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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

1.1 In a tug of war, each man shown in Fig. P1.1 exerts a force of 200 lbs. If the diameter of the rope is 1/2 inch, determine the axial stress in the rope.

Fig. P1.1

Solution

σ=? ------------------------------------------------------------

P=200 lbs

d=1/2 in

The free body diagram of the rope after an imaginary cut can be drawn as shown below. P=200 lbs

N

By equilibrium of forces we have: N = 200lbs 2 π 1 The area of cross-section of the rope is: A = --- ⎛⎝ ---⎞⎠ = 0.1963in 4 2 2

200 The average axial stress in the rope is: σ = N ---- = ---------------- = 1018.6psi or A

0.1963

σ = 1019 psi ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.2 A weight is being raised using a cable and a pulley as shown in Fig. P1.2. If the weight W = 200 lbs, determine the axial stress assuming (a) cable diameter is 1/8 inch. (b) cable diameter is 1/ 4 inch.

W

Fig. P1.2

Solution

(a) d=1/8 in σ =? (b) d=1/4 in σ =? ------------------------------------------------------------

W=200 lbs

The free body diagram of the weight is shown below. N

W

By equilibrium of forces we have: N = W 2 π 1 (a) The area of cross-section is: A = --- ⎛⎝ ---⎞⎠ = 0.01227in 4 8 2

200 The average axial stress in the rope is: σ = N ---- = ------------------- = 16297psi or A

0.01227

σ = 16.3ksi ( T )

2 π 1 (b) The area of cross-section is: A = --- ⎛⎝ ---⎞⎠ = 0.04909in 4 4 2

200 - = 4070psi or The average axial stress in the rope is: σ = N ---- = -----------------A

0.04909

σ = 4.1ksi ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.3 The cable shown in Fig. P1.3 has a diameter of 1/5 inch. If the maximum stress in the cable must be limited to 4 ksi (T), what is the maximum weight that can be lifted?

W

Fig. P1.3

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1-1

M. Vable

Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

σ max ≤ 4ksi ( T )

d=1/5 in

January 2014

Wmax = ?

-----------------------------------------------------------The free body diagram of the weight is shown below. N

W

By equilibrium of forces we have: N = W –3 2 π 1 The area of cross-section of the rope is: A = --- ⎛⎝ ---⎞⎠ = 31.41 ( 10 )in 4 5 2

3 W The average axial stress is: σ = N ---- = ----------------------------- ≤ 4 ( 10 ) or W ≤ 125.66lb or –3

A

31.41 ( 10 )

W max = 125.6lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.4 The weight shown in Fig. P1.4 is W = 250 lbs. If the maximum stress in the cable must be limited to 5 ksi (T), determine the minimum diameter of the cable to the nearest 1/16 inch.

W

Fig. P1.4

Solution

σ max ≤ 5ksi ( T )

W = 250 lbs.

dmin = ? to nearest 1/16th of an inch

-----------------------------------------------------------The free body diagram of the weight is shown below. N

W

By equilibrium of forces we have: N = W 3 W The average axial stress is: σ = N ---- = --------------------- ≤ 5 ( 10 ) or d ≥ 0.252in or 2

A

( πd ⁄ 4 )

5 d min = ------ inch 16

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.5 A 6 kg light shown in Fig. P1.5 is hanging from the ceiling by wires of diameter of 0.75 mm. Determine the tensile stress in the wires AB and BC. 2m A

A m

2 .5

m

2 .5

B C Light

Fig. P1.5

Solution

σAB = ? σBC = ? ------------------------------------------------------------

m = 6 kg.

d = 0.75 mm

The weight is: W = mg = 58.86N

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1-2

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The following free body diagrams can be drawn. (a)

NBC

(b)

NAB

1m

NAB θθ

2.5 m θ

Light Light W = 58.86 N

W = 58.86 N

o 1 θ = asin ⎛ -------⎞ = 23.6 ⎝ 2.5⎠

By force equilibrium in the y-direction in Figs. (a) and (b) we have the following. N BC = W 2N AB cos θ = W

1

W N AB = --------------------- = 0.5456W 2 cos 23.6

or

2

2 –6 2 π The area of cross-section is: A = --- ( 0.00075 ) = 0.4418 ( 10 )m

4

N

58.86 BC The average axial stress in BC is: σ BC = ---------- = -------------------------------- = 133.2 ( 10 ) or σ BC = 133.2MPa ( T ) 6

–6

A

0.4418 ( 10 )

N AB 6 ( 0.5456 ) ( 58.86 ) = The average axial stress in AB is: σ AB = ---------- = --------------------------------------72.7 ( 10 ) or σ AB = 72.7MPa ( T ) –6

A

0.4418 ( 10 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.6 A 8 kg light shown in Fig. P1.6 is hanging from the ceiling by wires. If the tensile stress in the wires cannot exceed 50 MPa, determine the minimum diameter of the wire to the nearest tenth of a millimeter. 2m A

A m

2.5 m

2.5

B C Light

Fig. P1.6

Solution

m = 8 kg.

σ max ≤ 50MPa

dmin = ? nearest tenth of a millimeter

-----------------------------------------------------------The weight is: W = mg = 58.86N The following free body diagrams can be drawn. (a)

NBC

(b)

NAB

NAB θθ

1m

2.5 m θ

Light Light W = 58.86 N

W = 58.86 N

o 1 θ = asin ⎛ -------⎞ = 23.6 ⎝ 2.5⎠

By force equilibrium in the y-direction in Figs. (a) and (b) we have the following. N BC = W 2N AB cos θ = W

or

1

W N AB = ---------------------- = 0.5456W 2 cos 23.6

2

We see that the internal force in BC will be more. As the area of cross-section is the same, the maximum axial stress will be in BC.

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

N

6 –3 78.48 BC - = --------------------- ≤ 50 ( 10 ) or d ≥ 1.41 ( 10 )m or The average axial stress in the rope is: σ max = ---------2

A

( πd ⁄ 4 )

Rounding to the nearest tenth of a millimeter that satisfy the inequality, we obtain: d min = 1.5 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.7 Wires of diameter 0.5 mm are to be used for hanging lights such as shown in Fig. P1.7. If the tensile stress in the wires cannot exceed 80 MPa, determine the maximum mass of the light that can be hung using these wires. 2m A

A m

2.5 m

2.5

B C Light

Fig. P1.7

Solution

d = 0.5 mm

σ max ≤ 80MPa

mmax = ?

-----------------------------------------------------------The weight is: W = mg = 58.86N The following free body diagrams can be drawn. (a)

NBC

(b)

NAB

NAB θθ

1m

2.5 m θ

Light Light W = 58.86 N

W = 58.86 N

o 1 θ = asin ⎛ -------⎞ = 23.6 ⎝ 2.5⎠

By force equilibrium in the y-direction in Figs. (a) and (b) we have the following. N BC = W 2N AB cos θ = W

or

1

W N AB = ---------------------- = 0.5456W 2 cos 23.6

2

We see that the internal force in BC will be more. As the area of cross-section is the same, the maximum axial stress will be in BC. 2 –6 2 π The area of cross-section is: A = --- ( 0.0005 ) = 0.1963 ( 10 )m

4

6 m ( 9.81 ) BC - = -------------------------------- ≤ 80 ( 10 ) or m ≤ 1.601kg or m max = 1.6kg The average axial stress is: σ max = ---------–6

N

A

0.1963 ( 10 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.8 A 3 kg picture is hung using a wire of diameter of 3 mm as shown in Fig. P1.8. What is the average normal stress in the wire. 54o

Fig. P1.8

Solution

m= 3 kg

σ=? -----------------------------------------------------------d = 10 mm

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The free body diagram of the picture frame is shown in Fig. (a). By equilibrium of forces in the y-direction we have the following equation. 2N sin 54 = W = 3g

N = 0.618 ( 3 ) ( 9.81 ) = 18.189 N

or

1

N N

(a)

54o

54o

W = 3g 2 –6 2 π The area of cross-section is: A = --- ( 0.003 ) = 7.0686 ( 10 )mm

4

6 2 18.189 The average axial stress in the wire is: σ = N ---- = -------------------------------- = 2.573 ( 10 )N ⁄ m or

A

–6

7.0686 ( 10 )

σ = 2.57MPa ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.9 A 5 kg picture is hung using a wire as shown in Fig. P1.9. If the tensile stress in the wires cannot exceed 10MPa, determine the minimum diameter of the wire to the nearest millimeter. 54o

Fig. P1.9

Solution

σ max ≤ 10MPa

m= 5kg

dmin = ? nearest millimeter.

-----------------------------------------------------------The free body diagram of the picture frame is shown in Fig. (a). By equilibrium of forces in the y-direction we have the following equation. 2N sin 54 = W = 5g

or

N = N = 0.618 ( 5 ) ( 9.81 ) = 30.315 N

1

N N

(a)

54o

54o

W = 5g 6 –3 30.315 The average axial stress in the wire is: σ = N ---- = --------------------- ≤ ( 10 ) ( 10 ) or d ≥ 1.96 ( 10 ) m or 2

A

( πd ⁄ 4 )

The minimum diameter to the nearest miliimeter that satisfy the inequality is:

d min = 2 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.10 Wires of diameter 16 mils are used for hanging picture such as shown in Fig. P1.10. If the tensile stress in the wire cannot exceed 750 psi, determine the maximum weight of the picture that can be hung using these wires. One mil equals 0.001 inch. 54o

Fig. P1.10

Solution

d= 0.016 inch

σ max ≤ 750psi

Wmax =?

-----------------------------------------------------------The free body diagram of the picture frame is shown in Fig. (a). By equilibrium of forces in the y-direction we have the following equation.

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

2N sin 54 = W

or

January 2014

N = 0.618W

1

N N

(a)

54o

54o

W = 3 lb 2 –3 2 π The area of cross-section is: A = --- ( 0.016 ) = 0.201 ( 10 )in

4

0.618W The average axial stress is: σ max = N ---- = ----------------------------- ≤ 750 or W ≤ 0.244lb or –3 A

0.201 ( 10 )

W max = 0.24lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.11 For swim meets, diving boards are raised to lean against the left wall using a cable and pulley as shown in Fig. P1.11. Determine the axial stress in the cable in terms of the length L of the diving board, γ the specific weight per unit length of the diving board, diameter d of the cable and the angles θ and α shown in Fig. P1.11.

Bo ing v i D

α ard

θ

Fig. P1.11

Solution

σ = f(L,γ,d,θ,α)=? ------------------------------------------------------------

The free body diagram of the diving board is shown below. N L/2 α L/2 Ox

θ W=γL Oy

By equilibrium of moment about point O we obtain the following. L ( γL ) ⎛ --- cos θ⎞ – ( N sin α )L = 0 ⎝2 ⎠

or

γL cos θ N = -----------------2 sin α

γL cos θ ⁄ ( 2 sin α ) ) or The average axial stress is: σ = N ---- = (---------------------------------------------A

2

( πd ⁄ 4 )

2γL cos θ σ = --------------------2 πd sin α

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.12 A hollow circular column supporting a building is attached to a metal plate and bolted into concrete foundation as shown in Fig. P1.12. The column outside diameter is 100 mm and an inside diameter is 75 mm.The metal plate dimensions are 200 mm x200 mm x 10 mm. The load P is estimated at 800 kN. Determine (a) the compressive stress in the column. (b) average bearing stress between the metal plate and

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

the concrete. P

Metal

Concrete

Fig. P1.12

Solution

di = 75 mm plate: 200 mm x200 mm x 10 mm 800 kNσcol = ?σb= ?

do = 100 mm

-----------------------------------------------------------The following free body diagrams can be drawn. P (a) (b)

P

Metal

N1

N2

By equilibrium of forces in Figs. (a) and (b) we obtain the following equations. N1 = P

and

N2 = P

1

2 2 –3 π The area of cross-section of the column is: A col = --- ( 0.1 – 0.075 ) = 3.436 ( 10 )

mm

4

3

2

800 ( 10 ) 1 The average normal stress in the column is: σ col = ---------- = ----------------------------- = 232.8 ( 10 )N ⁄ m or N

6

2

–3

A col

3.436 ( 10 )

σ col = 232.8MPa ( C )

The area of cross-section of the plate is: A plate = ( 0.2 ) ( 0.2 ) = 0.04 m

2

3

N2 6 2 800 ( 10 ) The average bearing stress is: σ b = -------------- = ---------------------- = 20 ( 10 )N ⁄ m or A plate

( 0.04 )

σ b = 20MPa ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.13 A hollow circular column supporting a building is attached to a metal plate and bolted into concrete foundation as shown in Fig. P1.13. The column outside diameter is 4 inches and an inside diameter is 3.5 inches.The metal plate dimensions are 10 in x 10 in x 0.75 inch. If the allowable average compressive stress in the column is 30 ksi and the allowable average bearing stress in concrete is 2 ksi, determine the maximum load P that can be applied to the column.. P

Metal

Concrete

Fig. P1.13

Solution

do = 4 inch

di = 3.5 inch

σ col ≤ 30ksi

σ b ≤ 2ksi

plate: 10 in x 10 in x 0.75 P=?

------------------------------------------------------------

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

The following free body diagrams can be drawn. P (a) (b)

January 2014

P

Metal

N1

N2

By equilibrium of forces in Figs. (a) and (b) we obtain the following equations. N1 = P

and

N2 = P

1

2 2 π 2 The area of cross-section of the column is: A col = --- ( 4 – 3.5 ) = 2.945 in

4

N

P 1 The average normal stress in the column is: σ col = ---------- = ------------- ≤ 30 or A col

2.945

The area of cross-section of the plate is: A plate = ( 10 ) ( 10 ) = 100 in

2

N

P 2 The average bearing stress is: σ b = -------------- = --------- ≤ 2 or A plate

P ≤ 88.35kips

P ≤ 200kips

100

The maximum value of P that satisfies the two inequalities above is:

P = 88.3kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.14 A hollow square column supporting a building is attached to a metal plate and bolted into concrete foundation as shown in Fig. P1.14. The column outside dimension is 120 mm x 120 mm and has a thickness of 10 mm. The load P is estimated at 600 kN. The metal plate dimensions are 250 mm x 250 mm x 15 mm. Determine (a) the compressive stress in the column. (b) the average bearing stress between the metal plate and the concrete. P

Metal

Concrete

Fig. P1.14

Solution

P= 600 kN

plate: 250 mm x 250 mm x 15 mm σcol = ?

σ b= ?

-----------------------------------------------------------Fig(a) below shows the geometry of the cross-section. Figs. (b) and (c) show the free body diagrams. (a) 100 mm 100 mm

120 mm

(b)

P

P

(c)

Metal 120 mm

N1

N2

By equilibrium of forces in Figs. (a) and (b) we obtain the following equations. N1 = P

and

N2 = P

1 –3

The area of cross-section of the column is: A col = ( 0.12 ) ( 0.12 ) – ( 0.1 ) ( 0.1 ) = 4.4 ( 10 ) m

2

3

N1 6 2 ( 600 ) ( 10 ) The average normal stress in the column is: σ col = ---------- = --------------------------- = 136.4 ( 10 ) N ⁄ m or A col

–3

4.4 ( 10 )

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

σ col = 136.4MPa ( C ) –3

The area of cross-section of the plate is: A plate = ( 0.25 ) ( 0.25 ) = 62.5 ( 10 ) m

2

3

N2 6 2 ( 600 ) ( 10 ) The average bearing stress is: σ b = -------------- = --------------------------- = 9.6 ( 10 ) N ⁄ m or –3

A plate

62.5 ( 10 )

σ b = 9.6MPa ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.15 A column with cross-section shown in Fig. P1.15 supports a building. The column is attached to a metal plate and bolted into concrete foundation. The load P is estimated at 750 kN. The metal plate dimensions are 300 mm x 300 mm x 20 mm. Determine (a) the compressive stress in the column. (b) the average bearing stress between the metal plate and the concrete. P

160 mm 10 mm

10

m

m

16 0

m

m

10 mm

Metal Concrete

Fig. P1.15

Solution

P = 750 kN

plate: 300 x 300 x 20 mm

(a) σcol = ?

(b) σp = ?

-----------------------------------------------------------The area of cross-section is: Acol = (160) (10) + (160) (10) + (160) (10) = 4800 mm2 = 4.80 (10-3) m2 P

P

(b)

(a)

Metal N1 N

Fig. (a) shows the free body diagram of the column after an imaginary cut is made through it. By equilibrium of forces we have: N1 = P = 750 (10-3) N 3

N1 750 ( 10 ) 6 2 - = ----------------------- = 156.25 ( 10 )N ⁄ m or σ col = 156 MPa ( C ) The average axial stress is: σ col = ---------–3 A col

4.8 ( 10 )

The area of cross-section of the metal plate is: Aplate = (300) (300) = 90 (10-3) m2 Fig. (b) shows the free body diagram after an imaginary cut is made between the plate and the ground. By force equilibrium we have: N2 = P = 750 (103) N. 3

N2 6 2 ( 10 ) - = 750 The average bearing stress is: σ b = ----------------------------------- = 8.33 ( 10 )N ⁄ m σ b = 8.33 MPa ( C ) A plate

–3

90 ( 10 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.16

A 70 kg. person is standing on a bathroom scale that has a dimensions 150

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

mm x 100 mm x 40 mm. Determine the bearing stress between the scales and the floor.

Fig. P1.16

Solution

σ b= ? ------------------------------------------------------------

m = 70 kg

Scale: 150 x 100

The weight of the person is: W = mg = (70) (9.81) = 686.7 N By making an imaginary cut between the scales and the floor we obtain the following free body diagram. W

N By equilibrium of forces we obtain: N = W = 686.7N The area of cross-section is: A = (150) (100) = 15 (103) mm2 = 15.0 (10-3) m2 –3 2 N 686.7 = The average bearing stress is: σ b = ---- = --------------------45.78 ( 10 ) N ⁄ m σ b = 45.8 kPa ( C )

A

–3

15 ( 10 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.17 A 30 ft tall brick-chimney has an outside diameter of 3 ft. and a wall thickness of 4 inch. If the specific weight of the bricks is 80 lbs/ft3, determine the average bearing stress at the base of the chimney.

30 ft

Fig. P1.17

Solution

L = 30 ft.

γ = 80 lbs/ft3 -----------------------------------------------------------d = 3 ft

t = 4 in = 1/3 ft

σb = ?

2 2 3 π 2 π 2 The volume of the brick material is: V = --- ( d o – d i ) L = --- ( 3 – 2.33 ) ( 30 ) = 83.77 ft

4

4

The weight is: W = γ V = (80) (83.77) = 6702 lbs. By making an imaginary cut at the base of the chimney we obtain the following free body diagram. W

N

By force equilibrium we have: N = W = 6702 lbs 2 2 2 π 2 The area of cross-section is: A = = --- ( 3 – 2.33 ) = 2.792 ft = 402.1 in

4

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

N 6702 The average bearing stress is: σ b = ---- = ------------ = 16.7 psi A

January 2014

σ b = 16.7 psi ( C )

402.1

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.18 Determine the average bearing stress at the bottom of a block shown in Fig. P1.18 in terms of the specific weight γ, and the length dimensions a and h. a 10 h h

100 a

Fig. P1.18

Solution

σb = f(γ,a,h) = ? ------------------------------------------------------------

2 1 The volume of material is: V = --- ( 100a + a )h ( 10h ) = 505ah

2

The weight of the material is: W = γV = 505γah2 By making an imaginary cut at the base of the block, we obtain the following free body diagram. W

N

By force equilibrium we have: N = W = 505γah2 The area of cross-section is: A = (100 a) (10 h) = 1000 ah 2

N -------------------- = 0.505γh The average bearing stress is: σ b = ---- = 505γah A

σ = 0.505γh ( C )

1000ah

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.19 The Washington monument is an Obelisk with a hollow rectangular cross-section that tapers along its length. The approximation of the monument geometry is shown on the right in Fig. P1.19. The thickness at the base is 4.5 m. Assume that the thickness varies proportionately with the outside taper. The monument is constructed from marble and granite.Using a specific weight for these material as 28 kN/m3, determine the average bearing stress at the base of the monument. 10 m 10 m 169 m

17 m 17 m

Solution

Fig. P1.19

γ= 28 kN/m3 σb = ? ------------------------------------------------------------

t0 = 4.5 m

tL = 2.5 m

The variation of the taper can be found using the geometry shown in Fig. (a).The outer dimension ho and

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

inner dimension hi are linear functions of x and can be written as: 10 m

(a)

h ----o- = m 1 x + C 1 2

5m

and

hi ---- = m 2 x + C 2 2

1

ho 169 m

hi x 8m 17 m

Noting that @ x = 0, ho = 17 and hi = 8, we obtain: C1 = 8.5 and C2 = 4 Noting that @ x = 169, ho = 10 and hi = 5 we obtain the following equations: 5 = m1(169) + 8. 5 and 2.5 = m2(169) + 4 which yields: m1 = -3.5/169 and m2 =-1.5/169. We obtain the following equations for the outer and inner dimension: h o = ( 17 – 7x ⁄ 169 )

and

h i = ( 8 – 3x ⁄ 169 )

169

The volume of the monument material is: V =

∫

169 2 ( ho

–

2 h i ) dx

∫

=

0

1 7x-⎞ 3 ⎛ 169⎞ V = --- ⎛ 17 – -------– --------3⎝ 169⎠ ⎝ 7 ⎠

169 0

2

7x ⎞ 2 ⎛ 3x 2 ⎛ 17 – -------- – 8 – ---------⎞ dx or ⎝ ⎠ ⎝ 169 169⎠

0

1 169 3x-⎞ 3 ⎛ – ------------⎞ – --- ⎛ 8 – -------3⎝ 169⎠ ⎝ 3 ⎠

169

= 24223.3 m

3

0

The weight of the monument is: W = γV = 28 (10 ) (24223.3) = 678253 (103) N. The force at the base of the monument is equal to the weight of the monument i.e., N = W = 678253 (103)N The area of cross-section of the base is: A = 172 - 82 = 225 m2 3

3

3 2 678253 ( 10 ) The bearing stress is: σ b = N ---- = ----------------------------------- = 3014 ( 10 ) N ⁄ m or

A

225

σ b = 3 MPa ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.20 Show that the average compressive stress due to weight on a cross section at a distance x from the top of the wall in Figure P1.20(b) is half that of wall in Figure P1.20(a), thus confirming the wisdom of ancient Egyptians in building inward-sloping walls for the pyramids. (Hint: Using γ the specific weight of wall material, H the height of the wall, t the thickness of the wall, and L the length of the wall, calculate the average compressive stress at any crossection at a distance x from the top for the two walls.) . (b) (a)

x

x

H

H L

t Fig. P1.20(a) Straight wall (b) Inward sloping tapered wall.

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L t

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Solution

-----------------------------------------------------------We make imaginary cut at a distance of x units from the top and draw the FBD.

W1

W2

x

x L t

(a)

L b

N1

(b)

N2

By equilibrium we obtain: N 1 = W 1 = γ ( xt ) ( L )

1 N 2 = W 2 = γ ⎛ --- xt⎞ ( L ) ⎝2 ⎠

1

The normal stresses are N ( xt ) ( L )- = γx σ 1 = ------1 = γ-------------------Lt A1

2

1 γ ⎛ --- xb⎞ ( L ) ⎝2 ⎠ N2 1 σ 2 = ------ = --------------------------- = --- γx 2 Lb A2

3

Thus, 1 σ 2 = --- σ 1 2

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.21 The Great pyramid of Giza shown in Figure 1.14d has a base of 757.71ft x 757.71ft and a height of 480.96 ft. Assume an average specific weight of γ = 75 lb/ft3. Determine (a) the bearing stress at the base of the pyramid. (b) the average compressive stress at mid height. Solution a= 757.7 ft h= 480.96 ftγ = 75 lb/ft3 σ b = ? σH = ?

-----------------------------------------------------------Fig. (a) shows the geometry of the pyramid. The volume of the pyramid can be found as shown below. 2 6 3 1 2 1 V = --- a h = --- ( 757.7 ) ( 480.96 ) = 92.040 ( 10 ) ft 3 3

1

The weight of the pyramid is:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

6

January 2014

9

W = γV = ( 75 ) ( 92.040 ) ( 10 ) = 6.903 ( 10 )lb

2

(b)

(a) h

W

h/2

b b

a

a

N

Fig. (b) shows the free body diagram of the entire pyramid. The total bearing force is 9

N = W = 6.903 ( 10 )lb

3

The bearing stress at the base is: 9

6.903 ( 10 ) N- = -------------------------σ b = ---- = 12024 psi 2 2 a ( 757.7 )

4 σ b = 12024 psi ( C )

The sides of square at half height will be have the base side. Thus b=a/2. Thus the volume will be 1 2 1 2 h V H = --- ( b ) ⎛ ---⎞ = --- b h ⎝ 2⎠ 6 3

The weight of top half of pyramid is: γ 2 W H = γV H = --- b h 6

5

WH NH γ ( 75 ) ( 480.96 ) σ H = ------- = -------- = --- h = -------------------------------- = 6012psi 2 2 6 6 b b

6

The compressive stress will be:

σ H = 6012psi ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.22 The bent pyramid shown in Figure 1.14c has a base of 188 m x 188 m. The initial slopes of the sides is 54o27’44”. After a certain height the slope is 43o22’. The total height of the pyramid is 105 m. Assume an average mass density of 1200 kg/ m3. Determine the bearing stress at the base of the pyramid. Solution θ1= 54o27’44” θ2= 43o22’ g = 1200 kg/ m3 σ=?

------------------------------------------------------------

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1-14

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The schematic of the pyramid is shown below.

h3

h2

W

θ2 x2 h1 N

θ1 94 m

x1

Converting to degrees we obtain: 44 θ 1 = 54 + ⎛⎝ 27 + ------⎞⎠ ⁄ 60 = 54.4622° 60

θ 2 = 43 + ( 22 ) ⁄ 60 = 42.3667°

1

From geometry we obtain: h 1 + h 2 = 105 m

2

x 1 + x 2 = 94 m

3

h 1 = x 1 tan θ 1 = 1.4x 1

h 2 = x 2 tan θ 2 = 0.9446x 2

4

Substituting 1.4x 1 + 0.9446x 2 = 105 m

5

Solving Equations 3 and 5 we obtain x 1 = 35.59 m

x 2 = 58.41 m

6

h 1 = 49.82 m

h 2 = 55.17 m

7

From Equation 4 we obtain: From geometry: h 3 = 94 tan θ 1 = 131.6 m The total volume of the pyramid is: 2 2 1 1 1 1 = --- ( 188 ) ( 188 )h 3 – --- ( 2x 2 ) ( 2x 2 ) ( h 3 – h 1 ) + --- ( 2x 2 ) ( 2x 2 ) ( h 2 ) = --- [ 188 ( h 3 ) – ( 4x 2 ) ( h 3 – h 1 – h 2 )or 3 3 3 3 6 6 6 3 1 V = --- [ 4.651 ( 10 ) – 0.363 ( 10 ) ] = 1.429 ( 10 ) m 3

The total weight of the pyramid is: 6

9

W = γgV = ( 1200 ) ( 9.81 ) [ 1.429 ( 10 ) ] = 16.827 ( 10 )N

The bearing stress is 9

6 2 16.827 ( 10 ) N N σ = ---- = ---------------------------- = ------------------------------ = 0.476 ( 10 ) ( N ⁄ m ) A ( 188 ) ( 188 ) ( 188 ) ( 188 )

σ = 0.476 MPa ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.23 A steel bolt of 25 mm diameter passes through an aluminum sleeve of thickness 4 mm and outside diameter of 48 mm as shown in Figure P1.23. Determine the average normal stress in the sleeve if in the

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

assembled position the bolt has an average normal stress of 100 MPa (T). Rigid washers

Sleeve

300 mm 25 mm

25 mm

Fig. P1.23

Solution

σbolt = 100 MPa(T) ------------------------------------------------------------

dbolt = 25 mm dal = 48 mm

t= 4mm

σbolt = ?

The cross-sectional areas of the bolt and aluminum sleeve are: 2 –6 2 π 2 π A bolt = --- d bolt = --- ( 0.025 ) = 490.87 ( 10 )m 4 4

1

2 2 –6 2 π 2 π A al = --- d bolt = --- [ ( 0.048 ) – ( 0.04 ) ] = 552.92 ( 10 )m 4 4

2

The internal force in the bolt is: 6

–6

N bolt = σ bolt A bolt = 100 ( 10 ) [ 490.87 ( 10 ) ] = 49087N

(a) N al

N bolt

Fig. (a) shows the free body diagram. By equilibrium of forces we obtain: N al = N bolt = 49087N

3

The normal stress in aluminium sleeve is: N al 6 2 49087 σ al = ------- = -------------------------------- = 88.78 ( 10 )N ⁄ m –6 A al 552.92 ( 10 )

4 σ al = 88.78 MPa ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.24 The device shown in Fig. P1.24 is used for determining the shear strength of the wood. The dimensions of the wood block are 6 in x 8 in x 2 in. If the force required to break the wood block is 15 kips, determine the average shear strength of the wood. P

6 in

Wood

6 in

2 in

Fig. P1.24

Solution

Wood: 6 in x 8 in x 2 in. τ=? ------------------------------------------------------------

P = 15 kips

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The free body diagram of the wood that would shear off is shown below. P

6 in τ 2 in 2. in

By equilibrium of forces we have the following equation. P = τ(2)(6) or τ = P ⁄ 12 = 15 ⁄ 12 or

τ = 1.25ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.25 The device shown in Fig. P1.24 is used for determining the shear strength of the wood. The dimensions of the wood block are 6 in x 8 in x 1.5 in. Estimate the force P that should be applied to break the block if the shear strength of the wood is 1.2 ksi.. P

6 in

Wood

6 in

Solution

2 in

Fig. P1.25 τ ≤ 1.2ksi

Wood: 6 in x 8 in x 1.5 in.

P=?

-----------------------------------------------------------The free body diagram of the wood that would shear off is shown below. P

6 in τ 1.5 in 2 in

By equilibrium of forces we have the following equation. P = τ ( 6 ) ( 1.5 )

or

τ = P ⁄ 9 ≤ 1.2

P ≤ 10.8 kips or

or

P max = 10.8 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.26 The schematic of a punch and die arrangement shown Fig. P1.26 is used to punch out thin plate objects of different shapes. The cross-section of the punch and the die is a circle of diameter 1 inch. A force of P = 6 kips is applied to the punch. If the plate thickness t = 1/8 inch, what would be the average shear stress in the plate along the path of the punch. P Punch t

Plate Die

Die

Fig. P1.26

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M. Vable

Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

τ=? ------------------------------------------------------------

P = 6 kips

t = 1/8 inch

d = 1 inch

The free body diagram of the punch as it will go through the plate is shown below. P

τ By equilibrium of forces we have the following equation. P = τ ( πd ) ( t )

P 6 τ = -------- = -------------------------- = 15.28ksi or πdt π( 1)(1 ⁄ 8)

or

τ = 15.3ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.27 The schematic of a punch and die arrangement shown in Fig. P1.27 is used to punch out thin plate objects of different shapes. The cross-section of the punch and the die is a square of 10 mm x 10 mm. The plate shown has a thickness of t = 3 mm and shear strength of 200 MPa. What is the average force P needed to drive the punch through the plate. P Punch t

Plate Die

Die

Fig. P1.27

Solution

t = 3 mm

τ ≤ 200MPa

plate: 10 mm x 10 mm

P=?

-----------------------------------------------------------The free body diagram of the punch as it will go through the plate is shown below. By equilibrium of forces we have the following equation. 6

3

P = τ ( 4a ) ( t ) = ( 200 ) ( 10 ) ( 4 ) ( 0.01 ) ( 0.003 ) = 24 ( 10 )N or

P = 24kN

P a=10 mm a=10 mm

τ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.28 A schematic of a punch and die for punching washers is shown in Fig. P1.28. Determine the force P needed to punch out washers, in terms of the plate thickness t, plate shear strength τ, and the inner and outer diameter of the washers di and do.

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

P

Punch t

Plate

di

Die

Die do

Fig. P1.28

P = f(τ, t, di, do) = ?

Solution

-----------------------------------------------------------The surface area on which the shear stress will is the inner and outer cylindrical surface area punched through the thickness of the plate i.e., A = π ( d o + d i )t . By equilibrium of forces we have: P = τA or P = τπ ( d o + d i )t

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.29 The forces acting on a pin in a truss joint are shown. The diameter of the pin is 25 mm. Determine the maximum transverse shear stress in the pin. 50 kN

40 kN

o 30 kN 36.9

Fig. P1.29

τmax = ? ------------------------------------------------------------

Solution

d = 25 mm

2

–6

The area of cross-section of the pin is: A = π ( 0.025 ) ⁄ 4 = 490.87 ( 10 )m The following free body diagrams can be drawn. V1

(a)

(b)

30 kN

2

40 kN

V2

From the above free body diagram we see that the maximum shear force is: V2 = 40kN. The maximum 3

40 ( 10 ) = 81.48 ( 10 )N ⁄ m or shear stress is: τ max = -----2- = -------------------------------V

A

6

2

–6

490.87 ( 10 )

τ max = 81.5MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.30 A weight W = 200lbs is being raised using a cable and a pulley as shown. The cable diameter is 1/ 4 inch and the pin in the pulley has diameter of 3/8 inch. Determine the axial stress in the cable and shear stress in the pin assuming the pin is in double shear. 55o W

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1-19

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Fig. P1.30

Solution

σC = ? τp = ? ------------------------------------------------------------

W = 200lbs

dc = 1/4 in

dp = 3/8 in

The following free body diagrams can be drawn. (a)

T

Oy

(b)

(c)

Ox

55o

Oy

V1 Ox

T

W

V1

T

By equilibrium of forces in Figs. (a) and (b) we have the following. T = W

1

O x = T cos 55 = 0.5736W

2

O y – T – T sin 55 = 0

O y = T ( 1 + sin 55 ) = 1.8192W

or

2

–3

The area of cross-section of the cable is: A C = π ( 1 ⁄ 4 ) ⁄ 4 = 49.08 ( 10 )in

3

2

T 200 The average axial stress in the cable is: σ C = ------- = ----------------------------= 4074.4 psi or σ C = 4074. psi ( T ) AC

–3

49.08 ( 10 )

From equilibrium of forces in Fig. (c) we have. 2V =

2

2

Ox + Oy

4

Substituting W = 200 lbs in Eqs. (2) and (3) we obtain: Ox = 114.7 lbs and Oy = 363,8 lbs. Substituting these values into Eq. (4) we obtain V = 190.7 lbs. 2

–3

The area of cross-section of the pin is: A p = π ( 3 ⁄ 8 ) ⁄ 4 = 110.4 ( 10 )in

2

V 190.7 The average shear stress in the pin is: τ = ------ = ----------------------------- = 1727.4psi AP

τ = 1727psi

–3

110.4 ( 10 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.31 The cable in Fig. P1.30 has a diameter of 1/5 inch and the pin in the pulley has a diameter of 3/ 8 inch. If the maximum normal stress in the cable must be limited to 4 ksi (T) and the maximum shear stress in the pin is to be limited to 2 ksi, determine the maximum weight that can be lifted. The pin is in double shear. 55o W

Solution

dc = 1/5 in

dp = 3/8 in

Fig. P1.31 σ C ≤ 4ksi

τp = ? Wmax = ?

τ p ≤ 2ksi

-----------------------------------------------------------The following free body diagrams can be drawn. (a)

T

Oy

(b)

Ox

55o

(c)

Oy

V1 Ox

W

T

T

V1

By equilibrium of forces in Figs. (a) and (b) we have the following.

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

T = W

1

O x = T cos 55 = 0.5736W

2

O y – T – T sin 55 = 0

O y = T ( 1 + sin 55 ) = 1.8192W

or

2

–3

The area of cross-section of the cable is: A C = π ( 1 ⁄ 5 ) ⁄ 4 = 31.42 ( 10 )in

2

3 T W The average axial stress in the cable is: σ C = ------- = ----------------------------≤ 4 ( 10 )psi or –3

AC

3

31.42 ( 10 )

W ≤ 125.66lbs

The shear force in the pin can be found as follows: 2V =

2

2

2

O x + O y = W ( 0.5736 ) + ( 1.8192 )

2

or

2

–3

The area of cross-section of the pin is: A p = π ( 3 ⁄ 8 ) ⁄ 4 = 110.4 ( 10 )in

V = 0.9537W

4

2

3 0.9537W V- = ----------------------------The average shear stress in the pin is: τ = -----≤ 2 ( 10 )psi or W ≤ 231.5lb –3

AP

110.4 ( 10 )

The maximum value of W that satisfies both inequalities above is:

W max = 125.6lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.32 The manufacturer of the plastic carrier for dry-wall panels prescribes a maximum load P of 200 lbs. If the cross-sectional areas at section AA and BB are 1.3 in2 and 0.3 in2 respectively, determine the average shear stress at section AA and average normal stress at section BB at the maximum load P.

A A B B P

Fig. P1.32

Solution

τAA = ? ------------------------------------------------------------

P = 200 lbs.

AAA = 1.3

in2

AAA = 0.3 in2

σBB = ?

The free body diagrams of the plastic carrier after making the imaginary cuts though AA and BB are shown below. By force equilibrium we obtain: V A = P ⁄ 2 = 100lbs and N A = P ⁄ 2 = 100lbs VA 100 The average shear stress at section AA is: τ AA = ----------= --------- = 76.92psi or (a)

A AA

1.3

(b)

NB

τ AA = 76.9psi

NB

VA VA P P

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

NB 100 The average normal stress at section BB is: σ BB = ---------- = --------- = 333.33psi or A BB

0.3

January 2014

σ BB = 333.3 psi ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.33 A bolt passing through a piece of wood is shown in Fig. P1.33. Determine (a) the axial stress in the bolt. (b) the average shear stress in the bolt-head. (c) the average bearing stress between the bolt head and the wood. (d) the average shear stress in the wood. 3/8 in

1/4 in

P= 1.5 kips

3/4 in

1/2 in

Fig. P1.33

Solution

σbolt = ?

τbolt = ? σbear = ? τwood = ? ------------------------------------------------------------

The following free body diagrams can be drawn. (a) N

P 1/4 in

(d) τwood

(c)

(b)

σbear

P

τ

P P

3/4 in

3/4 in

3/8 in 1/2 in 2

–3

The area of cross-section of the bolt is: A bolt = π ( 1 ⁄ 4 ) ⁄ 4 = 49.09 ( 10 )in

2

By equilibrium of forces in Fig. (a) we obtain: N = P = 1.5 kips The average normal stress at section BB is: N1.5 σ bolt = -----------= ----------------------------- = 30.556ksi or –3 A bolt 49.09 ( 10 )

σ bolt = 30.56 ksi ( T )

By equilibrium of forces in Fig. (b) we obtain the following. τ [ π ( 1 ⁄ 4 ) ] ( 3 ⁄ 8 ) = P = 1.5

1.5 τ = --------------------- = 5.093ksi or ( 3π ⁄ 32 )

or

τ = 5.09 ksi

As per Fig(c), the area of cross-section over which the bearing stress acts is as given below. 2

2

A bear = π [ ( 3 ⁄ 4 ) – ( 1 ⁄ 4 ) ] ⁄ 4 = 0.3927in

2

By equilibrium of forces in Fig. (c) we obtain the following. σ bear A bear = P = 1.5

or

1.5 σ bear = ---------------- = 3.8197ksi or 0.3927

σ bear = 3.82 ksi ( C )

By equilibrium of forces in Fig. (d) we obtain the following. τ wood [ π ( 3 ⁄ 4 ) ] ( 1 ⁄ 2 ) = P = 1.5

or

1.5 τ wood = ------------------ = 1.273ksi or ( 3π ⁄ 8 )

τ wood = 1.27ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.34 A load of P = 10 kips is transferred by the the riveted joint shown in Figure P1.34. Determine (a) the average shear stress in the rivet. (b) the largest average normal stress in the members attached (c) the

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

largest average bearing stress stress between the pins and members. P

1 in

2 in

P

1 in

0.5 in

0.5 in

P

P

0.5 in 0.5 in

Fig. P1.34

Solution

τr = ? σmax = ? σb = ? ------------------------------------------------------------

P = 10 kips

We make imaginary cuts and obtain the FBD’s below. V

P

P

P

Nb

N

V By equilibrium we obtain: V = P --- = 5 kips 2

N = P = 10 kips

N b = P = 10 kips

1

The stresses1 are: V5 5 - = 25.46 ksi τ r = ----= ------------------------------ = --------------2 Ar 0.1963 ( π ⁄ 4 ) ( 0.5 )

2

10 - = 20 ksi σ = N ---- = ------------------A ( 1 ) ( 0.5 )

3

N 10 10 10 σ b = ------b = --------------------------- = ---------------------------------------- = ---------------- = 25.46 ksi ( π ⁄ 2 ) ( t )d r Ab ( π ⁄ 2 ) ( 0.5 ) ( 0.5 ) 0.3927

4

The answers are: τ r = 25.46 ksi

σ = 20 ksi ( T )

σ b = 25.46 ksi ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.35

3 A joint in a wooden structure is shown in Figure P1.35. The dimension h = 4 --- in and 8

1 d = 1 --- in . Determine the average normal stress on plane BEF and average shear stress on plane BCD. 8

1. In the calcuation of bearing stress, a better alternative is to use the projected bearing area (tdr) which assume a uniform pressure acting on the hole boundary.

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Assume plane BEF and the horizontal plane at AB are a smooth surfaces. 10 kips

30⬚

n

4i

A

F

h d

E D B C

Fig. P1.35

Solution

-----------------------------------------------------------The FBD’s are: 10 kips

10 kips

E,F 30o

A

B

30o

C VBCD NABC

A

NBEF

B NAB

By equilibrium we obtain: V BCD = 10 cos 30 = 8.66 kips

N BEF = 10 cos 30 = 8.66 kips

The stresses are: V BCD 8.66 τ BCD = -------------- = -------------------- = 0.4949 ksi A BCD ⎛ 4 3---⎞ ( 4 ) ⎝ 8⎠ N BEF 8.66 - = 1.924 ksi σ BEF = ------------- = ------------------A BEF ⎛ 1 1---⎞ ( 4 ) ⎝ 8⎠

The answers are: τ BCD = 0.4949 ksi

σ BEF = 1.924 ksi ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.36 A metal plate welded to an I-beam is sequrely fastened to the foundation wall using four bolts of 1/2 in diameter as shown in Figure P1.36. If P = 12 kips determine the normal and shear stress in the bolt.

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Assume the load is equally distibuted between the four bolts.

P 60o

Fig. P1.36

Solution

-----------------------------------------------------------The FBD is:

P 60o

2V 2N 2V 2N

By equilibrium we have: 4N = P sin 60

or

4V = P cos 60

or

12 sin 60 N = ⎛ --------------------⎞ = 2.598 kips ⎝ 4 ⎠

1

12 cos 60 V = ⎛ ---------------------⎞ = 1.5 kips ⎝ ⎠ 4

2

The area of the bolt crossection is: 2 π 1 2 A bolt = --- ⎛ ---⎞ = 0.1963 in ⎝ ⎠ 4 2

3

The stresses are: N 2.598- = 13.23 ksi σ av = ------------- = --------------A bolt 0.1963

V 1.5 - = 7.64 ksi τ av = ------------- = --------------A bolt 0.1963 σ av = 13.23 ksi ( T )

4 τ av = 7.64 ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.37 A metal plate welded to an I-beam is sequrely fastened to the foundation wall using four bolts as shown Figure P1.36. The allowable normal stress in the bolts is 100 MPa and the allowable shear stress is 70 MPa. Assume the load is equally distibuted between the four bolts. If the beam load P= 50 kN, determine the minimum diameter to the nearest millimeter of the bolts. Solution

-----------------------------------------------------------The FBD is:

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

P 60o

2V 2N 2V 2N

By equilibrium we have: 3

4N = P sin 60

or

50 ( 10 ) sin 60 3 N = ⎛ ----------------------------------⎞ = 10.825 ( 10 ) N ⎝ ⎠ 4

1

3

4V = P cos 60

or

50 ( 10 ) cos 60 3 V = ⎛ -----------------------------------⎞ = 6.25 ( 10 ) N ⎝ ⎠ 4

2

The stresses have to be less than the limiting values, hence 3

10.825 ( 10 ) 6 N σ av = ------------- = ------------------------------ ≤ 100 ( 10 ) A bolt A bolt

or

3

( 10 ) 6 V - = 6.25 ------------------------ ≤ 70 ( 10 ) τ av = -----------A bolt A bolt

or

–6

A bolt ≥ 108.25 ( 10 ) m –6

A bolt ≥ 89.29 ( 10 ) m

2

2

3

4

The diameter of the bolt can be found as –6 π 2 A bolt = --- d bolt ≥ 108.25 ( 10 ) 4

or

–3

d bolt ≥ 11.74 ( 10 ) m

5 d bolt = 12 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.38 A metal plate welded to an I-beam is sequrely fastened to the foundation wall using four bolts of 1/ 2 in diameter as shown Figure P1.36. The allowable normal stress in the bolts is 15 ksi and the allowable shear stress is 12 ksi. Assume the load is equally distibuted between the four bolts. Determine the maximum load P to the nearest pound the beam can support. Solution

-----------------------------------------------------------The FBD is:

P 60o

2V 2N 2V 2N

By equilibrium we have:

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

4N = P sin 60

or

P sin 60 N = ⎛ -----------------⎞ = 0.2165P ⎝ 4 ⎠

1

4V = P cos 60

or

P cos 60 V = ⎛ ------------------⎞ = 0.125P ⎝ 4 ⎠

2

The stresses have to be less than the limiting values, hence 3 0.2165P N - = -------------------------------- ≤ 15 ( 10 )psi σ av = -----------2 A bolt (π ⁄ 4)(1 ⁄ 2)

or

P ≤ 13603.9 lbs

3

3 0.125P V τ av = ------------- = --------------------------------- ≤ 12 ( 10 )psi 2 A bolt (π ⁄ 4)( 1 ⁄ 2)

or

P ≤ 18849.6 lbs

4

The maximum value of P is: P max = 13603 lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.39 An adhesively bonded joint in wood is fabricated as shown in Figure P1.39. The length of the overlap is L= 4 in. and the thickness of the wood is 3/8 in. Determine the average shear stress in the adhesive..

Fig. P1.39

Solution

-----------------------------------------------------------The FBD is:

V 10 kips

V

By equilibrium we have: V = 10 ------ = 5 kips 2

1

V- = --------------5 - = 0.3125 ksi τ av = --A (2 )(8)

2

The average shear stress in the adhesive is:

τ av = 312.5 psi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.40 A double lap joint adhesively bonds three pieces of wood as shown in Figure P1.40. The joints transmits a force of P= 20 kips and has the following dimensions: L = 3 in. , a = 8 in. and h = 2 in. Deter-

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1-27

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

mine the maximum average normal stress in the wood and the average shear stress in the adhesive. P

a

P/2

h/2

P/2

h h/2

L

Fig. P1.40

Solution

-----------------------------------------------------------The FBD is:

V 20 kips

V

N

10 kips

By equilibrium we have: V = 20 ------ = 10 kips 2

N = 10 kips

1

The average stressesare: V- = --------------10 - = --------------10 - = 0.4167 ksi τ av = --A (L)( a) (3 )(8)

2

10 - = --------------20 - = 2.5 ksi σ av = N ---- = --------------------A (a)( h ⁄ 2) (8)( 1) τ av = 416.7 psi

3 σ av = 2500 psi ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.41 The wood in the double lap joint of Figure P1.41 has a strength of 15 MPa in tension and the strength of the adhesive in shear is 2 MPa. The joint has the following dimensions: L = 75 mm, a =200 mm, and h = 50 mm. Determine the maximum force P the joint can transfer. P

a h/2

P/2 P/2

h h/2

L

Fig. P1.41

Solution

-----------------------------------------------------------The FBD is:

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

V P

V

N

P/2

By equilibrium we have: V = P⁄2

N = P⁄2

1

The average stresses must be less than the limiting values, hence 6 P⁄2 V P⁄2 τ av = ---- = ---------------- = ------------------------------- ≤ 2 ( 10 ) ( 0.075 ) ( 0.2 ) A (L)(a)

or

6 P⁄2 N P⁄2 σ av = ---- = ---------------------- = ------------------------------- ≤ 15 ( 10 ) ( 0.2 ) ( 0.025 ) A (a)( h ⁄ 2)

or

3

P ≤ 60 ( 10 ) N

2

3

P ≤ 150 ( 10 ) N

3

The maximum force P is P max = 60 kN

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

1.42 A wooden dowel of diameter d = 20 mm is used for constructing the double lap joint in Figure P1.42. The wooden members have a strength of 10 MPa in tension, the bearing stress between the dowel and the members is to be limited to 18 MPa, the shear strength of the dowel is 25 MPa. The joint has the following dimensions: L = 75 mm, a =200 mm, and h = 50 mm. Determine the maximum force P the joint can transfer. d P

a

P/2

h/2

P/2

h h/2

L

Fig. P1.42

Solution

-----------------------------------------------------------The FBD is:

V P

Nb

N P/2

V

P/2

By equilibrium we have: V = P⁄2

N = P⁄2

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Nb = P ⁄ 2

1

1-29

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The average stresses must be less than the limiting values1, hence 6 P⁄2 V P⁄2 τ av = ---- = --------------------- = ------------------------------------- ≤ 25 ( 10 ) –6 2 A ( 314.16 ) ( 10 ) ( π ⁄ 4 )d 6 P⁄2 N P⁄2 σ av = ---- = ---------------------- = ------------- ≤ 10 ( 10 ) 0.005 A (a)(h ⁄ 2)

3

P ≤ 15.707 ( 10 ) N

or

or

N 6 P⁄2 P⁄2 σ b = ------b = -------------------------------------- = ------------------------------------- ≤ 18 ( 10 ) –3 Ab ( π ⁄ 2)( d)(h ⁄ 2) ( 0.7854 ) ( 10 )

2

3

P ≤ 100 ( 10 ) N

3 3

P ≤ 28.27 ( 10 ) N

or

4

The maximum force P is P max = 15.7 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.43 A couple is using the chair lift shown in Figure P1.43 to see the Fall colors in Michigan’s Upper Peninsula. The pipes of the chair frame are 1/16 in thick. Assuming each person weighs 180 lb, determine the average normal stress at section AA and BB and average shear stress at section CC. 1/16 in 2 in A 2 in. 1.5 in.

B B

A C

1/16 in

C

1.5 in

80o

Section CC

Fig. P1.43

Solution

-----------------------------------------------------------The following FBD can be drawn. NA NB

180 lb

180 lb

180 lb

NB

VC

180 lb

180 lb

VC

180 lb

By equilibrium N A = 360 lb

2N B sin 80 = 360 lb

or

N B = 182.78 lb

V C = 180 lb

The cross-sectional areas are:

1. In the calcuation of bearing stress, a better alternative is to use the projected bearing area (dh/2) which assume a uniform pressure acting on the hole boundary.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

2 π 2 2 2 A A = --- 2 – ⎛ 2 – ------⎞ = 0.3804 in ⎝ 4 16⎠

January 2014

2 2 π 2 2 A B = --- 1.5 – ⎛ 1.5 – ------⎞ = 0.2823 in ⎝ 4 16⎠

2 2 1 π 2 2 A c = ( 2 ) ⎛ ------⎞ + --- 1.5 – ⎛ 1.5 – ------⎞ = 0.4073 in ⎝ 16⎠ 4 ⎝ 16⎠

The stresses are: NA 360 σ A = ------- = ---------------- = 946.4 psi 0.3804 AA

N 182.78 σ B = ------B- = ---------------- = 647.5 psi 0.2823 AB

V 180 τ C = ------C- = ---------------- = 441.9 psi 0.4073 AC σ A = 946.4 psi ( T )

σ B = 647.5 psi ( T )

τ C = 441.9 psi

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

1. 5

in

1.44 The axial force P = 12 kips acts on a rectangular member shown in Fig. P1.44. Determine the average normal and shear stress on the inclined plane AA.

A

P

2 in

P A

65o

Fig. P1.44

Solution

σAA= ?

τAA= ? -----------------------------------------------------------1. 5

in

The following free body diagram can be drawn after making an imaginary cut through section AA. t

A P 65o

25oh

2 in

A

65o

V

n 25o

N

By equilibrium of forces in the n-direction we obtain: N = P sin 65 = 12 sin 65 = 10.876kips . By equilibrium of forces in the t-direction we obtain: V = P cos 65 = 12 cos 65 = 5.0714kips . The area of cross-section of plane AA is: A AA = ( h ) ( 1.5 ) = ⎛⎝ --------------⎞⎠ ( 1.5 ) = 3.3107in . cos 25 2

2

N 10.876 The average normal stress on plane AA is: σ AA = ----------= ---------------- = 3.286ksi or σ AA = 3.286ksi ( T ) A AA

3.3107

V = 5.0714 The average shear stress on plane AA is: τ AA = -------------------------- = 1.532ksi or τ AA = 1.53ksi A AA

3.3107

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.45 A wooden axial member has a cross section of 2 in × 4 in. The member was glued along line AA and trasmits a force of P = ,80 kips as shown in Figure P1.45 Determine the average normal and shear

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

stress on plane AA. P

A

40⬚ A 4 in

Fig. P1.45

Solution

-----------------------------------------------------------We draw the FBD below: N

V h 40o

40o 40o

P By equilbrium N = P cos 40 = 61.283 kips

V = P sin 40 = 51.42 kips

1

From geometry the cross-sectional area is: 2 4 A AA = h ( 2 ) = ⎛ --------------⎞ ( 2 ) = ( 5.2216 ) ( 2 ) = 10.4433 in ⎝ cos 40⎠

2

The stresses are N 61.283 σ AA = ----------- = ------------------- = 5.87ksi A AA 10.4433

V 51.42 τ AA = ----------- = ------------------- = 4.924ksi A AA 10.4433 σ AA = 5.87ksi ( T )

3 τ AA = 4.924ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.46 Two rectangular bars of thickness 10 mm are loaded as shown. If the normal stress on plane AA is 180 MPa (C), determine the force F1 and the normal and shear stress on plane BB. 10 mm A 30 mmF1

Fig. P1.46

Solution

B

50 kN

60 mm 75o

A 50 kN

F3 B 65o

σAA= 180 MPa (C) F1 = ? σBB= 1? τBB= ? ------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Figures below show the free body diagrams after the imaginary cuts are made through AA and BB. t (a) NBB 0.01 n (b) 0.01

F1

0.03

75o

NAA

VBB

n

65o

F3

hB 0.06

h 75oVAA

t

65o

By equilibrium of forces in the n-direction in Fig. (a) we obtain: N AA = F 1 sin 75 = 0.9659F 1 . The area of cross-section of plane AA is: A AA = ( h ) ( 0.01 ) = ⎛⎝ -------------⎞⎠ ( 0.01 ) = 0.3106 ( 10 )m . sin 75 0.03

–3

0.9659F

N

2

6

AA 1 The average normal stress on plane AA is: σ AA = ----------= -------------------------------= 180 ( 10 ) or F 1 = 57, 882N or –3

A AA

0.3106 ( 10 )

F 1 = 57.9kN

By equilibrium of forces in Fig. P1.46 we obtain F 1 + F 3 – 100 = 0 or F 3 = 42.1 kN By equilibrium of forces in the n-direction in Fig. (b) we obtain: N BB = F 3 sin 65 = 42.1 sin 65 = 38.16kN By equilibrium of forces in the t-direction in Fig. (b) we obtain: V BB = F 3 cos 65 = 42.1 cos 65 =17.79kN The area of cross-section of plane BB is: A BB = ( h B ) ( 1.5 ) = ⎛⎝ --------------⎞⎠ ( 0.01 ) = 0.662 ( 10 )m . cos 65 0.06

–3

2

3

N BB 6 2 ( 38.16 ) ( 10 ) The average normal stress on plane BB is: σ BB = ---------- = -------------------------------- = 57.64 ( 10 )N ⁄ m or –3

A BB

0.662 ( 10 )

σ BB = 57.6MPa ( T ) 3

V BB 6 2 ( 17.79 ) ( 10 ) The average shear stress on plane AA is: τ BB = ---------- = -------------------------------- = 26.87 ( 10 )N ⁄ m or A BB

–3

0.662 ( 10 )

τ BB = 26.87MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.47 A butt joint is created by welding two plates to transmits a force of P = 250 kN as shown in Figure P1.47. Determine the average normal and shear stress on the plane AA of the weld. 900 mm

P

P A

50 mm

60o

60o

200 mm

A

Fig. P1.47

Solution

-----------------------------------------------------------We draw the FBD below: N 60o

P

h

V 60o

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

By equilbrium N = P sin 60 = 216.5 kN

V = P cos 60 = 125 kN

1

From geometry the cross-sectional area is: –3

–3 –3 2 50 ( 10 ) A AA = h ( 0.2 ) = ⎛ ----------------------⎞ ( 0.2 ) = [ 57.735 ( 10 ) ] ( 0.2 ) = 11.547 ( 10 )m ⎝ sin 60 ⎠

2

The stresses are 3

6 2 216.5 ( 10 ) N σ AA = ----------- = -------------------------------- = 18.75 ( 10 )N ⁄ m –3 A AA 11.547 ( 10 )

6 2 V 125 τ AA = ----------- = -------------------------------- = 10.83 ( 10 )N ⁄ m 3 –3 A AA 11.547 ( 10 )

σ AA = 18.75 MPa ( T )

τ AA = 10.83 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.48 A square tube of 1/4 in thickness is welded along the seam and used for transmiting a force of P = 20 kips as shown in Figure P1.48. Determine average normal and shear stress on the plane AA of the weld.

P 2.5 in

2.5 in

30o

Fig. P1.48

P

Solution

-----------------------------------------------------------We draw the FBD below: N

V h 30o

30o 30

o

P By equilbrium N = P cos 30 = 17.32 kips

V = P sin 30 = 10 kips

1

From geometry the cross-sectional area is: 2 2 2 2 2 2.5 2 2.5 – 2 ⁄ 4 2 A AA = h o – h i = ⎛ --------------⎞ – ⎛ -----------------------⎞ = ( 2.887 ) – ( 2.309 ) = 3 in ⎝ cos 30⎠ ⎝ cos 30 ⎠

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1-34

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The stresses are N 17.32 σ AA = ----------- = ------------- = 5.77 ksi A AA 3

V 10 τ AA = ----------- = ------ = 3.33 ksi A AA 3 σ AA = 5.77 ksi ( T )

3 τ AA = 3.33 ksi

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

a

1.49 (a) In terms of P, a, b and θ, determine the average normal and shear stress on the inclined plane AA shown in Fig. P1.49. (b) Plot the normal and shear stress as a function of θ and determine the maximum value of the normal and shear stress. (c) At what angles of the incline plane the maximum normal and maximum shear stress occur. A

P

b P

θ

A

Fig. P1.49

Solution

σ = f1(P, a, b, θ) τ = f2(P, a, b,q) θσ = ? when σ is max. θτ = ? when τ is max. ------------------------------------------------------------

By making an imaginary cut through AA we obtain the following free body diagram. P

N

a θ

b

t

n

h θ

V

By equilibrium of forces in the n-direction we obtain: N – P cos θ = 0 or N = P cos θ By equilibrium of forces in the t-direction we obtain: P sin θ + V = 0 or V = – P sin θ The area of the incline can be found as: A = ah = a ( b ⁄ cos θ ) P cos θ The average normal stress on the incline plane is: σ = N ---- = --------------------------

2 P σ = ------ cos θ ab

V – P sin θ The average shear stress on the incline plane is: τ = --- = --------------------------

–P τ = ------ sin θ cos θ ab

a ( b ⁄ cos θ )

A

A

a ( b ⁄ cos θ )

The minus sign in the shear stress expression indicates that is acts down the incline. The normal stress and shear stress are plotted below. From the plot it is seen that normal stress is maximum at θ = 0 and shear stress magnitude is maximum at θ = 45o. Thus the answers are:

θσ = 0

o

θ τ = 45

o

σ -----------------( P ⁄ ab ) τ -----------------( P ⁄ ab )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.50 An axial load is applied to a 1 inch diameter circular rod. The shear stress on section AA was found to be 20 ksi. The section AA is at 45o to the axis of the rod. Determine the applied force P and the average normal stress acting on section AA. P

A A

P

Fig. P1.50

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M. Vable

Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

d=1

January 2014

τA = 20ksi P=? σA = ? ------------------------------------------------------------

By making an imaginary cut through AA we obtain the following free body diagram. N

B P

45o o

d=1

45

The length BC can be found as: be found as:

n a t

V 45o

a = d ⁄ ( sin 45 ) = 1.414

C

. The area of the elliptic inclined section can

2 π π A = --- ad = --- ( 1.414 ) = 1.1107in 4 4

The shear force can be found as:

V = τ A A = ( 20 ) ( 1.1107 ) = 22.214kips

By equilibrium of forces in the t-direction we obtain: P sin 45 – V = 0

V 22.214 P = ------------- = ---------------sin 45 sin 45

or

P = 31.4 kips

By equilibrium of forces in the n-direction we obtain: or The average normal stress on the incline plane is:

N = P cos 45 = 31.4 cos 45 = 22.214kips

N 22.214 σ = ---- = ---------------- = 20 A 1.1107

σ = 20ksi ( T )

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Biceps muscle Bone

1.51 A simplified model of a child’s arm lifting weight is shown in Fig. P1.51. The area of cross-section of the biceps muscle is estimated as 2 in2. Determine the average normal stress in the muscle and the average shear force at the elbow joint A.

5 lb

Bone 7 in

6 in

A 2 in

Fig. P1.51

Solution

A = 2in.2

σ=? V=? ------------------------------------------------------------

The free body diagram of the fore arm is shown below. 5 lb.

N θ Ax 2

7

The angle θ can be found as θ = atan ( 6 ⁄ 2 ) = 71.56 By equilibrium of moment about point A we obtain:

Ay

6 θ 2

o

45 ( N sin θ ) ( 2 ) – ( 5 ) ( 9 ) = 0 or N = ------------------------ = 23.717lbs 2 sin 71.56 23.717 The average normal stress in the muscle is: σ = N ---- = ---------------- = 11.858 or A

2

σ = 11.9psi ( T )

By equilibrium of forces in the x-direction we obtain: A x = N cos θ = 23.717 cos 71.56 = 7.5lbs·

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

By equilibrium of forces in the y-direction we obtain: A y = 5 – N sin θ = 5 – 23.717 sin 71.56 = – 17.5lbs 2

V =

2

A x + A y = 19.04

V = 19lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.52 Fig. P1.52 shows a truss and the sequence of assembly of members at pin H, G, and C. All members of the truss have a cross-sectional area of 250 mm2 and all pins have a diameter of 15 mm. Determine (a) the axial stresses in members HA, HB, HG, and HC. (b) the maximum shear stress in pin H. G HC

F

H

GF

HG HA

300

A 3m

C 3m

4 kN

D

E

3m

2 kN

FC FG

GH

300

B

FE FD

GC

HB

Pin G

Pin H

Pin F

3m

3 kN

Fig. P1.52

σHA = ? σHB = ? σHG = ? σHC = ? (τH)max = ? ------------------------------------------------------------

Am = 250 mm2 dp = 15 mm

Solution

The free body diagram of the entire truss can be drawn as shown below.By equilibrium of moment about point E, we obtain: Ay(12) - 4(9) - 2(6) - 3(3) = 0 or Ay = 4.75 kN

1

G F

H

300

300

B

Ay

3m

C 3m

4 kN

Ex

D 3m

2 kN

3m

Ey

3 kN

By equilibrium of forces in the y-direction, we obtain: Ay + Ey - 4 - 2 - 3 = 0 or Ey = 4.25 kN

2

By equilbrium of forces in the x-direction, we obtain: Ex = 0

3

Making an imaginary cut through members HG, HC, and BC we obtain the free body diagram in Fig. (a). By equilibrium of moment about point C, we obtain: [(NHG sin30) + Ay](6) - 4(3) = 0 or NHG = (12.0 - 6(4.75)) / 6sin30 = -5.5 kN NHG

(a)

A y

H

B

NHC

300 B

NBC

(c)

NBH

(b)

NBC

30o 60o 60o

NBA 4 kN

3m

4 kN

NHG

NHA

NHB

NHC

N = -4 kN By equilibrium of moment about point A, we obtain:NHC sin30(6) + 4(3) = 0 or HC Fig. (b) is the free body diagram of joint B. By equilibrium of forces in the y-direction we obtain: NBH = 4 kN Fig. (c) represents the free body diagram of joint H. By equilibrium of forces in the x-direction we obtain:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

NHG cos30 + NHC cos30 - NHA sin60 = 0or (-5.5 - 4) cos 30 = NHA sin60 or NHA = -9.5 kN The axial stresses in each member can be found as shown below.

σHA = NHA/A = (-9.5)(103)/(250)(10-6) = -38(106) N/m2

σHA = 38 MPa (C)

σHB = NHB/A = (4)(103)/(250)(10-6) = 16(106) N/m2

σHB = 16 MPa (T)

σHG = NHG/A = (-5.5)(103)/(250)(10-6) = -22(106) N/m2

σHG = 22 MPa (C)

σHC = NHA/A = (-4)(103)/(250)(10-6) = -16(106) N/m2

σHC = 16 MPa (C)

(b) The free body diagram of pin H is shown in Fig. (d). By making imaginary cuts at different sections of the pin the free body diagrams shown in Figs. (e), (f), and (g) are obtained.

(e)

(d)

(f)

5.5 kN Pin H

60o HC HG

HB

HA

9.5 kN

HC HB

60o 4 kN

HA

(g)

V2y

V1 V2x

HA 9.5 kN

4 kN

V3

4 kN

9.5 kN

4 kN

60o

60o

V1 = 9.5 kN V2x - 9.5 sin60 = 0 V2y + 4 - 9.5 cos60 = 0

From Fig. (e) From Fig. (f)

V2 =

2

(4)

V2x = 8.22 kN V2y = 0.75 kN

2

(5)

V 2x + V 2y = 8.25kN

From Fig (g) V3 = 4 kN (6) From equations (4), (5) and (6), we see that the maximum shear force is 9.5kN, thus the maximum sheer stress will be between members HA and HB. V1 9.5 ( 10 3 ) ( τ H ) max = ---------- = ------------------------ = 53.76 ( 10 6 ) π · 2 A pin --- ( 0.015 ) 4

( τ H ) max = 53.76 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.53 Fig. P1.53 shows a truss and the sequence of assembly of members at pin H, G, and C. All members of the truss have a cross-sectional area of 250 mm2 and all pins have a diameter of 15 mm. Determine (a) the axial stresses in members FG, FC, FD, and FE (b) the maximum shear stress in pin F. G F

H

GF

HC HG

0

30

B

A 3m

4 kN

C 3m

D 3m

2 kN

HA

3m

Pin H

FC FG

GH

E

FE FD

GC

HB

300

Pin G

Pin F

3 kN

Fig. P1.53

Solution

σFG = ? σFC = ? σFD = ? σFE = ? ------------------------------------------------------------

Am = 250 mm2 dp = 15 mm

(τF)max = ?

The free body diagram of the entire truss can be drawn as shown below in Fig. (a) .By equilibrium of moment about point E, we obtain: Ay(12) - 4(9) - 2(6) - 3(3) = 0 or

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Ay = 4.75 kN

1

G

(a)

F

H

300

300

B

Ay

3m

C 3m

4 kN

Ex

D 3m

2 kN

3m

Ey

3 kN

By equilibrium of forces in the y-direction, we obtain: Ay + Ey - 4 - 2 - 3 = 0 or Ey = 4.25 kN

2

By equilbrium of forces in the x-direction, we obtain: Ex = 0

3

We can make an imaginary cut through members FG, FC, and DC to obtain the free body diagram shown in Fig. (b). NFG

(b)

(d)

(c) F

NDC C

NFG

NDF

NFC E

0

30

NDC

NDE

x

30o

D 3m

3m

3 kN

E

30o

y

3 kN

NFC

60o 60o NFD

NFE

By equilibrium of moment about point C in Fig. (b), we obtain: NFG sin30)(6) + (4.25)(6) - 3(3) = 0 or NFG = -5.5 kN

4

By equilibrium of moment about point E, we obtain:NFC sin30)(6) + 3(3) = 0 NFC = -3 kN

5

Fig. (c) is the free body diagram of joint D. By equilibrium of forces in the y-direction we obtain: NDF = 3 kN

6

Fig. (d) represents the free body diagram of joint F. By equilibrium of forces in the x-direction we obtain: NFE sin60 - NFG cos30 - NFC cos30 = 0 or NFE = -8.5 kN

7

The axial stresses in each member can be found as shown below. σFG = NFG / A = -5.5(103) / 250(10-6) = -22(106) N/m2

σFC = NFC / A = -3(103) / 250(10-6) = -12(106) N/m2 σFD = NFD / A = 3(103) / 250(10-6) = 12(106) N/m2 σFE = NFE / A = -8.5(103) / 250(10-6) = -34(106) N/m2 The answers are: σ FG = 22 MPa (C) ; σ FC = 12 MPa (C) ; σ FD = 12 MPa (T) ; σ FE = 34 MPa (C) The free body diagram of pin F is shown in Fig. (e). By making imaginary cuts at different sections of the

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

pin the free body diagrams shown in Figs. (f), (g), and (h) are obtained.

(e)

Pin F

(f)

3kN

5.5 kN

60o

60o

FD

(h) FE

60o

FE 60o

FC

(g)

5.5 kN

FD

8.5 kN

FG

FG

V1

3kN

FE o

8.5 kN

60

60o

V2x V2y

8.5 kN

V3

3kN

V1 = 5.5 kN

From Fig. (f) From Fig. (g)

8

V2x - 8.5 sin60 = 0 -V2y - 3 + 8.5 cos60 = 0 V2 =

2

V2x = 7.36 kN V2y = 1.25 kN

2

V 2x + V 2y = 7.466kN

9

From Fig (h) V3 = 8.5 kN 10 From equations (8), (9) and (10), we see that the maximum shear force is 8.5kN, thus the maximum sheer stress will be between members FD and FE. V3 8.5 ( 10 3 ) = 48.1 ( 10 6 ) ( τ F ) max = ---------- = -----------------------π · 2 A pin --- ( 0.015 ) 4

( τ F ) max = 48.1 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.54 Fig. P1.54 shows a truss and the sequence of assembly of members at pin H, G, and C. All members of the truss have a cross-sectional area of 200 mm2 and all pins have a diameter of 10 mm. Determine (a) the axial stresses in members GH, GC, and GF (b) the maximum shear stress in pin G G F

H

HC

GF

HG 300

B

A 3m

C 3m

4 kN

D 3m

2 kN

HA

E 3m

FC FG

GH Pin G

Pin H

Pin F

Fig. P1.54

3 kN

Am = 200mm2 dp = 10mmσGH = ?

σGC = ? σGF = ? ------------------------------------------------------------

Solution

FE FD

GC

HB

300

(τG)max = ?

The free body diagram of the entire truss can be drawn as shown below in Fig. (a) .By equilibrium of moment about point E, we obtain: Ay(12) - 4(9) - 2(6) - 3(3) = 0 or Ay = 4.75 kN G

(a)

(b)

H

300 Ay

NHG

300

3m

4 kN

C 3m

2 kN

Ex

D 3m

3m

3 kN

(c) G

H

F

B

1

Ey

A y

600 600

NHC

300 B

NBC

3m

NGF

NGH NGC

4 kN

Making an imaginary cut through members HG, HC, and BC we obtain the free body diagram in Fig. (b). By equilibrium of moment about point C, we obtain: [(NHG sin30) + Ay](6) - 4(3) = 0 or NHG = (12.0 - 6(4.75)) / 6sin30 = -5.5 kN

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

2

NHG = -5.5 kN = NGH

The free body diagram of the joint G can be drawn as shown in Fig. (c). By equilibrium of forces in the x-direction we obtain: -NGH sin60 + NGF sin60 = 0 or NGF = NGH = -5.5 kN

3

By equilibrium of forces in the y-direction we obtain: -NGH cos60 - NGF cos60 - NGC = 0 or NGC = 5.5 kN

4

The axial stresses in each member can be found as shown below. σGH = 27.5 MPa (C)

σGH = - 5.5(10-3) / 200(10-6) = - 27.5(106) N/m2

σGC = 27.5 MPa (T)

σGC = 5.5(10-3) / 200(10-6) = 27.5(106) N/m2

σGF = 27.5 MPa (C)

σGF = - 5.5(10-3) / 200(10-6) = - 27.5(106) N/m2

The answers are: σ GH = 27.5 MPa (C) ; σ GC = 27.5 MPa (T) ; σ GF = 27.5 MPa (C) The free body diagram of pin G is shown in Fig. (d). By making imaginary cuts at different sections of the pin the free body diagrams shown in Figs. (e) and (f) are obtained. By force equilibrium in Figs. (e) and (f) we obtain:V1 = 22 kN V2 = 22 kN Magnitude of shear stress is constant though the direction changes. The maximum shear stress is ( 5.5 ) ( 10 3 ) ( τ G ) max = -------------------------- = 70.03 ( 10 6 ) = 70.03MPa π --- ( 0.01 ) 2 4

(d)

Pin G

( τ G ) max = 70 MPa

V1

(e)

GF

GC

(f)

V2 GF

GH

GH

60o 5.5 kN o 5.5 kN 60

60o 5.5 kN

o 5.5 kN 60

5.5 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.55 The pin at C is has a diameter of 1/2 in. and is in double shear. The area of cross-sections for members AB and BC are 2 in2 and 2.5 in2. Determine the axial stress in member AB and the shear stress in pin C. B

80 lb./in

C

60 in

A

60o 66 in

Fig. P1.55

Solution

σAB = ? τc = ? ------------------------------------------------------------

dc = 1/2 in.

AAB = 2 in

2

ABC = 2.5 in2

The free body diagram of member BC can be drawn with the distributed force replaced by an equivalent

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

=

52 80

lb s

force as shown in Fig (a). (a) (8 0) (6 6)

(b)

NAB

B

33

January 2014

V V

Cx

33 Cy

60o Cx

C Cy

By equilibrium of moment about point C we obtain: (5280)(33) - NAB(66sin60) = 0 or N AB = 3048.4lbs

1

By equilibrium of forces in the x-direction we obtain: Cx + NAB - 5280cos30 = 0 or Cx = 1524.2 lbs.

2

By equilibrium of forces in the y-direction we obtain-Cy + 5280sin30 = 0 Cy = 2640 lbs.

3

N AB 3048.4 The axial stress is: σ AB = ---------- = ---------------- = 1524.2psi or A AB 2

σ AB = 1524 psi ( T )

The free body diagram in Fig. (b) shows pin C in double shear. By equilibrium of force we obtain: 2V =

2

2

C x + C y = 1524.2

V1524.2 The shear stress in pin C is: τ C = -----= ------------------------------ = 7763psi

τ C = 7763psi

2

AC

( π ( 0.5 ) ⁄ 4 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.56 All pins shown are in single shear and have a diameter of 40 mm. All members have square crosssections. Determine the maximum shear stress in the pins and the axial stress in member BD. 50 kN/m A

B 50 mm

3m

0m 20

m

C

150 mm

D E

2.5 m

2.5 m

Fig. P1.56

σBD = ? τmax = ? ------------------------------------------------------------

Solution

d = 40mm

The free body diagram of the entire structure can be drawn as shown in Fig. (a). By equilibrium of moment about point A we obtain: Ex(3) - 250(2.5) = 0 or Ex = 208.33 kN

(a)

250 kN

Ay Ax

B

Cy

(b) C

NBD

Cx

150 mm

50 mm 3m Ex 0 20

E mm

D

D Ex 2.5 m

2.5 m

E

2.5 m

2.5 m

By equilibrium of force in the x-direction we obtain:Ax = Ex = 208.33 kN or Ax = 208.33 kN

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

By equilibrium of force in the y-direction we obtain:Ay = 250 kN The free body diagram of member CE can be drawn as shown in Fig. (b) By equilibrium of moment about point C we obtain: NBD(2.5) - 208.33(3) = 0 or NBD = 250 kN 250 ( 10 ) - = 100 ( 10 6 ) N ⁄ m or σ BD = 100MPa ( T ) The axial stress in member BD is: σ BD = -----------------------------3

2

( 0.05 ) ( 0.05 )

By equilibrium of force in the x-direction in Fig. (b) we obtain: Cx = Ex = 208.33 kN or Cx = 208.33 kN By equilibrium of force in the x-direction in Fig. (b) we obtain: Cy = NBD = 250 kN or Cy = 250 kN 2

A x + A y2 = 325kN

The resultant force at A: F A = The resultant force at F C =

2

C x + C y2 = 325kN

The force on the pin B and D will equal NBD. Thus, pins A & C will have maximum shear. 2 325 ( 10 3 ) τ max = --------------------------------- = 258.6 ( 10 6 ) N ⁄ m 2 ( π ⁄ 4 ) ( 0.04 )

τ max = 259MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.57 A student athlete is lifting weight W = 36 lbs as shown in Figure P1.57a. The weight of the athelete is WA = 140lb. A model of the student pelvis and legs is shown in Figure P1.57b. The weight of legs and pelvis WL = 32 lb acts at the center of gravity G. Determine the normal stress in the erector spinae

8. 2

in 1.5 in

muscle that supports th trunk if the average muscle area at the time of lifting the weight is 1.75 in2. Erector spinae muscles Hip Joint A

in

G

10

B

45o

45o

WL

20 in

(W+WA)/2 3 in

Fig. P1.57

Solution

------------------------------------------------------------

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

We draw the FBD as below From equilibrium of moment about A Ax

N ( 1.5 ) + 32a – 88b = 0

1

a = 8.2 cos 45 = 5.798 in

2

b = 18.2 cos 45 + 3 = 15.869 in

3

From geometry

8. 2

in 1.5 in

N 45o

Ay

10

in

G

Substituting a and b we obtain

a B

45o

( 88 ) ( 15.869 ) – 32 ( 5.798 ) N = --------------------------------------------------------------- = 807.31 lb 1.5

32 lb

Thus, the stress in the muscle is: N 807.31 σ = ---- = ---------------- = 461.3 psi A 1.75

20 in

σ = 461.3 psi ( T )

88 lb 3 in

b

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.58 A student is exercising his shoulder muscles using a W = 15 lb dumbell as shown in Figure P1.58a. The model of the student arm is shown in Figure P1.58b. The weight of the arm of WA = 9 lb acts at the center of gravity G. Determine the average normal stress in the deltoid muscle if the average area of the muscle is 0.75 in2 at the time the weight is in the horizontal position. (a)

(b) G

A 12 in

B 6 in

scle id mu delto

O 6 in

WA

W

15o Shoulder joint

Fig. P1.58

Solution

-----------------------------------------------------------We draw the FBD as below N G

A

15o

B

12 in

6 in

6 in

WA

W

Ox

Oy

From equilibrium of moment about O N sin 15 ( 6 ) – ( 6 )W A – ( 24 )W = 0

or

( 6 ) ( 9 ) + ( 24 ) ( 15 )- = 266.6 lb N = ------------------------------------------6 sin 15

1

Thus, the stress in the muscle is: N 266.6 σ = ---- = ------------- = 354.7 psi A 0.75

2 σ = 354.7 psi ( T )

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

1.59 The bottom screw in a hook supports 60% of the load P, while the remaining 40% of P is carried by the top screw. The shear strength of the screws is 50 MPa. Develop a table for maximum load P that the hook can support for screw diameters that vary from 1 mm to 5 mm in steps of 1 mm.

P

Fig. P1.59

Solution

τ max ≤ 50MPa

P vs. d = ?

-----------------------------------------------------------The shear force on the bottom screw is Vbottom = 0.6P V bottom 6 0.6P - = --------------------- ≤ 50 ( 10 ) τ max = -----------------2 A ( πd ⁄ 4 )

or

6 P ----- ≤ 65.45 ( 10 ) 2 d

The table below list the values of d and P that satisfies the above inequality. d (mm)

P (Newton)

1

65

2

261

3

589

4

1047

5

1636

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.60 Determine the maximum force P that can be transferred by the riveted joint shown in Figure P1.60 if the limits apply: maximum normal stress in the attached members can be 30 ksi., maximum bearing stress between the pins and members can be 15 ksi, and the maximum shear stress in the rivet can be 20 ksi. P

1 in

2 in

1 in

P

0.5 in

P

0.5 in

P

0.5 in 0.5 in

Fig. P1.60

Solution

------------------------------------------------------------

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

We make imaginary cuts and obtain the FBD’s below. V

P

P

P

Nb

N

V By equilibrium we obtain: P V = --2

N = P

Nb = P

1

The stresses1 should be below the limiting values P⁄2 V P⁄2 τ r = ------ = ------------------------------ = ---------------- ≤ 20 ksi 2 0.1963 Ar ( π ⁄ 4 ) ( 0.5 ) P N σ = ---- = -------------------- ≤ 30 ksi ( 1 ) ( 0.5 ) A

or

or

P ≤ 7.852 kips

2

P ≤ 15 kips

N P P P σ b = ------b = -------------------------- = --------------------------------------- = ---------------- ≤ 15 ksi 0.3927 ( π ⁄ 2 ) ( t )d r ( π ⁄ 2 ) ( 0.5 ) ( 0.5 ) Ab

or

3 P ≤ 5.891 kips

4

The maximum value of P is P max = 5.8 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.61 A tire swing is suspended using three chains. Each chain makes an angle of 12o with the vertical. The chain is made from links as shown. For purpose of design, assume that more than one person may use the swing and hence the swing is to be designed to carry a weight of 500-lb. If the maximum average normal stress in the links is not to exceed 10 ksi, determine to the nearest 1/16 inch the diameter of the wire that should be used for constructing the links.

12o

Fig. P1.61

Solution

σ max ≤ 10ksi

W = 500-lb

d=?

to nearest 1/16th of an inch

-----------------------------------------------------------The free body diagram of the tire and the chain link is shown below. By force equilibrium in the y-direction we obtain: 3T cos12 = 500 or T = 170.4bs

1. In the calcuation of bearing stress, a better alternative is to use the projected bearing area (tdr) which assume a uniform pressure acting on the hole boundary.

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

By force equilibrium on the link we obtain: 2N =T

or

12o T

N = 85.195 lbs. T

(b)

(a)

January 2014

T

T N N

W = 500 lb 3 85.195 The average axial stress in the link is: σ = N ---- = --------------------- ≤ 10 ( 10 ) 2

A

( πd ⁄ 4 )

or

d ≥ 0.1041inch 1 d min = --- inch 8

The minimum d to the nearest 1/16th of an inch is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.62 Two cast iron pipes are held together by a bolt as shown. The outer diameters of the two pipes are 50 mm and 70 mm and the wall thickness of each pipe is 10 mm. The diameter of the bolt is 15 mm. What is the maximum force P this assembly can transmit if the maximum permissible stresses in the bolt and cast iron are 200 MPa in shear and 150 MPa in tension, respectively. P

P

Fig. P1.62

Solution

σ cast ≤ 150MPa

τ bolt ≤ 200MPa

Pmax = ?

d bolt = 15mm

-----------------------------------------------------------The smaller pipe will have higher normal stress. The free body diagrams of the smaller pipe and the pipe with bolts cut by imaginary sections are shown below. V (b) (c) (a) P N P

V

By force equilibrium in Fig. (a) we have N = P The area of cross-section of the cast iron pipe is minimum at the bolt hole as shown in Fig. (c).The crosssectional area is: Acast = π(0.052 - 0.032)/4 - 2(0.01)(0.015) = 0.9566(10-3)m2 The average normal stress in the cast iron pipe is as shown below. 6 P N σ cast = ------------- = -------------------------------- ≤ 150 ( 10 ) –3 A cast 0.9566 ( 10 )

or

3

P ≤ 143.49 ( 10 )Newtons

1

By force equilibrium in Fig. (b) we have2V = P or V = P/2 2 The area of cross-section of the bolt is: A = π(0.015) /4 = 176.7(10-6)m2 6 (P ⁄ 2) V τ bolt = ------------- = ---------------------------------- ≤ 200 ( 10 ) –6 A bolt ( 176.7 ) ( 10 )

or

3

P ≤ 70.68 ( 10 )Newtons

The maximum value of P that satisfies inequalities in Eqs. (1) and (2) is:

2

P max = 70.6kN

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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1-47

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

1.63 A normal stress of 20 ksi is to be transferred from one plate to another by riveting a plate on top. The shear strength of the 1/2 inch rivets used is 40 ksi. Assuming all rivets carry equal shear stress, determine the minimum even number of rivets that must be used. 1 in

8 in 1 in σ = 20 ksi

1 in

σ = 20 ksi

6 in

Fig. P1.63

Solution

σ = 20 ksi drivet = 1/2 τ rivet ≤ 40ksi 2n= number of rivets =? ------------------------------------------------------------

We can draw the following free body diagram. n rivets

V

V V

V V

1 in

V

V

8 in

σ = 20 ksi

V

By equilibrium of forces we have: nV = (20)(8)(1) or V= 160/n kips The area of cross-section of each rivet is: Arivet = (π/4)(1/2)2 = 0.1963 in2 The average stress in each rivet can be written as: 160 V τ rivet = -------------- = ------------------------ ≤ 40 n ( 0.1963 ) A rivet

or

160 n ≥ -------------------------------( 40 ) ( 0.1963 )

or

n ≥ 20.37 or n = 21 rivets 2n = 42 rivets

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.64 Two possible joining configurations are to be evaluated. The forces on the joint A in a truss were calculated and are as shown. The diameter of the pin is 20 mm. Determine which joint assembly is better by calculating the maximum shear stress in the pin for each case NC = 50 kNN = 67.32 kN B ND= 30 kN NA = 32.68 kN

NA = 32.68 kN

NB = 67.32 kN 30o

30o

Solution

Configuration 2

NC = 50 kN

Configuration 1

dp = 20mm

30o

ND= 30 kN

30o

configuration = ?

-----------------------------------------------------------Configuration 1 Fig. (a) below shows the free body diagram of the entire pin. Figs. (b), (c), and (d) show the free body diagrams after making the imaginary cut at different sections of the pin. From Fig. (b) V1 = 32.68 kN From Fig. (c)

V2x = 30 kN

V2y = 50 kN. The resultant shear force is: V 2 =

2

2

30 + 50 = 58.31kN

From Fig. (d) V3 = 30 kN. The maximum shear force is V2. Thus, the maximum shear stress is: 3 V 6 2 58.31 ( 10 ) ( τ 1 ) max = -----2- = ---------------------------- = 185.6 ( 10 ) N ⁄ m 2 A π ( 0.02 ) ⁄ 4

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1-48

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

(a) NA = 32.68 kN 30

(b)

NC = 50 kN

30o

o

(c)

NA = 32.68 kN

ND= 30 kN NB = 67.32 kN

January 2014

(d)

NC = 50 kN

ND= 30 kN

ND= 30 kN

30o

V3 V1

V2x V2y

Configuration 2 Fig. (e) shows the free body diagram of the entire pin. Figs. (f), (g), and (h) show the free body diagrams after making the imaginary cut at different sections of the pin.

(e)

NC = 50 kN

NA = 32.68 kN 30

(f)

NB = 67.32 kN 30o ND= 30 kN

(g)

NA = 32.68 kN 30

30o

o

V1

(h)

NB = 67.32 kN

o

ND= 30 kN

V3

ND= 30 kN

V2x V2y

From Fig (f)

V1 = 32.68 kN

From Fig (g)V2x = 28.3 kN V2y = 33.66 kN

The resultant shear force is: V 2 =

2

2

28.3 + 33.66 = 43.97kN

From Fig (h) V3 = 30 kN The maximum shear force is V2. Thus, the maximum shear stress is: 3 V 6 2 43.97 ( 10 ) ( τ 2 ) max = -----2- = ---------------------------- = 139.98 ( 10 ) N ⁄ m 2 A π ( 0.02 ) ⁄ 4

2

Comparing Eqs. (1) and (2) we see that configuration 2 has lower shear stress. Configuration 2 is better.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.65 Truss analysis showed the following forces at joint A. Determine the sequence in which the three members at joint A should be assembled, so that the shear stress in the pin is minimum. ND =22.94 kips NC = 40 kips 35o A

NB= 32.77 kips

Fig. P1.65

Solution

Sequence = ?

-----------------------------------------------------------There are 6 possible sequences. Three of the sequences are shown below. The remaining three are the mirror image of the three shown and hence will result in the same shear stress.

(b)

(a)

NC = 40 kips

ND =22.94 kips

2

30o

1

(c)

ND =22.94 kips 2

30o

1 NB= 32.77 kips

ND =22.94 kips

NC = 40 kips 2 1 NB= 32.77 kips

NC = 40 kips 30o NB= 32.77 kips

The shear force at the imaginary cuts 1 and 2 will equal to the force at the nearest end. Thus, if smaller forces are on the outside, then the shear stress will be smaller. Configuration shown in Fig.(c) will yield the smallest shear stress.

Member should be assembled as in Fig.(c).

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

1.66 An 8 in x 8 in reinforced concrete bar needs to be designed to carry a compressive axial force of 235 kips. The reinforcement is done using 1/2 inch round cast iron bars. Assuming normal stress in concrete to be a uniform maximum value of 3 ksi and in iron bars to be a uniform value of 20 ksi, determine the minimum number of iron bars that are needed. Solution N ≥ 235kips ds = 1/2 in. bar = 8in x 8in σconc = 3 ksi (C) σs = 20 ksi (C) nmin = number of steel bars = ?

-----------------------------------------------------------The cross-section of the reinforced concrete bar is shown. 8 8

The material area of steel is: As = n (π/4)(1/2)2 = 0.19635n in2 The material area of concrete is: Aconc = (8)(8) - 0.19635n The axial stress in steel is: σs = Ns /As = 20 ksi or Ns = 3.927n kips The axial stress in concrete is: σconc = Nconc/Aconc = 3 ksi or Nconc = (192 - 0.58905n) kips. The total axial force that the bar can carry is: N = N s + N conc = 192 + 3.3379n ≥ 235 or n ≥ 12.88 or n min = 13

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.67 A wooden axial member has a cross-section of 2 in x 4 in. The member was glued along the line AA as shown in Fig. P1.67. Determine the maximum force P that can be applied to the repaired axial member if the maximum normal stress in the glue cannot exceed 800 psi and the maximum shear stress in the glue cannot exceed 350 psi. P

A 40o A 4 in

Solution

Fig. P1.67 σ ≤ 800 psi

cross-section = 2in x 4in

τ ≤ 350 psi

Pmax = ?

-----------------------------------------------------------The free body diagram of a lower part of the wedge is shown below. A

V

N t

n

h 4 inch

40o 40o A

50o P

By equilibrium of forces in the n-direction we have: N = Psin50o = 0.7660P By equilibrium of forces in the t-direction we have: V = Pcos50o = 0.6428P The area of the incline plane is: A = 2h = 2(4/cos40) = 10.44 in2

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

0.766P The normal stress on the incline is: σ = N ---- = ----------------- ≤ 800 0.6428P V The shear stress on the incline is: τ = --- = -------------------- ≤ 350

P ≤ 5684lbs

or

10.44

A

P ≤ 10902lbs

or

10.44

A

January 2014

The maximum value of P that satisfies the above two inequalities is:

P max = 5684 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.68 .An adhesively bonded joint in wood is fabricated as shown in Fig. P1.68. The length of bonded region L = 5 inches. Determine the maximum force P the joint can support if the shear strength of adhesive is 300 psi and the wood strength is 6 ksi in tension. P

P

8 in 1 in 1 in

1 in L

Fig. P1.68

Solution

σ W ≤ 6000psi

L = 5 in.

τ glue ≤ 300psi

Pmax = ?

-----------------------------------------------------------Fig. (a) shows the free body diagram after making an imaginary cut though the wood.

(a)

(b)

P

V

P

8 in

N

1 in

V L/2

By equilibrium of forces we have: N = P P N - = --------------- ≤ 6000 The average normal stress in wood is: σ W = --------

or

(8 )(1 )

AW

P ≤ 48000 lbs

Fig. (b) shows the free body diagram after imaginary cuts are made through the glue on the top and bottom surface of the wood. By equilibrium of forces we have: 2V = P P V The average shear stress in the glue is: τ glue = ------------= ------------------------------- ≤ 300 A glue

( 2)(8)(5 ⁄ 2 )

or

P ≤ 12000 lbs

The maximum value of P that satisfies equations the two inequalities above is: P max = 12 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.69 The joint in Fig. P1.69 is to support a force P = 25 kips, what should be the length L of the bonded region if the adhesive strength in shear is 300 psi. P

P

8 in 1 in 1 in

1 in L

Solution

P = 25 kips

Fig. P1.69 τ glue ≤ 300psi

L =?

-----------------------------------------------------------The FBD is shown below. V V

P

8 in 1 in

L/2

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

from equilibrium: 2V = P = 25(103) or V = 12.5(103) lbs. 3

12.5 ( 10 ) V The average shear stress is: τ glue = ------------= ------------------------ ≤ 300

L ≥ 10.412 inch orL = 10.4 in

or

( 8 )( L ⁄ 2)

A glue

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.70 The normal stress in the members of the truss shown in Fig. P1.70 is to be limited to 160 MPa in tension or compression. All members have a circular cross-section. The shear stress in the pins is to be limited to 250 MPa. Determine (a) the minimum diameters to the nearest millimeter of members ED, EG, and EF. (b) the minimum diameter of pin E to the nearest millimeter and the sequence of assembly of members ED, EG, and EF. 4m

4m C

D

E

G

F

3m B 3m A

Fig. P1.70 σ ≤ 160MPa

Solution

dEP = ?

40 kN

65 kN H

dEF = ? τ E ≤ 250MPa

dEG = ?

dE = ?

-----------------------------------------------------------Figs. (a) and (b) shows the free body diagrams of joints F and E. By equilibrium of force in Fig. (a) in the y-direction we obtain: NEF = 65kN NEF

(a)

(b)

θ

40 kN

NFG

4

(c)

NED E

3

θ 5

NEG

F

NEF

65 kN

By equilibrium of force in Fig. (b) in the y-direction we obtain: NEGsinθ + NEF = 0 or NEG = -108.3 kN By equilibrium of force in Fig. (b) in the y-direction we obtain: NED + NEGcosθ= 0 or NED = 86.67kN N ED 86.67 ( 10 3 ) -) σ ED = -----------------------≤ 160 ( 10 6 ) ) - = ---------------------------2 2 πd ( ) ⁄ 4 ( πd ED ) ⁄ 4 ED

or

d ED ≥ 26.26 ( 10 )m or d ED = 27 mm

N EG 108.3 ( 10 3 ) ) - ≤ 160 ( 10 6 ) ) σ EG = ------------------------ = ---------------------------2 2 πd ( ) ⁄ 4 ( πd EG ) ⁄ 4 EG

or

d EG ≥ 29.36 ( 10 )m or d EG = 30 mm

N EF 65 ( 10 3 ) )σ EF = ----------------------≤ 160 ( 10 6 ) ) - = ----------------------2 2 πd ( ) ⁄ 4 ( πd EF ) ⁄ 4 EF

or

–3

–3

–3

d EF ≥ 22.7 ( 10 )m or d EF = 23 mm

There are 6 possible assembly sequences. Three of the sequences are shown below. The remaining three are mirror images and will result in the same shear stress. The shear force at the imaginary cuts 1 and 2 will equal to the force at the nearest end. Thus, if smaller forces are on the outside, then the shear stress will be smaller. Assembly in Fig. (b) will yield the smallest shear stress. The sequence of assembly shown in Fig. (b) should be used.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

108.3 kN

(d) 86.67 kN

86.67 kN

(e)

108.3 kN ED

EG EF

EG

ED 2 1

EF 1

(f)

January 2014

86.67 kN ED EF 108.3 kN EG 2 1

2

65 kN 65 kN

65 kN

In Fig. (b) the max shear force is at cut 2 and its value is: V = 86.67 kN 86.67 ( 10 3 ) ) V - ≤ 250 ( 10 6 ) ) τ E = -------------------- = ---------------------------2 2 πd ( ) ⁄ 4 ( πd E ) ⁄ 4 E

–3

d E ≥ 21.01 ( 10 )m or d E = 22 mm

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.71 The normal stress in the members of the truss shown in Fig. P1.71 is to be limited to 160 MPa in tension or compression. All members have a circular cross-section. The shear stress in the pins is to be limited to 110 MPa. Determine (a) the minimum diameters to the nearest millimeter of members CG, CD, and CB. (b) the minimum diameter of pin C to the nearest millimeter and the sequence of assembly of members CG, CD, and CB.. 4m

4m D

E

B

G

F

3m A

H

C 3m

40 kN

65 kN

Fig. P1.71

Solution

σ ≤ 160MPa

dCG = ?

dCB = ? τ E ≤ 250MPa

dCD = ?

dC = ?

-----------------------------------------------------------Figs. (a) and (b) shows the free body diagrams of joints D and C. D NCD (a) (b) (c) NCD

NED NDG

C

θ

NCB

(d)

4 θ

3

5

NCG

CD

86.67 kN

CG 2

CB 1

108.3 kN

From problem 1-72 we know NED = 86.67 kN By equilibrium of force in Fig. (a) in the x-direction we obtain: NCD = NED = 86.67 kN By equilibrium of force in Fig. (b) in the x-direction we obtain: NCG cosθ + NCD = 0 or NcG = -108.3 kN By equilibrium of force in Fig. (b) in the y-direction we obtain: NCB + NCG sinθ= 0 or NCB = 65.0kN N CG 108.3 ( 10 3 ) ) - ≤ 160 ( 10 6 ) ) σ CG = ------------------------ = ---------------------------2 2 πd ( ) ⁄ 4 ( πd CG ) ⁄ 4 CG

or

d CG ≥ 29.36 ( 10 )m or d CG = 30 mm

N CD 86.67 ( 10 3 ) ) - ≤ 160 ( 10 6 ) ) σ CD = ------------------------ = ---------------------------2 2 ( ) ⁄ 4 πd ( πd CD ) ⁄ 4 CD

or

d CD ≥ 26.26 ( 10 )m or d CD = 27 mm

N CB 65 ( 10 3 ) )σ CB = -----------------------≤ 160 ( 10 6 ) ) - = -----------------------2 2 ( πd CB ) ⁄ 4 ( πd CB ) ⁄ 4

or

–3

–3

–3

d CB ≥ 22.7 ( 10 )m or d CB = 23 mm

The configuration that will produce the smallest maximum shear stress will be the one with members car-

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

rying the smallest forces on the outside edge of the pin. Free body diagram of pin C is shown below in Fig. (d) . The sequence of assembly shown in Fig. (d) should be used. In Fig. (d) the max shear force is at cut 2 and its value is: V = 86.67 kN 86.67 ( 10 3 ) ) V τ C = --------------------= ----------------------------- ≤ 250 ( 10 6 ) ) 2 2 ( πd C ) ⁄ 4 ( πd C ) ⁄ 4

–3

d C ≥ 21.01 ( 10 )m

or

or d C = 22 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.72 Truss analysis showed the following forces at joint A. Determine the sequence in which the four members at joint A should be assembled to minimize the shear stress in the pin. NC = 27.58 kips

NB= 40 kips

ND = 25 kips

NE = 28.34 kips

65o A

Fig. P1.72

Solution

assembly sequence = ?

-----------------------------------------------------------There are factorial four i.e., twenty-four possible ways of assembling the four member. Twelve of the twenty-four sequence are mirror images of the other twelve. To consider the twelve assembly sequence, we consider the figure below, where a member may be in any of the four location shown. 4 3 2 1

The table below describe the location of the four members and the maximum shear stress location and value of maximum shear stress for each sequence. Sequence Number

Member Location 1

Member Location 2

Member Location 3

Member Location 4

Location of Maximum shear force between members

Maximum Shear Force (kips)

1

AB

AC

2

AB

AC

AD

AE

1 and 2

40

AE

AD

1 and 2

40

3

AC

AB

4

AC

AB

AD

AE

2 and 3

37.8

AE

AD

2 and 3

37.8

5

AB

AD

6

AB

AD

AE

AE

2 and 3

47.17

AE

AC

2 and 3

47.17

7

AD

AB

8

AD

AB

AC

AE

2 and 3

47.17

AE

AC

2 and 3

47.17

9

AB

AE

AC

AD

1 and 2

40

10

AB

AE

AD

AC

1 and 2

40

11

AE

AB

AC

AD

1 and 2

28.34

12

AE

AB

AD

AC

1 and 2

28.34

The maximum shear stress is smallest for sequence 11 and 12, thus we have the following conclusion.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The members should be assembled in the following sequence or the mirror image of the given sequence AE, AB, AD, AC or AE, AB, AD, AC 1.73 The post shown has a rectangular cross-section 2 in x 4 in. The length L of the post buried in ground is 12 inches and the shear strength of the soil is 2 psi. Determine the force P needed to pull the post out of the ground.

Post

P

L

Ground

Fig. P1.73

Solution

τ = 2 psi Pmin = ? ------------------------------------------------------------

Post 2 in x 4 in

L = 12 in

The following free body diagram can be drawn by making an imaginary cut around the surface of the post. P

τ 12 in 4 in 2 in

By equilibrium of forces the applied force should equal the average shear stress multiplied by the surface area as shown below. P P = ( τ ) [ 12 ( 2 + 4 + 2 + 4 ) ] = 144τ or τ = --------- ≤ 2 psi or 144

P ≤ 288

or

P min = 288 lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.74 The post shown in 1.73 has a circular cross-section of diameter 100 mm. The length L of the post buried in ground is 400 mm. It took a force of 1250 N to pull the post out of ground. What was the average shear strength of the soil?

Post

P

L

Ground

Fig. P1.74

Solution

d = 100 mm = 0.1 m

L = 400 mm = 0.4 m

P = 1250 N

τ=?

------------------------------------------------------------

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The following free body diagram can be drawn by making an imaginary cut around the surface of the post. P

τ

0.4

0.1

By equilibrium of forces the applied force should equal the average shear stress multiplied by the surface area as shown below. 2 1250 P = τ ( 0.4 ) [ π ( 0.1 ) ] = 0.04πτ or τ = -------------- = 9947 N ⁄ m or 0.04π

τ = 9947 Pa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.75 The cross-section of the post shown in 1.73 is an equilateral triangle with each side of dimension ‘a’. If the shear strength of the soil is τ, determine the force P needed to pull the post out of the ground in terms of τ, L, and a.

Post

P

L

Ground

Solution

P = f( τ , h , a ) = ?

-----------------------------------------------------------The following free body diagram can be drawn by making an imaginary cut around the surface of the post. P

L

τ

a

τ

a

By equilibrium of forces the applied force should equal the average shear stress multiplied by the surface area as shown below. P = τ ( 3a ) ( L ) or P = 3aLτ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.76 A force P = 10 lbs is applied to the handle of a hammer in an effort of pulling the nail out of the wood as shown in Fig. P1.76. The nail has a diameter of 1/8 inch. Determine the shear stress acting on the

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

nail. P

12 in

2 in

Fig. P1.76

Solution

τ=? ------------------------------------------------------------

P = 10 lbs

d = 118 in

L = 2 in

The following free body diagram can be created. P

12 in Cy Cx P1

2 in

By moment equilibrium about point C, we obtain: P 1 ( 2 ) = P ( 12 ) or P 1 = 6P = 60 lbs 2

The surface area of the nail in the wood is: A = LπD = ( 2 )π ( 1 ⁄ 8 ) = 0.7854 in The shear force acting on the nail is V = P1. The shear stress acting on the nail can be found as: V 60 τ = ---- = ---------------- = 76.39 psi or A 0.7854

τ = 76.4 psi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.77 Two cast iron pipes are adhesively bonded together over a length of 200 mm. The outer diameters of the two pipes are 50 mm and 70 mm and wall thickness of each pipe is 10 mm. The two pipes separated while transmitting a force of 100 kN. What was the shear stress in the adhesive just before the two pipes separated.

P

P

Fig. P1.77

Solution

(do)small = 50 mm (di)small = 30 mm P = 100 kN L = 200 mm τ=?

(do)big = 70 mm

(di)big = 50 mm

-----------------------------------------------------------The free body diagram of the smaller pipe is shown below. (do)small

τ

P

L

By equilibrium of forces we have:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

P = τ [ π ( d o ) small ]L

January 2014

3

3 3 100 ( 10 ) P τ = ----------------------------------- = ------------------------------- = 3183 ( 10 )N ⁄ m or τ = 3.18MPa [ π ( d o ) small ]L π ( 0.05 ) ( 0.2 )

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.78 Two cast iron pipes are adhesively bonded together over a length of 200 mm. The outer diameters of the two pipes are 50 mm and 70 mm and wall thickness of each pipe is 10 mm. The two pipes separated while transmitting a torque of 2 kN-m. What was the shear stress in the adhesive just before the two pipes separated? T

T

Fig. P1.78

Solution

(do)small = 50 mm (di)small = 30 mm T = 2 kN-m L = 200 mm τ=?

(do)big = 70 mm

(di)big = 50 mm

-----------------------------------------------------------The free body diagram of the smaller pipe and the differential circular surface area over is shown below. T

dA = (Rodθ)dx

τ

(do)small = 2Ro

dθ

dV = τ dA dx

L

By moment equilibrium about the axis of the pipe we have the following. L 2π

T =

∫ Ro

dV =

A

L

∫ Ro τ dA = ∫ ∫ Ro τ ( Ro dθ ) dx A

0

=

0

2 τR o

2π

∫ ∫ dθ 0

2

dx = τR o [ 2π ] ( L )

1

0

3

6 2 2 ( 10 ) T τ = ---------------- = ---------------------------------------- = 2.546 ( 10 )N ⁄ m 2 2 2πR o L 2π ( 0.025 ) ( 0.2 )

2 τ = 2.55MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.79 Two cast iron pipes are held together by a bolt as shown. The outer diameters of the two pipes are 50 mm and 70 mm and wall thickness of each pipe is 10 mm.The diameter of the bolt is 15 mm. The bolt broke while transmitting a torque of 2 kN-m. On what surface(s) did the bolt break? What was the shear stress in the bolt on the surface where it broke? T T

Fig. P1.79

Solution

T = 2 kN-m

(do)small = 50 mm

dbolt = 15 mm

τ=?

------------------------------------------------------------

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The free body diagram with the bolt broken is shown below. T

V (do)small = 2Ro V 2

By equilibrium of moment we have: T = ( 2R o ) [ V ] = ( 2R o ) [ ( τ ) ( πd bolt ⁄ 4 ) ] or 3

6 2 ( 2 ) ( 2 ) ( 10 ) 2T τ = --------------------------- = ------------------------------------------- = 226.3 ( 10 )N ⁄ m 2 2 ( πd bolt )R o π ( 0.015 ) ( 0.025 )

τ = 226.3MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.80 The can lid in Figure P1.80a gets punched on two sides AB and AC of an equilateral triangle ABC. Figure P1.80b is the top view showing relative location of the points. The thickness of the lid is t = 1/64 in. and the lid material can at most support a shear stress of 1800 psi. Assume a uniform shear stress during punching and point D acts like a pin joint. Use a= 1/2 in, b = 3 in and c =1/4 in. Determine the minimum force F that must be applied to the can opener. F

(b)

(a) C B

A

C

D E

B

C

a

a a

E

D A

A B

b c Fig. P1.80

Solution

-----------------------------------------------------------We draw the FBD as

τ

F C

2V

Dx

E

A

F

(a cos30)/2 Dy C Dx

Dy

E

A B

B

b c

b c

By equilibrium of moment about D we obtain: cos 30 cos 30 F ( b ) = 2V ⎛ c + a --------------⎞ = 2 [ ( τ ) ( a ) ( t ) ] ⎛ c + a --------------⎞ ⎝ ⎝ 2 ⎠ 2 ⎠ 1 1 1 0.866 F ( 3 ) = 2 ( 1800 ) ⎛ ---⎞ ⎛ ------⎞ ⎛ 1--- + --- -------------⎞ = 13.12 ⎝ 2⎠ ⎝ 64⎠ ⎝ 4 2 2 ⎠

or

or

F = 13.12 ------------- = 4.374 lb 3

1 F = 4.38 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.81 It is proposed to use 1/2 inch bolts in a 10 inch diameter coupling for transferring a torque of 100 in-kips from a 4 inch diameter shaft onto another. The maximum average shear stress in the bolts is to

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

be limited to 20 ksi. How many bolts and at what radius should the bolts be placed on the coupling? T

T

Solution

Fig. P1.81 dbolt = 1/2 inch T = 100 in-kips τ bolt ≤ 20ksi

n = number of bolts = ? r = ?

-----------------------------------------------------------The free body diagram of the coupling can be drawn after making an imaginary cut through the bolts as shown below. By equilibrium of moment about the shaft axis we obtain: T = nVr = 100(103) or V = 100(103)/(nr) The shear stress in the bolt can be written as given below. 3

100 ( 10 ) 3 V τ bolt = ---- = ------------------------------- ≤ 20 ( 10 ) 2 A ( nr ) ( πd ⁄ 4 )

5 nr ≥ ------------------------2 ( π0.5 ⁄ 4 )

or

nr ≥ 25.465

or

1

V V

T

r

V

V V V

1

As per Eq. 1 there are several for n and r. The minimum value of r is rmin = 2+0.25=2.25 corresponding to n = 11.3 i.e., n = 12. The maximum value of r is rmax = 5-0.25=4.75 corresponding to n = 5.36 i.e., n = 6. the table below gives all possible answers between n = 6 to n = 12. n (number of bolts)

6

7

8

9

10

11

r in inches

4.25

3.64

3.18

2.83

2.55

2.31

For minimizing machining and assembly cost 6 bolts should be used.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.82 A human hand can comfortably apply a torsional moment of 15 in-lbs. (a)What should be the breaking shear strength of a seal between a lid and the bottle assuming the lid has a diameter of 1.5 inches and a height of 1/2 inch? (b) If the same sealing strength as in part a is used on a lid that is 1 inch in diameter and 1/2 inch in height, what would be the torque needed to open the bottle?

Solution

(a) T = 15 in-lb (b) d = 1 inch

d = 1.5 inch h = 1/2 inch

h = 1/2 inch Τ=?

τ=?

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

The free body diagram of the bottle with lid removed is shown below. dθ r=d/2 h = 1/2 in

dV

T By moment equilibrium about the axis of the bottle we obtain: 2π

T =

∫

⎛ d---⎞ ( dV ) = ⎝ 2⎠

A

∫

2

d d ⎛d ---⎞ τh ⎛ ---⎞ ( dθ ) = ⎛ -----⎞ hτ ⎝ 2⎠ ⎝ 2⎠ ⎝ 4⎠

0

2π

∫

2

τ ( πd h ) dθ = -------------------2

1

0

(a) Substituting d= 1.5 inch, h= 0.5 inch, T= 15 in-lbs, into Eq. (1) we obtain the following. 2

τπ ( 1.5 ) ( 0.5 ) 15 = ---------------------------------2

(b)

or

τ = 8.49psi or

τ = 8.5 psi

15 = τ [π(1.5)2(.5)/2] τ = 8.49 psi Substituting d= 1.0 inch, h= 0.5 inch, τ= 8.49 psi, into Eq. (1) we obtain the following 2

( 1.0 ) ( 0.5 ) T = τπ --------------------------------2

or

T = 6.67 in – lbs or

T = 6.7 in – lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.83 The hand exert a force F on the handle of a bottle opener shown in Figure P1.83. Assume the average shear strength of the bond between the lid and the bottle is 10 psi. Determine the minimum force needed to open the bottle. Use t = 3/8 in. d = 2 1/2 in. and a = 4 in. d

a

t

F

Fig. P1.83

Solution

-----------------------------------------------------------We draw the FBDs as shown below

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Text =F(a+d/2) dθ r=d/2

t

dV

Text

Text

By equilibrium we obatin d T ext = F ⎛ a + ---⎞ = T ⎝ 2⎠

1

where the internal torque T is 2π

T =

∫

⎛ d---⎞ ( dV ) = ⎝ 2⎠

A

∫

2

d ⎞ ⎛d ---⎞ τt ⎛ d ---⎞ ( dθ ) = ⎛ ---- tτ ⎝ 2⎠ ⎝ 2⎠ ⎝ 4⎠

0

2π

∫

2

τ ( πd t ) dθ = ------------------2

2

0

By equating the two equations we obtain 2

τ ( πd t ) d F ⎛ a + ---⎞ = ------------------⎝ ⎠ 2 2

2

( 10 ) ( π ( 2.5 ) ( 0.375 ) ) 2.5 F ⎛ 4 + -------⎞ = ----------------------------------------------------⎝ ⎠ 2 2

or

36.8155 F = ------------------- = 7.012 lb 5.25

or 3 F = 7.1 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.84

Show the stress components of a point in plane stress on the square σ xx = 100MPa ( T )

τ xy = – 75 MPa

τ yx = – 75 MPa

σ yy = 85MPa ( T )

y

x

Fig. P1.84

Solution

-----------------------------------------------------------85 y

75 75

75

100

100

75

x

85

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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M. Vable

1.85

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Show the stress components of a point in plane stress on the square σ xx = 85MPa ( C )

τ xy = 75MPa

τ yx = 75MPa

σ yy = 100MPa ( T ) y

x

Fig. P1.85

Solution

-----------------------------------------------------------100 y

75

75

75 85

85

75

x 100

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.86

Show the stress components of a point in plane stress on.the square σ xx = 27ksi ( C )

τ xy = 18ksi

τ yx = 18ksi

σ yy = 85ksi ( T ) y

x

Fig. P1.86

Solution

-----------------------------------------------------------85 y

18 18

18

27

27

18

x 85

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.87

Show the stress components of a point in plane stress on.the square σ xx = 27ksi ( C )

τ xy = 18ksi

τ yx = 18ksi

σ yy = 85ksi ( T )

x

y

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Solution

-----------------------------------------------------------85 18

x 18

27

27 18 18

y 85

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.88

Show the non-zero stress components on the A,B, and C faces of the cube. σ xx = 70MPa ( T )

τ xy = – 40 MPa

τ xz = 0

τ yx = – 40 MPa

σ yy = 85MPa ( C )

τ yz = 0

τ zy = 0

τ zx = 0

σ zz = 0

.C .B y

.A

z

x

Fig. P1.88

Solution

-----------------------------------------------------------70 40 40 85 y

z

x

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.89

Show the non-zero stress components on the A,B, and C faces of the cube. σ xx = 70MPa ( T )

τ xy = – 40 MPa

τ yx = – 40 MPa

σ yy = 85MPa ( C ) τ zy = 0

τ zx = 0

τ xz = 0 τ yz = 0 σ zz = 0

.C .B .A x

z

y

Fig. P1.89

Solution

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

-----------------------------------------------------------85

40 40 z 70

x

y

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.90

Show the stress components of a point in plane stress on the stress element in polar coordinates σ rr = 125MPa ( T )

τ rθ = – 65 MPa

τ θr = – 65 MPa

σ θθ = 90MPa ( C )

y

r

θ

x

Fig. P1.90

Solution

-----------------------------------------------------------125

90 y

65

65

r

65 65 125 θ

90

x

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.91

Show the stress components of a point in plane stress on the stress element in polar coordinates σ rr = 125MPa ( T )

τ rθ = – 65 MPa

τ θr = – 65 MPa

σ θθ = 90MPa ( C )

y r θ

x

Fig. P1.91

Solution

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

-----------------------------------------------------------y r

θ

x

90 125

65 65

90 65

125

65

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.92

Show the stress components of a point in plane stress on the stress element in polar coordinates σ rr = 18ksi ( T )

τ rθ = – 12 ksi

τ θr = – 12 ksi

σ θθ = 25ksi ( C )

y r θ

x

Fig. P1.92

Solution

-----------------------------------------------------------18

25 12 y 12

12 12

r

18 θ

25

x

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.93

Show the stress components of a point in plane stress on the stress element in polar coordinates σ rr = 25ksi ( C )

τ rθ = 12ksi

τ θr = 12ksi

σ θθ = 18ksi ( T ) y r θ

x

Solution

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

-----------------------------------------------------------y r θ

18 12 12

x

25

18 25

12 12

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.94

Show the non-zero stress components on the A,B, and C faces of the cube. σ xx = 100MPa ( T )

τ xy = 200MPa

τ xz = – 125MPa

τ yx = 200MPa

σ yy = 175MPa ( C )

τ yz = 225MPa

τ zx = – 125 MPa

τ zy = 225MPa

σ zz = 150MPa ( C )

y .C .B

x

.A z

Fig. P1.94

Solution

-----------------------------------------------------------y

175 200 200

225

125 100 x

225 125 150 z

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.95

Show the non-zero stress components on the A,B, and C faces of the cube. σ xx = 90MPa ( T )

τ xy = 200MPa

τ xz = 0

τ yx = 200MPa

σ yy = 175MPa ( T )

τ yz = – 225 MPa

τ zx = 0

τ zy = – 225 MPa

σ zz = 150MPa ( C )

x .C .B

z

.A y

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Fig. P1.95

Solution

-----------------------------------------------------------x

90 200 150

200

z 225

225 175 y

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.96

Show the non-zero stress components on the A,B, and C faces of the cube. τ xy = 15ksi

τ xz = 0

τ yx = 15ksi

σ xx = 0

σ yy = 10ksi ( T )

τ yz = – 25 ksi

τ zx = 0

τ zy = – 25 ksi

σ zz = 20ksi ( C )

y

z .C .B

x

.A

Fig. P1.96

Solution

-----------------------------------------------------------y

z 10

15 25

15 25

x 20

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.97

Show the non-zero stress components on the A,B, and C faces of the cube. σ xx = 0 τ yx = – 15 ksi τ zx = 0

τ xy = – 15 ksi

τ xz = 0

σ yy = 10ksi ( C ) τ zy = 25ksi

τ yz = 25ksi σ zz = 20ksi ( T )

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

z

January 2014

x .C .B

y

.A

Fig. P1.97

Solution

-----------------------------------------------------------z

x

20

25

25 15 10

15

y

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.98 Show the non-zero stress components in the r, θ, and x cylindrical coordinate system on the A,B, and C faces of the stress element shown. σ rr = 150MPa ( T ) τ θr = – 100 MPa τ xr = 125MPa

τ rθ = – 100 MPa

τ rx = 125MPa

σ θθ = 160MPa ( T )

τ θx = 165MPa

τ xθ = 165MPa

σ xx = 145MPa ( C )

A

r

B

θ x

C

Solution

-----------------------------------------------------------r 150 125 125 100

145 x

100

165 165 160 θ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.99 Show the non-zero stress components in the r, θ, and x cylindrical coordinate system on the A,B, and C faces of the stress element shown σ rr = 10ksi ( C )

τ rθ = 22ksi

τ rx = 32ksi

τ θr = 22ksi

σ θθ = 0

τ θx = 25ksi

τ xr = 32ksi

τ xθ = 25ksi

σ xx = 20ksi ( T )

C A

Fig.P1.99

B

r

θ x

Solution

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

-----------------------------------------------------------x 25 20

22 r 32

θ 25

32 22 10

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.100 Show the non-zero stress components in the r, θ, and φ spherical coordinate system on the A,B, and C faces of the stress element shown. σ rr = 150MPa ( T )

τ rθ = 100MPa

τ rφ = 125MPa

τ θr = 100MPa

σ θθ = 160MPa ( C )

τ θφ = – 175 MPa

τ φr = 125MPa

τ φθ = – 175 MPa

σ φφ = 135MPa ( C )

z φ

r C y

θ

A

B

x

Solution

-----------------------------------------------------------θ

r

150 100

160

125 125

175 100

175 135

φ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.101 Show the non-zero stress components in the r, θ, and φ spherical coordinate system on the A,B, and C faces of the stress element shown. σ rr = 0 τ θr = – 18 ksi τ φr = 0

τ rθ = – 18 ksi σ θθ = 10ksi ( C ) τ φθ = 25ksi

z φ

τ rφ = 0ksi

C

τ θφ = 25ksi σ φφ = 20ksi ( T )

r y

B

θ

A

Fig.P1.101

x

Solution

-----------------------------------------------------------r

θ

18 φ

10 25 25

18

20

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.102 Show that the normal stress σxx on a surface can be replaced by equivalent internal normal force N

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

and internal bending moments My and Mz as given below and shown in Figure 1.102.

∫ σxx dA

N =

1.5a

A M y = – zσ xx dA A

∫

1.5b

∫

1.5c

M z = – yσ xx dA A N x y

My

O Mz

z

Fig. P1.102

Solution

-----------------------------------------------------------Consider a differential area dA on which the normal force is dN= σxx dA as shown below. x

dN = σxx dA y z

O

y

z

For static equivalency, the force in the x direction and the moments about the y and z axis for the above figure and figure Fig. P1.102 must be same. Thus, we obtain: N =

∫ dN A

=

∫ σxx dA

1

A

∫

∫

2

∫

∫

3

M y = – z dN = – zσ xx dA A A M z = – y dN = – yσ xx dA A A

Eq’s (1),(2), and (3) are the same as Eq’s 1.5a,1.5b, and 1.5c.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.103 The normal stress on a cross-section is given by σxx = a +b y, where, y is measured from the centroid of the cross-section. If A is the area of cross-section, Izz is the area moment of inertia about the z axis,

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

N and Mz are the internal axial force and internal bending moment given by Equations 1.5a. and 1.5c, show the following: ⎛M ⎞ ---- – ⎜ -------z-⎟ y σ xx = N A ⎝ I zz ⎠ We will see the above equation in combined axial and symmetric bending problem later in the book. Solution

1.6

------------------------------------------------------------

Substituting σxx = a +b y into Eq’s 1.5a and 1.5c, we obtain the following: N =

∫ ( a + by )dA

∫

∫

∫

∫

∫

A

A

A

A

A

2

= a dA + b ydA and M z = – y ( a + by )dA = – a ydA – b y dA

A

The first moment of area ∫ ydA is zero because the origin is at the centroid. Noting that A = ∫ dA and A

I zz =

∫y

2

dA

A

we obtain:

A

N = aA

a = N⁄A

or

and

M z = – bI zz

b = – ( M z ⁄ I zz )

or

Substituting the values of a and b in σxx = a +b y we obtain Eq. 1.6.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.104 The normal stress on a cross-section is given by σxx = a + b y + c z, where y and z are measured from the centroid of the cross-section. If A is the area of cross-section, Iyy, Izz, and Iyz are the area moment of inertia, N, My and Mz are the internal axial force and internal bending moments given by Equations 1.5a., 1.5b, and 1.5c, show the following: ⎛ M z I yy – M y I yz⎞ ⎛ M y I zz – M z I yz⎞ σ xx = N ---- – ⎜ ---------------------------------⎟ y – ⎜ ---------------------------------⎟ z 2 A ⎝ I I –I ⎠ ⎝ I I – I2 ⎠ yy zz

yz

yy zz

1.7

yz

The above equation is used in unsymmetrical bending of beams. Note that if either y or z is an axis of symmetric then Iyz =0. In such a case the above equation simplifies considerably. Solution

------------------------------------------------------------

Substituting σxx = a + b y + c z, into Eq’s 1.5a, 1.5b, and 1.5c, we obtain the following: N =

∫ ( a + by + cz )dA

∫

∫

∫

A

A

A

= a dA + b ydA + c zdA

A

∫

∫

∫

A

A

A

∫

∫

∫

A

A

A

∫

1 2

M y = – z ( a + by + cz )dA = – a zdA – b yzdA – c z dA

2

A 2

∫

M z = – y ( a + by + cz )dA = – a ydA – b y dA – c yzdA

3

A

The first moment of area ∫ ydA and ∫ zdA are zero because the origin is at the centroid. Noting that A

A =

∫ dA , A

A

and Izz = ∫ y 2 dA , I yy = ∫ z 2 dA , I yz = ∫ yz dA we obtain: A

A

A

N = aA

or

N a = ---A

4

M y = – bI yz – cI yy

5

M z = – bI zz – cI yz

6

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Solving Eq’s (5) and (6) for the constants b and c, we obtain: ⎛ M z I yy – M y I yz⎞ b = – ⎜ ---------------------------------⎟ ⎝ I yy I zz – I 2yz ⎠

⎛ M y I zz – M z I yz⎞ c = – ⎜ ---------------------------------⎟ ⎝ I yy I zz – I 2yz ⎠

Substituting the values of a, b, and c in σxx = a + by + cz, we obtain Eq. 1.7.

7

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------1.105 An infinitesimal element in plane stress is shown. Fx and Fy are the body forces acting at the point and have the dimensions of force per unit volume. By converting stresses into forces and writing equilibrium equations show ∂σ xx ∂τ yx + + Fx = 0 ∂x ∂y

1.8a

∂τ xy ∂σ yy + + Fy = 0 ∂x ∂y τ xy = τ yx σ

yy

+ τ

dy

σ

∂σ yy ∂y

xy

1.8c

dy

Fx Fy

∂τ

yx dy ∂y ∂τ xy τ + dx xy ∂ x ∂σ xx σ + dx xx ∂ x

τ yx +

xx

τ yx

O σ

Fig. P1.105

1.8b

yy dx

Solution

-----------------------------------------------------------Converting stresses into forces by multiplying by the areas of the plane on which they act and multiplying body force by the differential volume we obtain the following free body diagram below. By force Equilibrium in the x-direction we obtain: ⎛ σ + ∂σ xx dx⎞ ( dydz ) – ( σ ) ( dydz ) + ⎛ τ + ∂τ yx dy⎞ ( dxdz ) – ( τ ) ( dxdz ) + ( F ) ( dxdydz ) = 0 or xx yx x ⎝ xx ∂ x ⎠ ⎝ yx ∂ y ⎠ ∂σ xx ∂x

( dxdydz ) +

∂τ yx ∂y

( dxdydz ) + ( F x ) ( dxdydz ) = 0

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1-73

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 1

January 2014

Noting that (dxdydz) is a common factor and can be removed, we obtain Eq. 1.8a. ∂σ yy ⎞ ⎛ dy⎟ dxdz ⎜ σ yy + ∂y ⎝ ⎠

(τ

xy

∂τ ⎛ yx ⎞ dy⎟ ( dxdz ) ⎜ τ yx + ∂y ⎝ ⎠ ∂τ ⎛ xy ⎞ dx⎟ ( dydz ) ⎜ τ xy + ∂ x ⎝ ⎠

) ( dydz ) ( F ) ( dxdydz ) x

dy

(σ

xx

∂σ ⎛ xx ⎞ dx⎟ ( dydz ) ⎜ σ xx + ∂x ⎝ ⎠

) ( dydz ) ( F y ) ( dxdydz ) O

(σ

yy

(τ

yx

) ( dxdz )

) ( dxdz )

dx

By force equilibrium in the y-direction we obtain: ⎛ σ + ∂σ yy dy⎞ dxdz – ( σ ) ( dxdz ) + ⎛ τ + ∂τ xy dx⎞ ( dydz ) – ( τ ) ( dydz ) + ( F ) ( dxdydz ) = 0 yy xy y ⎝ yy ∂ y ⎠ ⎝ xy ∂ x ⎠ ∂σ yy ∂y

( dxdydz ) +

∂τ xy ∂x

( dxdydz ) + ( F y ) ( dxdydz ) = 0

2

We now consider moment equilibrium about point about the center of the differential element. The body forces do not produce any moment as these pass through the center. The forces from the normal stresses do not produce any moment as these also pass through the center of the differential element. The moment from the variation of shear stresses can be written as ∂τ ∂τ ⎛ τ + xy dx⎞ ( dydz ) ⎛ dx ------⎞ + ( τ xy ) ( dydz ) ⎛ dx ------⎞ – ⎛ τ + yx dy⎞ ( dxdz ) ⎛ dy ------⎞ – ( τ yx ) ( dxdz ) ⎛ dy ------⎞ = 0 ⎝ xy ∂ x ⎠ ⎝ 2⎠ ⎝ 2 ⎠ ⎝ yx ∂ y ⎠ ⎝ 2⎠ ⎝ 2⎠

Neglecting the terms containing the product of four differentials as these term tend to zero faster in the limit than the terms with product of three differentials. We obtain: ( τ xy ) ( dydz ) ( dx ) – ( τ yx ) ( dxdz ) ( dy ) = 0 . Removing the common term (dxdydz) we obtain Eq. 1.8a.

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

2.1 A 80 cm stretch cord is used to tie the rear of the canoe to the car hook as shown. In the stretched position the cord forms the side AB of the triangle shown. Determine the average normal strain in the stretch cord. 80 cm

B

B

C

132 cm

A

A Solution

Figure P2.1

ε=?

Lo=80cm

-----------------------------------------------------------The final length of the cord is: L f = AB =

2

2

AC + BC =

2

2

80 + 132 = 154.35cm

Lf – Lo – 80- = 0.9294 or The average normal strain is: ε = ---------------= 154.35 --------------------------80 Lo

ε = 0.9294 cm ⁄ cm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.2 The diameter of a spherical balloon changes from 250 mm to 252 mm. Determine the change in the average circumferential normal strain.

Figure P2.2

Solution

do=250mm

df=252mm

ε=?

-----------------------------------------------------------πd f – πd o 2 - = 0.008 or The average normal strain is: ε = ---------------------- = -------250 πd o

ε = 0.008 mm/mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.3 Two rubber bands are used for packing an air mattress for camping. The undeformed length of the rubber band is seven inches. Determine the average normal strain in the rubber bands if the diameter of the mattress is 4.1 inch at the section where the rubber bands are on the mattress.

Figure P2.3

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M. Vable

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

Solution

Lo=7in

January 2014

ε=?

d=4.1in

-----------------------------------------------------------The final length is: L f = πd = 4.1π = 12.88 Lf – Lo 12.88 – 7 The average normal strain is: ε = ---------------= ---------------------- = 0.8401 7 Lo

ε = 0.8401 in ⁄ in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.4 A canoe on top of a car is tied down using rubber stretch cords as shown. The undeformed length of the stretch cord is 40 inches. Determine the average normal strain in the stretch cord assuming that the path of the stretch cord over the canoe can be approximated as shown below.The average normal strain can be found as shown below. C

C B

B

B

17 in 12 in A

A

A 6 in

18 in

Solution

Lo=40 in

ε=?

-----------------------------------------------------------The figure below can be used for calculating the lengths of segments AB and BC as AB = BC =

2

2

2

6 + 12 = 13.416 and

2

5 + 12 = 13 C 5 in

B

17 in 12 in A 12 in 6 in

The final length of the cord is: L f = 2 ( AB + BC ) = 52.83 Lf – Lo – 40- = 0.3208 or The average normal strain is: ε = ---------------= 52.83 -----------------------40 Lo

ε = 0.321 in ⁄ in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.5 The cable between two poles shown in Figure P2.5 is taut before the two trafic lights are hung on it. The lights are placed symmertically at 1/3 the distance between the poles. Due to the weight of the trafic lights the cable sags as shown. Determine the average normal strain in the cable. 27 ft 15 in

Figure P2.5

Solution

------------------------------------------------------------

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2-2

M. Vable

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

The deformed shape is 9 ft A

D

1.25 ft C

B From geometry: AB = CD =

2

9 ft

9 ft

2

9 + 1.25 = 9.0864 ft

The initial length is: L o = 27 ft . The final length is: L f = AB + BC + CD = 2 ( 9.0864 ) + 9 = 27.173 ft . Lf – Lo –3 27.173 – 27 The average normal strain is: ε av = ---------------= ---------------------------- = 6.399 ( 10 ) 27 Lo

ε av = 6399 μft ⁄ ft

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.6 Due to the application of the forces, the displacement of the rigid plates in the x direction were observed as given below. Determine the axial strains in rods in sections AB, BC, and CD. u B = – 1.8 mm

u C = 0.7 mm

u D = 3.7 mm

x F1 B

A

εAB=?

C

εBC=?

F3 D F3

F2

F1 2.5 m

1.5 m

Solution

F2

2m

εCD=?

-----------------------------------------------------------The average normal strains in the rods can be found as shown below. uB – uA –3 ( – 1.8 – 0 ) - = -------------------------- = – 1.2 ( 10 ) ε AB = --------------------3 ( xB – xA ) ( 1.5 ) ( 10 )

or

mm ε AB = – 1200μ --------mm

uC – uB –3 0.7 – ( – 1.8 ) - = ----------------------------- = 1 ( 10 ) ε BC = --------------------3 ( xC – xB ) 2.5 ( 10 )

or

mm ε BC = 1000μ --------mm

uD – uC –3 3.7 – ( 0.7 ) = ε CD = ---------------------- = ------------------------1.5 ( 10 ) 3 ( xD – xC ) 2 ( 10 )

or

mm ε CD = 1500μ --------mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.7 Due to the application of the forces, the average normal strains in the bars were found to be as given below. Determine the movement of point at D with respect to the left wall. ε AB = – 800μ

ε BC = 600μ

ε CD = 1100μ

x F1 B

A 1.5 m

F2 C

F1 2.5 m

F3 D

F2

F3

2m

Figure P2.7

Solution

uD-uA=?

-----------------------------------------------------------The relative displacements of the ends of each rod can be found as shown below.

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M. Vable

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

–6

3

–6

3

January 2014

u B – u A = ε AB ( x B – x A ) = ( – 800 ) ( 10 ) ( 1.5 ) ( 10 ) = – 1.2mm

1

u C – u B = ε BC ( x C – x B ) = ( 600 ) ( 10 ) ( 2.5 ) ( 10 ) = 1.5mm –6

2

3

u D – u C = ε CD ( x D – x C ) = ( 1100 ) ( 10 ) ( 2 ) ( 10 ) = 2.2mm Adding (1),(2) and (3): u D – u A = 2.2 + 1.5 – 1.2 = 2.5mm or u D – u A = 2.5mm

3

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.8 Due to the application of the forces, the rigid plate in Figure P2.9 is observed to move 0.0236 inch to the right. Determine the average normal strain in bars A and B. Rigid Plate

P Bar A

Bar B

P 60 in

Solution

εA=?

24 in 0.02 in

εB=?

-----------------------------------------------------------The exaggerated deformed geometry can be drawn as shown below.

(a)

0.0236 in

δA

From the above figure we obtain the following. δ A = 0.0236 δ B = δ A – 0.02 = 0.0036 ( Contraction ) The average normal strain can be found as shown below. δ –6 0.0236 ε A = -----A- = ---------------- = 393.3 ( 10 ) 60 LA

in ε A = 393.3μ ----in

δ 0.0036- = – 150 ( 10 –6 ) ε B = -----B- = –-----------------24 LB

in ε B = – 150 μ ----in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.9 Due to the application of the forces, the average normal strain in bar A in Figure P2.9 was found to be 2500μ in/in. Determine the normal strain in bar B. Rigid Plate

P Bar A

Bar B

P 60 in

24 in 0.02 in

Figure P2.9

Solution

εA=2500μ

εB=?

-----------------------------------------------------------–6

The deformation of bar A is: δ A = ε A L A = ( 2500 ) ( 10 ) ( 60 ) = 0.150in

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M. Vable

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

The exaggerated deformed geometry can be drawn as shown below.

δA

(a)

δB

0.02 in

From Fig. (a) δ B = δ A – 0.02 = 0.13in ( Contraction ) δ –3 – 0.13 ε B = -----B- = ------------- = – 5.4167 ( 10 ) 24 LB

in ε B = – 5416.7 μ ----in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.10 Due to the application of the forces, the average normal strain in bar B in Figure P2.10 was found to be -4000μin/in. Determine the normal strain in bar A. Rigid Plate

P Bar A

Bar B

P 60 in

Solution

εB=−4000μ

εA=?

24 in 0.02 in

Figure P2.10

-----------------------------------------------------------–6

The deformation of bar B is: δ B = ε B L B = ( 4000 ) ( 10 ) ( 60 ) = 0.24in The exaggerated deformed geometry can be drawn as shown below.

δA

(a)

δB

0.02 in

From Fig. (a) δ A = δ B + 0.02 = 0.26in ( extension ) δ –3 0.26 ε A = -----A- = ---------- = 4.3333 ( 10 ) 60 LA

in ε A = 4333.3 μ ----in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.11 The rigid bar BD in Figure P2.11 pivots about support C. Due to the application of force P, point B moves upward by 0.06 in. If the length of bar A is 24 in., determine the average normal strain in bar A 25in

125 in D

Rigid

A

Figure P2.11

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C

B

P

2-5

M. Vable

Solution

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

δB= 0.06 inch

January 2014

εA= ?

LA = 24 inches

-----------------------------------------------------------We draw the exaggerated deformed shape as shown below.

δB

125 in Contraction

δD= δA

25in

Using similar triangle, the deformation equations can be written as given below. δ δA -------- = -----Bor δ A = 5δ B = 5 ( 0.06 ) = 0.3 contraction 125 25 The average normal strain can be found as shown below. δ 0.3 ε A = -----A- = – ⎛ -------⎞ = – 0.0125 ⎝ 24 ⎠ LA

1

ε A = – 0.0125 in ⁄ in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.12 Due to the application of force P, the average normal strain in bar A in Figure P2.12 was found to be - 6000 μ. If length of bar A is 36 inches determine the movement of point B. 25in

125 in Rigid

D

B

C

A

P

Figure P2.12

Solution

εA= -6000μ

δB= ?

LA = 36 inches

-----------------------------------------------------------–6

The deformation of A can be found as: δ A = ε A L A = ( 6000 ) ( 10 ) ( 36 ) = 0.216inch contraction From Eq.(1) in problem 2.9 we obtain: 25 δ B = ⎛ ---------⎞ δ A = 0.216 ------------- = 0.0432inch or δ B = 0.0432 in. upwards ⎝ 125⎠ 5

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.13

The rigid bar BD in Figure P2.13 pivots about support C. Due to the application of force P, point B imoves upward by 0.06 in. If the length of bar A is 24 in., determine the average normal strain in bar A.

0.04 in

Figure P2.13

Solution

δB= 0.06 inch

D

25in

125 in Rigid

C

P

A

LA = 24 inches

B

εA= ?

------------------------------------------------------------

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

Consider point D on the rigid bar. We draw the exaggerated deformed shape as shown below.

δB

125 in 0.04

δD

δA

25in Contraction

The deformation equations can be written as shown below. δ D = δ A + 0.04 Using similar triangle, the deformation equations can be written as given below. δ δ A + 0.04 δD δ -------- = -----Bor ---------------------- = -----B125 25 25 125 δ A = 5δ B – 0.04 = 5 ( 0.06 ) – 0.04 = = 0.26inch contraction The average normal strain can be found as shown below.

1

or

δ 0.26 ε A = -----A- = – ⎛ ----------⎞ = – 0.0108 ⎝ 24 ⎠ LA

2

ε A = – 0.0108 in ⁄ in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.14 Due to the application of force P, the average normal strain in bar A in Figure P2.14 was found to be - 6000 μ. If the length of bar A is 36 inches, determine the movement of point B.

0.04 in

D

25in

125 in Rigid

B

C

P

A

Figure P2.14

Solution

εA= -6000μ

δB= ?

LA = 36 inches

-----------------------------------------------------------–6

The deformation of A can be found as: δ A = ε A L A = ( 6000 ) ( 10 ) ( 36 ) = 0.216inch contraction From Eq.(2) in problem 2.11 we obtain: δ A + 0.04 0.216 + 0.04- = 0.0512inch or δ B = ---------------------- = ----------------------------5 5

δ B = 0.0512 in. upwards

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.15

The rigid bar BED in Figure P2.15 pivots about support C. Due to the application of force P, point B moves upward by 0.06 in. If the lengths of bars A and F are 24 in., determine the average normal strain in bars A and F. 125 in 0.04 in

Figure P2.15

Solution

δB= 0.06

D

Rigid

A

30 in E

F

LA = 24 inches LF = 24 inches

25in C

B P

εA= ?

εF= ?

------------------------------------------------------------

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2-7

M. Vable

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

Consider point D on the rigid bar. We draw the exaggerated deformed shape as shown below. 125 in 0.04

δD

δA

δB

30 in

δE=δF 25in Contraction

The deformation equations can be written as shown below. δ D = δ A + 0.04

1

Using similar triangle, the deformation equations can be written as given below. δ δ δ A + 0.04 δ δD δ -------- = -----E- = -----Bor ---------------------- = -----F- = -----B125 30 25 30 25 125 δ A = 5δ B – 0.04 = 5 ( 0.06 ) – 0.04 = = 0.26inch contraction and 30 6 δ F = ------ δ = --- ( 0.06 ) = 0.072inch contraction 25 B 5 The average normal strain can be found as shown below.

or

δ 0.26 ε A = -----A- = – ⎛ ----------⎞ = – 0.0108 ⎝ 24 ⎠ LA

ε A = – 0.0108 in ⁄ in

δ 0.072 ε F = -----F- = – ⎛ -------------⎞ = – 0.003 ⎝ 24 ⎠ LF

ε F = – 0.003 in ⁄ in

2

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.16 Due to the application of force P, the average normal strain in bar A was found to be - 5000 μ. If lengths of bars A and F are 36 inches determine the movement of point B and the average normal strain in bar F. 125 in 0.04 in

D

Rigid

E

LA = 36 inches δB= ? εF= ?

25in B

C

F

A

Solution

30 in

P

εA= - 5000 μ

LF = 36 inches

-----------------------------------------------------------–6

The deformation of A can be found as: δ A = ε A L A = ( 5000 ) ( 10 ) ( 36 ) = 0.180inch contraction Consider point D on the rigid bar. We draw the exaggerated deformed shape as shown below. 125 in 0.04

δA

δD

δB

30 in

δE=δF 25in Contraction

The deformation equations can be written as shown below. δ D = δ A + 0.04

1

Using similar triangle, the deformation equations can be written as given below. δ δ δD -------- = -----E- = -----B125 30 25

or

δ A + 0.04 δ δ ---------------------- = -----F- = -----B30 25 125

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or

2

2-8

M. Vable

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

δ A + 0.04 0.18 + 0.04 δ B = ---------------------- = --------------------------- = 0.044inch 5 5 30 6 δ F = ------ δ = --- ( 0.044 ) = 0.0528inch contraction 25 B 5 The average normal strain can be found as shown below.

January 2014

δ B = 0.044 in. upwards

δ –3 0.0528 ε F = -----F- = – ⎛ ----------------⎞ = – 1.4667 ( 10 ) ⎝ 36 ⎠ LF

ε F = – 1466.7μ in/in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.17 Due to the application of force P, the average normal strain in bar F in Figure P2.17 was found to be -2000μ. If the lengths of bars A and F are 36 in, determine the movement of point B and the average normal strain in bar A. 125 in 0.04 in

D

Rigid

25in

30 in E

B

C

F

A

P

Figure P2.17

Solution

LA = 36 inches δB= ? εA= ?

εF= - 2000 μ

LF = 36 inches

-----------------------------------------------------------–6

The deformation of F can be found as: δ F = ε F L F = ( 2000 ) ( 10 ) ( 36 ) = 0.072in contraction Consider point D on the rigid bar. We draw the exaggerated deformed shape as shown below. 125 in 0.04

δA

δD

δB

30 in

δE=δF 25in Contraction

The deformation equations can be written as shown below. δ D = δ A + 0.04

1

Using similar triangle, the deformation equations can be written as given below. δ δ δD -------- = -----B- = -----E125 25 30

or

δ A + 0.04 δ δ ---------------------- = -----B- = -----F25 30 125

or

2

25 δ B = ------ δ = 0.06 30 F δ B = 0.06 in. upwards 125 125 δ A + 0.04 = --------- δ = --------- ( 0.072 ) = 0.3 30 F 30 The average normal strain can be found as shown below.

or

δ –3 0.26 ε A = -----A- = – ⎛ ----------⎞ = – 7.222 ( 10 ) ⎝ ⎠ 36 LA

δ A = 0.26inch contraction

3

ε A = – 7222μin./in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.18

The rigid bar BD in Figure P2.18 pivots about support C. Due to the application of force P, point B moves left by 0.75 mm. If the length of bar A is 1.2 m, determine the average normal strain in bar A.

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

P

B

1.25 m D

C

Rigid 2.5 m A

Figure P2.18

Solution

δB= 0.75 mm

εA= ?

LA = 1.2 m

-----------------------------------------------------------We draw the exaggerated deformed shape as shown below.

δB θ 1.25 m

Contraction θ

δD = δA

C

2.5 m

Using similar triangle, the deformation equations can be written as given below. δB δA ------ = --------2.5 1.25

1

or

–3

δ A = 2δ B = 2 ( 0.75 ) = 1.5mm = 1.5 ( 10 )m contraction The average normal strain can be found as shown below. –3 δ –3 1.5 ( 10 ) ε A = -----A- = – ⎛ -----------------------⎞ = – 1.2 ( 10 ) ⎝ 1.2 ⎠ LA

ε A = – 1250μmm/mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.19 Due to the application of force P, the average normal strain in bar A in Figure P2.19 was found to be - 2000 μ. If lengths of bar A is 2 m determine the movement of point B. P

B

1.25 m D

Rigid 2.5 m A

C

Figure P2.19

Solution

εA= -2000μ

δB= ?

LA = 2 m

-----------------------------------------------------------–6

The deformation of A can be found as: δ A = ε A L A = ( 2000 ) ( 10 ) ( 2 ) = 4mm contraction From Eq.(1) in problem 2.15 we obtain: 1.25 4 δ B = ⎛ ----------⎞ δ A = --- = 2mm or δ B = 2mm to the left ⎝ 2.5 ⎠ 2

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.20 The rigid bar BD in Figure P2.20 pivots about support C. Due to the application of force P, point B moves left by 0.75 mm. If the length of bar A is 1.2 m, determine the average normal strain in bar A.

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2-10

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

P

B

1.25 m 1 mm

D

C

Rigid 2.5 m A

Figure P2.20

Solution

δB= 0.75 mm

εA= ?

LA = 1.2 m

-----------------------------------------------------------Consider point D on the rigid bar. We draw the exaggerated deformed shape as shown below. We can write the deformation equations as shown below. δD = δA + 1 The deformation equations can be written as given below. δB δD ------ = --------2.5 1.25 or

1

δA + 1 δB -------------- = --------1.25 2.5

2

–3

δ A = 2δ B – 1 = 2 ( 0.75 ) – 1 = 0.5mm = 0.5 ( 10 )m contraction

δB θ 1μμ Contraction

θ δD

δA

1.25 m

C

2.5 m

The average normal strain can be found as shown below. –3 δ 0.5 ( 10 ) –3 ε A = -----A- = – ⎛ -----------------------⎞ = – 0.4167 ( 10 ) ⎝ 1.2 ⎠ LA

ε A = – 416.7μ mm/mm [

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.21 Due to the application of force P, the average normal strain in bar A was found to be - 2000 μ. If length of bar A is 2 m, determine the movement of point B. P

B

1.25 m 1 mm

D

Rigid 2.5 m

C

A

Figure P2.21

Solution

εA= -2000μ

δB= ?

LA = 2 m

-----------------------------------------------------------–6

The deformation of A can be found as: δ A = ε A L A = ( 2000 ) ( 10 ) ( 2 ) = 4mm contraction From Eq.(2) in problem 2.17 we obtain:

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

1.25 5 δ B = ⎛ ----------⎞ ( δ A + 1 ) = --- = 2.5mm or ⎝ 2.5 ⎠ 2

January 2014

δ B = 2.5mm to the left

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.22

The rigid bar BD in Figure P2.22 pivots about support C. Due to the application of force P, point B moves left by 0.75 mm. If the lengths of bars A and F are 1.2 m, determine the average normal strains in bars A and F. P 0.45 m F 0.8 m

B E

D

1 mm

C

Rigid 2.5 m A

Figure P2.22

Solution

δB= 0.75 mm

LA = 1.2 m

εA= ?

LF = 1.2 m

εF= ?

-----------------------------------------------------------Consider point D on the rigid bar. We draw the exaggerated deformed shape as shown below. δB

Extension

δE

0.45 m

θ

0.80 m

2.5 m 1mm Contraction δA

θ

C

δD

We can write the deformation equations as shown below. δD = δA + 1 δF δB δD ------ = ------ = --------2.5 0.8 1.25

δA + 1 δB δF - = ---------------------- = -----0.8 1.25 2.5

or

1 or

2

–3

δ A = 2δ B – 1 = 2 ( 0.75 ) – 1 = 0.5mm = 0.5 ( 10 )m contraction and –3

δ F = 0.8δ B ⁄ 1.25 = 0.48 mm = 0.48 ( 10 )m extension The average normal strains can be found as shown below. –3 δ 0.5 ( 10 ) –3 ε A = -----A- = – ⎛ -----------------------⎞ = – 0.4167 ( 10 ) ⎝ 1.2 ⎠ LA

or

ε A = – 416.7μ mm ⁄ mm

–3 δ 0.48 ( 10 ) –3 ε F = -----F- = ⎛ --------------------------⎞ = 0.4 ( 10 ) ⎝ ⎠ 1.2 LF

or

ε F = 400μ mm ⁄ mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.23 Due to the application of force P in Figure P2.23, the average normal strain in bar A was found to be - 2500 μ. Bars A and F are 2 m long. Determine the movement of point B and average normal strain in bar F.

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

P 0.45 m F 0.8 m

B E

1 mm

D

January 2014

C

Rigid 2.5 m A

Figure P2.23

Solution

εA= - 2500 μ

LA = 2 m

δB= ?

LF = 2 m

εF= ?

-----------------------------------------------------------–6

The deformation of A can be found as: δ A = ε A L A = ( 2500 ) ( 10 ) ( 2000 ) = 5mm contraction Consider point D on the rigid bar. We draw the exaggerated deformed shape as shown below. δB

Extension

δE

0.45 m

θ

0.80 m

2.5 m 1mm Contraction δA

θ

C

δD

We can write the deformation equations as shown below. δ D = δ A + 1 = 6mm δF δB δD ------ = ------ = --------2.5 0.8 1.25 1.25 δ B = ⎛ ----------⎞ ( 6 ) = 3.0mm ⎝ 2.5 ⎠

1

δA + 1 δB δF - = ---------------------- = -----0.8 1.25 2.5

or

or

2

δ B = 3.0mm to the left

or –3

δ F = 0.8δ B ⁄ 1.25 = 1.92mm = 1.92 ( 10 )m extension The average normal strains can be found as shown below. –3 δ 1.92 ( 10 ) –3 ε F = -----F- = ⎛ --------------------------⎞ = 0.96 ( 10 ) ⎝ ⎠ 2.0 LF

ε F = 960μmm/mm

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.24 Due to the application of force P, the average normal strain in bar F in Figure P2.24 was found to be 1000 μ. Bars A and F are 2 m long. Determine the movement of point B and the average normal strain in bar A. B E

1 mm

Rigid

D

P 0.45 m F 0.8 m

C

2.5 m A

Solution

εF= 1000 μ

Figure P2.24

LA = 2 m

LF = 2 m

δB= ?

εA= ?

------------------------------------------------------------

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

–6

The deformation of F can be found as: δ F = ε F L F = ( 1000 ) ( 10 ) ( 2000 ) = 2mm extension Consider point D on the rigid bar. We draw the exaggerated deformed shape as shown below. δB

Extension

δE

0.45 m

θ

0.80 m

2.5 m 1mm Contraction δA

θ

C

δD

We can write the deformation equations as shown below. δD = δA + 1 δF δB δD ------ = ------ = --------2.5 0.8 1.25 1.25 δ B = ⎛ ----------⎞ ( 2 ) = 3.125mm ⎝ 0.8 ⎠

1

δA + 1 δB δF - = ---------------------- = -----0.8 1.25 2.5

or

or

2

δ B = 3.125mm to the left;

or –3

δ A = 2.5δ F ⁄ 0.8 – 1 = 5.25mm = 5.25 ( 10 )m contraction The average normal strains can be found as shown below. –3 δ – 5.25 ( 10 ) –3 ε A = -----A- = ⎛ ------------------------------⎞ = – 2.625 ( 10 ) ⎝ ⎠ 2.0 LA

ε A = – 2625 μmm/mm

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.25 Two bars of equal lengths of 400 mm are welded to rigid plates at right angles. The right angles between the bars and the plates are preserved as the rigid plates are rotated by an angle of ψ as shown in Figure P2.25. The distance between the bars is h = 50 mm. The average normal strains in bars AB and CD were determined as -2500 μ mm/mm and 3500 μ mm/mm, respectively. Determinethe radius of curvature R and the angle ψ. D

C

h A

ψ

ψ

B

R

Figure P2.25

Solution

-----------------------------------------------------------From geometry the arc lengths can be written as: AB = R ( 2ψ ) CD = ( R + h ) ( 2ψ ) = 2Rψ + 2hψ The orignal length is Lo = 400 mm. The average strain in each bar can be written and equated to the given values: 2Rψ – L – 400- = – 2500 ( 10 – 6 ) or ε AB = ----------------------o- = 2Rψ ------------------------400 Lo 2Rψ – 400 = – 1

1

2Rψ + 2hψ – L + ( 2 ) ( 50 )ψ – 400- = 3500 ( 10 – 6 ) or ε CD = ---------------------------------------o- = 2Rψ ------------------------------------------------------400 Lo

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

2Rψ + 100ψ – 400 = 1.4 Substituting Eq. 1 into Eq. 2 we obtain: 100ψ = 2.4 or ψ = 0.024 rads From Eq. 1 we obtain: 2R ( 0.024 ) = 399 or R = 8312.5 mm

2 3 4

ψ = 0.024 rads = 1.375°

The answers are:

R = 8312.5 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.26 Two bars of equal lengths of 30 in are welded to rigid plates at right angles. The right angles between the bars and the plates are preserved as the rigid plates are rotated by an angle of ψ= 1.25o as shown in Figure P2.26. The distance between the bars is h = 2 in. If the average normal strain in bar AB is -1500 μ determine the strain in bar CD. D

C

h A

ψ

ψ

B

R

Figure P2.26

Solution

-----------------------------------------------------------1.25° The rotation angle is: ψ = ------------- π = 0.0218 rads 180 From geometry the arc lengths can be written as: AB = R ( 2ψ ) CD = ( R + h ) ( 2ψ ) = 2Rψ + 2hψ The orignal length is Lo = 30 in. The average strain in each bar can be written and equated to the given values: 2Rψ – L –6 2Rψ – 30 ε AB = ----------------------o- = ----------------------- = – 1500 ( 10 ) 30 Lo 2Rψ + 2hψ – L –6 2Rψ + ( 2 ) ( 2 )ψ – 30 2Rψ – 30 4ψ 4 ( 0.0218 ) ε CD = ---------------------------------------o- = -------------------------------------------------- = ----------------------- + ------- = – 1500 ( 10 ) + ------------------------ or 30 30 30 30 Lo –6

–6

–6

ε CD = – 1500 ( 10 ) + 2908.8 ( 10 ) = 1408.8 ( 10 ) The answer is:

ε CD = 1408.8 μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.27 Two bars of equal lengths of 48 in. are welded to rigid plates at right angles. The right angles between the bars and the plates are preserved as the rigid plates are rotated by an angle of ψ as shown in Figure P2.27. The average normal strains in bars AB and CD were determined as -2000 μ in./in. and 1500 μ in./in., respectively. Determine the location h of a third bar EF that should be placed such that it has

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

zero normal strain. D

C E 4 in

h

F A

ψ

ψ

B

R

Figure P2.27

Solution

-----------------------------------------------------------From geometry the arc lengths can be written as: AB = R ( 2ψ ) CD = ( R + 4 ) ( 2ψ ) = 2Rψ + 8ψ EF = ( R + h ) ( 2ψ ) = 2Rψ + 2hψ The orignal length is Lo = 48 in. The average strain in each bar can be written and equated to the given values: 2Rψ – L – 48- = – 2000 ( 10 – 6 ) ε AB = ----------------------o- = 2Rψ ---------------------48 Lo

1

2Rψ + 8ψ – L –6 2Rψ + 8ψ – 48- = 2Rψ – 48- + 8ψ ε CD = ------------------------------------o- = --------------------------------------------------------------- = 1500 ( 10 ) or 48 48 48 Lo –6 –6 ψ – 2000 ( 10 ) + ---- = 1500 ( 10 ) or 6 The average strain in EF can be written and equated to zero to obtain:

ψ = 0.021 rads

2

2Rψ + 2hψ – L 2Rψ + 2hψ – 48- = 2Rψ – 48- + 2hψ ε EF = ---------------------------------------o- = --------------------------------------------------------------------- or 48 48 48 Lo –6 h ( 0.021 ) ε EF = – 2000 ( 10 ) + --------------------- = 0 24

or

h = 2.2857 in.

The answer is:

h = 2.286 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.28

A rectangular plastic plate deforms into the shaded shape. Determine the average shear strain at point A 0.84 mm

0.84 mm

350 mm A

Solution

600 mm

γA=?

-----------------------------------------------------------The shear strain represents the change of angle from the right angle at A as shown in the figure below. 0.84 mm

γA 350 mm A

From geometry, we obtain the magnitude of γA as shown below.

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

0.840 tan γ A = ------------350

or

January 2014

γ A = 0.0024 rads

1 γ A = 2400μrads

Shear strain is positive as angle decreases.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.29 A rectangular plastic plate deforms into the shaded shape. Determine the average shear strain at point A. 1.7 in 0.0051 in 3.5 in

0.0051 in

A

Solution

γA=?

-----------------------------------------------------------The shear strain represents the change of angle from the right angle at A as shown in the figure below.

1.7 in

γA

A

0.0051 in

From geometry, we obtain the magnitude of γA as shown below. 0.0051 tan γ A = ---------------1.7 Shear strain is negative as angle increases.

or

γ A = 0.003 rads

1 γ A = – 3000 μ rad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.30

A rectangular plastic plate deforms into the shaded shape. Determine the average shear strain at point A. 0.007 in

0.007 in

1.4 in A 3.0 in

Solution

γA=?

-----------------------------------------------------------The shear strain represents the change of angle from the right angle at A as shown in the figure below. 0.007 in

γA 1.4 in

A

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

From geometry, we obtain the magnitude of γA as shown below. 0.007 tan γ A = ------------1.4

γ A = 0.005 rads

or

1 γ A = – 5000 μrads

Shear strain is negative as angle increases.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.31 A rectangular plastic plate deforms into the shaded shape. Determine the average shear strain at point A. 0.65 mm

450 mm

0.65 mm A 250 mm

Solution

γA=?

-----------------------------------------------------------The shear strain represents the change of angle from the original right angle at A as shown in the figure below.

γ A 0.65 mm A 250 mm

From geometry, we obtain the magnitude of γA as shown below. 0.65 tan γ A = ---------250 Shear strain is positive as angle decreases.

or

γ A = 0.00026 rads

1 γ A = 260μ rad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.32

A rectangular plastic plate deforms into the shaded shape. Determine the average shear strain at point A. 0.0056 in 1.4 in 0.0042 in

A 3.0 in

Solution

γA=?

------------------------------------------------------------

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

The angles φ1 and φ2 are measures of the change of angle as shown in the figure below. 0.0056 in

φ2 1.4 in

φ1

A

0.0042 in

3.0 in

From geometry, we obtain the angles φ1 and φ2 as shown below. 0.0042 tan φ 1 = ---------------3.0

φ 1 = 0.0014 rads

or

1

0.0056 tan φ 2 = ---------------or φ 2 = 0.004 rads 1.4 The decrease in the angle from the original right angle at A represents positive shear strain, i.e., γ A = φ 1 + φ 2 = 0.0054 rads or γ A = 5400μ rad

2

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.33 A rectangular plastic plate deforms into the shaded shape. Determine the average shear strain at point A 0.6 mm

600 mm 350 mm 0.6 mm

A

Solution

γA=?

------------------------------------------------------------

The angles φ1 and φ2 are measures of the change of angle as shown in the figure below. 0.6 mm

600 mm

φ2 350 mm

φ1

A

0.6 mm

From geometry we obtain the angles φ1 and φ2 as shown below. 0.6 tan φ 1 = --------600

or

–3

φ 1 = 1.0 ( 10 ) rads

1

–3 0.6 tan φ 2 = --------or φ 2 = 1.714 ( 10 ) rads 350 The increase in angle from the original right angle at A represents negative shear strain, i.e., –3

γ A = – ( φ 1 + φ 2 ) = – 2.714 ( 10 ) rads or

2

γ A = – 2714 μ rad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.34

A thin triangular plate ABC forms a right angle at point A. During deformation, point A moves

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

vertically down by δA. Determine the average shear strain at point A. 8 in 65o C

25o

B

δ A = 0.005 in A

Solution

δA=0.005in

δA

γA=?

-----------------------------------------------------------The exaggerated deformed geometry can be drawn as shown below. 8 in D

B 25o

C 65o

h A φ1 φ2

From triangle ABC From triangle ABD

δA

AB = 8 cos 25 = 7.2505 BD = AB cos 25 = 6.5712

1 2

h = AB sin 25 = 3.0642 BC = BD + DC 6.5712 BD tan φ 1 = --------------- = ----------------------------------------( 3.0642 + 0.005 ) h + δA

or

3

DC = 1.4288

φ 1 = 1.1338rads

or

DC 1.4288 tan φ 2 = -------------- = ---------------------------------------or φ 2 = 0.4357rads h + δA ( 3.0642 + 0.005 ) The decrease in the angle from the original right angle at A represents positive shear strain, i.e., –3 π γ A = --- – ( φ 1 + φ 2 ) = 1.2963 ( 10 )rads 2

γ A = 1296μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.35 A thin triangular plate ABC forms a right angle at point A. During deformation, point A moves vertically down by δA. Determine the average shear strain at point A. 5 in

3i

A

n

C

B

δ A = 0.006 in

δA

Solution

δA= 0.006

γA= ?

------------------------------------------------------------

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

The exaggerated deformed geometry can be drawn as shown below. 5 in D

θ

B

C

3i n

h A

φ1φ2

δA

From triangle ABC From triangle ABD

AB =

2

2

cos θ = 0.8

5 –3 = 4

sin θ = 0.6

BD = AB cos θ = 3.2

1

h = AB sin θ = 2.4 BC = BD + DC BD 3.2 tan φ 1 = -------------- = ------------------------------h + δA ( 2.4 + 0.006 )

or

2 3

DC = 1.8

φ 1 = 0.9261rads

or

1.8 DC or φ 2 = 0.6423rads tan φ 2 = --------------- = -------------------------------( 2.4 + 0.006 ) h + δA The decrease in the angle from the original right angle at A represents positive shear strain, i.e., –3 π γ A = --- – ( φ 1 + φ 2 ) = 2.397 ( 10 )rads 2

γ A = 2397μrads

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.36 A thin triangular plate ABC forms a right angle at point A. During deformation, point A moves vertically down by δA that is given in each problem. Determine the average shear strain at point A. 1300 mm C

B m 500

δ A = 0.75 mm

A

m

δA

Solution

δA=0.75mm

γA= ?

-----------------------------------------------------------The exaggerated deformed geometry can be drawn as shown below. 1300 mm B

θ

C

D h

500 mm

A φ2

δA

φ1

From triangle ABC

AC =

2

2

1300 – 500 = 1200

5 cos θ = -----13

12 sin θ = -----13

From triangle ABD BD = AB cos θ = 192.308 h = AB sin θ = 461.538 BC = BD + DC

or

DC = 1107.69

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

BD 192.308 tan φ 1 = --------------- = ---------------------------------------h + δA ( 461.538 + 0.75 )

January 2014

φ 1 = 0.3942rads

or

DC 1107.69 tan φ 2 = --------------- = ----------------------------------------or φ 2 = 1.1754rads h + δA ( 461.538 + 0.75 ) The decrease in the angle from the original right angle at A represents positive shear strain, i.e., –3 π γ A = --- – ( φ 1 + φ 2 ) = 1.1529 ( 10 )rads 2

γ A = 1153μrads

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.37 A thin triangular plate ABC forms a right angle at point A. During deformation, point A moves horizontally by δA that is given in each problem. Determine the average shear strain at point A. 8 in 65o C

25o

B

δ A = 0.005 in A δA

Solution

δA=0.005

γA= ?

-----------------------------------------------------------The exaggerated deformed geometry can be drawn as shown below. 8 in D 25o

B

h

65o C

φ1 φ2 A δA

From Equations (1),(2) and (3) of problem 2.27: BD = 6.5712 BD – δ 6.5712 – 0.005 tan φ 1 = -------------------A- = -----------------------------------h 3.0642

h = 3.0642

DC = 1.4288

φ 1 = 1.13417rads

or

DC + δ 1.4288 + 0.005tan φ 2 = --------------------A- = ----------------------------------or φ 2 = 0.43766rads h 3.0642 The decrease in the angle from the original right angle at A represents positive shear strain, i.e., –3 γA = π --- – ( φ 1 + φ 2 ) = – 1.0337 ( 10 )rads 2

γ A = – 1034 μrads

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.38 A thin triangular plate ABC forms a right angle at point A. During deformation, point A moves horizontally by δA that is given in each problem. Determine the average shear strain at point A. 5 in

3i

A

n

C

B

δ A = 0.008 in

δA

Solution

δA=0.008

γA= ?

------------------------------------------------------------

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

The exaggerated deformed geometry can be drawn as shown below. 5 in D

φ2

φ1

C

h 3i n

B

A δA

From Equations (1), (2) and (3) in Problem 2-28: BD = 3.2 BD – δ 3.2 – 0.008 tan φ 1 = -------------------A- = --------------------------h 2.4

h = 2.4

DC = 1.8

φ 1 = 0.92609rads

or

CD + δ 1.8 + 0.008 tan φ 2 = --------------------A- = -------------------------or φ 2 = 0.64563rads h 2.4 The decrease in the angle from the original right angle at A represents positive shear strain, i.e., –3 π γ A = --- – ( φ 1 + φ 2 ) = – 0.9280 ( 10 )rads 2

γ A = – 928 μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.39 A thin triangular plate ABC forms a right angle at point A. During deformation, point A moves horizontally by δA that is given in each problem. Determine the average shear strain at point A in each problem 1300 mm B

C

m 500

A

m

Solution

δA=0.90 mm

δ A = 0.90 mm

δA

γA= ?

-----------------------------------------------------------The exaggerated deformed geometry can be drawn as shown below. 1300 mm B

θ

D C hφ

1

m 500 m

From triangle ABC

AC =

A

2

φ2

δA

2

1300 – 500 = 1200

5cos θ = ----13

sin θ = 12 -----13

From triangle ABD BD = AB cos θ = 192.308 h = AB sin θ = 461.538 BC = BD + DC BD + δ 192.308 + 0.90 tan φ 1 = --------------------A = -----------------------------------h 461.538

or

or

DC = 1107.69

1 2 3

φ 1 = 0.39645rads

CD – δ – 0.90tan φ 2 = --------------------A = 1107.69 ----------------------------------or φ 2 = 1.17572rads h 461.538 The decrease in the angle from the original right angle at A represents positive shear strain, i.e.,

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

–3 π γ A = --- – ( φ 1 + φ 2 ) = – 1.3717 ( 10 )rads 2

January 2014

γ A = – 1371.7 μrad

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.40 Bar AB is bolted to a plate along the diagonal as shown in Figure P2.40. The plate experiences an average strain in the x-direction ε xx = 500μ in ⁄ in . Determine the average normal strain in the bar AB. y B 5 in A

x 10 in Figure P2.40

Solution

-----------------------------------------------------------The deformed geometry can be draw as: B

θ

C B1

5 in θ

A

10 in From geometry we have: 5 tan θ = -----10 AB =

2

or

θ = 26.565°

2

5 + 10 = 11.1803 in –6

From the given strain we have: BB 1 = ( 10 )ε xx = 10 [ 500 ( 10 ) ] = 0.005 –3

From geometry: BC = BB 1 cos θ = 4.4721 ( 10 ) The normal strain in bar AB can be written as: –3

–3 4.4721 ( 10 ) BC ε AB = -------- = -------------------------------- = 0.4 ( 10 ) AB 11.1803 The answer is

ε AB = 400 μin. ⁄ in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.41 Bar AB is bolted to a plate along the diagonal as shown in Figure P2.41. The plate experiences an average strain in the x-direction ε yy = – 1200μ mm ⁄ mm . Determine the average normal strain in the

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

bar AB. y B

100 mm

Figure P2.41

x

A 45 mm

Solution

-----------------------------------------------------------The deformed geometry can be draw as: B C

θ

B1 100 mm

A 45 mm From geometry we have: 45tan θ = -------100 AB =

2

or

θ = 24.23°

2

45 + 100 = 109.66 mm –6

From the given strain we have: BB 1 = ( 100 )ε yy = 100 [ 120 ( 10 ) ] = 0.12 –3

From geometry: BC = BB 1 cos θ = 109.43 ( 10 ) The normal strain in bar AB can be written as: –3 ( –B C ) – 109.43 ε AB = ---------------- = ------------------- = – 998 ( 10 ) AB 109.66 The answer is

ε AB = – 998 μmm ⁄ mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.42 A right angle bar ABC is welded to a plate as shown in Figure P2.42. Points B are fixed. The plate experiences an average strain in the x-direction ε xx = – 1000 μ mm ⁄ mm . Determmine the average normal

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M. Vable

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

strain in AB. y

C

B

A

300 mm B

x

Figure P2.42

150 mm

450 mm

Solution

-----------------------------------------------------------The deformed shape can be drawn as y

C1 A

α

A1

C

B

D

300 mm

θ

α B

150 mm

450 mm

From geometry we have: 300 tan θ = --------450

θ = 33.69°

or

α = 90 – θ = 56.31°

150 - = 270.42 mm AB = --------------------cos 56.31

AB cos α = 150

1 2

–6

From the given strain we have: AA 1 = ( 150 ) ε xx = 150 [ 1000 ( 10 ) ] = 0.15 mm –3

From geometry: AD = AA 1 cos α = 83.205 ( 10 ) The normal strain in bar AB can be written as: –3

The answer is

–3 – AD -) = – 83.205 ( 10 ) = ε AB = (-------------------------------------------------– 307.7 ( 10 ) AB 270.42 ε AB = – 307.7 μmm ⁄ mm

3

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.43 A right angle bar ABC is welded to a plate as shown in Figure P2.43. Points B are fixed. The plate experiences an average strain in the x-direction ε xx = 700μ mm ⁄ mm . Determmine the average nor-

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M. Vable

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

mal strain in BC. y

C

B

A

300 mm B

x

Figure P2.43

150 mm

450 mm

Solution

-----------------------------------------------------------The deformed shape can be drawn as E C A1

B

A

θ

C1

300 mm

B θ

150 mm

450 mm

From geometry we have: 300 tan θ = --------450

θ = 33.69°

or

1

300 - = 540.83 mm BC = -------------------sin 56.31

BC sin θθ = 300

2

–6

From the given strain we have: CC 1 = ( 450 ) ε xx = 450 [ 700 ( 10 ) ] = 0.315 mm –3

From geometry: CE = CC 1 cos θ = 262.09 ( 10 ) The normal strain in bar AB can be written as: –3

–3 262.09 ( 10 ) ε AB = CE -------- = -------------------------------- = 0.4846 ( 10 ) BC 540.83

The answer is

3 ε AB = 484.6 μmm ⁄ mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.44 A right angle bar ABC is welded to a plate as shown in Figure P2.44. Points B are fixed. The plate experiences an average strain in the x-direction ε xx = – 800 μ mm ⁄ mm . Determmine the average shear

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

strain at point B in the bar. y

C

B

A

300 mm B

x

Figure P2.44

150 mm

450 mm

Solution

-----------------------------------------------------------The deformed shape can be drawn as C1 A

α D

A1

C

B E φ1

300 mm

φ2 α B

θ

150 mm

450 mm

From geometry we have: 300 tan θ = --------450

θ = 33.69°

or

α = 90 – θ = 56.31°

1

150 AB = ---------------------- = 270.42 mm cos 56.31 300 - = 540.83 mm BC = -------------------sin 56.31

AB cos α = 150 BC sin θ = 300

2 3

From the given strain we have: –6

AA 1 = ( 150 ) ε xx = 150 [ 800 ( 10 ) ] = 0.12 mm

4

–6

CC 1 = ( 450 ) ε xx = 450 [ 800 ( 10 ) ] = 0.360 mm

5

From geometry: –3

A 1 D = AA 1 sin α = 99.85 ( 10 )

–3

C 1 E = CC 1 sin θ = 199.69 ( 10 )

AD = AA 1 cos α = 66.56 ( 10 ) CE = CC 1 cos θ = 299.54 ( 10 )

–3

6

–3

7

–3 C1 E C1 E –3 199.69 ( 10 ) tan φ 1 = --------- = --------------------- = ------------------------------------------------------- = 0.3694 ( 10 ) – 3 BE BC – CE 540.83 – 299.54 ( 10 )

or

φ 1 = 0.3694 ( 10 ) rads 8

–3 A1 D A1 D –3 99.85 ( 10 ) tan φ 2 = ---------= --------------------- = ---------------------------------------------------- = 0.3692 ( 10 ) –3 BD AB – AD 270.42 – 66.56 ( 10 )

or

φ 2 = 0.3692 ( 10 ) rads

–3

Shear strain at B is: γ B = φ 1 + φ 2 = 0.7388 ( 10 ) or

–3

–3

9

γ B = 738.8 μrads

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.45

A right angle bar ABC is welded to a plate as shown in Figure P2.45. Points B are fixed. The plate

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M. Vable

Mechanics of materials, 2nd edition, solutions manual: Chapter 2

experiences an average strain in the y-direction

January 2014

ε yy = 800μ in ⁄ in . Determmine the average normal

strain in AB. y

2 in C

3.0 in

B

B

x

1.0 in

Figure P2.45

A

Solution

-----------------------------------------------------------We can draw the deformed geometry as: 2 in C1 C

3.0 in

B θ

B

α

1.0 in A α D

A1

From geometry we have: tan θ = 3--2

or

θ = 56.31°

α = 90 – θ = 33.69°

1 - = 1.8028 in. AB = -------------------sin 33.69

AB sin α = 1

1 2

–6

From the given strain we have: AA 1 = ( 1 ) ε yy = 800 ( 10 ) in. –6

From geometry: AD = AA 1 sin α = 443.76 ( 10 ) The normal strain in bar AB can be written as: –6

–6 443.76 ( 10 ) -------- = -------------------------------- = 246.15 ( 10 ) ε AB = AD AB 1.8028

The answer is

3 ε AB = 246.2 μin. ⁄ in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.46 A right angle bar ABC is welded to a plate as shown in Figure P2.46. Points B are fixed. The plate experiences an average strain in the y-direction ε yy = – 500 μ in ⁄ in . Determmine the average normal

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January 2014

strain in BC. y

2 in C

3.0 in

x

B

B

1.0 in

Figure P2.46

A

Solution

-----------------------------------------------------------We can draw the deformed geometry as: 2 in C E θ

C1

3.0 in

B θ

B

α

A1

1.0 in

A

From geometry we have: 3 tan θ = --2

θ = 56.31°

or

1

3 AC = --------------------- = 3.6056 in. sin 56.31

AC sin θ = 3

2

–3

From the given strain we have: CC 1 = ( 3 ) ε yy = 1.5 ( 10 ) in. –3

From geometry: CE = CC 1 sin θ = 1.2481 ( 10 ) The normal strain in bar AB can be written as: –3

–3 – 1.2481 ( 10 ) – CE ε AC = ----------- = ------------------------------------ = ( – 0.3462 ) ( 10 ) AC 3.6056

The answer is

3

ε AC = – 346.2 μin. ⁄ in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.47 A right angle bar ABC is welded to a plate as shown in Figure P2.47. Points B are fixed. The plate experiences an average strain in the y-direction ε yy = 600μ in ⁄ in . Determmine the average shear strain

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January 2014

at B in the bar. y

2 in C

3.0 in

x

B

B

1.0 in

Figure P2.47

A

Solution

-----------------------------------------------------------We can draw the deformed geometry as: 2 in C1

θ E

φ1

C

3.0 in

B θ

B

α

1.0 in

φ2

A α D

A1

3 From geometry we have: tan θ = --2 AB sin α = 1

or

θ = 56.31°

1 - = 1.8028 in. AB = -------------------sin 33.69

α = 90 – θ = 33.69°

BC sin θ = 3 –3

From the given strain we have: AA 1 = ( 1 ) ε yy = 0.6 ( 10 ) in.

3 - = 3.6056 in. BC = -------------------sin 56.31

1

–3

CC 1 = ( 3 ) ε yy = 1.8 ( 10 ) in.

From geometry: –3

A 1 D = AA 1 cos α = 0.4992 ( 10 )

–3

C 1 E = AA 1 cos α = 0.9985 ( 10 )

AD = AA 1 sin α = 0.3328 ( 10 ) CE = CC 1 sin θ = 1.4977 ( 10 )

–3

2

–3

3

–3 C1 E C1 E –3 0.9985 ( 10 ) tan φ 1 = --------- = --------------------- = ------------------------------------------------------- = 0.2768 ( 10 ) –3 BE BC + CE 3.6056 + 1.4977 ( 10 )

or

φ 1 = 0.2768 ( 10 ) rads 4

–3 A1 D A1 D –3 0.4992 ( 10 ) tan φ 2 = ---------= --------------------- = ------------------------------------------------------- = 0.2769 ( 10 ) – 3 BD AB + AD 1.8028 + 0.3328 ( 10 )

or

φ 2 = 0.2769 ( 10 ) rads 5

–3

Shear strain at B is: γ B = – ( φ 1 + φ 2 ) = – 0.5537 ( 10 ) or

–3

–3

γ B = – 553.7 μrads

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.48 The diagonals of two square form a right angle at point A. The two rectangles are pulled horizontally to a deformed shape shown by the colored lines. Determine the average shear strain at point A, if the

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

displacements of points A and B are as given in each problem. δA= 0.4 mm and δB= 0.8 mm. 300 mm

300 mm

300 mm A1

A

B

Solution

δA=0.4mm

δB=0.8mm

B1 δB

δA

γA= ?

-----------------------------------------------------------The exaggerated deformed geometry can be drawn as shown below. 300 mm D

300 mm C

E

φ1 φ2

300 mm

A1

A

B

δA

300 + δ tan φ 1 = CD -------- = --------------------AAD 300

B1 δB

1

( 600 + δ B ) – ( 300 + δ A ) DE ( CE – CD ) tan φ 2 = -------- = --------------------------- = --------------------------------------------------------- or AD AD 300 300 + δ B – δ A tan φ 2 = -------------------------------300 Substituting δ A = 0.4 and δ B = 0.8 in Equations(1) and (2)

2

We obtain 300.4tan φ 1 = -----------or φ 1 = 0.78606rads and tan φ 2 = 300.4 ------------or φ 2 = 0.78606rads 300 300 The decrease in the angle from the original right angle at A represents positive shear strain, i.e., –3 γA = π --- – ( φ 1 + φ 2 ) = – 1.332 ( 10 )rads 2

γ A = – 1332 μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.49 The diagonals of two square form a right angle at point A. The two rectangles are pulled horizontally to a deformed shape shown by the colored lines. Determine the average shear strain at point A, if the displacements of points A and B are as given in each problem. δA= 0.3 mm and δB= 0.9 mm. 300 mm

300 mm

300 mm A1

A δA

Solution

δA=0.3

δB=0.9

B

B1 δB

γA= ?

-----------------------------------------------------------Substituting δ A = 0.3 and δ B = 0.8 in Equations (1) and (2) in problem 2-33,

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

We obtain: 300.3 300.6 tan φ 1 = ------------or φ 1 = 0.78590rads and tan φ 2 = ------------or φ 2 = 0.78640rads 300 300 The decrease in the angle from the original right angle at A represents positive shear strain, i.e., –3 π γ A = --- – ( φ 1 + φ 2 ) = – 1.4987 ( 10 ) rads 2

γ A = – 1499 μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.50 The roller at P can slide only in the slot by the given amount. Determine the strains in bar AP by (a) finding the deformed length of AP without small-strain approximation, (b) using Equation 2.6, and (c) using Equation 2.7. δP = 0.25 mm

20

0m m

P

A 50o

δP = 0.25 mm (a) εAP= ? using geometry (b) εAP= ? using small strain scaler approach (c) εAP= ? using small strain vector approach.

Solution

-----------------------------------------------------------We draw the exaggerated deformed shape as shown below. δA

Extension

B o

P

50 130o δP

P1

Lf A

(a) From triangle APP1, using cosine rule we obtain 2

2

2

2

2

L f = AP + PP 1 – 2AP ( PP 1 ) cos 130 = 200 + 0.25 – 2 ( 200 ) ( 0.25 ) cos 130 = 200.16079 Lf – Lo –3 0.16079 The average normal strain is ε AP = ---------------= ------------------- = 0.80394 ( 10 ) or ε AP = 803.9μmm ⁄ mm 200 Lo (b) From triangle PBP1 δ AP = PB = PP 1 cos 50 = 0.25 cos 50 = 0.1607 δ AP –3 0.16070 ε AP = -------- = ------------------- = 0.80348 ( 10 ) 200 Lo

or

ε AP = 803.5μmm ⁄ mm

(c) The deformation vector can be written as: D = 0.25i The unit vector can be written as: i AP = cos 50i + sin 50j The deformation of the bar AP can be calculated as: δ AP = i AP ⋅ D = 0.25 cos 50 = 0.1607 δ AP –3 The average normal strain is ε AP = -------- = 0.16070 ------------------- = 0.80348 ( 10 ) or ε AP = 803.5μmm ⁄ mm 200 Lo

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.51 The roller at P can slide only in the slot by the given amount. Determine the strains in bar AP by (a) finding the deformed length of AP without small-strain approximation, (b) using Equation 2.6, and (c)

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

using Equation 2.7. δP

.2 5 =0

mm

P 20

0m m

30o

A 50o

δP = 0.25 mm (a) εAP= ?

Solution

using geometry

(b) εAP= ? using small strain scaler approach (c) εAP= ? using small strain vector approach.

-----------------------------------------------------------We draw the exaggerated deformed shape as shown below. B

δA

Extension

20o P 160o

δP

P1

Lf A

(a) From triangle APP1, using cosine rule we obtain 2

2

2

2

2

L f = AP + PP 1 – 2AP ( PP 1 ) cos 160 = 200 + 0.25 – 2 ( 200 ) ( 0.25 ) cos 160 = 200.23494 Lf – Lo –3 0.23494 The average normal strain is: ε AP = ---------------= ------------------- = 1.1747 ( 10 ) or ε AP = 1174.7μmm ⁄ mm; 200 Lo (b) From triangle PBP1: δ AP = PB = PP 1 cos 20 = 0.25 cos 20 = 0.23492 δ AP –3 0.23492 The average normal strain is: ε AP = -------- = ------------------- = 01.1746 ( 10 ) or ε AP = 1174.6μmm ⁄ mm; Lo 200 (c) The deformation vector can be written as: D = 0.25 ( cos 30i + sin 30j ) The unit vector can be written as i AP = cos 50i + sin 50j δ AP = i AP ⋅ D = 0.25 ( cos 30 cos 50 + sin 30 sin 50 ) = 0.25 cos 20 = 0.23492 δ AP –3 ε AP = -------- = 0.23492 ------------------- = 01.1746 ( 10 ) 200 Lo

ε AP = 1174.6μmm ⁄ mm;

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.52 A roller at P slides in a slot as shown. Determine the deformation in bar AP and bar BP using small strain approximation. B 110o δP = 0.25 mm

A P

Solution

δAP= ?

δBP= ?

Figure P2.52

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

-----------------------------------------------------------We draw the exaggerated deformed shape as shown below.

Contraction

Β

δBP

P2

Extension

70o P δAP=δP P1

Α

As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP1 represents the deformation of bar AP and PP2 represents the deformation of bar BP and are calculated as shown below. δ AP = δ P = 0.25

δ AP = 0.25mm extension

or

δ BP = δ P cos 70 = 0.08551 or

δ BP = 0.0855mm contraction

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.53 A roller at P slides in a slot as shown Determine the deformation in bar AP and bar BP using small strain approximation. B

A

60o P δP = 0.25 mm

SolutionδAP= ?

Figure P2.53

δBP= ?

-----------------------------------------------------------We draw the exaggerated deformed shape as shown below. Β

60o Α

Extension

δAP=δP o P1 P 60

δBP

Extension

P2

As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP1 represents the deformation of bar AP and PP2 represents the deformation of bar BP and are calculated as shown below. δ AP = δ P = 0.25 or δ BP = δ P cos 60 = 0.125 or

δ AP = 0.25 mm extension δ BP = 0.125 mm extension

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.54 A roller at P slides in a slot as shown. Determine the deformation in bar AP and bar BP small strain approximation.

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

B 75o

A

o

30

P

Figure P2.54

Solution

δAP= ?

δBP= ?

-----------------------------------------------------------We draw the exaggerated deformed shape as shown below. Β

Α

30o Extension 75o P δAP P3 ExtensionδBP δP P2 P1

As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP1 represents the deformation of bar AP and PP2 represents the deformation of bar BP and are calculated as shown below. δ AP = δ P cos 75 = 0.06470 or δ BP = δ P cos 30 = 0.2165 or

δ AP = 0.0647mm extension δ BP = 0.2165mm extension

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.55 A roller at P slides in a slot as shown in each problem. Determine the deformation in bar AP and bar BP using small strain approximation. δP = 0.02 inch A

P

40o

110o B

Figure P2.55

Solution

δAP= ?

δBP= ?

-----------------------------------------------------------We draw the exaggerated deformed shape as shown below. P1 Extension δP P3 o A δBP 70 40o P2 δAP P Extension 110o

B

As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP2 represents the deformation of bar AP and PP3 represents the deformation of bar BP and are calculated as shown below. δ AP = δ P cos 40 = 0.01532 or δ BP = δ P cos 70 = 0.00684 or

δ AP = 0.0153 in. extension δ BP = 0.0068 in. extension

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.56 A roller at P slides in a slot as shown. Determine the deformation in bar AP and bar BP in each problem by using small strain approximation.

B

A

B

250 250 P δP=0.01 in

Figure P2.56

Solution

δAP= ?

δBP= ?

-----------------------------------------------------------We draw the exaggerated deformed shape as shown below. B

A

250 δBP P2

δP

B

250 P Extension δBP δAP

P2

P1

As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP1 represents the deformation of bar AP and PP2 represents the deformation of bar BP and are calculated as shown below. δ AP = δ P = 0.01 or δ BP = δ P cos 25 = 0.00906 or

δ AP = 0.01 in. extension δ BP = 0.0091 in. extension

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.57 A roller at P slides in a slot as shown. Determine the deformation in bar AP and bar BP using small strain approximation. A 60o 50o

P

δP = 0.02 inch

20o B

Figure P2.57

Solution

δAP= ?

δBP= ?

------------------------------------------------------------

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

We draw the exaggerated deformed shape as shown below. Contraction 60o δAP P2 P

30o

δP

δBP

A

50o

P1 20o P3 20o Contraction B

As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP2 represents the deformation of bar AP and PP3 represents the deformation of bar BP and are calculated as shown below. δ AP = δ P cos 80 = 0.00347 or δ BP = δ P cos 20 = 0.01879 or

δ AP = 0.0035inch contraction δ BP = 0.0188inch contraction

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.58 A gap of 0.004 of an inch exists between the rigid bar and bar A before the load P is applied in Figure P2.58. The rigid bar is hinged at point C. Due to force P, the strain in bar A was found to be - 600 μ. Determine the strain in bar B. The lengths of bars A and B are 30 and 50 inches respectively.

B C

75

o

P 24 in 36 in A 60 in

Figure P2.58

Solution

εA=-500μ

LA=30in

εB=?

LB=50in

-----------------------------------------------------------–6

The deformation of A is: δ A = ( 600 ) ( 10 ) ( 30 ) = 0.018 in contraction 36 in C δA 24 in Contraction

60 in D δD A

E δB

75o Extension δE

75o

From geometry of the deformed shape we obtain: δ D = δ A + gap = 0.018 + 0.004 = 0.022in and δ δD ------ = -----E36 96

96 ∴δ E = ------ ( 0.022 ) = 0.05867in and δ B = δ E cos 15 = 0.05667 36

δ –3 The average normal strain in B is: ε B = -----B- = 0.05667 ------------------- = 1.133 ( 10 ) 50 LB

ε B = 1133μin./in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.59

A gap of 0.004 in exists between the rigid bar and bar A before the load P is applied in Figure

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

P2.58. The rigid bar is hinged at point C. Due to force P, the strain in bar B was found to be 1500 μ. Determine the strain in bar A. The lengths of bars A and B are 30 and 50 in, respectively. Solution εB=1500μ LA=30in LB=50in εA=?

-----------------------------------------------------------–6

The deformation of B is: δ B = ( 1500 ) ( 10 ) ( 50 ) = 0.075 in. extension 36 in

60 in

C

D δD

δA 24 in Contraction

A

75o

E δB

Extension δE 75o

From geometry of the deformed shape we obtain: δ B = δ E cos 15

or

0.075 δ E = -------------- = 0.07765 cos 15

δ δD 36 ------ = -----Eor δ D = ------ ( 0.07765 ) = 0.02912in 96 36 96 δ A = δ D – 0.004 = 0.02512in contraction The average normal strain in A is:

0.02512- = – 0.8372 ( 10 –3 ) ε A = –--------------------30

ε A = – 837.2 μin./in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.60 For the truss shown, the pin displacements in the x and y direction, given by u and v respectively were computed by FiniteElement Method and are as given. Determine the axial strain in members AB, BF, FG, and GB. u B = 12.6 mm

v B = – 24.48 mm

u C = 21.0 mm

v C = – 69.97 mm

u D = – 16.8 mm

v D = – 119.65 mm

u E = – 12.6 mm

v E = – 69.97 mm

u F = – 8.4 mm

v F = – 28.68 mm

A

y x

B

2m G

C

E

F 2m

2m

Figure P2.60

Solution

εAB=?

εBF=?

εFG=?

D 2m P

εGB=?

-----------------------------------------------------------1. Deformation vector calculations D AB = ( u B – u A )i + ( v B – v A )j = ( 12.6i – 24.48j )mm D BF = ( u F – u B )i + ( v F – v B )j D BF = ( – 8.4 – 12.6 )i + ( – 28.68 + 24.48 )j = ( – 21 i – 4.2j )mm D FG = ( u G – u F )i + ( v G – v F )j = ( 8.4i + 28.68j )mm D GB = ( u B – u G )i + ( v B – v G )j = ( 12.6i – 24.48j )mm 2. Unit vector calculations i AB = i

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

i BF = – j i FG = – i i GB = cos 45i + sin 45j = 0.707i + 0.707j 3. Deformation calculation δ AB = D AB ⋅ i AB = 12.6mm δ BF = D BF ⋅ i BF = 4.2mm δ FG = D FG ⋅ i FG = – 8.4 mm δ GB = ( 12.6 ) ( 0.707 ) – ( 24.48 ) ( 0.707 ) = – 8.4004 4.

Strain Calculation

δ AB –3 12.6 ε AB = -------- = ---------------- = 6.3 ( 10 ) 3 L AB 2 ( 10 )

ε AB = 6300μmm/mm

δ BF –3 4.2 = ε BF = -------- = ---------------2.1 ( 10 ) 3 L BF 2 ( 10 )

ε BF = 2100μmm/mm

δ FG –3 – 8.4 = ε FG = --------- = ---------------– 4.2 ( 10 ) 3 L FG 2 ( 10 )

ε FG = – 4200 μmm/mm

δ GB –3 – 8.4004 ε GB = --------- = ------------------------ = – 2.9698 ( 10 ) 3 L GB 2 2 ( 10 )

ε GB = – 2970 μmm/mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.61 For the truss shown, the pin displacements in the x and y direction given by u and v respectively were computed by FiniteElement Method and are as given. Determine the axial strain in members BC, CF, and FE. A

u B = 12.6 mm

v B = – 24.48 mm

u C = 21.0 mm

v C = – 69.97 mm

u D = – 16.8 mm

v D = – 119.65 mm

u E = – 12.6 mm

v E = – 69.97 mm

u F = – 8.4 mm

v F = – 28.68 mm

B

C

2m G

y

E

F 2m

2m

D 2m

x P

Solution

εBC=?

εCF=?

εFE=?

-----------------------------------------------------------1. Deformation vector calculation D BC = ( u C – u B )i + ( v C – v B )j = ( 21 – 12.6 )i + ( – 69.97 + 24.48 )j = ( 8.4i – 45.52j )mm D CF = ( u F – u C )i + ( v F – v C )j = ( – 8.4 – 21.0 )i + ( – 28.68 + 69.97 )j = ( 29.4i + 41.29j )mm D FE = ( u E – u F )i + ( v E – v F )j = ( – 12.6 + 8.4 )i + ( – 69.97 + 28.68 )j = ( – 4.2i – 41.29 j )mm 2. Unit vector calculations i BC = i i CF = – cos 45i – sin 45j = – 0.707i – 0.707j i FE = i

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

3. Deformation calculation δ BC = D BC ⋅ i BC = 8.4mm δ CF = D CF ⋅ i CF = ( 29.4 ) ( 0.707 ) – ( 41.29 ) ( 0.707 ) = – 8.408mm δ FE = D FE ⋅ i FE = – 4.2 mm 4. Strain Calculation δ BC –3 8.4 ε BC = --------- = ---------------- = 4.2 ( 10 ) 3 L BC 2 ( 10 )

or

mm ε BC = 4200μ --------mm

δ CF –3 – 8.408 ε CF = --------- = ------------------------ = 2.973 ( 10 ) 3 L CF 2 2 ( 10 )

or

mm ε CF = – 2973 μ --------mm

δ FE –3 – 4.2 ε FE = -------- = ---------------- = – 2.1 ( 10 ) 3 L FE 2 ( 10 )

or

mm ε FE = – 2100 μ --------mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.62 For the truss shown, the pin displacements in the x and y direction given by u and v respectively, were computed by FiniteElement Method and are as given. Determine the axial strain in members ED, DC, and CE. u B = 12.6 mm

v B = – 24.48 mm

u C = 21.0 mm

v C = – 69.97 mm

u D = – 16.8 mm

v D = – 119.65 mm

u E = – 12.6 mm

v E = – 69.97 mm

u F = – 8.4 mm

v F = – 28.68 mm

A

B

2m G

y x

C

E

F 2m

2m

D 2m P

Solution

εED=?

εDC=?

εCE=?

-----------------------------------------------------------1. Deformation vector calculation D ED = ( u D – u E )i + ( v D – v E )j = ( – 16.8 + 12.6 )i + ( – 119.65 + 69.97 )j or D ED = ( – 4.2 i – 49.98j )mm D DC = ( u C – u D )i + ( v C – v D )j = ( 21.0 + 16.8 )i + ( – 69.97 + 119.65 )j or D DC = ( 37.8i + 49.68j )mm D CE = ( u E – u C )i + ( v E – v C )j = ( – 12.6 – 21 )i + ( – 69.97 + 69.97 )j = – 33.6i mm 2. Unit vector calculations i ED = i i DC = – cos 45i + sin 45j = – 0.707i + 0.707j i CE = j 3. Deformation calculation δ ED = D ED ⋅ i ED = – 4.2m m δ DC = D DC ⋅ i DC = – ( 37.8 ) ( 0.707 ) + ( 49.68 ) ( 0.707 ) = 8.4004mm δ CE = D CE ⋅ i CE = 0 4. Strain Calculation

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

δ ED –3 – 4.2 ε ED = --------- = ---------------- = – 2.1 ( 10 ) 3 L ED 2 ( 10 )

or

ε ED = – 2100 μ mm/mm;

δ DC –3 8.4004 ε DC = --------- = ------------------------ = 2.970 ( 10 ) 3 L DC 2 2 ( 10 )

or

ε DC = 2970 μ mm/mm; ε CE = 0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.63 For the truss shown, the pin displacements in the x and y directions given by u and v respectively, were computed by Finite Element Method and are as given. Determine the axial strains in members AB, BG, GA, and AH. u B = 7.00 mm

v B = 1.500 mm

u C = 17.55 mm

v C = 3.000 mm

u D = 20.22 mm

v D = – 4.125 mm

u E = 22.88 mm

v E = – 32.250 mm

u F = 9.00 mm

v F = – 33.750 mm

u G = 7.00 mm

v G = – 4.125 mm

uH = 0

Solution

εBG=?

D

E

G

F

3m B

3m

y x

P2

P1

A

vH = 0

εAB=?

4m

4m C

H

Figure P2.63

εGA=?

εAH=?

-----------------------------------------------------------1. Deformation vector calculations D AB = ( u B – u A )i + ( v B – v A )j = ( 7i + 1.5j )mm D BG = ( u G – u B )i + ( v G – v B )j D BG = ( 7 – 7 )i + ( – 4.125 – 1.5 )j = – 5.625j mm D GA = ( u A – u G )i + ( v A – v G )j = ( – 7 i + 4.125 j )mm D AH = ( u H – u A )i + ( v H – v A )j = 0 2. Unit vector calculations i AB = j

i BG = i

– 4i – 3j i GA = --------------------= – 0.8i – 0.6j 2 2 3 +4 i AH = i 3. Deformation calculation δ AB = D AB ⋅ i AB = 1.5mm δ BG = D BG ⋅ i BG = 0 δ GA = D GA ⋅ i GA = ( – 7 ) ( – 0.8 ) + ( 4.125 ) ( – 0.6 ) = 3.125mm δ AH = 0 4. Strain Calculation

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δ AB –3 1.5 ε AB = -------- = ---------------- = 0.5 ( 10 ) 3 L AB 3 ( 10 )

January 2014

ε AB = 500 μ mm/mm;

or

ε BG = 0 δ GA –3 3.125 ε GA = --------- = ---------------- = 0.625 ( 10 ) 3 L GA 5 ( 10 )

ε GA = 625 μ mm/mm;

or

ε AH = 0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.64 For the truss shown, the pin displacements in the x and y directions given by u and v respectively, were computed by Finite Element Method and are as given. Determine the axial strains in members BC, CG, GB, and CD. 4m

4m C

D

E

G

F

3m B

y

3m

x

Solution

P2

P1

A

H

εBC=?

εCG=?

εGB=?

u B = 7.00 mm

v B = 1.500 mm

u C = 17.55 mm

v C = 3.000 mm

u D = 20.22 mm

v D = – 4.125 mm

u E = 22.88 mm

v E = – 32.250 mm

u F = 9.00 mm

v F = – 33.750 mm

u G = 7.00 mm

v G = – 4.125 mm

u

= 0

εCD=?

v

= 0

-----------------------------------------------------------1. Deformation vector calculations D BC = ( u C – u B )i + ( v C – v B )j = ( 17.55 – 7 )i + ( 3 – 1.5 )j = ( 10.55i + 1.5j )mm D CG = ( u G – u C )i + ( v G – v C )j = ( 7 – 17.55 )i + ( – 4.125 – 3 )j = ( – 10.55 i – 7.125j )mm D GB = ( u B – u G )i + ( v B – v G )j = ( 7 – 7 )i + ( 1.5 + 4.125 )j = ( 5.625j )mm D CD = ( u D – u C )i + ( v D – v C )j = ( 20.22 – 17.55 )i + ( – 4.125 – 3 )j = ( 2.67i – 7.125j )mm 2. Unit vector calculations i BC = j 4i – 3j i CG = ---------------- = 0.8i – 0.6j 5 i GB = – i i CD = i 3. Deformation calculation δ BC = D BC ⋅ i BC = 1.5mm δ CG = D CG ⋅ i CG = ( 0.8 ) ( – 10.55 ) + ( – 0.6 ) ( – 7.125 ) = – 4.165mm δ GB = D GB ⋅ i GB = 0 δ CD = 2.67mm 4. Strain Calculation δ BC –3 1.5 = ε BC = --------- = ---------------0.5 ( 10 ) 3 L BC 3 ( 10 )

or

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δ CG –3 – 4.165 ε CG = --------- = ---------------- = – 0.833 ( 10 ) 3 L CG 5 ( 10 )

January 2014

mm ε CG = – 833 μ --------mm

or

δ GB ε GB = --------- = 0 L GB

ε GB = 0

δ CD –3 2.67 ε CD = --------- = ---------------- = 0.6675 ( 10 ) 3 L CD 4 ( 10 )

mm ε CD = 667.5μ --------mm

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.65 For the truss shown, the pin displacements in the x and y directions given by u and v respectively, were computed by Finite Element Method and are as given. Determine the axial strains in members GF, FE, EG, and DE.. 4m

4m u B = 7.00 mm

v B = 1.500 mm

u C = 17.55 mm

v C = 3.000 mm

u D = 20.22 mm

v D = – 4.125 mm

u E = 22.88 mm

v E = – 32.250 mm

u F = 9.00 mm

v F = – 33.750 mm

u G = 7.00 mm

v G = – 4.125 mm

uH = 0

Solution

C

D

E

G

F

3m B

3m

y x

A

P2

P1 H

vH = 0

εGF=?

εFE=?

εEG=?

εDE=?

-----------------------------------------------------------1. Deformation vector calculations D GF = ( u F – u G )i + ( v F – v G )j = ( 9 – 7 )i + ( – 33.75 + 4.125 )j = ( 2i – 29.625 j )mm D FE = ( u E – u F )i + ( v E – v F )j = ( 22.88 – 9 )i + ( – 32.25 + 33.75 )j = ( 13.88i + 1.5j )mm D EG = ( u G – u E )i + ( v G – v E )j = ( 7 – 22.88 )i + ( – 4.125 + 32.25 )j = ( – 15.88i + 28.125j )mm D DE = ( u E – u D )i + ( v E – v D )j = ( 22.88 – 20.22 )i + ( – 32.25 + 4.125 )j or D DE = ( 2.66i – 28.125j )mm 2. Unit vector calculations i GF = i

i FE = j

– 4i – 3j i EG = -------------------- = – 0.8i – 0.6j 5 i DE = i 3. Deformation calculation δ GF = D GF ⋅ i GF = 2mm δ FE = D FE ⋅ i FE = 1.5mm δ EG = D EG ⋅ i EG = ( – 15.88 ) ( – 0.8 ) + ( 28.125 ) ( – 0.6 ) = – 4.171mm δ DE = 2.66mm 4. Strain Calculation δ GF –3 2 ε GF = --------- = ---------------= 0.5 ( 10 ) 3 L GF 4 ( 10 )

or

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ε GF = 500 μ mm/mm;

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

δ FE –3 1.5 ε FE = -------- = ---------------- = 0.5 ( 10 ) 3 L FE 3 ( 10 )

or

ε FE = 500 μ mm/mm; [

δ EG –3 – 4.171 ε EG = --------- = ---------------- = – 0.8342 ( 10 ) 3 L EG 5 ( 10 )

or

ε EG = – 834 μ mm/mm;

δ DE –3 2.66 ε DE = --------- = ---------------- = 0.665 ( 10 ) 3 L DE 4 ( 10 )

or

ε DE = 665 μ mm/mm;

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.66 Three poles are pin connected to a ring at P and to the supports on the ground. The ring slides on a vertical rigid pole by 2 inches, as shown. The coordinates of the four points are as given. Determine the normal strain in each bar due to the movement of the ring. z δP = 2 inch P (0.0, 0.0, 6.0) ft. C -2.0, -3.0, 0.0) ft.

B (-4.0, 6.0, 0.0) ft. y

x

A (5.0, 0.0, 0.0) ft.

Figure P2.66

1. Deformation vector calculations: Points A,B, and C are fixed. Point P moves vertically upwards. Thus, the deformation vectors for the three bars is given by: D AP = 2k D BP = 2k D CP = 2k 2. Unit vector calcultions: The unit vectors in the directions of the rod can be found as shown below. r AP = – 5.0i + 6k

r AP =

2

2

5 + 6 = 7.81

r AP i AP = ---------- = – 0.6402i + 0.7682k r AP r BP = 4.0i – 6j + 6k

r BP =

2

2

r BP =

2

4 + 6 + 6 = 9.381

2

2

2 3

r BP i BP = ---------- = 0.4264i – 0.6396i + 0.6396k r BP r CP = – 2.0 i – 3j + 6k

1

2

2 + 3 + 6 = 7.0ft

4 5 6

r CP i CP = ----------- = 0.2857i + 0.4286i + 0.8571k r CP

7

δ AP = D AP • i AP = ( 2 ) ( 0.7682 ) = 1.5364 inch

8

δ BP = D BP • i BP = ( 2 ) ( 0.6396 ) = 1.2792 inch

9

3. Deformation calculations

δ CP = D CP • i CP = ( 2 ) ( 0.8571 ) = 1.7142 inch 10 4. Strain calcultions: The lengths of each bar are the magnitudes of the position vectors in Eqs. 2, 4, and 6. The normal strains can be found as shown below.

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δ AP –3 1.5364 ε AP = -------- = -------------------------- = 16.393 ( 10 ) L AP ( 12 ) ( 7.81 )

ε AP = 16393μin./in.

δ BP –3 1.2792 ε BP = -------- = ----------------------------- = 11.363 ( 10 ) L BP ( 12 ) ( 9.381 )

ε BP = 11363μin./in.

δ CP –3 1.7142 ε CP = --------- = ------------------- = 20.407 ( 10 ) ( 12 ) ( 7 ) L CP

ε CP = 20407μin./in.

January 2014

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.67 A rectangle deforms into the colored shape shown in each problem. Determine the average values of strain components ε xx, εyy, and γxy at point A. 0.0042 in

y 0.0056 in

0.0042 in

1.4 in

x A 3.0 in

Solution

εxx=?

εyy=?

0.0036 in

γxy=?

-----------------------------------------------------------The average strains can be found as shown below. Δu 0.0036 ε xx = ------ = ---------------- = 0.0012 Δx 3

ε xx = 1200μin./in.

Δv 0.0042 ε yy = ------ = ---------------- = 0.003 Δy 1.4

ε yy = 3000μin./in.

Δu Δv 0.0056 0.0042 γ xy = ------ + ------ = ---------------- + ---------------- = 0.0054 Δy Δx 1.4 3

γ xy = 5400μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.68 A rectangle deforms into the colored shape shown in each problem. Determine the average values of strain components ε xx, εyy, and γxy at point A. y 0.45 mm

450 mm

0.30 mm

0.65 mm

x

A 0.032 mm

250 mm

Solution

εxx=?

εyy=?

γxy=?

-----------------------------------------------------------The average strains can be found as shown below. –3 Δu – 0.32 ε xx = ------ = ------------- = – 0.128 ( 10 ) Δx 250

mm ε xx = – 128μ --------mm

0.3- = – 0.6667 ( 10 – 3 ) ε yy = Δ -----v- = – --------Δy 450

mm ε yy = – 666.7μ --------mm

–3 γ xy = Δ -----u- + Δ -----v- = 0.45 ---------- + 0.65 ---------- = 3.6 ( 10 ) Δy Δx 450 250

γ xy = 3600μrad

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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January 2014

2.69 A rectangle deforms into the colored shape shown in each problem. Determine the average values of strain components ε xx, εyy, and γxy at point A. y 0.033 mm 0.006 mm 3 mm x 0.024 mm

A

0.009 mm 6 mm

Solution

εxx=?

εyy=?

γxy=?

-----------------------------------------------------------The average strains can be found as shown below. 0.009- = – 1.5 ( 10 – 3 ) ε xx = Δ -----u- = –--------------Δx 6

ε xx = – 1500 μ mm/mm;

0.006- = – 2.0 ( 10 –3 ) ε yy = Δ -----v- = – --------------Δy 3

ε yy = – 2000 μ mm/mm;

– 0.024 )- = 7 ( 10 – 3 ) γ xy = Δ -----u- + Δ -----v- = 0.033 ------------- + (-------------------Δy Δx 3 6

γ xy = 7000μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.70 Displacements u and v in the x and y directions respectively were measured by Moire Interferometry method at many points on a body. Displacements of four points on a body are as given. Determine the average values of strain components εxx, εyy, and γxy at point A shown. y u = 0.500μmm v = – 1.000 μmm A

u B = 1.125μmm

v B = – 1.3125 μmm

uC = 0

v C = – 1.5625 μmm

u D = 0.750μmm

v D = – 2.125 μmm

Solution

εxx=?

εyy=?

0.0005 mm

A

C

D

A

B

x

0.0005 mm

γxy=?

-----------------------------------------------------------The average strains can be found as shown below. –6 uB – uA –6 0.625 ( 10 ) ε xx = ----------------- = ----------------------------- = 1250 ( 10 ) xB – xA 0.0005 –6 vC – vA –6 – 0.5625 ( 10 ) ε yy = ----------------= ------------------------------------ = – 1125 ( 10 ) 0.0005 yC – yA

ε xx = 1250 μ mm/mm;

or

or

ε yy = – 1125 μ mm/mm;

u C – u A v B – v A – 0.5 ( 10 – 6 ) ( – 0.3125 ) ( 10 – 6 ) –6 γ xy = ----------------- + ----------------- = --------------------------- + ----------------------------------------- = – 1625 ( 10 ) or yC – yA xB – xA 0.0005 0.0005 γ xy = – 1625 μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.71 Displacements u and v in the x and y directions respectively were measured by Moire Interferometry method at many points on a body. Displacements of four points on a body are as given. Determine the

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January 2014

average values of strain components εxx, εyy, and γxy at point A shown. v A = – 0.3125μmm

u B = 1.500μmm

v B = – 0.5000μmm

u C = 0.250μmm

v C = – 1.125 μmm

u D = 1.250μmm

v D = – 1.5625 μmm

0.0005 mm

y u A = 0.625μmm

C

D

A

B

x

0.0005 mm

Figure P2.71

Solution

εxx=?

εyy=?

γxy=?

-----------------------------------------------------------The average strains can be found as shown below. –6 uB – uA –6 ( 1.5 – 0.625 ) ( 10 ) - = ------------------------------------------------ = 1750 ( 10 ) ε xx = ----------------xB – xA 0.0005

mm ε xx = 1750μ --------mm

–6 vC – vA –6 ( – 1.125 + 0.3125 ) ( 10 ) ε yy = ----------------= -------------------------------------------------------------- = – 1625 ( 10 ) yC – yA 0.0005

mm ε yy = – 1625 μ --------mm

u C – u A v B – v A ( 0.25 – 0.625 ) ( 10 – 6 ) ( – 0.5 + 0.3125 ) ( 10 – 6 ) γ xy = ----------------- + ----------------- = --------------------------------------------------- + -----------------------------------------------------yC – yA xB – xA 0.0005 0.0005

γ xy = – 1125 μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.72 Displacements u and v in the x and y directions respectively were measured by Moire Interferometry method at many points on a body. Displacements of four points on a body are as given. Determine the average values of strain components εxx, εyy, and γxy at point A shown. v A = – 0.5625 μmm

u B = 0.250μmm

v B = – 1.125 μmm

u C = – 1.250 μmm

v C = – 1.250 μmm

u D = – 0.375 μmm

v D = – 2.0625 μmm

y 0.0005 mm

u A = – 0.500 μmm

C

D

A

B

x

0.0005 mm

Figure P2.72

Solution

εxx=?

εyy=?

γxy=?

-----------------------------------------------------------The average strains can be found as shown below. –6 uB – uA –6 ( 0.25 + 0.5 ) ( 10 ) ε xx = ----------------- = --------------------------------------------- = 1500 ( 10 ) xB – xA 0.0005

ε xx = 1500 μ mm/mm;

–6 vC – vA –6 ( – 1.25 + 0.5625 ) ( 10 ) ε yy = ----------------= ----------------------------------------------------------- = – 1375 ( 10 ) 0.0005 yC – yA

ε yy = – 1375 μ mm/mm;

u C – u A v B – v A ( – 1.25 + 0.5 ) ( 10 – 6 ) ( – 1.125 + 0.5625 ) ( 10 – 6 ) –6 γ xy = ----------------- + ----------------- = -------------------------------------------------- + -------------------------------------------------------------- = – 2625 ( 10 ) yC – yA xB – xA 0.0005 0.0005 γ xy = – 2625 μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.73 Displacements u and v in the x and y directions respectively were measured by Moire Interferometry method at many points on a body. Displacements of four points on a body are as given. Determine the

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average values of strain components εxx, εyy, and γxy at point A shown. v A = – 1.125 μmm

u B = 1.250μmm

v B = – 1.5625 μmm

u C = – 0.375 μmm

v C = – 2.0625 μmm

u D = 0.750μmm

v D = – 2.7500 μmm

0.0005 mm

y u A = 0.250μmm

C

D

A

B

x

0.0005 mm

Figure P2.73

Solution

εxx=?

εyy=?

γxy=?

-----------------------------------------------------------The average strains can be found as shown below. –6 uB – uA –6 ( 1.25 – 0.25 ) ( 10 ) ε xx = ----------------- = ------------------------------------------------ = 2000 ( 10 ) 0.0005 xB – xA

ε xx = 2000 μ mm/mm;

–6 vC – vA –6 ( – 2.0625 + 1.125 ) ( 10 ) ε yy = ----------------= -------------------------------------------------------------- = – 1875 ( 10 ) yC – yA 0.0005

ε yy = – 1875 μ mm/mm;

u C – u A v B – v A ( – 0.375 – 0.25 ) ( 10 – 6 ) ( – 1.5625 + 1.125 ) ( 10 – 6 ) –6 γ xy = ----------------- + ----------------- = ------------------------------------------------------- + -------------------------------------------------------------- = – 2125 ( 10 ) 0.0005 0.0005 yC – yA xB – xA γ xy = – 2125μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.74 In a tapered circular bar that is hanging vertically, the axial displacement due to its weight was found and is given below. Determine the axial strain εxx at x = 24 inches. 2 –3 933.12 u ( x ) = – 19.44 + 1.44x – 0.01x – ------------------- ( 10 ) inches ( 72 – x )

Solution

εxx(24)=?

-----------------------------------------------------------The strain at any point x can be found as shown below. ε xx ( x ) =

–3 du 933.12 = 1.44 – 0.02x – ---------------------- ( 10 ) 2 dx ( 72 – x )

933.12 - ( 10 – 3 ) = 0.555 ( 10 – 3 ) or ε xx ( 24 ) = du = 1.44 – 0.02 ( 24 ) – -----------------------2 dx ( 72 – 24 )

in ε xx ( 24 ) = 555μ ----in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.75 In a tapered rectangular bar that is hanging vertically, the axial displacement due to its weight was found and is given below. Determine the axial strain εxx at x = 100 mm –6

2

–6

u ( x ) = [ 7.5 ( 10 )x – 25 ( 10 )x – 0.15 ln ( 1 – 0.004x ) ] mm

Solution

εxx(100)=?

-----------------------------------------------------------The strain at any point x can be found as shown below ε xx ( x ) =

–6 –6 du ( 0.15 ) ( 0.004 ) = 15 ( 10 )x – 25 ( 10 ) + ---------------------------------dx ( 1 – 0.004x )

–6 –6 ( 0.15 ) ( 0.004 ) ε xx ( 100 ) = – 15 ( 10 ) ( 100 ) + 25 ( 10 ) + ---------------------------------- = 0.0015 – 0.000025 + 0.001 ( 0.6 )

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January 2014

ε xx ( 100 ) = 2475μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.76 The axial displacement in a quadratic one-dimensional finite element is as given below. Determine the strain at Node 2. Node 2

Node 1

u2 u3 u1 u ( x ) = --------2 ( x – a ) ( x – 2a ) – ----2- ( x ) ( x – 2a ) + --------2 ( x ) ( x – a ) 2a a 2a

Node 3

x x2= a

x1= 0

x3= 2a

εxx(a)=?

Solution

-----------------------------------------------------------The strain at any point x can be found as shown below. u1 u3 u du ε xx ( x ) = = -------- [ ( x – 2a ) + ( x – a ) ] – ----2- [ ( x – 2a ) + x ] + -------- [ ( x – a ) + x ] 2 2 2 dx 2a 2a a u1 u2 u3 u1 u2 u3 u3 – u1 ε xx ( x ) = -------- ( 2x – 3a ) – ----- ( 2x – 2a ) + -------- ( 2x – a ) = -------- ( – a ) – ----- ( 0 ) + -------- ( a ) or ε xx ( a ) = ---------------2 2 2 2 2 2 2a 2a a 2a 2a a 2a

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.77

0.2 - . Determine Due to the applied load, the strain in the tapered bar due was found to be εxx = --------------------2 ( 40 – x )

the extension of the bar. 20 in x

Solution

P

u(20)=?

-----------------------------------------------------------The strain at any point x can be written as shown below. ε xx =

du 0.2 = ---------------------2 dx ( 40 – x )

1

0.2 - + C Method I : Integrating Eq. (1) we obtain: u ( x ) = -----------------1 ( 40 – x ) Now

u ( 0 ) = 0 therefore C 1 = – 0.2 ------- = – 0.005 40

0.2 0.2 u ( x ) = ------------------- – 0.005 or u ( 20 ) = ---------------------- – 0.005 = 0.005 or ( 40 – x ) ( 40 – 20 )

u ( 20 ) = 0.005in

-----------------------------------------------------------Method II :Integrating Eq. (1) from x = 0 to x = 20 we obtain the following. u ( 20 )

∫0

du =

∫

20

20

0.2 0.2 0.2 --------------------= ------------------ = 0.005 or dx = ------------------2 ( 40 – x ) 40 – 20 0 ( 40 – x ) 0

u ( 20 ) = 0.005in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.78

KL The axial strain in a bar of length L was found to be ε xx = ---------------------( 4L – 2x )

0≤x≤L

where K is a con-

stant for a given material, loading, and cross-section dimension. Determine the total extension in terms of K and L. Solution u(L)-u(0)=?

------------------------------------------------------------

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The strain at any point x can be written as shown below. KL ε xx = du = ---------------------dx ( 4L – 3x ) Integrating Eq. (1) from x = 0 to x = L, we obtain the following. u(L)

L

KL

- dx or u ( L ) – u ( 0 ) ∫u ( 0 ) du = ∫0 ---------------------( 4L – 3x )

1

L

KL KL KL = ----------- ln ( 4L – 3x ) = – ------- [ ln ( L ) – ln ( 4L ) ] = – ------- ( – ln 4 ) 3 ( –3 ) 2 0 u( L ) – u ( 0 ) = 0.4621KL

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.79

The axial strain in a bar due to its own weight of length L was found to be 3

8L ε xx = K 4L – 2x – ------------------------2 ( 4L – 2x )

0≤x≤L

where K is a constant for a given material and cross-section

dimensions. Determine the total extension in terms of K and L. Solution u(L)-u(0)=?

-----------------------------------------------------------The strain at any point x can be written as shown below. ε xx =

3

du 8L = K 4L – 2x – -------------------------2 dx ( 4L – 2x )

1

Integrating Eq. (1) from x = 0 to x = L, we obtain the following. u(L)

∫u ( 0 )

du =

∫

3

L

8L K 4L – 2x – ------------------------- dx or 2 0 ( 4L – 2x ) 2

3

L

3

3

2 2 8L 2 8L 8L 2x u ( L ) – u ( 0 ) = K 4Lx – -------- – ----------------------= K 4L – L – --------- + --------- = 2KL 2 ( 4L – 2x ) 0 4L 8L

u( L ) – u ( 0 ) = 2KL

2

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.80 A bar has a tapered and a uniform section securely fastened. Determine the total extension of the bar if the axial strain in each section is as given below. 3

1500 ( 10 ) ε = ------------------------- μ 0 ≤ x ≤ 750 mm ( 1875 – x ) ε = 1500 μ 750 mm ≤ x ≤ 1250 mm P

x

750 mm

Solution

500 mm

u(1250)=?

-----------------------------------------------------------The strain at any point x can be written as shown below. 3

1500 ( 10 ) ε xx = du = ------------------------- μ ( 1875 – x ) dx ε xx = du = 1500 μ dx

0 ≤ x ≤ 750 mm

750 mm ≤ x ≤ 1250 mm

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Integrating Eq. (1), we obtain:

∫

u ( 750 ) du u(0)

=

∫

–3

3 750 1500 ( 10 ) ------------------------- dx 0 ( 1875 – x )

u ( 1250 )

du =

∫750

1250

–6

( 1500 ) ( 10 ) dx

–3

= – 1500 ( 10 ) [ ln ( 1875 – 750 ) – ln ( 1875 ) ] = 766.24 ( 10 ) 3 –6

–3

1250

u ( 1250 ) –u ( 750 ) = 1500 ( 10 )x 750 = 750 ( 10 )

or

4

–3

u ( 1250 ) – u ( 0 ) = 1516.24 ( 10 )m

Adding Eq.(3) and Eq.(4) Noting

or –3

750

u ( 750 ) –u ( 0 ) = – 1500 ( 10 ) ln ( 1875 – x ) 0

∫u ( 750 )

January 2014

u ( 0 ) = 0 we obtain

u ( 1250 ) = 1.516mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.81 N axial bars are securely fastened together. Determine the total extension of the composite bar if the strain in the ith section is as given. εi = ai

Solution

xi – 1 ≤

P

x 1

≤ xi

2 xi-1 xi

i

N-1 N

uN=?

-----------------------------------------------------------The axial strain in the ith segment can be written as: du ε i = ------ = a i dx ui

Integraing Eq. (1) we obtain:

∫u

xi

du = i–1

∫x

a i dx or

xi – 1 ≤

≤ xi

1

ui – ui – 1 = ai ( xi – xi – 1 )

1≤i≤N

i–1

Adding the relative displacements over all segments, we obtain: N

N

∑ ( ui – ui – 1 ) i=1

Noting

=

N

N

N

∑ ai ( xi – xi – 1 ) or ∑ ui – ∑ ui – 1 i=1

i=1

=

N

∑ ai ( xi – xi – 1 ) or uN – u0 i=1

i=1

u 0 = 0 we obtain:

=

N

uN =

∑ ai ( xi – xi – 1 ) i=1

∑ ai ( xi – xi – 1 ) i=1

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.82 True strain (εT) is calculated from dεT = ( du ) ⁄ ( L o + u ) , where u is the deformation at any given instance and Lo is the original undeformed length. Thus the increment in true strain is the ratio of change in length at any instance to the length at that given instance. If ε represents engineering strain show that at any instance the relationship between true strain and engineering strain is given by the following: ε T = ln ( 1 + ε )

Solution

2.12

uN=?

-----------------------------------------------------------Integrating dε T = ( du ) ⁄ ( L o + u ) we obtain: ε T = ln ( L o + u )

εT

∫o

dε T =

u

- or ∫0 ------------------( Lo + u ) du

Lo + u u⎞ = ln ( L o + u ) – ln ( L o ) = ln ⎛ ---------------⎞ = ln ⎛ 1 + ----⎝ ⎝ Lo ⎠ L o⎠ 0 u

1

Total Deformation u By defination Engineering strain is given by: ε = ---------------------------------------------------- = ----- Substituting this defination into Eq. Original Length Lo (1), we obtain: ε T = ln ( 1 + ε )

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.83

The displacements in a body are given by: 2

2

–3

u = [ 0.5 ( x – y ) + 0.5xy ] ( 10 )mm

2

2

–3

v = [ 0.25 ( x – y ) – xy ] ( ( 10 )mm )

Determine the strains ε xx , ε yy , and γ xy at x = 5 mm and y = 7 mm.’

Solution

εyy=?

{εxx=?

γxy=?} at x =5mm and y=7 mm

-----------------------------------------------------------The strains at a point can be found as shown below. –3 –3 ε xx = ∂u = [ 0.5 ( 2x ) + 0.5y ] ( 10 ) = ( x + 0.5y ) ( 10 ) ∂x

1

–3 –3 ∂v = [ 0.25 ( – 2y ) – x ] ( 10 ) = ( – 0.5 y – x ) ( 10 ) ∂y

2

–3 –3 –3 ∂u ∂v + = [ 0.5 ( – 2 y ) + 0.5x ] ( 10 ) + [ 0.25 ( 2x ) – y ] ( 10 ) = γ xy = ( x – 2y ) ( 10 ) ∂y ∂x

3

ε yy = γ xy =

Substituting x = 5 and y = 7 in Eq.(1),(2) and (3). We obtain: –3

–3

ε xx = ( 5 + 3.5 ) ( 10 ) = 8.5 ( 10 ) –3

–3

ε yy = ( – 3.5 – 5 ) ( 10 ) = – 8.5 ( 10 ) –3

–3

γ xy = ( 5 – 14 ) ( 10 ) = – 9 ( 10 )

or

ε xx = 8500 μ mm/mm;

or

ε yy = – 8500 μ mm/mm; γ xy = – 9000 μrad

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.84 A metal strip is to be pulled and bent to conform to a rigid surface such that the length of strip between OA fits the arc OB of the surface. The equation of the surface y=f(x) and the length OA are as given below. Determine the average normal strain in the metal strip. f ( x ) = 0.04x

3⁄2

y

y = f(x)

inches and length OA = 9 inches.

A

O

Solution

B x

ε=?

L o = OA = 9

-----------------------------------------------------------1 ---

1 ---

2 3 2 The first derivative of y can be written as: dy = ( 0.04 ) ⎛ ---⎞ x = 0.06x ⎝ ⎠ 2 dx The length of the deformed rod can be written as 9

Lf =

∫ ds = ∫0

2

9

3 --2

9

dy 2 1 + ⎛ ⎞ dx = ( 1 + 0.0036x ) dx = ------------------------ ( 1 + 0.0036x ) or ⎝ d x⎠ 3 ( 0.0036 ) 0 0

∫

3 --2

L f = 185.19 ( 1 + ( 0.0036 ) ( 9 ) ) – 1 = 9.07275 Lf – Lo –3 0.07275 The average normal strain is: ε = ---------------= ------------------- = 8.083 ( 10 ) 9 Lf

ε = 8083μin./in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.85 A metal strip is to be pulled and bent to conform to a rigid surface such that the length of strip between OA fits the arc OB of the surface. The equation of the surface y=f(x) and the length OA are as

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

given belo. Determine the average normal strain in the metal strip. f ( x ) = 625 x

3⁄2

μmm

y

and length OA = 200 mm.

y = f(x)

B A

O

Solution

x

ε=?

L o = OA = 200mm

-----------------------------------------------------------1 ---

1 ---

2 –6 –3 2 3 The first derivative of y can be written as: dy = ⎛ ---⎞ ( 625 ) x ( 10 ) = 0.9375 ( 10 )x ⎝ 2⎠ dx The length of the deformed rod can be written as

Lf =

∫

2

200

ds =

200

dy 1 + ⎛ ⎞ dx = ⎝ d x⎠ 0

∫0

∫

2

3 --2

200

2 –6 2 L f = ------------------------------------------- ( 1 + ( 0.9375 ) ( 10 )x ) 2 –6 3 ( 0.9375 ) ( 10 ) 0 2

–6

1 + ( 0.9375 ) ( 10 )x dx or

or

3 --2

–6

L f = 711.11 ( 1 + ( 0.9375 ) ( 10 )200 ) – 1 = 200.00844 mm Lf – Lo –6 The average normal strain is: ε = ---------------= 42.19 ( 10 ) or Lf

mm ε = 42.2μ --------mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.86 A metal strip is to be pulled and bent to conform to a rigid surface such that the length of strip between OA fits the arc OB of the surface. The equation of the surface y=f(x) and the length OA are as given below. Determine the average normal strain in the metal strip. 3⁄2

y

f ( x ) = ( 0.04x – 0.005x ) inches and length OA = 9 inches. Use numerical integration.

y = f(x)

O

Solution

B A

x

ε=?

L o = OA = 9

-----------------------------------------------------------1 ---

1 ---

2 2 3 dy The first derivative of y can be written as: = --- ( 0.04 )x – 0.005 = 0.06x – 0.005 2 dx

The length of the deformed rod can be written as: L f = 1

Let

f(x) =

2

--⎛ ⎞ 2 1 + ⎜ 0.06x – 0.005⎟ , thus ⎝ ⎠

∫

1

2

--9 ⎛ ⎞ 2 dy 2 ds = 1 + ⎛ ⎞ dx = 1 + ⎜ 0.06x – 0.005⎟ dx ⎝ d x⎠ ⎝ ⎠ 0 0

∫

9

∫

9

Lf =

∫0 f ( x ) dx

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

Using numerical integration as described in Figure B2 with a Δxi=0.25, we obtain the following the table below. Numerical Integration Results from Spread Sheet. xi

f(xi)

Integral value

xi

f(xi)

Integral value

0

1.0000125

0.250040619

4.75

1.007877637

5.020277911

0.25

1.000312451

0.500167191

5

1.008307175

5.272408426

0.5

1.000700123

0.750392467

5.25

1.008736947

5.524646409

0.75

1.001102085

1.000719147

5.5

1.009166921

5.776991909

1

1.001511358

1.251148721

5.75

1.009597073

6.029444965

1.25

1.001925237

1.501682168

6

1.010027379

6.282005615

1.5

1.002342333

1.752320187

6.25

1.010457817

6.534673888

1.75

1.002761823

2.003063311

6.5

1.010888369

6.787449811

2

1.00318317

2.253911957

6.75

1.011319017

7.040333407

2.25

1.003605998

2.504866462

7

1.011749746

7.293324693

2.5

1.004030038

2.755927102

7.25

1.012180542

7.546423685

2.75

1.004455082

3.007094109

7.5

1.012611393

7.799630394

3

1.004880973

3.258367678

7.75

1.013042285

8.052944831

3.25

1.005307582

3.509747977

8

1.013473208

8.306367001

3.5

1.005734807

3.761235148

8.25

1.013904153

8.559896909

3.75

1.006162564

4.012829316

8.5

1.01433511

8.813534556

4

1.006590781

4.264530589

8.75

1.01476607

9.067279943

9

1.015197025

4.25

1.007019398

4.516339059

4.5

1.007448365

4.768254809

The deformed length from above table is: L f = 9.0672799 in . Lf – Lo –3 The average normal strain is: ε = ---------------= 0.0672799 ------------------------- = 7.4755 ( 10 ) or 9 Lf

ε = 7475.5μin./in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2.87 Measurements were made along the path of the stretch cord that is stretched over the canoe in problem 2.4 are shown in the table below. The y-coordinate was measured to the closest 1/32 inch. Between points A and B the cord path can be approximated by a straight line. Determine the average strain in the stretch cord if the original length of it is 40 inches. Use spread sheet and approximate each 2 inch xinterval by a straight line. xi

yi

xi

yi

0

17

10

15 (16/32)

2

16 (30/32)

12

14 (24/32)

4

16 (29/32)

14

13 (28/32)

6

16 (19/32)

xB = 16

yB = 12

8

16 (3/32)

xA = 18

yA = 0

C

yi B

17 in

12 in

xi

A

18 in

Solution

ε=?

Lo=40

-----------------------------------------------------------The deformed length of each 2 inch segment can be found using. Li =

2

( xi – xi – 1 ) + ( yi – yi – 1 )

2

2 ≤ i ≤ 10

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Mechanics of materials, 2nd edition, solutions manual: Chapter 2

January 2014

10

The final length of AC can be found as AC =

∑ Li . Using Spread sheet the following table can be constructed. i=2

Results of Calculation of the total length using spread sheet. xi

yi

0

17

2

16.9375

2.000976324

4

16.90625

2.000244126

6

16.59375

2.024266843

8

16.09375

2.061552813

10

15.5

2.086273966

12

14.75

2.136000936

14

13.875

2.18303115

16

12

2.741464025

0

12.16552506

18 Total Length

From the above table:

Li

29.39933524

AC = 29.399335

( AC ) – 40- = 0.46997 The average normal strain is: ε = 2--------------------------40

ε = 47%

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

-

3.1 3.5 A tensile test specimen having a diameter of 10 mm and a gage length of 50 mm was tested to fracture. The stress-strain curve from the tension test is shown in Figure P3.1. The lower plot is the expanded region OAB and associated with the strain values given in the lower scale. Solve the problems 3.1 through 3.5 using this graph. C

480

Upper Scale

D

Stress MPa

360 AB

E

A

B

Lower Scale

240

120

Figure P3.1

O 0.00

0.04

0.00

0.002

0.08

0.12

0.004 0.006 Strain mm/mm

0.16

0.20

0.008

0.010

3.1 Determine: (a) the ultimate stress; (b) the fracture stress; (c) the modulus of elasticity; (d) the proportional limit; (e) the offset yield stress at 0.2%; (f) the tangent modulus at stress level of 420 MPa; (g) the secant modulus at stress level of 420 MPa Solution

σult=? σfrac=? E=? Es = ? @σ = 420 MPa

σprop= ? σ yield= ? @εoffset = 0.2% ET = ? @σ = 420 MPa

------------------------------------------------------------

(a) By inspection

σult=480+30

σ ult = 510MPa

(b) By inspection

σfrac=480

σ frac = 480MPa 6

9 N 300 ( 10 ) E = ---------------------- = 150 ( 10 ) ------2 0.002 m

(c) From the lower plot (d) By inspection of the lower plot

or

E = 150GPa σ prop = 300MPa

0.2- = 0.002 (e) The offset strain is: ε offset = -------Starting from a strain value of 0.002, we draw a line parallel 100

to OA on the lower plot in Figure P3.1, intersecting the stress-strain curve at point E. The stress at point E is 300 MPa σ yield = 300MPa The stress value of 420 MPa corresponds to point D on the upper plot in Figure P3.1. We draw a tangent line to the stress strain curve at point D and calculate the slope as shown below: 6

9 2 ( 450 – 400 ) ( 10 ) E t = ------------------------------------------ = 2.5 ( 10 )N ⁄ m 0.02

E t = 2.5GPa

To find the secant modulus, we draw a line from point D to point O in Figure P3.1 and find the slope as shown below. 6

9 2 ( 420 – 0 ) ( 10 ) E s = ------------------------------------ = 6.46 ( 10 )N ⁄ m 0.065 – 0

E s = 6.5GPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.2

Determine the axial force acting on the specimen when it is extended by (a) 0.2 mm (b) 4.0 mm.

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Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

(a) P=? @ δ = 0.2 mm

d = 10 mm; Lo = 50 mm;

January 2014

(b) P=? @ δ = 4.0 mm

------------------------------------------------------------

The area of cross-section can be found as shown below. 2 π 2 A = --- d = 78.5398mm 4

1

δ 0.2 (a) The strain in the specimen can be found as: ε = ----- = ------- = 0.004 L0

50

σ = 300 MPa

From the lower plot in Fig. P3.8

6 –6 P σ = ---- or P = ( 300 ) ( 10 ) ( 78.5398 ) ( 10 ) = 23561.9 N or A δ 4.0 (b) The strain in the specimen can be found as: ε = ----- = ------- = 0.08 L0 50

σ = 420 + 30 = 450MPa

From the upper plot in Fig. P3.8 P σ = ---- or A

P = 23.56kN

6

–6

P = ( 450 ) ( 10 ) ( 78.5398 ) ( 10 ) = 35342.9 N or

P = 35.34kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.3 Determine the extension of the specimen when the axial force on the specimen is 33 kN. Solution d = 10 mm L0= 50 δ = ? @P = 33 kN ------------------------------------------------------------

The area of cross-section from Eq. 1 in problem 3.1 is A = 78.5398 mm2. 3

6 N 33 ( 10 ) P The stress in the specimen can be found as: σ = --- = ----------------------------------- = 420.1 ( 10 ) ------- = 420MPa 2

A

m

ε = ( 0.064 )0.005 = 0.065

From the upper plot in Fig. P3.8 δ ε = ------ or L0

–6

78.5398 ( 10 )

δ = ( 0.065 ) ( 50 ) = 3.25 or

δ = 3.25mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.4 Determine the total strain, the elastic strain, and the plastic strain when the axial force on the specimen is 33 kN. Solution

ε total = ?

ε elas= ?

ε plas= ?@ P = 33kN

------------------------------------------------------------

The area of cross-section from Eq. 1 in problem 3.1 is A = 78.5398 mm2. The normal stress in the speci6

6 N 33 ( 10 ) Pmen can be found as: σ = --= ----------------------------------- = 420.2 ( 10 ) ------- = 420.2MPa 2 –6

A

78.5398 ( 10 )

m

εtotal= 0.06 + 0.005 = 0.065

From upper plot in Fig. P3.8

ε total = 0.065

From problem 3.1, we know E = 150 GPa, therefore, the elastic strain can be found as shown below. 6

–3 420.2 ( 10 ) ε elas = --σ- = --------------------------- = 2.801 ( 10 ) 9 E 150 ( 10 )

ε elas = 0.0028

The plastic strain can be found as shown below. –3

–3

–3

ε plas = ε total – ε elas = 65 ( 10 ) – 2.801 ( 10 ) = 62.2 ( 10 )

ε plas = 0.062

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.5 After the axial load was removed, the specimen was observed to have a length of 54 mm. What was the maximum axial load applied to the specimen.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

Solution

Lf = 54 mm

January 2014

P=?

-----------------------------------------------------------Lf – L0 4 The plastic strain can be found as: ε plas = ----------------= ------ = 0.08 50 L0

Starting from a strain value of 0.08, we draw a line parallel to OA on the upper plot, intersecting the stressstrain curve at point C. The stress at point C is approximated as σ c = 470MPa . The area of cross-section from Eq. 1 in problem 3.1 is A = 78.5398 mm2. The applied force can be found as shown below –6

–6

P = σ c A = ( 470 ) ( 10 ) ( 78.5398 ) ( 10 ) = 36913.7N

P = 36.9kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.6—3.10 A tensile test specimen having a diameter of 5/8 inch and a gage length of 2 inch was tested to fracture. The stress-strain curve from the tension test is shown in Figure P3.6. The lower plot is the expanded region OAB and associated with the strain values given in the lower scale. Solve the problems 3.7 through 3.8 using this graph 3.6 Determine: (a) the ultimate stress; (b) the fracture stress; (c) the modulus of elasticity; (d) the proportional limit.(e)the offset yield stress at 0.1%; (f) the tangent modulus at the stress level of 72 kips; (g) the secant modulus at the stress level of 72 kips. 80

C D

Upper Scale

Stress ksi

60 B A 40

E

B

Lower Scale A

20

O 0.00

0.04

0.00

0.002

0.08

0.12

0.004 0.006 Strain in/in

0.16

0.20

0.0.008

0.010

Figure P3.6

Solution

σult=? σfrac=? E = ? Es = ? @σ = 72ksi

σprop= ? σ yield= ? @εoffset = 0.1% ET = ? @ σ = 72ksi

------------------------------------------------------------

(a) By inspection of Figure P3.6, we find: σult=80 ksi

σ ult = 80ksi

(b) By inspection of Figure P3.6, we find:σfrac=80 - 4=76 ksi

σ frac = 76ksi

50 – 0 (c) From the lower plot in Figure P3.6 E = ---------------------

E = 25, 000ksi

(d) By inspection of the lower plot in Figure P3.6, we obtain

σ prop = 50ksi

0.002 – 0

0.1 - = 0.001 . Starting from a strain value of 0.001, we draw a line par(e) The offset strain is: ε offset = -------100

allel to OA on the lower plot in Figure P3.6, intersecting the stress-strain curve at point E. The stress at point E is 54ksi. The yield stress is: σ yield = 54ksi

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

(f) The stress value of 72 ksi corresponds to point D. on the upper plot in Figure P3.6, we draw a line to the stress strain curve at point D and calculate the slope as shown below: 75 – 68 E t = --------------------------- = 325ksi or 0.06 – 0.02

E t = 325ksi

(g) To find the secant modulus, we draw a line from point D to point O in Figure P3.6, and find the slope as shown below: 72 – 0 E t = ------------------- = 1800ksi 0.04 – 0

E s = 1, 800ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.7 Determine the axial force acting on the specimen when it is extended by (a) 0.006 inch (b) 0.120 inch. Solution

5 d = --- inch 8

Lo = 2 inch

(a) P=? @ δ = 0.006 inch (b) P =? @δ = 0.120 inch

------------------------------------------------------------

The area of cross-section can be found as shown below. 2 π 2 A = --- d = 0.3068in 4

1

δ 0.006 (a)The normal strain in specimen can be found as: ε = ----- = ------------- = 0.003 L0

2

From the lower plot in Figure P3.6, the approximate value of stress is σ = 53 ksi.The applied load can be found as: P = σA = ( 53 ) ( 0.3068 ) = 16.26kips or P = 16.3kips δ 0.120 (b) The normal strain in specimen can be found as: ε = ----- = ------------- = 0.060 L0

2

From the upper plot in Figure P3.6, the approximate value of stress is σ = 75 ksi. he applied load can be found as: P = σA = ( 75 ) ( 0.3068 ) = 23.01kips P = 23.0kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.8 Determine the extension of the specimen when the axial force on the specimen is (a) 10 kips (b) 20 kips. Solution

5 d = --- inch 8

L0 = 2 inch

δ = ? @ P = 10 kips

δ = ? @ P = 20 kips

-----------------------------------------------------------The area of cross-section from Eq. 1 in problem 3.7 is: A=0.3068 in2 10 P (a)The normal stress in specimen can be found as: σ = --- = ---------------- = 32.59ksi A

0.3068

From the lower plot in Figure P3.6, the approximate value of strain is ε = 0.0013. The deformation can be found as: δ = εL 0 = ( 0.0013 ) ( 2 ) = 0.0026in or δ = 0.0026in P 20 (b) The normal stress in specimen can be found as: σ = --- = ---------------- = 65.19ksi A

0.3068

From the upper plot of Figure P3.6, the approximate value of strain is ε = 0.02. The deformation can be found as: δ = εL 0 = ( 0.02 ) ( 2 ) = 0.04in δ = 0.04in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.9 Determine the total strain, the elastic strain, and the plastic strain when the axial force on the specimen is 20 kips. Solution

ε total = ?

ε elas= ? ε plas= ? @ P = 20 kips ------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

The area of cross-section from Eq. 1 in problem 3.7 is: A=0.3068 in2 . The normal stress in the specimen P 20 can be found as σ = --- = ---------------- = 65.19ksi A

0.3068

From upper plot in Figure P3.6 we have: εtotal= 0.02

ε total = 0.0200

From problem 3-12, we know E = 25,000 ksi, therefore the elastic strain can be found as shown below. –3 65.19 = ε elas = --σ- = -------------2.608 ( 10 ) E 25000

ε elas = 0.0026

The plastic strain can be found as shown below. –3

–3

–3

ε plas = ε total – ε elas = 20 ( 10 ) – 2.608 ( 10 ) = 17.392 ( 10 )

ε plas = 0.0174

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.10 After the axial load was removed, the specimen was observed to have a length of 2.12 inch. What was the maximum axial load applied to the specimen. Solution Lf = 2.12 inch P=? -----------------------------------------------------------Lf – L0 0.12 The plastic strain is: ε plas = ----------------= ---------- = 0.06 . Starting from a strain value of 0.06, we draw a line 2

L0

parallel to OA on the upper plot in Figure P3.6, intersecting the stress-strain curve at point C. The stress at point C is approximated as σ c = 77ksi . The area of cross-section from Eq. 1 in problem 3.7 is: A=0.3068

in2. The applied load can be found as shown below. P = σ c A = ( 77 ) ( 0.3068 ) = 23.62kips

or

P = 23.6kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.11 A typical stress-strain graph for cortical bone is shown in Figure P3.11. Determine the following quantities. (a) Modulus of elasticity. (b) Proportional limit (c) Yield stress at 0.15% offset. (d)Secant modulus at stress level of 130 MPa.(e) Tangent modulus at stress level of 130 MPa (f) Permanent strain at stress level of 130MPa (g) If the shear modulus of the bone is 6.6 GPa, determine the Poisson’s ratio assuming the bone is isotropic. (h) Assuming the bone specimen was 200 mm long and had a material cross-sectional area of 250 mm2, what is the elongation of the bone when a 20 kN force is applied. 140 D E

120

F

Stress (MPa)

C

100 80 60

A

40 20 B 0 0.000 0.003

Figure P3.11

0.006

0.009

0.012

0.015

0.018

G 0.021

0.024

0.027

0.030

Strain

Solution

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

The construction for calculation of various quantities is shown on graph in red. (a) Using the origin and point A for the calculation we obtain: 6

9 2 ( 50 – 0 ) ( 10 ) E = --------------------------------- = 16.667 ( 10 ) N ⁄ m 0.003 – 0

E = 16.667 GPa

(b) The end of linear region is

σ prop = 110 MPa

(c) The 0.15% offset corresponds to point B. Drawing a line parallel to the linear region we intersect the curve at point C. The offset yield stress is: σ offset = 115 MPa (d) Point D corresponds to stress level of 130 MPa. Thus, 6

9 2 ( 130 – 0 ) ( 10 ) E S = ------------------------------------ = 6.190 ( 10 ) N ⁄ m 0.021 – 0

E S = 6.190 GPa

(e) Using triangle DEF we obtain: 6

9 2 ( 130 – 120 ) ( 10 ) E T = ------------------------------------------ = 0.952 ( 10 ) N ⁄ m 0.021 – 0.0105

E T = 0.952 GPa

(f) The total strain at stress level of 130 MPa corresponds to point G. Thus. 6

130 ( 10 ) ε plas = 0.021 – ------------------------------ = 0.0132 9 16.667 ( 10 )

ε plas = 0.0132

(g) We are given G = 6.6 GPa, thus E G = ------------------2(1 + ν)

or

E- – 1 = 16.667 ν = --------------------- – 1 = 0.263 2G 2 ( 6.6 )

ν = 0.263

3

6 2 20 ( 10 ) (h) The axial stress is σ = ------------------------ = 80 ( 10 ) N ⁄ m = 80 MPa . The specimen is in linear elastic region. –6

250 ( 10 ) 6

80 ( 10 ) The strain is: ε = σ --- = ------------------------------ = 0.0048 . Thus the elongation is: δ = εL = ( 0.0048 ) ( 200 ) = 0.96 mm E

9

16.667 ( 10 )

δ = 0.96 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.12 A 12 mm x 12 mm square metal alloy having a gage length of 50 mm was tested in tension. Draw the stress-strain curve and calculate the following quantities. (Use of Spread Sheet recommended). (a) modulus of elasticity. (b) the proportional limit. (c) the yield stress at 0.2 percent offset. (d) the tangent Modulus at stress level of 1400 MPa. (e) the secant modulus at stress level of 1400 MPa. (f) the plastic strain at stress level of 1400 MPa. Load (KN)

Change in Length (mm)

Load (KN)

Change in Length (mm)

Load (KN)

Change in Length (mm)

0.00

0.00

176.03

1.96

212.06

9.99 11.01

17.32

0.02

182.80

2.79

212.17

60.62

0.07

190.75

4.00

208.64

11.63 12.03

112.58

0.13

193.29

4.71

204.99

147.22

0.17

200.01

5.80

199.34

12.31 12.47 12.63

161.18

0.53

204.65

7.15

192.15

168.27

1.10

209.99

8.88

185.46 break

Solution

Lo = 50 mm

cross-section is 12 mm x 12 mm

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

------------------------------------------------------------

The area of cross-section is A = (12)(12) = 144 mm2=144(10-6) m2. Dividing the column of load by the area A and the column of change of length by the gage length Lo, we can obtain the columns of stress and strains on a spread sheet as shown in the table below. The stresses and strain can be plotted to obtain the stress-strain graph in figure (a). The data corresponding to first 5 points can be also plotted to get an enlarged view of region OAB as shown in figure (b). 1600

(a)

D

1400

Stress-Strain values in problem 3.17. Strain

Stress (MPa)

Strain

Stress (MPa)

0

0

0.0942

1342.29

0.0004

120.28

0.116

1388.96

0.0014

420.7

0.143

1421.18

0.0026

781.36

0.1776

1458.26

0.0034

1022.36

0.1998

1472.64

0.0106

1119.31

0.2202

1473.40

0.022

1168.54

0.2326

1448.89

0.0392

1222.43

0.2406

1423.54

0.0558

1269.44

0.2462

1384.31

0.08

1324.65

0.2494

1334.38

0.2526

1287.92

Stress (MPa)

1200 1000

B A

800 600 400 200 0 O 0.00

E 0.05

0.10

0.15

0.20

0.25

0.30

Strain

(b)

C

B

A

O

The modulus of elasticity is calculated as the average slope from the first three points as shown below: 1 120.28 – 0 420.7 – 120.28 E = --- ------------------------- + --------------------------------------- = 300560MPa 2 0.0004 0.0014 – 0.0004

E = 300Gpa

From the figure (b) and the table:

σ prop = 1022MPa

0.2- = 0.002 On figure (b), starting at a strain value of 0.002, we draw a line The offset strain is: ε offset = -------100

parallel to OA to intersect the stress-strain curve at point C.The stress value at C is approximated as the σ yield = 1060MPa yield stress The stress value of 1400 MPa, corresponds to point D. in Figure (a). We draw tangent to the stress-strain curve at point D and calculate the slope as shown below: 1433 – 1367 E t = --------------------------------------- = 1719MPa 0.1467 – 0.1083

E t = 1.72GPa

To find the secant modulus we draw a line from point D to point O. and find the slope as 1400 E s = ------------- = 11200MPa 0.125

E s = 11.2GPa

The total strain at point D is εtotal= 0.125. The elastic strain at point D can be found as: 6 σD –3 1400 ( 10 ) - = ------------------------- = 4.667 ( 10 ) The plastic strain can be found as shown below. ε elas = -----9 E 300 ( 10 )

ε plas = ε total – ε elas = 0.125 – 0.004667 = 0.1203

ε plas = 0.1203

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

3.13 A mild steel specimen of diameter 0.5 inches and a gage length of 2 inches was tested in tension. The results of the test are reported in the table below. Draw the stress-strain curve and calculate the following quantities. (Use of Spread Sheet recommended). (a) Modulus of Elasticity. (b) the proportional limit. (c) the yield stress at 0.05 percent offset. (d) the tangent Modulus at stress level of 50 ksi. (e) the secant Modulus at stress level of 50 ksi. (f) the plastic strain at stress level of 50 ksi. Load (103)lbs

Change in Length (10-3)in

Load (103)lbs

Change in Length (10-3)in

Load (103)lbs

Change in Length (10-3)in

0.00

0.00

9.32

59.06

12.55

245.93

Solution

3.11

1.28

9.86

70.85

12.70

283.47

7.24

2.96

10.40

84.23

12.77

316.36

7.50

3.06

10.82

97.85

12.84

363.10

7.70

8.76

11.18

112.10

12.04

385.34

7.90

19.05

11.72

140.40

11.44

396.03

8.16

28.70

11.99

161.21

10.71

406.42

8.46

37.73

12.27

192.65

9.96

414.72

8.82

47.18

12.41

214.22

break

Lo = 2 inch

d = 0.5 inch ------------------------------------------------------------

The area of cross-section is A = πd2/4 = 0.1963 in2.Dividing the column of load by the area A and the column of change of length by the gage length Lo, we can obtain the columns of stress and strains on a spread sheet as shown in the table below. The stress and strain can be plotted to obtain the stress-strain graph in figure (a). The data corresponding to first 5 points can be also plotted to get an enlarged view of region OAB as shown in figure (b). The modulus of elasticity is calculated as the average slope from the first three points as shown below: 1 15.84 – 0 36.87 – 15.84 E = --- ---------------------- + --------------------------------------------- = 24892.8ksi 2 0.00064 0.00148 – 0.00064

E = 24893ksi

From the figure (b) and the table:

σ prop = 36.9ksi

The offset strain is: ε offset = 0.05 ---------- = 0.0005 on figure (b), starting at a strain value of 0.0005, we draw a 100

line parallel to OA to intersect the stress-strain curve at point C. The stress value at C is approximated as

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

σ yield = 37.5ksi

the yield stress Stress-Strain values in problem 3.18

(a) D

Stress (ksi)

Strain

Stress (ksi)

0.00000

0.00

0.05605

56.94

0.00064

15.84

0.07020

59.69

0.00148

36.87

0.08061

61.06

0.00153

38.25

0.09633

62.49

0.00438

39.22

0.10711

63.20

0.00953

40.23

0.12297

63.92

0.01435

41.56

0.14174

64.68

0.01887

43.09

0.15818

65.04

0.02359

44.92

0.18155

65.39

0.02953

47.47

0.19267

61.32

0.03543

50.22

0.19802

58.26

0.04212

52.97

0.20321

54.55

0.04893

55.11

0.20736

50.73

Strain

B A

O

E

(b)

B A C

O

The stress value of 50ksi, corresponds to point D. in Figure (a). We draw tangent to the stress-strain curve at point D and calculate the slope as shown below: 52 – 46 E t = ------------------------------ = 400ksi 0.04 – 0.025

E t = 400ksi

To find the secant modulus we draw a line from point D to point O. and find the slope as 50 - = 1428.57 E s = -----------0.035

E s = 1428.6ksi σ

50 The total strain at point D is εtotal= 0.035. The elastic strain at point D is ε elas = ------D- = -------------- = 0.002 . The E

24893

plastic strain can be found as shown below. ε plas = ε total – ε elas = 0.035 – 0.002 = 0.033

ε plas = 0.033

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.14 A rigid bar AB of negligible weight is supported by cable of diameter 1/4 in as shown in Figure P3.14. The cable is made from a material that has a stress- strain curve shown in Figure P3.6. (a) Determine the extension of the cable when P = 2 kips (b) What is the permanent deformation in BC when the

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

load P is removed? 5 ft B

C

P

40o

Figure P3.14

A

Solution

-----------------------------------------------------------We can draw the following free body diagram: 5 ft B

NBC P

h 40o

Ay A

From geometry: tan 40 = 5---

Ax

h = 5.958 ft

h

By moment equilibrium about A: N BC h = P ( 5 )

5 N BC = ------------- ( 2 ) = 1.6782 kips 5.958

or

1

2 π 1 The crosssectional area is: A = --- ⎛⎝ ---⎞⎠ = 0.04909 in. 4 4 2

N BC 1.6782- = 34.188 ksi The normal stress is: σ BC = ---------- = -----------------A

0.04909

50 - = 25000 ksi From the graph in Figure P3.6, the modulus of elasticity is E = -----------0.002

σ BC –3 34.188 - = ---------------- = 1.367 ( 10 ) The normal strain is: ε BC = --------E

25000

–3

The deformation is: δ BC = ε BC L = ( 1.367 ) ( 10 ) ( 5 ) ( 12 ) = 0.082 in. The stress of 34.188 ksi is below the yield stress, hence no permanent deformation. The answers are: δ BC = 0.082 in. ; ( δ BC ) plastic = 0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.15 A rigid bar AB of negligible weight is supported by cable of diameter 1/4 in. as shown in Figure P3.15. The cable is made from a material that has a stress- strain curve shown in Figure P3.6. (a) Determine the extension of the cable when P =4.25 kips (b) What is the permanent deformation in the cable

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

when the load P is removed? 5 ft B

C

P

40o

Figure P3.15

A

Solution

-----------------------------------------------------------We can draw the following free body diagram: 5 ft B

NBC P

h 40o

Ay A

Ax

From geometry: tan 40 = 5 ⁄ h h = 5.958 ft By moment equilibrium about A: N BC h = P ( 5 )

or

5 N BC = ------------- ( 4.25 ) = 3.566 kips 5.958

1

2 π 1 The crosssectional area is: A = --- ⎛⎝ ---⎞⎠ = 0.04909 in. 4 4 2

N BC 3.566 The normal stress is: σ BC = ---------- = ------------------- = 72.6 ksi A

0.04909

From the graph in Figure P3.6, the modulus of elasticity is the normal strain at 72.6 ksi is: ε BC = 0.04 The deformation is: δ BC = ε BC L = ( 0.04 ) ( 5 ) ( 12 ) = 2.4 in. From the graph in Figure P3.6, the modulus of elasticity is E = 50 ⁄ 0.002 = 25000 ksi –3

The elastic strain is: ( ε BC ) elastic = σ BC ⁄ E = 72.6 ⁄ 25000 = 2.906 ( 10 ) –3

–3

The plastic strain is: ( ε BC ) plastic = ε BC – ( ε BC ) elastic = 0.04 – 2.906 ( 10 ) = 37.094 ( 10 ) –3

The plastic deformation is: ( δ BC ) plastic = ( ε BC ) plastic L = 37.094 ( 10 ) ( 5 ) ( 12 ) = 2.4 in. = 2.22 in. The answers are:

δ BC = 2.4 in. ; ( δ BC ) plastic = 2.22 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.16 A rectangular bar has an cross-sectional area of 2 in2 and an undeformed length of five inches as shown in Figure P3.16. When a load of P = 50,000 lbs is applied, the bar deforms to a position shown by the colored shape. Determine the Modulus of Elasticity and the Poisson’s Ratio of the material.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

P

P

2 in

January 2014

1.9996 in

5 in 5.005 in

Figure P3.16

Solution

A = 2 in2

L0 = 5 in

Lf = 5.005

P = 50,000lbs

E=?

ν=?

-----------------------------------------------------------Lf – L0 0.005 The longitudinal (axial) strain is: ε long = ---------------- = ------------- = 0.001 5

L0

P 50000 The axial stress is: σ = --- = --------------- = 25, 000psi A

2

3 σ 25000 The modulus of elasticity is: E = ----------- = --------------- = 25000 ( 10 )psi

ε long

0.001

E = 25, 000ksi

The transverse strain is: ε trans = – 0.0004 ---------------- = – 0.0002 2

– ε trans ( – 0.0002 ) The Poisson’s ratio is: ν = ---------------= – ------------------------ = 0.2 ε long

ν = 0.2

0.001

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.17 A force P= 20 kips is applied to the rigid plate that is attached to a square bar as shown in Figure P3.17. If the plate moves an amount of 0.005 inch, determine the Modulus of Elasticity. P 2 in

2 in

10 in

Figure P3.17

Solution

δ=0.005

P = 20 kips

E=?

-----------------------------------------------------------P 20 The axial stress is: σ = --- = ------ = 5ksi . A

4

The axial strain is: ε = 0.005 ------------- = 0.0005 10

5 The modulus of elasticity is: E = σ --- = ---------------- = 10, 000ksi ε

0.0005

E = 10, 000ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.18 A force P= 20 kips is applied to the rigid plate that is attached to a square bar as shown in Figure P3.18. If the plate moves an amount of 0.0125 inch, determine the Shear Modulus of Elasticity.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

P

2 in

10 in

2 in

Figure P3.18

Solution

P = 20 ksi

δ = 0.0125

A= 4

G=?

------------------------------------------------------------

The free body diagram of the rigid plate can be drawn as shown in figure a. By equilibrium of forces we obtain V = P. The shear stress on the rigid plate can be found as: τ = V ⁄ A = 20 ⁄ 4 = 5ksi . An exaggerated deformed geometry can be drawn as shown in figure b and the shear strain found as: tan γ ≈ γ = δ ⁄ 10 = 0.00125 δ

(a)

(b)

P

10

V

γ

5 The shear modulus of elasticity can be found as: G = τ-- = ------------------ = 4000ksi γ

0.00125

G = 4000ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.19 Two rubber blocks of length L and cross section dimension a x b are bonded to rigid plates as shown in Figure P3.19. Point A was observed to move downwards by 0.02 in. when the weight W = 900 lb was hung from the middle plate. Determine the shear modulus of elasticity using small strain approximation. Use L= 12 in., a = 3 in., and b = 2 in.

L

A a

Figure P3.19

a

W

Solution

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

The deformed shape and the free body diagram can be drawn as shown V

V

3 in. γ

3 in. γ

0.02 in. W = 900 lb

–3 From geometry we have: tan γ ≈ γ = 0.02 ---------- = 6.667 ( 10 )

3

From free body diagram we have: 2V = 900

or

V = 450 lb

V- = -----------------450 - = 18.75 psi The shear stress on the rubber block is: τ = --( 12 ) ( 2 )

A

18.75 The shear modulus is: G = τ-- = ----------------------------= 2812.5 γ

G = 2812.5 psi

–3

6.667 ( 10 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.20 Two rubber blocks with a shear modulus of 1.0 MPa and length L and cross section dimension a x b are bonded to rigid plates as shown in Figure P3.20. Using small strain approximation, determine the displacement of point A, if a weight of 500 N is hung from the middle plate. Use L = 200 mm, a = 45 mm, and b = 60 mm.

L

A a

Figure P3.20

a

W

Solution

-----------------------------------------------------------The deformed shape and the free body diagram can be drawn as shown V

V

0.045 m γ

0.045 m γ

δ W = 500 N

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

δ From geometry we have: tan γ ≈ γ = ------------

0.045

From free body diagram we have: 2V = 500

or

V = 250 N

2 V 250 The shear stress on the rubber block is: τ = --- = ---------------------------- = 20833 N ⁄ m

( 0.2 ) ( 0.06 )

A

The displacement can be found as: –6 τ 20833 δ γ = ---- = ---------------- = 20833 ( 10 ) = ------------6 G 0.045 1 ( 10 )

–6

δ = 937.5 ( 10 ) m or

or

δ = 0.9375 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.21 Two rubber blocks with a shear modulus of 750 psi and length L and cross section dimension a x b are bonded to rigid plates as shown in Figure P3.21. If the allowable shear stress in the rubber is 15 psi, and allowable deflection is 0.03 in., determine the maximum weight W that can be hung from the middle plate using small strain approximation. Use L = 12 in, a = 2 in, and b = 3 in.

L

A a

Figure P3.21

a

W

Solution

-----------------------------------------------------------The deformed shape and the free body diagram can be drawn as shown V

V

3 in. γ

3 in. γ

δ W

From geometry we have: tan γ ≈ γ = δ---

3 From free body diagram we have: 2V = W

or

V = W⁄2

The shear stress on the rubber block is: W V- = -----------------W ⁄ 2 - = ----- ≤ 15 psi τ = --48 A ( 12 ) ( 2 )

or

W ≤ 720 lb

1

The displacement can be found as: τ- = (------------------W ⁄ 48 )- = δ--γ = --G 750 3

or

W δ = --------------- ≤ 0.03 12000

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or

W ≤ 360 lb

2

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

Thus

January 2014

W max = 360 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.22 Two rubber blocks with a shear modulus of G, length L and cross section dimension a x b are bonded to rigid plates as shown in Figure P3.22. Obtain the shear stress in the rubber block and the displacement of point A in terms of G, L, W, a, and b.

L

A a

Figure P3.22

a

W

Solution

-----------------------------------------------------------The deformed shape and the free body diagram can be drawn as shown V

V

a

a γ

γ

δ W

From geometry we have: tan γ ≈ γ = δ-a

From free body diagram we have: 2V = W The shear stress on the rubber block is:

or

V = W ----2

V- = --------------W ⁄ 2- = --------Wτ = --A (L)(b) 2Lb

The displacement can be found as: τ- = --------W - = δ-- or γ = --G 2Lb a

Waδ = --------2Lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.23 A circular bar of length 200 mm and diameter 20 mm is subjected to a tension test. Due to an axial force of 77 kN, the bar is seen to elongate by 4.5 mm and the diameter is seen to reduce by 0.162 mm. Determine the Modulus of Elasticity and the Shear Modulus of Elasticity. Solution L0 = 200 mm d = 20 mm P = 77 kN δ = 4-5mm Δd = -0.162 mm E=? G=? ------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

2 2 –6 2 π 2 π The area of cross-section is: A = --- d = --- ( 20 ) = 314.16mm = 314.16 ( 10 )m

4

4

3

6 N 77 ( 10 ) P The axial stress is: σ = --- = -------------------------------- = 245.1 ( 10 ) ------2

A

–6

314.16 ( 10 )

m

δ 4.5 The longitudinal strain (axial) is: ε long = ----- = --------- = 0.0225 L0

200

0.162 ------- = – ------------- = – 0.0081 The transverse strain is: ε tran = Δd d

20

6

9 245.1 ( 10 ) σ The modulus of elasticity is: E = ----------- = --------------------------- = 10.893 ( 10 )

ε long

E = 10.9GPa

0.0225

– ε tran ( – 0.0081 ) The Poisson’s ratio is: ν = ------------- = – ------------------------ = 0.36 ( 0.0225 )

ε long

E 10.893 The shear modulus of elasticity is: G = ------------------- = ---------------------------- = 4.005GPa 2(1 + ν)

G = 4.0GPa

2 ( 1 + 0.36 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.24 A circular bar of length 6 inch and diameter of 1 inch is made from a material with a Modulus of Elasticity of E=30,000 ksi and a Poisson’s ratio of ν=1/3. Determine the change in length and diameter of the bar when a force of 20 kips is applied to the bar. Solution L0 = 6 d = 1 E= 30,000ksi ν=1/3 P = 20kips Δd = ? ΔL = ? -----------------------------------------------------------2 π 2 π 2 The area of cross-section is: A = --- d = --- ( 1 ) = 0.7854in 4 4 P 20 The axial stress is: σ = --- = ---------------- = 25.465ksi A

0.7854

25.465- = 0.8488 ( 10 – 3 ) The longitudinal strain is: ε long = σ --- = ----------------E

30, 000

–3

–3

The change of length is: ΔL = Lε long = ( 6 ) ( 0.8488 ) ( 10 ) = 5.093 ( 10 )

ΔL = 0.0051in

–3

–3

The transverse strain is: ε tran = – νε long = – [ 0.8488 ( 10 ) ] ⁄ 3 = – 0.2829 ( 10 ) –3

The change in diameter is Δd = ε tran d = ( 1 ) ( – 0.2829 ) ( 10 ) or

Δd = – 0.0002

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.25 A circular bar of length 400 mm and diameter of 20 mm is made from a material with a modulus of elasticity of E = 180 GPa and a Poisson’s ratio of ν = 0.32. Due to a force the bar is seen to elongate by 0.5 mm. Determine the change in diameter and the applied force. Solution L0 = 400 mm d = 20 mm E= 180 GPa ν=0.32 δ = 0.5 mm Δd = ? P=? -----------------------------------------------------------δ- = -------0.5- = 1.25 ( 10 – 3 ) The longitudinal strain is: ε long = ----L0

400 9

–3

6

The axial stress is: σ = Eε long = ( 180 ) ( 10 ) ( 1.25 ) ( 10 ) = 225 ( 10 )Pa 6

π 4

2

3

The applied force is: P = σA = 225 ( 10 ) --- ( 0.02 ) = 70.686 ( 10 )N or The transverse strain is:

–3

P = 70.7kN –3

ε tran = – νε long = – ( 0.32 ) ( 1.25 ) ( 10 ) = – 0.4 ( 10 )

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

–3

January 2014

–3

The change in diameter is: Δd = ( d )ε tran = ( 20 ) ( – 0.4 ) ( 10 ) = – 8 ( 10 ) or Δd = – 0.008mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.26 A 25 mm x 25 mm square bar is 500 mm long and is made from a material that has a Poisson’s ratio of 1/3. In a tension test, the bar is seen to elongate by 0.75 mm. Determine the percentage change in volume of the bar. Solution

A = 625 mm2

L0 = 500 mm

ν = 1/3

δ = 0.75 mm

% change in V= ?

------------------------------------------------------------

The original volume is: V0 = AL0 = 312.5(103) mm3. Let Δa represent the change in width and thickness. The volume of the deformed bar is as given below. V f = FinalVolume = ( 25 + Δa ) ( 25 + Δa ) ( L 0 + δ ) 1 δ 0.75 The longitudinal strain is: ε long = ----- = ---------- = 0.0015 L0

500

1 3

The transverse strain is: ε tran = – νε long = – --- ( 0.0015 ) = – 0.0005 We can find Δa = ( 25 )ε tran = – 0.0125mm . Substituting δ and Δa into Eq. 1, we obtain: V f = ( 25 – 0.0125 ) ( 25 – 0.0125 ) ( 500 + 0.75 ) = 312.65586 mm ΔV f 0 - × 100 = 0.04988 or The percentage change in volume is: -------- × 100 = -----------------V –V

V

3

0.05%

V0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.27 A circular bar of length 50 inch and diameter of 1 inch is made from a material with a Modulus of Elasticity of E = 28,000 ksi and a Poisson’s ratio of 0.32. Determine the percentage change in volume of the bar when an axial force of 20 kips is applied to the bar. Solution d = 1 in L0 = 15in ν = 0.32 Ε = 28,000ksi % change in V @P = 20 kips -----------------------------------------------------------P 20 The axial stress is: σ = --- = --------------------- = 25.465ksi A

2

π(1 ) ⁄ 4

–3 25.465 --- = ------------------ = 0.9095 ( 10 ) The longitudinal strain is: ε long = σ

E

28, 000

–3

The deformation is: δ = L 0 ε long = ( 50 ) ( 0.9095 ) ( 10 ) = 0.04547in –3

–3

The transverse strain is: ε tran = – νε long = – ( 0.32 ) ( 0.9095 ) ( 10 ) = – 0.2910 ( 10 ) –3

–3

The change in diameter is: Δd = ( d )ε tran = – ( 0.2910 ) ( 10 ) ( 1 ) = – 0.2910 ( 10 )in The original volume is: V 0 = AL 0 = ( 0.7854 ) ( 50 ) = 39.2699in

3

The volume of the deformed bar is as shown below. 2 –3 2 3 π π V f = --- ( 1 – Δd ) ( L 0 + δ ) = --- ( 1 – 0.291 ( 10 ) ) ( 50 + 0.04547 ) = 39.2827in 4 4

ΔV f 0 - × 100 = 0.0327 The percentage change in volume is: -------- × 100 = -----------------V –V

V0

V0

0.0327%

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.28

An aluminum rectangular bar has a cross-section of 25 mm x 50 mm and a length of 500 mm. The

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

Modulus of Elasticity of E = 70 GPa and a Poisson’s ratio of ν = 0.25. Determine the percentage change in the volume of the bar when an axial force of 300 kN is applied to the bar. Solution E = 70GPa L0 = 500mm ν = 0.25 % change in V @ P = 300 kN -----------------------------------------------------------3

6 N 300 ( 10 ) P The axial stress is: σ = --- = ---------------------------------- = 240 ( 10 ) ------2

A

( 0.05 ) ( 0.025 )

m

6

–3 240 ( 10 ) The longitudinal strain is: ε long = σ --- = ---------------------- = 3.4286 ( 10 ) 9

E

70 ( 10 )

–3

–3

The transverse strain is: ε tran = – νε long = – ( 0.25 ) ( 3.4286 ) ( 10 ) = – 0.85714 ( 10 ) –3

The deformation of the bar is: δ = L 0 ε long = ( 500 ) ( 3.4286 ) ( 10 ) = 1.7143mm The change in cross-sectional dimensions can be found as shown below. –3

–3

–3

–3

Δa = aε tran = ( 25 ) ( – 0.85714 ) ( 10 ) = – 21.429 ( 10 )mm Δb = bε tran = ( 50 ) ( – 0.85714 ) ( 10 ) = – 42.857 ( 10 )mm 3

The original volume is: V 0 = AL 0 = ( 1250 ) ( 500 ) = 625 ( 10 )mm

3 3

The volume of the deformed bar is: V f = ( a + Δa ) ( b + Δb ) ( L 0 + δ ) = 626.068 ( 10 )mm ΔV f 0 - × 100 = 0.1709 or The percentage change in volume is: -------- × 100 = -----------------V –V

V0

V0

3

0.1709%

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.29 A circular bar of length L and diameter of d is made from a material with a Modulus of Elasticity of E and a Poisson’s ratio of ν. Assuming small strain, show that the percentage change in the volume of 2

the bar when an axial force P is applied is given by 400P ( 1 – 2ν ) ⁄ ( Eπd ) . Note the percentage change is zero when ν = 0.5? Solution %change in V = f(L,d,E,ν,P) -----------------------------------------------------------4P P P The axial stress is: σ = --- = ---------------- = --------A

2

πd ⁄ 4

πd

2

4P 4νP --- = ------------- and the transverse strain is: ε tran = – νε long = – ------------The longitudinal strain is: ε long = σ E

Eπd

2

Eπd

2

4LP 4νP The change in length is: δ = Lε long = ------------ and the change in diameter is: Δd = dε tran = – ---------Eπd

2

Eπd

2

πd The original volume is: V 0 = --------- L . For small strain, the volume of the deformed bar can be found as 4

shown below. 2 π 2 π 2 π 2 π Δd 2 δ 2Δd δ 2Δd δ V f = --- ( d + Δd ) ( L + δ ) = --- d L ⎛ 1 + -------⎞ ⎛ 1 + ---⎞ ≈ --- d L ⎛ 1 + ----------⎞ ⎛ 1 + ---⎞ ≈ --- d L ⎛ 1 + ---------- + ---⎞ ⎝ ⎝ ⎝ 4 4 d⎠ ⎝ L⎠ 4 d ⎠⎝ L⎠ 4 d L⎠

The percentage change in volume is as shown below. Vf – V0 Δd- δ ⎞ 4P 8νP ⎛ ΔV --------⎞ ( 100 ) = ⎛ ------------------⎞ ( 100 ) = ⎛ 2 -----+ --- ( 100 ) = ⎛ – ------------- + -------------⎞ ( 100 ) or ⎝ ⎝ V⎠ ⎝ V0 ⎠ ⎝ d L⎠ 2 2⎠ Eπd Eπd

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400P ------------- ( 1 – 2ν ) 2 Eπd

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.30 A rectangular bar has a cross-sectional dimensions of a x b and a length L. The bar material has a Modulus of Elasticity of E and a Poisson’s ratio of ν. Assuming small strain, show that the percentage change in the volume of the bar when an axial force P is applied is given by 100P ( 1 – 2ν ) ⁄ ( Eab ) . Note the percentage change is zero when ν = 0.5? Solution %change in V = f(L,d,E,ν,P) -----------------------------------------------------------P P The axial stress is: σ = --- = -----A

ab

P νP --- = ---------- o and the transverse strain is: ε tran = – νε long = – ---------The longitudinal strain is: ε long = σ E

Eab

Eab

The change in length is: δ = Lε long = νLP ---------- . The change in cross-sectional dimensions can be found as Eab

shown below. Paν Δa = aε tran = – ---------Eab

Pbν Δb = bε tran = – ---------Eab The original volume is: V 0 = abL . The volume of deformed bar can be found as shown below. δ Δa Δb δ Δb V f = ( a + Δa ) ( b + Δb ) ( L + δ ) = abL ⎛ 1 + Δa ------⎞ ⎛ 1 + -------⎞ ⎛ 1 + ---⎞ ≈ ( abL ) ⎛ 1 + ------ + -------⎞ + ⎛ 1 + ---⎞ ⎝ ⎝ L⎠ a b⎠ ⎝ L⎠ b ⎠⎝ a ⎠⎝

Assuming small strain, the product terms of Δa, Δb, and δ can be neglected to obtain the following. Δa Δb δ V f ≈ ( abL ) ⎛ 1 + ------ + ------- + ---⎞ . The percentage change in volume can be found as shown below. ⎝ a b L⎠ Vf – V0 Δa Δb δ Pν Pν P ⎛ ΔV --------⎞ = ⎛⎝ -------------------⎞⎠ ( 100 ) = ⎛⎝ ------ + ------- + ---⎞⎠ ( 100 ) = ⎛⎝ – ---------- – ---------- + ----------⎞⎠ ( 100 ) or ⎝ V ⎠ ( 100 ) V0 a b L Eab Eab Eab P ( 100 ) ( 1 – 2ν ) ---------- % Eab

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.31 A rectangular bar has a cross-sectional area of 2 in.2 and an undeformed length of 5 in., as shown in Figure P3.16. When a load P = 50,000 lb is applied, the bar deforms to a position shown by the colored shape. What is the strain energy in the bar ? 2 in

P

P

1.9996 in

5 in 5.005 in

Figure P3.31

Solution

P=50,000lbs

A=2 in2

L=5 in

δ=0.005in

U=?

-----------------------------------------------------------The strain energy density for the bar can be written as: 3 1 1 P δ 1 50, 000 0.005 U 0 = --- σε = --- ⎛ ----⎞ --- = --- ⎛ ------------------⎞ ⎛ -------------⎞ = 12.5in – lbs ⁄ in 2 2 ⎝ A⎠ L 2⎝ 2 ⎠ ⎝ 5 ⎠ As the strain energy density is a constant, the total strain energy can be found as shown below: U =

∫ U0 dv

= U 0 V = U 0 AL = ( 12.5 ) ( 2 ) ( 5 ) = 125 or

U = 125in – lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.32

A force P = 20 kips is applied to a rigid plate that is attached to a square bar, as shown in Figure P3.32. The

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

plate moves a distance of 0.005 in. What is the strain energy in the bar? P

2 in

2 in

10 in

Figure P3.32

Solution

P=20 kips

A=4 in2

δ=0.005in

L=10 in

U=?

-----------------------------------------------------------The strain energy density for the bar can be written as: 3

3 1 1 P δ 1 20 ( 10 ) 0.005 U 0 = --- σε = --- ⎛ ----⎞ --- = --- ⎛ -------------------⎞ ⎛ -------------⎞ = 1.25in – lbs ⁄ in 2 2 ⎝ A⎠ L 2⎝ 4 ⎠ ⎝ 10 ⎠

As the strain energy density is a constant, the total strain energy can be found as shown below: U =

∫ U0 dV

= U 0 V = U 0 AL = ( 1.25 ) ( 4 ) ( 10 ) = 50

U = 50in – lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.33 A force P = 20 kips is applied to a rigid plate that is attached to a square bar, as shown in Figure P3.17. The plate moves a distance of 0.0125 in. Assume line AB remains straight.What is the strain energy in the bar? 2 in 2 in B

P

10 0 in

Figure P3.33

Solution

P=20 kips

2

A=4 in

L=10 in

δ=0.0125in

U=?

-----------------------------------------------------------The strain energy density for the bar can be written as: 3

3 1 1 P δ 1 20 ( 10 ) 0.0125 U 0 = --- σε = --- ⎛ ----⎞ --- = --- ⎛ -------------------⎞ ⎛ ----------------⎞ = 3.125in – lbs ⁄ in 2 2 ⎝ A⎠ L 2⎝ 4 ⎠ ⎝ 10 ⎠

As the strain energy density is a constant, the total strain energy can be found as shown below: U =

∫ U0 dv

= U 0 V = U 0 AL = ( 3.125 ) ( 4 ) ( 10 ) = 125

U = 125 in.-lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.34 A circular bar of length L and diameter of d is made from a material with a modulus of elasticity of E and a Poisson’s ratio of ν. In terms of the given variables, what is the linear strain energy in the bar of problem when axial load P is applied to the bar.

Solution

U=f(L,d,E,ν,p)=?

-----------------------------------------------------------The strain energy density for the bar can be written as:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

2 2 2 1 1 σ 1⎛ P ⎞ 1 p 8 U 0 = --- σε = --- ⎛ ------⎞ = --- ⎜ ----------⎟ = --- --------------------- = --------2 2 2⎝ E ⎠ 2 ⎝ EA 2⎠ 2 πd 2 2 π E E ⎛ ---------⎞ ⎝ 4 ⎠

As the strain energy density is a constant, the total strain energy can be found as shown below: U =

∫

⎛ 8P 2 ⎞ πd 2 U 0 dv = U 0 V = U 0 AL = ⎜ ---------------⎟ ⎛ ---------⎞ L ⎝ π 2 Ed 4⎠ ⎝ 4 ⎠

2

2P L U = ------------2 πEd

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.35 A rectangular bar has a cross-sectional dimensions of a x b and a length L. The bar material has a Modulus of Elasticity of E and a Poisson’s ratio of ν. In terms of the given variables, what is the linear strain energy in the bar of problem when axial load P is applied to the bar.

Solution

U=f(L,a,b,E,ν,p)=?

-----------------------------------------------------------The strain energy density for the bar can be written as: 2 2 2 1 1 σ 1⎛ P ⎞ 1 P U 0 = --- σε = --- ⎛ ------⎞ = --- ⎜ ----------⎟ = --- --------------2 2⎝ E ⎠ 2 ⎝ EA 2⎠ 2 Ea 2 b 2

As the strain energy density is a constant, the total strain energy can be found as shown below: U =

∫ U0 dv

2 ⎛ ⎞ P = U 0 V = U 0 AL = ⎜ -----------------------⎟ ( ab ) ( L ) ⎝ 2 ( Ea 2 b 2 )⎠

2

P L U = ------------2Eab

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.36

For the material that has the stress-strain curve shown in Figure P3.3, determine

(a) Modulus of Resilience, use proportional limit as an approximation for yield point. (b) Strain energy density at a stress level of 420 MPa. (c) Complimentary strain energy density at a stress level of 420 MPa. (d) Modulus of Toughness. Solution

Modulus of Resilience=?

Uo= ? and Uo= ? @ σ = 420 MPa

Modulus of Toughness=?

-----------------------------------------------------------The proportional limit corresponds to point A. By inspection of the lower plot σ prop = 300MPa . 6

9 N 300 ( 10 ) From the lower plot E = ---------------------- = 150 ( 10 ) ------- = 150GPa 2 0.002 m

Thus, the strain at proportional limit (at A1) is εprop = 300(106) /[(150)(109)] = 0.002. All other points are approximate points chosen to represent the stress strain curve by a series of straight line. The areas under the stress strain

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

curve are calculated as shown on the right of the figure. 510 480

D

Stress MPa

420

( 300 ) ( 10 6 ) ( 0.002 ) AOA 1 = ---------------------------------------------- = 300 ( 10 3 ) 2

C

C2

360 340 A 300

E

( 300 + 340 ) ( 10 6 ) ( 0.038 ) AA 1 BB 1 = ------------------------------------------------------------- = 12160 ( 10 3 ) 2

B

( 340 + 420 ) ( 10 6 ) ( 0.025 ) BB 1 CC 1 = ------------------------------------------------------------= 9500 ( 10 3 ) 2

240

· ( 420 + 480 ) ( 10 6 ) ( 0.0325 ) CC 1 DD 1 = ---------------------------------------------------------------- = 14625 ( 10 3 ) 2

120

B1 O A1 0.00 0.04 0.002

C1 D1 0.08 0.12 0.065 0.0975

E1 0.16

0.20

( 480 + 510 ) ( 10 6 ) ( 0.0425 ) DD 1 EE 1 = ---------------------------------------------------------------- = 21038 ( 10 3 ) 2

Strain mm/mm

The yield point corresponds to point A. The area of the triangle OAA1 represents Modulus of Resilience.

Modulus of Resilience is 300 kN-m/m3 Point C is at 420 MPa. The strain energy density at point C is the area AOA1 plus areas AA1BB1 and BB1CC1. Thus the strain energy density at C is

U O = ( 300 + 12160 + 9500 ) ( 10 3 ) N – m ⁄ m

3

or 3

U O = 21, 960 kN – m ⁄ m . The complementary strain energy density at C can be found by subtracting UO from the area of the rectangle OC2CC1. Thus,

3

3

U O = ( 420 ) ( 10 6 ) ( 0.065 ) – 21, 960 ( 10 ) = 5340 ( 10 ) N – m ⁄ m

3

or 3

U O = 5, 340 kN – m ⁄ m . The ultimate stress corresponds to point E on the graph. The area underneath the curve can be calculated as the sum of the areas shown on the right of the graph. The total sum of the areas is: 57,623(103) N-m/m3. Thus,

Modulus of Toughness is 57,623 kN-m/m3

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.37

For the material that has the stress-strain curve shown in Figure P3.6, determine

(a) Modulus of Resilience, use proportional limit as an approximation for yield point. (b) Strain energy density at a stress level of 72 ksi. (c) Complimentary strain energy density at a stress level of 72ksi. (d) Modulus of Toughness. Solution

Modulus of Resilience=?Uo= ?

Uo= ? at 72 ksi

Modulus of Toughness=?

------------------------------------------------------------

The proportional limit corresponds to point A. The stress at problem limit is σprop = 50 ksi and the modulus of elasticity is 25,000 ksi. Thus, the strain at proportional limit (at A1) is εprop = 50/25,000 = 0.002. All other points are approximate points chosen to represent the stress strain curve by a series of straight line. The areas under the stress strain curve are calculated as shown on the right of the figure.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

3

The area of the triangle OAA1 represents Modulus of Resilience Modulus of Resilience is 50 in-lbs/in

Stress ksi

80 78 72 66 60

D

E

( 50 ) ( 10 3 ) ( 0.002 ) AOA 1 = ------------------------------------------- = 50 2

C

C2 B

( 50 + 66 ) ( 10 3 ) ( 0.018 ) AA 1 BB 1 = ------------------------------------------------------- = 1044 2

A

50 40

( 66 + 72 ) ( 10 3 ) ( 0.02 ) BB 1 CC 1 = ---------------------------------------------------- = 1380 2

20

( 72 + 78 ) ( 10 3 ) ( 0.04 ) CC 1 DD 1 = ---------------------------------------------------= 3000 2 O A1 B1 C1 0.00 0.04 0.002

D1 E1 0.08 0.12

0.16

0.20

78 + 80 ) ( 10 3 ) ( 0.02 ) = 1580 DD 1 EE 1 = (---------------------------------------------------2

Strain in/in

Point C is at 72 ksi. The strain energy density at point C is the area AOA1 plus areas AA1BB1 and BB1CC1. Thus the strain energy density at C is U O = ( 50 + 1044 + 1380 )

3

in – lbs ⁄ in or U O = 2474 in.-lbs/ in.

3

.

The complementary strain energy density at C can be found by subtracting UO from the area of the rectangle 3

3

3

OC2CC1. Thus, U O = ( 72 ) ( 10 ) ( 0.04 ) – 2474 = 406 in – lbs ⁄ in or U O = 406 in.-lbs/ in. . The ultimate stress corresponds to point E on the graph. The area underneath the curve can be calculated as the sum of the areas shown on the right of the graph. The total sum of the areas is: 7054 in-lbs/in3. Thus,

Modulus of Toughness is 7054in-lbs/in3

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.38

For the metal alloy given in Problem 3.12, determine

(a) Modulus of Resilience, use proportional limit as an approximation for yield point. (b) Strain energy density at a stress level of 1400 MPa. (c) Complimentary strain energy density at a stress level of 1400 MPa. (d) Modulus of Toughness. Solution

Modulus of Resilience=?

Uo= ? and

Uo= ? @σ = 1400 MPa

Modulus of Toughness=?

-----------------------------------------------------------The proportional limit corresponds to point A in the figure below. The stress at proportional limit is σprop = 1022 MPa and the modulus of elasticity is 300 GPa. Thus, the strain at proportional limit (at A1) is

εprop = 1022(106) /[(300)(109)] = 0.0034. All other points are approximate points chosen to represent the stress strain curve by a series of straight line. The areas under the stress strain curve are calculated as shown on the right of the figure. 3

The area of the triangle OAA1 represents Modulus of Resilience. Modulus of Resilience is 1734 kN-m/m Point E is at 1400 MPa. The strain energy density at point E is the area AOA1 through DD1EE1. Thus the strain 3

energy density at E is U O = ( 1734 + 10775 + 43920 + 65850 + 34588 ) ( 10 3 ) N – m ⁄ m or

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

3

U O = 157 MN – m ⁄ m .

E2

D

( 1022 ) ( 10 6 ) ( 0.0034 ) AOA 1 = ----------------------------------------------------- = 1734 ( 10 3 ) 2

F

E

( 1022 + 1133 ) ( 10 6 ) ( 0.01 ) AA 1 BB 1 = ----------------------------------------------------------------- = 10775 ( 10 3 ) 2

C B 1022 A

( 1133 + 1267 ) ( 10 6 ) ( 0.0366 ) BB 1 CC 1 = ----------------------------------------------------------------------- = 43920 ( 10 3 ) 2 ( 1267 + 1367 ) ( 10 6 ) ( 0.05 ) CC 1 DD 1 = ----------------------------------------------------------------- = 65850 ( 10 3 ) 2 A1 B1

C1

F1

D1 E1

( 1367 + 1400 ) ( 10 6 ) ( 0.025 ) DD 1 EE 1 = -------------------------------------------------------------------- = 34588 ( 10 3 ) 2

0.0034 0.0134

( 1400 + 1467 ) ( 10 6 ) ( 0.075 ) EE 1 FF 1 = -------------------------------------------------------------------- = 107512 ( 10 3 ) 2 The complementary strain energy density at C can be found by subtracting UO from the area of the rectangle OE2EE1. Thus,

6

6

U O = ( 1400 ) ( 10 6 ) ( 0.125 ) – 157 ( 10 ) = 18 ( 10 ) N – m ⁄ m

3

or

U O = 18 MN – m ⁄ m

3

. The ultimate stress corresponds to point E on the graph. The area underneath the curve can be calculated as the sum of the areas shown on the right of the graph. The total sum of the areas is: 264(106) N-m/m3. Thus,

Modulus of Toughness is 264 MN-m/m3

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.39

For the mild steel given in Problem 3.13, determine

(a) Modulus of Resilience, use proportional limit as an approximation for yield point. (b) Strain energy density at a stress level of 50 ksi. (c) Complimentary strain energy density at a stress level of 50 ksi. (d) Modulus of Toughness. Solution

Uo= ? and Uo= ? @ σ = 50 ksi

Modulus of Resilience=? Modulus of Toughness=?

------------------------------------------------------------

The proportional limit corresponds to point A in the figure below.The stress at proportional limit is σprop = 36.9 ksi and the modulus of elasticity is 24,893 ksi. Thus, the strain at proportional limit (at A1) is εprop = 36.9/ 24,893 = 0.00148. All other points are approximate points chosen to represent the stress strain curve by a series of straight line. The areas under the stress strain curve are calculated as shown on the right of the figure. The area of the triangle OAA1 represents Modulus of Resilience. Modulus of Resilience is 27.3 in-lbs/in3 Point B is at 50 ksi. The strain energy density at point B is the area AOA1 plus area AA1BB1. Thus the strain energy 3

3

density at B is U O = ( 27.31 + 1521 ) in – lbs ⁄ in or U O = 1548.3 in.-lbs/ in. . The complementary strain energy density at B can be found by subtracting UO from the area of the rectangle 3

3

OB2BB1. Thus, U O = ( 50 ) ( 10 ) ( 0.035 ) – 1548.3 = 201.7 in – lbs ⁄ in or

U O = 201.7 in.-lbs/ in.

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.

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D

E

F

( 36.9 ) ( 10 3 ) ( 0.00148 ) AOA 1 = ------------------------------------------------------ = 27.31 2

G

C B2

January 2014

( 36.9 + 50 ) ( 10 3 ) ( 0.035 ) AA 1 BB 1 = ----------------------------------------------------------- = 1521 2

B

A

( 50 + 56 ) ( 10 3 ) ( 0.05 ) BB 1 CC 1 = ---------------------------------------------------- = 2650 2

36.9

( 56 + 60 ) ( 10 3 ) ( 0.02 ) CC 1 DD 1 = ---------------------------------------------------- = 1160 2 A1

B1 C1 D1

0.00148

0.035

E1

F1

( 60 + 64 ) ( 10 3 ) ( 0.05 ) DD 1 EE 1 = ---------------------------------------------------= 3100 2

( 64 + 66 ) ( 10 3 ) ( 0.05 ) DD 1 EE 1 = ---------------------------------------------------- = 3250 2 The ultimate stress corresponds to point G on the graph. The area underneath the curve can be calculated as the sum of the areas shown on the right of the graph. The total sum of the areas is: 11708.3 in-lbs/in3. Thus,

Modulus of Toughness is 11,708 in-lbs/in3

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.40

The roller at P slides in the slot by an amount δP = 0.25 mm due to the force F. Member AP has an

area of cross-section of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. If the roller moves by the amount given, determine the force F. F

20

0m

m

P

A 50o Figure P3.40

Solution

δP = 0.25 mm E = 200 GPa

A = 100 mm2

F=?

------------------------------------------------------------

1.Strain Calculation: Assuming small strain, the average normal strain of bar AP is given by Eq.(2) in problem 2.35 ε AP = 803.5μ mm ⁄ mm 2.Stress Calculation: By Hooke’s law we can find the normal stress as shown below. 9

–6

6

σ AP = Eε AP = ( 200 ) ( 10 ) ( 803.5 ) ( 10 ) = 160.7 ( 10 )N ⁄ m

2

(T)

3.Internal Force Calculation: The internal normal force can be found as shown below. –6

6

3

N AP = Aσ AP = ( 100 ) ( 10 ) ( 160.7 ) ( 10 ) = 16.07 ( 10 )N = 16.07kN ( T )

4. External Force Calculation: We draw the free body diagram of the rigid bar as shown below. F 50o NA

R Tensile

By force equilibrium in the x-direction, we obtain the following.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

F = N AP cos 50 = ( 16.07 ) cos 50 = 10.33 or

January 2014

F = 10.3 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.41

The roller at P slides in the slot by an amount δP = 0.25 mm due to the force F. Member AP has an

area of cross-section of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. If the roller moves by the amount given, determine the force F. F

P 20

0m

m

30o

A 50o

Figure P3.41

Solution

δP = 0.25 mm E = 200 GPa

A = 100 mm2

F=?

------------------------------------------------------------

1.Strain Calculation: Assuming small strain, the average normal strain of bar AP is given by E q.(2) in problem 2.36 as ε AP = 1174.6μ mm ⁄ mm 2.Stress Calculation: By Hooke’s law we can find the normal stress as shown below. 9

–6

6

σ AP = Eε AP = ( 200 ) ( 10 ) ( 1174.6 ) ( 10 ) = 234.9 ( 10 )N ⁄ m

2

(T)

3.Internal Force Calculation: The internal normal force can be found as shown below. –6

6

3

N AP = Aσ AP = ( 100 ) ( 10 ) ( 234.9 ) ( 10 ) = 23.49 ( 10 )N = 23.49kN

4. External Force Calculation: We draw the free body diagram of the rigid bar as shown below. F

y

x

20o

R Tensile

NAP

By force equilibrium in the x-direction in Fig.(a), we obtain the following. F = N AP cos 20 = ( 23.49 ) cos 20 = 22.073

F = 22.1kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.42

A roller slides in a slot by the amount δP = 0.01 in. in the direction of the force F. Both bars have an

.

area of cross-section of A = 0.2 in2 and a modulus of elasticity E = 30,000 ksi. Bar AP and BP have lengths of LAP= 8 in. and LBP= 10 in. respectively. Determine the applied force F. B 110o F

A P

Figure P3.42

Solution

δP = 0.01 in LAP= 8 in.

E = 30,000 ksi LBP= 10 in.

A = 0.2 in F=?

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

1. Strain calculation: We draw the exaggerated deformed shape as shown in Fig. (a) Β

(a)

2.052

(b)

Compressive Tensile

110o 70o

δBP

Contraction

P2

Extension

70o P δAP=δP P1

Α

F

7.5 R

As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP1 represents the deformation of bar AP and PP2 represents the deformation of bar BP and are calculated as shown below. δ AP = δ P = 0.01in extension δ BP = δ P cos 70 = 0.00342in contraction δ AP –3 ε AP = --------- = 0.01 ---------- = 1.25 ( 10 )extension 8 L AP δ BP –3 ε BP = --------- = 0.00342 ------------------- = 0.342 ( 10 )contraction 10 L BP

2.Stress calculation: –3

σ AP = Eε AP = ( 30000 ) ( 1.25 ) ( 10 ) = 37.5ksi ( T ) –3

σ BP = Eε BP = ( 30000 ) ( 0.342 ) ( 10 ) = 10.26ksi ( C )

3.Internal force Calculation: N AP = Aσ AP = ( 37.5 ) ( 0.2 ) = 7.5kips ( T ) N BP = Aσ BP = ( 10.26 ) ( 0.2 ) = 2.052kips ( C )

4. External force Calculation: We draw the free body diagram of the rigid bar as shown in Fig. (b) . By force equilibrium in the x-direction we obtain the following. F – 7.5 – 2.052 cos 70 = 0 or F = 8.2 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.43

A roller slides in a slot by the amount δP = 0.25 mm in the direction of the force F. Both bars have

.

an area of cross-section of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. Bar AP and BP have lengths of LAP= 200 mm and LBP= 250 mm respectively. Determine the applied force F. B

A

60o P F

Figure P3.43

Solution

δP = 0.25 mm

A = 100 mm2

LAP= 200 mm

LBP= 250 mm F=?

------------------------------------------------------------

1.Strain Calculation: Assuming small strain, the deformation of the bars can be found as in problem 2-38. δ AP = 0 and δ BP = 0.2165mm extension

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ε AP = 0

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

and

January 2014

δ BP –3 0.2165 - = ---------------- = 0.866 ( 10 ) extension ε BP = ---------LBP 250

2. Stress calculation: By Hooke’s law we can find the normal stress as shown below 9

–3

6

2

σ AP = 0 and σ BP = Eε BP = ( 200 ) ( 10 ) ( 0.866 ) ( 10 ) = 173.2 ( 10 )N ⁄ m ( T )

3. Internal force Calculation: 6

–6

3

N AP = 0 and N BP = Aσ BP = ( 173.2 ) ( 10 ) ( 100 ) ( 10 ) = 17.32 ( 10 )N = 17.32kN ( T )

4. External Force Calculation We draw the free body diagram as shown below. 17.32 kN

Tensile

60o

R F

By force equilibrium in the y-direction, we obtain the F = 17.32 sin 60 = 14.999kN F = 15 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.44

A roller slides in a slot by the amount δP = 0.25 mm in the direction of the force F. Both bars have

an area of cross-section of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. Bar AP and BP have lengths of LAP= 200 mm and LBP= 250 mm respectively. Determine the applied force F. B A

75o

o

30

P F Figure P3.44

Solution

δP = 0.25 mm LAP= 200 mm

A = 100 mm2 F=?

E = 200 GPa LBP= 250 mm

------------------------------------------------------------

1.Strain Calculation: Assuming small strain, the deformation of the bars can be found as in problem 2-39. δ AP = 0.0647mm extension and δ BP = 0.21651mm extension δ AP δ BP –3 –3 0.0647 0.21651 ε AP = ----------- = ---------------- = 0.3235 ( 10 ) extension and ε BP = ---------- = ------------------- = 0.866 ( 10 ) extension LAP 200 LBP 250

2.Stress calculation: By Hooke’s law we can find the normal stress as shown below 9

–3

6

2

6

2

σ AP = Eε AP = ( 200 ) ( 10 ) ( 0.3235 ) ( 10 ) = 64.7 ( 10 )N ⁄ m ( T ) 9

–3

σ BP = Eε BP = ( 200 ) ( 10 ) ( 0.866 ) ( 10 ) = 173.2 ( 10 )N ⁄ m ( T )

3.Internal force Calculation: 6

–6

3

N AP = Aσ AP = ( 64.7 ) ( 10 ) ( 100 ) ( 10 ) = 6.47 ( 10 )N = 6.47kN ( T ) 6

–6

3

N BP = Aσ BP = ( 173.2 ) ( 10 ) ( 100 ) ( 10 ) = 17.32 ( 10 )N = 17.32kN ( T )

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

4. External Force Calculation: We draw the free body diagram of the roller as shown below. Tensile

NB

30o

75o

R

NA Tensile

F

By force equilibrium in the y-direction, we obtain the following. F = 17.32 cos 30 + 6.47 cos 75 = 12.674 or

F = 16.7kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.45 A little boy is shooting paper darts at his friends using a rubber band that has an un-stretched length of 7 inches. The piece of rubber band between point A and B is pulled to form two sides AC and CB of a triangle as shown. Assume the same normal strain in AC and CB, and the rubber band around the thumb and forefinger is a total of one inch. The area of cross-section of the band is (1/128) in2 and the rubber has Modulus of Elasticity E= 150 psi. Determine the approximate force F and the angle θ at which the paper dart leaves the boy’s hand. 3.2 in A

C

A

C

F

B

θ

2.5 2.9 in

B

Figure P3.45

Solution

E =150 psi

A = 1/128 in

θ=?

L0= 7 in

F=?

------------------------------------------------------------

1.Strain Calculation: The stretched length is: L f = 3.2 + 2.9 + 2.5 + 1 = 9.6in Lf – L0 – 7- = 0.37143 extension ε = ----------------= 9.6 --------------7 L0

2.Stress calculation: By Hooke’s law: σ = Eε = ( 150 ) ( 0.37143 ) = 55.71psi ( T ) 1 128

3.Internal force Calculation: N = Aσ = ( 55.71 ) --------- = 0.43527lbs ( T ) 4. External Force Calculation.The following free body diagram of the dart can be created

F

y

C θ

N

2

2

2

3.2 + 2.9 – 2.5 cos β = -----------------------------------------2 ( 3.2 ) ( 2.9 )

β

or

β = 48.08

0

N

x By equilibrium of forces in the x-direction F cos θ – N sin β = 0 or F cos θ = 0.43527 ( sin 48.08 ) = 0.32387

1

By equilibrium of forces in the y-direction. F sin θ – N – N cos β = 0 F sin θ = 0.43527 ( 1 + cos 48.08 ) = 0.72608

Dividing Eq(2) by Eq(1) we obtain tan θ = 2.24187 or

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2 θ = 65.96

°

3-30

M. Vable

From Eq(1)

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

0.32387 F = ---------------------- = 0.795lb cos 65.96

January 2014

F = 0.795lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.46 Three poles are pin connected to a ring at P and to the supports on the ground. The coordinates of the four points are as given. All poles have an area of cross-section of A=1 in2. and a Modulus of Elasticity E= 10,000 ksi. If under the action of force F the ring at P moves vertically by an amount δP = 2 inch determine the force F. z F P (0.0, 0.0, 6.0) ft. B

C

(-4.0, 6.0, 0.0) ft.

(-2.0, -3.0, 0.0) ft.

y A

x

(5.0, 0.0, 0.0) ft.

Figure P3.46

Solution

δP= 2 in

E =10,000ksi

A = 1in2

F=?

------------------------------------------------------------

1. Strain Calculation: From the solution of problem 2-50, we have: ε AP = 16393μin ⁄ in

ε BP = 11364μin ⁄ in

ε CP = 20407μin ⁄ in

2.Stress calculation: –6

σ AP = Eε AP = ( 10000 ) ( 16393 ) ( 10 ) = 163.93ksi ( T ) –6

σ BP = Eε BP = ( 10000 ) ( 11364 ) ( 10 ) = 113.64ksi ( T ) –6

σ CP = Eε CP = ( 10000 ) ( 20407 ) ( 10 ) = 204.07ksi ( T ) 3.Internal force Calculation: N AP = Aσ AP = 163.93kips ( T ) N BP = Aσ BP = 113.64kips ( T ) and N CP = Aσ CP = 204.07kips ( T )

4. External Force Calculation. The free body diagram of the collar can be drawn as shown below. F Ry

Rx

D NB

NC NA

N AP + N BP + N CP + Fk + R = 0

1

Noting that the unit vectors in problem 2-50 were directed towards P and the internal forces are directed away from P, we can write

N BP

N AP = – N AP i AP = ( – 163.93 ) ( – 0.6402iˆ + 0.7682kˆ ) = –N i = ( – 113.64 ) ( – 0.4264iˆ – 0.6396jˆ + 0.6396kˆ ) BP BP

2 3

N CP = – N CP i CP = ( – 204.07 ) ( 0.2857iˆ + 0.4286jˆ + 0.8571kˆ )

4

R = R x ˆi + R y ˆj

5

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

Substituting Eq’s (2),(3),(4) and (5) into Eq(1) and considering the coefficient of unit vector k we obtain: – 163.93 ( 0.7682 ) – 113.64 ( 0.6396 ) – 204.07 ( 0.8571 ) + F = 0 or F = 373.52 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.47 A gap of 0.004 inch exists between the rigid bar and bar A before the force F is applied. The rigid bar is hinged at point C. Due to force F the strain in bar A was found to be - 500 μ in/in. The lengths of bar A and B are 30 and 50 inches respectively. Both bars have an area of cross-section A= 1 in2 and Modulus of Elasticity E = 30,000 ksi. Determine the applied force F. B C F 24 in

75 36 in

o

A 60 in

Figure P3.47

Solution

gap= 0.004

EA=EB =30,000ksi

LA= 30in

LB= 50in

εΑ=-500μ in/in

AA=AB= 1in2

------------------------------------------------------------

1.Strain Calculation: From the solution of problem 2-43, we have: ε B = 978.8μ –6

2.Stress calculation: σ A = Eε A = ( 30000 ) ( – 500 ) ( 10 ) = – 15k si σ A = 15ksi ( C ) –6

σ B = Eε B = ( 30000 ) ( 978.8 ) ( 10 ) = 29.364ksi

σ B = 29.364ksi ( C )

3.Internal force Calculation: N A = Aσ A = 15kips ( C ) N B = Aσ B = 29.364kips ( T ) 4. External Force Calculation: The free body diagram of the rigid bar can be drawn as shown below. NB

Tension Cy Cx 24 in

75o 60 in

36 in

F

Compressive NA

By moment equilibrium about point C, we obtain: 24F = ( 36 ) ( 15 ) + ( 29.364 ) ( sin 75 ) ( 96 ) or F = 136 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.48 The cable between two poles shown in Figure P3.48 is taut before the two traffic lights are hung on it. The lights are placed symmetrically at 1/3 the distance between the poles. The cable has a diameter of 1/ 16 in and a modulus of elasticity of 28,000 ksi. Determine the weight of the traffic lights if the cable sags as shown. 27 ft 10 in

Figure P3.48

Solution

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

The deformed shape and free body diagram are as shown 108 in A

108 in

108 in

θ

θ

10 in.

N

N D

A

B

θ

C

C

B

D

θ

W

W

From geometry: AB = CD =

2

2

108 + 10 = 108.46 in.

10 tan θ = --------or θ = 5.29° 108 The initial length is: L o = 27 ft = 324 in. . The final length is: L f = AB + BC + CD = 2 ( 108.46 ) + 108 = 324.9239 in. . Lf – Lo –3 324.9239 – 324 The average normal strain is: ε av = ---------------- = ------------------------------------- = 2.8517 ( 10 ) 324 Lo –3

The average normal stress is: σ av = Eε av = ( 28000 ) ( 2.8517 ) ( 10 ) = 79.84 ksi –3 π 1 2 The internal force in the cable is: N = σ av A = 79.84 --- ⎛ ------⎞ = 244.97 ( 10 ) ksi = 244.97 psi 4 ⎝ 16⎠

From FBD we have: 2N sin θ = 2W

or

W = 244.97 sin 5.29 = 22.58 lb

W = 22.6 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.49 A steel bolt (Es= 200 GPa) of 25 mm diameter passes through an aluminum (Eal = 70 GPa) sleeve of thickness 4 mm and outside diameter of 48 mm as shown in Figure P3.49. Due to the tightening of the nut the rigid washers move towards each other by 0.75 mm. (a) Determine the average normal stress in the sleeve and the bolt. (b) What is the extension of the bolt? Sleeve

Rigid washers

300 mm

Figure P3.49 25 mm

25 mm

Solution

-----------------------------------------------------------The crosssectional areas are: 2 2 –3 2 π A Al = --- ( 0.048 – 0.04 ) = 0.5529 ( 10 ) m 4

2 –3 2 π A s = --- ( 0.025 ) = 0.4909 ( 10 ) m 4

δ Al –3 0.75 The average normal strain in aluminum is: ε Al = -------- = ---------- = 2.5 ( 10 ) L Al

300

9

–3

6

2

The average normal stress in aluminum is: σ Al = Eε Al = ( 70 ) ( 10 ) ( 2.5 ) ( 10 ) = 175 ( 10 ) N/m ( C )

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

6

January 2014

–3

6

The internal force in aluminum is: N Al = σ Al A = 175 ( 10 ) ( 0.5529 ) ( 10 ) = 96.76 ( 10 ) ( C ) The free body diagram is: Rigid washers

Sleeve

NAl

Ns

300 mm 25 mm

25 mm

6

By equilibrium: N S = N Al = 96.76 ( 10 ) ( T ) 6

N 6 2 96.76 ( 10 ) The average normal stress in steel is: σ S = ------S- = -------------------------------= 197.1 ( 10 ) N/m ( T ) A

–3

0.4909 ( 10 )

σ –3 197.1 ( 10 ) The average normal strain in steel is: ε S = -----S- = -------------------------- = 0.9856 ( 10 ) 6

E

9

( 200 ) ( 10 )

–3

–3

The elongation of steel bolt is: δ S = L S ε S = ( 300 + 25 + 25 ) ( 0.9856 ) ( 10 ) = 344.9 ( 10 ) mm σ Al = 175 MPa ( C ) ; σ S = 197.1 MPa ( T ) ; δ S = 0.345 mm

The answers are:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.50 The pins in the truss shown are displaced by u and v in the x and y direction respectively, as given. All rods in the truss have an area of cross-section A= 100 mm2 and a Modulus of Elasticity E= 200 GPa. Determine the external forces P1 and P2. u A = – 4.6765 mm

vA = 0

u B = – 3.3775 mm

v B = – 8.8793 mm

u C = – 2.0785 mm

v C = – 9.7657 mm

u D = – 1.0392 mm

v D = – 8.4118 mm

u E = 0.0000 mm

v E = 0.0000 mm

u F = – 3.260 0 mm

v F = – 8.4118 mm

u G = – 2.5382 mm

v G = – 9.2461 mm

u H = – 1.5500 mm

v H = – 8.8793 mm

Solution

A=100 mm2

P3

y x

P5

H

A

300 3m

P2

G

P4

P1

F

300 B

C 3m

D 3m

E 3m

Figure P3.50

E = 200GPa

P 1 = ? P2 = ?

------------------------------------------------------------

1. Strain Calculation in all members at joint F. Deformation vector calculation. D FC = ( u C – u F )iˆ + ( v C – v F )jˆ = ( – 2.0785 + 3.26 )iˆ + ( – 9.7657 + 8.4118 )jˆ = ( 1.1815iˆ – 1.3539jˆ ) mm D FG= ( u G – u F )iˆ + ( v G – v F )jˆ = ( – 2.5382 + 3.26 )iˆ + ( – 9.2461 + 8.4118 )jˆ =( 0.7218iˆ – 0.8343jˆ )mm D FE = ( u E – u F )iˆ + ( v E – v F )jˆ = ( 3.26iˆ – 8.4118jˆ )mm D FD = ( u D – u F )iˆ + ( v D – v F )jˆ = ( – 1.0392 + 3.26 )iˆ + ( – 8.4118 + 8.4118 )jˆ = ( 2.2208iˆ )mm

Unit vector calculation. ˆi ˆ ˆ ˆ ˆ FC = – cos 30i – sin 30j = – 0.866i – 0.5j

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

ˆi ˆ ˆ ˆ ˆ FG = – cos 30i – sin 30j = – 0.866i – 0.5j ˆi ˆ ˆ ˆ ˆ FE = – cos 30i – sin 30j = – 0.866i – 0.5j ˆi ˆ FD = – j

Deformation calculation. δ FC = i FC ⋅ D FC = ( – 0.866 ) ( 1.1815 ) + ( – 0.5 ) ( – 1.3539 ) = – 0.34626mm δ FG = i FG ⋅ D FG = ( – 0.866 ) ( 0.7218 ) + ( 0.5 ) ( – 0.8343 ) = – 1.04225mm δ FE = i FE ⋅ D FE = ( – 0.866 ) ( 3.26 ) + ( – 0.5 ) ( 8.4118 ) = – 1.38274mm δ FD = i FD ⋅ D FD = 0

Strains. L FC = L FG = L FE = 31 cos 30 = 3.4641m –3 0.34626 ε FC = – ------------------------------ = – 0.1 ( 10 ) 3 3.4641 ( 10 )

–3 1.04225 ε FG = – ------------------------------ = – 0.3 ( 10 ) 3 3.4641 ( 10 )

–3 1.38274 ε FE = – ------------------------------ = – 0.4 ( 10 ) 3 3.4641 ( 10 )

ε FD = 0

2. Stress Calculation: 9 –3 6 6 N σ FC = Eε FC = ( 200 ) ( 10 ) ( – 0.1 ) ( 10 ) = – 20 ( 10 )orσ FC = 20 ( 10 ) ------- ( C ) 2 m 9 –3 6 6 N σ FG = Eε FG = ( 200 ) ( 10 ) ( – 0.3 ) ( 10 ) = – 60 ( 10 )orσ FC = 60 ( 10 ) ------- ( C ) 2 m 9 –3 6 6 N σ FE = Eε FE = ( 200 ) ( 10 ) ( – 0.4 ) ( 10 ) = – 80 ( 10 )orσ FC = 80 ( 10 ) ------- ( C ) 2 m

σ FD = 0

3.Internal Force calculation. 6

–6

6

–6

6

–6

N FC = σ FC A = ( 20 ) ( 10 ) ( 100 ) ( 10 ) = 2000N = 2kN ( C ) N FG = σ FG A = ( 60 ) ( 10 ) ( 100 ) ( 10 ) = 6000N = 6kN ( C ) N FE = σ FE A = ( 80 ) ( 10 ) ( 100 ) ( 10 ) = 8000N = 8kN ( C ) N FD = 0

4.External Force Calculation. We draw the free body diagram of joint F as shown below. P2

6kN

P1 F

2 kN

300

300

8 kN

By force equilibrium in the x-direction, we obtain

– P 1 – 8 cos 30 + 2 cos 30 + 6 cos 30 = 0

or P 1 = 0

By force equilibrium in the y-direction, we obtain

– P 2 + 2 sin 30 + 8 sin 30 – 6 sin 30 = 0

or P 2 = 2 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.51 The pins in the truss shown are displaced by u and v in the x and y direction respectively, as given. All rods in the truss have an area of cross-section A= 100 mm2 and a Modulus of Elasticity E=

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

200 GPa.Determine the external force P3 u A = – 4.6765 mm

vA = 0

u B = – 3.3775 mm

v B = – 8.8793 mm

u C = – 2.0785 mm

v C = – 9.7657 mm

u D = – 1.0392 mm

v D = – 8.4118 mm

u E = 0.0000 mm

v E = 0.0000 mm

u F = – 3.260 0 mm

v F = – 8.4118 mm

u G = – 2.5382 mm

v G = – 9.2461 mm

u H = – 1.5500 mm

v H = – 8.8793 mm A=100 mm2

Solution

P3

y x P5

H

A

300 3m

P2

G

P4

P1

F

300 B

C 3m

D 3m

E 3m

Figure P3.51

E = 200GPa

P3 = ?

------------------------------------------------------------

1. Strain Calculation in all members at joint G. (i) Deformation vector calculation. D GF = ( u F – u G )iˆ + ( v F – v G )jˆ = ( – 3.26 + 2.5382 )iˆ + ( – 8.4118 + 9.2461 )jˆ = ( – 0.7218 ˆi + 0.8343jˆ )mm D GH = ( u H – u G )iˆ + ( v H – v G )jˆ = ( – 1.55 + 2.5382 )iˆ + ( – 8.8793 + 9.2461 )jˆ = ( 0.98820iˆ + 0.3668jˆ )mm D GC = ( u C – u G )iˆ + ( v C – v G )jˆ = ( – 2.0785 + 2.5382 )iˆ + ( – 9.7657 + 9.2461 )jˆ = ( 0.4597iˆ – 0.5196jˆ )mm

(ii) Unit vector calculation. ˆi ˆ ˆ ˆ ˆ GF = cos 30i – sin 30j = 0.866i – 0.5j ˆi ˆ ˆ ˆ ˆ GH = – cos 30i – sin 30j = – 0.866i – 0.5j ˆi ˆ GC = – j

(iii) Deformation calculation. δ GF = i GF ⋅ D GF = ( 0.866 ) ( – 0.7218 ) + ( – 0.5 ) ( 0.8343 ) = – 1.0423mm δ GH = i GH ⋅ D GH = ( – 0.866 ) ( 0.9882 ) + ( – 0.5 ) ( 0.3668 ) = – 1.03918mm δ GC = i GC ⋅ D GC = 0.5196mm

(iv) Strains. L GF = L GN = L GC = 3 ⁄ ( cos 30 ) = 3.4641m –3 1.0423 ε GF = – ----------------------------- = – 0.3 ( 10 ) 3 3.4641 ( 10 )

1.03918 = – ( – 3 ) ε GH = – ----------------------------0.3 10 3 3.4641 ( 10 ) –3 0.5196 ε GC = – ----------------------------- = 0.15 ( 10 ) 3 3.4641 ( 10 )

2.Stress Calculation: 9 –3 6 6 N σ GF = Eε GF = ( 200 ) ( 10 ) ( – 0.3 ) ( 10 ) = – 60 ( 10 )orσ GF = 60 ( 10 ) ------- ( C ) 2 m 9 –3 6 6 N σ GH = Eε GH = ( 200 ) ( 10 ) ( – 0.3 ) ( 10 ) = – 60 ( 10 )orσ GH = 60 ( 10 ) ------- ( C ) 2 m

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

9 –3 6 6 N σ GC = Eε GC = ( 200 ) ( 10 ) ( 0.15 ) ( 10 ) = – 30 ( 10 )orσ GC = 30 ( 10 ) ------- ( T ) 2 m

3.Internal Force calculation. 6

–6

N GF = σ GF A = ( 60 ) ( 10 ) ( 100 ) ( 10 ) = 6000N = 6kN ( C ) 6

–6

6

–6

N GH = σ GH A = ( 60 ) ( 10 ) ( 100 ) ( 10 ) = 6000N = 6kN ( C ) N GC = σ GC A = ( 30 ) ( 10 ) ( 100 ) ( 10 ) = 3000N = 3kN ( T )

4. External Force Calculation. We draw the free body diagram of the joint G as shown below. P3 G 0

60 6kN

600 6kN 3kN

By equilibrium of forces in y-direction – P 3 + 6 cos 60 + 6 cos 60 – 3 = 0

or

P 3 = 3 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.52 The pins in the truss shown are displaced by u and v in the x and y direction respectively, as given. All rods in the truss have an area of cross-section A= 100 mm2 and a Modulus of Elasticity E= 200 GPa.Determine the external force P4 and P5. vA = 0

u B = – 3.3775 mm

v B = – 8.8793 mm

u C = – 2.0785 mm

v C = – 9.7657 mm

u D = – 1.0392 mm

v D = – 8.4118 mm

u E = 0.0000 mm

v E = 0.0000 mm

u F = – 3.260 0 mm

v F = – 8.4118 mm

u G = – 2.5382 mm

v G = – 9.2461 mm

u H = – 1.5500 mm

v H = – 8.8793 mm

Solution

A=100 mm2

P3

y

u A = – 4.6765 mm

E = 200GPa

x

P2

G

P4 P5

H

P1

F

300

300 A

B 3m

C 3m

D 3m

E 3m

Figure P3.52

P4 = ?

P5 = ?

------------------------------------------------------------

1. Strain Calculation in all members at joint H. (i) Deformation vector calculation. D HA = ( u A – u H )iˆ + ( v A – v H )jˆ = ( – 4.6765 + 1.55 )iˆ + ( 8.8793 )jˆ = ( – 3.1265iˆ + 8.8793jˆ ) D HG = ( u G – u H )iˆ + ( v G – v H )jˆ = ( – 2.5382 + 1.55 )iˆ + ( – 9.2461 + 8.8793 )jˆ or D HG = ( – 0.9882iˆ – 0.3668jˆ ) D HC = ( u C – u H )iˆ + ( v C – v H )jˆ = ( – 2.0785 + 1.55 )iˆ + ( – 9.7657 + 8.8793 )jˆ = ( – 0.5285iˆ – 0.8864jˆ ) D HB = ( u B – u H )iˆ + ( v B – v H )jˆ = ( – 3.375 + 1.55 )iˆ + ( – 8.8793 + 8.8793 )jˆ = – 1.825iˆ

(ii) Unit vector calculation. ˆi ˆ ˆ ˆ ˆ HA = – cos 30i – sin 30j = – 0.866i – 0.5j ˆi ˆ ˆ ˆ ˆ HG = cos 30i + sin 30j = 0.866i + 0.5j ˆi ˆ ˆ ˆ ˆ HC = cos 30i – sin 30j = 0.866i – 0.5j

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

ˆi ˆ HB = – j

(iii) Deformation calculation. δ HA = i HA ⋅ D HA = ( – 0.866 ) ( – 3.1265 ) + ( – 0.5 ) ( 8.8793 ) = – 1.73202mm δ HG = i HG ⋅ D HG = ( 0.866 ) ( – 0.9882 ) + ( 0.5 ) ( – 0.3668 ) = – 1.03918mm δ HC = i HC ⋅ D HC = ( 0.866 ) ( – 0.5285 ) + ( – 0.5 ) ( – 0.8864 ) = – 0.01449mm δ HB = i HB ⋅ D HB = 0

(iv) Strains. L HA = L HG = L HC = 3 ⁄ cos 30 = 3.4641m δ HA –3 1.73202 ε HA = ---------- = – ------------------------------ = – 0.5 ( 10 ) 3 L HA 3.4641 ( 10 )

δ HG –3 1.03918 ε HG = ---------- = – ------------------------------ = – 0.3 ( 10 ) 3 L HG 3.4641 ( 10 )

δ HC –3 0.01449 = – ε HC = ---------= – ----------------------------0.004 ( 10 ) 3 L HC 3.4641 ( 10 )

ε HD = 0

2.Stress Calculation: 9

–3

9

–3

6 N σ HA = 100 ( 10 ) ------- ( C ) 2 m

6

σ HA = Eε HA = ( 200 ) ( 10 ) ( – 0.5 ) ( 10 ) = – 100 ( 10 ) or

6 N σ HG = 60 ( 10 ) ------- ( C ) 2 m

6

σ HG = Eε HG = ( 200 ) ( 10 ) ( – 0.3 ) ( 10 ) = – 60 ( 10 ) or 9

–3

6

σ HC = Eε HC = ( 200 ) ( 10 ) ( – 0.004 ) ( 10 ) = – 0.8 ( 10 )

6 N σ FC = 0.8 ( 10 ) ------- ( C ) 2 m

or

σ HB = 0

3.Internal Force calculation. 6

–6

N HA = σ HA A = ( 100 ) ( 10 ) ( 100 ) ( 10 ) = 10000N = 10kN ( C ) 6

–6

6

–6

N HG = σ HG A = ( 60 ) ( 10 ) ( 100 ) ( 10 ) = 6000N = 6kN ( C ) N HC = σ HC A = ( 0.8 ) ( 10 ) ( 100 ) ( 10 ) = 80N = 0.08kN ( C ) N HB = 0

4. External Force Calculation. We draw the free body diagram of the joint H as shown below. P4 P5

H

10 kN

By equilibrium of forces in x-direction:

600

6 kN

600

0.08 kN

– P 5 – 6 sin 60 + 10 sin 60 – 0.8 sin 60 = 0

or P 5 = 3.39 kN

By equilibrium of forces in y-direction – P 4 + 10 cos 60 – 6 cos 60 + 0.08 cos ( 60 ) = 0 or P 4 = 2.04 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.53 A joint in a wooden structure is to be designed for a factor of safety of 3. If the average failure stress in shear on surface BCD is 1.5 ksi and if the average failure bearing stress on surface BEF is 6 ksi, determine the smallest dimensions h and d to the nearest 1/16th of an inch.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

10 kips

30o

4 in A

F

E

h

d

D

B C

Figure P3.53

Solution

K=3

(τf)BCD=1.5ksi

(σf)BEF=6ksi

h=?

d=? to the nearest 1/16 inch

------------------------------------------------------------

The following free body diagrams can be constructed. (a)

10 kips

10 kips

(b)

30o

E,F A B

30o

C,D VBCD

E A B

NBEF

N2

N1

From Fig.(a), by force equilibrium in the x-direction we obtain: – V BCD + 10 cos 30 = 0 or V BCD = 10 cos 30 = 8.66kips V BCD The average shear stress is: τ BCD = -------------- = 8.66 ---------A BCD

( τ f ) BCD K = ------------------τ BCD

or

1.5 ------------------------( 8.66 ⁄ 4h )

4h

or

h = 4.33

3 h = 4 ---- in 8

From Fig.(b), by force equilibrium in the x-direction we obtain: 10 cos 30 – N BEF = 0 or N BEF = 8.66kips (C) N

8.66 BEF The average normal stress is: σ BEF = ------------- = ---------- (C) A BEF

( σ f ) BEF K = ------------------σ BEF

or

6 ------------------------ = 3 ( 8.66 ⁄ 4d )

4d

or

d = 1.0825 or

1 d = 1 ---- in 8

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.54 A 125 kg light is hanging from a ceiling by a chain as shown in Figure P3.54. The links of the chain are loops made from a thick wire. Determine the minimum diameter of the wire to the nearest millimeter for a factor of safety of 3. The normal failure stress for the wire is 180MPa

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Figure P3.54

Solution

m=125 kg

K=3

σfail= 180 MPa d=? nearest millimeter

------------------------------------------------------------

The following free body diagram can be drawn by making an imaginary cut through the two wires in a link of the chain. NN

125g=1226.2

By force equilibrium 2N = 1226.2 or N = 613.125N 780.65 σ fail 613.125 The average normal stress is: σ = N ---- = --------------------- = ---------------- ≤ ----------2 A

6

180 ( 10 ) 780.65 ---------------- = ---------------------2 3 d

( πd 2 ⁄ 4 )

d

–3

or d ≥ 3.61 ( 10 )m

3

or

d = 4 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.55

A light is hanging from a ceiling by a chain as shown in Figure P3.55. The links of the chain are

1 loops made from a thick wire with a diameter of --- in. The normal failure stress for the wire is 25 ksi. For a 8

factor of safety of 4, determine the maximum weight of the light to the nearest lb.

Solution

W=?

K=4

σfail=25ksi

Figure P3.55

d= 1/8

------------------------------------------------------------

The following free body diagram can be drawn by making an imaginary cut through the two wires in a link

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

of the chain. NN

W

By force equilibrium 2N = W

or N = 0.5W

σ fail 0.5W 0.5W The average normal stress is: σ = N ---- = --------------------- = --------------------------------- = 40.74W ≤ ----------A

( πd 2 ⁄ 4 )

(π( 1 ⁄ 8)2 ⁄ 4)

4

3

25 ( 10 ) 40.74W ≤ ------------------- or W ≤ 153.4 lb 4

or

W max = 153 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.56 Determine the maximum weight W that can be suspended using cables as shown in Figure P3.56 for a factor of safety of 1.2. The cable’s fracture stress is 200 MPa and the diameter is 10 mm.

22o

37o

W

Figure P3.56

Solution

K=1.2

σfail=200MPa

d=10mm

Wmax=?

------------------------------------------------------------

By making imaginary cuts through the cables, the following free body diagram can be drawn. NAB

NAC 22o

37o A

W

By force equilibrium in the x-direction: – N AB cos 37 + N AC cos 22 = 0 or N AB = 1.16096N BC

1

By force equilibrium in the y-direction N AB sin 37 + N BC sin 22 – W = 0 or ( 1.16096 sin 37 + sin 22 )N BC = W or N BC = 0.9317W

2

N AB = 1.0817W

3

From Eq. (1) π The area of cross-section is: A AB = A BC = --- ( 10 ) 2 = 78.54mm 2 = 78.54 ( 10 – 6 )m 2 4

The average normal stress is: σ fail N AB 200 ( 10 6 ) 1.0817W σ AB = ---------- = ------------------------------ = 0.01377W ( 10 –6 ) ≤ ----------- or 0.01377W ( 10 6 ) ≤ ---------------------1.2 K A AB 78.54 ( 10 –6 ) W ≤ 12.103 ( 10 3 )N

4

σ fail N BC 200 ( 10 6 ) 0.9317W σ BC = ---------- = ------------------------------ = 0.01186W ( 10 6 ) ≤ ----------- or 0.01186W ( 10 6 ) ≤ ---------------------- or – 6 1.2 K A BC 78.54 ( 10 )

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

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W ≤ 14.049 ( 10 3 )N

5

The maximum weight must satisfy the inequalities in Eq(4) and (5) i.e.,

W max = 12.1 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.57 The cable in Figure P3.57 has a fracture stress of 30 ksi and is used for suspending the weight W = 2500 lb. For a factor of safety of 1.25, determine the minimum diameter of the cables to the nearest 1 ------- in. that can be used. 16

22o

37o

W

Solution

K=1.25

Figure P3.57

σfail=30 ksi

dmin=?

W= 2500lb

------------------------------------------------------------

By making imaginary cuts through the cables, the following free body diagram can be drawn. NAB

NAC 22o

37o A

W

By force equilibrium in the x-direction: – N AB cos 37 + N AC cos 22 = 0 or 1

N AB = 1.16096N BC

By force equilibrium in the y-direction N AB sin 37 + N BC sin 22 – W = 0 or ( 1.16096 sin 37 + sin 22 )N BC = W or N BC = 0.9317W = 2329.25lb

2

N AB = 2704.17lb

3

From Eq. (1) π The area of cross-section is: A AB = A BC = --- ( d ) 2 4

The average normal stress is: N AB 3443.05 σ fail 3443.05 30 ( 10 3 ) 2704.17σ AB = ---------- = -------------------= ------------------- ≤ ----------- or ------------------- ≤ ------------------2 2 1.25 K A AB d d2 ( π ⁄ 4 )d d ≥ 0.3788in N BC 2965.7 σ fail 2329.25σ BC = ---------- = -------------------= ---------------- ≤ ----------2 K A BC d2 ( π ⁄ 4 )d

or

4 30 ( 10 3 )

2965.7 ---------------- ≤ ------------------1.25 d2

or

d ≥ 0.3515in

5

7 The minimum diameter must satisfy Eq. 4. To the nearest 1/16 the diameter is: d min = ------ in. 16

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.58 An adhesively bonded joint in wood is fabricated as shown. For a factor of safety of 1.25, determine the minimum overlap length L and dimension h to the nearest 1/8th inch. The shear strength of adhesive is 400 psi and the wood strength is 6 ksi in tension.

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10 kips

January 2014

10 kips

8 in h

h

h L

Figure P3.58

Solution

K = 1.25 τfail = 400psi σfail = 6ksi Lmin = ? nearest1/8 inch, hmin = ? nearest 1/8 inch ------------------------------------------------------------

The following free body diagrams can be drawn. V

(a)

10 kips

(b)

10 kips

N

V

From Fig.(a): 2V = 10

V = 5kips From Fig.(b) N = 10kips 10 σ fail 10 6 N------ ≤ ---------The average normal stress is: σ = ----= ------ ≤ ----------or 8h K 8h 1.25 8h or

h ≥ 0.2604 or

or

3 h min = --- in. 8 τ

5 fail V The average shear stress is: τ = -----------------= ------- ≤ ---------8(L ⁄ 2)

4L

K

5- --------0.4-----≤ 4L 1.25

or

or

h ≥ 3.906 L min = 4 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.59 A joint in a truss has the configuration shown. Determine the minimum diameter of the pin to the nearest millimeter for a factor of safety of 2.0. The pin’s failure stress in shear is 300 MPa. ND = 50 kN

NC= 30 kN NA = 32.68 kN NB = 67.32 kN 30o

30o

Figure P3.59

Solution

τfail=300MPa

K=2

dmin=? nearest mm

------------------------------------------------------------

By making imaginary cuts through the pin between the truss members, the following free body diagrams can be obtained. V1

(a)

50 kN

(b)

30 kN

32.68 kN

(c)

30 kN

V2x V3 V2y

From Fig.(a) V 1 = 32.68kN

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From Fig.(b) V 2x = 30kN

V 2y = 50kN

V2 =

January 2014

30 2 + 50 2 = 583kN

From Fig.(c) V 3 = 30kN τ

5831 ( 10 ) fail The maximum shear force is V 2 . Thus, the maximum shear stress is: τ max = -----2- = ------------------------≤ ---------- or 2 V

A

( 4 ) ( 58.31 ) ( 10 3 ) 300 ( 10 6 ) ---------------------------------------- ≤ ---------------------- or d 2 ≥ 0.4949 ( 10 –3 ) 2 πd 2

3

( πd ⁄ 4 )

K

d ≥ 0.02225 ( 10 – 3 )m or d min = 23mm

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.60

The shear stress on the cross-section of the wire of a helical spring is given by 2 τ = K ( 8PC ⁄ ( πd ) ) where, P is the force on the spring, d is the diameter of the wire from which the spring is constructed, C is called the spring index given by the ratio C = D / d, D is the diameter of coiled spring, and K is called the Wahl factor given by

4C – 1 0.615 K = ---------------- + ------------4C – 4 C

. The spring is to be designed to

resist a maximum force of 1200 N and must have a factor of safety of 1.1 in yield. The shear stress in yield is 350 MPa. Make a table for admissible values of D for values of d that range 8mm ≤ d ≤ 16 mm in steps of 2 mm.

P

D

P

d

Figure P3.60

Solution

P=1200N

τyield=350MPa

K=1.1

D=? for ( 8mm ≤ d ≤ 16mm )

-----------------------------------------------------------τ yield The allowable maximum stress is: τ max = ------------= 350 --------- = 318.18MPa K

1.1

( KC ) ( 8 ) ( 1200 ) = 318.18 ( 10 6 ) From the given expression of shear stress: τ = -------------------------------------πd 2

or KC = 0.1041d 2 ( 10 6 ), where d is in meters or KC = 0.1041d 2 , where d is in millimeters let a = 0.1041d 2

1

4C – 1 Substituting for K: ⎛⎝ ----------------⎞⎠ C + 0.615 = a or C 2 + ( 0.365 – a )C – ( 0.615 – a ) = 0 or 4C – 4 ( a – 0.365 ) + ( a – 0.365 ) 2 + 4 ( 0.615 – a ) C = -------------------------------------------------------------------------------------------------------2 Let C 1 and C 2 be the two roots. Then D 1 = C 1 d and D 2 = C 2 d . The values of D 1 & D 2 are tabulated

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below. d (mm)

a

C1

C2

D1 (mm)

D2 (mm)

40.9

9.5

8

6.662

5.115

1.182

10

10.410

8.951

1.094

89.5

10.9

12

14.990

13.566

1.060

162.8

12.7

14

20.404

18.997

1.042

266.0

14.6

16

26.650

25.254

1.031

404.1

16.5

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.61 Two cast-iron pipes are held together by a steel bolt, as shown in Figure P3.61. The outer diameters of the two pipes are 2 in and 2 3 ⁄ 4 in and the wall thickness of each pipe is 1/4 in. The diameter of the bolt is 1/2 in. The yield strength of cast iron is 25 ksi in tension and steel is 15 ksi in shear. What is the maximum force P to the nearest pound this assembly can transmit for a factor of safety of 1.2. P

P

Figure P3.61

Solution

-----------------------------------------------------------The cross-sectional area of the inner pipe and bolt are: 2 2 π 2 A iron = --- ( 2 – 1.5 ) = 1.3744 in 4

2 π 1 2 A bolt = --- ⎛ ---⎞ = 1.9635 in ⎝ ⎠ 4 2

2

The free body diagrams are: Vbolt Niron

P

P

Vbolt By equilibrium we have: N iron = P

V bolt = P --2

The normal stress in inner pipe is: σ yield N iron P σ iron = ------------ = ---------------- ≤ -------------1.3744 1.2 A iron

or

25 P ≤ ------- 1.3744 1.2

or

P ≤ 28.63 kips

3

or

15 P ≤ ------- 1.9635 1.2

or

P ≤ 4.908 kips

4

The shear stress in bolt is: τ yield N bolt P⁄2 τ bolt = ------------ = ---------------- ≤ ------------1.9635 1.2 A bolt

The answer is:

P max = 4.9 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.62 A coupling of diameter 250-mm is assembled using 6 bolts of diameter 12.5 mm as shown in Figure P3.62. The holes for the bolts are drilled with center on a circle of diameter 200 mm. A factor of safety of 1.5 for the assembly is desired. If the shear strength of the bolts is 300 MPa, determine the maximum

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torque that can be transfered by the coupling.

T

T

Figure P3.62

Solution

------------------------------------------------------------

2 –3 2 π The crosssectional area of the bolt is: A bolt = --- ( 0.0125 ) = 0.1227 ( 10 ) m .

4

The free body diagram is V

T R

By equilibrium we obtain: T = 6 ( VR ) = 0.6V

or

TV = -----0.6

The shear stress in the bolt is: 3 6 300 V - = T τ = -------------------------------------------------------- = 13.58T ( 10 ) ≤ --------- ( 10 ) –3 1.5 A bolt 0.6 [ 0.1227 ( 10 ) ]

or

The answer is:

3

T ≤ 14.727 ( 10 ) N-m T max = 14.72 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.63 A circular rod of diameter 15 mm is acted upon by a distributed force p(x) that has the units of kN / m as shown. The Modulus of Elasticity of the rod is 70 GPa. Determine the distributed force p(x) if the displacement u(x) in the x-direction is as given below. x is measured in meters. 2

–6

u ( x ) = 30 ( x – x ) ( 10 )m

p(x)

x Figure P3.63

Solution

d = 15 mm

E = 70 GPa

p(x) = ?

-----------------------------------------------------------–6 du 1.Strain Calculation: ε xx = ------ = 30 ( 1 – 2x ) ( 10 ) dx 9

–6

6

N

2.Stress Calculation: σ xx = Eε xx = ( 70 ) ( 10 ) ( 30 ) ( 1 – 2x ) ( 10 ) = 2.1 ( 1 – 2x ) ( 10 ) ------2m

3.Internal Force Calculation: 6 2 π N = σ xx A = 7.0 ( 10 ) ( 1 – 2x ) ⎛ ---⎞ ( 0.015 ) = 371.1 ( 1 – 2x )Newtons ⎝ 4⎠

4. External Force Calculations: A free body diagram of a differential element can be drawn as shown

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

below. N

N+dN

p(x) dx

dx

By balancing forces in the x -direction we obtain: ( N + dN ) + p ( x )dx – N = 0 or dN p ( x ) = – ⎛ ⎞ = – 371.1 ( – 2 ) or ⎝dx ⎠

p ( x ) = 742.2 kN/m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.64 A circular rod of diameter 15 mm is acted upon by a distributed force p(x) that has the units of kN / m as shown. The Modulus of Elasticity of the rod is 70 GPa. Determine the distributed force p(x) if the displacement u(x) in the x-direction is as given below. x is measured in 2

3

–6

meters. u ( x ) = 50 ( x – 2x ) ( 10 )m

p(x)

x

Figure P3.64

Solution

d =15mm

E = 70G Pa

p(x) = ?

-----------------------------------------------------------2 –6 1.Strain Calculation: ε xx = du ------ = 50 ( 2x – 6x ) ( 10 )

dx

9

2

–6

6

2

2.Stress Calculation: σ xx = Eε xx = ( 70 ) ( 10 ) ( 50 ) ( 2x – 6x ) ( 10 ) = 7.0 ( 10 ) ( x – 3x )N ⁄ m

2

6 2 π 2 2 3.Internal Force Calculation: N = σ xx A = 7.0 ( 10 ) ( x – 3x ) ⎛⎝ ---⎞⎠ ( 0.015 ) = 1237 ( x – 3x )Newtons 4

4.External Force Calculation: A free body diagram of a differential element can be drawn as shown below N

p(x) dx

N+dN

dx

By balancing forces in the x -direction we obtain: ( N + dN ) + p ( x )dx – N = 0 or dN p ( x ) = – ⎛ ⎞ = – 1237 ( 1 – 6x )Newtons ⁄ m ⎝dx ⎠

p ( x ) = – 1.237 ( 1 – 6x ) kN/m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.65

Due to the action of the forces, the displacement in the x-direction was found to be 2

( 60x + 80xy – x y ) u = ------------------------------------------------ in . The modulus of elasticity of the beam is 30,000 ksi. Determine the statically 180000

equivalent internal normal force N and the internal bending moment Mz acting at point O at a section at x = 20 inch.The cross-section of the beam is shown on the right. Assume some unknown shear stress is present at the cross-section.

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y

January 2014

P2 lbs. P1 lbs.

x

y z

20 in

20 in

3 in O

Cross-section z

2 in

Figure P3.65

Solution

E=30000ksi

N=? and Mz =? @ x=20

-----------------------------------------------------------du 60 + 80y – 2xy ) 1.Strain Calculation: ε xx = ------ = (----------------------------------------dx 180000 3 ( 60 + 80y – 2xy ) ( 30 + 40y – xy ) 2.Stress Calculation: σ xx = Eε xx = 30 ( 10 ) ------------------------------------------ = -------------------------------------- ksi 180000 3 30 + 20y 30 + 40y – 20y ) = -------------------- ksi at x=20, σ xx = (----------------------------------------3 3

3. Internal forces and moments: The stress distribution can be replaced by an equivalent force at the centroid of the distribution and then replaced by an equivalent force and moment at O. 2 in

y

y

20 ksi

N O 0.5 in z

3 in

Mz

N O

z

1 N = --- ( 20 ) ( 3 ) ( 2 ) = 60kips or 2

N = 60 kips

M z = N ( 0.5 ) = 30in – kips or

M z = 30 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.66

Assume that the stress-strain curve after yield stress in Problem 3.12 is described by the quadratic 2 equation σ = a + bε + cε . (a) Determine the coefficients a, b, and c by Least Square Method. (b) Find the tangent modulus of elasticity at a stress level of 1400 MPa. Solution a = ? b = ? c = ? Uo = ? @ε = 0.18 ET = ? @σ = 1400 MPa ------------------------------------------------------------

Using the results of stresses and strains in problem 3.12 after the proportional limit and the Least Square Method, we obtain the value of the values of constants a, b, and c on a spread sheet as shown in the table

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

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below. Strain xi

Stress (MPa) fi

xi2

xi3

xi4

xi*fi

xi2*fi 0.13

1

0.0106

1119.31

0.0001

0.0000

0.0000

11.86

2

0.0220

1168.54

0.0005

0.0000

0.0000

25.71

0.57

3

0.0392

1222.43

0.0015

0.0001

0.0000

47.92

1.88

4

0.0558

1269.44

0.0031

0.0002

0.0000

70.83

3.95

5

0.0800

1324.65

0.0064

0.0005

0.0000

105.97

8.48

6

0.0942

1342.29

0.0089

0.0008

0.0001

126.44

11.91

7

0.1160

1388.96

0.0135

0.0016

0.0002

161.12

18.69

8

0.1430

1421.18

0.0204

0.0029

0.0004

203.23

29.06

9

0.1776

1458.26

0.0315

0.0056

0.0010

258.99

46.00

10

0.1998

1472.64

0.0399

0.0080

0.0016

294.23

58.79

11

0.2202

1473.40

0.0485

0.0107

0.0024

324.44

71.44

12

0.2326

1448.89

0.0541

0.0126

0.0029

337.01

78.39

13

0.2406

1423.54

0.0579

0.0139

0.0034

342.50

82.41

14

0.2462

1384.31

0.0606

0.0149

0.0037

340.82

83.91

15

0.2494

1334.38

0.0622

0.0155

0.0039

332.79

83.00

16

0.2526

1287.92

0.0638

0.0161

0.0041

325.33

82.18

bij & ri

2.3798

21540.14

0.4730

0.1034

0.0236

3309.21

660.77

Cij

0.0005

0.0072

0.0224

0.1533

0.5288

1.90

D

0.000770195

ai

1062.06

4493.16

-12993.12

The values of the constants are: a = 1062.1MPa b = 4493.3MPa c = – 12993.1MPa . At σ = 1400 MPa we can write the following quadratic equation: 2 1062.1 + 4493.3 ε – 12993.1 ε = 1400 1 The root value for which Equation (1) is satisfied is: ε1400=0.11052. The tangent modulus is the slope of the stress-strain curve in the non-linear region and can be written as ET =

dσ = b + 2cε . Substituting ε1400=0.11052 and the values of b and c, we obtain the Tangent Modulus dε

as given below. E T = 4493.3 + 2 ( – 12993.1 ) ( 0.11052 ) = 1621.3MPa

E T = 1.621 GPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.67

Assume that the stress-strain curve after yield stress in Problem 3.13 is described by the quadratic 2 equation σ = a + bε + cε . (a) Determine the coefficients a, b, and c by Least Square Method. (b) Find the Tangent Modulus of Elasticity at a stress level of 50 ksi. Solution a=? b=? c=? Uo = ? @ ε = 0.15 ET = ? @ σ = 50 ksi ------------------------------------------------------------

Using the results of stresses and strains in problem 3.13 after the proportional limit and the Least Square Method described in Appendix B, we obtain the value of the values of constants a, b, and c on a spread

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sheet as shown in the table below. Strain xi

Stress (ksi) fi

xi2

xi3

xi4

xi*fi

xi2*fi

1

0.00438

39.22

0.000

0.000

0.000

0.172

0.001

2

0.00953

40.23

0.000

0.000

0.000

0.383

0.004

3

0.01435

41.56

0.000

0.000

0.000

0.596

0.009

4

0.01887

43.09

0.000

0.000

0.000

0.813

0.015

5

0.02359

44.92

0.001

0.000

0.000

1.060

0.025

6

0.02953

47.47

0.001

0.000

0.000

1.402

0.041

7

0.03543

50.22

0.001

0.000

0.000

1.779

0.063

8

0.04212

52.97

0.002

0.000

0.000

2.231

0.094

9

0.04893

55.11

0.002

0.000

0.000

2.697

0.132

10

0.05605

56.94

0.003

0.000

0.000

3.191

0.179

11

0.0702

59.69

0.005

0.000

0.000

4.190

0.294

12

0.08061

61.06

0.006

0.001

0.000

4.922

0.397

13

0.09633

62.49

0.009

0.001

0.000

6.020

0.580

14

0.10711

63.2

0.011

0.001

0.000

6.769

0.725

15

0.12297

63.92

0.015

0.002

0.000

7.860

0.967

16

0.14174

64.68

0.020

0.003

0.000

9.168

1.299

17

0.15818

65.04

0.025

0.004

0.001

10.288

1.627

18

0.18155

65.39

0.033

0.006

0.001

11.872

2.155

19

0.19267

61.32

0.037

0.007

0.001

11.815

2.276

20

0.19802

58.26

0.039

0.008

0.002

11.537

2.284

21

0.20321

54.55

0.041

0.008

0.002

11.085

2.253

22

0.20736

50.73

0.043

0.009

0.002

10.519

2.181

bij & ri

2.04273

1202.06

0.29666296

0.050327768

0.009120184

120.3690455

17.60174158

Cij

0.000

0.004

0.015

0.113

0.501

2.354

D

0.001

ai

35.995

455.956

-1756.978

The values of the constants are: a = 36.0ksi b = 456.0ksi c = – 1757.0ksi . At σ = 1400 MPa we can write the following quadratic equation: 2 36.0 + 456.0 ε – 1757.0 ε = 50 1 The root value for which Equation (1) is satisfied is: ε50=00.03558 The tangent modulus is the slope of the stress-strain curve in the non-linear region and can be written as E T = dσ = b + 2cε . Substituting ε50=00.03558 and the values of b and c, we obtain the tangent modulus dε

as given below. E T = 455.96 + 2 ( – 1756.96 ) ( 0.03558 ) = 330.94ksi

E T = 330.9ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.68 Marks were made on the cord used for tying the canoe on top of the car in Example 3.5. These marks were made every 2 in. to produce a total of 20 segments. The stretch cord is symmetric with respect to the top of the canoe. The starting point of the first segment is on the carrier rail of the car and the end point of the tenth segment is on the top of the canoe. The measured length of each segment is as shown in Table 3.64. Determine (a) the tension in the cord of each segment; (b) the force exerted by the cord on the carrier of the car. Use the modulus of elasticity E = 510 psi and the diameter of the stretch cord as 0.5 in:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

Table 3.64.

Solution

Lo= 2 in

do= 1/2 in

E = 510 psi

Segment Number

Deformed Length inches

1

3.4

2

3.4

3

3.4

4

3.4

5

3.4

6

3.4

7

3.1

8

2.7

9

2.3

10

2.2

Ti = ?

P=?

------------------------------------------------------------

The strain εi in the ith segment of column three in table below is given by: ε i = ( ΔL i ) ⁄ L 0 , where ΔLi is the deformed length in column two and Lo= 2 in. Segment Number

Deformed Length. ΔLi (inches)

Strain

(psi)

Internal Tension Ti (lbs)

1

3.4

0.70

357

70.10

2

3.4

0.70

357

70.10

3

3.4

0.70

357

70.10

4

3.4

0.70

357

70.10

5

3.4

0.70

357

70.10

6

3.4

0.70

357

70.10

7

3.1

0.55

280.5

55.08

8

2.7

0.35

178.5

35.05

9

2.3

0.15

76.5

15.02

10

2.2

0.10

51

10.01

εi

Stress

σi

The stress σi in the ith segment in column four is given by σ i = Eε i . Multiplying the column of strain by E = 510 psi, we obtain the column of stress. The area of each segment Ai is 2

2

A i = ( π0.5 ) ⁄ 4 = 0.19635 in . The internal tension Ti in the ith segment is given by: T i = σ i A i . Multiply-

ing the column of stress by the area Ai we obtain the column of internal tension. The force P exerted on the car carrier by the cord is given by the tension in segment 1. P = 70.1 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.69 Marks were made on the cord used for tying the canoe on top of the car in Example 3.5. These marks were made every 2 in. to produce a total of 20 segments. The stretch cord is symmetric with respect to the top of the canoe. The starting point of the first segment is on the carrier rail of the car and the end point of the tenth segment is on the top of the canoe. The measured length of each segment is as shown in Table 3.64. Determine (a) the tension in the cord of each segment; (b) the force exerted by the cord on the carrier of the car. Use the diameter of the stretch cord as (1/2) inch and the following equation for stressstrain curve:

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⎧ ( 1020ε – 1020ε 2 ) psi σ = ⎨ ⎩ 255 psi Solution

Lo= 2 in

do= 1/2 in

Ti = ?

January 2014

ε < 0.5 ε ≥ 0.5 P=?

------------------------------------------------------------

The strain calculation is shown in the table below. 2 The stress σi in the for segments 8,9, and 10 is found using the equation σ i = 1020 ε i – 1020 ε i , as the strains in these segments is less than 0.5. For segments 1 through 7, the stress is 255 psi as the strain is greater than 0.5.Multiplying the column of strain by E = 510 psi, we obtain the column of stress. 2

2

The area of each segment Ai is A i = ( π0.5 ) ⁄ 4 = 0.19635 in . The internal tension Ti in the ith segment is given by: T i = σ i A i . Multiplying the column of stress by the area Ai we obtain the column of internal tension. Deformed Length. ΔLi (inches)

Strain

(psi)

Internal Tension Ti (lbs)

1

3.4

2

3.4

0.70

255.00

50.07

0.70

255.00

50.07

3 4

3.4

0.70

255.00

50.07

3.4

0.70

255.00

50.07

5

3.4

0.70

255.00

50.07

6

3.4

0.70

255.00

50.07

7

3.1

0.55

255.00

50.07

8

2.7

0.35

232.05

45.56

9

2.3

0.15

130.05

25.54

10

2.2

0.10

91.80

18.02

Segment Number

εi

Stress

σi

The force P exerted on the car carrier by the cord is given by the tension in segment 1. P = 50.1 lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.70 Marks were made on the cord used for tying the canoe on top of the car in Example 3.5. These marks were made every 2 in. to produce a total of 20 segments. The stretch cord is symmetric with respect to the top of the canoe. The starting point of the first segment is on the carrier rail of the car and the end point of the tenth segment is on the top of the canoe. The measured length of each segment is as shown in Table 3.64. Determine (a) the tension in the cord of each segment; (b) the force exerted by the cord on the carrier of the car.Use Poisson’s ratio of ν=1/2 and initial diameter of 1/2 inch and calculate the diameter in deformed position for each segment. Use stress-strain relationship given in problem 3.69. Solution Lo= 2 in do= 1/2 in ν=1/2 Ti = ? P=? ------------------------------------------------------------

The longitudinal strain εi and the corresponding stress si are calculated as in problem 3.69 and are shown in the table below. The transverse strain is -νεi. The change in diameter is thus -νεi do. The deformed diameter di of the ith segment is d i = d 0 – νε i d 0 . Substituting the values of ν=1/2 and do= 1/2 in. the entries in column five can be found using d i = ( 1 – 0.5ε i ) ⁄ 2 . 2

The area of each segment Ai is A i = πd i ⁄ 4 . The internal tension Ti in the ith segment is given

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

by: T i = σ i A i . Multiplying the column of stress by the area Ai we obtain the column of internal tension. Deformed Length. ΔLi (inches)

Strain

(psi)

Deformed Diameter di (inches)

Internal Tension Ti (lbs)

1

3.4

2

3.4

0.70

255.00

0.33

21.15

0.70

255.00

0.33

21.15

3 4

3.4

0.70

255.00

0.33

21.15

3.4

0.70

255.00

0.33

21.15

5

3.4

0.70

255.00

0.33

21.15

6

3.4

0.70

255.00

0.33

21.15

7

3.1

0.55

255.00

0.36

26.32

8

2.7

0.35

232.05

0.41

31.01

9

2.3

0.15

130.05

0.46

21.85

10

2.2

0.10

91.80

0.48

16.27

Segment Number

εi

Stress

σi

The force P exerted on the car carrier by the cord is given by the tension in segment 1. P = 21.2 lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.71 Write the Generalized Hooke’s Law for isotropic material in cylindrical coordinates (r, θ, z). Solution ------------------------------------------------------------

The Generalized Hooke’s Law for isotropic material in cylindrical coordinates (r, θ, z) can be written as shown below. ε rr = [ σ rr – ν ( σ θθ + σ zz ) ] ⁄ E ε θθ = [ σ θθ – ν ( σ zz + σ rr ) ] ⁄ E

γ rθ = τ rθ ⁄ G γ θz = τ θz ⁄ G

ε zz = [ σ zz – ν ( σ rr + σ θθ ) ] ⁄ E

γ zr = τ zr ⁄ G

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.72 Write the Generalized Hooke’s Law for isotropic material in spherical coordinates (r, θ, φ). Solution ------------------------------------------------------------

The Generalized Hooke’s Law for isotropic material in cylindrical coordinates (r, θ, φ) can be written as shown below. ε rr = [ σ rr – ν ( σ θθ + σ φφ ) ] ⁄ E ε θθ = [ σ θθ – ν ( σ φφ + σ rr ) ] ⁄ E

γ rθ = τ rθ ⁄ G γ θφ = τ θφ ⁄ G

ε φφ = [ σ φφ – ν ( σ rr + σ θθ ) ] ⁄ E

γ φr = τ φr ⁄ G

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.73 The stresses and two material constants are given, calculate εxx, εyy, γxy, εzz, and σzz (a) assuming plane stress, and (b) assuming plane strain. σ xx = 100 MPa ( T )

Solution

εxx=?

σ yy = 150 MPa ( T )

εyy=?

γxy=?

εzz=?

τ xy = – 125 MPa

E = 200 GPa

σzz=? (a) plane stress

ν = 0.32

(b) plane strain

-----------------------------------------------------------E 200 G = -------------------- = ------------------ = 75.76GPa 2(1 + ν) 2 ( 1.32 )

(a) Assuming Plane Stress

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

6 6 σ xx – ν ( σ yy + σ zz ) –3 [ 100 – 0.32 ( 150 ) ] ( 10 ) 52 ( 10 ) ε xx = ---------------------------------------------- = ---------------------------------------------------------- = ---------------------- = 0.26 ( 10 ) 9 9 E 200 ( 10 ) 200 ( 10 )

ε xx = 260μ

6 6 σ yy – ν ( σ xx ) –3 118 ( 10 ) [ 150 – 0.32 ( 100 ) ] ( 10 ) ε yy = ------------------------------- = ---------------------------------------------------------- = ---------------------- = 0.59 ( 10 ) 9 9 E 200 ( 10 ) 200 ( 10 )

ε yy = 590μ

6 τ xy –3 – 125 ( 10 ) γ xy = ------- = --------------------------- = – 1.65 ( 10 ) 9 G 75.76 ( 10 )

γ xy = – 1650 μ

6 6 σ zz – ν ( σ xx + σ yy ) –3 0.32 ( 250 ) ( 10 ) 80 ( 10 ) ε zz = ---------------------------------------------- = – -------------------------------------- = ---------------------- = – 0.4 ( 10 ) 9 9 E 200 ( 10 ) 200 ( 10 ) ------------------------------------------------------------

(b) Assuming Plane Strain

ε zz = – 400 μ ε zz = 0

σ zz – ν ( σ xx + σ yy ) 0 = ---------------------------------------------E 6

6

∴σ zz = ν ( σ xx + σ yy ) = 0.32 ( 250 ) ( 10 ) = 80 ( 10 )

σ zz = 80 MPa (T)

6 σ xx – ν ( σ yy + σ zz ) –3 [ 100 – 0.32 ( 230 ) ] ( 10 ) ε xx = ---------------------------------------------- = ---------------------------------------------------------- = 0.132 ( 10 ) 9 E 200 ( 10 )

ε xx = 132μ

6 σ yy – ν ( σ xx + σ zz ) –3 [ 150 – 0.32 ( 180 ) ] ( 10 ) ε yy = ---------------------------------------------- = ---------------------------------------------------------- = 0.462 ( 10 ) 9 E 200 ( 10 )

ε yy = 462μ γ xy = – 1650 μ

Same as in part (a)

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

3.74 The stresses and two material constants are given, calculate εxx, εyy, γxy, εzz, and σzz (a) assuming plane stress, and (b) assuming plane strain. σ xx = 225 MPa ( C )

Solution

εxx=?

σ yy = 125 MPa ( T )

εyy=?

γxy=?

εzz=?

τ xy = 150 MPa

σzz=?

E = 70 GPa

(a) plane stress

G = 28 GPa

(b) plane strain

-----------------------------------------------------------E G = -------------------2(1 + ν)

9

9 70 ( 10 ) ∴28 ( 10 ) = -------------------2(1 + ν)

ν = 0.25

(a) Assuming Plane Stress

σ zz = 0

6 σ xx – ν ( σ yy + σ zz ) –3 [ – 225 – 0.25 ( 125 ) ] ( 10 ) ε xx = ---------------------------------------------- = -------------------------------------------------------------- = – 3.661 ( 10 ) 9 E 70 ( 10 )

ε xx = – 3661 μ

6 σ yy – ν ( σ xx ) –3 [ 125 – 0.25 ( – 225 ) ] ( 10 ) ε yy = ------------------------------- = ------------------------------------------------------------- = 2.589 ( 10 ) 9 E 70 ( 10 )

ε yy = 2589μ

6 τ xy –3 150 ( 10 ) - = ---------------------- = 5.357 ( 10 ) γ xy = ------9 G 28 ( 10 )

γ xy = 5357μrad

6 σ zz – ν ( σ xx + σ yy ) –3 0.25 ( – 225 + 125 ) ( 10 ) ε zz = ---------------------------------------------- = – ---------------------------------------------------------- = 0.357 ( 10 ) 9 E 70 ( 10 )

ε zz = 357μ

------------------------------------------------------------

(b) Assuming Plane Strain

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

σ zz – ν ( σ xx + σ yy ) 0 = ---------------------------------------------E 6

6

∴σ zz = ν ( σ xx + σ yy ) = 0.25 ( – 225 + 125 ) ( 10 ) = – 25 ( 10 )

σ zz = 25MPa ( C )

6 σ xx – ν ( σ yy + σ zz ) –3 [ – 225 – 0.25 ( 125 – 25 ) ] ( 10 ) ε xx = ---------------------------------------------- = -------------------------------------------------------------------------- = – 3.571 ( 10 ) 9 E 70 ( 10 )

ε xx = – 3571 μ

6 σ yy – ν ( σ xx + σ zz ) –3 [ 125 – 0.25 ( – 225 – 25 ) ] ( 10 ) ε yy = ---------------------------------------------- = -------------------------------------------------------------------------- = 2.679 ( 10 ) 9 E 70 ( 10 )

ε yy = 2679μ

Same as in part (a)

γ xy = 5357μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.75 The stresses and two material constants are given, calculate εxx, εyy, γxy, εzz, and σzz (a) assuming plane stress, and (b) assuming plane strain. σ xx = 22 ksi ( C )

Solution

σ yy = 25 ksi ( C )

εxx=?

εyy=?

γxy=?

τ xy = – 15 ksi

εzz=?

σzz=?

E = 30, 000 ksi

ν = 0.3

(a) plane stress

(b) plane strain

-----------------------------------------------------------6

6 30 ( 10 ) E G = -------------------- = ------------------------- = 11.54 ( 10 )psi 2(1 + ν) 2 ( 1 + 0.3 )

(a) Assuming Plane Stress

σ zz = 0

3 σ xx – ν ( σ yy + σ zz ) –3 [ – 22 – 0.3 ( – 25 ) ] ( 10 ) ε xx = ---------------------------------------------- = --------------------------------------------------------- = – 0.483 ( 10 ) 6 E 30 ( 10 )

ε xx = – 483 μ

3 σ yy – ν ( σ xx ) –3 [ – 25 – 0.3 ( – 22 ) ] ( 10 ) ε yy = ------------------------------- = --------------------------------------------------------- = – 0.613 ( 10 ) 6 E 30 ( 10 )

ε yy = – 613 μ

3 τ xy –3 – 15 ( 10 ) γ xy = ------- = --------------------------- = – 1.3 ( 10 ) 6 G 11.54 ( 10 )

γ xy = – 1300μ

3 σ zz – ν ( σ xx + σ yy ) –3 0.3 ( – 22 – 25 ) ( 10 ) ε zz = ---------------------------------------------- = – ------------------------------------------------- = 0.47 ( 10 ) 6 E 30 ( 10 )

ε zz = 470μ

-----------------------------------------------------------ε zz = 0

(b) Assuming Plane Strain σ zz – ν ( σ xx + σ yy ) 0 = ---------------------------------------------E 3

3

∴σ zz = ν ( σ xx + σ yy ) = 0.3 ( – 22 – 25 ) ( 10 ) = ( – 14.1 ) ( 10 )

σ zz = 14.1 ksi (C)

3 σ xx – ν ( σ yy + σ zz ) –3 [ – 22 – 0.3 ( – 25 – 14.1 ) ] ( 10 ) ε xx = ---------------------------------------------- = -------------------------------------------------------------------------- = – 0.342 ( 10 ) 6 E 30 ( 10 )

ε xx = – 342 μ

3 σ yy – ν ( σ xx + σ zz ) –3 [ – 25 – 0.3 ( – 22 – 14.1 ) ] ( 10 ) ε yy = ---------------------------------------------- = -------------------------------------------------------------------------- = – 0.472 ( 10 ) 6 E 30 ( 10 )

ε yy = – 472 μ

Same as in part (a)

γ xy = – 1300μ

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

3.76 The stresses and two material constants are given, calculate εxx, εyy, γxy, εzz, and σzz (a) assuming plane stress, and (b) assuming plane strain. σ xx = 15 ksi ( T )

Solution

σ yy = 12 ksi ( C )

εxx=?

εyy=?

γxy=?

τ xy = – 10 ksi

εzz=?

σzz=?

E = 10, 000 ksi

G = 3900 ksi

(a) plane stress

(b) plane strain

-----------------------------------------------------------E G = -------------------2(1 + ν)

3

3 10000 ( 10 ) ∴3900 ( 10 ) = ---------------------------2(1 + ν)

ν = 0.282

(a) Assuming Plane Stress

σ zz = 0

3 σ xx – ν ( σ yy + σ zz ) –3 [ 15 – 0.282 ( – 12 ) ] ( 10 ) ε xx = ---------------------------------------------- = -------------------------------------------------------- = 1.838 ( 10 ) 6 E 10 ( 10 )

ε xx = 1838μ

3 σ yy – ν ( σ xx ) –3 [ – 12 – 0.282 ( – 15 ) ] ( 10 ) ε yy = ------------------------------- = --------------------------------------------------------------- = – 1.623 ( 10 ) 6 E 10 ( 10 )

ε yy = – 1623 μ

3 τ xy –3 – 10 ( 10 ) γ xy = ------- = ----------------------- = – 2.564 ( 10 ) 6 G 3.9 ( 10 )

γ xy = – 2564μ

3 σ zz – ν ( σ xx + σ yy ) –3 0.282 ( – 15 – 12 ) ( 10 ) ε zz = ---------------------------------------------- = – ------------------------------------------------------- = – 0.085 ( 10 ) 6 E 30 ( 10 )

ε zz = – 85μ

-----------------------------------------------------------ε zz = 0

(b) Assuming Plane Strain σ zz – ν ( σ xx + σ yy ) 0 = ---------------------------------------------E 3

∴σ zz = ν ( σ xx + σ yy ) = 0.282 ( 15 – 12 ) ( 10 ) = 846

σ zz = 846 psi (T)

3 σ xx – ν ( σ yy + σ zz ) –3 [ 15 – 0.282 ( – 12 + 0.846 ) ] ( 10 ) ε xx = ---------------------------------------------- = -------------------------------------------------------------------------------- = 1.815 ( 10 ) 6 E 10 ( 10 )

ε xx = 1815μ

3 σ yy – ν ( σ xx + σ zz ) –3 [ – 12 – 0.282 ( 15 + 0.846 ) ] ( 10 ) ε yy = ---------------------------------------------- = ------------------------------------------------------------------------------- = – 1.647 ( 10 ) 6 E 10 ( 10 )

ε yy = – 1647 μ

Same as in part (a)

γ xy = – 2564μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.77 The stresses and two material constants are given, calculate εxx, εyy, γxy, εzz, and σzz (a) assuming plane stress, and (b) assuming plane strain. σ xx = 300 MPa ( C )

Solution

εxx=?

σ yy = 300 MPa ( T )

εyy=?

γxy=?

εzz=?

τ xy = 150 MPa

σzz=?

G = 15 GPa

(a) plane stress

ν = 0.2

(b) plane strain

-----------------------------------------------------------E G = -------------------2(1 + ν)

9 9 E ∴15 ( 10 ) = ------------------------- E = 36 ( 10 ) 2 ( 1 + 0.2 )

i.e., E = 36GPa

(a) Assuming Plane Stress

σ zz = 0

6 σ xx – ν ( σ yy + σ zz ) –3 [ – 300 – 0.2 ( 300 ) ] ( 10 ) ε xx = ---------------------------------------------- = ----------------------------------------------------------- = – 10 ( 10 ) 9 E 36 ( 10 )

ε xx = – 10 000μ

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

6 σ yy – ν ( σ xx + σ zz ) –3 [ 300 – 0.2 ( – 300 ) ] ( 10 ) ε yy = ---------------------------------------------- = ---------------------------------------------------------- = 10 ( 10 ) 9 E 36 ( 10 )

ε yy = 10000μ

6 τ xy –3 150 ( 10 ) - = ---------------------- = 10 ( 10 ) γ xy = ------9 G 15 ( 10 )

γ xy = 10000μrad

6 σ zz – ν ( σ xx + σ yy ) 0.2 ( 300 – 300 ) ( 10 ) ε zz = ---------------------------------------------- = – -------------------------------------------------- = 0 9 E 36 ( 10 )

ε zz = 0

-----------------------------------------------------------ε zz = 0

(b) Assuming Plane Strain σ zz – ν ( σ xx + σ yy ) 0 = ---------------------------------------------E

σ zz = 0

6

∴σ zz = ν ( σ xx + σ yy ) = 0.2 ( 300 – 300 ) ( 10 ) = 0

As σzz is zero, the results are the same as in plane stress.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.78 The stresses and two material constants are given, calculate εxx, εyy, γxy, εzz, and σzz (a) assuming plane stress, and (b) assuming plane strain. σ xx = 100

psi ( C )

Solution

εxx=?

σ yy = 150 psi ( T )

εyy=?

γxy=?

εzz=?

τ xy = 100

psi

σzz=?

E = 2000 psi

G = 800 psi

(a) plane stress

(b) plane strain

-----------------------------------------------------------E G = ------------------2(1 + ν)

2000 ∴800 = ------------------2(1 + ν)

ν = 0.25

(a) Assuming Plane Stress

σ zz = 0

σ xx – ν ( σ yy + σ zz ) ( – 100 ) – 0.25 ( 150 ) ε xx = ---------------------------------------------- = --------------------------------------------- = – 0.06875 2000 E

ε xx = – 0.06875

σ yy – ν ( σ xx + σ zz ) – 0.25 ( – 100 )- = 0.0875 ε yy = ---------------------------------------------- = 150 --------------------------------------E 2000 τ xy 100 - = --------- = 0.125 γ xy = ------G 800

ε yy = 0.0875 γ xy = 0.125

σ zz – ν ( σ xx + σ yy ) ( – 100 + 150 )- = – 0.00625 ε zz = ---------------------------------------------- = – 0.25 ------------------------------------------E 2000 ------------------------------------------------------------

ε zz = – 0.00625

(b) Assuming Plane Strain

ε zz = 0

σ zz – ν ( σ xx + σ yy ) 0 = ---------------------------------------------E ∴σ zz = ν ( σ xx + σ yy ) = 0.25 ( – 100 + 150 ) = 12.5

σ zz = 12.50psi ( T )

σ xx – ν ( σ yy + σ zz ) – 100 – 0.25 ( 150 + 12.5 )- = – 0.0703 ε xx = ---------------------------------------------- = -----------------------------------------------------------E 2000 σ yy – ν ( σ xx + σ zz ) – 0.25 ( – 100 + 12.5 )- = 0.0430 ε yy = ---------------------------------------------- = 150 --------------------------------------------------------E 2000

Same as in part (a)

ε xx = – 0.0703 ε yy = 0.08594 γ xy = 0.125

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

3.79 The strains and two material constants are given. Calculate σxx, σyy, τxy, σzz, and εzz assuming the point is in plane stress. ε xx = 500 μ

Solution

ε yy = 400 μ

σxx=? σyy=? τxy=?

γ xy = – 300 μ

σzz=? εzz=?

E = 200 GPa

ν = 0.32

plane stress

-----------------------------------------------------------E G = -------------------2(1 + ν)

9

9 15 ( 10 ) G = ------------------- = 75.76 ( 10 ) 2 ( 1.32 )

σ zz = 0

Assuming Plane Stress

σ xx – ν ( σ yy + σ zz ) 9 –6 3 N ε xx = ---------------------------------------------- or σ xx – ν ( σ yy + σ zz ) = Eε xx = 200 ( 10 )500 ( 10 ) = 100000 ( 10 ) ------2 E m ∴σ xx – ν ( σ yy + σ zz ) = 100MPa

1

Similarly σ yy – ν ( σ xx + σ zz ) 9 –6 3 N ε yy = ---------------------------------------------- or σ yy – ν ( σ xx + σ zz ) = Eε yy = 200 ( 10 )400 ( 10 ) = 80000 ( 10 ) ------2 E m σ yy – ν ( σ xx + σ zz ) = 80MPa

2 2

Multiplying Eq.(2) by ν and adding to Eq.(1), we obtain: σ xx ( 1 – ν ) = 100 + ν80 100 + 0.32 ( 80 ) 125.6 ∴σ xx = ------------------------------------- = ---------------- = 139.929 2 0.8976 1 – 0.32

3 σ xx = 139.93MPa ( T )

Substituting Eq.(3) in Eq.(2) σ yy = 80 + ν ( 139.93 ) = 124.78

σ yy = 124.78MPa ( T )

σ zz – ν ( σ xx + σ yy ) 6 –3 0.32 ε zz = ---------------------------------------------- = – ---------------------- ( 139.9 + 124.8 ) ( 10 ) = – 0.423 ( 10 ) 9 E 200 ( 10 )

ε zz = – 423μ

9

–6

3

τ xy = Gγ xy = 75.76 ( 10 ) ( – 300 ) ( 10 ) = 22727 ( 10 )

τ xy = – 22.7MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.80 The strains and two material constants are given. Calculate σxx, σyy, τxy, σzz, and εzz assuming the point is in plane stress. ε xx = 2000 μ

Solution

ε yy = – 1000 μ

σxx=? σyy=? τxy=?

γ xy = 1500 μ

σzz=? εzz=?

E = 70 GPa

G = 28 GPa

plane stress

-----------------------------------------------------------E G = ------------------2(1 + ν)

9

9 70 ( 10 ) 28 ( 10 ) = -------------------2(1 + ν)

ν = 0.25 σ zz = 0

Assuming Plane Stress

σ xx – ν ( σ yy + σ zz ) 9 –6 6 N ε xx = ---------------------------------------------- or σ xx – νσ yy = Eε xx = 70 ( 10 ) ( 2000 ) ( 10 ) = 140 ( 10 ) ------- or 2 E m σ xx – νσ yy = 140MPa

1

Similarly

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

σ yy – ν ( σ xx + σ zz ) ε yy = ---------------------------------------------- or E

9 –6 6 N σ yy – νσ xx = Eε yy = 70 ( 10 ) ( – 1000 ) ( 10 ) = – 70 ( 10 ) ------2 m σ yy – νσ xx = – 70 MPa

January 2014

or 2

2

Multiplying Eq.(1) by ν and adding to Eq.(2) we obtain: σ yy ( 1 – ν ) = – 70 + ν ( 140 ) or – 70 + 0.25 ( 140 ) – 35 σ yy = ----------------------------------------- = ---------------- = – 37.33 2 0.9375 1 – 0.25

3 σ yy = 37.3MPa ( C )

Substituting Eq.(3) into Eq.(1) σ xx = 140 + ν ( – 37.33 ) = 130.67

σ xx = 130.7MPa ( T )

σ zz – ν ( σ xx + σ yy ) 6 –3 0.25 ε zz = ---------------------------------------------- = – ------------------- ( 130.67 – 37.33 ) ( 10 ) = – 0.333 ( 10 ) 9 E 70 ( 10 )

ε zz = – 333.3μ

9

–6

6

τ xy = Gγ xy = 28 ( 10 ) ( 1500 ) ( 10 ) = 42 ( 10 )

τ xy = 42MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.81 The strains and two material constants are given. Calculate σxx, σyy, τxy, σzz, and εzz assuming the point is in plane stress. ε xx = – 800 μ

Solution

ε yy = – 1000 μ

γ xy = – 500 μ

σxx=? σyy=? τxy=?

σzz=? εzz=?

E = 30, 000 ksi

ν = 0.3

plane stress

-----------------------------------------------------------E - = -------------30000- = 11538ksi G = ------------------2(1 + ν) 2 ( 1.3 ) σ zz = 0

Assuming Plane Stress

σ xx – ν ( σ yy + σ zz ) 3 –6 ε xx = ---------------------------------------------- or σ xx – ν ( σ yy + σ zz ) = Eε xx = 30 ( 10 ) ( – 800 ) ( 10 ) = – 24ksi E ∴σ xx – ν ( σ yy + σ zz ) = – 24ksi

1

Similarly σ yy – ν ( σ xx + σ zz ) 3 –6 ε yy = ---------------------------------------------- or σ yy – ν ( σ xx + σ zz ) = Eε yy = 30 ( 10 ) ( – 1000 ) ( 10 ) = – 30ksi E σ yy – ν ( σ xx + σ zz ) = – 30ksi

2

2

Multiplying Eq.(1) by ν and adding to Eq.(2) we obtain: σ yy ( 1 – ν ) = – 24ν – 30 – 0.3 ( 24 ) – 30 ∴σ yy = --------------------------------= – 40.88ksi 2 1 – 0.3

3 σ yy = 40.9ksi ( C )

Substituting Eq.(3) in Eq.(1) σ xx = – 24 + ν ( – 40.88 ) = – 36.26

σ xx = 36.26ksi ( C )

σ zz – ν ( σ xx + σ yy ) –3 0.3 ε zz = ---------------------------------------------- = – ------------------- ( – 36.36 – 40.88 ) = 0.771 ( 10 ) 3 E 30 ( 10 )

in ε zz = 771μ ----in

–6

τ xy = Gγ xy = 11538 ( – 500 ) ( 10 ) = – 5.769

τ xy = – 5.77ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.82

The strains and two material constants are given. Calculate σxx, σyy, τxy, σzz, and εzz assuming the

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point is in plane stress. ε xx = 1500 μ

Solution

ε yy = – 1200 μ

γ xy = – 10 00 μ

σxx=? σyy=? τxy=?

σzz=? εzz=?

E = 10, 000 ksi

G = 3900 ksi

plane stress

-----------------------------------------------------------E 10000 G = -------------------- or 3900 = -------------------2(1 + ν) 2(1 + ν)

∴ν = 0.282 σ zz = 0

Assuming Plane Stress

σ xx – ν ( σ yy + σ zz ) 3 –6 ε xx = ---------------------------------------------- or σ xx – ν ( σ yy + σ zz ) = Eε xx = 10 ( 10 ) ( 1500 ) ( 10 ) = 15ksi E ∴σ xx – ν ( σ yy + σ zz ) = 15ksi

1

Similarly σ yy – ν ( σ xx + σ zz ) 3 –6 ε yy = ---------------------------------------------- or σ yy – ν ( σ xx + σ zz ) = Eε yy = 10 ( 10 ) ( – 1200 ) ( 10 ) = – 12ksi E σ yy – ν ( σ xx + σ zz ) = – 12ksi

2

2

Multiplying Eq.(1) by ν and adding to Eq.(2) we obtain: σ yy ( 1 – ν ) = 15ν – 12 0.282 ( 15 ) – 12 ∴σ yy = ----------------------------------- = – 8.44 2 1 – ( 0.282 )

3 σ yy = 8.44ksi ( C )

Substituting Eq.(3) in Eq.(1) σ xx = 15 + ν ( – 8.44 ) = 15 + 0.282 ( – 8.44 ) = 12.619

σ xx = 12.62ksi ( T )

σ zz – ν ( σ xx + σ yy ) 0.282 ε zz = ---------------------------------------------- = – ------------------- ( 12.619 – 8.44 ) = – 0.1178 3 E 10 ( 10 )

ε zz = – 118 μ

–6

τ xy = Gγ xy = 3900 ( – 1000 ) ( 10 ) = – 3.9

τ xy = – 3.9ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.83 The strains and two material constants are given. Calculate σxx, σyy, τxy, σzz, and εzz assuming the point is in plane stress. ε xx = – 2000 μ

Solution

ε yy = 2000 μ

γ xy = 1200 μ

σxx=? σyy=? τxy=?

σzz=? εzz=?

G = 15 GPa

ν = 0.2

plane stress

-----------------------------------------------------------E - or 15 = -----------------------E - or G = ------------------2(1 + ν) 2 ( 1 + 0.2 )

E = 36.0GPa σ zz = 0

Assuming Plane Stress

σ xx – ν ( σ yy + σ zz ) 9 –6 ε xx = ---------------------------------------------- or σ xx – ν ( σ yy + σ zz ) = Eε xx = 36 ( 10 ) ( – 2000 ) ( 10 ) = – 72MPa or E σ xx – ν ( σ yy + σ zz ) = – 72MPa

1

Similarly σ yy – ν ( σ xx + σ zz ) 9 –6 ε yy = ---------------------------------------------- or σ yy – ν ( σ xx + σ zz ) = Eε yy = 36 ( 10 ) ( 2000 ) ( 10 ) = 72MPa or E σ yy – ν ( σ xx + σ zz ) = 72MPa

2

2

Multiplying Eq.(1) by ν and adding to Eq.(2) σ yy ( 1 – ν ) = – 72ν – 72 or

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– 0.2 ( 72 ) + 72 σ yy = ---------------------------------- = 60.1 2 1 – 0.2

January 2014

or σ yy = 60MPa ( T )

3

Substituting Eq.(3) in Eq.(1) σ xx = – 72 + ν ( σ yy ) = – 72 + 0.2 ( 60.1 ) = – 59.98 or

σ xx = 60MPa ( C )

σ zz – ν ( σ xx + σ yy ) 0.2 ε zz = ---------------------------------------------- = – ------------------- ( – 60 + 60 ) = 0 9 E 36 ( 10 )

ε zz = 0

9

–6

6

τ xy = Gγ xy = 15 ( 10 ) ( 1200 ) ( 10 ) = 18 ( 10 )

τ xy = 18MPa

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

3.84 The strains and two material constants are given. Calculate σxx, σyy, τxy, σzz, and εzz assuming the point is in plane stress. ε xx = 50 μ

Solution

ε yy = 75 μ

γ xy = – 25 μ

σxx=? σyy=? τxy=?

E = 2000 psi

σzz=? εzz=?

G = 800 psi

plane stress

-----------------------------------------------------------E 2000 G = -------------------- or 800 = -------------------2(1 + ν) 2(1 + ν)

∴ν = 0.25 σ zz = 0

Assuming Plane Stress

σ xx – ν ( σ yy + σ zz ) –6 ε xx = ---------------------------------------------- or σ xx – ν ( σ yy + σ zz ) = Eε xx = 2000 ( 50 ) ( 10 ) = 0.1psi E ∴σ xx – ν ( σ yy + σ zz ) = 0.1psi

1

Similarly σ yy – ν ( σ xx + σ zz ) –6 ε yy = ---------------------------------------------- or σ yy – ν ( σ xx + σ zz ) = Eε yy = 2000 ( 75 ) ( 10 ) = 0.15psi E σ yy – ν ( σ xx + σ zz ) = 0.15psi

2

2

Multiplying Eq.(1) by ν and adding to Eq.(2) we obtain: σ yy ( 1 – ν ) = 0.1ν + 0.15 0.25 ( 0.1 ) + 0.15 ∴σ yy = --------------------------------------- = 0.1867 2 1 – 0.25

3 σ yy = 0.187psi ( T )

Substituting Eq.(3) in Eq.(1) σ xx = 0.1 + ν ( 0.1867 ) = 0.1 + 0.25 ( 0.1867 ) = 0.1467 σ zz – ν ( σ xx + σ yy ) –6 0.25 ε zz = ---------------------------------------------- = – ------------ ( 0.1467 + 0.1867 ) = – 41.7 ( 10 ) 2000 E –6

τ xy = Gγ xy = 800 ( – 25 ) ( 10 ) = – 0.02

σ xx = 0.147psi ( T ) ε zz = – 41.7 μ τ xy = – 0.02 psi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.85 The cross-section of the wooden piece that is visible in the picture is 40 mm x 25 mm. The clamped length of the wooden piece in the vice is 125 mm. The Modulus of Elasticity of wood is E = 14 GPa and a Poisson’s ratio of 0.3. The jaws of the vice exert a uniform pressure of 3.2 MPa on the wood. Determine the average change of length of the wood. .

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40 mm

Figure P3.85

Solution

E=14GPa

ν=0.3

σxx=3.2MPa(C)

ΔL=?

-----------------------------------------------------------We assume σ yy = 0 and σ zz = 0 σ zz – ν ( σ xx + σ yy ) – ν σ xx ε zz = ---------------------------------------------- = -------------E E 6

–3 ( 0.3 ) ( – 3.2 ) ( 10 ) ε zz = – ------------------------------------------ = 0.06857 ( 10 ) 9 14 ( 10 ) –3

3

ΔL = ε zz L = 0.06857 ( 10 ) ( 125 ) = 8.571 ( 10 )

ΔL = 0.0086mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.86 A thin plate (E=30,000 ksi; ν = 0.25) under the action of uniform forces deforms to the shaded position. Assuming plane stress determine the average normal stresses in the x and y direction. y

0.005 in

5 in x 10 in

Figure P3.86

Solution

E=30000ksi

Assuming plane stress

ν=0.25

σxx=?

σyy=?

-----------------------------------------------------------σ zz = 0

0.005 ε xx = – ------------- = – 0.0005 or σ xx – νσ yy = Eε xx = 30000 ( – 0.0005 ) 10 σ xx – νσ yy = Eε xx = – 15ksi ε yy = 0

1

σ yy – νσ xx = 0

or

2

2

Substituting Eq.(2) into Eq.(1) we obtain: σ xx ( 1 – ν ) = – 15psi or – 15 σ xx = --------------------- = – 16ksi 2 1 – 0.25 From Eq.(2) we obtain: σ yy = 0.25 ( – 16 ) = – 4ksi

σ xx = 16ksi ( C ) σ yy = 4ksi ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.87 A thin plate (E=30,000 ksi; ν = 0.25) is subjected to a uniform stress σ = 10 ksi as shown in Figure P3.87. Assuming plane stress, determine (a) the average normal stress in the y direction (b) the contraction

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of the plate in the x direction. y

σ

5 in

x 10 in Figure P3.87

Solution

ν=0.25

E=30000ksi

σxx=-10 ksi

σyy=?

δx=?

------------------------------------------------------------

Assuming plane stress σ zz = 0 ε yy = 0

σ yy – νσ xx = 0 or

or

σ yy = νσ xx = 0.25 ( – 10 ) = – 2.5ksi

1 σ yy = 2.5ksi ( C )

We can write σ xx – νσ yy –3 ( – 10 ) – ( 0.25 ) ( – 2.5 ) ε xx = -------------------------- = --------------------------------------------------- = – 0.3125 ( 10 ) E 30, 000

2

–3

δ x = ε xx ( 10 ) = – 3.125 ( 10 )

3 δ x = – 0.0031in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.88 A rubber (ER=300 psi and νR = 0.5) rod of diameter dR =4 in is placed in a steel (rigid) tube dS =4.1 in as shown in Figure P3.88. What is the smallest value of P that can be applied so that the space between the rubber rod and the steel tube would close. y

x

z

P

Figure P3.88

dR dS

Solution

-----------------------------------------------------------The normal strain in radial direction (y & z) is: dS – dR – 4.0- = 0.025 ε yy = ε zz = ----------------- = 4.1 -------------------4.0 dR

1

Just closes implies: σ yy = σ zz = 0 By generalized Hooke’s law: σ yy – ν ( σ xx + σ zz ) – νσ xx ε yy = ---------------------------------------------- = -------------E E

or

Eε yy ( 300 ) ( 0.025 ) - = – -------------------------------- = – 15 psi σ xx = – ----------ν 0.5

2

The compresive internal force is equal to applied force. We obtain:

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–N P σ xx = ------- = – -------------------------- = – 15 2 A (π ⁄ 4)(4 )

or

January 2014

3

P = 188.49 lb

The answer is:

P = 188.5 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.89 A rubber (ER= 2.1GPa and νR = 0.5) rod of diameter dR = 200 mm is placed in a steel (rigid) tube dS = 204 mm as shown in Figure P3.89. If the applied force is P = 10 kN, determine the average normal stress in the y and z direction. y

x

z

P

Figure P3.89

dR dS

Solution

-----------------------------------------------------------The compresive internal force is equal to applied force. We obtain: 3

6 2 10 ( 10 ) –N σ xx = ------- = – --------------------------------- = – 31.83 ( 10 ) N/m 2 A ( π ⁄ 4 ) ( 0.02 )

1

The normal strain in radial direction (y & z) is: dS – dR 204 – 200 ε yy = ε zz = ----------------- = ------------------------ = 0.02 200 dR

2

The normal stress in y and a direction is same by symmetry. By generalized Hooke’s law: 6

σ yy – ν ( σ xx + σ zz ) σ yy ( 1 – ν ) – νσ xx 0.5σ yy – 0.5 [ – 31.83 ( 10 ) ] ε yy = ---------------------------------------------- = -------------------------------------------= ----------------------------------------------------------------- = 0.02 9 E E 2.1 ( 10 ) 6

0.5σ yy = [ 42 – 0.5 ( 31.83 ) ] ( 10 )

6

σ yy = 52.17 ( 10 ) N ⁄ m

or

or

2

3

σ yy = σ zz = 52.17 MPa(T)

The answer is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.90 A 2in x 2 in square with a circle inscribed is stressed as shown Figure P3.90. The plate material has a Modulus of Elasticity of E = 10,000 ksi and a Poisson’s ratio ν = 0.25. Assuming plane stress, determine the major and minor axis of the ellipse formed due to deformation. 10 ksi

20 ksi

Figure P3.90

Solution

E=10000ksi

ν=0.25

a = major axis=?

b = minor axis=?

-----------------------------------------------------------We are given: σ xx = 20ksi and σ yy = – 10 ksi and Plane stress ∴σ zz = 0

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σ xx – ν ( σ yy + σ zz ) –3 20 – 0.25 ( – 10 ) ε xx = ---------------------------------------------- = ------------------------------------ = 2.25 ( 10 ) E 10000 σ yy – ν ( σ xx + σ zz ) –3 – 10 – 0.25 ( 20 ) ε yy = ---------------------------------------------- = -------------------------------------- = – 1.5 ( 10 ) E 10000

The circle extends in x direction and contracts in y direction –3

a = 2 + Δa = 2.0045in

a = 2.0045in

–3

b = 2 + Δb = 1.9970in

b = 1.9770in

Δa = ε xx ( 2 ) = 4.5 ( 10 ) Δb = ε yy ( 2 ) = – 3 ( 10 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.91 A 2in x 2 in square with a circle inscribed is stressed as shown Figure P3.91. The plate material has a Modulus of Elasticity of E = 10,000 ksi and a Poisson’s ratio ν = 0.25. Assuming plane stress, determine the major and minor axis of the ellipse formed due to deformation. 10 ksi

20 ksi

Figure P3.91

Solution

E=10000ksi

ν=0.25

a = major axis =?

b = minor axis =?

-----------------------------------------------------------We are given: σ xx = – 20 ksi and σ yy = – 10 ksi and plane stress ∴σ zz = 0 σ xx – ν ( σ yy + σ zz ) – 20 – 0.25 ( – 10 )- = – 1.75 ( 10 – 3 ) ε xx = ---------------------------------------------- = ---------------------------------------E 10000 σ yy – ν ( σ xx + σ zz ) –3 – 10 – 0.25 ( – 20 ) ε yy = ---------------------------------------------- = ----------------------------------------- = – 0.5 ( 10 ) E 10000

The circle contracts more in x direction than in y direction –3

Δa = ε xx ( 2 ) = – 3.5 ( 10 ) –3

Δb = ε yy ( 2 ) = – 1 ( 10 )

a = 2 + Δa = 1.9965in

a = 1.9965in

b = 2 + Δb = 1.999in

b = 1.999in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.92 A 50 mm x 50 mm square with a circle inscribed is stressed as shown Figure P3.92. The plate material has a Modulus of Elasticity of E = 70 GPa and a Poisson’s ratio ν = 0.25. Assuming plane stress, determine the major and minor axis of the ellipse formed due to deformation. 280 MPa

154 MPa

Figure P3.92

Solution

E=70GPa

ν=0.25

a= major axis =?

b = minor axis =?

-----------------------------------------------------------We are given: σ xx = 154MPa and σ yy = 280MPa and plane stress ∴σ zz = 0

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6 σ xx – ν ( σ yy + σ zz ) –3 [ 154 – 0.25 ( 280 ) ] ( 10 ) ε xx = ---------------------------------------------- = -------------------------------------------------------- = 1.2 ( 10 ) 9 E 70 ( 10 ) 6 σ yy – ν ( σ xx + σ zz ) –3 [ 280 – 0.25 ( 154 ) ] ( 10 ) ε yy = ---------------------------------------------- = -------------------------------------------------------- = 3.45 ( 10 ) 9 E 70 ( 10 )

The circle elongates more in x direction than in y direction –3

Δa = ε xx ( 50 ) = 60 ( 10 ) –3

Δb = ε yy ( 50 ) = 172.5 ( 10 )

a = 50 + Δa = 50.06mm

a = 50.06mm

b = 50 + Δb = 50.1725mm

b = 50.1725mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.93 A rectangle inscribed on a thin aluminum (10,000 ksi, ν = 0.25) plate is observed to deform into the colored shape shown in Figure P3.93. Assuming plane stress, determine the average stress components σ xx, σ yy, and τ xy . 0.0035 in

y 0.0042in

0.0048 in

1.4 in

x A 3.0 in 0.0036 in

Figure P3.93

Solution

σxx=?

σyy=?

σxy=?

------------------------------------------------------------

The average strains can be found as shown below. ε xx = Δ -----u- = 0.0036 ---------------- = 0.0012 Δx 3

1

Δv 0.0035 ε yy = ------ = ---------------- = 0.0025 Δy 1.4

2

Δu Δv 0.0042 0.0048 γ xy = ------ + ------ = ---------------- + ---------------- = 0.0046 Δy Δx 1.4 3

3

E 10000 G = -------------------- = ------------------ = 4000ksi 2(1 + ν) 2 ( 1.25 )

4

The shear modulus is

Assuming Plane Stress: σ zz = 0 σ xx – ν ( σ yy + σ zz ) ε xx = ---------------------------------------------- or E

σ xx – νσ yy = Eε xx = ( 10000 ) ( 0.0012 ) = 12ksi σ xx – νσ yy = 12ksi

5

Similarly σ yy – ν ( σ xx + σ zz ) ε yy = ---------------------------------------------- or σ yy – νσ xx = Eε yy = ( 10000 ) ( 0.0025 ) = 25ksi E σ yy – νσ xx = 50ksi

6

2

Multiplying Eq.(5) by ν and adding to Eq.(6) we obtain: σ yy ( 1 – ν ) = ( 12 )ν + 25 ( 12 ) + ( 25 ) σ yy = 0.25 -------------------------------------- = 29.87ksi 2 1 – 0.25

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σ yy = 29.87ksi ( T )

Substituting Eq.(7) in Eq.(5) σ xx = 12 + ν ( 29.87 ) = 19.47ksi

σ xx = 19.47ksi ( T )

τ xy = Gγ xy = ( 4000 ) ( 0.0046 ) = 18.4ksi

τ xy = 18.4ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.94 A rectangle inscribed on an steel (E = 210 GPa ν = 0.28) plate is observed to deform into the colored shape shown in Figure P3.94. Determine the average stress components σ xx, σ yy, and τ xy . y 0.09 mm

450 mm

0.060 mm

0.10 mm

x

A 0.075 mm

250 mm

Figure P3.94

The average strains can be found as shown below. 0.075- = – 0.0003 ε xx = Δ -----u- = – --------------Δx 250

1

0.6- = – 0.000133 ε yy = Δ -----v- = – --------Δy 450

2

Δu Δv 0.09 0.1 γ xy = ------ + ------ = ---------- + --------- = 0.0006 Δy Δx 450 250

3

E - = ----------------210 - = 82.03GPa G = ------------------2(1 + ν) 2 ( 1.28 )

4

The shear modulus is

Assuming Plane Stress: σ zz = 0 σ xx – ν ( σ yy + σ zz ) 9 ε xx = ---------------------------------------------- or σ xx – νσ yy = Eε xx = ( 210 ) ( 10 ) ( – 0.0003 ) = – 63 MPa E σ xx – νσ yy = – 63 MPa

5

Similarly σ yy – ν ( σ xx + σ zz ) 9 ε yy = ---------------------------------------------- or σ yy – νσ xx = Eε yy = ( 210 ) ( 10 ) ( – 0.000133 ) = – 28 MPa E σ yy – νσ xx = – 28 MPa

Multiplying Eq.(5) by ν and adding to Eq.(6) we obtain:

6

2

σ yy ( 1 – ν ) = ( – 63 )ν + ( – 28 )

0.28 ( – 63 ) + ( – 28 ) σ yy = --------------------------------------------- = – 49.52MPa 2 1 – 0.28

7 σ yy = 49.52MPa ( C )

Substituting Eq.(7) in Eq.(5)

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σ xx = – 63 + ν ( – 49.52 ) = – 76.87 MPa

January 2014

σ xx = 76.87MPa ( C )

9

τ xy = Gγ xy = ( 82.03 ) ( 10 ) ( 0.0006 ) = 49.2MPa

τ xy = 49.2MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.95 A 5-ft mean diameter spherical steel (E = 30,000 ksi, ν = 0.28) tank has a wall thickness of 3/4 in. Determine the increase in the mean diameter when the the gas pressure inside the tank is 600 psi. Solution

-----------------------------------------------------------The hoop stress is: pR ( 600 ) ( 30 ) σ θθ = σ φφ = ------- = ------------------------- = 24000 psi = 24 ksi t (3 ⁄ 4)

1

The radial stress is approximated as zero. From Hooke’s law we obtain: σ θθ – ν ( σ φφ + σ rr ) –3 24 – 0.28 ( 24 ) ε θθ = --------------------------------------------- = --------------------------------- = 0.576 ( 10 ) E 30000

2

We can write: –3 πΔd Δd ε θθ = ---------- = ------- = 0.576 ( 10 ) πd 60

or

Δd = 0.0346 in

3

Δd = 0.0346 in

The answer is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.96 A steel (E = 210 GPa ν = 0.28) cylinder of mean diameter of 1 m and wall thickness of 10 mm has gas at 250 kPa. Determine the increase in the mean diameter due to gas pressure. Solution

-----------------------------------------------------------The hoop stress is: 3

6 2 ( 250 ) ( 10 ) ( 0.5 ) σ θθ = pR ------- = ---------------------------------------- = 12.5 ( 10 ) N ⁄ m t ( 0.01 )

1

The axial stress is: 3

6 2 ( 250 ) ( 10 ) ( 0.5 ) pR σ xx = ------- = ---------------------------------------- = 6.25 ( 10 ) N ⁄ m 2t 2 ( 0.01 )

2

The radial stress is approximated as zero. From Hooke’s law we obtain: 6 σ θθ – ν ( σ xx + σ rr ) –6 [ 12.5 – 0.28 ( 6.25 ) ] ( 10 ) ε θθ = --------------------------------------------- = ------------------------------------------------------------- = 51.19 ( 10 ) 9 E 210 ( 10 )

3

–6 πΔd Δd ε θθ = ---------- = ------- = 51.19 ( 10 ) πd 1

4

We can write: or

–6

Δd = 51.19 ( 10 ) m Δd = 0.052 mm

The answer is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.97 Derive the following relation of normal stresses in terms of normal strain from the generalized Hooke’s Law. E σ xx = [ ( 1 – ν )ε xx + νε yy + νε zz ] -------------------------------------( 1 – 2ν ) ( 1 + ν ) E σ yy = [ ( 1 – ν )ε yy + νε zz + νε xx ] -------------------------------------( 1 – 2ν ) ( 1 + ν )

3.16

E σ zz = [ ( 1 – ν )ε zz + νε xx + νε yy ] -------------------------------------( 1 – 2ν ) ( 1 + ν )

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Solution ------------------------------------------------------------

Adding the three equations for normal strains in the generalized Hooke’s Law. 1 – 2ν ε xx + ε yy + ε zz = --------------- ( σ xx + σ yy + σ zz ) E E ∴( σ xx + σ yy + σ zz ) = ⎛ ---------------⎞ ( ε xx + ε yy + ε zz ) ⎝ 1 – 2ν⎠

1

From generalized Hooke’s Law. σ xx ν 1+ν ν - – --- ( σ + σ zz ) = σ xx ⎛ ------------⎞ – --- ( σ xx + σ yy + σ zz ) ε xx = -------⎝ E ⎠ E E E yy

2

Substituting Eq. (1) in Eq. (2) we obtain 1+ν ν E ε xx = σ xx ⎛ ------------⎞ – ⎛ ---⎞ ⎛ ---------------⎞ ( ε xx + ε yy + ε zz ) ⎝ E ⎠ ⎝ E⎠ ⎝ 1 – 2ν⎠ 1+ν ν ν ν ∴σ xx ⎛⎝ ------------⎞⎠ = ε xx ⎛⎝ 1 + ---------------⎞⎠ + --------------- ε yy + --------------- ε zz E 1 – 2ν 1 – 2ν 1 – 2ν E ∴σ xx = -------------------------------------- [ ε xx ( 1 – ν ) + νε yy + νε zz ] ( 1 + ν ) ( 1 – 2ν )

3

From generalized Hooke’s Law. σ yy ν 1+ν ν ε yy = -------- – --- ( σ + σ zz ) = σ yy ⎛ ------------⎞ – --- ( σ xx + σ yy + σ zz ) ⎝ E ⎠ E E E xx

4

Substituting Eq. (1) in Eq. (4) we obtain ν 1+ν ν E 1+ν ν ν ε yy = σ yy ⎛ ------------⎞ – ⎛ ---⎞ ⎛ ---------------⎞ ( ε xx + ε yy + ε zz ) or σ yy ⎛ ------------⎞ = --------------- ε xx + ε ⎛ 1 + ---------------⎞ + --------------- ε zz ⎝ E ⎠ ⎝ E⎠ ⎝ 1 – 2ν⎠ ⎝ E ⎠ 1 – 2ν yy ⎝ 1 – 2ν⎠ 1 – 2ν E ∴σ yy = -------------------------------------- [ νε xx + ( 1 – ν )ε yy + νε zz ] ( 1 + ν ) ( 1 – 2ν )

5

From generalized Hooke’s Law. σ zz ν ν 1+ν ε zz = ------- – --- ( σ + σ yy ) = σ zz ⎛ ------------⎞ – --- ( σ xx + σ yy + σ zz ) ⎝ E ⎠ E E E xx

6

Substituting Eq. (1) in Eq. (6) we obtain 1+ν ν E ε zz = σ zz ⎛ ------------⎞ – ⎛ ---⎞ ⎛ ---------------⎞ ( ε xx + ε yy + ε zz ) ⎝ E ⎠ ⎝ E⎠ ⎝ 1 – 2ν⎠ 1+ν ν ν ν ∴σ zz ⎛ ------------⎞ = --------------- ε xx + --------------- ε yy + ⎛ 1 + ---------------⎞ ε zz ⎝ ⎝ E ⎠ 1 – 2ν 1 – 2ν 1 – 2ν⎠ E ∴σ zz = -------------------------------------- [ νε xx + νε yy + ( 1 – ν )ε zz ] ( 1 + ν ) ( 1 – 2ν )

7

Eqs. (3), (5), and (7) are the same as Eq. (3.14)

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.98

For a point in plane stress show E σ xx = [ ε xx + νε yy ] ------------------2 (1 – ν )

E σ yy = [ ε yy + νε xx ] ------------------2 (1 – ν )

3.17

Solution ------------------------------------------------------------

Plane Stress ∴σ zz = 0 From generalized Hooke’s Law.

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σ xx ν - – --- ( σ + σ zz ) ε xx = -------E E yy

or

σ xx – νσ yy = Eε xx

1

σ yy ν ε yy = -------- – --- ( σ + σ zz ) E E xx

or

σ yy – νσ xx = Eε yy

2

2

Multiplying Eq. (2) by ν and adding to Eq. (1) we obtain: σ xx ( 1 – ν ) = E ( ε xx + νε yy ) E ∴σ xx = ------------------- ( ε xx + νε yy ) 2 (1 – ν )

3 2

Multiplying Eq. (1) by ν and adding to Eq. (2) we obtain: σ yy ( 1 – ν ) = E ( ε yy + νε xx ) E ∴σ yy = ------------------- ( ε yy + νε xx ) 2 (1 – ν )

4

Eq. (3) and (4) are the same as given in Eq. 3.15

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.99

For a point in plane stress show ν ε zz = – ⎛ ------------⎞ ( ε xx + ε yy ) ⎝ 1 – ν⎠

3.18

Solution ------------------------------------------------------------

Plane Stress ∴σ zz = 0 Adding Eq. (1) and (2) in Problem 3.82 we obtain: σ xx ( 1 – ν ) + σ yy ( 1 – ν ) = E ( ε xx + ε yy ) E ∴σ xx + σ yy = ---------------- ( ε xx + ε yy ) (1 – ν)

1

From generalized Hooke’s Law σ zz ν ν E - – --- ( σ + σ yy ) = 0 – ⎛ ---⎞ ⎛ ----------------⎞ ( ε xx + ε yy ) ε zz = ------⎝ E⎠ ⎝ ( 1 – ν )⎠ E E xx ν ∴ε zz = – ⎛ ----------------⎞ ( ε xx + ε yy ) ⎝ ( 1 – ν )⎠

2

Eq. (2) is the same as given in Eq. 3.16

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.100

Using Eqs. 3.17 and 3.18 solve for σxx, σyy, and εzz in Problem 3.79.

Solution

σxx=?

σyy=? and

εzz=?

-----------------------------------------------------------–6 9 N Substituting ε xx = 500 ( 10 ) , ε yy = 400 ( 10 ) , E = 200 ( 10 ) ------2- and ν = 0.32 into Eqs. 3.15 and 3.16 m –6

we obtain the following. –6

9

–6

9

( 10 )200 ( 10 ) 6 N σ xx = [ 500 + ( 0.32 ) ( 400 ) ] --------------------------------------- = 139.9 ( 10 ) ------2 2 1 – 0.32 m ( 10 )200 ( 10 ) 6 N σ yy = [ 400 + ( 0.32 ) ( 500 ) ] --------------------------------------- = 124.77 ( 10 ) ------2 2 1 – 0.32 m –6 –6 0.25 ε zz = – ------------------- ( 500 + 400 ) ( 10 ) = – 423.5 ( 10 ) 1 – 0.25

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σ xx = 139.9MPa ( T ) σ yy = 124.8MPa ( T ) ε zz = – 423.5μ

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-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.101

Using Eqs. 3.17 and 3.18 solve for σxx, σyy, and εzz in Problem 3.80.

Solution

σxx=?

σyy=? and

εzz=?

-----------------------------------------------------------–6 9 N Substituting ε xx = – 3000 ( 10 ) , ε yy = 1500 ( 10 ) , E = 70 ( 10 ) ------2- and ν = 0.25 into Eqs. 3.15 and 3.16 m –6

we obtain the following. –6

9

–6

9

( 10 )70 ( 10 ) 6 N σ xx = [ – 3000 + ( 0.25 ) ( 1500 ) ] ------------------------------------ = – 196 ( 10 ) ------2 2 m 1 – 0.25

σ xx = 196MPa ( C )

( 10 )70 ( 10 ) 6 N σ yy = [ 1500 + ( 0.25 ) ( – 3000 ) ] ------------------------------------ = 56 ( 10 ) ------2 2 m 1 – 0.25

σ yy = 56MPa ( T )

–6 –6 0.25 ε zz = – ------------------- ( – 3000 + 1500 ) ( 10 ) = 500 ( 10 ) 1 – 0.25

ε zz = 500μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.102

Using Eqs. 3.17 and 3.18 solve for σxx, σyy, and εzz in Problem 3.81.

Solution

σxx=?

σyy=? and

εzz=?

-----------------------------------------------------------–6

–6

Substituting ε xx = – 800 ( 10 ) , ε yy = – 1000 ( 10 ) , E = 30000ksi and ν = 0.3 into Eqs. 3.15 and 3.16 we obtain the following. –6

( 10 )30000 σ xx = [ – 800 + ( 0.3 ) ( – 1000 ) ] ------------------------------- = – 36.26ksi 2 1 – 0.3 –6

σ xx = 36.3ksi ( C )

( 10 )30000 σ yy = [ – 1000 + ( 0.3 ) ( – 800 ) ] ------------------------------- = – 40.88ksi 2 1 – 0.3

σ yy = 40.9ksi ( C )

–6 –6 0.3 ε zz = – ---------------- ( – 800 – 1000 ) ( 10 ) = 771.4 ( 10 ) 1 – 0.3

ε zz = 771.4μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.103

Using Eqs. 3.17 and 3.18 solve for σxx, σyy, and εzz in Problem 3.82.

Solution

σxx=?

σyy=? and

εzz=?

-----------------------------------------------------------–6

–6

Substituting ε xx = 1500 ( 10 ) , ε yy = – 1200 ( 10 ) , E = 10000ksi and ν = 0.282 into Eqs. 3.15 and 3.16 we obtain the following. –6

( 10 )10000 σ xx = [ 1500 + ( 0.282 ) ( – 1200 ) ] ------------------------------- = 12.62ksi 2 1 – 0.282 –6

( 10 )10000 σ yy = [ – 1200 + ( 0.282 ) ( 1500 ) ] ------------------------------- = – 8.44ksi 2 1 – 0.282 –6 –6 0.282 ε zz = – ---------------------- ( 1500 + ( – 1200 ) ) ( 10 ) = – 117.8 ( 10 ) 1 – 0.282

σ xx = 12.6ksi ( T ) σ yy = 8.44ksi ( C ) ε zz = – 118μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.104 Using Eqs. 3.17 and 3.18 solve for σxx, σyy, and εzz in Problem 3.83. Solution σxx=? σyy=? and εzz=?

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-----------------------------------------------------------–6

–6

9

2

Substituting ε xx = – 2000 ( 10 ) , ε yy = 2000 ( 10 ) , E = 36 ( 10 )N ⁄ m and ν = 0.2 into Eqs. 3.15 and 3.16 we obtain the following. –6

9

–6

9

( 10 ) ( 36 ) ( 10 ) 6 N σ xx = [ – 2000 + ( 0.2 ) ( 2000 ) ] ----------------------------------------- = – 60 ( 10 ) ------2 2 m 1 – 0.2

σ xx = 60MPa ( C )

( 10 ) ( 36 ) ( 10 ) 6 N σ yy = [ 2000 + ( 0.2 ) ( – 2000 ) ] ----------------------------------------- = 60 ( 10 ) ------2 2 m 1 – 0.2

σ yy = 60MPa ( T )

–6 0.2 ε zz = – ---------------- ( – 2000 + 2000 ) ( 10 ) = 0 1 – 0.2

ε zz = 0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.105

Using Eqs. 3.17 and 3.18 solve for σxx, σyy, and εzz in Problem 3.84.

Solution

σxx=?

σyy=? and

εzz=?

-----------------------------------------------------------–6

–6

Substituting ε xx = 50 ( 10 ) , ε yy = 75 ( 10 ) , E = 2000psi and ν = 0.25 into Eqs. 3.15 and 3.16 we obtain the following. –6

( 10 )2000 σ xx = [ 50 + ( 0.25 ) ( 75 ) ] ---------------------------- = 0.1467psi 2 1 – 0.25

σ xx = 0.147psi ( T )

–6

( 10 )2000 σ yy = [ 75 + ( 0.25 ) ( 50 ) ] ---------------------------- = 0.187psi 2 1 – 0.25

σ yy = 0.187psi ( T )

–6 –6 0.25 ε zz = – ------------------- ( 50 + 75 ) ( 10 ) = – 41.67 ( 10 ) 1 – 0.25

ε zz = – 41.67μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.106

For a point in plane strain show (1 + ν) ε xx = [ ( 1 – ν )σ xx – νσ yy ] ----------------E

(1 + ν) ε yy = [ ( 1 – ν )σ yy – νσ xx ] ----------------E

3.19

Solution ------------------------------------------------------------

Plane strain ∴ε zz = 0 σ

ν E E ∴σ zz = ν ( σ xx + σ yy )

zz --From Generalized Hooke’s Law: ε zz = ------- – ( σ xx + σ yy ) = 0 or

1

From Generalized Hooke’s Law 2 σ xx ν σ xx ν (1 – ν ) ν(1 + ν) ε xx = -------- – --- ( σ yy + σ zz ) = -------- – --- ( σ yy + νσ xx + νσ yy ) = ------------------- σ xx – -------------------- σ yy E E E E E E (1 + ν) ∴ε xx = ( ( 1 – ν )σ xx – νσ yy ) ----------------E

2

From Generalized Hooke’s Law 2 σ yy ν σ yy ν (1 – ν ) ν(1 + ν) - – --- ( σ xx + σ zz ) = -------- – --- ( σ xx + νσ xx + νσ yy ) = ------------------- σ yy – -------------------- σ xx ε yy = -------E E E E E E (1 + ν) ∴ε yy = ( ( 1 – ν )σ yy – νσ xx ) ----------------E

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Eqs. (2) and (3) are same as Eq. 3.17

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.107

For a point in plane strain show E σ xx = [ ( 1 – ν )ε xx + νε yy ] -------------------------------------( 1 – 2ν ) ( 1 + ν )

E σ yy = [ ( 1 – ν )ε yy + νε xx ] -------------------------------------( 1 – 2ν ) ( 1 + ν )

3.20

Solution ------------------------------------------------------------

Plane strain ∴ε zz = 0 Multiplying Eq. (1) in Problem 3.89 by ( 1 – ν ) and Eq. (2) by ν and adding 2 2 2 2 (1 + ν) (1 + ν) ( 1 – ν )ε xx + νε yy = ( ( 1 – ν ) – ν )σ xx ----------------- = ( 1 – 2ν + ν – ν )σ xx ----------------E E

E ∴σ xx = [ ( 1 – ν )ε xx + νε yy ] -------------------------------------( 1 – 2ν ) ( 1 + ν )

1

Multiplying Eq. (1) in Problem 3.89 by ν and Eq. (2) by ( 1 – ν ) and adding 2 2 2 2 (1 + ν) (1 + ν) νε xx + ( 1 – ν )ε yy = ( ( 1 – ν ) – ν )σ yy ----------------- = ( 1 – 2ν + ν – ν )σ yy ----------------E E

E ∴σ yy = [ νε xx + ( 1 – ν )ε yy ] -------------------------------------( 1 – 2ν ) ( 1 + ν )

2

Eq. (1) and (2) are the same as Eq. 3.18

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.108 A differential element subjected to only normal strains is shown in Figure P3.108. The ratio of change in volume (ΔV) to the original volume (V) is called the volumetric strain (εV) or dilation. For small strain, prove y

ε V = ΔV ⁄ V = ε xx + ε yy + ε zz

(1+εxx) Δx

3.21

(1+εyy) Δy Δy

Δx

Δz

Figure P3.108

x (1+εzz) Δz

z

Solution -----------------------------------------------------------The original volume is: V = ( Δx ) ( Δy ) ( Δz )

The change of volume is: ΔV = ( 1 + ε xx ) Δx ( 1 + ε yy ) Δy ( 1 + ε zz ) Δ( z ) – ( Δx Δy Δz ) ∴ΔV = [ 1 + ε xx + ε yy + ε zz + ε xx ε yy + ε yy ε zz + ε zz ε xx + ε xx ε yy ε zz – 1 ] ( Δx Δy Δz )

For small strain, the quadratic and the cubic terms can be neglected i.e., ΔV = [ ε xx + ε yy + ε zz ] ( Δx Δy Δz ) ΔV ε v = -------- = ε xx + ε yy + ε zz V

1

Eq. 1 is same as Eq. 3.19

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.109

Prove p = – Kε v

σ xx + σ yy + σ zz p = – ⎛ ---------------------------------------⎞ ⎝ ⎠ 3

E K = ----------------------3 ( 1 – 2ν )

3.22 where, K is called the bulk modulus and p is called the hydrostatic pressure because at a point in fluid the

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January 2014

normal stresses in all directions are equal to -p. Note that at ν = 1/2 there is no change in volume regardless of the value of stresses. Such materials are called incompressible materials. Solution ------------------------------------------------------------

Adding the three equations for normal strain in generalized Hooke’s law we obtain ( 1 – 2ν ) ε xx + ε yy + ε zz = -------------------- ( σ xx + σ yy + σ zz ) E 3 ( 1 – 2ν )p Substituting σ xx + σ yy + σ zz = – 3p and ε xx + ε yy + ε zz = ε v we obtain: ε v = –----------------------------E

–E ∴p = ----------------------- ε v 3 ( 1 – 2ν )

1

Eq. (1) is the same as Eq. 3.20

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.110 The stresses at a point on free surface of an orthotropic material are as given. Also given are the material constants. Using Equation (3.23), solve for the strains εxx, εyy, and γxy. σ xx = 5ksi ( C )

σ yy = 8ksi ( T )

E x = 7500ksi

τ xy = 6ksi ν xy = 0.3

G xy = 1250ksi

σ yy ν xy ε yy = -------– -------- σ xx

τ xy γ xy = --------

E y = 2500ksi σ xx ν yx ε xx = -------– -------- σ yy Ex

Solution

εxx= ?

Ey

Ey

εyy = ?

Ex

G xy

ν yx

ν

xy -------- = ------Ey Ex

(3.23)

γxy = ?

-----------------------------------------------------------ν xy 0.3 - = ν yx -------- = ----------------7500 Ex Ey

Substituting the given value of stresses and material constants in Eq. 3.21, we obtain the following –3 0.3 – 5 - – ----------- ( 8 ) = – 0.9987 ( 10 ) ε xx = ----------7500 7500

ε xx = – 998.7 μ

–3 0.3 ( 8 )- – ----------- ( – 5 ) = 3.4 ( 10 ) ε yy = ----------2500 7500 –3 6 γ xy = ------------ = 4.8 ( 10 ) 1250

ε yy = 3400μ γ xy = 4800μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.111 The stresses at a point on free surface of an orthotropic material are as given. Also given are the material constants. Using Equation (3.23), solve for the strains εxx, εyy, and γxy. σ xx = 25ksi ( C )

σ yy = 5ksi ( C )

E x = 25000ksi

Solution

E y = 2000ksi

εxx= ?

τ xy = – 8 ksi ν xy = 0.32

εyy = ?

G xy = 1500ksi

γxy = ?

-----------------------------------------------------------ν yx ν xy 0.32 -------- = --------------- = ------Ex Ey 25000

Substituting the given value of stresses and material constants in Eq. 3.21, we obtain the following –3 0.32 ( – 25 ) ε xx = --------------- – --------------- ( – 5 ) = – 0.936 ( 10 ) 25000 25000

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ε xx = – 936μ

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January 2014

–3 0.3 ( –5 ) ε yy = ------------ – --------------- ( – 25 ) = – 2.18 ( 10 ) 2000 25000

ε yy = – 2180 μ

–3 –8 γ xy = ------------ = – 5.333 ( 10 ) 1500

γ xy = – 5333μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.112 The stresses at a point on free surface of an orthotropic material are as given. Also given are the material constants. Using Equation (3.23), solve for the strains εxx, εyy, and γxy. σ xx = 200 MPa ( C ) E x = 53 GPa

σ yy = 80 MPa ( C )

E y = 18 GPa

εxx= ?

Solution

τ xy = – 54 MPa

ν xy = 0.25

εyy = ?

G xy = 9 GPa

γxy = ?

-----------------------------------------------------------ν yx ν xy 0.25 -------- = ------------------- = ------9 Ex Ey 53 ( 10 )

Substituting the given value of stresses and material constants in Eq. 3.21, we obtain the following 6

6 –3 0.25 ( – 200 ) ( 10 ) ε xx = ------------------------------ – ------------------- ( – 80 ) ( 10 ) = – 3.396 ( 10 ) 9 9 53 ( 10 ) 53 ( 10 )

ε xx = – 3396μ

6

6 –3 0.25 ( – 80 ) ( 10 ) ε yy = --------------------------- – ------------------- ( – 200 ) ( 10 ) = – 3.501 ( 10 ) 9 9 53 ( 10 ) 18 ( 10 )

ε yy = – 3501μ

6

–3 – 54 ( 10 ) γ xy = ----------------------- = – 6 ( 10 ) 9 9 ( 10 )

γ xy = – 6000μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.113 The stresses at a point on free surface of an orthotropic material are as given. Also given are the material constants. Using Equation (3.23), solve for the strains εxx, εyy, and γxy. σ xx = 300 MPa ( T ) E x = 180 GPa

σ yy = 50 MPa ( T ) E y = 15 GPa

εxx= ?

Solution

εyy = ?

τ xy = 60 MPa

ν xy = 0.28

G xy = 11 GPa

γxy = ?

-----------------------------------------------------------ν xy ν yx 0.28 -------- = ---------------------- = ------9 Ex Ey 180 ( 10 )

Substituting the given value of stresses and material constants in Eq. 3.21, we obtain the following 6

6 –3 0.28 300 ( 10 ) ε xx = ---------------------- – ---------------------- ( 50 ) ( 10 ) = 1.589 ( 10 ) 9 9 180 ( 10 ) 180 ( 10 ) 6

6 –3 0.28 ( 50 ) ( 10 ) ε yy = ------------------------ – ---------------------- ( 300 ) ( 10 ) = 2.867 ( 10 ) 9 9 15 ( 10 ) 180 ( 10 ) 6

–3 60 ( 10 ) γ xy = ------------------- = 5.454 ( 10 ) 9 11 ( 10 )

ε xx = 1589μ ε yy = 2867μ γ xy = 5454μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.114

The strains at a point on free surface of an orthotropic material is given in each problem. Also

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given are the material constants. Using Equation (3.23), solve for the stresses σxx, σyy, and τxy. ε xx = – 1000μ

ε yy = 500μ

E x = 7500ksi

E y = 2500ksi

σxx= ?

Solution

γ xy = – 250μ ν xy = 0.3

σyy = ?

G xy = 1250ksi

τxy = ?

-----------------------------------------------------------ν yx ν xy –6 0.3 -------- = ------- = ------------ = 40 ( 10 ) Ex Ey 7500 –6 –6 1 1 1 1 ------ = ------------ = 133.33 ( 10 ) and ------ = ------------ = 400 ( 10 ) Ex Ey 7500 2500

Substituting the given strain values and the constants above in Eqs. 3.21 we obtain the following –6

–6

–6

ε xx = 133.33 ( 10 )σ xx – 40 ( 10 )σ yy = – 1000 ( 10 ) –6

–6

–6

ε yy = 400 ( 10 )σ yy – 40 ( 10 )σ xx = 500 ( 10 )

or

133.33σ xx – 40σ yy = – 1000

1

or

400σ yy – 40σ xx = 500

2

Solving Eqs. (1) and (2), we obtain σ xx = – 7.34ksi

σ xx = 7.34ksi ( C )

σ yy = 0.515ksi

σ yy = 0.515ksi ( T ) –6

τ xy = ( 1250 ) ( – 250 ) ( 10 ) = – 0.3125ksi

τ xy = – 0.3125ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.115 The strains at a point on free surface of an orthotropic material is given in each problem. Also given are the material constants. Using Equation (3.23), solve for the stresses σxx, σyy, and τxy. ε xx = – 750μ E x = 25000ksi

ε yy = – 250 μ

ν xy = 0.32

E y = 2000ksi

σxx= ?

Solution

γ xy = 400μ

σyy = ?

G xy = 1500ksi

τxy = ?

-----------------------------------------------------------ν yx ν xy –6 0.32 -------- = ------- = --------------- = 12.8 ( 10 ) 25000 Ex Ey 1 - = 40 ( 10 –6 ) and ----1 - = 500 ( 10 –6 ) 1- = -------------1- = --------------25000 2000 Ex Ey

Substituting the given strain values and the constants above in Eqs. 3.21 we obtain the following –6

–6

–6

–6

–6

–6

ε xx = 40 ( 10 )σ xx – 12.8 ( 10 )σ yy = – 750 ( 10 ) ε yy = 500 ( 10 )σ yy – 12.8 ( 10 )σ xx = – 250 ( 10 )

or

40σ xx – 12.8σ yy = – 750

1

or

500σ yy – 12.8σ xx = – 250

2

Solving Eqs. (1) and (2), we obtain σ xx = – 19.07ksi

σ xx = 19.07ksi ( C )

and σ yy = – 0.988k si

σ yy = 0.99ksi ( C ) –6

τ xy = ( 1500 ) ( 400 ) ( 10 ) = 0.6ksi

τ xy = 0.6ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.116 The strains at a point on free surface of an orthotropic material is given in each problem. Also given are the material constants. Using Equation (3.23), solve for the stresses σxx, σyy, and τxy.

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ε xx = 1500μ

ε yy = 800μ

E x = 53 GPa

Solution

January 2014

γ xy = 600μ ν xy = 0.25

E y = 18 GPa

σxx= ?

σyy = ?

G xy = 9 GPa

τxy = ?

-----------------------------------------------------------ν yx ν xy – 12 0.25 -------- = -------- = ------------------- = 4.717 ( 10 ) 9 Ex Ey 53 ( 10 ) – 12 – 12 1 1 1- = 1 ----------------------- = 18.868 ( 10 ) and ------ = ------------------- = 55.556 ( 10 ) 9 9 Ex E y 53 ( 10 ) 18 ( 10 )

Substituting the given strain values and the constants above in Eqs. 3.21 we obtain the following ε xx = 18.868 ( 10

– 12

ε yy = 55.556 ( 10

)σ xx – 4.717 ( 10

– 12

– 12

)σ yy – 4.717 ( 10

–6

)σ yy = 1500 ( 10 )

– 12

–6

)σ xx = 800 ( 10 )

or

18.868σ xx – 4.717σ yy = 1500MPa 1

or

55.556σ yy – 4.717σ xx = 800MPa

2

Solving Eqs. (1) and (2), we obtain σ xx = 84.90MPa ( T ) σ yy = 21.60MPa ( T ) 9 –6 3 N τ xy = 9 ( 10 ) ( 600 ) ( 10 ) = 5400 ( 10 ) ------2 m

τ xy = 5.4MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.117 The strains at a point on free surface of an orthotropic material is given in each problem. Also given are the material constants. Using Equation (3.23), solve for the stresses σxx, σyy, and τxy. ε xx = 1500μ

ε yy = – 750 μ

E x = 180 GPa

Solution

γ xy = – 450 μ ν xy = 0.28

E y = 15 GPa

σxx= ?

σyy = ?

G xy = 11 GPa

τxy = ?

-----------------------------------------------------------ν yx ν xy – 12 0.28 -------- = ------- = ---------------------- = 1.555 ( 10 ) 9 Ex Ey 180 ( 10 ) – 12 – 12 1 1 1 1 ------ = ---------------------- = 5.555 ( 10 ) and ------ = ------------------- = 66.667 ( 10 ) 9 9 Ex E y 180 ( 10 ) 15 ( 10 )

Substituting the given strain values and the constants above in Eqs. 3.21 we obtain the following ε xx = 5.555 ( 10

– 12

)σ xx – 1.555 ( 10

– 12

)σ yy = 1500 ( 10 )

–6

ε yy = 66.667 ( 10

– 12

)σ yy – 1.555 ( 10

– 12

)σ xx = – 750 ( 10 )

–6

1

or

5.555σ xx – 1.555σ yy = 1500MPa

or

66.667σ yy – 1.555σ xx = – 750MPa 2

Solving Eqs. (1) and (2), we obtain σ xx = 268.60MPa

σ xx = 268.6MPa ( T )

and σ yy = – 4.98MPa

σ yy = 4.98MPa ( C )

9 –6 3 N τ xy = 11 ( 10 ) ( – 450 ) ( 10 ) = – 4950 ( 10 ) ------2 m

τ xy = – 4.95M Pa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.118 Using Equation (3.23), show that on a free surface of an orthotropic material, the following relationships can be written.

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E y ( ε yy + ν xy ε xx ) σ yy = -----------------------------------------( 1 – ν yx ν xy )

E x ( ε xx + ν yx ε yy ) σ xx = -----------------------------------------( 1 – ν yx ν xy )

3.24

Solution -----------------------------------------------------------ν yx 1 ------ – ------Ex E y σ xx ε Eq. 3.21 in matrix form can be written as: = xx σ yy ε yy ν xy 1 – ------- -----Ex Ey ν ν

(1 – ν ν )

1 xy yx xy yx The determinant (D) of the matrix on the left hand side yields: D = -----------– ---------------- = -----------------------------Ex Ey

Ex Ey

Ex Ey

Using Kramer’s rule, we can write ν yx ε xx – ------Ey 1 ε yy -----Ey ⎛ ε xx ν yx ⎞ E x E y σ xx = ------------------------------ or σ xx = ⎜ ------- + -------- ε yy⎟ -------------------------- or D ⎝ Ey Ey ⎠ 1 – ν xy ν yx

Ex σ xx = -------------------------- ( ε xx + ν yx ε yy ) 1 – ν xy ν yx

Similarly 1----ε E x xx ν xy – ------- ε E x yy ⎛ ε yy ν xy ⎞ E x E y σ yy = ------------------------------ or σ yy = ⎜ ------- + -------- ε xx⎟ -------------------------- or D ⎝ Ex Ex ⎠ 1 – ν xy ν yx

Ey σ yy = -------------------------- ( ε yy + ν xy ε xx ) 1 – ν xy ν yx

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.119 A steel bar is axially loaded as shown. Determine the factor of safety for the bar if yielding is to be avoided. The normal yield stress for steel is 30 ksi. Use stress concentration factor chart in Section C.4 in Appendix. 0.5 in 10 kips

10 kips

5 in

1 in

Figure P3.119

Solution

σyield=30ksi

K=? ------------------------------------------------------------

10 - = 4ksi The nominal axial stress is: σ nom = ------------------( 5 ) ( 0.5 )

From Figure C.1, for the nominal stress and d--- = 1--- = 0.2 the approximate value of stress concentration h

5

3 + 3.25 )- = 3.125 factor is K gross = (----------------------2 σ max = K gross σ nom = ( 3.125 ) ( 4 ) = 12.5

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The factor of safety can be found as shown below σ yield 30 K = -------------= ---------- = 2.4 12.5 σ max

K = 2.4

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.120 as:

The stress concentration factor for a stepped flat tension bar with shoulder fillets was determined 2r 2r 2 2r 3 K conc = 1.970 – 0.384 ⎛⎝ -----⎞⎠ – 1.018 ⎛⎝ -----⎞⎠ + 0.430 ⎛⎝ -----⎞⎠ H H H

r r L-⎞ ⎛ ----⎞ > ⎛ 1 + 2 ---⎞ and ⎛ --> 5.784 – 1.89 -----⎞ . The above equation is valid only if ⎛⎝ H ⎝ H⎠ ⎝ d⎠ d )⎠ d⎠ ⎝

L

r

t P

P d

H r

Figure P3.120

The nominal stress is P/(dt). Make a chart for the stress concentration factor vs. (H/d) for the following values of (r/d): 0.2, 0.4, 0.6, 0.8, 1.0. Use of spread sheet is recommended. Solution ------------------------------------------------------------

The equation for Kconc can be re-written as: 2(r ⁄ d) 2(r ⁄ d) 2 2(r ⁄ d) 3 K conc = 1.970 – 0.384 ⎛⎝ -----------------⎞⎠ – 1.018 ⎛⎝ -----------------⎞⎠ + 0.430 ⎛⎝ -----------------⎞⎠ (H ⁄ d) (H ⁄ d) (H ⁄ d)

1

Using r/d as a parameter and Eq.1, the values of Kconc can be found for various values of H/d on a spread ----⎞ > ⎛ 1 + 2 --r-⎞ does not hold can be eliminated as shown in the sheet. Those values for which the inequality ⎛⎝ H d⎠ ⎝ d⎠

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table below. H/d

r/d=0.2

1.4

1.787

1.6

1.817

r/d=0.4

1.8

1.839

1.636

2.0

1.856

1.681

r/d=0.6

2.2

1.869

1.716

1.527

2.4

1.880

1.745

1.577

r/d=0.8

2.6

1.888

1.768

1.618

1.448

2.8

1.896

1.787

1.652

1.498

r/d=1.0

3.0

1.902

1.803

1.681

1.541

1.389

3.2

1.907

1.817

1.706

1.577

1.437

3.4

1.911

1.829

1.727

1.609

1.479

3.6

1.915

1.839

1.745

1.636

1.516

3.8

1.919

1.848

1.761

1.660

1.549

4.0

1.922

1.856

1.775

1.681

1.577

4.2

1.925

1.863

1.787

1.700

1.603

4.4

1.927

1.869

1.798

1.716

1.626

4.6

1.929

1.875

1.808

1.731

1.646

H/d

r/d=0.2

r/d=0.4

r/d=0.6

r/d=0.8

r/d=1.0

4.8

1.931

1.880

1.817

1.745

1.664

5.0

1.933

1.884

1.825

1.757

1.681

5.2

1.935

1.888

1.832

1.768

1.696

5.4

1.936

1.892

1.839

1.778

1.710

5.6

1.938

1.896

1.845

1.787

1.723

5.8

1.939

1.899

1.851

1.796

1.734

6.0

1.940

1.902

1.856

1.803

1.745

6.2

1.941

1.904

1.861

1.810

1.755

6.4

1.942

1.907

1.865

1.817

1.764

6.6

1.943

1.909

1.869

1.823

1.772

6.8

1.944

1.911

1.873

1.829

1.780

7.0

1.945

1.913

1.876

1.834

1.787

7.2

1.946

1.915

1.880

1.839

1.794

7.4

1.946

1.917

1.883

1.844

1.800

7.6

1.947

1.919

1.886

1.848

1.806

7.8

1.948

1.920

1.888

1.852

1.812

8.0

1.948

1.922

1.891

1.856

1.817

The values in the table can be plotted as shown in the graph below. r/d = 0.2 r/d = 0.4

r/d = 0.6 r/d = 0.8 r/d = 1.0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.121 Determine the maximum normal stress in the stepped flat tension bar shown in Figure P3.120 for the following data: P = 9 kips, H= 8 inches, d = 3 inches, t = 0.125 inches, and r = 0.625 inches. Solution σmax = ? -----------------------------------------------------------P- = ------------------------9 The nominal stress is: σ nom = ---- = 24ksi and ---r- = 0.625 ------------- = 0.07813 dt

( 3 ) ( 0.125 )

H

8

Substituting (r/H) in the expression for Kconc given in problem 3.110, we obtain K conc = 1.97 – 0.384 [ ( 2 ) ( 0.07813 ) ] – 1.018 [ ( 2 ) ( 0.07813 ) ] 2 + 0.430 [ ( 2 ) ( 0.07813 ) ] 3 = 1.8868 σ max = K conc σ nom = ( 1.8868 ) ( 24 ) = 45.283 or

σ max = 45.3ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.122 An aluminum stepped tension bar is to carry a load P = 56 kN. The normal yield stress of aluminum is 160 MPa. The bar has H = 300 mm, d = 100 mm, t =10 mm in Figure P3.120. For a factor of safety of 1.6, determine the minimum value r of the fillet radius if yielding is to be avoided.

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Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

σyield=160MPa K=1.6

P=56kN t=10mm

H=300mm r=?

January 2014

d=100mm

-----------------------------------------------------------σ yield 160 = --------- = 100MPa The allowable maximum stress is: σ max = -------------12

1.6

56 ( 10 3 )

N P - = ---------------------------- = 56 ( 10 6 ) ------- = 56MPa The nominal stress is: σ nom = ---2 ( 0.1 ) ( 0.01 )

dt

m

σ max H 100 The stress concentration factor is: K cone = ------------ = --------- = 1.785 and ---- = 3 σ nom

d

56

From the graphs in problem 3.110, for H/d=3 and Kconc=1.785, the approximate value of r/d=0.4 Thus r = 0.4d = ( 0.4 ) ( 100 ) = 40mm or r = 40mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.123

The stress concentration factor for a flat tension bar with U-shaped notches was determined as: 4r 2 4r 4r 3 K conc = 3.857 – 5.066 ⎛ -----⎞ + 2.469 ⎛ -----⎞ – 0.258 ⎛ -----⎞ ⎝ H⎠ ⎝ H⎠ ⎝ H⎠

t

P

2r d 2r

H

P

2r

Figure P3.123

The nominal stress is P/(Ht). Make a chart for the stress concentration factor vs. (r/d) for the following values of (H/d): 1.25, 1.50, 1.75, 2.0. Use of spread sheet is recommended. Solution ------------------------------------------------------------

The equation for Kconc can be re-written as: 4(r ⁄ d) 4(r ⁄ d) 2 4(r ⁄ d) 3 K conc = 3.857 – 5.066 ⎛⎝ -----------------⎞⎠ + 2.469 ⎛⎝ -----------------⎞⎠ – 0.258 ⎛⎝ -----------------⎞⎠ (H ⁄ d) (H ⁄ d) (H ⁄ d)

1

Using H/d as a parameter and Eq.1, the values of Kconc can be found for various values of r/d on a spread sheet as shown in the table below.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

r/d

H/d=1.25

H/d=1.50

H/d=1.75

H/d=2.00

r/d

H/d=1.25

H/d=1.50

H/d=1.75

H/d=2.00

0.05

3.109

3.225

3.310

3.375

0.18

1.709

1.966

2.173

2.341

1.632

1.890

2.101

2.274

0.06

2.974

3.109

3.208

3.284

0.19

0.07

2.843

2.996

3.109

3.195

0.20

1.558

1.818

2.032

2.209

1.489

1.749

1.966

2.146

0.08

2.718

2.886

3.012

3.109

0.21

0.09

2.597

2.780

2.917

3.024

0.22

1.424

1.683

1.901

2.084

1.363

1.619

1.839

2.024

0.10

2.480

2.677

2.825

2.940

0.23

0.11

2.368

2.577

2.735

2.859

0.24

1.306

1.558

1.778

1.966

1.252

1.501

1.720

1.909

0.12

2.261

2.480

2.648

2.780

0.25

0.13

2.158

2.387

2.563

2.702

0.26

1.203

1.445

1.664

1.854

1.157

1.393

1.610

1.801

0.14

2.060

2.296

2.480

2.626

0.27

0.15

1.966

2.209

2.400

2.552

0.28

1.114

1.343

1.558

1.749

1.076

1.297

1.509

1.699

1.041

1.252

1.461

1.651

0.16

1.876

2.125

2.322

2.480

0.29

0.17

1.790

2.044

2.246

2.410

0.30

The values in the table can be plotted as shown in the graph below. H/d=2.0 H/d=1.75 H/d=1.5

H/d=1.25

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.124 Determine the maximum normal stress in the flat tension bar shown in Figure P3.123 for the following data: P =150 kN , H = 300 mm, r = 15 mm, t = 5 mm. Solution σmax=? -----------------------------------------------------------N P150 ( 10 3 ) The nominal stress is: σ nom = ----= ------------------------------- = 100 ( 10 6 ) ------- = 100MPa 2 Ht

15- = 0.05 ---r- = -------H 300

or

( 0.3 ) ( 0.005 )

m

r 4 ---- = 0.2 H

Substituting(4r/H) in the expression for Kconc given in problem 3.113, we obtain K conc = 3.857 – 5.066 ( 0.2 ) + 2.469 ( 0.2 ) 2 – 0.258 ( 0.2 ) 3 = 2.9405 σ max = K conc σ nom = ( 2.9405 ) ( 100 ) = 294.05 or

σ max = 294MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.125 A steel tension bar with U-shaped notches of the type shown in Figure P3.123, is to carry a load P = 18 kips. The yield stress of steel is 30 ksi. The bar has H = 9 in., d= 6 in. t =0.25 in. For a factor of safety of 1.4, determine the value of r if yielding is to be avoided. Solution P=18kips σyield=30ksi H=9in. t=0.25in. d=6in., K=1.4 r=? ------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

σ yield 30 = ------- = 21.43ksi σ max = -------------K 1.4 P 18 σ nom = ------ = ----------------------- = 8ksi Ht ( 9 ) ( 0.25 ) σ max 21.43 K conc = ------------ = ------------- = 2.68 8 σ nom H 9 ---- = --- = 1.5 d 6

From the graph in problem 3.113, for H/d=1.5 and Kconc=2.68, the approximate value of r/d=0.1 ∴r = 0.1d = 0.6in or r = 0.6inch

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.126 An iron rim (α = 6.5 μ/oF) of diameter 35.98 inches is to be placed on a wooden cask of diameter 36 inches. Determine the minimum temperature increase needed to slip the rim onto the cask. Solution

α=6.5 μ/oF

di=35.98 in

ΔT=?

df=36in

------------------------------------------------------------

All stresses are zero. Thus, the total radial normal strain = thermal strain ( df ⁄ 2 ) – ( di ⁄ 2 ) 36 – 35.98 ε = -------------------------------------- = αΔT or ------------------------- = ( 6.5 ) ( 10 – 6 )ΔT or ΔT = 85.518°F or 35.98 ( di ⁄ 2 )

ΔT = 85.52°F

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.127

The temperature in steel (Es = 200 GPa, and αs = 12.0 μ/oC) aluminum (E = 72 GPa, and

α = 23.0 μ/oC is increased by 60oC. Determine the angle by which the pointer rotates from the vertical position.

Aluminum 50 mm Steel 450 mm

Figure P3.127

Solution

Es=200GPa

αs=12.0 μ/0C

αa1=23.0 μ/0C ΔT=60

Ea1=72GPa

θ=?

------------------------------------------------------------

The thermal strain and the corresponding deformation in each rod can be found as shown below ε a1 = α a1 ΔT = ( 23 ) ( 60 ) ( 10 – 6 ) = 1.38 ( 10 – 3 ) and δ a1 = ε a1 L a1 = ( 1.38 ) ( 10 – 3 ) ( 450 ) = 0.621mm ε s = α s ΔT = ( 12 ) ( 60 ) ( 10 –6 ) = 0.72 ( 10 – 3 ) and δ s = ε s L s = ( 0.72 ) ( 10 – 3 ) ( 450 ) = 0.324mm

The deformed geometry of the assembly can be drawn as shown below. δal Aluminum θ

50 mm

Steel δS

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

δ al – δ s 0.621 – 0.324 ∴ tan θ = ----------------- = --------------------------------- or 50 50

θ = 0.34°

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.128

Solve Problem 3.73 if the temperature decrease is 25o C. Use α = 11.7 μ/oC

Solution

ΔT=-25oC α=11.7 μ/oC {εxx=?, εyy=?, γxy=?, εzz=?, σzz=?} for plane stress and plane strain ------------------------------------------------------------

(a) Plane stress σ zz = 0 N m

N m

N m

Substituting σ xx = 100 ( 10 6 ) ------2- , σ yy = 150 ( 10 6 ) ------2- , σ zz = 0, E = 200 ( 10 9 ) ------2- and ν = 0.32 into Eq’s(3-27a), (3-27b), and(3-27c) we obtain the following: [ 100 – 0.32 ( 150 ) ] ( 10 6 ) ε xx = ---------------------------------------------------------- + ( 11.7 ) ( – 25 ) ( 10 –6 ) = – 32.5 ( 10 – 6 ) or 200 ( 10 9 )

ε xx = – 32.5 μ

[ 150 – 0.32 ( 100 ) ] ( 10 6 )- + 11.7 ( – 25 ) ( 10 – 6 ) = 297.5 ( 10 – 6 ) or ε yy = --------------------------------------------------------200 ( 10 9 )

ε yy = 297.5μ

0.32 ( 100 + 150 ) ( 10 6 )- + 11.7 ( – 25 ) ( 10 –6 ) = – 692.5 ( 10 – 6 ) or ε zz = –-------------------------------------------------------200 ( 10 9 )

ε zz = – 692.5μ

Shear strain is same as in problem 3.63

γ xy = – 1650μ

(b) Plane strain ε zz = 0 σ

– 0.32 ( 100 + 150 ) ( 10 6 )

N m

zz From Eq(3-27c) 0 = ------------------------------------------------------------------- + 11.7 ( – 25 ) ( 10 – 6 ) or σ zz = 138.5 ( 10 6 ) ------- or 2

200 ( 10 9 )

σ zz = 138.5MPa ( T )

From Eq’s(3-27a) and(3-27b) 100 – 0.32 ( 150 + 138.5 ) ] ( 10 6 )- + ( 11.7 ) ( – 25 ) ( 10 –6 ) = – 254.1 ( 10 – 6 ) or ε xx = [----------------------------------------------------------------------------200 ( 10 9 )

ε xx = – 254.1 μ

[ 150 – 0.32 ( 100 + 138.5 ) ] ( 10 6 ) ε yy = ------------------------------------------------------------------------------ + ( 11.5 ) ( – 25 ) ( 10 –6 ) = 75.9 ( 10 – 6 ) or 200 ( 10 9 )

ε yy = 75.9μ

Shear strain value as in part(a)

γ xy = – 1650μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.129

Solve Problem 3.74 if the temperature increase is 50o C. Use α = 23.6 μ/oC

Solution

ΔT=50o C. α=23.6μ {εxx=?, εyy=?, γxy=?, εzz=?, σzz=?} for plane stress and plane strain ------------------------------------------------------------

(a) Plane stress σ zz = 0 N m

N m

N m

Substituting σ xx = – 225 ( 10 6 ) ------2- , σ yy = 125 ( 10 6 ) ------2- , σ zz = 0, E = 70 ( 10 9 ) ------2- and ν = 0.25 into Eq’s(3-27a), (3-27b), and(3-27c) we obtain the following: [ – 225 – 0.25 ( 125 ) ] ( 10 6 ) ε xx = --------------------------------------------------------------- + ( 23.6 ) ( 50 ) ( 10 – 6 ) = – 2481 ( 10 – 6 ) or 70 ( 10 9 )

ε xx = – 2481μ

[ 125 – 0.25 ( 225 ) ] ( 10 6 ) ε yy = ---------------------------------------------------------- + 23.6 ( 50 ) ( 10 – 6 ) = 3769 ( 10 –6 ) or 70 ( 10 9 )

ε yy = 3769μ

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

– 0.25 ( – 225 + 125 ) ( 10 6 ) ε zz = ------------------------------------------------------------- + 23.6 ( 50 ) ( 10 – 6 ) = 1537 ( 10 – 6 ) or 70 ( 10 9 )

ε zz = 1537μ

Shear strain is same as in problem 3.64

γ xy = 5357μ

(b) Plane strain ε zz = 0 σ

– 0.25 ( – 225 + 125 ) ( 10 6 )

N m

zz - + 23.6 ( – 50 ) ( 10 – 6 ) or σ zz = – 107.6 ( 10 6 ) ------- or From Eq(3-27c): 0 = -----------------------------------------------------------------------2

70 ( 10 9 )

σ zz = 107.6MPa ( C )

From Eq’s(3-27a) and(3-27b) [ – 225 – 0.25 ( 125 + 107.6 ) ] ( 10 6 ) ε xx = ---------------------------------------------------------------------------------- + ( 23.6 ) ( 50 ) ( 10 – 6 ) = – 2096.4 ( 10 – 6 ) or 70 ( 10 9 )

ε xx = – 2096 μ

[ 125 – 0.25 ( – 225 – 107.6 ) ] ( 10 6 ) ε yy = ---------------------------------------------------------------------------------- + ( 23.6 ) ( 50 ) ( 10 – 6 ) = 4153.6 ( 10 –6 ) or 70 ( 10 9 )

ε yy = 4154μ

Shear strain value as in part(a)

γ xy = 5357μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.130

Solve problem 3.81 if the temperature increase is 40o F-. Use α = 6.5 μ/oF

Solution

ΔT=40oF

α=6.5μ/οF

σxx=?, σyy=?, τxy=?, εzz=?

for plane stress

------------------------------------------------------------

For Plane stress: σ zz = 0 . Eqs. (3-27a) and (3-27b) can be rewritten with σzz=0 σ xx – νσ yy = E ( ε xx – αΔT ) = ( 30, 000 ) ( – 800 – 6.5 ( 40 ) ) ( 10 – 6 ) or σ xx – νσ yy = – 31.8ksi

1

6.5 ( 40 ) ) ( 10 – 6 )

σ yy – νσ xx = E ( ε yy – αΔT ) = ( 30, 000 ) ( – 1000 – or σ yy – νσ xx = – 37.8ksi

Solving Eq(1) and (2) with ν=0.3, we obtain:

2

σ xx = – 47.4ksi and σ yy = – 52.02ksi or σ xx = 47.4ksi ( C ) and σ yy = 52.02ksi ( C )

0 – 0.3 ( – 47.4 – 52.02 )- + ( 40 ) ( 6.5 ) ( 10 – 6 ) = 1254.2 or From Eq(3-27c): ε zz = -------------------------------------------------------

ε zz = 1254μ

Shear stress is same as in problem 3.81.

τ xy = – 5.77ksi

30, 000

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.131

Solve problem 3.82 if the temperature decrease is 100o F. Use α = 12.8 μ/oF

Solution

ΔT=-100

α=12.8μ/oF

σxx=?, σyy=?, τxy=?, εzz=?

for plane stress

------------------------------------------------------------

For plane stress σ zz = 0 Eqs. (3-27a) and (3-27b) can be rewritten with σzz=0 σ xx – νσ yy = E ( ε xx – αΔT ) = ( 10, 000 ) ( 1500 – 12.8 ( – 100 ) ) ( 10 –6 ) or σ xx – νσ yy = 27.8ksi σ yy – νσ xx = E ( ε yy – αΔT ) = 10, 000 ( – 1200 – 12.8 ( – 100 ) ) ( 10 – 6 ) or σ yy – νσ xx = 0.8ksi

Solving Eq’s(1) and (2) with ν=0.282 we obtain:

1

2

σ xx = 30.45ks ( T ) and σ yy = 9.39ksi ( T )

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

– 0.282 ( 30.45 + 9.39 ) From Eq(3-27c): ε zz = 0--------------------------------------------------------- + 12.8 ( – 100 ) ( 10 –6 ) = – 2403 ( 10 –6 ) or ε zz = – 2403μ ( 10, 000 )

τ xy = – 3.9ksi

Shear stress is same as in problem 3.82

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.132

Solve problem 3.83 if the temperature decrease is 75 o C. Use α = 26.0 μ/oC

Solution

ΔT=-75oF

α=26 μ/oC

σxx=? σyy=? τxy=?

εzz=?

for plane stress

------------------------------------------------------------

For plane stress σ zz = 0 Eq’s(3-27a) and (3-27b) can be rewritten with σzz=0 σ xx – νσ yy = E ( ε xx – αΔT ) = ( 36 ) ( 10 9 ) ( – 2000 – 26 ( – 75 ) ) ( 10 – 6 )

σ xx – νσ yy = – 1.8MPa

1

σ yy – νσ xx = E ( ε yy – αΔT ) = 36 ( 10 9 ) ( – 2000 – 26 ( – 75 ) ) ( 10 –6 ) or σ yy – νσ xx = 142.2MPa Solving Eqs. (1) and (2) with ν=0.2 we obtain: σ xx = 27.75MPa and σ yy = 147.75MPa

2

or

σ xx = 27.8MPa ( T ) and σ yy = 147.8MPa ( T ) – 0.2 ( 27.75 + 147.75 ) ( 10 6 -) + 26 ( – 75 ) ( 10 –6 ) = – 2925 ( 10 –6 ) or ε = – 2925μ From Eq(3-27c) ε zz = 0----------------------------------------------------------------------zz ( 36 ) ( 10 9 )

τ xy = 18MPa

Shear stress is same as in problem 3.83

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.133 A plate (E=30,000ksi; ν = 0.25 ; α = 6.5x10-6/oF) cannot expand in the y direction and can expand at most by 0.005 inch in the x direction as shown. Assuming plane stress determine the average normal stresses in the x and y direction due to a uniform temperature increase of 100oF. y

5 in x 10 in

0.005 in

Figure P3.133

Solution

E=30,000ksi

ν=0.25

δx=0.005

ΔT=100 F o

α=6.5 μ/oF

εyy=0,

σxx=?

σyy=?

-----------------------------------------------------------δ The normal strain in the x-direction is: ε xx = -----x- = 0.005 ------------- = 0.005 = 500 ( 10 –6 ) 10

10

Assuming plane stress σ zz = 0 , Eq’s(3-27a) and (3-27b) can be rewritten as σ xx – νσ yy = E ( ε xx – αΔT ) = ( 30, 000 ) [ 500 – ( 6.5 ) ( 100 ) ] ( 10 –6 )

or

σ yy – νσ xx = E ( ε yy – αΔT ) = 30, 000 [ 0 – ( 6.5 ) ( 100 ) ] ( 10 – 6 ) or Solving Eqs. (1) and (2) we obtain: σ xx = – 10ksi and σ yy = – 22ksi or

σ xx – 0.25σ yy = – 4.5ksi σ yy – νσ xx = – 19.5ksi

1 2

σ xx = 10ksi ( C ) and σ yy = 22ksi ( C )

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

3.134 Derive the following relation of normal stresses in terms of normal strain from Equations (3.28a), (3.28b), and (3.28c). E EαΔT σ xx = [ ( 1 – ν )ε xx + νε yy + νε zz ] -------------------------------------- – -------------------( 1 – 2ν ) ( 1 + ν ) ( 1 – 2ν ) E EαΔT σ yy = [ ( 1 – ν )ε yy + νε zz + νε xx ] -------------------------------------- – -------------------( 1 – 2ν ) ( 1 + ν ) ( 1 – 2ν )

3.29

E EαΔT σ zz = [ ( 1 – ν )ε zz + νε xx + νε yy ] -------------------------------------- – -------------------( 1 – 2ν ) ( 1 + ν ) ( 1 – 2ν )

Solution ------------------------------------------------------------

Adding Equations (3.28a), (3.28b), and (3.28c) we obtain: 1 – 2ν ε xx + ε yy + ε zz = --------------- ( σ xx + σ yy + σ zz ) + 3αΔT E E 3 ( EαΔT ) ∴σ xx + σ yy + σ zz = --------------- ( ε xx + ε yy + ε zz ) – -----------------------1 – 2ν 1 – 2ν σ

1

ν E

xx - – --- ( σ xx + σ yy + σ zz ) + αΔT . Substituting Eq(1) we Equation (3.28c) can be written as ε xx = ( 1 + ν ) --------

E

obtain: 3ν 1+ν ν ε xx = ------------ σ xx – --------------- ( ε xx + ε yy + ε zz ) + --------------- αΔT + αΔT or 1 – 2ν E 1 – 2ν ν 3ν + 1 – 2ν 1 + νν ν ----------σ = ε xx ⎛ 1 + ---------------⎞ + --------------- ε yy + --------------- ε zz – αΔT ⎛ ----------------------------⎞ or ⎝ ⎠ ⎝ 1 – 2ν 1 – 2ν ⎠ E xx 1 – 2ν 1 – 2ν E EαΔT σ xx -------------------------------------- [ ( 1 – ν )ε xx + νε yy + νε zz ] – --------------2 ( 1 + ν ) ( 1 – 2ν ) 1 – 2ν 1+ν ν Eq(3-27b) can be written as: ε yy = ------------ σ yy – --- ( σ xx + σ yy + σ zz ) + αΔT . Substituting Eq.(1) and solving E E

for σyy we obtain: E σ yy = -------------------------------------- [ νε xx + ( 1 – ν )ε yy + νε zz ] – EαΔT --------------( 1 + ν ) ( 1 – 2ν ) 1 – 2ν

3

1+ν ν Eq(3-27c) can be written as ε zz = ------------ σ zz – --- ( σ xx + σ yy + σ zz ) + αΔT . Substituting Eq.(1) and solving for E

E

σzz we obtain E EαΔT σ zz = -------------------------------------- [ νε xx + νε yy + ( 1 – ν )ε zz ] – --------------( 1 + ν ) ( 1 – 2ν ) 1 – 2ν

4

Eqs. (2),(3) and (4) are the same as Eq.(3-31)

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.135

For a point in plane stress show E EαΔT σ xx = [ ε xx + νε yy ] ------------------- – ---------------2 ( (1 – ν ) 1 – ν)

E EαΔT σ yy = [ ε yy + νε xx ] ------------------- – ---------------2 ( (1 – ν ) 1 – ν)

3.30

Solution ------------------------------------------------------------

For plane stress(σzz=0) Eqs. (3-27a) and (3-27b) can be written as σ xx – νσ yy = E ( ε xx – αΔT ) σ yy – νσ xx = E ( ε yy – αΔT )

1 2

Multiplying Eq.(2) by ν and adding it to Eq.(1) we obtain: σ xx ( 1 – ν 2 ) = E [ ε xx + νε yy – αΔT ( 1 + ν ) ] or

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

E EαΔT σ xx = -------------- ( ε xx + νε yy ) – --------------1–ν 1 – ν2

3

Multiplying Eq.(1) by ν and adding to Eq.(2) we obtain: σ yy ( 1 – ν 2 ) = E [ νε xx + ε yy – αΔT ( 1 + ν ) ] or E EαΔT σ yy = -------------- ( ε yy + νε xx ) – --------------2 1–ν 1–ν

4

Eqs.(3) and (4) are same as Eq(3-32)

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.136

For a point in plane stress show (1 + ν) ν ε zz = – ⎛ ------------⎞ ( ε xx + ε yy ) + ----------------- αΔT ⎝ 1 – ν⎠ (1 – ν)

3.31

Solution ------------------------------------------------------------

Adding Eqs.(3) and(4) an problem 3.124, we obtain E ( EαΔT )- or σ xx + σ yy = -------------- [ ( 1 + ν )ε xx + ( 1 + ν )ε yy ] – 2----------------------1–ν 1 – ν2 E 2 ( EαΔT ) σ xx + σ yy = ------------ ( ε xx + ε yy ) – -----------------------1–ν 1–ν

5

ν E

Eq.(3-27c) for plane stress can be written as: ε zz = – --- ( σ xx + σ yy ) + αΔT . Substituting Eq.(1) ν 2ν ε zz = – ------------ ( ε xx + ε yy ) + ⎛ ------------ + 1⎞ αΔT or ⎝1 – ν ⎠ 1–ν (1 + ν) ν ε zz = – ------------ ( ε xx + ε yy ) + ----------------- αΔT 1–ν 1–ν

6

Eq.(2) is the same as Eq.(3-33)

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.137 A machine component is made from a steel alloy that has an S-N curve shown in Figure 3.36. Estimate the service life of the component if the peak stress is reversed at the rate shown. (a) 40 ksi at 200 cycles per minute. (b) 36 ksi at 250 cycles per minute. (c) 32 ksi at 300 cycles per minute. Solution

T=? for the three cases ------------------------------------------------------------

(a) From figure 3.34 for steel at 40ksi, the number of cycles to failure is n=400,000 , 000- = 2, 000 minutes or T = 400 -------------------200

T = 33.33hours

(b) At 36ksi, the number of cycles to failure is n=2(106) ( 10 6 ) = 8000 minutes T = 2---------------250

T = 133.33hours

(c) The stress value of 32 ksi is below the endurance limit and the component won’t fail. T = ∞

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.138 A machine component is made from a aluminum alloy that has an S-N curve shown in Figure 3.36. What should be the maximum permissible peak stress in MPa for the following situations: (a) 17 hours of service at 100 cycles per minute. (b) 40 hours of service at 50 cycles per minute. (c) 80 hours of service at 20 cycles per minute.

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Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

σmax= ? for the three cases ------------------------------------------------------------

(a) T=17 hours = 1020 minutes and n = number of cycles=(100)(1020)=102,000 From figure 3-34 for aluminum, the peak stress value at 100,000 is 182 MPa and the peak stress value at 2000 100, 000 σ max = 182MPa

200,000 is 168 MPa. By linear interpolation we obtain a peak stress= 182 – --------------------- ( 14 ) = 181.7 or (b) T=40 hours =2400 minutes n=(2400)(50)=120,000 20, 000 100, 000

By linear interpolation, the peak stress value= 182 – --------------------- ( 14 ) = 177.2 or

σ max = 177MPa

(c)T=80 hours =4800 min and n=(4800)(20)=96,000 cycles The peak stress value between 96,000 cycles &100,000 cycles changes slightly and is approximated at 183MPa. σ max = 183MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.139 A uni-axial stress acts on an aluminum plate with a hole as shown. The aluminum has an S-N curve shown in Figure 3.36. Predict the number of cycles the plate could be used if d = 3.2 inches and the far field stress σ = 6 ksi. σ

σ 8 inch

d

Figure P3.139

Solution

σ=6ksi

d=3.2

H=8

n=?

-----------------------------------------------------------d- = 3.2 From Figure C.1, for --------- = 0.4 , the stress concentration factor Kgross=3.75 H

8

The maximum normal stress is: σ max = K gross σ = ( 3.75 ) ( 6 ) = 22.5ksi From figure 3.34 for alumina, the number of cycles at 22.5ksi are estimated at 400,000 n = 400, 000cycles

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.140 A uni-axial stress acts on an aluminum plate with a hole as shown. The aluminum has an S-N curve shown in Figure 3.36. Determine the maximum diameter of the hole to the nearest 1/8th of an inch, if the predicted service life of half-million cycle is desired for a uniform far field stress σ=6 ksi. σ

σ 8 inch

d

Figure P3.140

Solution

n=500,000 cycles

σ= 6 ksi

H=8

dmax= ? to the nearest 1/8 inch

-----------------------------------------------------------From figure 3.34 for aluminum, the peak stress at 500,000 cycles is σ max = 22ksi

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January 2014

σ max 22 = ------ = 3.67 The stress concentration factor is: K gross = -----------σ

6

d d - = 0.34 for Kgross=3.5 and ---- = 0.4 for Kgross=3.75, by linear interpolation the value From figure C.1, --H H 0.4 – 0.34 of ---d- = 0.34 + ------------------------ ( 3.67 – 3.5 ) or ---d- = 0.38 or d = ( 0.38 ) ( 8 ) = 3.04 or H

3.75 – 3.5

H

d max = 3.0 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.141 A uni-axial stress acts on an aluminum plate with a hole as shown. The aluminum has an S-N curve shown in Figure 3.36. Determine the maximum far field stress σ if the diameter of the hole is 2.4 inch and a predicted service life of three-quarters of a million is desired. σ

σ 8 inch

d

Figure P3.141

Solution

d=2.4 H=8

n=750,000 cycles

σ=?

------------------------------------------------------------

From figure 3.34 for aluminum, the peak stress at 750,000 cycles is estimated as σmax=21ksi 3.25 + 3.5 From Figure C.1, for ---d- = 0.3 the stress concentration factor is estimated K gross = ⎛⎝ ------------------------⎞⎠ = 3.375 2 H σ

21 max The nominal normal stress is: σ = --------------- = ------------- = 6.222 K gross

3.375

σ = 6.2 ksi(T)

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.142 Bronze has a yield stress of σyield = 18 ksi in tension, yield strain of εyield = 0.0012, ultimate stress of σult = 50 ksi and the corresponding ultimate strain of εult = 0.50. Determine the material constants and plot the resulting stress-strain curve for (a) elastic-perfectly plastic model (b) linear strain hardening model (c) and the non-linear power law model. Solution

σyield = 18 ksi εyield = 0.0012 σult = 50 ksi Constants in the three material models=?

εult = 0.50

------------------------------------------------------------

We have coordinates of three points on the curve: Po (σ0 = 0.00, ε0 = 0.000), P1 (σ1 = 18.0, ε1 = 0.0012), and P2 (σ2 = 50, ε2 = 0.5). Using this data we can find the various constants in the material models. (a) The modulus of elasticity E is slope between points Po and P1 and can be found as shown below. σ1 – σ0 18 - = ---------------- = 15, 000 ksi E 1 = ----------------0.0012 ε1 – ε0

After yield stress, the stress is a constant. The stress strain behavior can be written as shown below. 15, 000ε ksi ε ≤ 0.0012 σ = ⎛ ⎝ 18 ksi 0.0012 ≤ ε ≤ 0.5

1

(b) In the linear strain hardening model the slope of the straight line before yield stress is as calculated in part (a). After the yield stress, the slope of the line can be found from the coordinates of point P1 and P2 . σ2 – σ1 50 – 18 E 2 = ----------------- = ------------------------------ = 64.15 ksi 0.5 – 0.0012 ε2 – ε1

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

The stress strain behavior can be written as shown below. ε ≤ 0.0012

15, 000ε ksi σ = ⎛ ⎝ ( 18 + 64.15 ( ε – 0.0012 ) ) ksi

0.0012 ≤ ε ≤ 0.5

2

n

(c) The two constants E and n in σ = Eε can be found by substituting the coordinates of the two point P1 and P2 , to generate the following two equations: 18 = E ( 0.0012 ) 50 = E ( 0.5 )

n

3

n

4

Dividing Eq. 4 by Eq. 3 and taking the logarithm of both sides, we can solve for ‘n’ as shown below: 50 0.5 n ln ⎛ ----------------⎞ = ln ⎛ ------⎞ ⎝ 18⎠ ⎝ 0.0012⎠

or

n ln ( 416.67 ) = ln ( 2.778 )

or

n = 0.1694

50 Substituting the value of n in Eq. 4, we can obtain the value of E as E = ------------------------ = 56.2 ksi ( 0.5 )

0.1694

We can now write the stress-strain equations for the power law model as shown below: 0.1694

⎛ 56.2ε ksi σ = ⎜ 0.1694 ⎝ – 56.2 ( – ε ) ksi

ε≥0 ε0 ε0

y0

The plots of normal strain and stress are as shown below. y (mm)

y (mm) 150

150 125

125 12.5 15

15 12.5

εxx (μ)

250 259.3 σxx(MPa)

259.3 250

125

125

150

150

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3.155 A rectangular beam has the dimensions as shown in Figure P3.152. The normal strain due to bending about the z-axis was found to vary as εxx = – 0.01y with y measured in meters. For the given material model in the problem. Write an expressions for normal stress σxx as a function of y and plot the σxx distribution across the cross-section. The beam material has a stress strain relationship given by 2

σ = ( 200ε – 2, 000ε ) MPa

Solution

εxx=-0.01y where y is in meters

σxx = f(y) = ?

------------------------------------------------------------

As the material behavior in tension and compression is the same, we have: 2

⎧ ( 200ε – 2, 000ε )MPa σ xx = ⎨ ⎩ ( 200ε + 2, 000ε 2 )MPa

ε>0 ε0

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 3

January 2014

The plots of normal strain and stress are as shown below y (mm) 150

y (mm) 150 125 12.5 15 15 12.5

125 εxx (μ)

246.9 304.5

304.5246.9

125 150

σxx (kPa)

125 150

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

4.1 Aluminum (E = 30,000 ksi) bars are welded to rigid plates as shown. All bars have an area of cross-section of 0.5 in2. Due to the applied forces the rigid plates at A, B, C, and D displaced in the xdirection without rotating by the amount given below. Determine the applied forces F1, F2, F3, and F4. F1

u A = – 0.0100 in

u B = 0.0080 in

u C = – 0.0045 in

u D = 0.0075 in

A

Solution

E = 30,000 ksi

B x

F1

Figure P4.1

A = 0.5in

F1 = ?

F4 C

F2

36 in

2

F3

F2

D

F3

50 in

F4 36 in

F2 = ?

F3 = ? F4 = ?

-----------------------------------------------------------1. Deformation calculations u D – u C = 0.0075 – ( – 0.0045 ) = 0.0120 in

u C – u B = – 0.0045 – 0.0080 = – 0.0125 in

u B – u A = 0.0080 – ( – 0.0100 ) = 0.0180 in

2. Strain calculations uD – uC ε DC = ------------------ = 0.012 ------------- = 0.3333 ( 10 – 3 ) 36 xD – xC

uC – uB – 0.0125- = – 0.2500 ( 10 – 3 ) ε CB = -----------------= -----------------50 xC – xB

uB – uA 0.0180 ε BA = ------------------ = ---------------- = 0.5000 ( 10 – 3 ) 36 xB – xA

3. Stress calculations σ DC = ( 30 ) ( 10 3 ) ( 0.3333 ( 10 –3 ) ) = 10

or

σ DC = 10 ksi ( T )

σ CB = ( 30 ) ( 10 3 ) ( – 0.25 ) ( 10 –3 ) = – 7.5

or

σ CB = 7.5 ksi ( C )

σ BA = ( 30 ) ( 10 3 ) ( 0.50 ) ( 10 – 3 ) = 15

or

σ BA = 15 ksi ( T )

4. Internal force calculations N DC = ( 10 ) ( 0.5 ) = 5 kips ( T )

N CB = ( 7.5 ) ( 0.5 ) = 3.75 kips ( C )

N BA = ( 15 ) ( 0.5 ) = 7.5 kips ( T )

5. External force calculations By making imaginary cuts through segments AB, BC, and CD the following free body diagrams can be obtained. (a)

F1

NAB A

(b)

NBC A

NAB

F1

F2

F1 B

NBC F1

(c)

F4

NCD NCD

D

F2

F4

By equilibrium of forces in figure (a) 2F 1 = 2N BA

By equilibrium of forces in figure (b) – 2F 1 + 2F 2 – 2N CB = 0 or F 2 = F 1 + N CD = 7.5 + 3.75

F 1 = 7.5 kips F 2 = 11.25 kips

By equilibrium of forces in figure (c) 2F 4 = 2N DC

By equilibrium of forces on the entire bar – 2F 1 + 2F 2 – 2F 3 + 2F 4 = 0 or F 3 = F 2 + F 4 – F 1

F 4 = 5 kips F 3 = 8.75 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.2 Brass bars between section A and B, aluminum bars between B and C, and steel bars between section C and D, are welded to rigid plates. The rigid plates displace in the x-direction without rotating by the

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following amounts: uB= -1.8 mm, uC = 0.7 mm, and uD = 3.7 mm. Determine the external forces F1, F2 and F3using the properties given in the adjoining Table. x Brass

Aluminum

F1

Steel

Modulus of Elasticity

70 GPa

100 GPa

200 GPa

Diameter

30 mm

25 mm

20 mm

A

F2

F3

C

B

F3

F2 2m

F1 2.5 m

1.5 m

D

Figure P4.2

Solution

F1 = ?

F2 = ?

F3 = ?

-----------------------------------------------------------1. Deformation calculations u D – u C = 3.7 – 0.7 = 3.0 mm

u C – u B = 0.7 + 1.8 = 2.5 mm

u B – u A = – 1.8 – 0 = – 1.8 mm

2. Strain calculations uD – uC 3 ε DC = ------------------ = ---------------- = 1.5 ( 10 –3 ) xD – xC 2 ( 10 3 )

uC – uB 2.5 ε CB = -----------------= --------------------- = 1.0 ( 10 – 3 ) xC – xB 2.5 ( 10 3 )

uB – uD – 1.8 - = ------------------------ = – 1.2 ( 10 – 3 ) ε BD = -----------------xB – xD 1.5 ( 10 –3 )

3. Stresses 2

(T)

2

(T)

σ DC = E DC ε DC = ( 200 ) ( 10 9 ) ( 1.5 ) ( 10 – 3 ) = 300 ( 10 6 ) N ⁄ m σ BC = E BC ε BC = ( 100 ) ( 10 9 ) ( 1.0 ) ( 10 – 3 ) = 100 ( 10 6 ) N ⁄ m σ AB = E AB ε AB = ( 70 ) ( 10 9 ) ( – 1.2 ) ( 10 – 3 ) = – 84 ( 10 6 )

2

or σ AB = 84 ( 10 6 ) ( N ⁄ m ) ( C )

4. Internal forces π N DC = σ DC A DC = ( 300 ) ( 10 6 ) --- ( 0.02 ) 2 = 94.25 ( 10 3 ) N = 94.25 kN ( T ) 4 π N BC = σ BC A BC = ( 100 ) ( 10 6 ) --- ( 0.025 ) 2 = 49.09 ( 10 3 ) N = 49.09 kN ( T ) 4 π N AB = σ AB A AB = ( 84 ) ( 10 6 ) --- ( 0.03 ) 2 = 59.38 ( 10 3 ) N = 59.38 kN ( C ) 4

5. External forces F3 (a) NCD

(b)

D NCD

F2

NBC C

F3

NBC

F3

(c)

NAB B

D F2

F1

F3

NAB

F2 C

F1

F3 D

F2

F3

By equilibrium of forces in figure (a) 2F 3 = 2N CD

By equilibrium of forces in figure (b) 2F 3 – 2F 2 – 2N BC = 0 or F 2 = F 3 – N BC = 94.25 – 49.09 = 45.16 or

F 3 = 94.3 kN F 2 = 45.2 kN

By equilibrium of forces in figure (c) 2F 3 – 2F 2 – 2F 1 + 2N AB = 0 or F 1 = F 3 – F 2 + N AB = 94.25 – 45.16 + 59.38 = 108.47 F 1 = 108.5 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.3 Ends of four circular steel (E = 200 GPa) bars are welded to a rigid plate. The other ends of the bars are built into the wall. Due to the action of the external force F the rigid plate moves to the right by

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

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0.1 mm without rotating. If the bars have a diameter of 10 mm, determine the applied force F. F

Rigid Plate

F

2.5 m

1.5 m

Figure P4.3

Solution

E = 200 GPa

d = 10 mm

δ = 0.1 mm to the right

F=?

-----------------------------------------------------------1. Deformation calculation. Fig. (a) shows an exaggerated deformed geometry. (a)

δA = δ

(b)

NA

F

NB NB

NA

F δB = δ

From the deformed geometry in figure (a) we obtain the following. δ A = δ = 0.1 mm extension δ B = δ = 0.1 mm contraction

2. Strain calculation δA 0.1 ε A = ------ = --------------------- = 40 ( 10 – 6 ) extension LA 2.5 ( 10 3 ) δB 0.1 ε B = ------ = --------------------- = 66.67 ( 10 – 6 ) contraction LB 1.5 ( 10 3 )

3. Stress calculation σ A = E A ε A = ( 200 ) ( 10 9 ) ( 40 ) ( 10 – 6 ) = 8.0 ( 10 6 ) ( N ⁄ m 2 ) ( T ) σ B = E B ε B = ( 200 ) ( 10 9 ) ( 66.67 ) ( 10 – 6 ) = 13.333 ( 10 6 ) ( N ⁄ m 2 ) ( C )

4. Internal force calculation π A A = A B = --- ( 0.01 ) 2 = 78.54 ( 10 – 6 ) m 2 4 N A = σ A A A = ( 8.0 ) ( 10 6 ) ( 78.54 ) ( 10 – 6 ) = 628.31 N ( T ) N B = σ B A B = ( 13.333 ) ( 10 6 ) ( 78.54 ) ( 10 – 6 ) = 1047.2 N ( T )

5. External force calculation. Fig. (b) shows the free body diagram of the rigid plate. By force equilibrium 2F – 2N A – 2N B = 0 or F = N A + N B = 1675.5

F = 1676 N

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.4 Rigid plates are securely fastened to bars A and B as shown. A gap of 0.02 inch exists between the rigid plates before the forces are applied. After the application of the forces the normal strain in bar A was found to be 500 μ. The area of cross-section and Modulus of Elasticity for the each bar are as follows: AA= 1 in2, EA=10,000 ksi, AB= 0.5 in2, EB=30,000 ksi. Determine the applied forces F, assuming that the rigid plates do not rotate.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

Bar A

F

January 2014

Rigid Plates Bar B Bar B

Bar A

F 60 in

24 in 0.02 in

Figure P4.4

Solution

AA = 1 in2 ε A = 500μ

EA = 10,000ksi F=?

AB = 0.5in2

EB = 30,000ksi

-----------------------------------------------------------1. Deformation calculation. Fig. (a) shows an exaggerated deformed geometry. (a)

δA

(b) NA

F

NB

NA δB

NB

F

0.02 in

δ A = ε A L A = ( 500 ) ( 10 –6 ) ( 60 ) = 0.03 in

From deformed shape in Fig. (a) δ B = δ A – 0.02 = 0.01 in contraction 2. Strain calculation ε A = 500 ( 10 –6 ) extension δB 0.01 ε B = ------ = ---------- = 416.7 ( 10 – 6 ) contraction LB 24

3. Stress calculations σ A = E A ε A = 10 ( 10 3 ) ( 500 ) ( 10 – 6 ) = 5 ksi ( T ) σ B = E B ε B = 30 ( 10 3 ) ( 416.7 ) ( 10 – 6 ) = 12.5 ksi ( C )

4. Internal force calculations N A = σ A A A = ( 5 ) ( 1 ) = 5.0 kips ( T ) N B = σ B A B = ( 12.5 ) ( 0.5 ) = 6.25 kips ( C )

5. External force calculation. Fig.(b) shows the free body diagram of the rigid plate. By force equilibrium 2F – 2N A – 2N B = 0 or F = N A + N B = 5.0 + 6.25 or F = 11.25 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.5 The strain at a cross-section shown in Figure P4.5 of an axial rod is assumed to have a uniform value of ε xx = 200 μ . (a) Plot the stress distribution across the laminated cross-section. (b) Determine the equivalent internal axial force N and its location from the bottom of the cross-section. Use

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January 2014

Ealuminium = 100 GPa, Ewood = 10 GPa, and Esteel = 200 GPa. 80 mm

y

Aluminum

z Wood

10 mm

100 mm

x FigureP4.5

Solution

εxx = 200μ σxx = ?

10 mm

Steel

Eal = 100GPa N=?

Ew = 10GPa yN = ?

Es = 200GPa

------------------------------------------------------------

The normal stress in steel, wood and aluminum can be found using Hooke’s Law as shown below. ( σ xx ) s = E s ε xx = ( 200 ) ( 10 9 ) ( 200 ) ( 10 –6 ) = 40 ( 10 6 ) ( N ⁄ m 2 ) = 40 MPa ( σ xx ) w = E s ε xx = ( 10 ) ( 10 9 ) ( 200 ) ( 10 –6 ) = 2 ( 10 6 ) ( N ⁄ m 2 ) = 2 MPa ( σ xx ) a1 = E s ε xx = ( 100 ) ( 10 9 ) ( 200 ) ( 10 – 6 ) = 20 ( 10 6 ) ( N ⁄ m 2 ) = 20 MPa

The stress distribution across the cross-section is as shown below. (b) 5 mm (c) (a) 20 MPa Aluminum

Wood

(d) Nal

NW

2 MPa 40 MPa

60 mm

N yN

NS

Steel

O

5 mm

O

The stress distribution in Fig. (b) can be replaced by equivalent internal normal forces shown in Fig. (c) that act at the centroid of each distribution. The internal normal forces, can be found as shown below. N s = ( σ xx ) s A s = 40 ( 10 6 ) ( 0.08 ) ( 0.01 ) = 32 ( 10 3 ) N = 32kN N w = ( σ xx ) w A w = 2 ( 10 6 ) ( 0.08 ) ( 0.1 ) = 16 ( 10 3 ) N = 16kN N A1 = ( σ xx ) A1 A A1 = 20 ( 10 6 ) ( 0.08 ) ( 0.01 ) = 16 ( 10 3 ) N = 16kN

Equating the forces in Figs. (c) and (d) we obtain: N = N s + N w + N A1 = 64kN

ANS.

N = 64kN ( T )

Equating moment about point O in Figs. (c) and (d), we obtain N ( y N ) = N s ( 0.005 ) + N w ( 0.06 ) + ( 16 ) ( 0.115 ) or ( 64 )y N = ( 32 ) ( 0.005 ) + ( 16 ) ( 0.006 ) + ( 16 ) ( 0.115 )

or

y N = 2.96 ---------- = 46.25 ( 10 – 3 ) m 64 ANS. y N = 46.25 mm

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

4.6 A reinforced concrete bar is constructed by embedding 2 in x 2 in square iron rods in concrete as shown in Figure P4.6. Assuming uniform strain of ε xx = – 1500 μ in the cross-section (a) Plot the stress distribution across the cross-section (b) Determine the equivalent internal axial force N. Use

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

Eiron = 25,000 ksi and Econcrete = 3,000 ksi.

y

24in

z

4 in iron rods

x

4 in

FigureP4.6

4 in

εxx = -1500μ

Solution

24in

4 in

Ei = 25,000ksi

σxx = ?

Ec = 3,000ksi

N=?

------------------------------------------------------------

The normal stress in iron and concrete can be found using Hooke’s Law as shown below. ( σ xx ) i = E i ε xx = 25 ( 10 3 ) ( – 1500 ) ( 10 –6 ) = – 37.5 ksi = 37.5 ksi ( σ xx ) c = E c ε xx = 3.0 ( 10 3 ) ( – 1500 ) ( 10 –6 ) = – 4.5 ksi = 4.5 ksi

The normal stress distribution across the cross-section is as shown in Fig. (b): (a)

(b)

(c)

4.5 ksi

(d)

37.5 ksi

2Ni N

Nc 2Ni

The stress distribution in Fig. (b) can be replaced by equivalent internal normal forces shown in Fig. (c) that act at the centroid of each distribution. The internal normal forces can be found as shown below. A i = ( 2 ) ( 2 ) = 4 in 2

and

A c = ( 24 ) ( 24 ) – 4 ( 4 ) = 560 in 2

N i = ( σ xx ) i A i = ( 37.5 ) ( 4 ) = 150 kips ( C ) N c = ( σ xx ) c A c = ( 4.5 ) ( 560 ) = 2520 kips ( C )

Equating forces in Figs. (c) and (d) we obtain: N = 4N i + N c = 4 ( 150 ) + 2520 = 3120 kips

ANS.

N = 3120 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.7 A crane is lifting a mass of 1000 kg. The mass of the iron ball at B is 25 kg. A single cable runs between A and B that has a diameter of 25 mm. Two cables are between the B and C, each cable has a diameter of 10 mm. Determine the axial stresses in the cables. A

B

C

Figure P4.7

-----------------------------------------------------------Solution

m1 = 1000kg σAB=?

m2 = 25kg σBC=?

dAB = 25mm

dBC = 10mm

-----------------------------------------------------------By making imaginary cuts between A and B and between B and C, we obtain the following free body dia-

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grams. (a)

NAB

January 2014

NBC NBC

(b)

m2g m1g m1g

By force equilibrium in figure (a) N AB = m 1 g + m 2 g = ( 1025 ) ( 9.81 ) = 10.055 ( 10 3 ) N N AD 10.055 ( 10 3 ) σ AB = ----------= ------------------------------- = 20.483 ( 10 6 ) N ⁄ m 2 or A AD π ( 0.025 ) 2 ⁄ 4

σ AB = 20.5 MPa ( T )

By force equilibrium in figure (b) 2N BC = m 1 g or

1000 ) ( 9.81 )- = 4.905 ( 10 3 ) N N BC = (------------------------------2

N BC 4.905 ( 10 3 ) σ BC = ---------- = ---------------------------- = 62045 ( 10 6 ) ( N ⁄ m 2 ) or A BC π ( 0.01 ) 2 ⁄ 4

σ BC = 62.5 MPa ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.8 The counter-weight in a lift bridge has 12 cables on the left and 12 cables on the right. Each cable has an effective diameter of 0.75 inch, a length of 50 feet, Modulus of Elasticity of 30,000 ksi, and an ultimate strength of 60 ksi. (a) If the counter-weight is100 kips, determine the factor of safety for cable. (b) What is the extension of each cable when the bridge is being lifted?

Set of 12 Cables Counter-weight

Figure P4.8

Solution

d = 0.75in 2 sets of 12 cables

L = soft W = 100 ksi

σ ult = 60 ksi

E = 30,000ksi ksaftey = ?

-----------------------------------------------------------The free body diagram of the counter-weight is shown below. 12NC

12NC

100 kips

By force equilibrium 24N C = 100 or

N C = 4.167 kips

N

4.167 The axial stress is: σ C = ------C- = --------------------------- = 9.43 ksi AC

π ( 0.75 ⁄ 2 ) 2

σ ult 60 The factor of safety is: k saftey = -------- = ---------- = 6.361 or σc

9.43

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

σ –3 9.43 - = 0.3143 ( 10 ) The axial strain is: ε C = -----c = -------------E

30000

–3

The deformation of the cable is: δ C = ε C L = ( 0.3143 ) ( 10 ) ( 50 ) ( 12 ) = 0.1886 in or δ C = 0.19 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.9 Draw the axial force diagrams and then check your results by finding internal forces in each segment AB, BC, and CD by making imaginary cut and drawing free body diagrams.The axial rigidity of the bar is EA = 8,000 kips. Determine the movement of section at D with respect to section at A. Solution NAB = ? NBC = ? NCD = ? EA = 8,000 kips uD-uA = ?

-----------------------------------------------------------We draw a template as shown in figure (a) and write the template equation. 25 kips (a) (b) F ext ----------2

10 kips

20 kips A

B

N2

N1

20 in

(c) Axial force diagram N kips

+ F ext

10

D

30 kips 50 in

20 in

N2 = N1

C

25 kips

F ext ----------2 Template Equation

30 kips

20

20

10

x 40

40

We start in the imaginary extension where the internal axial force is zero. If the external forces in figure (b) are in the direction shown on the template in figure (a), then we add the value as per the template equation, otherwise we subtract. In this manner we otain the axial force diagram shown in figure (c). By making imaginary cuts between A and B, between B and C, and between C and D, we obtain the free body diagrams shown in figures (d), (e), and (f), respectively. 25 kips (d) (e) (f) NAB

10 kips

NBC

10 kips A

A

NCD

B

20 kips D

25 kips

By equilibrium of forces in figure (d), we obtain: N AB = 10 kips

(1)

By equilibrium of forces in figure (d), we obtain: N BC = – 40 kips

(2)

By equilibrium of forces in figure (d), we obtain: N CD = 20 kips

(3)

The value of internal forces in Eq’s (1), (2) and (3) are the same as obtained from the axial force diagram in figure (c). N AB ( x B – x A ) 10 ) ( 20 )- = 0.025 in u B – u A = ---------------------------------- = (--------------------EA 8000

1

N BC ( x C – x B ) ( – 40 ) ( 50 ) u C – u B = ---------------------------------- = ------------------------- = – 0.250 in EA 8000

2

N CD ( x D – x C ) 20 ) ( 20 )- = 0.050 in u D – u C = ----------------------------------= (--------------------EA 8000

3

Adding Eq’s (1), (2), and (3), we obtain: u D – u A = 0.025 – 0.25 + 0.05 = – 0.175 in

or

u D – u A = – 0.175 in

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

4.10 Draw the axial force diagrams and then check your results by finding internal forces in each segment AB, BC, and CD by making imaginary cut and drawing free body diagrams. The axial rigidity of the bar is EA = 80,000 kN. Determine the movement of section at C with respect to section at A Solution NAB = ? NBC = ? NCD = ? EA = 80,000 kN uC = ?

-----------------------------------------------------------By force equilibrium of the entire axial member, we obtain the reaction force at A as shown below. or

R A – 150 + 90 – 70 = 0

R A = 130 kN

We draw a template as shown in figure (a) and write the template equation. 75 kN (a) (b) F ext ----------2

2

RA

= N

1

+ F ext

C

75 kN

(c) Axial force diagram N kN

70 kN

B

N2

N1

N

A

45 kN

45 kN 0.5 m

0.25 m

F ext ----------2 Template Equation

D

0.25 m

20

20

x 70

70 130

130

We start in the imaginary extension where the internal axial force is zero. If the external forces in figure (b) are in the direction shown on the template in figure (a), then we add the value as per the template equation, otherwise we subtract. In this manner we obtain the axial force diagram shown in figure (c). By making imaginary cuts between A and B, between B and C, and between C and D, we obtain the free body diagrams shown in figures (d), (e), and (f), respectively. 75 kN (d) (e) (f) NAB NBC

A RA=130 kN

RA=130 kN

A

NCD

B

70 kN D

75 kN

By equilibrium of forces in figure (d), we obtain: N AB = – 130 kN

(1)

By equilibrium of forces in figure (d), we obtain: N BC = 20 kN

(2)

By equilibrium of forces in figure (d), we obtain: N CD = – 70 kN

(3)

The value of internal forces in Eq’s (1), (2) and (3) are the same as obtained from the axial force diagram in figure (c). N AB ( x B – x A ) –3 ( – 130 ) ( 0.25 ) u B – u A = ---------------------------------- = -------------------------------- = – 0.4063 ( 10 ) m EA 80, 000

1

N BC ( x C – x B ) 20 ) ( 0.5 )- = 0.125 ( 10 – 3 ) m u C – u B = ---------------------------------- = (---------------------EA 80, 000

2

–3

–3

Adding Eq’s (1) and (2) we obtain: u C – u A = ( – 0.4063 + 0.125 ) ( 10 ) = – 0.28125 ( ( 10 ) m ) Point A is fixed to the wall, i.e., uA = 0, thus we obtain:

u C = – 0.281 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.11 Draw the axial force diagrams and then check your results by finding internal forces in each segment AB, BC, and CD by making imaginary cut and drawing free body diagrams. The axial rigidity of the bar is EA = 2,000 kips. Determine the movement of section at B. Solution NAB = ? NBC = ? NCD = ? EA = 2,000 kips uB= ?

------------------------------------------------------------

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January 2014

By force equilibrium of the entire axial member, we obtain the reaction force at D as shown below. or

1.5 + 4 – 8 + R D = 0

R D = 2.5 kips

We draw a template as shown in figure (a) and write the template equation 2 kips

(b)

(a) F ext ----------2

A

B

C

+ F ext

RD

60 in

25 in

N2 = N1

D

4 kips

2 kips

N2

N1

4 kips

1.5 kips

20 in

(c) Axial force diagram F ext ----------N 2 kips Template Equation

2.5

2.5

x

1.5

1.5

5.5

5.5

We start in the imaginary extension where the internal axial force is zero. If the external forces in figure (b) are in the direction shown on the template in figure (a), then we add the value as per the template equation, otherwise we subtract. In this manner we obtain the axial force diagram shown in figure (c). By making imaginary cuts between A and B, between B and C, and between C and D, we obtain the free body diagrams shown in figures (d), (e), and (f), respectively. 2 kips (d) (e) (f) N N AB

1.5 kips

A

1.5 kips

BC

A

B

NCD

D

RD=2.5 kips

2 kips

By equilibrium of forces in figure (d), we obtain: N AB = – 1.5 kips

(1)

By equilibrium of forces in figure (d), we obtain: N BC = – 5.5 kips

(2)

By equilibrium of forces in figure (d), we obtain: N CD = 2.5 kips

(3)

The value of internal forces in Eq’s (1), (2) and (3) are the same as obtained from the axial force diagram in figure (c). N BC ( x C – x B ) ( – 5.5 ) ( 60 ) u C – u B = ---------------------------------- = -------------------------- = – 0.165 in 2000 EA

1

N CD ( x D – x C ) 2.5 ) ( 20 -) = 0.025 in u D – u C = ----------------------------------= (---------------------EA 2000 Adding Eq’s (1) and (2) we obtain: u D – u B = – 0.165 + 0.025 = – 0.140 in

Point D is fixed to the wall,i.e., uD = 0, thus we obtain:

2

u B = 0.14 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.12 Draw the axial force diagrams and then check your results by finding internal forces in each segment AB, BC, and CD by making imaginary cut and drawing free body diagrams. The axial rigidity of the bar is EA = 50,000 kN. Determine the movement of section at D with respect to section at A. Solution NAB = ? NBC = ? NCD = ? EA = 50,000 kN uD-uA = ?

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

We draw a template as shown in figure (a) and write the template equation, 90 kN

60 kN

(b)

(a)

100kN

200 kN

B

A

F ext ----------2

C 60 kN

N2

N1

F ext ----------2 N2 = N1

+ F ext

90 kN 0.6 m

0.4 m

(c) Axial force diagram N kN

D

0.4 m 200

200 20

20

x

Template Equation 100

100

We start in the imaginary extension where the internal axial force is zero. If the external forces in figure (b) are in the direction shown on the template in figure (a), then we add the value as per the template equation, otherwise we subtract. In this manner we obtain the axial force diagram shown in figure (c). By making imaginary cuts between A and B, between B and C, and between C and D, we obtain the free body diagrams shown in figures (d), (e), and (f), respectively. 60 kN (d) (e) (f) N NAB 100 kN

NBC

A

A

100 kN

CD

200 kN

D

B 60 kN

By equilibrium of forces in figure (d), we obtain: N AB = – 100 kN

(1)

By equilibrium of forces in figure (d), we obtain: N BC = 20 kN

(2)

By equilibrium of forces in figure (d), we obtain: N CD = 200 kN

(3)

The value of internal forces in Eq’s (1), (2) and (3) are the same as obtained from the axial force diagram in figure (c). N AB ( x B – x A ) –3 ( – 100 ) ( 0.4 ) u B – u A = ---------------------------------- = ----------------------------- = – 0.80 ( 10 ) m EA 50, 000

1

N BC ( x C – x B ) 20 ) ( 0.6 )- = 0.24 ( 10 – 3 ) m u C – u B = ---------------------------------- = (---------------------50, 000 EA

2

N CD ( x D – x C ) –3 ( 200 ) ( 0.4 ) u D – u C = ----------------------------------= -------------------------- = 1.60 ( 10 ) m EA 50, 000

3

Adding Eq’s (1),(2), and (3) we obtain: –3

–3

u C – u A = ( – 0.80 + 0.24 + 1.60 ) ( 10 ) = 1.04 ( ( 10 ) m ) or

u C = 1.04 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.13 Three segments of 4 in x 2 in rectangular wooden bars (E = 1600 ksi) are secured together with rigid plates and subjected to axial forces as shown. Determine: (a) the movement of rigid plate at D with respect to plate at A. (b) the maximum axial stress. .

22.5 kips 4 in

15 kips A

27.5 kips

B

30 in

C 22,5 kips 27.5 50 in

D

25 kips

30 in

Figure P4.13

Solution

E = 1600 ksi

A = (4)(2)=8 in2

uD-uA = ?

σmax = ?

-----------------------------------------------------------We draw a template as shown in figure (a) and write the template equation. Using the template and tem-

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

plate equation, we draw the axial force diagram as shown in figure (b) F ext (b) Axial force diagram (a) ----------2

25

N2

N1

Template Equation

F ext ----------2 N2 = N1

N kips

25

15

15 A

B

C

30

+ F ext

D

x

30

From figure (b) the internal axial forces are: N AB = 15 kips , N BC = – 30 kips , and N CD = 25 kips . N AB ( x B – x A ) ( 15 ) ( 30 ) u B – u A = ---------------------------------- = ------------------------- = 0.0352 in EA ( 1600 ) ( 8 )

1

N BC ( x C – x B ) ( – 30 ) ( 50 ) u C – u B = ---------------------------------- = ------------------------- = – 0.1172 in EA ( 1600 ) ( 8 )

2

N CD ( x D – x C ) ( 25 ) ( 30 )- = 0.0586 in u D – u C = ----------------------------------= -----------------------EA ( 1600 ) ( 8 )

3

Adding Eq’s (1), (2), and (3), we obtain: or

u D – u A = 0.0352 – 0.1172 + 0.0586 = – 0.0234 in

u D – u A = – 0.0234 in.

With all segments having the same area of cross-section, the maximum axial stress will be in the segment with the maximum internal axial force, i.e., segment BC. The axial stress in segment BC can be found as shown below. N BC – 30 - = --------- = – 3.75 ksi σ max = ---------A 8

σ max = 3.75 ksi ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.14 Aluminum bars (E = 30,000 ksi) are welded to rigid plates, as shown in Figure P4.1. All bars have a cross-sectional area of 0.5 in2. The applied forces are F1 = 8 kips; F2 = 12 kips; and F3= 9 kips. Determine (a) the displacement of the rigid plate at D with respect to the rigid plate at A. (b) the maximum axial stress in the assembly. F1

F2 A

F1

Figure P4.14

F3

B x 36 in

F4 C

F2

D

F3

F4

50 in

36 in

Solution

-----------------------------------------------------------By equilibrium of the entire bar: – 2 F 1 + 2F 2 – 2F 3 + 2F 4 = 0

or

F 4 = 8 + 9 – 12 = 5 kips

1

We can draw the axial force diagram as shown F (b) Axial force diagram ext (a) ----------2

N1

Template Equation

16

N2

F ext ----------2 N = N 2 1

N kips

16 B

A

8

10

10 C 8

D

x

+ F ext

From figure (b) the internal axial forces are: N AB = 16 kips , N BC = – 8 kips , and N CD = 10 kips .

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

N AB ( x B – x A ) –3 ( 16 ) ( 36 ) u B – u A = ---------------------------------- = ---------------------------------------- = 19.2 ( 10 ) in EA ( 30000 ) ( 0.5 ) ( 2 )

1

N BC ( x C – x B ) –3 ( – 8 ) ( 50 ) u C – u B = ---------------------------------- = ---------------------------------------- = – 13.33 ( 10 ) in EA ( 30000 ) ( 0.5 ) ( 2 )

2

N CD ( x D – x C ) –3 ( 25 ) ( 36 ) u D – u C = ----------------------------------= ---------------------------------------- = 12 ( 10 ) in EA ( 30000 ) ( 0.5 ) ( 2 )

3

Adding Eq’s (1), (2), and (3), we obtain: –3

–3

u D – u A = ( 19.2 – 13.33 + 12 ) ( 10 ) = 17.87 ( 10 ) in

4

With all segments having the same area of cross-section, the maximum axial stress will be in the segment with the maximum internal axial force, i.e., segment AB. The axial stress in segment AB can be found as shown below. N AB 16 - = -------------------- = 16 ksi σ max = ---------A ( 0.5 ) ( 2 )

The answers are:

σ max = 16 ksi ( T )

u D – u A = 0.01787 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.15 Brass bars between sections A and B, aluminum bars between sections B and C, and steel bars between sections C and D are welded to rigid plates, as shown in Figure P4.2. The properties of the bars are given in The applied forces are F1 = 90 kN; F2 = 40 kN; and F3= 70 kN. Determine (a) the displacement of the rigid plate at D.(b) the maximum axial stress in the assembly Table P4.15

x Brass

Modulus of 70 GPa

Aluminum

Steel

100 GPa

200 GPa

25 mm

20 mm

A

30 mm

F2 C

D F2

F1 2.5 m

1.5 m

elasticity Diameter

F1 B

F3 F3

2m

Figure P4.15

Solution

-----------------------------------------------------------We can find the reactions at the wall from equilibrium of the entire free body diagram. 2R A – 2F 1 – 2F 2 + 2F 3 = 0 RA RA

x A 1.5 m

F1 C F1 2.5 m

R A = 90 + 40 – 70 = 60 kN

F ext ----------2

F2

B

or

D F2

F3 F3

N2

N1

F ext ----------2

2m

Template Equation

N2 = N1

The internal forces are: N AB = – 120 kN

1 140

140 N kN

A

60 C

60

D

x

B 120

120

+ F ext

N BC = 60 kN

N CD = 140 kN

2 –3 2 2 –3 2 π π The crosssectional areas are: A AB = --- ( 0.03 ) = 0.706 ( 10 ) m ; A BC = --- ( 0.025 ) = 0.4909 ( 10 ) m ;

4

4

2 –3 2 π A CD = --- ( 0.02 ) = 0.3142 ( 10 ) m 4

The relative displacements are: N AB ( x B – x A ) –3 ( – 120 ) ( 1.5 ) u B – u A = ----------------------------------= ------------------------------------------------------ = – 1.819 ( 10 ) m 9 –3 E AB A AB 70 ( 10 ) ( 0.706 ) ( 10 )

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

N BC ( x C – x B ) –3 ( 20 ) ( 0.6 ) u C – u B = ---------------------------------- = ------------------------------------------------------------- = 1.528 ( 10 ) m 9 – 3 E BC A BC 100 ( 10 ) ( 0.4909 ) ( 10 )

2

N CD ( x D – x C ) –3 ( 200 ) ( 0.4 ) u D – u C = ----------------------------------= ------------------------------------------------------------- = 2.228 ( 10 ) m 9 – 3 E CD A CD 200 ( 10 ) ( 0.3142 ) ( 10 )

3

–3

–3

Adding we obtain: u D – u A = ( – 1.819 + 1.528 + 2.228 ) ( 10 ) = 1.937 ( 10 ) m The stresses in each bar are: 3 3 N AB N BC 6 2 6 2 60 ( 10 ) – 120 ( 10 ) σ AB = ---------- = ----------------------------- = – 169.7 ( 10 ) N ⁄ m ; σ BC = ---------- = -------------------------------- = 61.1 ( 10 ) N ⁄ m – 3 – 3 A AB A BC 0.706 ( 10 ) 0.4909 ( 10 ) 3 N CD 6 2 140 ( 10 ) σ CD = ---------- = -------------------------------- = 222.8 ( 10 ) N ⁄ m –3 A CD 0.3142 ( 10 )

The answers are:

σ max = 222.8 MPa ( T )

u D = 1.937 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.16 A solid circular steel (Es = 30,000 ksi) rod BC is securely attached to two hollow steel rods AB and CD as shown. Determine: (a) the angle of displacement of section at D with respect to section at A. (b) the maximum axial stress in the axial member. 210 kips

60 kips

A

60 kips

100 kips

210 kips

Figure P4.16

C

4 in.

B

24 in.

50 kips

100 kips

2 in.

D

50 kips 24 in.

36 in.

Solution

------------------------------------------------------------

2 2 π 2 π 2 π 2 The crosssectional areas are: A AB = A CD = --- ( 4 ) – --- ( 2 ) = 9.428 in ; A BC = --- ( 4 ) = 12.566 in ;

4

4

4

We can draw the axial force diagram as shown F (b) Axial force diagram ext (a) ----------2

N1

Template Equation

300

N2

F ext ----------2 N2 = N1

N kips

B

100

100

A

120

C 8

D

x

+ F ext

From figure (b) the internal axial forces are: N AB = – 120 kips , N BC = 300 kips , and N CD = 100 kips . N AB ( x B – x A ) –3 ( – 120 ) ( 24 ) u B – u A = ---------------------------------- = -------------------------------------- = – 10.186 ( 10 ) in EA ( 30000 ) ( 9.428 )

1

N BC ( x C – x B ) ( 300 ) ( 36 ) - = 28.678 in u C – u B = ---------------------------------- = ---------------------------------------EA ( 30000 ) ( 12.566 )

2

N CD ( x D – x C ) –3 ( 100 ) ( 24 ) u D – u C = ----------------------------------= -------------------------------------- = 8.488 ( 10 ) in EA ( 30000 ) ( 9.428 )

3

Adding Eq’s (1), (2), and (3), we obtain:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

–3

January 2014

–3

u D – u A = ( – 10.186 + 28.678 + 8.488 ) ( 10 ) = 26.95 ( 10 ) in

4

The axial stress in segments AB & BC can be found as shown below. N AB – 120 σ AB = ---------- = ------------- = – 12.732 ksi 9.428 A AB

N BC 300 σ BC = ---------- = ---------------- = 23.87 ksi 12.566 A BC

The answers are:

σ max = 23.87 ksi ( T )

u D – u A = 0.027 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.17 Two circular steel bars (E=30,000 ksi, ν = 0.3) of diameter 2 inch are securely connected to an aluminum bar (E=10,000 ksi, ν = 0.33) of diameter 1.5 inch. Determine (a) the displacement of section at C with respect to the wall. (b) the maximum change in the diameter. 5 kips 17.5 kips RA

Solution

A steel

B C aluminum

25 kips

D

steel 5 kips 17.5 kips 15 in 25 in

Figure P4.17

40 in

EAB = ECD = 30,000 ksi dAB = dCD = 2.0 in

EBC = 10,000 ksi dBC = 1.5 in

νAB =νCD =0.3 νBC = 0.33 uC-uA = ? Δdmax = ?

-----------------------------------------------------------By force equilibrium of the entire axial member, we obtain the reaction force at A as shown below. or

R A – 10 – 35 + 25 = 0

R A = 20 kips

We draw a template as shown in figure (a) and write the template equation. Using the template and template equation, we draw the axial force diagram as shown in figure (b) F (a) (b) Axial force diagram ext ----------2

N

2

= N

1

+ F ext

25

25

N2

N1

N kips

F ext ----------2 Template Equation

B 10

A 20

D

C

x

10

20

From figure (b) the internal axial forces are: N AB = – 20 kips , N BC = – 10 kips , and N CD = 25 kips . 2

2

2

The area of cross-sections are: A AB = A CD = π ( 2 ) ⁄ 4 = 3.142 in , and A BC = π ( 1.5 ) ⁄ 4 = 1.767 in

2

N AB ( x B – x A ) ( – 20 ) ( 40 ) - = – 8.488 ( 10 – 3 ) in u B – u A = ---------------------------------- = ---------------------------------------EA ( 30, 000 ) ( 3.142 )

1

N BC ( x C – x B ) –3 ( – 10 ) ( 15 ) u C – u B = ---------------------------------- = ----------------------------------------- = – 8.488 ( 10 ) in EA ( 10, 000 ) ( 1.767 )

2

Adding Eq’s (1) and (2) –3

–3

u C – u A = ( – 8.488 – 8.488 ) ( 10 ) = ( – 16.98 ) ( 10 ) in

or

u C – u A = – 0.017 in.

The maximum change in diameter will occur in that segment in which the axial stress is maximum. As area of cross-sections for AB and CD are the same, we conclude that CD will have a larger stress as the axial force is larger. We calculate the axial stress in BC and CD as shown below: N BC N CD ( – 10 ) ( 25 ) σ BC = ---------- = -------------- = – 5.659 ksi and σ CD = ---------- = ------------- = 7.958 ksi 1.767 3.142 A BC A CD

The axial strain in BC and CD is:

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January 2014

σ BC σ CD –3 –3 ( – 5.659 ) ( 7.958 ) ε BC = --------- = --------------------- = ( – 0.5659 ) ( 10 ) and ε CD = ---------= ------------------ = 0.2653 ( 10 ) E BC E CD 10, 000 30, 000

The transverse strain in BC and CD is: –3 –3 · ε BC ) T = – ν BC ε BC = – ( 0.33 ) ( – 0.5659 ) ( 10 ) = 0.1867 ( 10 and –3

–3

( ε CD ) T = – ν CD ε CD = – ( 0.3 ) ( 0.2653 ) ( 10 ) = – 0.07959 ( 10 ).

The change in diameter in BC and CD is: –3

–3

–3

–3

Δd BC = ( 1.5 ) ( 0.1867 ) ( 10 ) = ( 0.2801 ) ( 10 ) in andΔd CD = ( 2 ) ( – 0.07959 ) ( 10 ) = ( – 0.1592 ) ( 10 ) in Δd max = 0.00028 in.

The maximum change in diameter occurs in BC. It’s value is: or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.18 Two cast iron pipes (E = 100 GPa) are adhesively bonded together as shown. The outer diameters of the two pipes are 50 mm and 70 mm and wall thickness of each pipe is 10 mm. Determine the displacement of end B with respect to end A. 500 mm 20 kN

150 mm 400 mm C

A

D B

20 kN

Figure P4.18

Solution

E = 100GPa (dDB)i = 30 mm

(dAC)o = 70 mm (dCD)o = 70 mm

(dAC)i = 50 mm (dCD)i = 30 mm

(dDB)o = 50 mm uB-uA = ?

-----------------------------------------------------------The internal normal force in each segment is: N AC = N CD = N DB = 20 kN . The area of material crosssections are as shown below. π [ ( 0.07 ) 2 – ( 0.05 ) 2 ] A AC = -------------------------------------------------- = 1.885 ( 10 –3 ) m 2 4 [ ( 0.05 ) 2 – ( 0.03 ) 2 ]- = 1.257 ( 10 –3 ) m 2 A DB = π ------------------------------------------------4 [ ( 0.07 ) 2 – ( 0.03 ) 2 ]- = 3.142 ( 10 –3 ) m 2 A CD = π ------------------------------------------------4

The relative displacement of the ends of each segment can be written as shown below. N AC ( x C – x A ) ( 20 ) ( 10 3 ) ( 0.5 ) u C – u A = ----------------------------------= -------------------------------------------------------------- = 0.05305 ( 10 – 3 ) m E AC A AC ( 100 ) ( 10 9 ) ( 1.885 ) ( 10 – 3 )

1

N CD ( x D – x C ) ( 20 ) ( 10 3 ) ( 0.15 ) u D – u C = ----------------------------------= -------------------------------------------------------------= 0.00955 ( 10 – 3 ) m E CD A CD ( 100 ) ( 10 9 ) ( 3.142 ) ( 10 – 3 )

2

N BD ( x B – x D ) ( 20 ) ( 10 3 ) ( 0.4 ) u B – u D = ----------------------------------= -------------------------------------------------------------= 0.06364 ( 10 – 3 ) m E BD A BD ( 100 ) ( 10 9 ) ( 1.257 ) ( 10 – 3 )

3

Adding Eq’s (1), (2), and (3) u B – u A = ( 0.05305 + 0.00955 + 0.06364 ) ( 10 – 3 ) = 0.12624 ( 10 – 3 ) m or

u B – u A = 0.126 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.19

The tapered bar shown in Figure P4.19 has a cross-sectional area that varies with x as A = K ( 2 L – 0.25 x ) 2 . Determine the elongation of the bar in terms of P, L, E and K.

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x

A

January 2014

P

B L

Figure P4.19

Solution

uB-uA = ?

-----------------------------------------------------------We can write: N AB P du = ---------------------= ---------------------------------------------------dx E AB A AB E [ K ( 2L – 0.25x ) 2 ]

1

Integrating we obtain: L

∫0 du

P L dx = -------- --------------------------------EK 0 ( 2L – 0.25x ) 2

∫

L

P 1 u = -------- -----------------------------------------EK 0.25 ( 2L – 0.25x ) 0

2

P 1 1 = -------- -------------------- – ----------EK 0.4375L 0.5L

3 P u = 0.2857 -----------EKL

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.20

The tapered bar shown in Figure P4.19 has a cross-sectional area that varies with x as A = K ( 4 L – 3 x ) . Determine the elongation of the bar in terms of P, L, E and K.

A

x

B

P

L

Solution

uB - uA = ?

-----------------------------------------------------------We can write: N AB P du = ------------------------------------- = ---------------------dx EK ( 4L – 3x ) E AB A AB

Integrating we obtain: L

∫0

L

du =

P

dx

- ⋅ -----------------------∫0 ------EK ( 4L – 3x )

or

L P P P 1 u = ----------- [ ln 4L – 3x 0 ] = – ----------- [ ln L – ln 4L ] = – ----------- ln --3EK 3EK 3EK 4

4 P u = 0.4621 -------EK

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.21 A tapered and un-tapered solid circular steel bars (E=30,000 ksi) are securely fastened to a solid circular aluminum bar (E=10,000 ksi). The un-tapered steel bar has a diameter of 2 inches. The aluminum bar has a diameter of 1.5 inches. The tapered bars diameter varies from 1.5 inches to 2 inches.(a) Determine the displacement of section at C with respect to the section at A (b) Determine the maximum axial stress in the bar. .

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

10 kips

20 kips

40 kips A

January 2014

D 60 kips

B C steel aluminum steel 20 kips 10 kips 60 in 40 in 15 in

Figure P4.21

Solution

EAB = ECD = 30,000 ksi dAB = 2.0 in uC-uA = ?

EBC = 10,000 ksi dBC = 1.5 in σmax = ?

dCD varies from 1.5 in to 2 in

-----------------------------------------------------------We draw a template as shown in figure (a) and write the template equation. Using the template and template equation, we draw the axial force diagram as shown in figure (b) F ext (b) Axial force diagram (a) ----------60 60 N1

2 N2

F ext ----------2

N kips

Template Equation N

2

= N

1

40

40 20

+ F ext

20 B

A

C

D

x

From figure (b) the internal axial forces are: N AB = 40 kips , N BC = 20 kips , and N CD = 60 kips . N AB ( x B – x A ) –3 ( 40 ) ( 40 ) u B – u A = ---------------------------------- = ------------------------------------------- = 16.976 ( 10 ) in 2 E AB A AB ( 30, 000 ) ( π2 ⁄ 4 )

1

N BC ( x C – x B ) –3 ( 20 ) ( 15 ) u C – u B = ---------------------------------- = -----------------------------------------------= 16.976 ( 10 ) in 2 E BC A BC ( 10, 000 ) ( π1.5 ⁄ 4 )

2

Adding Eq’s (1), (2), and (3), we obtain: –3

–3

u C – u A = ( 16.976 + 16.976 ) ( 10 ) = 33.952 ( 10 )in

or

u C – u A = 0.034 in

The axial stress in CD will be maximum where the area of cross-section is minimum. Thus, the maximum stress in CD will be just after C, where the diameter is 1.5 in.This stress will be greater than the stress in BC as the diameter there is also 1.5 but axial force is smaller. N CD 60 - = ------------------------- = 33.95 σ max = ---------2 A ( π1.5 ⁄ 4 )

σ max = 33.95 ksi ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.22 The column shown has a length L, Modulus of elasticity E, specific weight γ, and radius a. Determine the contraction of the column in terms of L, E, γ, and a. x

Solution

uB - uA = ?

-----------------------------------------------------------The distributed force is product of specific weight and the area of cross-section and it acts in the negative x-direction: p = – γA = – γπa 2

1

We can write:

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

dN ------- = – p = γπa 2 dx

2

N = γπa 2 x + c 1

3

Integrating we obtain: at x = L

N = 0 We obtain:

c 1 = – γπa 2 L N = γπa 2 x – γπa 2 L

4

γπa 2 γ du N ------ = -------- = ------------- ( x – L ) = --- ( x – L ) 2 E dx EA Eπa

5

We can write:

Integrating we obtain:

∫

uB

γ L du = --- ( x – L ) dx or E 0 uA

∫

γ x2 u B = --- ⎛ ---- – Lx⎞ ⎠ E⎝ 2

L

6 0

– γL 2 u B = -----------2E

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.23 The column shown has a length L, Modulus of elasticity E, specific weight γ, and length a as the side of an equilateral triangle. Determine the contraction of the column in terms of L, E, γ, and a. x

Solution

uB - uA = ?

-----------------------------------------------------------The area of cross-section is: 1 3 A = --- ab = ------- a 2 2 4

1

The distributed force is product of specific weight and the area of cross-section and it acts in the negative x-direction: 3 p = – γ ⎛ ------- a 2⎞ ⎝ 4 ⎠

2

3 dN ------- = – p = ------- γa 2 4 dx

3

3 N = ------- γa 2 x + c 1 4

4

We can write:

Integrating we obtain:

at x = L

3 N = 0 we obtain: c 1 = – ------- γa 2 L 4 3 N = ------- γa 2 ( x – L ) 4

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4-19

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

We can write: N 3 γπa 2 γ du ------ = -------- = ------- ----------------- ( x – L ) = --- ( x – L ) EA 4 E dx 3 E ------- a 2 4

6

Integrating we obtain:

∫

L

uB

γ du = --- ( x – L )dx or E uA

∫ 0

γ x2 u B = --- ⎛ ----- – Lx⎞ ⎠ E⎝ 2

L

7 0

– γL 2 u B = -----------2E

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.24 The columns shown has a length L, Modulus of elasticity E, specific weight γ, and cross-sectional area A. Determine the contraction of the column in terms of L, E, γ, and A.

x

Solution

uB - uA = ?

-----------------------------------------------------------The distributed force is product of specific weight and the area of cross-section and it acts in the negative x-direction: p = – γA

1

dN- = – p = γA -----dx

2

N = γAx + c 1

3

N = 0 we obtain: c 1 = – γAL N = γA ( x – L )

4

We can write:

Integrating we obtain: at x = L We can write:

N du ------ = -------- = γA ( x – L ) EA dx

5

Integrating we obtain: uB

∫u

A

du = γA

L

∫0 ( x – L ) dx or

γA 2 u B – u A = -------- ⎛ x----- – Lx⎞ ⎠ EA ⎝ 2

L 0

γ –L = ------- ⎛ ------⎞ 2E ⎝ 2 ⎠

6 γL 2 u B = – --------2E

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

2 4.25 On the truncated cone of Example 4.7 a force of P = ( γπr L ⁄ 5 ) is also applied as shown. Determine the total elongation of the cone due to its weight and the applied force. (Hint: use superposition) 5r

x R(x)

L

r P

Solution

Figure P4.25

δ=?

------------------------------------------------------------

Let the total elongation δ= δ W+ δP, where δW is the elongation due to weight and δP is the elongation due to the force P. From example 4.6 we have: 7 γL 2 δ W = ------ --------30 E r R ( x ) = --- ( 5L – 4x ) L

and

or

1 r2 A ( x ) = π ------ ( 5L – 4x ) 2 L2

2

The free body diagram at any location x is: N

P

γπr 2 L By equilibrium of forces we obtain: N = P = -------------5

We can write: du N γπr 2 L γL 3 ------ = -------- = -------------------------------------------- = ---------------------------------2 dx EA 5E ( 5L – 4x ) 2 r 5Eπ ------ ( 5L – 4x ) 2 L2

3

Integrating from x = 0 to x = L we can obtain δP as shown below δP =

uL

∫u

du = 0

∫

L

3 γL 3 1 - ----------------------1 --------------------------------- dx = γL --------- ------------or ⋅ 2 5E – ( – 4 ) ( 5L – 4x ) 0 5E ( 5L – 4x ) 0 L

γL 3 1 1 γL 2 - --- – ------- = --------δ P = --------20E L

5L

4

25E

Adding Equations (1) and (3), we obtain the total elongation as shown below 7 γL 2 γL 2 41 γL 2 δ = ------ --------- + ---------- = --------- --------30 E 150 E 25E

5 γL 2 δ = 0.273 --------E

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.26

A 20 feet tall, thin-hollow tapered tube of uniform wall thickness 1/8 inch is used for a light pole

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

in a parking lot. The mean diameter at the bottom is 8 inches and at the top it is 2 inches. The weight of the lights on top of the pole is 80 lbs. The pole is made of aluminum alloy with a specific weight of 0.1 lb/in3 and Modulus of Elasticity E = 11,000 ksi. Determine (a) the maximum axial stress. (b) the contraction of the pole. (Hint: Approximate the area of cross-section as A = πDt, where D is the mean diameter and t is the thickness of the thin-walled pole.)

Tapered pole

Figure P4.26

Solution

L = 20ft

t = 118in

γ = 0.1 lb ⁄ in 3

db = 8in

E = 11,000ksi

dt = 2in

WLight = 80lbs

G = 8000ksi

σ max =

δ=?

-----------------------------------------------------------The geometry off the pole in two-dimension is as shown below. P 2 in

x

R(x)

20 ft

8 in

Let R(x) be the outer radius of the pole at any x. R ( x ) = a + bx

At x = 0

1

R(x) = 1 inch . Thus: a = 1

At x = 20 ft = 240 in

1R(x) = 4 inch, thus 4 = 1 + b ( 240 ) or b = ----80

( 80 + x ) R ( x ) = -------------------- in 80

2

For thin-walled tubes the area of cross-section can be approximated as A = 2πRt or ( 80 + x ) 1 A ( x ) = ( 2π ) -------------------- ⎛ ---⎞ = 9.817 ( 10 3 ) ( 80 + x ) 80 ⎝ 8⎠

3

With gravity being the same direction as x p ( x ) = γA = 0.9817 ( 10 – 3 ) ( 80 + x )

4

dN ------- = – p = – 0.9817 ( 10 – 3 ) ( 80 + x ) dx

5

N = – 0.4909 ( 10 – 3 ) ( 80 + x ) 2 + c 1

6

We can write:

Integrating we obtain

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

N = -80lbs, thus – 80 = – 0.4909 ( 10 –3 ) ( 80 ) 2 + c 1 or c 1 = – 76.858

At x = 0

N = – 0.4909 ( 10 – 3 ) ( 80 + x ) 2 – 76.858

7

The maximum axial force will be at the bottom of the pole i.e., x = 240 in N max = – 0.4909 ( 10 –3 ) ( 320 ) 2 – 76.858 = – 127.12lbs

8

The area of cross-section at the bottom is: A b = 9.817 ( 10 3 ) ( 320 ) = 3.1416 in 2

9

N max – 127.12 σ max = ------------ = ------------------- = – 54.78 Ab 3.1416

10

The maximum axial force is:

σ max = 54.8 psi ( C )

We can write: du N – 0.4909 ( 10 – 3 ) ( 80 + x ) 2 – 76.858 ------ = -------- = ---------------------------------------------------------------------------------- or dx EA 11 ( 10 6 )L9.817 ( 10 –3 ) ( 80 + x ) du 0.7117 ( 10 – 3 )------ = – 0.4545 ( 10 – 9 ) ( 80 + x ) – -------------------------------dx 80 + x

11

Integrating from x = 0 to x = 240 u 240

∫u

du = – 0.4545 ( 10 – 9 )

0

240

∫0

( 80 + x ) dx – 0.7117 ( 10 – 3 )

240

∫0

dx --------------- or 80 + x

240

( 80 + x ) 2 u 240 – u 0 = – 0.4545 ( 10 – 9 ) ---------------------2 0

240

– 0.7117 ( 10 – 3 ) ln ( 80 + x ) 0

= – 1.205 ( 10 – 3 )

12

Now u240 = 0 as the point is fixed to the ground and we obtain the contraction δ = u 0 = 1.205 ( 10 – 3 ) δ = 0.0012 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.27 Determine the contraction of a column due to its own weight. The specific weight γ, the Modulus of Elasticity E, the length L, and the radius R are as given. γ = 0.28 lb ⁄ in

3

L = 120 in

E = 3, 600 ksi R =

240 – x x R(x)

where, R and x are in inches.

L

Figure P4.27

Solution

δ=?

-----------------------------------------------------------Gravity is in the negative x-direction hence p = – γA or p = – γπR 2 = – 0.28π ( 240 – x ) . We can write: dN ------- = – p = 0.28π ( 240 – x ) dx

1

Integrating from x = L = 120, where N = 0 to any local x we obtain N

∫0

x

dN =

∫120 0.2817 ( 240 – x ) dx or

x ( 0.28π ) N = – ------------------- ( 240 – x ) 2 120 = – 0.14π [ ( 240 – x ) 2 – 120 2 ] 2

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

We can write: du = N = –----------------------------------------------------------------0.14π [ ( 240 – x ) 2 – 120 2 ] ------------dx EA ( 3 ⋅ 6 ) ( 10 6 )π ( 240 – x )

du 120 2 ------ = – 3.889 ( 10 – 6 ) ( 240 – x ) – ----------------dx 240 – x

or

3

Integrating from x = 0 to x = 120 u 120

∫u

du = – 3.889 ( 10 – 6 )

0

120

120 2 ( 240 – x ) – ---------------------- dx or ( 240 – x )

∫0

– ( 240 – x ) 2 u 120 – u 0 = – 3.889 ( 10 – 6 ) ---------------------------- + 120 2 ln ( 240 – x ) 2

4

Now point at x = 0 is fixed to the ground, i.e., u0 = 0 and we obtain the following. – 120 2 240 2 u 120 = – 3.889 ( 10 – 6 ) -------------- + ----------- + 120 2 ln ( 240 – x ) or 2 2 u 120 = – 45.18 ( 10 –3 ) inches

5 δ = 0.045 in

The top point moves in negative x-direction, hence it is in contraction

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.28 Determine the contraction of a column due to its own weight. The specific weight γ, the Modulus of Elasticity E, the length L, and the radius R are as given.

γ = 24 kN ⁄ m

3

E = 25 GPa

L = 10 m R = 0.5e where, R and x are in meters.

x R(x)

– 0.07x

L

Figure P4.28

δ=?

Solution

-----------------------------------------------------------Gravity is in the negative x-direction, hence p = – γA = – γπR 2 = – 24 ( 10 3 )π ( 0.5e –0.07x ) 2 or p = – 18.85 ( 10 3 )e 0.14x . We can write: dN ------- = – p = 18.85 ( 10 3 )e –0.14x dx

1

Integrating from x = L = 10 m, where N = 0 to any location x, we obtain N

∫0

x

dN =

∫10 ( 18.85π ) ( 10 3 )e –0.14x dx or N =

e – 0.14x – 18.85π ( 10 3 ) ⎛ ----------------⎞ ⎝ 0.14 ⎠

x

= – 134.64 ( 10 3 ) ( e – 0.14x – e – 1.4 )

2

10

We can write: du N- = –-------------------------------------------------------------------------134.64 ( 10 3 ) ( e – 0.14 – 0.2466 )- = – 6.857 ( 10 – 6 ) ( 1 – 0.2466e 0.14x ) ------ = ------dx EA 25 ( 10 9 )π [ ( 0.5 ) 2 e –0.14x ]

3

Integrating from x = 0 to x = 10 m u 10

∫u

0

du = – 6.857 ( 10 –6 )

10

∫0

( 1 – 0.2466e 0.14 ) dx

0.14x u 10 – u 0 = – 6.857 ( 10 – 6 ) ⎛ x – 0.2466e -------------------------------⎞ ⎝ ⎠ 0.14

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10

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

Point x = 0 is fixed to the ground i.e., u0 = 0 u 10 = – 6.857 ( 10 – 6 ) [ 10 – 1.761 ( e 1.4 – 1 ) ] = – 31.67 ( 10 – 6 )

5 δ = 0.317 mm

The top point moves in negative x-direction, hence it is contraction

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.29 The frictional force per unit length on a cast iron pipe being pulled from the ground varies as a quadratic function as shown. Determine the force F needed to pull the pipe out of ground and the elongation of the pipe before the pipe slips in terms of the modulus of elasticity E, area of cross-section A, length L and the maximum value of frictional force fmax. F x 2

L

x f = f max -----2L Figure P4.29

Solution

δ=?

F=?

-----------------------------------------------------------The distributed force is: p ( x ) = f max ( x 2 ⁄ L 2 ) . We can write: dN ------- = – p ( x ) = – f max ( x 2 ⁄ L 2 ) dx

1

N

Integrating from x = L, where N = 0, to any location x, we obtain

∫0 dN

x

= – f max

x2

- dx ∫L ----L2

or

x

f max – f max x 3 N = ------------------- = ----------- ( L 3 – x 3 ) 3L 2 3L 2 L

At x = 0

2 f max L F = -------------3

N = F; Thus,

We can write: f max du N------ = ------= ------------------ ( L 3 – x 3 ) dx EA 3EAL 2

3

Integrating from x = 0 to x = L

∫

uL

f max L du = ------------------ ( L 3 – x 3 ) dx or 3EAL 2 0 u0

∫

L f max f max 4 4 u L – u 0 = ------------------ ⎛ L 3 x – x-----⎞ = ------------------ ⎛ L 4 – L ------⎞ ⎝ ⎠ ⎝ 2 2 4 4⎠ 3EAL 3EAL 0

f max L 2 u L – u 0 = ----------------4EA

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.30 The spare wheel in an automobile is stored under the vehicle and raised and lowered by a cable. The wheel has a mass of 25 kg. The ultimate strength of the cable is 300 MPa and an effective Modulus of Elasticity of E = 180 GPa. The length of the hanging cable length is 36 cm. (a) For a factor of safety of four, determine to the nearest millimeter, the minimum diameter of the cable if failure due to rupture is to

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

be avoided. (b) What is the maximum extension of the cable for the answer in part (a).

Figure P4.30

Solution

σult = 300 MPa E = 180 GPa dmin = ? nearest millimeter d = uL-u0 = ?

m = 25 kg ksaftey = 4

L0 = 36 cm

-----------------------------------------------------------The free body diagram of the tire is as shown below. N

25g

N = 25g = 245.25 Newtons

1

The maximum allowable axial stress is: σ ult σ max = --------------- = 300 --------- = 75MPa 4 k saftey

2

The axial stress can be written as: 245.25 312.26 N σ = ---- = --------------------- ≤ σ max or ---------------- ≤ 75 ( 10 6 ) or d 2 ≥ 4.163 ( 10 – 6 ) or 2 A ( πd ⁄ 4 ) d2 d ≥ 2.04 ( 10 –3 ) m

3

To the nearest millimeter the diameter that satisfies the inequality is

d min = 3 mm

The extension is NL ( 245.25 ) ( 0.36 ) δ = -------- = --------------------------------------------- = 0.0694 ( 10 –3 ) m EA ( 180 ) ( π0.003 2 ⁄ 4 )

4 δ = 0.069 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.31 A adhesively bonded joint in wood (E = 1800 ksi) is fabricated as shown. If the total elongation of the joint between A and D is to be limited to 0.05 inches, determine the maximum axial force F that can be applied. F A

1 in 1 in x 36 in

C

B 5 in

4 in

F D

1 in

36 in

Figure P4.31

Solution

E = 1800 ksi

δ ≤ 0.05

Fmax = ?

-----------------------------------------------------------The internal axial force in all segments of the joint is N = F. The area of cross-sections of the segments are:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

A AB = ( 4 ) ( 1 ) = 4 in 2

A BC = 3 ( 4 ) ( 1 ) = 12 in 2

January 2014

A CD = ( 4 ) ( 1 ) = 4 in 2

The relative displacement of the ends of each segment can be written as shown below. N AB ( x B – x A ) F ( 36 ) u B – u A = ----------------------------------= ------------------------- = 5 ( 10 – 3 )F ( 1800 ) ( 4 ) EA AB

1

N BC ( x C – x B ) F(5) u C – u B = ---------------------------------- = ---------------------------- = 0.2318 ( 10 – 3 )F ( 1800 ) ( 12 ) EA BC

2

N CD ( x D – x C ) F ( 36 ) u D – u C = ----------------------------------= ------------------------- = 5 ( 10 – 3 )F ( 1800 ) ( 4 ) EA CD

3

Adding equations (1), (2) and (3)

u D – u A = F ( 5 + 0.2318 + 5 ) ( 10 –3 ) = 10.2318 ( 10 – 3 ) ≤ 0.05 or F ≤ 4.8869 kips or F max = 4886 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.32 A 5 feet long hollow rod is to transmit an axial force of 30 kips. The outer diameter of the rod must be 6 inches to fit existing attachments. The relative displacement of the two ends of the shaft is limited to 0.027 inches. The axial rod can be made of steel or aluminum. The Modulus of Elasticity E, the allowable axial stress σallow, and the specific weight γ, are given in Table 4.1. Determine the maximum inner diameter in increments of 1/8 inch of the lightest rod that can be used for transmitting the axial force and the corresponding weight. Table 4.1.Material Properties in Problem 4.32 E ksi

σallow ksi

γ lbs./in3

Steel

30,000

24

0.285

Aluminum

10,000

14

0.100

Material

Solution

L = 5 ft P=30 kips di= ? nearest 1/8th inch,

do= 6 in Δu = 0.024 W=? for lightest rod

-----------------------------------------------------------We calculate the area of cross-section for steel to satisfy the stress and deformation requirements. ( 30 ) ( 60 ) ( Δu ) S = -------------------------------- ≤ 0.027 3 ( 30 ) ( 10 )A S ( 30 ) σ S = ---------- ≤ 24 AS

A S ≥ 2.22 in

or

A S ≥ 1.25 in

or

2

1

2

2

Using similar calculations for Aluminum we obtain the following limits on AAl. ( 30 ) ( 60 ) ( Δu ) Al = ----------------------------- ≤ 0.027 3 10 ( 10 )A Al ( 30 ) σ Al = ---------- ≤ 14 A Al

or

or

2

A Al ≥ 6.67 in A Al ≥ 2.143 in

2

3

2

4 2

Thus if A S ≥ 2.22 in then it will meet both conditions in Eq’s (1) and (2). Similarly if A Al ≥ 6.67 in then it will meet both conditions in Eq’s (3) and (4). The internal diameters DS and DAl can be found as shown below:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

2 π 2 A S = --- ( 6 – D S ) ≥ 2.22 4 2 π 2 A Al = --- ( 6 – D Al ) ≥ 6.67 4

January 2014

D S ≤ 5.759 in 5

D Al ≤ 5.245 in

Rounding downwards to the closest 1/8th inch, we obtain: D S = 5.75 in

D Al = 5.125 in

We can find the weight of each material by taking the product of the specific weight and the volume of a hollow cylinder as shown below: 2 π 2 W S = ( 0.285 ) --- ( 6 – 5.75 ) ( 60 ) = 39.45 lbs 4

2 π 2 W Al = ( 0.1 ) --- ( 6 – 5.125 ) ( 60 ) = 45.87 lbs 4

The rod should be of steel. 3 4

The inside diameter should be d i = 5 --- inch . The corresponding weight is W S = 39.5 lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.33 A hitch for an automobile is to be designed for pulling a maximum load of 3,600 lbs. A solidsquare-bar fits into a square-tube, and is held in place by a pin as shown. The allowable axial stress in the bar is 6 ksi, the allowable shear stress in the pin is 10 ksi, and the allowable axial stress in the steel tube is 12 ksi. To the nearest 1/16th of an inch, determine the minimum cross-sectional dimensions of the pin, the bar and the tube. Negect stress concentration.(Note: Pin is in double shear) Square Tube Pin

Square Bar

Figure P4.33

Solution

σ b ≤ 6 ksi

P = 3,600lbs σ s = 12 ksi

dp = ?

τ p = 10 ksi

ab = ?

double shear

bs = ? nearest 1116 inch

-----------------------------------------------------------The two-dimensional picture of the hitch and the cross-section of the tube are shown below.

bs

P

P

ab

dp

ab ab

dp

bs

bs

As pin is in double shear: 2V = P or V = 180 V 1800 - ≤ 10, 000 d p2 ≥ 0.2292 The shear stress is: τ p ≥ ------ = ----------------------2 Ap

( πdp ⁄ 4 )

d p ≥ 0.4787 inch

d p = 0.5 in

The axial force in bar is N = P = 3600. The minimum area of cross-section in the bar is is at the pin hole. 3600 The axial stress is: σ b = --------------------------- ≤ 6000 or ab ( ab – dp )

a b2 – 0.5a b ≥ 0.6 or a b ≥ 1.063

1 a b = 1 --- in 8

The axial force in the steel tube is Ns = P = 3600. The minimum area of cross-section in the tube is at the pin hole.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

3600 The axial stress is: σ s = --------------------------------------------------------------------------- ≤ 12, 000 or b s2 – 0.5b s – ( 1.125 ) ( 1.125 – 0.5 ) ≥ 0.3 or [ ( bs ) ( bs – dp ) – ( ab ) ( ab – dp ) ]

b s2 – 0.5b s – 1.003125 ≥ 0

5 b s = 1 ------ in 16

b s ≥ 1.282

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.34 An axial rod has a constant axial rigidity EA and is acted upon by a distributed axial force p(x). If at section A the internal axial force is zero show that the relative displacement of section at B with respect to displacement of section at A is given by 1 u B – u A = -------EA

Solution

N(xA)= 0

xB

∫ ( x – xB )p ( x ) dx

4.15

xA

uB - uA = ?

-----------------------------------------------------------We can write: dN = –p ( x ) dx

1

Integrating from x = xA to any location x, we obtain the following. N

x

∫

dN = –

NA = 0

x

∫ p ( x ) dx

N(x) = –

or

xA

∫ p ( x ) dx

2

xA

(x) ------------ from xA to xB, we obtain the following. Integrating the equation du = N dx

EA

uB

xA

∫ du uA

=

∫

N(x) ------------ dx EA

x

1 u B – u A = -------EA

or

xA

∫ N ( x ) dx

3

xA

Integrating by parts, the above equation can be written as x xB 1 u B – u A = -------- xN ( x ) x – EA A

dN

∫ x d x dx

4

xA

Imposing the limits and substituting Eq. (1), we obtain the following. x

1 u B – u A = -------- x B N ( x B ) – N ( x A ) – EA

∫ x ( –p ( x ) ) dx

5

xA

Substituting Eq. (2) at x = xB and noting that N(xA)= 0, we obtain xB ⎧ xB ⎫ ⎪ ⎪ 1 u B – u A = -------- x B ⎨ – p ( x ) dx ⎬ – 0 + xp ( x ) dx EA ⎪ ⎪ xA ⎩ xA ⎭

∫

∫

6

Noting that xB is a constant and can be taken inside the integral, we obtain: 1 u B – u A = -------EA

xB

∫ ( x – xB )p ( x ) dx

7

xA

Eq. (7) is same as Eq. 4.15.

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

4.35 A composite laminated bar made from n materials is shown in Figure P4.35. Ei and Ai are the modulus of elasticity and cross sectional area of the ith material. (a) If Assumptions from 1 through 5 are valid, show that the stress ( σ xx ) i in the ith material is given Equation 4.16a, where N is the total internal force at a cross section. (b) If Assumptions 7 through 9 are valid, show that relative deformation u 2 – u 1 is given by Equation 4.16b.(c) Show that for E1=E2=E3....=En=E Equations (4.16a) and (4.16b) give the same results as Equations (4.8) and (4.10). NE i ( σ xx ) i = --------------------

y

(4.16a)

n

∑ Ej Aj j

1

N ( x2 – x1 ) u 2 – u 1 = --------------------------

En

x

z

Ei

(4.16b)

n

∑ Ej Aj j

E2 E1

Figure P4.35

1

Solution

-----------------------------------------------------------The axial strain is: du (x) dx

ε xx =

1

The axial stress is: σ xx = E

du (x) dx

2

The internal axial force can be written as: N =

∫ σxx dA A

=

du

∫ E d x dA A

N =

=

du du E dA = dx dx

∫

∫ E1 dA + ∫ E2 dA +

A

A1

du [ E A + E2 A2 + dx 1 1

⋅

du ⋅ + En An ] = dx

+

⋅

+

⋅

+

A2

∫ En dA

or

An n

∑ Ej Aj

or

j=1

du = -------------------N dx n Ej Aj

3

∑

For the ith material as ( σ xx ) i = E i

du , where dx

j

1

( σ xx ) is the axial stress in the ith material, we obtain: i

N ( σ xx ) i = E i --------------------n Ej Aj

4

∑

j=1

Equation (4) is same as Equation (4.16a). Assuming all quantities on right hand side of Equation (3) are constant, then we can write: u2 – u1 du = ---------------N - = -------------------dx x2 – x1 n Ej Aj

or

N ( x2 – x1 ) u 2 – u 1 = -------------------------

∑

j

1

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n

∑ Ej Aj j

1

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

Equation (5) is same Equation (4.16b). If E1=E2=E3....=En=E then n

n

∑

Ej Aj = E

j=1

∑ Aj

= EA

6

j=1

Substituting Equation (6) into Equation (4) and (5) we obtain: N N ( σ xx ) i = E -------- = ---EA A

7

N ( x2 – x1 ) u 2 – u 1 = -------------------------EA

8

Equations (7) and (8) are same as Equations (4.8) and (4.10)

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.36 The stress-strain relationship for a non-linear material is given by the power law σ = Eε n . If all assumptions except Hooke’s law are valid, show 1 --n N ⎛ ⎞ Δu = -------- L ⎝ EA⎠

4.16

and the axial stress σxx is given by Equation 4.8, where N is the internal axial force, L is the length of the rod, A is the area of cross-section, and Δu is the relative displacement of the rod.

Solution

σ = Eε

n

σxx = ?

Δu = ?

-----------------------------------------------------------From kinematics we have ε xx =

du ( x ) . Substituting this in the given stress-strain relationship, we obtain. dx σ xx = E

n du (x) dx

1

The normal stress and internal normal force are related as N =

∫ σxx dA

2

A

Substituting Eq. (1) into Eq. (2) and assuming homogenous material we obtain the following: N =

∫

E

n n du du ( x ) dA = E (x) dx dx

A

∫

dA = EA

n du ( x ) or dx

0 n du N ( x ) = -------dx EA

3 σ xx = N ---A

Substituting Eq. (3) into Eq. (1), we obtain: Eq. (2) can be re-written as: du N ( x ) = -------dx EA

1 --n

4

Assuming all quantities on the right in Eq (4) are constant over the length of the shaft, then du = Δu ------- . dx

L

1 ---

Substituting this in Eq. (4) we obtain:

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N n Δu = -------- L EA

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.37 Determine the elongation of a rotating bar in terms of the rotating speed ω, density ρ, length L, area of cross-section A, and Modulus of Elasticity E. (Hint: the body force per unit volume is ρω2x). ω x L

Figure P4.37

Δu = f(ω, ρ, L, A, E) = ?

Solution

-----------------------------------------------------------The free body diagram of a differential element is shown below. N+dN

N

(ρω2x) Adx dx

By equilibrium N + dN – N = – ρw 2 xAdx or dN = ρw 2 Axdx

At x = L N

∫0

dN =

∫

1

N = 0 . Integrating Eq (1) from x = L to any location x we obtain x

x

x2 – ρw 2 Ax dx or N – 0 = ρw 2 A ----or 2 L L ρw 2 N = ---------- A ( L 2 – x 2 ) 2

2

ρw 2 2 du N- = --------- ( L – x2 ) ------ = ------2E dx EA

3

We have the following.

Integrating Eq (3) from x = 0 to x = L we obtain: ρw 2 x3 Δu = u L – u 0 = ---------- ⎛ L 2 x – -----⎞ ⎝ 2E 3⎠

L 0

∫

uL

ρw 2 L du = ---------- ( L 2 – x 2 ) dx or 2E 0 u0

∫

ρw ------⎞ or Δu = ---------- ⎛⎝ L 3 – L 2E 3⎠ 2

ρw 2 L 3 Δu = ----------------3E

3

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.38

Consider the dynamic equilibrium of differential element shown in Figure P4.38. Where N is the

internal force, ρ is the density, A is the area of cross-section and N

2

∂u ∂t

2

N+dN

is acceleration. =

Figure P4.38 dx

dx

2 ρA ∂ u dx 2 ∂t

By substituting for N from Equation 4.8 into the dynamic equilibrium equation derive the following wave equation: 2 2 2 ∂ u ∂ u = c 2 2 ∂x ∂t

c =

E --ρ

4.17

The material constant c is the velocity of propagation of sound in the material. Solution

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

-----------------------------------------------------------2

By equilibrium of forces in Fig. P4.36 we obtain: N + dN – N = ρA

∂ u ∂t

2

or

2

∂ u dN ------- = ρA 2 dx ∂t

1

∂u N = EA ----∂x

2

We can write

2

∂ u ∂u Substituting Eq(2) into Eq(1), we obtain ∂ ⎛⎝ EA -----⎞⎠ = ρA 2 ∂x ∂x ∂t 2

2

∂ u

2

2

∂ u

∂ u Assuming EA is not a function of x we obtain the following. EA 2 = ρA 2 or ∂ u = c 2 2 2 ∂x ∂t ∂x ∂t

Above equation is same as Eq(4-17)

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.39 Show by substitution that the functions f(x - ct) and g(x + ct) satisfy the wave Equation 4.17. Solution

-----------------------------------------------------------∂f ∂y ∂f Consider the solution: u = f ( x – ct ) . Substituting y = x – ct we obtain: ∂-----u = ----- ----- = ----. We obtain the ∂x

∂y ∂x

∂y

following: 2

∂ u ∂x

2

2

=

∂f ⎞ ∂ ⎛ ----∂f ⎞ ∂-----y ∂ ⎛ ----∂ f = = 2 ∂ y ⎝ ∂y⎠ ∂x ∂ x ⎝ ∂y⎠ ∂y

3

∂f ∂y ∂f We can also write: ∂-----u = ----- ----- = – c ----- . We obtain the following. ∂t

∂y ∂t

∂y

2

2

∂ ∂f ∂ ⎛ ----∂f ∂ f –c ⎞ ( –c ) = c 2 = ⎛ – c -----⎞ = 2 2 ∂ t ⎝ ∂y⎠ ∂ y ⎝ ∂y⎠ ∂y ∂t ∂ u

4

Substituting Eq.(2) and (3) in Eq. 4-17, we observe that the equation is identically satisfied, hence f is a solution of Eq. 4-17. 2

2

∂g ∂ ∂g ∂y Consider now u = g ( x + ct ) . Substituting z = x + ct we obtain: ∂ u = ∂ ⎛⎝ -----⎞⎠ = ⎛⎝ -----⎞⎠ ----- = ∂ g and ∂x

∂t

2

∂ x ∂z

∂ z ∂z ∂ x

∂z

2

2

2

∂ u

2

=

∂ g ∂ ⎛ ∂g⎞ ∂ ⎛ ∂g⎞ c ----- = c ----- ( c ) = c 2 2 ∂ t ⎝ ∂z⎠ ∂ z ⎝ ∂z ⎠ ∂z

which on substitution satisfies Eq 4-17. Hence g(x + ct) is a

solution of Eq 4-17.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.40

The strain displacement relationship for large axial strain is given by ε xx =

du 2 du 1 + --- ⎛ ⎞ d x 2 ⎝ d x⎠

4.18

where we recognize that as u is only a function of x and hence the strain from Equation 4.18 is uniform across the cross-section. For linear, elastic, homogenous material show:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

du = dx

and the axial stress σxx is given by Equation 4.1. Solution

January 2014

2N- – 1 1 + ------EA

4.19

-----------------------------------------------------------By Hooke’s Law du 1 du 2 σ xx = Eε xx = E ------ + --- ⎛ ------⎞ dx 2 ⎝ dx⎠

By static equivalency: N =

∫A σxx dA

=

∫E A

1

du 2 du 1 ------ + --- ⎛ ------⎞ dA ⎝ dx 2 dx⎠

du Assuming material is homogenous across the cross-section i.e. E is a constant, and nothing that ------ is a dx

2

1 du function of x and does not change across the cross-section, we obtain. N = E du ------ + --- ⎛ ------⎞ A or dx 2 ⎝ dx⎠ du 2 du 1 N ------ + --- ⎛ ------⎞ = -------⎝ 2 dx⎠ dx EA

2

---- as given by Eq. 4-9. Substituting Eq(2) into Eq(1) we obtain σ xx = E ⎛⎝ --------⎞⎠ = N EA A N

du N du 2 Eq. (3) can be re-written as ⎛⎝ ------⎞⎠ + 2 ⎛⎝ ------⎞⎠ – 2 -------- = 0 . dx EA dx 1 2N N Solving the quadratic, we obtain du ------ = --- ⎛ – 2 ± 4 + 4 ⎛ --------⎞ ⎞ or du ------ = ± 1 + 2 ⎛ --------⎞ – 1 ⎝ EA⎠ ⎠ ⎝ EA⎠ 2⎝ dx dx du The positive root is the admissible root, because with the negative root ------ is always negative and less that dx

-1 irrespective of the value of N i.e. applied forces with a positive root we obtain Eq 4-19.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.41 Table 4.2shows the measured radii of a solid tapered rod at several point along the axis. The rod is made of aluminum (E = 100 GPa) and has a length of 1.5 meters. Determine (a) elongation of the rod using numerical integration. (b) the maximum axial stress in the rod. R(x) 500 kN B

A x

Figure P4.41

Table 4.2.Data for Problem 4.41. x (m)

R(x) (mm)

x (m)

R(x) (mm)

0.0

100.6

0.8

60.1

0.1

92.7

0.9

60.3

0.2

82.6

1.0

59.1

0.3

79.6

1.1

54.0

0.4

75.9

1.2

54.8

0.5

68.8

1.3

54.1

0.6

68.0

1.4

49.4

0.7

65.9

1.5

50.6

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Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

E = 100 GPa

xA= 0

Δu = uB-uA = ?

xB= 1.5 m

January 2014

σmax= ?

-----------------------------------------------------------The internal axial force is N = 500 kN. We have the following. 3

–6

( 500 ) ( 10 ) 5 ( 10 ) du N = -------- = --------------------------------------------------- = ------------------9 2 2 dx EA ( 100 ) ( 10 ) ( πR ( x ) ) πR ( x )

1

5 In Eq. (1), R(x) is measured in meters. Eq. (1) can be rewritten as du = ----------------- , where now R(x) is in mil2

dx

πR ( x )

limeters. Integrating from xA= 0 to xB= 1.5, we obtain the following: x B = 1.5

uB

∫ du

=

uA

∫ xA = 0

1.5

5 ------------------ dx 2 πR ( x )

Δu =

or

∫ 0

5 ------------------ dx 2 πR ( x )

2

2

Representing f ( x ) = 1.5915 ⁄ [ R ( x ) ] and using numerical integration as described in Appendix B, we obtain the value of the integral on a spread sheet as shown in the table below. From the results in the table we see the elongation of the bar is: Δu = 0.60 mm The maximum stress will occur at the cross-section where area is the smallest, i.e., just before point B and can be found as shown below. 3

6 2 500 ( 10 ) N σ max = ------- = --------------------------- = 62.2 ( 10 ) N ⁄ m or 2 AB π ( 0.0506 )

σ max = 62.2 MPa ( T )

x (m)

R(x) (mm)

f(x)

Δu (10-3)(m)

x (m)

R(x) (mm)

f(x)

Δu (10-3)(m)

0.00

100.60

0.1573

0.0171

0.80

60.10

0.4406

0.2731

0.10

92.70

0.1852

0.0380

0.90

60.30

0.4377

0.3178

0.20

82.60

0.2333

0.0623

1.00

59.10

0.4557

0.3678

0.30

79.60

0.2512

0.0886

1.10

54.00

0.5458

0.4216

0.40

75.90

0.2763

0.1193

1.20

54.80

0.5300

0.4753

0.50

68.80

0.3362

0.1533

1.30

54.10

0.5438

0.5351

0.60

68.00

0.3442

0.1888

1.40

49.40

0.6522

0.5988

0.70

65.90

0.3665

0.2292

1.50

50.60

0.6216

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.42 Let the radius of the tapered rod in problem 4.41 be represented by the equation R ( x ) = a + bx . Using the data in Table 4.2 determine the constant a and b by the least-square method and then find the elongation of the rod by analytical integration. Solution E = 100 GPa a = ? b = ? xA= 0 xB= 1.5 m Δu = uB-uA = ?

-----------------------------------------------------------We develop the equations for the least square method for a linear representation as described in Appendix B. This difference between the radius value and the value obtained by substituting x = xi in the equation R ( x ) = a + bx is the error ei that can be written as given below. e i = R i – R ( x i ) = R i – ( a + bx i ) N

We define the error E as E =

1

∑ ei . The error E is minimized with respect to coefficients a, and b to gen2

i=1

erate a set of linear algebraic equations as shown below.

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∂E = 0 ∂a ∂E = 0 ∂b

or

∂e i

∑ 2ei ∂ a ""

or

∑

2e i

January 2014

= 0

or

∑ 2 [ Ri – ( a + bxi ) ] [ –1 ]

= 0

= 0

or

∑ 2 [ Ri – ( a + bxi ) ] [ –xi ]

= 0

∂e i ∂b

The above equations on the right can be rearranged and written in matrix form as shown below:

∑ xi ⎧⎨ 2 ∑ xi ∑ xi ⎩ N

⎧ a ⎫ = ⎪ ⎬ ⎨ b ⎭ ⎪ ⎩

⎫

∑ Ri ⎪⎬ ∑ xi Ri ⎪⎭

⎧ b 11 b 12 ⎧ a ⎫ ⎪ r1 ⎨ ⎬ = ⎨ b 21 b 22 ⎩ b ⎭ ⎪ r2 ⎩

or

⎫ ⎪ ⎬ ⎪ ⎭

2

The coefficients of the b-matrix and the r-vector can be determined by comparison to the matrix form of the equations on the left. The coefficients a and b can be determined by Cramer’s rule. Let D represent the determinant of the b matrix. By Cramer’s rule, we replace the first column in the matrix of b’s by the right hand side and find the determinant of the so constructed matrix and divide by D. Thus, the coefficients a, and b can be written as shown below. D = b 11 b 22 – b 12 b 21 a =

r 1 b 22 – r 2 b 12 r 1 b 12 ⁄ D = ------------------------------D r 2 b 22

b =

3 r 2 b 11 – r 1 b 21 b 11 r 1 ⁄ D = ------------------------------D b 21 r 2

4

The given data and Eq. (2) through (4) can be put in a spread sheet and the coefficients a and b can be found as shown in the table below as: a = 90.226 mm b = – 30.593 xi (m)

Ri (mm)

xi2

xi (m)

Ri (mm)

xi2

xi*Ri

10

0.90

60.30

0.8100

54.270

xi*Ri

1

0.00

100.60

0.0000

0.000

2

0.10

92.70

0.0100

9.270

11

1.00

59.10

1.0000

59.100

3

0.20

82.60

0.0400

16.520

12

1.10

54.00

1.2100

59.400

4

0.30

79.60

0.0900

23.880

13

1.20

54.80

1.4400

65.760

30.360

14

1.30

54.10

1.6900

70.330

15

1.40

49.40

1.9600

69.160

5

0.40

75.90

0.1600

6

0.50

68.80

0.2500

34.400

7

0.60

68.00

0.3600

40.800

16

1.50

50.60

2.2500

75.900

8

0.70

65.90

0.4900

46.130

bij & ri

12.0000

1076.50

12.4000

703.360

9

0.80

60.10

0.6400

48.080

D

54.4

a and b

90.226

-30.593

Thus R(x) = 90.226-30.593x, where x is in meters and R is in millimeters. Substituting this in Eq. (2) of problem 4.41 and integrating, we obtain the elongation as shown below. 1.5

1.5

–3 5 5 1 ----------------------------------------------------- dx = ⎛ ---⎞ -------------------------------------------------------------------= 0.5968 ( 10 ) m ⎝ π⎠ ( 30.593 ) ( 90.226 – 30.593x ) 2 π ( 90.226 – 30.593x ) 0 0 Thus the elongation of the bar is: Δu = 0.60 mm

Δu =

∫

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.43 Table 4.4 shows the value of distributed axial force at several point along the axis of a hollow steel (E = 30,000 ksi) rod. The rod has a length of 36 inches, an outside diameter of 1 inch, and an inside diameter of 0.875 inch. Determine (a) the displacement of the end A using numerical integration. (b) the maxi-

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

mum axial stress in the rod. p(x) lb/in

A

B

x

Figure P4.43

Table 4.4.

Solution

x (inches)

p (x) lbs./in

x (inches)

p (x) lbs./in

0

260

21

-471

3

106

24

-598

6

32

27

-645

9

40

30

-880

12

-142

33

-1035

15

-243

36

-1108

18

-262

E = 30,000 ksi

do= 1 inch

σmax= ?

di = 0.875 inch uA = ?

-----------------------------------------------------------The equilibrium equation dN + p ( x ) = 0 can be integrated from point A, where N = 0 to any location x, to dx

obtain the internal axial force as a function of x as shown below. N(x)

N(x) =

The area of cross-section is A =

2 π ( do

–

∫

x

∫

dN = –

NA = 0 2 di ) ⁄ 4 =

p ( x ) dx

1

xA = 0 2

0.1841in

.

–9 du = N ( x )- = N(x) -------------------------------------------------------- = 181.08 ( 10 )N ( x ) 6 dx EA ( 30 ) ( 10 ) ( 0.1841 )

2

Integrating from point A i.e., xA =0 to point B, i.e., xB =36. We note that point B is fixed to the wall, hence uB= 0. we obtain the following integral: uB = 0

x B = 36

∫

∫

uA

du =

36 –9

[ 181.08 ( 10 )N ( x ) ] dx

or

∫

–9

u A = – [ 181.08 ( 10 )N ( x ) ] dx

xA = 0

The axial stress σxx can be found as shown below: N σ xx = N ---- = --------------A 0.1841

3

0

4

Eq. (1) can be numerically integrated on a spread sheet to obtain N(xi), the value of internal force at any xi. Then, Eq (3) can be numerically integrated to obtain the elongation. Eq. (4) can be used to find the axial stress at various xi and the maximum value chosen by inspection. These calculations can be done on a spread sheet as shown in the table below. The negative sign for the displacement uA implies it moves in the negative x-direction, which means it displaces to the left. The answers are: u A = 0.0176 in. to the left and σ max = 73.7 ksi ( T )

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

xi (inches)

p(xi) (lbs./in)

N(xi) (lbs.)

[uA-u(xi+1)](10-3) (inches)

σxx(xi)

0

260

0.00

0.149

0.000

3

105.5

-548.25

0.503

-2.978

6

32.0

-754.50

0.942

-4.098

9

39.5

-861.75

1.368

-4.681

12

-142.0

-708.00

1.596

-3.846

15

-242.5

-131.25

1.462

-0.713

18

-262.0

625.50

0.824

3.398

21

-470.5

1724.25

-0.548

9.366

24

-598.0

3327.00

-2.862

18.072

27

-644.5

5190.75

-6.303

28.195

30

-880.0

7477.50

-11.145

40.617

33

-1034.5

10349.25

-17.640

36

-1108.0

13563.00

(ksi)

56.215 73.672

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.44 Let the distributed force p(x) in problem 4.43 be represented by the equation p ( x ) = a + bx + cx2 . Using the data in Table 4.4 determine the constant a, b, and c by the least-square method and then find the displacement of section at A by analytical integration. Solution a=? b=? c=? uA = ?

-----------------------------------------------------------The equilibrium equation dN + p ( x ) = 0 can be integrated from point A, where N = 0 to any location x, to dx

obtain the internal axial force as a function of x as shown below. N(x)

N(x) =

∫

x

∫

dN = –

NA = 0

The area of cross-section is A =

x

p ( x ) dx = –

xA = 0 2 2 π ( do – di )

∫

2

3

2 bx [ a + bx + cx ] dx = – ax + -------+ cx -------2 3

1

xA = 0 2

⁄ 4 = 0.1841in . 2

3

–9 bx du = N ( x )- = N(x) cx -------------------------------------------------------- = ( – 181.08 ) ( 10 ) ax + -------- + -------6 dx EA 2 3 ( 30 ) ( 10 ) ( 0.1841 )

2

Integrating from point A i.e., xA =0 to point B, i.e., xB =36. We note that point B is fixed to the wall, hence x B = 36

uB = 0

∫

uB= 0. we obtain the following integral:

du =

uA 2

3

3

xA = 0 4

ax bx cx u A = ( 181.08 ) ( 10 ) ⎛ -------- + -------- + --------⎞ ⎝ 2 6 12 ⎠ –9

∫

2

–9 bx cx ( – 181.08 ) ( 10 ) ax + -------- + -------- dx or 2 3

36

2

3

4

– 9 a ( 36 ) b ( 36 ) c ( 36 ) = ( 181.08 ) ( 10 ) ---------------- + ---------------- + ---------------2 6 12 0

3

Using the Least Square Method described in Appendix B, we obtain the value of the values of constants a, b, and c on a spread sheet as shown in the table below. The values of the constants are: a = 224.40 b = – 23.60 c = – 0.40 Substituting the values of the above constants in Eq. (3), we obtain the displacement of point A

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

2

3

January 2014

4

– 9 ( 224 ) ( 36 ) – 23.60 ) ( 36 ) - + ------------------------------( – 0.40 ) ( 36 ) - = – 0.017 or u = 0.017in to the left u A = ( 181.08 ) ( 10 ) --------------------------- + (---------------------------------A 2 6 12

xi

p(xi)

xi2

xi3

xi4

x*pi

xi2*pi

1

0.0

260.0

0.0

0.000E+00

0.000E+00

0.000E+00

0.000E+00

2

3.0

105.5

9.0

2.700E+01

8.100E+01

3.165E+02

9.495E+02

3

6.0

32.0

36.0

2.160E+02

1.296E+03

1.920E+02

1.152E+03

4

9.0

39.5

81.0

7.290E+02

6.561E+03

3.555E+02

3.200E+03

5

12.0

-142.0

144.0

1.728E+03

2.074E+04

-1.704E+03

-2.045E+04

6

15.0

-242.5

225.0

3.375E+03

5.063E+04

-3.638E+03

-5.456E+04

7

18.0

-262.0

324.0

5.832E+03

1.050E+05

-4.716E+03

-8.489E+04

8

21.0

-470.5

441.0

9.261E+03

1.945E+05

-9.881E+03

-2.075E+05

9

24.0

-598.0

576.0

1.382E+04

3.318E+05

-1.435E+04

-3.444E+05

10

27.0

-644.5

729.0

1.968E+04

5.314E+05

-1.740E+04

-4.698E+05

11

30.0

-880.0

900.0

2.700E+04

8.100E+05

-2.640E+04

-7.920E+05

12

33.0

-1034.5

1089.0

3.594E+04

1.186E+06

-3.414E+04

-1.127E+06

13

36.0

-1108.0

1296.0

4.666E+04

1.680E+06

-3.989E+04

-1.436E+06 -4.531E+06

bij & ri

234.0

-4945.0

5850.0

1.643E+05

4.918E+06

-1.513E+05

Cij

1.783E+09

1.897E+08

4.216E+06

2.971E+07

7.666E+05

2.129E+04

D

3.453E+09

ai

224.40

-23.60

-0.40

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.45 A rigid bar is hinged at C. The Modulus of Elasticity of the bar A is E= 30,000 ksi, the area of cross-section is A = 1.25 in2, and the length is 24 inches. Determine the applied force F if point B moves upward by 0.002 inches. D

C Rigid

B 125 in 25in

A

Solution

A = 1.25 in2

E = 30,000 ksi

F

δB = 0.002 in

L = 24 inch

-----------------------------------------------------------We draw the free body diagram of the rigid bar as shown in Fig. (a) (a) Cy (b) C D

x

Rigid

B

δD= δA

125 in NA

Compressive

25in

δB

125 in

Contraction 25in

F

By taking moment equilibrium about point C in Fig.(a), we obtain the following. 125N A = 25F

or

F = 5N A

1

We draw the exaggerated deformed shape as shown in Fig. (b) and write the deformation equations using

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

similar triangles. δ δA -------- = -----B125 25

or

δ A = 5δ B = 5 ( 0.002 ) = 0.01in

2

From Eq 4.27, we have the following: NA L N A ( 24 ) δ A = ---------- = ------------------------------------- = 0.01 EA ( 30, 000 ) ( 1.25 )

N A = 15.625 kips

or

3

Substituting Eq. (3) into Eq. (1), we obtain the force F as shown below. F = 5 ( 15.625 ) = 78.125kips or

F = 78.1 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.46 A rigid bar is hinged at C. The Modulus of Elasticity of the bar A is E= 30,000 ksi, the area of cross-section is A = 1.25 in2, and the length is 24 inches. Determine the applied force F if point B moves upward by 0.002 inches. C

D

B

Rigid

0.004 in

125 in 25in

A

Solution

A = 1.25 in2

E = 30,000 ksi

F

δB = 0.002 in

L = 24 inch

-----------------------------------------------------------We draw the free body diagram of the rigid bar as shown in Fig. (a) Cy (a) (b) Rigid

0.004

B

Compressive

δD

δA

125 in NA

25in

δB

125 in

Cx

D

25in Contraction

F

By taking moment equilibrium about point C in Fig.(a), we obtain the following. 125N A = 25F

F = 5N A or We draw the exaggerated deformed shape as shown in Fig. (b) and write the deformation equations. δ D = δ A + 0.04

δ δD -------- = -----B125 25

δ A = 5δ B – 0.004 = 5 ( 0.002 ) – 0.004 = 0.006in or

1

2

From Eq 4.27, we have the following: NAL N A ( 24 ) δ A = ---------- = ------------------------------------- = 0.006 EA ( 30, 000 ) ( 1.25 )

or

Substituting Eq. (3) into Eq. (1), we obtain the force F as shown below. F = 5 ( 9.375 ) = 46.875kips or

N A = 9.375 kips

3

F = 46.9 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.47 A rigid bar is hinged at C. The Modulus of Elasticity of the bar A is E= 100 GPa, the area of crosssection is A = 15 mm2, and the length is 1.2 m. Determine the applied force F if point B moves to the left by 0.75 mm.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

F

B

1.25 m

A

Solution

C

Rigid

D

2.5 m

A = 15 mm2

E = 100 GPa

δB = 0.75 mm

L = 1.2 m

-----------------------------------------------------------We draw the free body diagram of the rigid bar as shown in Fig. (a) B (b) F (a)

δB

1.25 m Rigid

D

C

Contraction

Cx

θ

2.5 m NA

θ

2.5 m

1.25 m C

δD = δA

Compressive Cy

By taking moment equilibrium about point C in Fig.(a), we obtain the following. 2.5N A = 1.25F

F = 2N A or We draw the exaggerated deformed shape as shown in Fig. (b) and write the deformation equations. δB δA ------ = --------2.5 1.25

1

–3

δ A = 2δ B = 2 ( 0.75 ) = 1.5mm = 1.5 ( 10 )m

or

2

From Eq 4.27, we have the following: NAL N A ( 1.2 ) –3 δ A = ---------- = ------------------------------------------------------ = 1.5 ( 10 ) 9 –6 EA ( 100 ) ( 10 ) ( 15 ) ( 10 )

3

N A = 1.875 ( 10 ) Newtons = 1.875kN

or

Substituting Eq. (3) into Eq. (1), we obtain the force F as shown below. F = 2 ( 1.875 ) = 3.75kN or

3

F = 3.75 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.48 A rigid bar is hinged at C. The Modulus of Elasticity of the bar A is E= 100 GPa, the area of crosssection is A = 15 mm2, and the length is 1.2 m. Determine the applied force F if point B moves to the left by 0.75 mm. B

F 1.25 m

1 mm

D

Rigid

C

2.5 m A

Solution

E = 100 GPa

A = 15 mm2

L = 1.2 m

δB = 0.75 mm

------------------------------------------------------------

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

We draw the free body diagram of the rigid bar as shown in Fig. (a) B (a) F (b)

δB θ

1.25 m D

Rigid

C

2.5 m Cy Compressive

NA

2.5 m

1mm Contraction δA

Cx

January 2014

θ

1.25 m C

δD

By taking moment equilibrium about point C in Fig.(a), we obtain the following. 2.5N A = 1.25F F = 2N A or We draw the exaggerated deformed shape as shown in Fig. (b) and write the deformation equations. δB δD ------ = --------2.5 1.25

δD = δA + 1

1

–3

δ A = 2δ B – 1 = 2 ( 0.75 ) – 1 = 0.5mm = 0.5 ( 10 )m

or

2

From Eq 4.27, we have the following: NA L N A ( 1.2 ) –3 δ A = ---------- = ------------------------------------------------------ = 0.5 ( 10 ) 9 –6 EA ( 100 ) ( 10 ) ( 15 ) ( 10 )

3

N A = 0.625 ( 10 ) Newtons = 0.625kN

or

3

Substituting Eq. (3) into Eq. (1), we obtain the force F as shown below. F = 2 ( 0.625 ) = 1.25kN or

F = 1.25 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.49 The roller at P slides in the slot due to the force F = 20 kN. Member AP has an area of cross-section of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. Determine the displacement of the roller. P

20

0m m

F

A 50o

Solution

A = 100 mm2

E = 200 GPa

δP = ?

F=20 kN

-----------------------------------------------------------We draw the free body diagram of the rigid bar as shown in Fig. (a) (a)

Extension

(b)

F

δA P

50o NA

P2 50o δP

P1

R Tensile A

By force equilibrium in the x-direction in Fig.(a), we obtain the following. N A cos 50 = F

N A = 31.11kN or 1 We draw the exaggerated deformed shape as shown in Fig. (b). As per small strain approximation, we need component of PP1 in the original direction of AP, i.e., PP2 represents the deformation of bar AP and is calculated as shown below. δ A = δ P cos 50

or

δ P = 1.556δ A

2

From Eq 4.27, we have the following:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

3 NAL –3 ( 31.11 ) ( 10 ) ( 0.2 ) δ A = ---------- = ---------------------------------------------------------- = 0.311 ( 10 )m = 0.311mm 9 –6 EA ( 200 ) ( 10 ) ( 100 ) ( 10 )

3

Substituting Eq. (3) into Eq. (2), we obtain the displacement of roller as shown below. δ P = 1.556δ A = ( 1.556 ) ( 0.311 ) = 0.484mm or δ P = 0.48 mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.50 The roller at P slides in the slot due to the force F = 20 kN. Member AP has an area of cross-section of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. Determine the displacement of the roller.

0m m

P

F

20

30o

A 50o

Solution

A = 100 mm2

E = 200 GPa

δP = ?

F=20 kN

-----------------------------------------------------------We draw the free body diagram of the rigid bar as shown in Fig. (a) (b)

(a) y

F

x

20o

P

20o

P2

δA

Extension

δP

P1

R NA

Tensile

A

By force equilibrium in the x-direction in Fig.(a), we obtain the following. N A cos 20 = F

N A = 21.28kN

or

1

We draw the exaggerated deformed shape as shown in Fig. (b). As per small strain approximation, we need component of PP1 in the original direction of AP, i.e., PP2 represents the deformation of bar AP and is calculated as shown below. δ A = δ P cos 20

or

δ P = 1.064δ A

2

From Eq 4.27, we have the following: 3 NA L –3 ( 21.28 ) ( 10 ) ( 0.2 ) δ A = ---------- = ---------------------------------------------------------- = 0.2128 ( 10 )m = 0.2128mm 9 – 6 EA ( 200 ) ( 10 ) ( 100 ) ( 10 )

Substituting Eq. (3) into Eq. (2), we obtain the displacement of roller as shown below. δ P = 1.064δ A = ( 1.064 ) ( 0.2128 ) = 0.2264mm or

3

δ P = 0.23mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.51 A rigid bar is hinged at C. The Modulus of Elasticity of the bar A is E= 30,000 ksi, the area of cross-section is A = 1.25 in2 and the length is 24 inches. Determine the axial stress in bar A and the displacement of point B on the rigid bar.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

B

C

D Rigid

0.004 in

125 in 25in

A

Solution

A = 1.25 in2 σA = ?

E = 30,000 ksi L = 24 inch

F = 50 kips

δD = ?

-----------------------------------------------------------We draw the free body diagram of the rigid bar as shown in Fig. (a) (a) (b) Cy Cx

D Rigid

0.004

B

δA

125 in NA

Compressive

25in

δB

125 in

δD

25in Contraction

F

By taking moment equilibrium about point C in Fig.(a), we obtain the following. 125N A = 25F

N A = F ⁄ 5 = 10kips

or

N 10 The axial stress in member A is: σ A = ------A- = --------- = 8.0 ksi or A

1

σ A = 8.0 ksi ( C )

1.25

We draw the exaggerated deformed shape as shown in Fig. (b) and write the deformation equations. δ δD -------- = -----B125 25

δ D = δ A + 0.004

( δ A + 0.004 ) δ B = -----------------------------5

or

2

From Eq 4.27, we have the following: NA L ( 10 ) ( 24 ) δ A = ---------- = -------------------------------------- = 0.0064 in EA ( 30, 000 ) ( 1.25 )

3

Substituting Eq. (3) into Eq. (2), we obtain the displacement δB as shown below. 0.0064 + 0.004 )- = 0.00208in δ B = (---------------------------------------5

δ B = 0.0021 in

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.52 A rigid bar is hinged at C. The Modulus of Elasticity of the bar A is E= 30,000 ksi, the area of cross-section is A = 1.25 in2 and the length is 24 inches. Determine the axial stress in bar A and the displacement of point B on the rigid bar. M= 2500 in-kips C

D Rigid

0.004 in

B 125 in

A

Solution

E = 30,000 ksi

A = 1.25 in2

25in

L = 24 inch

σA = ?

δD = ?

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

We draw the free body diagram of the rigid bar as shown in Fig. (a) (a) Cy M= 2500 in-kips (b) Cx

D Rigid

B

δA 25in

Compressive

δB

125 in

0.004

125 in NA

January 2014

δD

25in Contraction

F

By taking moment equilibrium about point C in Fig.(a), we obtain the following. 125N A = 2500

N A = 20kips

or

N 20 - = 16.0 ksi The axial stress in member A is: σ A = ------A- = --------A

σ A = 16.0 ksi ( C )

or

1.25

1

We draw the exaggerated deformed shape as shown in Fig. (b) and write the deformation equations. δ δD -------- = -----B125 25

δ D = δ A + 0.004

( δ A + 0.004 ) δ B = -----------------------------5

or

2

From Eq 4.27, we have the following: NA L ( 20 ) ( 24 ) - = 0.0128 in δ A = ---------- = ------------------------------------( 30, 000 ) ( 1.25 ) EA

3

Substituting Eq. (3) into Eq. (2), we obtain the displacement δB as shown below. ( 0.0128 + 0.004 ) δ B = ----------------------------------------- = 0.00336in 5

δ B = 0.0034 in

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.53 A steel (E = 30,000 ksi, ν = 0.28) rod passes through a copper ( E = 15,000 ksi, ν = 0.35) tube as shown in Figure P4.53. The steel rod has a diameter of 1/2 in. and the tube has an inside diameter of 3/ 4 in. and a thickness of 1/8 in. If the applied load is P = 25 kips, determine: (a) the movement of point A (b) the change in diameter of the steel rod. 24 in.

C P

B

A

Figure P4.53

16 in.

Solution

-----------------------------------------------------------2 2 π 1 π 2 π 3 The crosssectional areas are: A S = --- ⎛⎝ ---⎞⎠ = 0.19635 in ; A cu = --- ( 1 ) – --- ⎛⎝ ---⎞⎠ = 0.3461 in ; 4 2 4 4 4 We can draw the free body diagrams as shown 2

P

A

2

NS

NCu

NS B

From figure above the internal axial forces are: N S = P = 2.5 kips and N Cu = – N S = – 2.5 kips .

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

N Cu ( x B – x A ) –3 ( – 2.5 ) ( 16 ) u B – u C = ---------------------------------= ----------------------------------------- = – 7.761 ( 10 ) in ( 15000 ) ( 0.3461 ) E Cu A Cu

1

NS ( xA – xB ) ( 2.5 ) ( – 24 ) u B – u A = ------------------------------ = -------------------------------------------- = – 10.186 in ( 30000 ) ( 0.19635 ) ES AS

2

Adding Eq’s (1) and (2) we obtain:

–3

–3

u A – u C = ( – 7.761 – 10.186 ) ( 10 ) = ( – 17.947 ) ( 10 ) in

3

The axial stress and strain in steel can be found as shown below. σ –3 12.732 ε S = ------S- = ---------------- = 0.4244 ( 10 ) AS 30000

N 2.5 σ S = ------S- = ------------------- = 12.732 ksi AS 0.19635

The transverse strain and change of diameter can be found as: –3

–3

–3

( ε S ) tran = – ν S ε S = – 0.28 ( 0.4244 ) ( 10 ) = – 0.1188 ( 10 )

Noting that point C is fixed, the answers are:

Δd = d ( ε S ) tran = 0.059 ( 10 ) in –3

Δd = 0.059 ( 10 ) in.

u c = 0.0018 in. to left

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.54 A rigid bar ABC is supported by two aluminum cables (E = 10,000 ksi) with a diameter of 1/2 in. as shown in Figure P4.54. The bar is horizontal before the force is applied. Determine the angle of rotation of the bar from the horizontal when a force P= 5 kips is applied

5 ft

P

Figure P4.54

B

A

C

3 ft

5 ft

Solution

-----------------------------------------------------------2 π 1 The crosssectional area is: A = --- ⎛⎝ ---⎞⎠ = 0.19635 in 4 2 2

We can draw the free body diagram and the deformed shape as shown below: N1 N2 (a) (b) 8 ft =96 in. P

3 ft

δ1

θ

δ2

5 ft

By equilibrium of moment about C and force balance in y-direaction we obtain: ( 8 )N 1 = ( 5 ) ( P )

or

5(5) N 1 = ----------- = 3.125 kips 8

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

N1 + N2 = P

or

January 2014

N 2 = 5 – 3.125 = 1.875 kips

2

The deformations are N1 L –3 ( 3.125 ) ( 5 ) ( 12 ) δ 1 = ---------= ----------------------------------------------- = 95.49 ( 10 ) in EA ( 10, 000 ) ( 0.19635 )

3

N2 L –3 ( 1.875 ) ( 5 ) ( 12 ) = δ 2 = ---------= ---------------------------------------------57.29 ( 10 ) in EA ( 10, 000 ) ( 0.19635 )

4

–3 –3 δ1 – δ2 –3 95.49 ( 10 ) – 57.29 ( 10 ) tan θ = ---------------- = ----------------------------------------------------------------- = 0.3978 ( 10 ) 96 96

5

From geometry we obtain:

θ = 0.0004 rads CCW

The angle of rotation is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.55 Two rigid beams are supported by four axial steel (E = 210 GPa) rods of diameter 10 mm as shown in Figure P4.55. Determine the angle of rotation of the bars from the horizontal no load position when a force of F = 5 kN is applied

2m

A

B 1.2 m

1.6 m

D

C 1.2 m

4m

F

Figure P4.55 2m

1.2m

4.56 Two rigid beams are supported by four axial steel (E = 210 GPa, σyield = 210 MPa) rods of diameter 20 mm as shown in Figure P4.55. For a factor of safety of 1.5, determine the maximum value of force F that can be applied without causing any rod to yield. 4.57 A rigid bar ABC is supported by two aluminum cables (E = 10,000 ksi) with a diameter of 1/2 in. as shown in Figure P4.57. Determine the extensions of cables CE and BD when a force P= 5 kips is applied. E

C

5 ft

P D

B 40o

5 ft Figure P4.57

A

Solution

------------------------------------------------------------

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

From geometry: CE tan 40 = ---------------------( 10 ) ( 12 )

or

( 10 ) ( 12 ) cos 40 = ---------------------AC

or

CE BD = -------- = 50.34 in. 2

1

AC BC = AB = -------- = 78.32 in. 2

2

CE = 100.69 in. AC = 156.65 in.

We can draw the following free body diagram and deformed shape: C

δCE

NCE

P

C

δBD

NBD

B

B

40o

Ay Ax

A

A

By moment equilibrium about A: 120N CE + 60N BD = 156.65P = 783.25 kips

3

By similar triangles in deformed shape: δ BD δ CE --------- = --------AC AB

or

δ CE = 2δ BD

4

2 π 1 The crosssectional area is: A = --- ⎛⎝ ---⎞⎠ = 0.1964 in. 4 2 The deformation in each bar can be written as: 2

N CE ( 100.69 ) N CE L CE δ CE = --------------------= ---------------------------------------EA ( 10000 ) ( 0.1964 )

or

δ CE = 51.28 ( 10 )N CE

N BD ( 50.34 ) N BD L BD δ BD = ---------------------= ---------------------------------------EA ( 10000 ) ( 0.1964 )

or

δ BD = 25.64 ( 10 )N BD

–3

–3

5 6

Substituting Eqs. 5 and 6 into Eq. 4 we obtain: –3

–3

51.28 ( 10 )N CE = 2 [ 25.64 ( 10 )N BD ]

or

N CE = N BD

7

From Eq. 3 we obtain: 120N CE + 60N CE = 783.25 kips

N CE = 4.34 kips

N BD = 4.34 kips

8 9

The deformations from Eqs. 5 and 6 are: –3

δ CE = 51.28 ( 10 ) ( 4.34 ) = 0.222 in. –3

δ BD = 25.64 ( 10 ) ( 4.34 ) = 0.111 in.

The answers are:

10 11

δ CE = 0.222 in. ; δ BD = 0.111 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.58 A rigid bar ABC is supported by two aluminum cables (E = 10,000 ksi) as shown in Figure P4.58. The yield stress of aluminum is 40 ksi. If the applied force P = 10 kips, determine the minimum diameter

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

of cables CE and BD to the nearest 1/16 in. E

C

5 ft

P D

B 40o

5 ft Figure P4.58

A

Solution

-----------------------------------------------------------From geometry: CE tan 40 = ---------------------( 10 ) ( 12 )

or

( 10 ) ( 12 ) cos 40 = ---------------------AC

or

CE BD = -------- = 50.34 in. 2

1

AC BC = AB = -------- = 78.32 in. 2

2

CE = 100.69 in. AC = 156.65 in.

We can draw the following free body diagram and deformed shape: C

δCE

NCE

P B

C

δBD

NBD

B

40o

Ay A

Ax

A

By moment equilibrium about A: 120N CE + 60N BD = ( 156.65 sin 40 )P = 1006.9 kips

3

By similar triangles in deformed shape: δ BD δ CE --------- = --------AC AB

or

δ CE = 2δ BD

4

The deformation in each bar can be written as: N CE ( 100.69 ) N CE L CE δ CE = --------------------= -------------------------------EA EA

5

N BD L BD N BD ( 50.34 ) δ BD = ---------------------= ----------------------------EA EA

6

Substituting Eqs. 5 and 6 into Eq. 4 we obtain: N BD ( 50.34 ) N CE ( 100.69 ) ------------------------------- = 2 -----------------------------EA EA

or

N CE = N BD

7

From Eq. 3 we obtain: 120N CE + 60N CE = 1006.9 kips

N CE = 5.594 kips

8

The normal stress in each bar is the same as internal force and area is the same, hence

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

N CE 5.594 - = -------------- ≤ 40 ksi σ CE = ---------A π 2 --- d 4

2 4 5.594 d ≥ ⎛ ---⎞ ⎛ -------------⎞ ⎝ π⎠ ⎝ 40 ⎠

or

January 2014

d ≥ 0.422 in.

or

7 d min = ------ in. 16

The answer is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.59 A rigid bar ABC is supported by two aluminum cables (E = 10,000 ksi) with a diameter of 1/2 in. as shown in Figure P4.57. The yield stress of aluminum is 40 ksi. Determine the maximum force P to the nearest pound that can be applied. E

C

5 ft

P D

B 40o

5 ft Figure P4.59

A

Solution

-----------------------------------------------------------From geometry: CE tan 40 = ---------------------( 10 ) ( 12 )

or

10 ) ( 12 )cos 40 = (--------------------AC

or

CE BD = -------- = 50.34 in. 2

1

BC = AB = AC -------- = 78.32 in. 2

2

CE = 100.69 in. AC = 156.65 in.

We can draw the following free body diagram and deformed shape: C

δCE

NCE

P B

C

δBD

NBD

B

40o

Ay A

Ax

A

By moment equilibrium about A: 120N CE + 60N BD = 156.65P

3

By similar triangles in deformed shape: δ BD δ CE --------- = --------AC AB

or

δ CE = 2δ BD

4

2 π 1 The crosssectional area is: A = --- ⎛⎝ ---⎞⎠ = 0.1964 in. 4 2 2

The deformation in each bar can be written as:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

N CE ( 100.69 ) N CE L CE δ CE = --------------------= ---------------------------------------EA ( 10000 ) ( 0.1964 )

or

δ CE = 51.28 ( 10 )N CE

N BD ( 50.34 ) N BD L BD δ BD = ---------------------= ---------------------------------------EA ( 10000 ) ( 0.1964 )

or

δ BD = 25.64 ( 10 )N BD

–3

5

–3

6

Substituting Eqs. 5 and 6 into Eq. 4 we obtain: –3

–3

51.28 ( 10 )N CE = 2 [ 25.64 ( 10 )N BD ]

or

N CE = N BD

7

From Eq. 3 we obtain: 120N CE + 60N CE = 156.65P

N CE = 0.868P kips

8 9

N BD = 0.868P kips

The normal stress in each bar is the same as internal force and area is the same, hence N CE 0.868P- ≤ 40 ksi σ CE = ---------- = -------------------A [ 0.1964 ]

P ≤ 9.0507 kips

or

P ≤ 9050.7 lb

or

The answer is:

P max = 9050 lb

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.60 A force F= 20 kN is applied to the roller that slides inside a slot. Both bars have an area of crosssection of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. Bar AP and BP have lengths of LAP= 200 mm and LBP= 250 mm respectively. Determine the displacement of the roller and axial stress in bar A .

B

110o

A

F P

F=20 kN E = 200 GPa A = 100 mm2 LAP= 200 mm LBP= 250 mm δP = ?

Solution

σA = ?

-----------------------------------------------------------We draw the free body diagram of the rigid bar as shown in Fig. (a) (a) NB (b)

Β

Compressive Tensile

o

110

70o

Contraction

F

NA

δB

Α

R

P2

Extension

70o P

δA=δP

P1

By force equilibrium in the x-direction in Fig.(a), we obtain the following. N A + N B cos 70 = F

or

3

N A + 0.342N B = 20 ( 10 )

1

We draw the exaggerated deformed shape as shown in Fig. (b). As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP1 represents the deformation of bar AP and PP2 represents the deformation of bar BP and are calculated as shown below. δA = δP δ B = δ P cos 70 = 0.342δ P

2 3

From Eq 4.27, we have the following:

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N A ( 0.2 ) NA LA δ A = -------------- = --------------------------------------------------------- = δP 9 –6 EA ( 200 ) ( 10 ) ( 100 ) ( 10 )

January 2014

6

or

N A = 100δ P ( 10 ) 4

N B ( 0.25 ) NB LB δ B = -------------- = --------------------------------------------------------- = 0.342δ P 9 –6 EA ( 200 ) ( 10 ) ( 100 ) ( 10 )

6

or

N B = 27.36δ P ( 10 )

Substituting Eqs. (4) and (5) into Eq. (1), we obtain the displacement δP as shown below. 6

3

[ ( 100δ P ) + 0.342 ( 27.36δ P ) ] ( 10 ) = 20 ( 10 )

5

–3

δ P = 0.1829 ( 10 )m

or

6

δ P = 0.18 mm –3

6

3

Substituting Eq.(6) into Eq. (4) we obtain: N A = ( 100 ) ( 0.1829 ) ( 10 ) ( 10 ) = 18.29 ( 10 ) Newtons 3

N 6 18.29 ( 10 ) The axial stress in member A is: σ A = ------A- = ----------------------------- = 182.9 ( 10 ) or A

–6

( 100 ) ( 10 )

σ A = 183 MPa ( T ) )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.61 A force F= 20 kN is applied to the roller that slides inside a slot. Both bars have an area of crosssection of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. Bar AP and BP have lengths of LAP= 200 mm and LBP= 250 mm respectively. Determine the displacement of the roller and axial stress in bar A. B

60o P

30o A

Solution

F

A = 100 mm2 δP = ?

F=20 kN E = 200 GPa LAP= 200 mm LBP= 250 mm

σA = ?

-----------------------------------------------------------We draw the free body diagram of the rigid bar as shown in Fig. (a) By force equilibrium in the y-direction in Fig.(a), we obtain the following. N A sin 30 + N B sin 60 = F

(a)

60o

R

30o NA

Contraction

F Compressive

1

Β

(b)

NB Tensile

3

0.5N A + 0.866N B = 20 ( 10 )

or

Α

δA

P 60o δB P3 δ P o P2 60 P1

Extension

We draw the exaggerated deformed shape as shown in Fig. (b) As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP3 represents the deformation of bar AP and PP2 represents the deformation of bar BP and are calculated as shown below. δ A = δ P cos 60 = 0.5δ P

δ B = δ P sin 60 = 0.866δ P

2

We have the following:

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N A ( 0.2 ) NA LA δ A = -------------- = --------------------------------------------------------- = 0.5δ P 9 –6 EA ( 200 ) ( 10 ) ( 100 ) ( 10 )

or

N B ( 0.25 ) NB LB δ B = -------------- = --------------------------------------------------------- = 0.866δ P 9 –6 EA ( 200 ) ( 10 ) ( 100 ) ( 10 )

or

January 2014

6

N A = 50δ P ( 10 ) 3 6

N B = 69.28δ P ( 10 )

Substituting Eqs. (3) and (4) into Eq. (1), we obtain the displacement δP as shown below. 6

3

[ 0.5 ( 50δ P ) + 0.866 ( 69.2δ P ) ] ( 10 ) = 20 ( 10 )

4

–3

δ P = 0.2353 ( 10 )m

or

5

δ P = 0.24 mm –3

6

3

Substituting Eq.(6) into Eq. (4) we obtain: N A = ( 50 ) ( 0.2353 ) ( 10 ) ( 10 ) = 11.76 ( 10 ) Newtons 3

N 6 11.76 ( 10 ) The axial stress in member A is: σ A = ------A- = ----------------------------- = 117.6 ( 10 ) or –6

A

( 100 ) ( 10 )

σ A = 118MPa ( C ) )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.62 A force F= 20 kN is applied to the roller that slides inside a slot. Both bars have an area of crosssection of A = 100 mm2 and a Modulus of Elasticity E = 200 GPa. Bar AP and BP have lengths of LAP= 200 mm and LBP= 250 mm respectively. Determine the displacement of the roller and axial stress in bar A

B

75o

30o

A

P F

Solution

A = 100 mm2 δP = ?

F=20 kN E = 200 GPa LAP= 200 mm LBP= 250 mm

σA = ?

-----------------------------------------------------------We draw the free body diagram of the rigid bar as shown in Fig. (a) (a) (b) Tensile

75o

30o

Tensile

Α R

NA

Β

NB

F

30o Extension 75o P δA P3 ExtensionδB δP P2 P1

By force equilibrium in the y-direction in Fig.(a), we obtain the following. N A cos 75 + N B cos 30 = F

or

3

0.2588N A + 0.8660N B = 20 ( 10 )

1

We draw the exaggerated deformed shape as shown in Fig. (b). As per small strain approximation, we need component of PP1 in the original direction of AP and BP, i.e., PP1 represents the deformation of bar AP and PP2 represents the deformation of bar BP and are calculated as shown below. δ A = δ P cos 75 = 0.2588δ P δ B = δ P cos 30 = 0.8660δ P

2 3

From Eq 4.27, we have the following:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

N A ( 0.2 ) NA LA δ A = -------------- = --------------------------------------------------------- = 0.2588δ P 9 –6 EA ( 200 ) ( 10 ) ( 100 ) ( 10 )

or

N B ( 0.25 ) NB LB δ B = -------------- = --------------------------------------------------------- = 0.866δ P 9 –6 EA ( 200 ) ( 10 ) ( 100 ) ( 10 )

or

January 2014

6

N A = 25.88δ P ( 10 ) 4 6

N B = 69.28δ P ( 10 ) 5

Substituting Eqs. (4) and (5) into Eq. (1), we obtain the displacement δP as shown below. 6

3

[ 0.2588 ( 25.88δ P ) + 0.866 ( 69.2δ P ) ] ( 10 ) = 20 ( 10 )

–3

δ P = 0.3002 ( 10 )m

or

6

δ P = 0.30 mm –3

6

3

Substituting Eq.(6) into Eq. (4) we obtain: N A = ( 25.88 ) ( 0.3002 ) ( 10 ) ( 10 ) = 7.769 ( 10 ) Newtons 3

N 6 7.769 ( 10 ) The axial stress in member A is: σ A = ------A- = ----------------------------- = 77.69 ( 10 ) or A

–6

( 100 ) ( 10 )

σ A = 77.7 MPa ( T ) )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.63 An aluminum (E = 70 GPa, σyield = 280 MPa, ν = 0.28) wire of diameter 0.5 mm is to hang two flower pots of equal mass as shown in Figure P4.63. (a) Determine the maximum mass of the pots to the nearest gram that can be hung if yielding is to be avoided in all wires. (b) For the maximum mass what is the percentage change in the diameter of the wire BC.. 225 mm

600 mm

A 200 mm 340 mm

Flower Pot Flower Pot

Figure P4.63

4.64 An aluminum (E = 70 GPa, σyield = 280 MPa, ν = 0.28) wire is to hang two flower pots of equal mass of 5 kg as shown in Figure P4.64. Determine the minimum diameter of the wires to the nearest 1/10 of a millimeter if yielding is to be avoided in all wires. 225 mm

600 mm

A 200 mm 340 mm

Flower Pot Flower Pot

Figure P4.64

4.65 An aluminum (Eal=10,000 ksi and νal= 0.25) and steel (Es=30,000 ksi and νs= 0.28) hollow cylinder are securely fastened to a rigid plate as shown in Figure P4.65. Both cylinders are made from sheet metal thickness 1/8 inch. The outer diameters of aluminum and steel cylinders are 4 inch and 3 inch respectively. The load P = 20 kips in Figure P4.65. Determine (a) the displacement of the rigid plate. (b) the change in diameters of each cylinder.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

Aluminum 40 in P

P

Steel

30 inch

Figure P4.65

Solution

νal = 0.25 t = 1/8 in δp = ?

Eal = 10,000ksi νs = 0.28 P = 20 kips

Es = 30,000ksi (dal)0 = 4in Δds = ?

(ds)0 = 3in Δdal = ?

-----------------------------------------------------------We draw the free body diagram of the rigid plate as shown in Fig.(a). Nal Tensile

(a)

(b)

P

P

Contraction

δP δal

δS

Extension

Compressive NS

The inner diameter for each cylinder can be found as shown: ( d s ) i = ( d s ) 0 – 2t = 2.75in and ( d al ) i = ( d al ) 0 – 2t = 3.75in The area of the cross section can be found as 2 2 π 2 A s = --- ( 3 – 2.75 ) = 1.129in 4

1

2 2 π 2 A al = --- ( 4 – 3.75 ) = 1.522in 4

2

We draw the free body diagram of the rigid plate as shown in Fig (a). By force equilibrium in Fig. (a) we obtain the following. N al + N s = 2P = 40kips

3

We draw the exaggerated deformed shape as shown in Fig.(b) and write the deformation equations as shown below. δs = δp δ al = δ p

4 5

From Eq 4.27, we have the following N al ( 40 ) N al L al –3 δ al = --------------- = ------------------------------------- = 2.628 ( 10 )N al = δ p 3 E al A al 10 ( 10 ) ( 1.522 )

or

N al = 0.3804δ p ( 10 )

N s ( 30 ) Ns Ls –3 δ s = ----------- = ------------------------------------- = 0.8857 ( 10 )N s = δ p 3 Es As 30 ( 10 ) ( 1.129 )

or

N s = 1.129δ p ( 10 )

3

3

6

7

Substituting Eq.(6) and Eq. (7) into Eq. (3) we obtain the displacement of the plate δp as shown below 3

–3

( 0.3804 + 1.129 )δ p ( 10 ) = 40 or δ p = 26.5 ( 10 )in or –3

δ p = 0.0265 in. –3

From Eq’s(4) and (5) we obtain δ al = 26.5 ( 10 )in and δ s = 26.5 ( 10 )in .The axial strains are

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

–3 –3 δ δ al –3 –3 26.5 ( 10 ) 26.5 ( 10 ) ε s = ----s- = -------------------------- = 0.8833 ( 10 ) and ε al = ------ = -------------------------- = 0.6625 ( 10 ) Ls L al 30 40

Noting that steel cylinder contracts and Aluminum cylinder extends in the axial direction, we obtain the change in diameter as shown below. –3

–3

Δd s = ( d s ) 0 [ – ν s ( – 0.8833 ) ( 10 ) ] = 0.7420 ( 10 ) or –3

Δd s = 0.00074 in.

–3

Δd al = ( d al ) 0 [ – ν al ( 0.6625 ) ( 10 ) ] = – 0.6625 ( 10 ) or

Δd al = – 0.00066 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.66 An aluminum (Eal=10,000 ksi and νal= 0.25) and steel (Es=30,000 ksi and νs= 0.28) hollow cylinder are securely fastened to a rigid plate as shown in Figure P4.65. Both cylinders are made from sheet metal thickness 1/8 inch. The outer diameters of aluminum and steel cylinders are 4 inch and 3 inch respectively. The allowable stresses in aluminum and steel are 10 ksi and 25 ksi, determine the maximum force P that can be applied to the assembly shown in Figure P4.65. Solution Eal = 10,000ksi νal = 0.25 Es = 30,000ksi νs = 0.28 t = 1/8 in (dal)0 = 4in (ds)0 = 3in σ al ≤ 10ksi

σ s ≤ 25ksi

Pmax = ?

-----------------------------------------------------------From Eq. (3), (6) and (7) in problem 4-64, we have N al + N s = 2P

8 3

N al = 0.3804δ p ( 10 )

9

3

N s = 1.129δ p ( 10 )

10 3

Substituting Eq.(2) and (3) into Eq. (1) we obtain: [ 0.3804 + 1.129 ]δ p ( 10 ) = 2P or –3

δ p = 1.325P ( 10 )

11

We obtain the internal forces as N al = 0.504P and N s = 1.496P The axial stresses in each cylinder can be found using the areas in Eq.(1) and (2) in the problem 4-64 and maximum value of P calculated as shown below N al 0.504P σ al = ------- = ----------------- ≤ 10 1.522 A al N 1.496P σ s = ------s = ----------------- ≤ 25 1.129 As

or or

The maximum value of P that satisfies Eq’s (5) and (6) is

P ≤ 30.19kips

12

P ≤ 18.87kips

13 P max = 18.8 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.67 A gap of 0.004 inch exists between the rigid bar and bar A before the force P is applied. The rigid bar is hinged at point C. The lengths of bar A and B are 30 and 50 inches respectively. Both bars have an area of cross-section A= 1 in2 and Modulus of Elasticity E = 30,000 ksi. If P = 100 kips in Figure P4.67, determine the axial stresses in bars A and B.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

B C

P

o

75

24 in 36 in A

60 in

Figure P4.67

Solution

EA = EB = 30,000ksi AA = AB = 1in2

LA = 30 in P = 100 kips

LB = 50 in σA = ?

σB = ?

-----------------------------------------------------------Assume gap closes at Equilibrium. The free body diagram of the rigid body as shown in Fig(a). NB (a) (b) 36 in

C

Cy

75o

Tensile

Cx

75o

δA

24 in 24 in

60 in D

F

δB

δE

Extension

75o

Contraction

60 in

36 in

δD

E

Compressive NA

By moment equilibrium about point C in Fig.(a) we obtain: F ( 24 ) – N A ( 36 ) – N B sin 75 ( 96 ) = 0

or

36N A + 92.73N B = 24P = 2400

1

We draw the exaggerated deformed shape as shown in Fig (b) and write deformation equations δ δD ------ = -----E36 96 δ A = δ D – 0.004

2 3

δ B = δ E sin 75

4

N A ( 30 ) NA LA –3 δ A = -------------- = -------------------------------= N A ( 10 ) 3 EA AA ( 30 ) ( 10 ) ( 1 )

5

N B ( 50 ) NB LB –3 δ B = -------------- = -------------------------------= 1.667N B ( 10 ) 3 EB AB ( 30 ) ( 10 ) ( 1 )

6

From Eq.4.27

From Eq’s (5) and (3) we obtain. –3

N A ( 10 ) = δ D – 0.004

3

N A = 10 δ D – 4

or

7

From Eq’s (2),(3) and (6) we obtain. –3 96 δ B = 1.667N B ( 10 ) = ⎛ ------ δ D⎞ sin 75 ⎝ 36 ⎠

3

N B = 0.647δ D ( 10 )

or

8

Substituting Eq’s (7) and (8) into Eq. (1), we obtain the following 3

3

3

36 [ 10 ( δ D ) – 4 ] + 92.73 [ 0.647δ D ( 10 ) ] = 2400 or 96δ D ( 10 ) = 2544 or δ D = 0.0265in

9

As δD is greater than the gap, the assumption of gap closing is correct. From Eq.(7) and (8) we obtain. 3

3

N A = ( 10 ) ( 0.0265 ) – 4 = 22.5kips and N B = ( 0.647 ) ( 0.0265 ) ( 10 ) = 17.15kips

The axial stress can be found as

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

NA 22.5 σ A = ------- = ---------- or AA 1

σ A = 22.5ksi ( C )

N 17.15 σ B = ------B- = ------------- or AB 1

σ B = 17.2ksi ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.68 A gap of 0.004 inch exists between the rigid bar and bar A before the force P is applied in Figure P4.67. The rigid bar is hinged at point C. The lengths of bars A and B are 30 and 50 inches respectively. Both bars have an area of cross-section A= 1 in2 and Modulus of Elasticity E = 30,000 ksi. If the allowable normal stress in the bars is 20 ksi in tension or compression, determine the maximum force P that can be applied to the assembly shown in Figure P4.67. Solution EA = EB = 30,000ksi LA = 30 in LB = 50 in 2 AA = AB = 1in σ A ≤ 20ksi σ B ≤ 20ksi Pmax = ?

-----------------------------------------------------------From Eq’s (1), (7) and (8) in problem 4-66, we have the following 36N A + 92.73N B = 24P

1

3

N A = ( 10 )δ D – 4

2

3

N B = 0.647δ D ( 10 )

3

Substituting Eq.(2) and (3) into Eq.(1), we obtain the following 3

3

3

36 [ ( 10 )δ D – 4 ] + 92.73 [ 0.647δ D ( 10 ) ] = 24P = 96 ( 10 )δ D = 24P + 144 or –3

δ D = ( 0.25P + 1.5 ) ( 10 )

4

N A = 0.25P – 2.5

5

N B = 0.1618P + 0.9705

6

From Eq.(2) and (3) we obtain.

The axial stresses can be found and using the limiting values, the limit on P can be found as shown below NA σ A = ------- = ( 0.25P – 2.5 ) ≤ 20 AA

or

N σ B = ------B- = ( 0.1618P + 0.9705 ) ≤ 20 AB

The maximum value of P that satisfies Eq’s(7) and (8) is

or

P ≤ 90kips

7

P ≤ 117.6kips

8

P max = 90 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.69 In Figure P4.69, a gap exists between the rigid bar and rod A before the force F is applied. The rigid bar is hinged at point C. The lengths of bar A and B are 1 m and 1.5 m respectively and the diameters are 50 mm and 30 mm respectively. The bars are made of steel with a Modulus of Elasticity E = 200 GPa and Poisson’s ratio is 0.28. If F=75 kN in Figure P4.69, determine (a)the deformation of the two bars. (b) the change in the diameters of the two bars.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

A

F 0.4 m P

January 2014

C

0.0002 m

0.9 m Rigid 40o B

Figure P4.69

Solution

Es = 200GPa dA = 50 mm

ν = 0.28 dB = 30mm

LA = 1 m δA = ?

LB = 1.5m δB = ?

F = 75 kN ΔdA = ?

ΔdB = ?

-----------------------------------------------------------The area of cross-sections are. 2 –3 2 π A A = --- ( 0.05 ) = 1.963 ( 10 )m 4

1

2 –3 2 π A B = --- ( 0.03 ) = 0.7068 ( 10 )m 4

2

We assume the gap is closed at equilibrium. We can draw the free body diagram of bar as shown in Fig. (a) NA

(a)

F 0.4 m

Compressive

(b)

Contraction

Cy

P

Cx

0.9 m

δD

δB

40o

40o

C NB

δA

50o

0.4 m

0.9 m

Extension

Tensile

By moment equilibrium about point C, we obtain: F ( 0.4 ) – N A ( 0.9 ) – ( N B sin 40 ) ( 0.9 ) = 0 or 3

0.9N A + 0.5785N B = 0.4F = 30 ( 10 )

3

We draw the exaggerated deformed shape and write the deformation equations as shown below δ A = δ D – 0.0002 δ B = δ D cos 50 = 0.6428δ D

4 5

From Eq. (4.27) NA ( 1 ) NA LA –9 δ A = -------------- = -------------------------------------------------------------= 2.547N A ( 10 ) 9 – 3 EA AA ( 200 ) ( 10 ) ( 1.963 ) ( 10 )

6

N B ( 1.5 ) NBLB –9 δ B = -------------- = ----------------------------------------------------------------= 10.611N B ( 10 ) 9 – 3 EB AB ( 200 ) ( 10 ) ( 0.7068 ) ( 10 )

7

From Eq’s (6) and (4) we obtain. 6

N A = ( 392.6δ D – 0.0785 ) ( 10 )

8

From Eq’s (5) and (7) we obtain. 6

N B = 60.576δ D ( 10 )

9

Substituting Eq’s (8) and (9) into Eq (3), we obtain 6

6

3

0.9 ( 392.6δ D – 0.0785 ) ( 10 ) + 0.5785 ( 60.576 )δ D ( 10 ) = 30 ( 10 ) or

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January 2014

–3

388.4δ D = 0.10065or, δ D = 0.2591 ( 10 )m

As δD is greater than the gap, the assumption of gap closing is correct, From Eq’s (4) and (5) we obtain –3

–3

–3

δ A = 0.059 mm

δ B = 0.2591 ( 0.6428 ) ( 10 ) = 0.1666 ( 10 )m

δ B = 0.166 mm

δ A = 0.2591 ( 10 ) – 0.2 ( 10 ) = 0.0591 ( 10 )m –3

–3

The axial strains in A and B can be found as –3 δA δB –3 –3 0.166 ( 10 ) ε A = ------ = 0.0591 ( 10 ) and ε B = ------ = ----------------------------- = 0.1106 ( 10 ) LA LB 1.5

Noting that A contracts and B extends in the axial direction, we obtain the change in the diameter as shown below –3

–3

Δd A = 0.00083 mm

–3

Δd B = – 0.00093 mm

Δd A = d A [ – ν ( – 0.0581 ) ( 10 ) ] = 0.828 ( 10 )mm –3

Δd B = d B [ – ν ( 0.1106 ) ( 10 ) ] = – 0.929 ( 10 ) mm

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.70 In Figure P4.69, a gap exists between the rigid bar and rod A before the force F is applied. The rigid bar is hinged at point C. The lengths of bar A and B are 1 m and 1.5 m respectively and the diameters are 50 mm and 30 mm respectively. The bars are made of steel with a Modulus of Elasticity E = 200 GPa and Poisson’s ratio is 0.28. In Figure P4.69, the allowable axial stresses in bars A and B are 110 MPa and 125 MPa, respectively. Determine the maximum force F that can be applied. Solution Es = 200 GPa LA = 1 m LB = 1.5 m dA = 50 mm dB = 30 mm ν = 0.28

σ A ≤ 110MPa

σ B ≤ 125MPa

Fmax = ?

-----------------------------------------------------------From Eq’s (3), (8) and (9) in problem (4-68) we have the following. 0.9N A + 0.5785N B = 0.4F

1 6

N A = ( 392.6δ D – 0.0785 ) ( 10 )

2

6

N B = 63.581δ D ( 10 )

3

Substituting Eq’s (2) and (3) into Eq. (1) we obtain. 6

–6

[ 0.9 ( 392.6δ D – 0.0785 ) + ( 0.5785 ) ( 63.581 )δ D ] ( 10 ) = 0.4F + 390.1δ D = 0.4F ( 10 ) + 0.07065 or –9

–3

δ D = 1.025F ( 10 ) + 0.1811 ( 10 )

4

From Eq. (2) and (3) we obtain. –3

3

N A = 402.4F ( 10 ) – 7.400 ( 10 ) –3

5

3

N B = 65.17F ( 10 ) – 11.514 ( 10 )

6

Dividing the internal force in Eq’s (5) and (6) by the areas given by Eq’s (1) and (2) in the problem 4-68 we obtain the following. –3 3 NA 6 402.4F ( 10 ) – 7.4 ( 10 ) σ A = ------- = ------------------------------------------------------------ ≤ 110 ( 10 ) –3 AA 1.963 ( 10 ) –3 3 N 65.17F ( 10 ) + 11.514 ( 10 ) 6 σ B = ------B- = ---------------------------------------------------------------------- ≤ 125 ( 10 ) – 3 AB 0.7068 ( 10 )

The maximum value of F that satisfies Eq’s (7) and (8) is

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or

or

3

F ≤ 555.0 ( 10 )Newtons

3

F ≤ 1179 ( 10 )Newtons

7

8

F max = 555kN

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.71 A rectangular aluminum bar (E = 10,000 ksi) a steel bar (E = 30,000 ksi), and a brass bar (E = 15,000 ksi) form a structure as shown in Figure P4.71. All bars have the same thickness of 0.5 inch. A gap of 0.02 inch exists before the load of P = 15 kips is applied to the rigid plate. Assume that the rigid plate does not rotate. If P = 15 kips in Figure P4.71, determine (a) the axial stress in steel. (b) the displacement of the rigid plate with respect to the right wall. 0.02 inch P 2 in

Brass

2 in

Steel

Aluminum

6 in

P

30 in

60 in

Figure P4.71

Solution

Eal = 10,000ksi

Es = 30,000ksi

Ebl = 15,000ksi

t = 0.5 in

P = 15 kips

σs =

δp = ?

-----------------------------------------------------------The areas of cross sections of various material can be found as shown below. A br = ( 2 ) ( 0.5 ) = 1in

2

A s = ( 2 ) ( 0.5 ) = 1in

2

A al = ( 6 ) ( 0.5 ) = 3in

2

Assume that the gap closes at equilibrium. We can draw the free body diagram of the rigid plate as shown in Fig. (a) (a)

δP=δal

(b)

δbr

Nal

Brass Compressive

Extension

Contraction

P

Nbr

Aluminum Steel

NS

P

Tensile δS

Contraction

By force equilibrium in Fig (a) N al + N br + N s = 2P = 30kips

1

We draw the exaggerated deformed shape as shown in Fig. (b) and write the deformation equation as shown below. δ al = δ p δ s = δ p – 0.02

2

δ br = δ p – 0.02

4

3

From Eq. (4-27) N al ( 60 ) N al L al –3 δ al = --------------- = -------------------------------= 2N al ( 10 ) = δ p 3 E al A al ( 10 ) ( 10 ) ( 3 )

or

N al = 0.5δ p ( 10 )

N s ( 30 ) Ns Ls –3 δ s = ----------- = -------------------------------= N s ( 10 ) = δ p – 0.02 3 Es As ( 30 ) ( 10 ) ( 1 )

or

N s = ( 10 )δ p – 20

N br ( 30 ) N br L br –3 δ br = ---------------- = -------------------------------= 2N br ( 10 ) = δ p – 0.02 3 E br A br ( 15 ) ( 10 ) ( 1 )

or

N br = 0.5 ( 10 )δ p – 10

3

3

3

5

6

7

Substituting Eq. (5), (6) and (7) into Eq. (1)

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3

3

3

3

0.5δ p ( 10 ) + ( 10 )δ p – 20 + 0.5 ( 10 )δ p – 10 = 30 or 2.0δ p ( 10 ) = 60

or

As δp > 0.02 the assumption gap closes is correct. 3

January 2014

–3

δ p = 30 ( 10 )inch δ p = 0.03 in.

–3

From Eq. (6) N s = ( 10 ) ( 30 ) ( 10 ) – 20 = 10kips N The axial stress in steel is σ s = ------s = 10 ------

σ s = 10 ksi ( C )

1

As

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.72 A rectangular aluminum bar (E = 10,000 ksi) a steel bar (E = 30,000 ksi), and a brass bar (E = 15,000 ksi) form a structure as shown in Figure P4.71. All bars have the same thickness of 0.5 inch. A gap of 0.02 inch exists before the load of P = 15 kips is applied to the rigid plate. Assume that the rigid plate does not rotate. Determine the maximum load P in Figure P4.71, if the allowable axial stresses in brass, steel, and aluminum are 8 ksi, 15 ksi, and 10 ksi, respectively. Solution Eal = 10,000ksi Es = 30,000ksi Ebl = 15,000ksi σ br ≤ 8ksi

t = 0.5 in

σ s ≤ 15ksi

σ al ≤ 10ksi

-----------------------------------------------------------From Eq’s (1), (5), (6) and (7) in problem 4-70, we have N al + N s + N br = 2P

1

3

N al = 0.5 ( 10 )δ p

2

3

N s = ( 10 )δ p – 20

3

3

N br = 0.5 ( 10 )δ p – 10

4

Substituting Eq. (2), (3) and (4) into Eq(1) we obtain the following. 3

3

3

0.5 ( 10 )δ p + ( 10 )δ p – 20 + 0.5 ( 10 )δ p – 10 = 2P or 3

–3

2.0δ p ( 10 ) = 2P + 30, or, δ p = P ( 10 ) + 0.015

5

Substituting Eq. (5) into Eq(2), (3) and (4) we obtain N al = 0.5P + 7.5

Ns = P – 5

N br = 0.5P – 2.5

The axial stress in each material can be found and the limits on P establish as shown below. N al 0.5P + 7.5 σ al = ------- = ------------------------- ≤ 10ksi 3 A al N P–5 σ s = ------s = ------------ ≤ 15ksi 1 As N br 0.5P – 2.5 σ br = -------= ------------------------ ≤ 8ksi 1 A br

or

P ≤ 45kips

6

P ≤ 20kips

or or

The maximum value of P that satisfies the three inequalities is

7

P ≤ 21kips

8 P max = 20 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.73 Bar A and bar B have a cross-sectional area of 400 mm2and a Modulus of Elasticity E = 200 GPa. A gap exists between bar A and the rigid bar before the force F is applied as shown in Figure P4.73. The applied force F= 10 kN in, determine (a) the axial stress in bar B and (b) the deformation of bar A.

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F 0.0005 m 1.5m

2m

A 5m C 3m B 2m

Figure P4.73

Solution

AA = AB = 400 mm2

EA = EB = 200 GPa

δA = ? σ B =

F = 10 kN

-----------------------------------------------------------Assume the gap closes at equilibriums, we draw the free body diagram of the rigid bar as shown in Fig(a) F

(a) 2m

(b)

δA Contraction

2m

Compressive

δD

NA Cy

5m

5m Cx

3m

3m NB

δB Contraction

Compressive

From moment equilibrium about point C, we obtain. NA ( 5 ) + NB ( 3 ) – F ( 7 ) = 0

3

5N A + 3N B = 7F = 70 ( 10 )

or

1

We draw an exaggerated deformed shape as shown in Fig(b) and write the deformation equations δB δD 3 ------ = ------ or, δ B = --- δ D 5 3 5

2

δ A = δ D – 0.0005

3

N ( 1.5 )

N L

–9

A A A From Eq. 4-27 we have δ A = -------------- = --------------------------------------------------------- = 18.75N A ( 10 ) = δ D – 0.0005 or

EA AA

9

–6

( 200 ) ( 10 ) ( 400 ) ( 10 )

6

N A = ( 53.33δ D – 0.02667 ) ( 10 )

4

NB ( 2 ) NBLB –9 3 δ B = -------------- = --------------------------------------------------------- = 25N B ( 10 ) = --- δ D 9 –6 5 EB AB ( 200 ) ( 10 ) ( 400 ) ( 10 )

6

N B = 24δ D ( 10 )

or

5

Substituting Eq’s (4) and (5) into Eq. (1) we obtain 6

6

3

( 5 ) ( 53.33δ D – 0.02667 ) ( 10 ) + ( 3 ) ( 24 ) ( δ D ) ( 10 ) = 70 ( 10 ) or 6

338.65δ D ( 10 ) = 203.33

or

–3

δ D = 0.6004 ( 10 )m

6

As δD > 0.0005 the assumption that gap closes is correct. By substituting Eq.(6) into Eq. (3) we obtain the –3

–3

deformation of A as δ A = ( 0.6004 – 0.5 ) ( 10 ) = 0.1004 ( 10 )m or –3

δ A = 0.1 mm 6

3

Substituting Eq. (3) into Eq. (5) we obtain N B = ( 24 ) ( 0.6004 ) ( 10 ) ( 10 ) = 14.41 ( 10 )Newtons

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3

N 6 2 ( 10 ) The axial stress in B is σ B = ------B- = 14.41 --------------------------- = 36.03 ( 10 )N ⁄ m or AB

January 2014

σ B = 36 MPa ( C )

–6

400 ( 10 )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.74 Bar A and bar B have a cross-sectional area of 400 mm2and a Modulus of Elasticity E = 200 GPa. A gap exists between bar A and the rigid bar before the force F is applied as shown in Figure P4.73. Determine the maximum force F that can be applied if the allowable stress in member B is 120 MPa(C) and the allowable deformation of bar A is 0.25 mm. Solution

AA = AB = 400 mm2

EA = EB = 200 GPa Pmax = ? σ B ≤ 120MPa ( C )

δ A ≤ 0.25mm

-----------------------------------------------------------From Eq’s (1), (4) and (5) in problem 4-72, we have 5N A + 3N B = 7F

1 6

N A = ( 53.33δ D – 0.02667 ) ( 10 )

2

6

N B = 24δ D ( 10 )

3

Substituting Eq. (6) and (2) into Eq (1). 6

6

5 ( 53.33δ D – 0.02667 ) ( 10 ) + ( 3 ) ( 24 )δ D ( 10 ) = 7F

–6

338.65δ D = 7F ( 10 ) + 0.1333 or

or –9

–3

δ D = 20.67F ( 10 ) + 0.3936 ( 10 )

4

From Eq. (3) in problem 4-72 –3

–9

–3

–3

δ A = δ D – 0.5 ( 10 ) = 20.67F ( 10 ) – 0.1064 ( 10 ) ≤ 0.25 ( 10 )

3

F ≤ 17.24 ( 10 )newtons

or

5

3

Substituting Eq. (4) into Eq. (3): N B = 0.496F + 9.446 ( 10 ) The axial stress in B is 3 N 0.496F + 9.446 ( 10 ) 6 σ B = ------B- = -------------------------------------------------- ≤ 120 ( 10 ) –6 AB 400 ( 10 )

F ≤ 77.73

or

The maximum value of F that satisfies Eq(5) and (6) is

6

F max = 17.2kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.75 A rectangular steel (E = 30,000 ksi, ν=0.25) bar of 0.5 inch thickness has a gap of 0.01 inch between the section at D and a rigid wall before the forces are applied. Assuming that the applied forces are sufficient to close the gap determine: (a) the movement of section at C with respect to the left wall.(b) the change in the depth d of segment CD. 12.5 kips B A

17.5 kips

12.5 kips 18 in

17.5 kips 24 in

C

D

d =3 in

36 in 0.01 in

Figure P4.75

Solution

E = 30,000

ν = 0.25

t = 0.5

δc = ?

ΔdCD

-----------------------------------------------------------Assume gap is closed at equilibrium. We make imaginary cuts through segments AB, BC and CD and draw free body diagram as shown below. By force equilibrium in Figs. (a), (b) and (c) we obtain the following.

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(a) RA

A

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

(b) NAB Tensile

January 2014

N AB = R A N BC = R A – 25

1

N CD = 60 – R A

3

2

(c)

12.5 kips B A

12.5 kips B A

17.5 kips

C

NBC

RA

NCD

Tensile

RA 12.5 kips

12.5 kips

17.5 kips

Compressive

The total extension of the bar is equal to the gap. Noting that segment AB and BC will extend and segment CD will contract, we obtain the equation for total extension as δ AB + δ BC – δ CD = 0.01

4

N AB L AB ( R A ) ( 18 ) –3 δ AB = ---------------------= -------------------------------------------- = 0.40R A ( 10 ) 3 EA ( 30 ) ( 10 ) ( 0.5 ) ( 3 )

5

N BC L BC ( R A – 25 ) ( 24 ) –3 δ BC = --------------------- = -------------------------------------------- = 0.533 ( R A – 25 ) ( 10 ) 3 EA ( 30 ) ( 10 ) ( 0.5 ) ( 3 )

6

N CD L CD ( 60 – R A ) ( 36 ) –3 δ CD = ---------------------= -------------------------------------------- = 0.80 ( 60 – R A ) ( 10 ) 3 EA ( 30 ) ( 10 ) ( 0.5 ) ( 3 )

7

From Eq (4-27)

–3

Substituting Eq. (5), (6) and (7) into Eq. (4): [ 0.4R A + 0.533 ( R A – 25 ) – 0.8 ( 60 – R A ) ] ( 10 ) = 0.01 or 1.733R A = 71.325

or

R A = 41.157kips

8

The movement of rigid plate at C is the extension of segment, –3

–3

δ C = δ AB + δ BC = [ ( 0.4 ) ( 0.533 ) ( 41.157 – 25 ) ] ( 10 ) = 25.07 ( 10 )inch or

δ C = 0.025 in.

δ CD –3 0.8 ( 60 – 41.157 ) ( 10 ) The axial strain in CD is ε CD = ---------= --------------------------------------------------------- = 0.4187 ( 10 ) –3

L CD

36

Noting that εCD is contraction, the change in the depth can be found as shown below. –3

–3

Δd CD = – ν ( – 0.4187 ) ( 10 ) ( 3 ) = 0.314 ( 10 ) or

Δd CD = 0.00031 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.76 Three plastic members of equal cross-section are shown in Figure P4.76. Member B is smaller than member A by 0.5 mm. A distributed force is applied to the rigid plate, which moves downwards without rotating. The modulus of elasticity for members A and B are 1.5 GPa and 2.0 GPa, respectively. The distributed force w = 20 MPa in Figure P4.76. Determine the axial stress in each member.

w 20 mm 0.5 mm

A

10 mm B

A

80 mm

100 mm

Figure P4.76

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Solution

ΕΒ = 2.0GPa

EA = 1.5GPa

σA =

w = 20 MPa

January 2014

σB =

-----------------------------------------------------------Assume gap closes at equilibrium, we can draw the free body diagram of the rigid body after replacing the distributed force by an equivalent force as shown in Fig (a). By equilibrium of forces in Fig (a), we obtain. 3

2N A + N B = 0.002w = 40 ( 10 )N ⁄ m (0.1)(0.02)W

(a)

NB Compressive

1

(b) Contraction δA

NA

2

δB

Contraction

NA

We can draw an exaggerated deformed geometry as shown in Fig (b) and write the deformation equations δ A = δ B + 0.0005

2

N A ( 0.08 ) NA LA –6 δ A = -------------- = -------------------------------------------------------- = 0.2667 ( 10 )N A 9 EA AA ( 1.5 ) ( 10 ) ( 0.01 ) ( 0.02 )

3

N B ( 0.08 ) NBLB –6 δ B = -------------- = -------------------------------------------------------- = 0.2000 ( 10 )N B 9 EB AB ( 2.0 ) ( 10 ) ( 0.01 ) ( 0.02 )

4

From Eq. (4-27)

Equations (3) and (4) can be written in terms of δA as shown below 6

N A = 3.75 ( 10 )δ A

5

6

N B = 5.00 ( 10 ) ( δ A – ( 0.0005 ) )

6 6

3

Substituting Eq. (5) and (6) into Eq (1), we obtain [ 2 ( 3.75δ A ) + 5 ( δ A – 0.0005 ) ] ( 10 ) = 40 ( 10 ) or 12.5δ A = 0.0425

or

δ A = 0.0034

7

The internal forces in A and B are 6

3

2

6

3

N A = ( 3.75 ) ( 0.0034 ) ( 10 ) = 12.75 ( 10 )N ⁄ m and N B = ( 5.0 ) ( 0.0034 – 0.0005 ) ( 10 ) = 14.5 ( 10 )N ⁄ m

2

The axial stress in the bar A and B are 3 NA 6 2 ( 12.75 ) ( 10 ) σ A = ------- = -------------------------------- = 63.75 ( 10 )N ⁄ m AA ( 0.01 ) ( 0.02 )

σ A = 63.75 MPa ( C )

3 N 6 2 ( 14.5 ) ( 10 ) σ B = ------B- = ------------------------------- = 72.5 ( 10 )N ⁄ m AB ( 0.01 ) ( 0.02 )

σ B = 72.5 MPa ( C )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.77 Three plastic members of equal cross-section are shown in Figure P4.76. Member B is smaller than member A by 0.5 mm. A distributed force is applied to the rigid plate, which moves downwards without rotating. The modulus of elasticity for members A and B are 1.5 GPa and 2.0 GPa, respectively.The allowable stresses in members A and B in Figure P4.76 are 50 MPa and 30 MPa. Determine the maximum intensity of the distributed force that can be applied to the rigid plate. Solution EA = 1.5GPa ΕΒ = 2.0GPa wmax = ?

σ A ≤ 50MPa

σ B = 30MPa

------------------------------------------------------------

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January 2014

From Eq’s (1), (5) and (6) in problem 4-75, we have 2N A + N B = 0.002w

1

6

N A = 3.75 ( 10 )δ A

2

6

N B = 5.0 ( 10 ) ( δ A – 0.0005 )

3

Substituting Eq. (2) and (3) into Eq(1) we obtain 6

[ 2 ( 3.75 )δ A + 5 ( δ A – 0.0005 ) ] ( 10 ) = 0.002w

–9

–3

12.5δ A = 2w ( 10 ) + 2.5 ( 10 )

or –9

–3

–3

3

δ A = 0.16w ( 10 ) + 0.2 ( 10 )

4

Substituting Eq. (4) into Eq’s (2) and (3), we obtain N A = 0.6w ( 10 ) + 0.75 ( 10 ) –3

5

3

N B = 0.8w ( 10 ) – 1.5 ( 10 )

6

The axial stress can be found and limits on N obtained as shown below –3 3 NA 0.6w ( 10 ) + 0.75 ( 10 ) 6 2 σ A = ------- = ----------------------------------------------------------- ≤ 50 ( 10 )N ⁄ m ( 0.01 ) ( 0.02 ) AA –3 3 N 0.8w ( 10 ) – 1.5 ( 10 ) 6 2 σ B = ------B- = ------------------------------------------------------- ≤ 30 ( 10 )N ⁄ m ( 0.01 ) ( 0.02 ) AB

or

or

The maximum value of w that satisfies Eq.(7) and (8) is

6

w ≤ 17.92 ( 10 )N ⁄ m 6

w ≤ 9.375 ( 10 )N ⁄ m

2

2

7

8

w max = 9.4MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.78 Figure P4.78 shows an aluminum rod (E = 70 GPa, ν = 0.25) inside a steel (E = 210 GPa, ν = 0.28) tube. The aluminium rod is slightly longer than the steel tube and has a diameter of 40 mm. The steel tube has an inside diameter of 50 mm and is 10 mm thick. If the applied load P = 200 kN, determine: (a) the axial stresses in aluminium rod and steel tube. (b) the change in diameter of aluminium. P

250 mm

Aluminum

0.15 mm

Figure P4.78

4.79 Figure P4.79 shows an aluminium rod (E = 70 GPa, σyield = 280 MPa) inside a steel (E = 210 GPa, σyield = 210 MPa) tube. The aluminium rod is slightly longer than the steel tube and has a diameter of 40 mm. The steel tube has an inside diameter of 50 mm and is 10 mm thick. What is the maxi-

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January 2014

mum force P that can be applied without yielding either material. P

250 mm

Aluminum

0.15 mm

Figure P4.79

4.80 A rigid bar ABCD hinged at one end and is supported by two aluminum cables (E = 10,000 ksi) with a diameter of 1/4 in. as shown in Figure P4.80. The bar is horizontal before the force is applied. Determine the angle of rotation of the bar from the horizontal when a force P= 10kips is applied.

5 ft

P D B

A 3 ft

C 5 ft

5 ft

Figure P4.80

4.81 A suspended walkway is modelled as a rigid bar and supported by steel (E =30,000 ksi) rods as shown in Figure P4.81. The rods have a diameter of 2 in. and the nut has a contact area with the bottom of the walkway is 4 in.2 The weight of the walk per unit length is w = 725lb/ft. Determine (a) the axial stress

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January 2014

in the steel rods. (b) the average bearing stress between the nuts at A and D and the walkway. Ceiling C

B

36ft

w

D

A

10ft

10 ft

20 ft

Figure P4.81

4.82 An aluminum (EAl = 70 GPa) circular bar of and a steel (E = 200 GPa) tapered circular bar are securely attached to a rigid plate on which axial forces are applied as shown in Figure P4.82. Determine (a) the displacement of the rigid plate (b) the maximum axial stress in steel. 50 kN 50 mm

A aluminum

B

C

steel

100 mm

50 Kn 2.0 m

0.5 m

Figure P4.82

Solution

Eal = 70GPa

Es = 200GPa

δB = ?

( σ S ) max =

-----------------------------------------------------------The radius of the bar BC varies linearly and R ( x ) = a + bx

1

At x=0.5m; R=0.025m and x=2.5m; R=0.05. Using these values we obtain the following equations a + b ( 0.5 ) = 0.025

2

a + b ( 2.5 ) = 0.050

3

Solving the above two equations, we obtain. –3

–3

a = 18.75 ( 10 )andb = 12.5 ( 10 )

Thus, the area of cross-section in segment be varies as 2

–6

A = π ( 18.75 + 12.5x ) ( 10 )

4

We make imaginary cuts in segment AB and BC to obtain free body diagrams shown below (a)

A RA

NAB Compressive

50 kN

(b)

NBC A

B

RA

Tensile 50 kN

By force equilibrium we obtain.

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January 2014

N AB = R A

5

3

N BC = 100 ( 10 ) – R A

6

The total extension of the bar is zero. Thus, the contraction of segment AB is equal to extension of segment BC δ AB = δ BC

7

R A ( 0.5 ) N AB L AB –9 δ AB = ---------------------= ------------------------------------------------ = 3.638R A ( 10 ) 9 2 E AB A AB ( 70 ) ( 10 )π ( 0.025 )

8

From Eq. (4-27) we have

We can find the extension of bar BC as shown below. 3

N BC [ 100 ( 10 ) – R A ] du ------ = --------------------- = ---------------------------------------------------------------------------------------9 2 –6 dx E BC A BC ( 200 ) ( 10 )π ( 18.75 + 12.5x ) ( 10 )

Integrating from B to C, we obtain the extension as shown below 3 3 2.5 [ 100 ( 10 ) – R A ] 2.5 [ 100 ( 10 ) – R A ] 1 1 dx ------------------------------------------------------------------------------- ------------------ ------------------------------------δ BC = du = ----------------------------------------- = or 3 3 2 uB ( 200π ) ( 10 ) 0.5 ( 18.75 + 12.5x ) ( 200π ) ( 10 ) ( – 12.5 ) ( 18.75 + 12.5x ) 0.5

∫

uA

∫

3

–9

δ BC = 2.546 [ ( 100 ) ( 10 ) – R A ] ( 10 )

9

Substituting Eq’s (8) and (9) into Eq. (7), we obtain the following. –9

3

–9

3.638R A ( 10 ) = 2.546 [ 100 ( 10 ) – R A ] ( 10 )

3

R A = 41.17 ( 10 )Newtons

or

10

The movement of the rigid plate is the same as contraction of segment AB. From Eq. (8) we obtain. 3

–9

–3

δ B = ( 3.638 ) ( 41.17 ) ( 10 ) ( 10 ) = 0.1498 ( 10 )m

δ B = 0.15 mm to the left

3

From Eq. (6) and (10): N BC = 58.83 ( 10 )Newtons . The maximum axial stress in the segment BC will be just after B, as the area of the cross-section is smallest there. 3

6 2 58.83 ( 10 ) ( σ S ) max = --------------------------- = 29.96 ( 10 )N ⁄ m 2 π ( 0.025 )

( σ S ) max = 30 MPa ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.83 A rigid bar hinged at point O has a force P applied to it as shown. Bars A and B are made of steel (E = 30,000 ksi). The cross-sectional areas of the bars A and B are AA = 1 in 2 and AB = 2 in 2 respectively. If the allowable deflection at point C is 0.01 inch and the allowable stress in the bars is 25 ksi, determine the maximum force P that can be applied. P 24 in

30 in rigid

42 in

0.005in

C O

36 in A

48 in B

Figure P4.83

Solution

E = 30000 δ C ≤ 0.01in

AA = 1in2 σ A ≤ 25ksi

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AB = 2 in2 σ B ≤ 25ksi

Pmax = ?

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

-----------------------------------------------------------Assume gap closes. We draw the free body diagram of the rigid bar as shown in Fig (a). P Oy 30 in (a) (b) δ 24 in

30 in

42 in

42 in

δA

C 24 in

Ox NA

D 0.005in

C

δB

δD

Contraction

Compressive NB

By moment equilibrium about point O, we obtain. N A ( 30 ) + N B ( 72 ) – P ( 24 ) = 0

or

30N A + 72N B = 24P

1

We draw an exaggerated deformed shape as shown in Fig. (b) and write the deformation Equations. δ δ δC ------ = -----A- = -----D24 30 72

2

δ D = δ B + 0.005

3

30 5 δ A = ------ δ C = --- δ C 24 4

4

72 δ B = ------ δ C – 0.005 = 3δ C – 0.005 24

5

Eq. (2) and (3) can be rewritten as

From Eq. (4-27) N A ( 36 ) NA LA 5 δ A = -------------- = --------------------------- = --- δ C 4 EA AA ( 30000 ) ( 1 )

3

N A = 1.0417δ C ( 10 )

or

N B ( 48 ) NB LB δ B = -------------- = --------------------------- = 3δ C – 0.005 EBAB ( 30000 ) ( 2 )

6

3

N B = 3.75δ C ( 10 ) – 6.25

or

7

Substituting Eq. (6) and (7) into Eq (1). We obtain 3

3

30 ( 1.0417δ C ) ( 10 ) + 72 ( 3.75δ C ( 10 ) – 6.25 ) = 24P

3

301.25 ( 10 )δ C = 24P + 450

or

–3

δ C = ( 0.0791P + 1.4938 ) ( 10 )

8

Noting the limit on δC, we obtain one limit on P as –3

( 0.0797P + 1.4938 ) ( 10 ) ≤ 0.01

P ≤ 106.73kips

or

9

From Eq. (6) and (7). we obtain: N A = ( 0.083P + 1.556 )kips and N B = ( 0.2988P – 0.6483 )kips NA 0.083P + 1.556 σ A = ------- = ------------------------------------- ≤ 25 1 AA

or

N 0.2988P – 0.6483 σ B = ------B- = ------------------------------------------ ≤ 25 2 AB

The maximum value of P that satisfies Eq. (9), (11) and (12) is

or

P ≤ 282.5

10

P ≤ 169.5

11 P max = 106.7kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.84 The structure at the base of a crane is modeled by the pin connected structure shown. The allowable axial stresses in members AC and BC are 15 ksi, and the Modulus of Elasticity is 30,000 ksi. To ensure adequate stiffness at the base, the displacement of pin C in the vertical direction to be limited to

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January 2014

0.1 inches. Determine the minimum area of cross-sections for members AC and BC. F = 75 kips

55o

C

C 12 ft

52o

A

B

Figure P4.84

σ AC ≤ 15ksi

Solution

A σ BC = 15ksi

ν C ≤ 0.1in

B

AAC = ?

ABC = ?

-----------------------------------------------------------We can draw the free body diagram as shown in Fig.(a) F = 75 kips (a) (b) 52o Tensile NAC

C

vc

55o

uc

C

12 ft

Compressive NBC

A

52o

B

By force equilibrium in x-direction in Fig (a) N BC cos 52 = 75 cos 55

or

N BC = 69.87kips

N BC 69.87 σ BC = ---------- = ------------- ≤ 15 A BC A BC

1

2

or

A BC ≥ 4.658in

or

N AC = 116.49kips

2

By force equilibrium in y-direction in Fig (b) N AC = N BC sin 52 + 75 sin 55 N AC 116.49 σ AC = ---------- = ---------------- ≤ 15 A AC A AC

or

A AC ≥ 7.766in

3

2

4

The length LBC = 12/sin52 = 15.23 ft. The deformation of each bar can be found as shown below N AC L AC 0.3354 2 ( 69.87 ) ( 12 ) ( 12 ) δ AC = ---------------------= ---------------------------------------- = ---------------- in A AC EA AC ( 30000 )A AC

5

N BC L BC 0.7097 2 ( 116.49 ) ( 15.23 ) ( 12 ) δ BC = --------------------- = --------------------------------------------------- = ---------------- in A BC EA BC ( 30000 )A BC

6

Let the pin move by uC and vC in the x and y-direction, respectively. Assuming small strain. δ AC = v C

7

Substituting Eq. (5) into Eq. (7). we obtain 0.3354 v C = ---------------- ≤ 0.1 A AC

A AC ≥ 3.35in

2

8

The maximum area of cross section that satisfies Eq.(2), (4) and (8) are A AC = 7.77 in

2

and

A BC = 4.66 in

2

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.85 The landing wheel of the plane in Figure P4.85 is modeled as shown to the right. Pin at C is in double shear and has an allowable shear stress of 12 ksi The allowable axial stress for link BC is 30 ksi. Determine the diameter of pin C and the effective area of cross-section of link BC.

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(Note the following: Attachments at A and B are approximated by pins to simplify analysis. There are two links represented by BC, one on either side of the hydraulic cylinder that we are modeling as a single link with an effective area cross-section that is to be determined so that the free body diagram is two-dimensional) A

B B

A

12 in 36.5o

C

C 2 in

32 in

D

D 20o

18 kips

Figure P4.85

Solution

σ BC ≤ 30ksi

τ C ≤ 12ksi

dc = ?

ABC = ?

-----------------------------------------------------------We draw the free body diagram of bar ACD. as shown in Fig. (a) Ay Compressive Ax

NBC 36.5o

C

12 in

2 in 32 in

D 20o 18 kips

By moment equilibrium about point A. 18 sin 20 ( 44 ) – ( N BC cos 36.5 ) ( 12 ) + ( N BC sin 36.5 ) ( 2 ) = 0 N

32.03 BC The axial stress in BC is: σ BC = ---------- = ------------- ≤ 30 A BC

A BC

or

270.9 N BC = ------------- = 32.03kips 8.457 2

A BC ≥ 1.068in or

or

A BC = 1.1in

2

The pin being in double sheer will have a shear or force Vc = NBC/2=16.015 V

16.015 - ≤ 12 The shear stress in pin C is: τ C = ------C- = -------------------2 AC

( πd ⁄ 4 )

or

2

d ≥ 1.703in or

d = 1.3in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.86 The allowable normal stress in the stepped axial rod shown in Figure P4.86 is 20 ksi. If F = 10kips, determine the smallest fillet radius that can be used at section B. Use stress concentration graphs given in Section C.4.2.

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2 in 1 in F

A

B

C

Figure P4.86

σ max ≤ 20ksi

Solution

r=?

-----------------------------------------------------------10 The axial stress in BC can be found as: σ BC = -------------------- = 12.73ksi 2

π(1 ) ⁄ 4

σ max = k conc σ BC = 12.73k conc ≤ 20

k conc ≤ 1.57

or

1

From Fig. C.2, we obtain the approximate value of r/d corresponding to D/d = 2 and kconc = 1.5 r --- = 0.26 d

or

r = ( 0.26 ) ( 1 ) = 0.26inch

r = 0.26 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.87 The fillet radius in the stepped circular rod shown is 6 mm. Determine the maximum axial force F that can act on the rigid wheel if the allowable axial stress is 120 MPa, and modulus of elasticity is 70 GPa.Use stress concentration graphs given in Section C.4.2. 60 mm A

B

0.9 m

48 mm F C F 0.75 m

D

1.0 m

Figure P4.87

σ max ≤ 120 MPa

Solution

r = 6 mm

E = 70GPa

Fmax = ?

-----------------------------------------------------------The area of cross-sections of various segments can be found as shown below. 2 –3 2 π A AB = --- ( 0.06 ) = 2.83 ( 10 )m 4

1

2 –3 2 π A BC = A CD = --- ( 0.048 ) = 1.81 ( 10 )m 4

2

6 - = 0.125 The ratio’s --r- = ----d

D 60 ---- = ------ = 1.25 d 48

and

48

From Fig. C-2. we obtain the stress concentration factor as K conc = 1.7

3

We made imaginary cuts in segments AB, BC and CD and draw free body diagram as shown below (a) RA

A

NAB

(b)

Tensile

RA

A

B

NBC Tensile

F

(c) RA

A

B

C F

NCD Compressive

By force equilibrium we obtain N AB = R A N BC = R A

1

N CD = 2F – R A

3

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January 2014

The total extension of the bar is zero. δ AB + δ BC – δ CD = 0

4

R A ( 0.9 ) N AB L AB –6 δ AB = ---------------------= --------------------------------------------------= 4.543R A ( 10 ) 9 –3 E AB A AB 70 ( 10 ) ( 2.83 ) ( 10 )

5

R A ( 0.75 ) N BC L BC –6 δ BC = --------------------- = --------------------------------------------------= 5.919R A ( 10 ) 9 – 3 E BC A BC 70 ( 10 ) ( 1.81 ) ( 10 )

6

N CD L CD ( 2F – R A ) ( 1 ) –6 δ CD = ---------------------= --------------------------------------------------= 7.893 ( 2F – R A ) ( 10 ) 9 – 3 E CD A CD 70 ( 10 ) ( 1.81 ) ( 10 )

7

From Eq. (4-27) we have

Substituting Eq’s (5), (6) and (7) into Eq. (4) –6

[ 4.543R A + 5.919R A – 7.893 ( 2F – R A ) ] ( 10 ) = 0

15.786F R A = -------------------- = 0.86F 18.355

or

The nominal axial stress in each member can be found as N AB 0.86F σ AB = ---------- = -------------------------- = 303.9F –3 A AB 2.83 ( 10 )

8

N BC 0.86F σ BC = ---------- = -------------------------- = 475.1F –3 A BC 1.81 ( 10 )

9

N CD 2 – 0.86 )F σ CD = ---------- = (--------------------------- = 629.8F –3 A CD 2.83 ( 10 )

10

( σ BC ) max = k conc σ BC = ( 1.7 ) ( 475.1F ) = 807.7F

11

The maximum stress in BC is As the maximum stress in BC is the largest, we use to find maximum F as shown below. 6

807.7F ≤ 120 ( 10 )

or

3

F ≤ 148.6 ( 10 )Newtons or

F max = 148.6 kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.88 The fillet radius is 0.2 mm in the stepped steel circular rod shown in Figure P4.86. What should be the peak value of the cyclic load F to ensure a service life of half-million cycles. Use the SN-curve shown in Figure 3.36. Solution

n = 0.5(106) steel

r = 0.2

Fmax = ?

-----------------------------------------------------------From Figure 3.36 the peak stress for half a million cycles is σ max = 40ksi For r/d = 0.2 and d/D = 2, the stress concentration factor is Kconc = 1.65 The nominal stress in BC is as shown below. σ max 40 F σ BC = ------------- = --------= --------------------2 1.65 K conc π(1) ⁄ 4

or

F = 19.04kips or

F max = 19 kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.89 The aluminum axial rods in Figure P4.87 are subjected to a cyclic force F. Determine the peak value of the force F to ensure a service life of one-million cycle. Use the SN-curves shown in Figure 3.36 and a modulus of elasticity E = 70 GPa. Solution

n = 106 Aluminum

Fmax = ?

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

From Figure 3.36, the peak stress for a million cycles in Aluminum is approximately σ max = 147MPa From Eq. (11) in problem 4-84, we have 6

( σ BC ) max = 807.7F ≤ 147 ( 10 )

3

F ≤ 181.998 ( 10 )Newtons

or

F max = 181.9kN

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.90 During assembly of the structure, a misfit between bar A and the attachment of the rigid bar was found as shown. If bar A is pulled and attached, determine the initial stress introduced due to the misfit. The modulus of elasticity of the circular bars A and B is E= 10,000 ksi and the diameter is 1 inch.

A

B 80 in

60 in 40 in 0.05 in

60 in C

Figure P4.90

Solution

E = 10,000ksi

σA =

d = 1inch

σB =

-----------------------------------------------------------We draw the free body diagram of rigid bar as shown in Fig. (a) NB (b) (a) N A

Tensile Cy 40 in

Cx

Extension 60 in

δA 0.05 in δD

60 in

C δB 40 in

Extension

By moment equilibrium about point C, we obtain 0.4N A = 0.6N B

1

We can draw the exaggerated deformed geometry as shown in Fig.(b), and write the deformation equations δB δD ------ = -----0.4 0.6

δ B = 1.5δ D

or

2

δ A = 0.05 – δ D

3

From Eq. 4-27 · N A ( 60 ) NA LA –3 δ A = -------------- = ------------------------------------------ = 7.639N A ( 10 ) = 0.05 – δ D EA AA 2 π ( 10, 000 ) ⎛ ---⎞ ( 1 ) ⎝ 4⎠ · N B ( 80 ) NB LB –3 δ B = -------------- = ------------------------------------------------ = 10.186N B ( 10 ) = 1.5δ D 2 EB AB ( 10, 000 ) ( π ( 1 ) ⁄ 4 )

or

or

N A = 6.545 – 130.91δ D

4

N B = 147.26δ D

5

Substituting Eq. (4) and (5) into Eq. (1) 0.4 ( 6.545 – 130.91δ D ) = 0.6 ( 147.26δ D )

From Eq. (4) and (5) we obtain: N A = 4.11kips

or and

2.618- = 18.6 ( 10 – 3 ) δ D = --------------140.72 N B = 2.74kips

6

The initial axial stresses are

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January 2014

NA 4.11 σ A = ------- = -------------------------- = 5.233ksi 2 AA (π(1) ⁄ 4)

σ A = 5.2ksi ( T )

N 2.74 σ B = ------B- = -------------------------- = 3.488ksi 2 AB (π(1) ⁄ 4)

σ B = 3.5ksi ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.91 Bar A was manufactured 2 mm less than bar B due to an error. The attachment of these bars to the rigid bar would cause a misfit of 2 mm. Calculate the initial stress in each assembly. Which of the two assembly configuration you would recommend? Use modulus of elasticity of E = 70 GPa and diameter of the circular bars as 25 mm.

A

B

2m

2m

2mm

2 mm

C

1.5 m

E = 70GPa

C

1.5 m

1.5 m

Figure P4.91

Assembly 1

Solution

A

B

d = 25mm

1.5 m Assembly 2

σA =

assembly = ?

σB =

-----------------------------------------------------------2 –3 π The area cross-section is A = --- ( 0.025 ) = 0.4909 ( 10 )

4

Assembly 1 Analysis We can draw the free body diagram of rigid bar as shown in Fig.(a) (a)

NB

NA

(b) Compressive

Tensile

Cy

Extension

Cx

δA

2mm 1.5 m

1.5 m

Contraction δD

C

δB 1.5 m

1.5 m

By moment equilibrium about point C N A ( 3 ) = N B ( 1.5 )

or

2N A = N B ( 1 )

1

We can draw an exaggerated deformed geometry and write deformation equations δB δD δ ------ = -----or δ B = -----D3 1.5 2 δ A = 0.002 – δ D

2 3

From Eq. (4-27) NA ( 2 ) NA LA –9 δ A = -------------- = -------------------------------------------------------------= 58.2N A ( 10 ) 9 –3 EA AA ( 70 ) ( 10 ) ( 0.4909 ) ( 10 )

4

Substituting Eq. (3) into Eq. (4), we obtain 3

6

N A = 34.36 ( 10 ) – 17.18δ D ( 10 )

5

NB LB –9 δ B = -------------- = 58.2N B ( 10 ) EB AB

6

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

Substituting Eq. (2) into Eq. (6), we obtain 6

N B = 8.591δ D ( 10 )

7

Substituting Eq. (5) and (7)into Eq. (1), we obtain 3

6

6

[ 34.36 – 36 ( 10 ) – 17.18δ D ( 10 ) ] = 8.591δ D ( 10 )

–3 –3 68.72 δ D = ------------- ( 10 ) = 1.60 ( 10 42.95

or

3

8

3

From Eq. (4) and (5): N A = 6.872 ( 10 ) and N B = 13.75 ( 10 ) The initial axial stresses are 3 NA 6 6.872 ( 10 ) σ A = ------- = ------------------------------------- = 14.0 ( 10 ) –3 AA ( 0.4909 ) ( 10 )

σ A = 14.0 MPa ( T )

3 N 6 13.75 ( 10 ) σ B = ------B- = ------------------------------------- = 28.0 ( 10 ) –3 AB ( 0.4909 ) ( 10 )

σ B = 28.0 MPa ( C )

Assembly 2 Analysis We can draw the free body diagram of rigid bar as shown in Fig.(c) (a) NB Compressive NA (b) Tensile Cy Cx Contraction 1.5 m

δB

Extension δA 2mm

δD

1.5 m 1.5 m

C

1.5 m

By moment equilibrium about point C N B ( 3 ) = 1.5N A

or

N A = 2N B

9

We can draw an exaggerated deformed geometry and write deformation equations δD δB ------ = -----or δ B = 2δ D 3 1.5 δ A = 0.002 – δ D

10 11

Substituting Eq. (11) into Eq. (4) we obtain 3

6

N A = 34.36 ( 10 ) – 17.18δ D ( 10 )

12

Substituting Eq. (10) into Eq. (6) we obtain 6

N B = 34.36δ D ( 10 )

13

Substituting Eq. (12) and (13) into Eq. (9) we obtain 3

6

3

6

[ 34.36 ( 10 ) – 17.18δ D ( 10 ) ] = 2 ( 34.36 )δ D ( 10 )

–3 34.36 ( 10 ) δ D = ------------------------------ = 0.4 ( 10 ) 6 85.908 ( 10 )

or 3

From Eq. (12) and (13) we obtain: N A = 27.49 ( 10 )

or

3

N B = 13.74 ( 10 )

The axial stresses are 3 NA 6 27.49 ( 10 ) σ A = ------- = -------------------------------- = 56 ( 10 ) – 3 AA 0.4909 ( 10 )

σ A = 56 MPa ( T )

3 N 13.74 ( 10 ) σ B = ------B- = -------------------------------- = 27.989 –3 AB 0.4909 ( 10 )

σ B = 28 MPa ( C )

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

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Assembly 1 is preferred as the initial stress in member A is significantly smaller.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.92 A steel bolt is passes through a aluminum sleeve. After assembling the unit by finger tightening the nut (no deformation) the nut is given 1/4 turn. If the pitch of the threads is 3.0 mm, determine the initial axial stress developed in the sleeve and the bolt. The modulus of elasticity for steel and copper are Est = 200 GPa, Eal = 70 GPa, respectively the area of cross-sections areAst = 500 mm2 Aal= 1100 mm2. Rigid washers

Sleeve

Bolt

25 mm

25 mm

300 mm

Figure P4.92

Solution

p =0.3 mm

n=1/4

Est = 200 GPa

Ast = 500 mm2

Aal= 1100 mm2

Eal = 70 GPa

-----------------------------------------------------------We can make an imaginary cut through the sleeve and the bolt and draw the free body diagram shown in Fig. (a). 0.75 mm (a) (b) δal δs Nal

Ns P

P1

Nal represent the equivalent axial force that acts on the entire circulation cross-section of the sleeve By force equilibrium in Fig(a) N al = N s

1

Consider point P on the bolt. As the nut is tightened, point P moves to point P1, and the distance PP1, represents the extension of the bolt. Relative to point P1(bolt) the nuts moves by an amount (n)(p) = 0.75mm. The deformation equation is –3

δ al + δ s = 0.75 ( 10 )m

2

From Eq. (4-27). · N al ( 0.3 ) N al L al –9 δ al = ---------------- = ---------------------------------------------------------- = 3.896N al ( 10 ) 9 – 6 E al A al ( 70 ) ( 10 ) ( 1100 ) ( 10 ) · N s ( 0.3 + 0.025 + 0.025 ) Ns Ls –9 δ s = ----------- = ---------------------------------------------------------= 3.5N s ( 10 ) 9 –6 Es As ( 200 ) ( 10 ) ( 500 ) ( 10 )

3

4

Substituting Eq. (3),(4) and (1) into Eq. (2) we obtain –9

–3

( 3.896 + 3.5 )N al ( 10 ) = 0.75 ( 10 )

or

3

3

N al = 101.4 ( 10 )Newtons and N s = 101.4 ( 10 )Newtons

The axial stresses are

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

3 N al 6 101.4 ( 10 ) σ al = ------- = ---------------------------- = 92.2 ( 10 ) –6 A al 1100 ( 10 )

σ al = 92.2 MPa ( C )

3 N 6 101.4 ( 10 ) σ s = ------s = --------------------------- = 202.8 ( 10 ) – 6 As 500 ( 10 )

σ s = 202.8 MPa ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.93 The rigid bar is horizontal when the unit is put together by finger tightening the nut. The pitch of the threads is 0.125 inch. Develop a table in steps of quarter turn of the nut that can be used for prescribing the pre-tension in bar B. Maximum number of quarter turns is limited by the yield stress. 15 in

5 in

Bar A

Bar B

Modulus of Elasticity

10,000 ksi

30,000 ksi

Yield Stress

24 ksi

30 ksi

Area of cross-section

0.5 in

2

0.75 in

rigid A

B

50 in

2

Figure P4.93

Solution

σA = ?

p = 0.125 inch n = number of quarter turns = ?

σB = ?

-----------------------------------------------------------We can draw the free body diagram of the rigid plate as shown in Fig. (a). (a) (b) 5 in 15 in (c) δA

15 in D

Extension

Cy Tensile NA

δD

5 in

δB

Extension

δD

P1 P

np/4

NB

By moment equilibrium about point C N A ( 5 ) = 15N B

or

N A = 3N B

1

Consider point D on the rigid plate. Using deformed geometry in Fig(b), we can write δ δA ------ = -----D5 15

or

δ δ A = -----D3

2

Point P on the bolt moves to point P1 as the bolt material is pulled through the nut. The distance PP1 in the Fig (c) represents the extension of the bolt. The nut moves relative to the bolt. i.e. with respect to point P1 by the amount nP/4. The deformed equation is thus nP δ B + δ D = -----4

δ B = 0.03125n – δ D

3

· N A ( 50 ) NA LA δ δ A = -------------- = ------------------------------------ = 0.01N A = -----D3 EA AA 3 ( 10 ) ( 10 ) ( 0.5 )

4

or

From Eq. (4-27)

or

N A = 33.33δ D

· N B ( 50 ) NB LB –3 δ B = -------------- = --------------------------------------- = 2.222 ( 10 )N B = 0.03125n – δ D 3 EB AB ( 30 ) ( 10 ) ( 0.75 ) or

N B = 14.0625n – 450δ D

5 6 7

Substituting Eq. (5) and (7) into eq. (1) we obtain.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

42.1875n δ D = ---------------------- = 0.0305n 1383.33 From Eq. (5) and (7) we obtain: N A = 1.0166nkips and N B = 0.3388n 33.33δ D = 3 ( 14.0625n – 450δ D )

or

8

The axial stresses are NA 1.0166n σ A = ------- = ------------------- = 2.033n 0.5 AA

9

N 0.3388n σ B = ------B- = ------------------- = 0.4518n AB 0.75

10

The stress in A exceeds material yield stress when n = 12. The initial stress for 11 quarter turns are given in the table below. n σA σB 1

2.03

0.45

2

4.07

0.90

3

6.10

1.36

4

8.13

1.81

5

10.17

2.26

6

12.20

2.71

7

14.23

3.16

8

16.26

3.61

9

18.30

4.07

10

20.33

4.52

11

22.36

4.97

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.94 The increase in temperature varies as ΔT = TL ( x 2 ⁄ L2 ) . Determine the axial stress and movement of a point at x=L/2 in terms of length L, Modulus of elasticity E, area of cross-section A, coefficient of thermal expansion α and the increase of temperature at the end TL x L

⎛ x2⎞ ΔT = T L ⎜ ------⎟ ⎝ L 2⎠

Solution

σ xx =

L u ⎛ ---⎞ = ⎝ 2⎠

-----------------------------------------------------------As the right side is free to expand and there are no external forces.

σ xx = 0

2

x ------ = αΔT = αT L -----The axial strain is: ε xx = du 2 dx

L

Integrating from x=0, where displacement is zero to x=L/2 we obtain L u ⎛ ---⎞ ⎝ 2⎠

∫0

L --2

∫

L ---

2

x du = αT L ------ dx 2 0 L

or

3

αT L 3 2 αT L L L u ⎛ ---⎞ = ---------- x 0 = ---------------⎝ 2⎠ 2 2 3L 24L

αT L L L u ⎛ ---⎞ = -------------⎝ 2⎠ 24

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.95 The increase in temperature varies as ΔT = TL ( x 2 ⁄ L2 ) . Determine the axial stress and movement of a point at x=L/2 in terms of length L, Modulus of elasticity E, area of cross-section A, coefficient of ther-

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

mal expansion α and the increase of temperature at the end TL x L

Solution

2

2

ΔT = T L ( x ⁄ L )

σ xx =

u(L ⁄ 2) =

-----------------------------------------------------------We make an imaginary cut and draw free body diagram as shown below. N

RL

By force equilibrium we obtain N = RL

1

R σ xx = N ---- = ------LA A

2

σ xx ------ = -------- + αΔT The axial strain is: ε xx = du dx

RL 2 2 du ------ = ------- + αT L ( x ⁄ L ) Integrating we obtain. dx EA

or

E

3

RL αT L x u ( x ) = -------- x + ---------------+ C1 2 EA 3L

Noting u(0) = 0.

3

C1 = 0. Noting that u(L) = 0, we obtain the following. 3

R L L αT L L u ( L ) = ---------- + ----------------- = 0 2 EA 3L

αT L EA R L = – -----------------3

or

4 EαT L σ xx = --------------- ( C ) 3

From Eq. (2) we obtain. αT

αT x

3

L Substituting Eq. (4) and C1 = 0 in Eq (3): u ( x ) = – ----------L x + ---------------at x=L/2, we obtain

3

3L

2

3

αT L L αT L L αT L L L u ⎛ ---⎞ = – -------------- + ----------------- = – -------------⎝ 2⎠ 2 6 8 24L

αT L L L u ⎛ ---⎞ = – -------------⎝ 2⎠ 8

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.96

The tapered bar shown has a cross-sectional area that varies with x as A = K ( L – 0.5 x ) 2 . If the

increase in temperature of the bar varies as ΔT = TL ( x2 ⁄ L 2 ) . Determine the axial stress at mid point in terms of the length L, Modulus of elasticity E, area of cross-section parameter K, coefficient of thermal expansion α, and the increase of temperature at the end TL

x

L

Figure P4.96

Solution

A = K(L-0.5x)2

2

2

ΔT = T L ( x ⁄ L )

σ xx ( L ⁄ 2 ) =

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

We make an imaginary cut and draw a free body diagram as shown below. R

N

By force equilibrium N = R

1

The axial stress and strain can be written as shown below. N R σ xx = ---- = -------------------------------2 A K ( L – 0.5x )

2

2 σ xx x du R ε xx = ------ = -------- + αT L ------ + αΔT = ----------------------------------2 2 dx E L EK ( L – 0.5x )

3

Integrating from x=0 to x=L and noting that displacement at both points are zero, we obtain the following

∫

2

3

αT L x ⎞ αT L x ⎞ ⎛ R R du = ⎜ ------------------------------------ + ----------------⎟ dx = ⎜ ---------------------------------------------- + ----------------⎟ 2 2 u0 = 0 ⎝ ( 0.5EK ) ( L – 0.5x ) 0 ⎝ EK ( L – 0.5x ) 2 L ⎠ 3L ⎠ uL = 0

∫

L⎛

L

= 0 or 0

αT L L R 1 - – --1--------------------- ---------+ -------------- = 0 ( 0.5EK ) 0.5L L 3 αT L L 2R ----------= -------------EKL 3

2

EKαT L L R = – ------------------------6

or

4

Substituting Eq. (4) in Eq. (2) and evaluating the stress at x=L/2 EαT L R R σ xx = --------------------------------- = ------------------------- = – -------------------2 2 2 L ⎛ ⎞ K ( 0.75L ) 6 ( 0.75 ) -K L – 0.5 ⎝ 2⎠

8 σ xx = ------ EαT L ( C ) 27

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.97 Three metallic rods are attached to a rigid plate as shown. The temperature of the rods is lowered oF after the forces are applied. Assuming the rigid plate does not rotate, determine the movement of by 100 the rigid plate Area

6500 lbs

α

in2

E ksi

10 -6/ oF

Aluminum

4

10,000

12.5

Steel-1

4

30,000

6.6

Steel-2

12

30,000

6.6

Aluminum Steel -2 Steel-1 6500 lbs 100 inch

100 inch

Figure P4.97

ΔT = – 100°F

Solution

δP =

-----------------------------------------------------------We can draw the free body diagram of the rigid plate as shown in Fig.(a). (a)

Nal

6500 lbs Ns2

δP

(b)

Tensile Compressive Ns1

6500 lbs

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δs1 = δal

δs2

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

By force equilibrium in Fig (a) N a1 + N s1 + N s2 = 13000lbs = 13kips

1

We can draw an exaggerated deformed geometry as shown in Fig.(b) and write the deformation equations δ a1 = δ P

δ s1 = δ P

δ s2 = δ P

2

Noting that the deformation shown in Fig (b) are considered positive we have the following N a1 ( 100 ) –6 δ a1 = --------------------------- – ( 100 ) ( 12.5 ) ( 10 ) ( 100 ) = δ P 3 10 ( 10 ) ( 4 ) N s1 ( 100 ) –6 δ s1 = --------------------------- – ( 100 ) ( 6.6 ) ( 10 ) ( 100 ) = δ P 3 30 ( 10 ) ( 4 ) N s2 ( 100 ) –6 δ s2 = ------------------------------ + ( 100 ) ( 6.6 ) ( 10 ) ( 100 ) = δ P 3 30 ( 10 ) ( 12 )

or

N a1 = 400δ P + 50

3

or

N s1 = 1200δ P + 79.2

4

or

N s2 = 3600δ P – 237.6

5

Substituting Eq’s (3), (4) and (5) into Eq. (1) we obtain ( 400 + 1200 + 3600 )δ P + 50 + 79.2 – 237.6 = 13

–3 121.4 δ P = ------------- = 23.35 ( 10 ) 5200 δ P = 0.0233 in. to the right

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.98 Solve problem 4.92 assuming that in addition to turning the nut the temperature of the assembled unit is raised by 40oC. The coefficients of thermal expansion for steel and aluminum are αst = 12 μ / oC,

αal = 22.5 μ / oC. ΔT = 40°C

Solution

α st = 12μ ⁄ °C

α al = 22.5μ ⁄ °C

-----------------------------------------------------------From Eq. (1) and (2) in problem 4-89 N al = N s

1 –3

δ al + δ s = 0.75 ( 10 )m

2

Noting that the deformation in Eq. (2) assumes for steel as positive in extension and for aluminum as positive in contraction, we can modify the deformation equations (3) and (4) of problem 4-89 as shown below. –9

–6

δ al = 3.896N al ( 10 ) – ( 40 ) ( 22.5 ) ( 10 ) ( 0.3 ) –9

or

–6

δ s = 3.5N s ( 10 ) – ( 40 ) ( 12 ) ( 10 ) ( 0.3 + 0.025 + 0.025 )

–9

–6

δ al = 3.896N al ( 10 ) – 270 ( 10 ) or

–9

3 –6

δ s = 3.5N s ( 10 ) – 168 ( 10 )

4

Substituting Eq’s (3), (4) and (1) into Eq. (2) –9

–6

–3

( 3.896 + 3.5 ) ( N al ) ( 10 ) + ( – 270 + 168 ) ( 10 ) = 0.75 ( 10 ) or –3

3 0.852 ( 10 ) N al = ----------------------------- = 115.2 ( 10 )Newtons –9 7.396 ( 10 )

5

3

N s = 115.2 ( 10 )Newtons

6

The axial stresses are 3 N al 6 115.2 ( 10 ) σ al = ------- = ---------------------------- = 104.7 ( 10 ) – 6 A al 1100 ( 10 )

σ al = 104.7MPa ( C )

3 N 6 115.2 ( 10 ) σ s = ------s = --------------------------- = 230.4 ( 10 ) –6 As 500 ( 10 )

σ s = 230.4 MPa ( T )

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January 2014

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.99 Determine the axial stress in bar A of problem 4.93 assuming that the nut is turned 1 full turn and the temperature of bar A is decreased by 80oF. The coefficients of thermal expansion for bar A is αst = 22.5 μ / oF. Solution

ΔTA = - 80oF

p = 0.125 inch n = 4 quarter turns σA = ? σB = ?

αst = 22.5 μ / oF

-----------------------------------------------------------From Eqs. (1) (2) and (3) in problem 4.93 we have: N A ( 5 ) = 15N B

or

δ δA ------ = -----D5 15 δ B + δ D = nP -----4

N A = 3N B

1

δ δ A = -----D3

or

2

δ B = 0.03125n – δ D

or

3

Noting that the deformation in Eq. (4) of problem 4.93 assumes for bar A as positive in extension, we can modify the deformation in Eq. (4) of problem 4.93 as shown below. · δ N A ( 50 ) –6 –3 δ A = ------------------------------------- – ( 80 ) ( 22.5 ) ( 10 ) ( 50 ) = 0.01N A – 90 ( 10 ) = -----D3 3 ( 10 ) ( 10 ) ( 0.5 ) or

4 5

N A = 33.33δ D + 9

There is no temperature change in bar B, hence from Eq. (7) of problem 4.93 we have the following with n = 4. N B = 14.0625 ( 4 ) – 450δ D = 56.25 – 450δ D

or

6

Substituting Eq. (5) and (6) into Eq. (1) we obtain. 33.33δ D + 9 = 3 ( 56.25 – 450δ D )

or

159.75 δ D = ------------------- = 0.1155 1383.33

7

From Eq. (5) we obtain: N A = 12.849kips NA The axial stress in A is: σ A = ------- = 12.849 ---------------- = 25.70ksi AA

σ A = 25.70ksi ( T )

0.5

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.100 Fifty rivets of 10 mm diameter are used to attach caps at each end on a 1000 mm diameter cylinder. The wall of the cylinder is 10 mm thick and the gas pressure is 200 kPa. Determine the hoop stress and the axial stress in the cylinder and the shear stress in each rivet.

Figure P4.100

Solution

n = 50 dr = 10 mm

d = 1000 mm

t = 10 mm p = 200kPa σθθ = ?

τr =?

-----------------------------------------------------------The hoop stress can be found as shown below: 3

6 2 ( 200 ) ( 10 ) ( 0.500 ) σ θθ = pr ----- = ---------------------------------------------- = 10 ( 10 ) N ⁄ m t ( 0.01 )

σ θθ = 10MPa ( T )

The shear force on each rivet is Vr=τrAr, where Ar is the area of cross-section of each rivet. The free body

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January 2014

diagram of the cap can be drawn as shown below. Vr=τrAr p

By force equilibrium we obtain: 2

p ( πr ) = ( 50 ) ( τ r A r )

3

2

6 2 ( 200 ) ( 10 )π ( 0.5 ) τ r = ---------------------------------------------- = 40 ( 10 ) N ⁄ m or τ r = 40MPa 2 ( 50 )π ( 0.01 ⁄ 2 )

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.101 A pressure tank 15 feet long and 40 inch diameter is to be fabricated from a 1/2 inch thick sheet. A 15 feet long, 8 inch wide and 1/2 inch thick plate is bonded onto the tank to seal the gap. What is the shear stress in the adhesive when the pressure in the tank is 75 psi? Assume uniform shear stress over the entire inner surface of the attaching plate.

Figure P4.101

Solution

L = 15 ft.

d = 40 inch

t = 0.5 inch

W = 8 inch

p = 500 psi

τa = ?

-----------------------------------------------------------( 75 ) ( 20 ) ----- = ---------------------- = 3000psi The hoop stress can be found as σ θθ = pr t

( 0.5 )

.

We make two imaginary cuts, one through the adhesive and another one through the tank on the other side to produce the free body diagram shown below. σθθ p W/2

τa L

By force equilibrium we obtain: τ a ( W ⁄ 2 ) ( L ) + σ θθ ( L ) ( t ) – ( p ) ( L ) ( d ) = 0 ( p ) ( d ) – σ θθ ( t ) ( 40 ) – ( 3000 ) ( 0.5 )- = 375psi τ a = ------------------------------------- = 75 --------------------------------------------------(W ⁄ 2) (8 ⁄ 2)

τ a = 375 psi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.102 A 5 feet diameter spherical tank has a wall thickness of 3/4 inch. If the maximum normal stress is not to exceed 10 ksi, determine the maximum permissible pressure. Solution d = 5 ft. t = 0.75 inch σ max ≤ 10 ksi pmax = ?

-----------------------------------------------------------The maximum pressure can be found as shown below. p ( 5 ) ( 12 ) σ = pr ----- = ---------------------- ≤ 10 0.75 t

or

p ≤ 0.125ksi or

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p max = 125psi

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.103 In a spherical tank of radius 500 mm and a thickness of 40 mm, a hole of 50 mm diameter is drilled and then plugged using adhesive of shear strength 1.2 MPa to form a safety pressure release mechanism.Determine the maximum allowable pressure and the corresponding hoop stress in the tank material.

Figure P4.103

Solution

r = 500 mm

t = 40 mm

τ a ≤ 1.2MPa

dplug = 50 mm

pmax = ?

σ=?

-----------------------------------------------------------We can draw the free body diagram of the plug as shown below. τa p 2

By force equilibrium we obtain p ( πd plug ⁄ 4 ) = ( τ a ) ( πd plug ) ( t ) or pd plug 6 p ( 0.05 ) ( τ a ) = --------------- = ----------------------- ≤ 1.2 ( 10 ) ( 4 ) ( 0.04 ) 4t

6

p ≤ 3.84 ( 10 )

or

or

p max = 3.8 MPa

The hoop stress can be found as shown below. 6

6 2 3.84 ( 10 ) ( 0.5 ) σ = pr ----- = ------------------------------------- = 48 ( 10 ) N ⁄ m t ( 0.04 )

σ = 48 MPa ( T )

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.104 A 20 inch diameter pressure cooker is to be designed for a 15 psi pressure. The allowable normal stress in the cylindrical pressure cooker is to be limited to 3 ksi. Determine the minimum wall thickness of the pressure cooker. A 1/2 lbs. weight on top of the nozzle is used to control the pressure in the cooker. Determine the diameter ‘d’ of the nozzle.

Figure P4.104

Solution

d = 20 inch

p = 15 psi

σ max ≤ 3 ksi

tmin = ?

W = 0.5 lb

dnoz = ?

-----------------------------------------------------------The hoop stress in the cylinder is greater than the axial stress and is used to determine the minimum thickness as shown below. 3 ( 15 ) ( 10 ) σ θθ = pr ----- = ---------------------- ≤ 3 ( 10 ) t t

or

t ≥ 0.05 in

or

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t min = 0.05in

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

January 2014

We can draw the free body diagram of the weight on the nozzle as shown below. W = 0.5 lb

p

By force equilibrium we obtain: d noz =

2 ------ = πp

2 pπ ( d noz

⁄ 4 ) = 0.5 or

2 -------------- = 0.206 or π ( 15 )

d noz = 0.206 in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.105 The cylindrical gas tank shown is made from a sheet metal that is 8 mm thick and must be designed to sustain a maximum normal stress of 100 MPa. Develop a table of maximum permissible gas pressure and the corresponding diameter of the tank in steps of 100mm between diameter values of 400 mm to 900 mm.

Figure P4.105

Solution

σ max ≤ 100MPa

t = 8 mm

p vs. d

-----------------------------------------------------------The hoop stress in the cylinder is greater than the axial stress and is used to determine the values of p and d. Assuming d is measured in millimeters and p in MPa, we can write the hoop stress expression as shown below: 6

–3

( p ) ( 10 ) ( d ⁄ 2 ) ( 10 ) 6 σ θθ = pr ----- = ----------------------------------------------------- ≤ 100 ( 10 ) 0.008 t

or pd ≤ 1600

1

Eq. 1 can be used to determine the values of p and d as shown in the table below. All pressure values are rounded downwards, d (mm

p (MPa)

400

4.00

500

3.20

600

2.66

700

2.28

800

2.00

900

1.77

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.106 A pressure tank 15 feet long and 40 inch diameter is to be fabricated from a 1/2 inch thick sheet. A 15 feet long, 8 inch wide and 1/2 inch thick plate is to be used for sealing the gap by using two rows of 90 rivets each. If the shear strength of the rivets is 36 ksi and the normal stress in the tank is to be limited to 20

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January 2014

ksi, determine the maximum pressure and the minimum diameter of the rivets that can be used.

Figure P4.106

Solution

L = 15 ft.

d = 40 inch

t = 0.5 inch

nrivets=90

σ max ≤ 20 ksi

τ riv ≤ 36 ksi

Wplate = 8 inch tplate = 0.5 inch pmax = ? (driv )min= ?

-----------------------------------------------------------The hoop stress in the cylinder is greater than the axial stress and is used to determine the maximum pressure as shown below. 3 ( p ) ( 20 ) pr σ θθ = ----- = ------------------- ≤ 20 ( 10 ) 0.5 t

p ≤ 500psi

or

or p max = 500psi We make imaginary cuts through the rivets and also cut through the tank on the other side to produce the free body diagram shown below.

τr

W/2

σθθ

p

L

2

By force equilibrium we obtain: n riv τ r ( πd riv ) ⁄ 4 + σ θθ ( L ) ( t ) – ( p ) ( L ) ( d ) = 0 or ( p ) ( L ) ( d ) – ( σ θθ ) ( L ) ( t ) 3 ( 500 ) ( 15 ) ( 12 ) ( 40 ) – ( 20000 ) ( 15 ) ( 12 ) ( 0.5 ) τ riv = ----------------------------------------------------------- = ------------------------------------------------------------------------------------------------------------ ≤ ( 36 ) ( 10 ) or 2 2 ( 90 )π ( d riv ⁄ 4 ) n riv π ( d riv ⁄ 4 ) 3

1800 ( 10 ) d riv ≥ -------------------------- or d riv ≥ 0.841inch or 3 810π ( 10 )

( d riv ) min = 0.85in

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4.107 A pressure tank 5 m long and a mean diameter of 1 m is to be fabricated from a 10 mm thick sheet. A 5 m-long, 200 mm wide, 10-mm-thick plate is to be used for sealing the gap by using two rows of 100 rivets each. The shear strength of the rivets is 300 MPa and the yield strength of the tank material is 200 MPa. Determine the maximum pressure and the minimum diameter of the rivets to the nearest millimeter that can be used for a factor of safety of 2.

Figure P4.107

Solution

L=5m d=1m ksafety= 2 σ yield = 200MPa

t = 0.01m τ riv

W = 0.2 m = 300MPa p=?

nriv=100 dr= ?

------------------------------------------------------------

The hoop stress can be found as

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 4

200 pr p ( 0.5 ) σ θθ = ----- = --------------- = 50p ≤ --------2 t ( 0.01 )

p ≤ 2MPa .

or

January 2014

p max = 2 MPa

We make imaginary cuts through the rivets and also cut through the tank on the other side to produce the free body diagram shown below.

τr

W/2

p

σθθ

L

2

By force equilibrium we obtain: n riv τ r ( πd riv ) ⁄ 4 + σ θθ ( L ) ( t ) – ( p ) ( L ) ( d ) = 0 or 6 6 6 ( p ) ( L ) ( d ) – ( σ θθ ) ( L ) ( t ) ( 4 ) ( 10 ) ( 5 ) ( 1 ) – ( 200 ) ( 10 ) ( 5 ) ( 0.01 ) 300 6 10 ( 10 ) τ riv = ----------------------------------------------------------- = ----------------------------------------------------------------------------------------------- ≤ --------- ( 10 ) or d riv ≥ ----------------------------2 2 6 2 ( 100 )π ( d riv ⁄ 4 ) n riv π ( d riv ⁄ 4 ) 3750π ( 10 )

or d riv ≥ 0.02913m or

( d riv ) min = 30 mm

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

5.1 A pair of 48 inch long bars and a pair of 60 inch long bars are symmetrically attached to a rigid disc at a radius of 2 inch at one end and built into the wall at the other end. The shear strain at point A due to a twist of the rigid disc was found to be 3000 μ rads. Determine the magnitude shear strain at point D. T A

B

A

C

D

D

C

B

60 in

48 in

Figure P5.1

Solution

γA= 3000 μ r= 2 inch. γD= ? ------------------------------------------------------------

We can draw an exaggerated deformed geometry as shown below. A

γA A

B

C

D

γD

B1 C1 φ C B

48 in

D

60 in

From triangle ABB1 and small strain approximation, we have the following: BB tan γ A = ----------1 AB

BB –6 γ A = ----------1 = 3000 ( 10 ) 48

or

or

BB 1 = 0.144 in

(1)

Note that BB1= CC1. From triangle DCC1 we have the following: CC BB tan γ D = ----------1 = ----------1 CD CD

or

–3 0.144 γ D = ------------- = 2.400 ( 10 ) or 60

γ D = 2400 μrad

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.2 If the four bars in problem 5.1 are made from material that has a Shear Modulus of 12,000 ksi., determine the applied torque T on the rigid disc.The area of cross-section of all bars is 0.25 in2. Solution

A = 0.25in2

G = 12,000 ksi

T=?

-----------------------------------------------------------–3

–3

From problem 5.1 γ A = 3.000 ( 10 ) and γ D = 2.400 ( 10 ) . The shear stresses in bars AB and CD can be found from Hooke’s Law as shown below. –3

τ AB = Gγ A = ( 12, 000 ) ( 3.000 ) ( 10 ) = 36 ksi –3

τ CD = Gγ D = ( 12, 000 ) ( 2.400 ) ( 10 ) = 28.8 ksi

(1) (2)

The internal shear forces in the bars can be found as shown below: V AB = τ AB A = ( 36 ) ( 0.25 ) = 9.0 kips V CD = τ CD A = ( 28.8 ) ( 0.25 ) = 7.2 kips

(3) (4)

We can make imaginary cuts through the bars and draw the free body diagram of the rigid disc as shown

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

below. T

VAB

VCD

C

B

2 in 2 in

C

B

VAB

VCD

By equilibrium of moment about the axis of the rigid disc, we obtain: T = ( 2 ) ( V AB ) ( 2 ) + ( 2 ) ( V CD ) ( 2 ) = ( 4 ) ( V AB ) + ( 4 ) ( V CD ) T = ( 4 ) ( 9 ) + ( 4 ) ( 7.2 ) = 64.8 or

or

(5)

T = 64.8 in – kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.3 If the bars AB in problem 5.1 are made of aluminum with a Shear Modulus of Gal = 4,000 ksi and bars CD are made from bronze with a Shear Modulus of Gbr=6,500 ksi, determine the applied torque T on the rigid disc.The area of cross-section of all bars is 0.25 in2 Solution

GAB= 4,000 ksi

A = 0.25in2

GCD= 6,500 ksi

T=?

-----------------------------------------------------------–3

–3

From problem 5.1 γ A = 3.000 ( 10 ) and γ D = 2.400 ( 10 ) . The shear stresses in bars AB and CD can be found from Hooke’s Law as shown below. –3

τ AB = G AB γ A = ( 4, 000 ) ( 3.000 ) ( 10 ) = 12.0 ksi

(1)

–3

τ CD = G CD γ D = ( 6, 500 ) ( 2.400 ) ( 10 ) = 15.6 ksi

(2)

The internal shear forces in the bars can be found as shown below: V AB = τ AB A = ( 12.0 ) ( 0.25 ) = 3.0 kips

(3)

V CD = τ CD A = ( 15.6 ) ( 0.25 ) = 3.9 kips

(4)

From Eq. (5) in problem 5.2, we have: T = ( 4 ) ( V AB ) + ( 4 ) ( V CD ) = ( 4 ) ( 3 ) + ( 4 ) ( 3.9 ) = 27.6

or

T = 27.6 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.4 Three pairs of bars are symmetrically attached to rigid discs at the radii shown. The discs were observed to rotate by angles φ 1 = 1.5° , φ2 = 3.0° , and φ3 = 2.5° in the direction of the applied torques T1, T2, and T3 respectively. The shear modulus of the bars is 40 ksi and the area of cross-section is 0.04 in2. Determine the applied torques. T2

T1

A

B

C

T3

D

E

F

D

1.5 in E

F

2in 1.25 in A

B

25 in

C 40 in

30in

Figure P5.4

Solution

φ 1 = 1.5°

φ 2 = 3.0°

G = 40 ksi

A = 0.04 in2

φ 3 = 2.5°

T1 = ?

T2 = ?

T3 = ?

------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

Converting the given angles from degrees to radians we obtain: 1.5° φ 1 = ⎛ ----------⎞ π = 0.0262rads ⎝ 180 ⎠

3.0° φ 2 = ⎛ ----------⎞ π = 0.0524rads ⎝ 180 ⎠

2.5° φ 2 = ⎛ ----------⎞ π = 0.0436rads ⎝ 180 ⎠

We can draw an exaggerated deformed geometry as shown below. B1 A γ B AB A

D2

C1 γ CD C C φ1

B

D

φ E12 γEF

D1 D

C

25 in

F1 F φ 3

E

F2 F

E

40 in

30in

The magnitude of shear strains can be found using small strain approximation as shown below. BB tan γ AB = ----------1 AB

1.25φ ( 0.0262 -) = 1.31 ( 10 – 3 ) γ AB = ---------------1- = 1.25 ------------------------------25 25

or

D2 D1 tan γ CD = ------------C1 D F2 F1 tan γ EF = ----------E1 F2

D 2 D + DD 1 2φ 1 + 2φ 2 –3 2 ( 0.0262 + 0.0524 ) γ CD = ---------------------------- = ----------------------- = ----------------------------------------------- = 3.93 ( 10 ) CD CD 40

or

F 2 F + FF 1 1.5φ 2 + 1.5φ 3 –3 2 ( 0.0524 + 0.0436 ) γ EF = ------------------------ = -------------------------------- = ----------------------------------------------- = 4.8 ( 10 ) EF EF 30

or

We note that the angle between the axial direction and the tangent direction, increases for AB and EF and decreases for CD. Using the sign convention that positive shear strain result in decrease in angle from right angle, we write the shear strains with the following signs. –3

–3

γ AB = – 1.31 ( 10 )

–3

γ CD = 3.93 ( 10 )

γ EF = – 4.8 ( 10 )

(1)

The shear stress in each bar can be found using Hooke’s law. –3

–3

τ AB = Gγ AB = ( 40 ) ( – 1.31 ) ( 10 ) = – 52.4 ( 10 )ksi = – 52.4 psi –3

(2)

–3

τ CD = Gγ CD = ( 40 ) ( 3.93 ) ( 10 ) = 157.2 ( 10 )ksi = 157.2 psi –3

(3)

–3

τ EF = Gγ EF = ( 40 ) ( – 4.80 ) ( 10 ) = – 192.0 ( 10 )ksi = – 192.0 psi

(4)

The internal shear forces in the bars can be found as shown below: V AB = τ AB A = ( – 52.4 ) ( 0.04 ) = – 2.096 lbs V CD = τ CD A = ( 157.2 ) ( 0.04 ) = 6.288 lbs

(5)

V EF = τ EF A = ( – 192.0 ) ( 0.04 ) = – 7.680 lbs

(7)

(6)

We can make imaginary cuts through the bars and draw free body diagrams as shown.The direction of VEF is determined by inspection to oppose the torque T3.The direction of VCD is opposite to VEF and direction of VAB is in the same direction as VEF as per the signs obtained above. T3

(a)

(b)

T2

F VEF 1.5 in 1.5 in

VEF

F

T3

VCD D 2.0 in 2.0 in D

E

E

T1

(c)

F

VAB A

F

1.25 in A 1.25 in V AB

B

B

T2 D

C

C

D

E

T3 F

E

F

VCD

From moment equilibrium in Fig. (a), we obtain the following. T 3 = ( 2 ) ( V AB ) ( 1.5 ) = ( 3 ) ( 7.628 ) = 23.04 or

T 3 = 23.0 in.-lbs

From moment equilibrium in Fig. (b), we obtain the following.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

T 2 – T 3 – ( 2 ) ( V CD ) ( 2.0 ) = 0

or

T 2 = 23.04 + ( 4 ) ( 6.288 ) = 48.192 or

January 2014

T 2 = 48.2 in.-lbs

From moment equilibrium in Fig. (c), we obtain the following. T 1 – T 2 + T 3 + ( 2 ) ( V AB ) ( 1.25 ) = 0 or T 1 = 48.2 – 23.04 + ( 2.5 ) ( 2.096 ) = 30.4 or T 1 = 30.4 in.-lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.5 A circular shaft of radius r and length Δx has to rigid discs attached at each end. If the rigid discs are rotated as shown. Determine the shear strain γ at point A in terms of r, Δx, and Δφ assuming line AB remains straight, where. Δφ = φ 2 – φ1 . φ1

A

φ2

B

Δx

FigureP5.5

Solution

γ=f(r, Δx, Δφ)=? ------------------------------------------------------------

We can draw an exaggerated deformed geometry as shown below. A A1

φ1

B B2

γ

φ2

B1 Δx

From triangle AB2B1 and small strain approximation, we have the following: B2 B1 tan γ = ------------A1 B2

or

BB 1 – BB 2 BB 1 – AA 1 rφ 2 – rφ 1 γ = --------------------------- = --------------------------- = --------------------- or AB AB Δx

Δφ γ = r ------Δx

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.6 A hollow circular shaft made from hard rubber has an outer diameter of 4 inch and an inner diameter of 1.5 inch. The shaft is fixed to the wall on the left end and the rigid disc on the right hand is twisted as shown. The shear strain at point A which is on the outside surface was found to be 4000 μ rads. Determine the shear strain at point C which is on the inside surface and the angle of rotation. Assume lines AB and CD remain straight during deformation.

A

C

B

D

φ

36 in

Figure P5.6

Solution

γA = 4000 μ rads γC = ? ------------------------------------------------------------

do = 4 inch

di = 1.5 inch

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

We can draw the exaggerated deformed geometry as shown below. D

γC

C A

γA

D1 B

φ φ

B1 36 in

From triangle ABB1 and small strain approximation, we have the following: BB ( d o ⁄ 2 )φ –6 4 ⁄ 2 )φ tan γ A ≈ γ A = ----------1 = -------------------or 4000 ( 10 ) = (----------------- or φ = 0.072 rads 36 AB AB

From triangle CDD1 and small strain approximation, we have the following: DD ( d i ⁄ 2 )φ –3 ( 1.5 ⁄ 2 ) ( 0.072 ) tan γ C ≈ γ C = ----------1- = ------------------= -------------------------------------- = 1.500 ( 10 ) or CD AB 36

γ C = 1500 μ rads

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.7

The magnitude of shear strain in each of the segment of the stepped shaft shown were found to be , γ CD = 2500 μrad , and γEF = 6000 μrad . The radius of section AB is 150 mm, of section CD is 70 mm, and section EF is 60 mm. Determine the angle by which each of the rigid discs was rotated. γ AB = 3000 μrad

φ1 A

B

φ2 D

C

1.8 m

2m

φ3

E

F

1.2m

Figure P5.7

Solution

γ AB = 3000 μrads

rAB = 150 mm φ1 = ?

γ CD = 2500 μrads

γ EF = 6000 μrads

rCD = 70 mm φ2 = ?

rEF = 60 mm φ3 = ?

-----------------------------------------------------------We can draw the exaggerated deformed geometry as shown below. φ1

A

γAB

B B1

2m

φ2

C C1

γCD

D1

E1

D D2

E

1.8 m

φ3 F2 γEF F F1

1.2m

The magnitude of shear strains can be found using small strain approximation as shown below. r AB φ 1 ( 0.150 )φ BB –6 - = -----------------------1- = 3000 ( 10 ) or tan γ AB ≈ γ AB = ----------1 = -------------AB AB 2

φ 1 = 0.0400 rad

D2 D1 D 2 D + DD 1 r CD φ 1 + r CD φ 2 0.07 ( 0.04 + φ 2 ) –6 tan γ CD ≈ γ CD = ------------- = ---------------------------- = ------------------------------------ = ------------------------------------- = 2500 ( 10 ) C1 D2 CD CD 1.8

or

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φ 2 = 0.0243 rad;

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

F2 F1 F 2 F + FF 1 r EF φ 2 + r EF φ 3 0.06 ( 0.0243 + φ 3 ) –6 tan γ CD ≈ γ CD = ----------- = ------------------------ = ---------------------------------- = ------------------------------------------- = 6000 ( 10 ) E1 F2 EF EF 1.2

φ 3 = 0.0957 rad

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.8 A hollow aluminum (G= 26 GPa) shaft cross section is shown in Figure P5.8. The shear strain γxθ in polar coordinates at the section is γxθ = – 0.06ρ where ρ is in meters. Determine the equivalent internal torque acting at the cross-section. Use di= 30 mm and do = 50 mm. ρ

θ x

di do Figure P5.8

Solution

G = 26 GPa

γ xθ = – 0.06ρ

di= 30 mm

do = 50 mm

T=?

-----------------------------------------------------------From Hooke’s law we can write: 9

9

τ xθ = Gγ xθ = ( 26 ) ( 10 ) ( – 0.06ρ ) = – 1.5 6ρ ( 10 ) N ⁄ m

2

(1) (2)

The differential area dA is the area of a ring of radius ρ and thickness dρ i.e., dA = ( 2πρ )dρ . We can write: 0.025

T =

0.025

∫ ρ τxθ ( 2πρ )dρ 0.015

=

∫ 0.015

4

0.025

9 ρ ρ ( – 1.56 ρ ) ( 10 ) ( 2πρ )dρ = ( – 3.12π ) ----( 10 ) or 4 0.015 9

4 4 9 – 3.12π T = ⎛ -----------------⎞ ( 0.025 – 0.015 ) ( 10 ) ⎝ 4 ⎠

(3) T = – 833.1 N-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.9 A hollow aluminum (G= 26 GPa) shaft cross section is shown in Figure P5.9. The shear strain γxθ in polar coordinates at the section is γxθ = 0.05ρ where ρ is in meters. Determine the equivalent internal torque acting at the cross-section. Use di= 40 mm and do = 120 mm. Solution

G = 26 GPa

γ xθ = 0.05ρ

di= 40 mm

do = 120 mm

T=?

-----------------------------------------------------------From Hooke’s law we can write: 9

9

τ xθ = Gγ xθ = ( 26 ) ( 10 ) ( 0.05ρ ) = 1.3ρ ( 10 ) N ⁄ m

2

The differential area dA is the area of a ring of radius ρ and thickness dρ i.e., dA = ( 2πρ )dρ . We can write:

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(1) (2)

5-6

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

θ

ρ

x

di do Figure P5.9 0.06

T =

0.06

∫ ρ τxθ ( 2πρ )dρ 0.02

=

∫ 0.02

4

0.06

9 ρ ρ ( 1.3ρ ) ( 10 ) ( 2πρ )dρ = ( 2.6π ) ----( 10 ) 4 0.02 9

or

4 4 9 2.6π T = ⎛ -----------⎞ ( 0.06 – 0.02 ) ( 10 ) ⎝ 4 ⎠

(3) T = 26.1 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.10 A hollow brass (GB = 6500 ksi) shaft and a solid steel (GS = 13,000 ksi) shaft are securely fastened to form a composite shaft as shown. The shear strain γxθ in polar coordinates at the section is γ xθ = 0.001ρ where ρ is in inches. Determine the equivalent internal torque acting at the cross-section. Use dB= 4 in and dS = 2 in. ρ

θ x

Steel Brass dS dB

Figure P5.10

Solution

GB = 6500 ksi γ xθ = 0.001ρ

GS = 13,000 ksi) dB= 4 in dS = 2 in.

T=?

-----------------------------------------------------------From Hooke’s law we can write: ( τ xθ ) steel = G S γ xθ = ( 13, 000 ) ( 0.001ρ ) = 13ρ ksi

(1)

( τ xθ ) brass = G B γ xθ = ( 6, 500 ) ( 0.001ρ ) = 6.5ρ ksi

(2)

The shear stress across the cross-section can be written as shown below. ⎧ 13ρ ksi τ xθ = ⎨ ⎩ 6.5ρ ksi

0.0 ≤ ρ < 1.0 1.0 < ρ ≤ 2.0

(3)

The differential area dA is the area of a ring of radius ρ and thickness dρ ι.e., dA = ( 2πρ )dρ . We can write

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

2

T =

1

∫ ρ τxθ ( 2πρ )dρ = ∫ 0

0

2

4

January 2014

1

4

2

ρ ρ ρ ( 13ρ ) ( 2πρ )dρ + ρ ( 6.5ρ ) ( 2πρ )dρ = ( 26π ) ----- + ( 13π ) ----4 0 4 1

∫

or

1

4 26π 13π 4 T = ⎛ ---------⎞ + ⎛ ---------⎞ ( 2 – 1 ) = 173.57 in – kips ⎝ 4 ⎠ ⎝ 4 ⎠

(4) T = 173.6 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.11 A hollow brass (GB = 6500 ksi) shaft and a solid steel (GS = 13,000 ksi) shaft are securely fastened to form a composite shaft as shown. The shear strain γxθ in polar coordinates at the section is γ xθ = – 0.0005 ρ where ρ is in inches. Determine the equivalent internal torque acting at the cross-section. Use dB= 6 in and dS = 4 in. ρ

θ x

Steel Brass dS dB

Figure P5.11 .

Solution

GB = 6500 ksi

GS = 13,000 ksi) γxθ = – 0.0005 ρ

dB= 6 in

dS = 4 in.

-----------------------------------------------------------From Hooke’s law we can write: ( τ xθ ) brass = G B γ xθ = ( 6, 500 ) ( – 0.0005 ρ ) = – 3.25 ρ ksi

(1)

( τ xθ ) steel = G S γ xθ = ( 13, 000 ) ( – 0.0005 ρ ) = – 6.5 0ρ ksi

(2)

The shear stress across the cross-section can be written as shown below. ⎧ – 3.25 ρ ksi τ xθ = ⎨ ⎩ – 6.5 0ρ ksi

0.0 ≤ ρ < 2.0

(3)

2.0 < ρ ≤ 3.0

The differential area dA is the area of a ring of radius ρ and thickness dρ ι.e., dA = ( 2πρ )dρ . We can write: 3

T =

2

∫ ρ τxθ ( 2πρ )dρ = ∫ 0

0

3

4

2

4

3

ρ ρ ρ ( – 6.5 0ρ ) ( 2πρ )dρ + ρ ( – 3.25 ρ ) ( 2πρ )dρ = ( – 13 π ) ----- + ( – 6.5 π ) ----- or 4 0 4 2

∫ 2

4 – 13 π 4 – 6.5 π 4 T = ⎛ ------------⎞ ( 2 – 0 ) + ⎛ --------------⎞ ( 3 – 2 ) = – 495.19 in – kips ⎝ 4 ⎠ ⎝ 4 ⎠

(4)

T = – 495.2 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.12 A hollow brass (GB = 6500 ksi) shaft and a solid steel (GS = 13,000 ksi) shaft are securely fastened to form a composite shaft as shown. The shear strain γxθ in polar coordinates at the section is γ xθ = 0.002ρ where ρ is in inches. Determine the equivalent internal torque acting at the cross-section. Use dB= 3 in and dS = 1 in.

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

θ

ρ

x Steel Brass dS dB

Figure P5.12

GB = 6500 ksi GS = 13,000 ksi) γxθ = 0.002ρ

Solution

dB= 3 in

dS = 1 in.T = ?

-----------------------------------------------------------From Hooke’s law we can write: ( τ xθ ) steel = G S γ xθ = ( 13, 000 ) ( 0.002ρ ) = 26ρ ksi

(1)

( τ xθ ) brass = G B γ xθ = ( 6, 500 ) ( 0.002ρ ) = 13ρ ksi

(2)

The shear stress across the cross-section can be written as shown below. 0.0 ≤ ρ < 0.5

⎧ 26ρ ksi τ xθ = ⎨ ⎩ 13ρ ksi

(3)

0.5 < ρ ≤ 1.5

The differential area dA is the area of a ring of radius ρ and thickness dρ ι.e., dA = ( 2πρ )dρ . We can write: 1.5

T =

0.5

∫ ρ τxθ ( 2πρ )dρ

=

0

1.5

∫ ρ ( 26ρ ) ( 2πρ )dρ + ∫ 0

0.5

4

0.5

4

1.5

ρ ρ ρ ( 13ρ ) ( 2πρ )dρ = ( 52π ) ----+ ( 26π ) ----or 4 0 4 0.5

4 4 4 52π 26π T = ⎛ ---------⎞ ( 0.5 – 0 ) + ⎛ ---------⎞ ( 1.5 – 0.5 ) = 104.65 in – kips ⎝ 4 ⎠ ⎝ 4 ⎠

(4) T = 104.65 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.13 A hollow titanium (GTi = 36 GPa) shaft and a hollow Aluminum (GAl = 26 GPa) shaft are securely fastened to form a composite shaft as shown. The shear strain γxθ in polar coordinates at the section is γ xθ = 0.04ρ where ρ is in meters. Determine the equivalent internal torque acting at the cross-section Use di= 50 mm, dAl= 90 mm and dTi = 100 mm. Titanium

ρ

Aluminum

θ x

di dAl dTi

Solution

GTi = 36 GPa di= 50 mm

GAl = 26 GPa γ xθ = 0.04ρ dAl= 90 mm dTi = 100 mm

T=?

-----------------------------------------------------------From Hooke’s law we can write: 9

9

( τ xθ ) Al = G Al γ xθ = ( 26 ) ( 10 ) ( 0.04ρ ) = 1.04ρ ( 10 ) N ⁄ m

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2

(1)

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

9

9

( τ xθ ) Ti = G Ti γ xθ = ( 36 ) ( 10 ) ( 0.04ρ ) = 1.44ρ ( 10 ) N ⁄ m

January 2014

2

(2)

The shear stress across the cross-section can be written as shown below. 9

2

⎧ 1.04ρ ( 10 ) ( N ⁄ m ) τ xθ = ⎨ ⎩ 1.44ρ ( 10 9 ) N ⁄ m 2

0.025 ≤ ρ < 0.045

(3)

0.045 < ρ ≤ 0.050

The differential area dA is the area of a ring of radius ρ and thickness dρ ι.e., dA = ( 2πρ )dρ . We can write: 0.050

0.045

∫ ρ τxθ ( 2πρ )dρ

T =

0.025

=

∫

0.05 9

ρ ( 1.04ρ ) ( 10 ) ( 2πρ )dρ +

0.025 4

0.045

4

∫

9

ρ ( 1.44ρ ) ( 10 ) ( 2πρ )dρ or

0.045 0.050

9 ρ ρ T = ( 2.08π ) ----+ ( 2.88π ) ----( 10 ) or 4 0.025 4 0.045 4 4 4 4 9 3 2.08π 2.88π T = ⎛ --------------⎞ ( 0.045 – 0.025 ) + ⎛ --------------⎞ ( 0.05 – 0.045 ) ( 10 ) = 10.92 ( 10 ) N – m ⎝ 4 ⎠ ⎝ 4 ⎠

(4)

T = 10.9 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.14 A circular shaft made from elastic - perfectly plastic material has a torsional shear stress distribution across the cross section shown in Figure P5.14. Determine the equivalent internal torque.I

τxθ 24 ksi

ρ 0.3 in. 0.3 in.

Figure P5.14

5.15

A solid circular shaft of 3-in. diameter has a shear strain at a section in polar coordinates of γxθ =

ρ is the radial coordinate measured in inches. The shaft is made from an elastic–perfectly 2ρ × plastic material, which has a yield stress τyield = 18 ksi and a shear modulus G = 12,000 ksi. Determine the equivalent internal torque. (See problem 3.144). Solution γxθ=0.002ρ τyield=18ksi G=12,000 ksi T = ? Elastic perfectly plastic 10-3, where

-----------------------------------------------------------18 - = 0.0015 = 0.002ρ yield or ρ yield = 0.75inch where, ρyield The strain at yield point is: γ yield = ----------------12, 000

is the radial coordinate of elastic-plastic boundary. For 0 < ρ < 0.75 , τ xθ = Gγ xθ = ( 12000 ) ( 0.002ρ ) = 24ρ . The stress distribution can be written as: ⎧ 24ρ ksi τ xθ = ⎨ ⎩ 18 ksi

0 < ρ < 0.75in 0.75in < ρ < 1.5in

1

The internal torque can be found as shown below.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

∫ ρτxθ dA

T =

=

A

January 2014

∫ ρτxθ dA + ∫ ρτxθ dA AE

AP

TE

2

TP

The differential area dA is the area of a ring of radius ρ and thickness dρ ι.e. dA = ( 2πρ )dρ . Substituting the stress distribution and performing integration we obtain the internal torque 0.75

∫

TE =

0 1.5

3

∫

TP =

4

0 75

0.75

ρ ( ρ ) ( 24ρ ) ( 2πρ ) dρ = ( 48π ) ----4 0

= 11.93 in – kips

1.5

ρ ( ρ ) ( 18 ) ( 2πρ ) dρ = ( 36π ) ----= 111.33 in – kips 3 0.75

T = T E + T P = 123.26 in – kips or

T = 123.3 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.16

A solid circular shaft of 3-in. diameter has a shear strain at a section in polar coordinates of γxθ =

2ρ × 10 , where ρ is the radial coordinate measured in inches.The shaft is made form a bilinear material as shown in Figure 3.40. The material has a yield stress τyield = 18 ksi and shear moduli G1 = 12,000 ksi and G2 = 4800 ksi. Determine the equivalent internal torque.(See problem 3.145). -3

γxθ=0.002ρ

Solution

τyield=18ksi

G1=12,000ksi

G2=4,800ksi

τxθ= f(ρ) = ? T = ?

-----------------------------------------------------------18 - = 0.0015 = 0.002ρ yield or ρ yield = 0.75inch where, ρyield The strain at yield point is: γ yield = ----------------12, 000

is the radial coordinate of elastic-plastic boundary. Before yield point we have: τ1=G1γxθ=(12,000)(0.002ρ)=24ρ τ

18 yield The strain at yield point is: γ yield = ------------= ------------------ = 0.0015 12, 000

G1

After yield point we can write: τ 2 = τ yield + G 2 ( γ xθ – γ yield ) = 18 + 4800 ( 2ρ ( 10 – 3 ) – 0.0015 ) = 10.8 + 9.6ρ The stress distribution can be written as: 0 < ρ < 0.75in

⎧ 24ρ ksi τ xθ = ⎨ ⎩ ( 10.8 + 9.6ρ )ksi

0.75in < ρ < 1.5in

1

The internal torque can be found as shown below.

∫ ρτxθ dA

T =

A

=

∫ ρτxθ dA + ∫ ρτxθ dA AE

TE

AP

2

TP

The differential area dA is the area of a ring of radius ρ and thickness dρ ι.e. dA = ( 2πρ )dρ . Substituting the stress distribution and performing integration we obtain the internal torque 0.75

TE =

∫ 0 1.5

TP =

∫

4

0.75

ρ ( ρ ) ( 24ρ ) ( 2πρ ) dρ = ( 48π ) ----4 0

= 11.93 in – kips 3

4

ρ ρ ( ρ ) ( 10.8 + 9.6ρ ) ( 2πρ ) dρ = ( 2π ) ⎛ 10.8 ----- + 9.6 -----⎞ ⎝ 3 4⎠

0.75

T = T E + T P = 150.29 in – kips or

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1.5

= 138.37 in – kips 0.75

T = 150.3 in.-kips

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.17 A solid circular shaft of 3-in. diameter has a shear strain at a section in polar coordinates of γxθ = -3 2ρ × 10 , where ρ is the radial coordinate measured in inches.The shaft material has a stress–strain relationship given by τ = 243γ 0 .4 ksi. Determine the equivalent internal torque. (See problem 3.146). Solution γxθ= 0.002ρ T=?

-----------------------------------------------------------The stress distribution is: τ xθ = 243γ xθ = 243 ( 0.002ρ ) 0.4 or

τ xθ = 20.23ρ 0.4 ksi

0 ≤ ρ ≤ 1.5

The differential area dA is the area of a ring of radius ρ and thickness dρ ι.e. dA = ( 2πρ )dρ . The internal torque can be found as shown below. 1.5

T =

∫ ρτxθ dA

=

A

∫

( ρ ) ( 20.23ρ 0.4 ) ( 2πρ ) dρ

0

3.4

1.5

ρ = ( 40.46π ) --------3.4 0

= 148.39 in – kips or T = 148.4 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------A solid circular shaft of 3-in diameter has a shear strain at a section in polar coordinates of γxθ = 2ρ

5.18

× 10 , where ρ is the radial coordinate measured in inches. The shaft material has a stress–strain relation-3

ship given by τ = 12,000γ − 120,000γ 2 ksi. Determine the equivalent internal torque. (See problem 3.147). Solution

γxθ=0.002ρ

2

τ = ( 12, 000γ – 120, 000γ ) ksi

T=?

-----------------------------------------------------------2

The shear stress distribution is: τ xθ = ( 12, 000γ xθ – 120, 000γ xθ ) = ( 12, 000 ) ( 0.002ρ ) – 120, 000 ( 0.002ρ ) 2 or τ xθ = ( 24ρ – 0.48ρ 2 )ksi

0 ≤ ρ ≤ 1.5

1

The differential area dA is the area of a ring of radius ρ and thickness dρ ι.e. dA = ( 2πρ )dρ . The internal torque can be found as shown below. 1.5

T =

∫ ρτxθ dA A

=

∫

4

5

ρ ρ ( ρ ) ( 24ρ – 0.48ρ ) ( 2πρ ) dρ = ( 2π ) ⎛ 24 ----- – 0.48 -----⎞ ⎝ 4 5⎠ 2

0

or

1.5

= 186.27 in – kips 0

T = 186.3 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.19 The torsional shear stress at point A on a solid circular homogenous cross-section was found to be τA= 120 MPa. Determine the maximum torsional shear stress on the cross-section. A 300

60 mm 100 mm

FigureP5.19

5.20

The torsional shear strain at point A on a homogenous circular section shown in Figure P5.20 was

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

found to be 900 μ rads. Using a shear modulus of elasticity of 4000 ksi, determine the torsional shear stress at point B. B A 550

300

1.5 in. 2.5 in.

Figure P5.20

5.21 An aluminum shaft (Gal= 28 GPa) and a steel shaft (GS=82 GPa) are securely fastened to form composite shaft with a cross section shown in Figure P5.21. If the maximum torsional shear strain in aluminum is 1500 m rads, determine the maximum torsional shear strain in steel. Steel Aluminum

60 mm 100 mm

FigureP5.21

5.22 An aluminum shaft (Gal= 28 GPa) and a steel shaft (GS=82 GPa) are securely fastened to form composite shaft with a cross section shown in Figure P5.21. If the maximum torsional shear stress in aluminum is 21 MPa, determine the maximum torsional shear stress in steel. 5.23 Determine the direction of shear stress at points A and B (a) by inspection, and (b) by using the sign convention for internal torque and the subscripts. Report your answer as a positive or negative τxy. B

y

T

A B

x

x A x

Solution

-----------------------------------------------------------By Inspection: The reaction torque at the wall will be counter-clockwise with respect to the x-axis. Thus, the segment of the shaft near the wall would rotate counter-clockwise, that is point A would move downwards and point B upwards. To oppose this imaginary motion, the shear stress at point A will be upward and at point B it will be downwards as shown in Figs (a) and (b). The shear stress on the rest of the surfaces

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

can be drawn using the observation that a symmetric pair of shear stress points towards the corner or away from the corner. TI (d) (b) y (c) (a) y y T

θ

A

x

B

x

x

By Subscripts: We can make an imaginary cut through the cross-section containing points A and B and draw the free body diagram as shown in Fig. (c). the internal torque is drawn as per our sign convention. By moment equilibrium we obtain T I = – T . Thus from the torsional stress formula we obtain τ xθ < 0 . Noting that the outward normal of the cross-section is in the positive x-direction, a negative τxθ requires the stress to be in the negative θ-direction as shown in Fig. (d). In the x-y coordinate system we obtain: ( τ xy ) A > 0

( τ xy ) B < 0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.24 Determine the direction of shear stress at points A and B (a) by inspection, and (b) by using the sign convention for internal torque and the subscripts. Report your answer as a positive or negative τxy

y

A

x T

A B x x

B

Solution

-----------------------------------------------------------By Inspection: The reaction torque at the wall will be counter-clockwise with respect to the x-axis. Thus, the segment of the shaft near the wall would move counter-clockwise, that is point A would move upwards and point B would downwards. To oppose this imaginary motion, the shear stress at point A will be downwards and at point B it will be upwards as shown in Figs (a) and (b). The shear stress on the rest of the surfaces can be drawn using the observation that a symmetric pair of shear stress points towards the corner or away from the corner. TI (d) (b) (c) y y (a) T

y

θ

x

A

x

B x

By Subscripts: We can make an imaginary cut through the cross-section containing points A and B and draw the free body diagram as shown in Fig. (c). the internal torque is drawn as per our sign convention. By moment equilibrium we obtain T I = T . Thus from the torsional stress formula we obtain τ xθ > 0 .

Noting that the outward normal of the cross-section is in the positive x-direction, a positive τxθ requires the stress to be in the negative θ-direction as shown in Fig. (d). In the x-y coordinate system we obtain: ( τ xy ) A < 0

( τ xy ) B > 0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.25 Determine the direction of shear stress at points A and B (a) by inspection, and (b) by using the sign convention for internal torque and the subscripts. Report your answer as a positive or negative τxy

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

x A

T

y x A B

x B

Solution

-----------------------------------------------------------By Inspection: Due to the applied torque, the segment of the shaft containing the points A and B would rotate clockwise, that is point A would move downwards and point B upwards. To oppose this imaginary motion, the shear stress at point A will be upward and at point B it will be downwards as shown in Figs (a) and (b). The shear stress on the rest of the surfaces can be drawn using the observation that a symmetric pair of shear stress points towards the corner or away from the corner. y (d) y (b) (c) (a) y T

x A

T

θ

TI

x

x

B

By Subscripts: We can make an imaginary cut through the cross-section containing points A and B and draw the free body diagram as shown in Fig. (c). the internal torque is drawn as per our sign convention. By moment equilibrium we obtain T I = – T . Thus from the torsional stress formula we obtain τ xθ < 0 .

Noting that the outward normal of the cross-section is in the negative x-direction, a negative τxθ requires the stress to be in the positive θ-direction as shown in Fig. (d). In the x-y coordinate system we obtain: ( τ xy ) A < 0

( τ xy ) B > 0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.26 Determine the direction of shear stress at points A and B (a) by inspection, and (b) by using the sign convention for internal torque and the subscripts. Report your answer as a positive or negative τxy x A

T

y x B A

x B

Solution

-----------------------------------------------------------By Inspection: Due to the applied torque, the segment of the shaft containing the points A and B would rotate in the direction of the torque, that is point A would move upwards and point B would move downwards. To oppose this imaginary motion, the shear stress at point A will be downward and at point B it will be upwards as shown in Figs (a) and (b). The shear stress on the rest of the surfaces can be drawn using the observation that a symmetric pair of shear stress points towards the corner or away from the corner. y (d) y y (b) (c) (a) T

x

x A

θ

TI x

A

B

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B

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

By Subscripts: We can make an imaginary cut through the cross-section containing points A and B and draw the free body diagram as shown in Fig. (c). The internal torque is drawn as per our sign convention. By moment equilibrium we obtain T I = – T . Thus from the torsional stress formula we obtain τ xθ < 0 . Noting that the outward normal of the cross-section is in the negative x-direction, a negative τxθ requires the stress to be in the positive θ-direction as shown in Fig. (d). In the x-y coordinate system we obtain: ( τ xy ) A > 0

( τ xy ) B < 0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.27 The two shaft shown have the same cross-sectional area A, show that the ratio of polar moment of inertia of the hollow shaft to the solid shaft is as given below:

2 J hollow + 1------------------------------ = α 2 J Solid α –1

RH

RS

αRH

Figure P5.27

Solution

-----------------------------------------------------------We can calculate the radius RH and RS in terms of the cross-sectional area A as shown below 2

A H = π [ ( αR H ) 2 – R H ] = A 2

A S = πR S = A

2 A R H = ----------------------2 π(α – 1)

or or

(1)

2 A R S = ---π

(2)

The polar area moment of inertia for a hollow shaft with inside radius Ri and outside radius Ro is

π J = --- ( R o4 – R i4 ) . For the hollow shaft Ro = αRH and Ri = RH, while for solid shaft Ro=RS and Ri=0. Sub2

stituting these values and using Eqs (1) and (2), we obtain the two polar area moments shown below: 2 ⎛ α 2 + 1⎞ A 2 4 4 4 π 4 A π π 4 J H = --- [ ( αR H ) – ( R H ) ] = --- ( α – 1 ) ( R H ) = --- ( α – 1 ) ⎛ -----------------------⎞ = ⎜ ---------------⎟ ⎛ ------⎞ 2 ⎝ ⎠ 2 2 2 ⎝ α 2 – 1 ⎠ ⎝ 2π⎠ π(α – 1)

π π A 2 A2 J S = --- R S4 = --- ⎛ ----⎞ = ------⎝ ⎠ 2 2 π 2π

(3)

(4)

Dividing Eq. (3) by Eq. (4) we obtain: ⎛ α 2 + 1⎞ ⎛ A 2⎞ ⎜ --------------⎟ -----⎝ α 2 – 1⎠ ⎝ 2π⎠ or JH ------ = ---------------------------------JS A 2 ⁄ ( 2π )

JH ⎛ α 2 + 1⎞ ------ = ⎜ ---------------⎟ JS ⎝ α 2 – 1⎠

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.28

Show that for a thin tube of thickness t and center-line radius R, the polar moment of inertia can be

approximated by J = 2πR 3 t . By thin tube we imply t < ( R ⁄ 10 ) Solution

-----------------------------------------------------------The polar moment of inertia of a hollow shaft in terms of outer radius Ro and inner radius Ri can be written as shown below:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

4 π 4 J = --- ( R o – R i ) = 2 Substituting R o = R + t ⁄ 2 and R i 2

January 2014

2 2 2 2 π 2 π 2 --- ( R o + R i ) ( R o – R i ) = --- ( R o + R i ) ( R o + R i ) ( R o – R i ) 2 2 = R – t ⁄ 2 in the equation above we obtain: 2

(1)

2

2 2 t π π 2 t t J = --- ⎛ R + Rt + ---- + R – Rt + ----⎞ ( R + t ⁄ 2 + R – t ⁄ 2 ) ( R + t ⁄ 2 – R + t ⁄ 2 ) = --- ⎛ 2R + ----⎞ ( 2R ) ( t ) ⎠ ⎝ 2 2⎝ 4 4 2⎠

(2)

Noting that the t2 term is two orders of magnitude smaller than the R2 term and hence can be neglected to obtain the following approximated expression: 3 2 π J ≈ --- ( 2R ) ( 2R ) ( t ) or J = 2πR t 2

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.29 (a) Draw the torque diagram in Figure P5.29. (b) Check the values of internal torque by making imaginary cuts and drawing free-body diagrams. (c) Determine the rotation of the rigid wheel D with respect to the rigid wheel A if the torsional rigidity of the shaft is 90,000 kips · in.2 10 in-kips 50 in-kips A 60 in-kips

B

20 in-kips

20 in

C

x

36 in

D

Figure P5.29

30 in

φD - φA = ? ------------------------------------------------------------

GJ = 90,000 kips-in2

Solution

The torque diagram can be drawn using the template and template equation shown. (b) Torque Diagram

T1

(a) Template

Text

T 10 in-kips A

T2

T 2 = T 1 + T ext

20

10

20 x

B

D

C

40

40

By making imaginary cuts in segments AB, BC, and CD, the following free body diagrams can be drawn. The internal torque is drawn as per our sign convention. TCD 10 in-kips (c) 10 in-kips (d) (e) 50 in-kips

TAB

20 in-kips

A

A

B

TBC D

By equilibrium of moment about the shaft axis in Figs. (c), (d), and (e), we obtain: T AB = 10in – kips T BC – 10 + 50 = 0

or

(1)

T BC = – 40in – kips

(2)

T CD = 20in – kips

(3)

The value of internal torques in Eqs. (1), (2), and (3) are the same as on the torque diagram in Fig. (b). The relative rotation of the ends of each segment can be found as shown below: T AB ( x B – x A ) 10 ) ( 20 )- = 2.222 ( 10 – 3 ) rads φ B – φ A = ---------------------------------- = (--------------------GJ 90, 000

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5-17

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

T BC ( x C – x B ) – 40 ) ( 36 -) = – 16.0 ( 10 – 3 ) rads φ C – φ B = --------------------------------- = (-----------------------GJ 90, 000

(2)

T CD ( x C – x B ) –3 ( 20 ) ( 30 ) φ D – φ C = ---------------------------------- = ---------------------- = 6.667 ( 10 ) rads GJ 90, 000

(3)

Adding Eqs. (1), (2), and (3) we obtain: –3

–3

φ D – φ A = ( 2.222 – 16.0 + 6.667 ) ( 10 ) = – 7.111 ( 10 ) rads or φ D – φ A = 0.00711 rads CW

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.30 (a) Draw the torque diagram in Figure P5.30. (b) Check the values of internal torque by making imaginary cuts and drawing free-body diagrams. (c) Determine the rotation of the rigid wheel D with respect to the rigid wheel A if the torsional rigidity of the shaft is 1,270 kN-m2. 20 kN-m 18 kN-m A 12 kN-m 10 kN-m

B 0.4 m

C

FigureP5.30

1.0 m

D

x

0.5 m

φD - φA = ? ------------------------------------------------------------

GJ = 1,270 kN-m2

Solution

The torque diagram can be drawn using the template and template equation shown. (a) Template

T1

(b) Torque Diagram

Text

T2

20

20

T kN-m

T 2 = T 1 + T ext

A

2

2

B

C

x D 10

10

By making imaginary cuts in segments AB, BC, and CD, the following free body diagrams can be drawn. The internal torque is drawn as per our sign convention. 20 kN-m

(c)

20 kN-m

(d)

18 kN-m

TAB A

TCD

(e) A

10 kN-m

TBC B

D

By equilibrium of moment about the shaft axis in Figs. (c), (d), and (e), we obtain: T AB = 20 kN – m T BC – 20 + 18 = 0

or

(1)

T BC = 2 kN – m

(2)

T CD = – 10 kN – m

(3)

The value of internal torques in Eqs. (1), (2), and (3) are the same as on the torque diagram in Fig. (b). The relative rotation of the ends of each segment can be found as shown below: T AB ( x B – x A ) –3 ( 20 ) ( 0.4 ) φ B – φ A = ---------------------------------- = ----------------------- = 6.299 ( 10 ) rads GJ 1, 270

(1)

T BC ( x C – x B ) 2 ) ( 1 )- = 1.575 ( 10 – 3 ) rads φ C – φ B = --------------------------------- = (--------------GJ 1, 270

(2)

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

T CD ( x C – x B ) – 10 ) ( 0.5 )- = – 3.937 ( 10 – 3 ) rads φ D – φ C = ---------------------------------- = (------------------------GJ 1, 270

(3)

Adding Eqs. (1), (2), and (3) we obtain: –3

–3

φ D – φ A = ( 6.299 + 1.575 – 3.937 ) ( 10 ) = 3.937 ( 10 ) rads or

φ D – φ A = 0.00394 rads CCW

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.31 The shaft in Figure P5.31 is made of steel (G = 80 GPa) and has a diameter of 150 mm. Determine: (a) the rotation of the rigid wheel D; (b) the magnitude of the torsional shear stress at point E and show it on a stress cube (Point E is on the top surface of the shaft.); (c) the magnitude of maximum torsional shear strain in the shaft. 150 kN-m A B

90 kN-m

E

TA

70 kN-m 0.25 m

C 0.5 m

Figure P5.31

0.3 m

x Solution

D

φD = ? (τxθ)E = ? ------------------------------------------------------------

G = 80 GPa

γmax= ?

d = 150 mm

The wall reaction torque TA can be found from the free body diagram of the entire shaft in Fig. 5.22. By equilibrium of moments about the shaft axis, we obtain: T A – 150 + 90 – 70 = 0

or

T A = 130 kN – m

(1)

The torque diagram can be drawn using the template and template equation shown. T1

(a) Template

Text

(b) Torque Diagram T 130 kN-m

T2

T 2 = T 1 + T ext

130 70 B

A

70

C

20

D

x

20

The internal torques are: T AB = 130 kN – m

T BC = – 20 kN – m

T CD = 70 kN – m

(2)

The polar moment of the shaft cross-section can be found as: 4 –6 4 π J = ------ ( 0.15 ) = 49.701 ( 10 ) m 32

(3)

The torsional rigidity can be found as: 9

–6

6

2

3

GJ = ( 80 ) ( 10 ) ( 49.701 ) ( 10 ) = 3.976 ( 10 )N – m = 3.976 ( 10 )kN – m

2

(4)

The relative rotation of the ends of each segment can be found as shown below: T AB ( x B – x A ) –3 ( 130 ) ( 0.25 ) φ B – φ A = ---------------------------------- = ----------------------------- = 8.174 ( 10 ) rads 3 GJ 3.976 ( 10 )

(5)

T BC ( x C – x B ) –3 ( – 20 ) ( 0.5 ) = – φ C – φ B = --------------------------------- = -------------------------2.515 ( 10 ) rads 3 GJ 3.976 ( 10 )

(6)

T CD ( x C – x B ) ( 70 ) ( 0.3 )- = 5.282 ( 10 –3 ) rads φ D – φ C = ---------------------------------- = -------------------------3 GJ 3.976 ( 10 )

(7)

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

Adding Eqs. (5), (6), and (7) we obtain: –3

–3

φ D – φ A = ( 8.174 – 2.515 + 5.282 ) ( 10 ) = 10.94 ( 10 ) rads

Noting that point A is fixed to the wall, hence φA = 0, we obtain:

φ D = 0.0109 rads CCW

The shear stress at point E can be found as shown below. 3 T BC ρ E 6 2 ( – 20 ) ( 10 ) ( 0.075 ) ( τ xθ ) E = ---------------- = ---------------------------------------------- = – 30.18 ( 10 ) N ⁄ m or –6 J 49.701 ( 10 )

( τ xθ ) E = – 30.2 MPa

The outward normal is in the positive x-direction, thus the shear stress must be in the negative θ-direction at point E as shown below. 150 kN-m A 30.2 MPa

Eθ

B

x

x

The maximum shear stress will be in segment AB as the internal torque is maximum and the diameter is the same for the entire shaft. The maximum shear stress can be found as shown below: 3 T AB ρ max 6 2 130 ( 10 ) ( 0.075 ) τ max = ----------------------= ----------------------------------------- = 196.2 ( 10 ) N ⁄ m –6 J 49.701 ( 10 )

(8)

The maximum shear strain can be found by using Hooke’s law as shown below: 6 τ max –3 196.2 ( 10 ) γ max = ----------= --------------------------- = 2.452 ( 10 ) 9 G 80 ( 10 )

(9) γ max = 2452μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.32 The shaft in Figure P5.31 is made of aluminum (G = 4000 ksi) and has a diameter of 4 in. Determine: (a) the rotation of the rigid wheel D; (b) the magnitude of the torsional shear stress at point E and show it on a stress cube (Point E is on the bottom surface of the shaft.); (c) the magnitude of maximum torsional shear strain in the shaft. TA 80 in-kips A 40 in-kips

B

15 in-kips C

E 32 in

25 in

x

D

Solution

20 in

Figure P5.32

φD = ? (τxθ)E = ? ------------------------------------------------------------

G = 4000 ksi

d = 4 in

γmax= ?

The wall reaction torque TA can be found from the free body diagram of the entire shaft in Fig. 5.23. By equilibrium of moments about the shaft axis, we obtain: T A – 80 + 40 + 15 = 0

or

T A = 25 in – kips

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5-20

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

The torque diagram can be drawn using the template and template equation shown. (a) Template

T1 Text

25

(b) Torque Diagram

25

A T in-kips

B

D

C

x 15

T2 T 2 = T 1 + T ext

55

15

55

The internal torques are: T AB = 25 in – kips

T BC = – 55 in – kips

T CD = – 15 in – kips

(2)

The polar moment of the shaft cross-section can be found as: 4 π 4 J = ------ ( 4 ) = 25.133 in 32

(3)

The torsional rigidity can be found as: 3

GJ = ( 4000 ) ( 25.133 ) = 100.53 ( 10 )kips – in

2

(4)

The relative rotation of the ends of each segment can be found as shown below: T AB ( x B – x A ) –3 ( 25 ) ( 20 ) φ B – φ A = ---------------------------------- = ----------------------------------- = 4.974 ( 10 ) rads 3 GJ ( 100.53 ) ( 10 )

(5)

T BC ( x C – x B ) –3 ( – 55 ) ( 32 ) φ C – φ B = --------------------------------- = ----------------------------------- = – 17.507 ( 10 ) rads 3 GJ ( 100.53 ) ( 10 )

(6)

T CD ( x C – x B ) –3 ( – 15 ) ( 25 ) = φ D – φ C = ---------------------------------- = ---------------------------------– 3.730 ( 10 ) rads 3 GJ ( 100.53 ) ( 10 )

(7)

Adding Eqs. (5), (6), and (7) we obtain: –3

–3

φ D – φ A = ( 4.974 – 17.507 – 3.730 ) ( 10 ) = – 16.263 ( 10 ) rads

Noting that point A is fixed to the wall, hence φA = 0, we obtain:

(8)

φ D = 0.0163 rads CW;

The shear stress at point E can be found as shown below. T BC ρ E ( – 55 ) ( 2 ) - = ---------------------- = – 4.377 ksi ( τ xθ ) E = ---------------J 25.133

(9) ( τ x θ ) E = – 4.4 ksi

The outward normal is in the positive x-direction, thus the shear stress must be in the negative θ-direction at point E as shown below. 80 in-kips A θ x

B

4.4 ksi x E

The maximum shear stress will be in segment BC as the internal torque is maximum and the diameter is the same for the entire shaft. The maximum shear stress value is the same as the shear stress at point E. The maximum shear strain can be found by using Hooke’s law as shown below: τ max –3 – 4.377 γ max = ----------= ---------------- = – 1.094 ( 10 ) G 4, 000

(10) γ max = – 1094 μ

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

5.33 Two circular steel shafts (G =12,000 ksi) of diameter 2 in. are securely connected to an aluminum shaft (G =4,000 ksi) of diameter 1.5 in. as shown in Figure P5.33. Determine (a) the rotation of section at D with respect to the wall, and (b) the maximum shear stress in the shaft. 12 in.-kips

A

B

25 in.-kips

C aluminum

steel

D steel

Figure P5.33 15 in.

40 in.

15 in.-kips

25 in.

Solution

-----------------------------------------------------------The polar moment of inertias are: 4 4 π J AB = J CD = ------ ( 2 ) = 1.5708 in 32

4 4 π J BC = ------ ( 1.5 ) = 0.497 in 32

(1)

From the free body diagram below we can find the reation torque at the wall and draw the torque diagram. T A = 12 + 25 – 15 = 22 in-kips TA

22

12 in.-kips 25 in.-kips 15 in.-kips

A

B steel

C aluminum

40 in.

15 in.

(2) 10

T in-kips A

D steel

B

x C 15

D

25 in.

From the torque diagram the internal torques are: T AB = 22 in-kips

T BC = 10 in-kips

T CD = – 15 in-kips

(3)

T AB ( x B – x A ) –3 ( 22 ) ( 40 ) φ B – φ A = ---------------------------------- = ----------------------------------------- = 46.68 ( 10 ) rads ( 12000 ) ( 1.5708 ) G AB J AB

(4)

T BC ( x C – x B ) ( 10 ) ( 15 ) - = 75.45 ( 10 –3 ) rads φ C – φ B = --------------------------------- = ---------------------------------( 4000 ) ( 0.497 ) G BC J BC

(5)

T CD ( x C – x B ) –3 ( – 15 ) ( 25 ) φ D – φ C = ---------------------------------- = ----------------------------------------- = – 19.89 ( 10 ) rads ( 12000 ) ( 1.5708 ) G CD J CD

(6)

The relative rotations of each section can be found as:

Adding we obtain: –3

–3

φ D – φ A = ( 46.68 + 75.45 – 19.89 ) ( 10 ) = 102.2 ( 10 ) rads

(7)

The maximum torsional shear stress in each segment can be found as:

The answers are:

T AB ( ρ AB ) max 22 ( 1 ) τ AB = ----------------------------------= ---------------- = 14.01 ksi 1.5708 J AB

(8)

T BC ( ρ BC ) max 10 ) ( 0.75 )- = 15.09 ksi τ BC = ---------------------------------- = (------------------------1.5708 J BC

(9)

T AB ( ρ AB ) max ( – 15 ) ( 1 ) τ CD = ----------------------------------= ---------------------- = – 9.55 ksi J AB 1.5708

(10)

φ D – φ A = 0.102 rads CCW ; τ max = 15.09 ksi

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

5.34 A solid circular steel (Gs = 12,000 ksi) shaft BC is securely attached to two hollow steel shafts AB and CD as shown. Determine: (a) the angle of rotation of section at D with respect to section at A. (b) the magnitude of maximum torsional shear stress in the shaft (c) the torsional shear stress at point E and show it on a stress cube. Point E is on the inside bottom surface of CD. 420 in-kips

120in-kips B

A

100 in-kips

200 in-kips C

2 in 4 in

D E

24 in

36 in

24 in

Figure P5.34

Solution

φD - φA= ? τmax= ? (τxθ)E = ? ------------------------------------------------------------

G = 12,000 ksi

The polar moment of the cross-sections can be found as shown below. 4 4 π 4 J AB = J AB = ------ ( 4 – 2 ) = 23.562 in 32

4 π 4 J BC = ------ ( 4 ) = 25.133 in 32

(1)

The torque diagram can be drawn using the template and template equation shown. (a) Template T1

Text

(b) Torque Diagram T2

T in-kips

120

120

A

B

C 100

T 2 = T 1 + T ext 300

D 100

x

300

From Fig. (b) we have: T AB = 120 in – kips T BC = – 300 in – kips T CD = – 100 in – kips The relative rotation of the ends of each segment can be found as shown below: T AB ( x B – x A ) –3 ( 120 ) ( 24 ) φ B – φ A = ---------------------------------- = -------------------------------------------- = 10.186 ( 10 ) rads ( 12, 000 ) ( 23.562 ) G AB J AB

(2)

T BC ( x C – x B ) ( – 300 ) ( 36 ) - = – 35.810 ( 10 – 3 ) rads φ C – φ B = --------------------------------- = ------------------------------------------( 12, 000 ) ( 25.133 ) G BC J BC

(3)

T CD ( x D – x C ) –3 ( – 100 ) ( 24 ) φ D – φ C = ---------------------------------- = -------------------------------------------- = – 8.488 ( 10 ) rads ( 12, 000 ) ( 23.562 ) G CD J CD

(4)

Adding Eqs. (2), (3), and (4) we obtain: –3

–3

φ D – φ A = ( 10.186 – 35.810 – 8.488 ) ( 10 ) = – 34.11 ( 10 ) rads

φ D – φ A = 0.0341 rads CW

The maximum torsional shear stress will be in BC and can be found as shown below. T BC ρ max ( 300 ) ( 2 ) τ max = ---------------------- = ---------------------- = 23.87 ksi 25.133 J BC

τ max = 23.87 ksi

The torsional shear stress at point E can be found as shown below. T CD ρ E – 100 ) ( 1 )- = – 4.244 ksi or ( τ xθ ) E = ---------------- = (-----------------------23.562 J CD

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( τ xθ ) E = – 4.24 ksi

5-23

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

Figure below shows the shear stress on a stress cube. 200 in-kip s

θ

C

4.24 ksi x E

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.35 A steel shaft (G=80GPa) is subjected to the torques shown. Determine (a) the rotation of section A with respect to no load position. (b) the torsional shear stress at point E and show it on a stress cube. Point E is on the surface of the shaft. TD

x 160 kN.m C

80 kN.m 300 mm

500 mm

D

120 kN.m

A

B

E

3.0 m 2.5 m

2.0 m

Figure P5.35

φA -= ? (τxθ)E = ? ------------------------------------------------------------

Solution

G=80GPa

The polar moment of the cross-sections can be found as shown below. 4 –3 4 π J AB = J BC = ------ ( 0.3 ) = 0.7952 ( 10 ) mm 32

4 –3 4 π J CD = ------ ( 0.5 ) = 6.1359 ( 10 ) mm 32

(1)

The wall reaction torque TD can be found from the free body diagram of the entire shaft in Fig. 5.23. By equilibrium of moments about the shaft axis, we obtain: T D – 80 + 120 – 160 = 0

or

T D = 120 kN – m

(2)

The torque diagram can be drawn using the template and template equation shown. (a) Template

Text T1

T2

(b) Torque Diagram

120

120 80

T kN-m

80 x

A

T 2 = T 1 + T ext

From Fig. (b) we have: T AB = 80 kN – m T BC = – 40 kN – m

40

D

C 40

B

T CD = 120 kN – m

The relative rotation of the ends of each segment can be found as shown below: 3 T AB ( x B – x A ) –3 ( 80 ) ( 10 ) ( 2 ) φ B – φ A = ---------------------------------- = -------------------------------------------------------------- = 2.515 ( 10 ) rads 9 – 3 G AB J AB ( 80 ) ( 10 ) ( 0.7952 ) ( 10 )

(3)

3 T BC ( x C – x B ) –3 ( – 40 ) ( 10 ) ( 2.5 ) φ C – φ B = --------------------------------- = -------------------------------------------------------------- = – 1.572 ( 10 ) rads 9 – 3 G BC J BC ( 80 ) ( 10 ) ( 0.7952 ) ( 10 )

(4)

3 T CD ( x D – x C ) –3 ( 120 ) ( 10 ) ( 3.0 ) φ D – φ C = ---------------------------------- = -------------------------------------------------------------- = 0.7334 ( 10 ) rads 9 – 3 G CD J CD ( 80 ) ( 10 ) ( 6.1359 ) ( 10 )

(5)

–3

–3

Adding Eqs. (3), (4), and (5) we obtain: φ D – φ A = ( 2.515 – 1.572 + 0.7334 ) ( 10 ) = 1.676 ( 10 ) rads –3

Note that point D is fixed to the wall, hence φD is zero. Thus, φ A = – 1.676 ( 10 ) rads

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

φ A = 1676 μ rads CW

or The torsional shear stress at point E can be found as shown below. 3 T AB ρ E 6 2 ( 80 ) ( 10 ) ( 0.150 ) ( τ xθ ) E = ---------------- = ------------------------------------------- = 15.09 ( 10 ) N ⁄ m or – 3 J AB ( 0.7952 ) ( 10 )

( τ xθ ) E = 15.1 MPa

Figure below shows the shear stress at point E on a stress cube. 120 kN.m x B

θ E

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.36 The radius of the tapered circular shaft varies from 200 mm at A to 50 mm at B. The shear modulus of the material is G = 40 GPa. Determine: (a) the angle of rotation of wheel C with respect to the fixed end. (b) the magnitude of maximum torsional shear strain in the shaft. 10 kN-m

2.5 kN-m 50 mm C

B

A

2m

7.5 m

Figure P5.36

Solution

φC = ? γmax = ? ------------------------------------------------------------

RA= 200 mm

RB= 50 mm

G = 40 GPa

We note that R in segment AB is a linear function of x i.e., R ( x ) = a + bx . The constants can be found as shown below. R ( x = 0 ) = R A = a = 0.2

R ( x = 7.5 ) = R B = 0.2 + b ( 7.5 ) = 0.05

and

or

R ( x ) = 0.2 – 0.02x

b = – 0.02 (1)

The polar moment of the cross-sections can be calculated as shown below. 4 π J AB = --- ( 0.2 – 0.02x ) 4 m 2

–6 4 π J BC = --- ( 0.05 ) 4 = 9.817 ( 10 ) m 2

(2)

We can make imaginary cuts in segment AB and BC and obtain the free body diagrams shown in Fig.(a) and Fig. (b). The internal torques are drawn as per our sign convention. From moment equilibrium in Figs. (a) and (b) we obtain: T AB = – 7.5kN – m

(a)

TAB

10 kN-m

B

2.5kN-m

T BC = 2.5kN – m

(b)

TBC

C

(3) 2.5 kN-m

C

The relative rotation of the ends of segment AB can be found as shown below 3 T AB – ( 7.5 ) ( 10 ) ⎛ dφ⎞ = -------------------- = --------------------------------------------------------------------⎝ d x⎠ AB G AB J AB 9 π ( 40 ) ( 10 ) --- ( 0.2 – 0.02x ) 4 2

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

Integrating from point A to point B, we obtain: φB

∫ φA

–6

15 ( 10 ) d φ = – --------------------40π

7.5

∫

–6

0.1194 ( 10 ) 1 dx ---------------------------------- = – -------------------------------- ---------------------------------4 ( – 3 ) ( – 0.02 ) ( 0.2 – 0.02x )3 ( 0.2 – 0.02x )

0

7.5

or 0

–6 –3 1 1 φ B – φ A = – ( 1.989 ) ( 10 ) ----------------- – ----------------- = – 15.668 ( 10 ) rads 3 3 ( 0.05 ) ( 0.20 )

(5)

The relative rotation 3 T BC ( x C – x B ) –3 ( 2.5 ) ( 10 ) ( 2 ) φ C – φ B = --------------------------------- = ----------------------------------------------------------- = 12.732 ( 10 )rads 9 – 6 G BC J BC ( 40 ) ( 10 ) ( 9.817 ) ( 10 )

(6)

Adding Eqs (5) and (6), we obtain the rotation of section at C with respect to section at A. Noting that the section at A is built into the wall we obtain the rotation of section at C. –3

–3

φ C – φ A = ( – 15.668 + 12.732 ) ( 10 ) = – 2.934 ( 10 ) rads or

φ C = 0.0029 rads CW

The maximum shear stress will be on a cross-section just left of B. Just left of section at B we have: J AB = 9.817 ( 10–6 ) m4 and ρ max = 0.05m . The magnitude of maximum torsional shear stress in the shaft can be found as shown below. 3 T AB ρ max 6 2 ( 7.5 ) ( 10 ) ( 0.05 ) τ max = ---------------------- = ----------------------------------------- = 38.199 ( 10 ) N ⁄ m –6 J AB 9.817 ( 10 )

(7)

From Hooke’s law the maximum shear strain can be found, as shown below. 6 τ max –3 38.199 ( 10 ) γ max = ----------= ------------------------------ = 0.955 ( 10 ) 9 G ( 40 ) ( 10 )

γ max = 955 μ

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------– ax

5.37 The radius of the tapered shaft in Figure P5.37 varies as R = Ke . Determine the rotation of the section at B in terms of the applied torque Text, length L, shear modulus of elasticity G, and geometric parameters K and a.

Text ⴢm

x Figure P5.37

Solution

R = Ke

φB = f(Text, L, G, r) = ? τmax = g(Text, L, G, r) = ? ------------------------------------------------------------

– ax

The polar moment of the cross-section can be calculated as shown below. π 4 – 4ax J AB = --- K e 2

(1)

We can make imaginary cut in segment AB and obtain the following free body diagram. The internal

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5-26

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

torque is drawn as per our sign convention. TAB

Text

B

From moment equilibrium we obtain: T AB = T ext The relative rotation of the ends of segment AB can be found as shown below 4ax

T ext 2T ext e T AB ⎛ dφ⎞ = -------------------- = --------------------------------- = ----------------------⎝ d x⎠ AB 4 G AB J AB π 4 –4ax πGK G --- K e 2

(2)

Integrating from point A to point B, we obtain: φB

∫ φA

L

L

2T ext 4ax ⎛ 2T ext ⎞ e 4ax dφ = --------------- e dx = ⎜ ---------------⎟ ---------4 ⎝ πGK 4⎠ 4a 0 πGK

∫

or

0

⎛ T ext ⎞ 4aL φ B – φ A = ⎜ ---------------------⎟ [ e – 1 ] or ⎝ 2πaGK 4⎠

⎛ T ext ⎞ 4aL φ B = ⎜ ---------------------⎟ [ e – 1 ] CCW ⎝ 2πaGK 4⎠

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.38 The radius of the tapered shaft shown in Figure P5.37 varies as R = r ( 2 – 0.25x ⁄ L ) . In terms of Text, L, G, and r, determine: (a) the rotation of the section at B; (b) the magnitude of maximum torsional shear stress in the shaft. Solution

φB = f(Text, L, G, r) = ? τmax = g(Text, L, G, r) = ? ------------------------------------------------------------

R = r ( 2 – 0.25x ⁄ L )

The polar moment of the cross-section can be calculated as shown below. 4

4

πr πr J AB = -------- ( 2 – 0.25 x ⁄ L ) 2 = --------- ( 2L – 0.25 x ) 2 2 2 2L

(1)

We can make imaginary cut in segment AB and obtain the following free body diagram. The internal torque is drawn as per our sign convention. TAB

Text

B

From moment equilibrium we obtain: T AB = T ext The relative rotation of the ends of segment AB can be found as shown below 2

T ext 2T ext L T AB ⎛ dφ⎞ = -------------------- = -----------------------------------------------------= ---------------------------------------------⎝ d x⎠ AB 4 G AB J AB 4 2 πGr ( 2L – 0.25x ) πr G --------- ( 2L – 0.25x ) 2 2 2L

(2)

Integrating from point A to point B, we obtain: φB

2 L

∫

∫

2

L ⎛ 2T ext L ⎞ ⎛ 2T ext L ⎞ 1 1 dx dφ = ⎜ --------------------⎟ ---------------------------------- = ⎜ --------------------⎟ ------------------------------ ------------------------------⎝ πGr 4 ⎠ ( 2L – 0.25x ) 2 ⎝ πGr 4 ⎠ ( – 1 ) ( – 0.25 ) ( 2L – 0.25x ) 0 φA 0

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

2

⎛ 8T ext L ⎞ ⎛ T ext L⎞ 1 1 φ B – φ A = ⎜ --------------------⎟ ------------------- – ----------- = 0.1819 ⎜ --------------⎟ 4 ⎝ πGr ⎠ ( 1.75L ) ( 2L ) ⎝ Gr 4 ⎠

January 2014

⎛ T ext L⎞ φ B = 0.1819 ⎜ --------------⎟ CCW ⎝ Gr 4 ⎠

or

The maximum shear stress will be on a cross-section just left of B. Just left of section at B we have: R = ( r ) ( 2 – 0.25 ) = 1.3229r . The magnitude of maximum torsional shear stress in the shaft can be found as shown below. T ext ( 1.3229r ) T AB ρ max τ max = ---------------------- or - = -----------------------------------4 J AB π ( 1.3229r ) ⁄ 2

0.275T ext τ max = ----------------------3 r

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.39 The external torque on a drill bit varies as a quadratic function to a maximum intensity of q in-lb/in as shown. If the drill bit diameter is d, its length L, and modulus of rigidity G, determine (a) the maximum shear stress on the drill bit. (b) the relative rotation of the end of the drill bit with respect to the chuck. L x

⎛ x2 ⎞ A q ⎜ -----2-⎟ in – lb ⁄ in ⎝L ⎠

B

Figure P5.39

Solution

φB -φA= f(q, L, G, r) = ? τmax = g(q, L, G, r) = ? ------------------------------------------------------------

The distributed torque on the drill bit is counter-clockwise with respect to the x-axis, thus we can substitute ⎛ x2⎞ t ( x ) = q ⎜ -----2-⎟ ⎝L ⎠

in the equilibrium equation and integrate to obtain the following: ⎛ x2⎞ dT + q ⎜ ------⎟ = 0 dx ⎝ L 2⎠

or

⎛ x3 ⎞ T = – q ⎜ ---------⎟ + c ⎝ 3L 2⎠

(1)

Where, c is an integration constant. At point B i.e., at x = L the internal torque should be zero as there is no concentrated applied torque at B. Substituting this in Eq. (1) we can find the integration constant as shown below. ⎛ L3 ⎞ T ( x = L ) = – q ⎜ ---------⎟ + c = 0 ⎝ 2L 2⎠

or

c = qL ------3

(2)

Substituting Eq (2) into Eq (1) and simplifying, we obtain: 3 3 q T = --------- ( L – x ) 2 3L

(3)

------- . The maximum The maximum internal torque will exist at x = 0. From Eq. (3) we obtain: T max = qL 3

shear stress can be found as shown below. T AB ρ max ( qL ⁄ 3 ) ( d ⁄ 2 ) 16qL- or τ max = ---------------------- = ------------ = --------------------------------4 3 J [ πd ⁄ 32 ] 3πd

τ max = 16qL ------------3 3πd

The relative rotation can be found as shown below.

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

3 q 3 ------- ( L – x ) dφ 3L = -----------------------------dx G ( πd 4 ⁄ 32 )

(4)

Integrating from A to B we obtain: φB

32q ⎞ . d φ = ⎛ ----------------------∫ ⎝ 2 4⎠ φA

3πGL d

xB = L xA

∫= 0 ( L

L

4

3

3 32q x φ B – φ A = ⎛ ------------------------⎞ ⎛ L x – -----⎞ ⎝ 2 4⎠ ⎝ 4⎠ 3πGL d 0

or

3

– x ) dx

2

8qL φ B – φ A = ⎛ --------------⎞ CCW ⎝ πGd 4⎠

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.40 A circular solid shaft is acted upon by torques as shown. Determine the rotation of the rigid wheel A with respect to the fixed end C in terms of q, L, G and J. TA = 2qL in-lb

TB = qL in-lb q in-lb/in

A

C

B

x 0.5 L

0.5 L

Figure P5.40

φA = f(q, L, G, J) = ? ------------------------------------------------------------

Solution

We can make imaginary cuts in segments AB and BC to obtain the following free body diagrams. (a)

TA = 2qL in-lb

TA = 2qL in-lb

(b)

q in-lb/in TAB

TB = qL in-lb

q in-lb/in TBC

A

A

x

B

x

From moment equilibrium in Fig (a) we obtain: T AB – 2qL + qx = 0

or

T AB = 2qL – qx

(1)

From moment equilibrium in Fig (b) we obtain: T BC – 2qL + qx + qL = 0

or

T BC = qL – qx

(2)

The relative rotation for the ends of segments AB can be found by integrating from point A to point B as shown below T AB 2qL – qx⎛ dφ⎞ = -------------------- = --------------------⎝ d x⎠ AB G J GJ AB AB

φB

or

∫ dφ φA

0.5L

=

∫

2qL – qx⎞ ⎛ --------------------- dx or ⎝ GJ ⎠

0 2

1 qx ⎞ φ B – φ A = ------- ⎛ 2qLx – -------GJ ⎝ 2 ⎠

0.5L 0

7 qL 2 = --- ⎛ ---------⎞ 8 ⎝ GJ ⎠

(3)

The relative rotation for the ends of segments BC can be found by integrating from point A to point B as shown below.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

T BC qL – qx ⎛ dφ⎞ = -------------------- = ------------------⎝ d x⎠ BC G J GJ BC BC

φC

or

∫ dφ φB

January 2014

L

∫

=

– qx ⎛ qL -------------------⎞ dx or ⎝ GJ ⎠

0.5L L

2

1 qx φ C – φ B = ------- ⎛ qLx – --------⎞ GJ ⎝ 2 ⎠

0.5L

1 qL 2 = --- ⎛ ---------⎞ 8 ⎝ GJ ⎠

(4)

Adding Eqs. (3) and (4), we obtain the rotation of section at C with respect to section at A. Noting that section at C is built into the wall we obtain the rotation of section at A as shown below. qL 2 φ C – φ A = ⎛ ---------⎞ or ⎝ GJ ⎠

qL 2 φ A = ⎛ ---------⎞ CW ⎝ GJ ⎠

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.41 A thin steel (G = 12,000 ksi) tube of 1/8 inch thickness has a mean diameter of 6 inches and a length of 36 inches. What is the maximum torque the tube can transmit if the allowable shear stress in is 10 ksi and the allowable relative rotation of the two ends is 0.015 rads? Solution: G = 12,000 ksi t = 1/8 inch d = 6 inch L = 36 inch τ max ≤ 10ksi Δφ ≤ 0.015rads Tmax =?

-----------------------------------------------------------The outer and inner diameters of the tube are: do = 6 + (1/8) = 6.125 in. and di = 6 - (1/8) = 5.875 in. 4 4 π The polar moment of inertia is: J = ------ [ ( 6.125 ) – ( 5.875 ) ] = 21.215 in4

32

The maximum stress is T ( d o ) ⁄ 2 T ( 6.125 ⁄ 2 ) τ max = --------------------- = ----------------------------- ≤ 10 J 21.215

or

T ≤ 69.27in – kips

(1)

The relative rotation of the two ends is T ( 36 ) TL Δφ = ------- = ----------------------------------------- ≤ 0.015 ( 12000 ) ( 21.211 ) GJ

or

T ≤ 106.1in – kips

The maximum torque that satisfies Eq. (1) and (2) is

(2)

T = 69.2 in-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.42 Determine the maximum torque that can be applied on a 2 inch diameter solid aluminum (G = 4,000 ksi) shaft if the allowable shear stress is 18 ksi and the relative rotation over 4 feet of the shaft is to be limited to 0.2 rads. Solution: d = 2 inch. G = 4,000 ksi L = 4 feet τ max ≤ 18ksi Δφ ≤ 0.2rads Tmax =?

------------------------------------------------------------

4 4 π The polar moment of inertia is: J = ------ ( 2 ) = 1.5708 in

32

The maximum shear stress is T( 1) ( d ⁄ 2 ) = --------------- ≤ 18 τ max = T -----------------1.5708 J

or

T ≤ 28.27in – kips

(1)

The relative rotation of the two ends is T ( 4 ) ( 12 ) TL Δφ = ------- = -------------------------------------- ≤ 0.2 ( 4000 ) ( 1.5708 ) GJ

or

The maximum torque that satisfies Eq(1) and (2) is

T ≤ 26.18in – kips

(2)

T = 26.1 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.43 A hollow steel (G = 80 GPa) shaft with an outside radius of 30 mm is to transmit a torque of 2700 N-m. The allowable shear stress is 120 MPa and the allowable relative rotation over 1 m is 0.1 rad.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

Determine the maximum permissible inner radius to the nearest millimeter. Solution: G = 80 GPa ro = 3.0 mm T = 2700 m L = 1 m. τ max ≤ 120MPa Δφ ≤ 0.1rads (ri)max =?nearest mm.

-----------------------------------------------------------The maximum shear stress is: –3 Tr ( 2700 ) ( 30 ) ( 10 ) 6 τ max = -------o- = -------------------------------------------- ≤ 120 ( 10 ) J J

–9

J ≥ 675 ( 10 )m

or

4

(1)

The relative rotation of the two ends is 2700 ( 1 ) TL Δφ = ------- = -------------------------------- ≤ 0.1 9 GJ ( 80 ) ( 10 ) ( J )

–9

J ≥ 337.5 ( 10 )m

or

4

(2)

If J is given by Eq(1) then it satisfies Eq(2). –9 π 4 4 Thus: --- ( r o – r i ) ≥ 675 ( 10 )

2

or

4 4 1350 –9 r i ≤ ( 0.03 ) – ------------ ( 10 ) π

r i ≤ 0.0248m

or

( r i ) max = 24mm

The maximum inner radius to the nearest millimeter is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.44 A 5 feet long hollow shaft is to transmit a torque of 200 in-kips. The outer diameter of the shaft must be 6 inches to fit existing attachments. The relative rotation of the two ends of the shaft is limited to 0.05 rads. The shaft can be made of steel or aluminum. The shear modulus of elasticity G, the allowable shear stress τallow, and the specific weight γ, are given in Table 5.1. Determine the maximum inner diameter to the nearest 1/8 inch of the lightest shaft that can be used for transmitting the torque and the corresponding weight. Table 5.1.Material Properties. Material

G ksi

τallow ksi

γ lbs./in3

Steel

12,000

18

0.285

Aluminum

4,000

10

0.100

Solution

L = 5 ft. T= 200 in-kips (di)max = ? nearest 1/8 inch

do = 6 inch W=?

Δφ ≤ 0.05 rads

-----------------------------------------------------------The relative rotation of the ends and maximum shear stress in steel can be found and the following limits on JS, the polar moment of steel can be obtained. ( 200 ) ( 5 ) ( 12 ) ( Δφ ) S = --------------------------------- ≤ 0.05 3 12 ( 10 )J S ( 200 ) ( 3 ) ( τ max ) S = ---------------------- ≤ 18 JS

or

J S ≥ 20 in J S ≥ 33.33 in

or

4

(1)

4

(2)

Using similar calculations for Aluminum shaft we obtain the following limits on JAl. ( 200 ) ( 5 ) ( 12 ) ( Δφ ) Al = --------------------------------- ≤ 0.05 ( 4000 )J Al ( 200 ) ( 3 ) ( τ max ) Al = ---------------------- ≤ 10 J Al

or or

J Al ≥ 60 in J Al ≥ 60 in

4

4

(3) (4)

Thus if J S ≥ 33.33 in 4 it will meet both conditions in Eqs (1) and (2). Similarly if J Al ≥ 60 in4 it will meet both conditions in Eqs. (3) and (4). The internal diameters dS and dAl can be found as shown below:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

4 4 π 4 J S = ------ ( 6 – d S ) ≥ 33.33 in 32

or

4 4 π 4 J Al = ------ ( 6 – d Al ) ≥ 60 in 32

or

January 2014

d S ≤ 5.561 in

(5)

d Al ≤ 5.116 in

(6)

Rounding downwards to the closest 1/8 inch we obtain: d S = 5.5 in or d Al = 5.0 in (7) We can find the weight of each material by taking the product of the material density and the volume of a hollow cylinder as shown below: 2 π 2 W S = ( 0.285 ) ⎛ ---⎞ ( 6 – 5.5 ) ( 60 ) = 77.224 lbs ⎝ 4⎠

(8)

2 π 2 W Al = ( 0.1 ) ⎛ ---⎞ ( 6 – 5.0 ) ( 60 ) = 51.8 lbs ⎝ 4⎠

(9)

From Eqs 8 and 9 we see that the aluminum shaft is lighter. Thus, the answer is: ( d i ) max = 5.0 in.

W = 51.8 lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.45 A 100-hp motor is driving a pulley and belt system as shown in Figure 5.45. If the system is to operate at 3,600 rpm, determine the minimum diameter of the solid shaft AB to the nearest 1/8 inch if the allowable torsional stress in the shaft is 10 ksi.

A

B

Figure P5.45

Solution: P = 100 hpw = 3600 rpm τ max ≤ 10ksi

dmin =?

nearest 1/8 inch.

-----------------------------------------------------------The rotational speed in rads/sec can be found as: ω = ( 3600 ) ( 2π ) ⁄ 60 = 376.99 ( rad ) ⁄ ( sec ) The torque on the shaft can be found as shown below 100 ) ( 6600 ) P- = (------------------------------ = 1751in – lbs T = --( 376.99 ) ω

(1)

The maximum torsion and shear stress in terms of T can be found and from it the maximum torque can be obtained as shown below. 3 16 1751 T( d ⁄ 2) τ max = ------------------ = ------ ------------ ≤ 10 ( 10 ) π d3 π 4 ------ d 32

or

d ≥ 0.9625

The minimum diameter to the nearest inch 1/8 inch

d min = 1.0 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.46 The bolts used in the coupling for transferring power in problem 5.45 have an allowable strength of 12 ksi. Determine the minimum number of 1/4 inch diameter bolts that must be placed at a radius of 5/ 8 inch.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

A

January 2014

B

Figure P5.46

Solution:

τ bolt ≤ 12ksi

dbolt = 1/4 inch

R = 5/8 inch

nmin =?

-----------------------------------------------------------From Eq(1) in problem 5-38, we have T = 1751 in - kips 4 π 2 The area of cross section of a bolt is: A = --- ( d bolt ) = 0.4909in

4

The following force body diagram can be drawn

T

V = τbolt R

Each bolt resists a shear force of V. By moment equilibrium about the shaft axis we obtain T = nVR = ( nR ) ( τ bolt ) ( A )

or

3 1751 T - = -----------------------------------------τ bolt = ----------≤ 12 ( 10 ) n ( 5 ⁄ 8 ) ( 0.04909 ) nRA

The minimum number of bolts needed are

or

n ≥ 4.76

n min = 5

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.47 A 20 kW motor drives 3 gears that are rotating at a frequency of 20 Hz. Gear A next to the motor transfers 8 kW of power. Gear B which is in the middle transfers 7 kW of power. Gear c which is at the far end from the motor transfers the remaining 5 kW of power. A single steel solid shaft connecting the motors to all three gears is to be used. The steel used has a yield strength in shear of 145 MPa. Assuming a factor of safety of 1.5, what is the minimum diameter of the shaft to the nearest millimeter that can be used if failure due to yielding is to be avoided? What is the magnitude of maximum torsional shear stress in the shaft segment between gears A and B. Solution: P = 20 kW f = 20 Hz. PA = 8kW PB = 7KN PC = 5 kW τyield = 145 MPa k = 1.5 dmin =? to the nearest millimeter τAB =?

-----------------------------------------------------------The torque delivered by the motor and the torques transferred by the gears can be found as shown below. 3

( 20 ) ( 10 ) PT = -------= ------------------------ = 159.2N – m 2πf ( 2π ) ( 20 ) 3 Pb ( 7 ) ( 10 ) T b = -------- = ---------------------- = 55.7N – m 2πf ( 2π ) ( 20 )

3 Pa ( 8 ) ( 10 ) T a = -------- = ---------------------- = 63.7N – m 2πf ( 2π ) ( 20 ) 3 Pc ( 5 ) ( 10 ) T c = -------- = ---------------------- = 39.8N – m 2πf ( 2π ) ( 20 )

The shaft with the torques are shown in Fig (a) and the associated torque diagram shown in Fig(b). From the torque diagram the internal torques are:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

T MA = 159.2N – m

T MB = 95.5N – m

T MB = 39.8N – m

(a) T = 159.2 N-m

TA = 63.7 N-m TB = 55.7 N-m

TC = 39.8 N-m

(b)

T 159.2

159.2

95.5 M

A

B

January 2014

C M

A

95.5 39.8 B

39.8 C

x

The maximum torsional shear stress in the shaft will be in segment MA 6 T MA ( d ⁄ 2 ) 16 ( 159.2 ) 145 ( 10 ) τ max = -------------------------- = ------ ------------------ ≤ ---------------------4 π d3 1.5 ( πd ) ⁄ 32

d ≥ 0.0203m

or

The minimum diameter is

d min = 21mm

The maximum torsional shear stress in segment AB is T AB ( d ⁄ 2 ) 16 T AB 16 ( 95.5 )- = 52.5 ( 10 6 ) or τ AB = ------------------------- = ------ ---------- = ----------------------3 4 3 π d ( πd ) ⁄ 32 π ( 0.021 )

τ AB = 52.5MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.48 A circular shaft has a constant torsional rigidity GJ and is acted upon by a distributed torque t(x). If at section A the internal torque is zero show that the relative rotation of section at B with respect to rotation of section at A is given by xB

1 φ B – φ A = ------GJ

∫ ( x – xB )t ( x ) dx

5.14

xA

Solution

φB - φA = ? ------------------------------------------------------------

T(xA) = 0

From equilibrium equation, we have dT = – t ( x ) dx

(1)

Integrating from x = xA to any location x, we obtain the following. T

x

∫

dT = –

TA = 0

x

∫ t ( x ) dx

T(x) = –

or

xA

∫ t ( x ) dx

(2)

xA

( x ) from x to x , we obtain the following. Integrating the equation dφ = T ----------A B dx

GJ

φB

xA

∫ dφ φA

=

∫

T (x) ----------dx GJ

x

or

1 φ B – φ A = ------GJ

xA

∫ T ( x ) dx

(3)

xA

Integrating by parts, the above equation can be written as x xB 1 φ B – φ A = ------- xT ( x ) x – GJ A

dT

∫ x d x dx

(4)

xA

Imposing the limits and substituting Eq. (1), we obtain the following. x

1 φ B – φ A = ------- x B T ( x B ) – T ( x A ) – GJ

∫ x ( –t ( x ) ) dx

(5)

xA

Substituting Eq. (2) at x = xB and noting that T(xA) = 0, we obtain

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

xB ⎧ xB ⎫ ⎪ ⎪ 1 φ B – φ A = ------- x B ⎨ – t ( x ) dx ⎬ – 0 + xt ( x ) dx GJ ⎪ ⎪ xA ⎩ xA ⎭

∫

∫

(6)

Noting that xB is a constant and can be taken inside the integral, we obtain: xB

1 φ B – φ A = ------GJ

∫ ( x – xB )t ( x ) dx

(7)

xA

Eq. (7) is same as Eq. 5.14.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.49 A composite shaft made from n materials is shown in Figure P5.49. Gi and Ji are the shear modulus of elasticity and polar moment of inertia of the ith material. (a) If Assumptions from 1 through 6 are valid, show that the stress ( τ xθ ) i in the ith material is given Equation 5.16a, where T is the total internal torque at a cross section. (b) If Assumptions 8 through 10 are valid, show that relative rotation φ 2 – φ 1 is given by Equation 5.16b.(c) Show that for G1=G2=G3....=Gn=G. Equations (5.16a) and (5.16b) give the same results as Equations (5.10) and (5.12). Gn Gi ρ T ( τ x θ ) i = -----------------------n Gj Jj

(5.16a)

T ( x2 – x1 ) φ 2 – φ 1 = -----------------------n Gj Jj

(5.16b)

∑ j=1

G2 G1

∑ j=1

Figure P5.49

Solution

-----------------------------------------------------------From kinematics we have: γ xθ = ρ

dφ (x) dx

(1)

From Hooke’s law, we obtain. dφ (x) dx

(2)

T = ∫ ρ τxθ dA

(3)

τ xθ = Gρ

The shear stress and internal torque are related as A

Assuming homogenous material, and noting that dA = 2πρ dρ, we obtain the following:

T = ∫ ρ τ xθ dA = ∫ ρ ⎛⎝ Gρ dφ ( x )⎞⎠ ( 2πρ ) dA = dφ ∫ Gρ 2 dA or dx dx A

T =

dφ dx

∫ G1 ρ A1

2

dA +

∫ G2 ρ A2

A 2

dA +

⋅

+

A

⋅

+

∫ Gn ρ

2

dA =

dφ [ G J + G2 J2 + dx 1 1

⋅ +

⋅ + G N J N ] or

An

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

T =

January 2014

n

dφ dx

∑ Gj Jj

(4)

j=1

dφ T = -------------------dx n Gj Jj

(5)

∑

j

We can write for the ith material

1

( τ xθ ) i = G i ρ

dφ dx

(6)

Substituting Equation (E5), we obtain:

T ( τ xθ ) i = G i ρ -------------------------

(7)

n

∑ Gj Jj j=1

φ2 – φ1 If the right hand side in Equation (7) is constant between x1 and x2 and we can write dφ = ---------------- and dx

x2 – x1

obtain the following. T ( x2 – x1 ) φ 2 – φ 1 = -------------------------

(8)

n

∑ Gj Jj j 1 Equations (7) and (8) are same as Equations (5.16a) and (5.16b). If G1=G2=G3....=Gn=G, then n

∑ j=1

n

Gj Jj = G

∑ Jj

= GJ

(9)

j=1

Substituting Equation (9) into Equations (7) and (8) we obtain:

T T ρ( τ xθ ) i = Gρ ------- = -------

(10)

T ( x2 – x1 ) φ 2 – φ 1 = ------------------------GJ

(11)

GJ

J

Equations (10) and (11) are same as Equations (5.10) and (5.12).

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.50 A circular solid shaft of radius R is made from a non-linear material that has a shear stress-shear strain relationship given by τ = Gγ0.5. Assume that the kinematic assumptions are valid and shear strain varies linearly with the radial distance across the cross-section. Determine the maximum shear stress and the rotation of section at B in terms of external torque Text, radius R, material constant G, and length Text A

B L FigureP5.50

5.51 A hollow circular shaft is made from a non-linear materials that has the following shear stress-shear strain relation τ = Gγ2. Assume that the kinematic assumptions are valid and shear strain varies lin-

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

early with the radial distance across the cross-section. In terms of internal torque T, material constant G, and R, obtain formulas for (a) the maximum shear stress τ max and (b) the relative rotation φ 2 – φ 1 of two cross-sections at x1 and x2.

R 2R

Figure P5.51

5.52

A solid circular shaft of radius R, and length L, is twisted by an applied torque T. The stress-strain

relationship for a non-linear material is given by the power law τ = Gγn . If all assumptions except Hooke’s law are applicable, show that the maximum shear stress in the shaft and the relative rotation of the two ends are as given below. 1 ---

( n + 3 )T n Δφ = ----------------------------- L (3 + n) 2πGR

T(n + 3) τ max = --------------------3 2πR

Substitute n = 1 in the above formulas and show that we obtain the same results as from Equations (5.9) and (5.10). τ = Gγ

Solution

n

τmax = ? Δφ = ? ------------------------------------------------------------

From kinematics we have: γ xθ = ρ

dφ (x) dx

(1)

Substituting this in the given stress-strain relationship, we obtain. τ xθ = Gρ

n

n dφ (x) dx

(2)

The shear stress and internal torque are related as

T = ∫ ρ τxθ dA

(3)

A

Assuming homogenous material, and noting that dA = 2πρ dρ, we obtain the following: R

T =

∫ 0

n⎫ n ⎧ n dφ dφ ρ ⎨ Gρ (x) ( x ) ⎬( 2πρ ) dρ = 2πG dx dx ⎩ ⎭

dφ (x) dx

n

R

∫

ρ

n+2

dρ = 2πG

n+3 n R dφ ( x ) ⎛ --------------⎞ or ⎝ n + 3⎠ dx

0

T(n + 3) = ------------------------n+3 2πGR

(4)

Substituting Eq. (4) into Eq. (2), we obtain: n T( n + 3) τ xθ = Gρ ------------------------n+3 2πGR

(5)

The maximum torsional shear stress is at ρ = R:

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

n T(n + 3) τ max = GR ------------------------- or n+3 2πGR

T(n + 3) τ max = --------------------3 2πR

Eq. (4) can be re-written as: 1 ---

T(n + 3) n dφ = ------------------------n+3 dx 2πGR

(6)

Δφ- . SubAssuming all quantities on the right in Eq (4) are constant over the length of the shaft, then dφ = -----dx

( n + 3 )T Δφ = ----------------------------(3 + n) 2πGR

stituting this in Eq. (4) we obtain:

L

1 --n

L

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.53 The internal torque T and the displacements of a point on a cross-section of a non-circular shaft are as given below. y

u = ψ ( y, z ) y

dVy = τxy dA z dVz = τxz dA

x

z

w = xy x

z

T =

T

5.15a

dφ dx

5.15b

dφ dx

5.15c

v = – xz

y

dφ dx

∫ ( yτxz – zτxy ) dA

5.16

A

Figure P5.53 Torsion of non-circular shafts.

where u, v, and w are the displacements in the x, y, and z, respectively, dφ is the rate of twist and is considered constant, dx ψ ( x, y ) is called the warping function and describes the movement of points out of the plane of

cross-section. Using Equations 2.12a and 2.12f, and the Hooke’s law show that the shear stresses for non-circular are given by: dφ τ xy = G ⎛ ∂ψ – z⎞ ⎝∂y ⎠ dx

5.17a

dφ τ xz = G ⎛ ∂ψ + y⎞ ⎝∂z ⎠ dx

5.17b

Solution

-----------------------------------------------------------From Eq. 2.12d, we have γ xy = ∂u + ∂v . Substituting Eqs. (5.15a) and (5.15b), we obtain the following. ∂y

∂x

dφ dφ dφ ∂ψ γ xy = ∂ ψ ( y, z ) + ∂ – xz = –z d x d x dx ∂y ∂y ∂x

(1)

From Hooke’s law, we have τ xy = Gγ xy . Substituting Eq. (1), we obtain Eq. (5.17a). From Eq. 2.12f, we have γ xz = ∂w + ∂u . Substituting Eqs. (5.15a) and (5.15c), we obtain the following. ∂x

γ xz =

∂z

dφ dφ dφ ∂ ∂ ∂ψ xy + ψ ( y, z ) = y+ d x d x dx ∂x ∂z ∂z

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(2)

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

From Hooke’s law, we have τ xz = Gγ xz . Substituting Eq. (2), we obtain Eq. (5.17b). -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.54

Show that for circular shafts ( ψ ( x, y ) = 0 ) the Equations in problem 5.53 reduce to Equation 5.7.

Solution

-----------------------------------------------------------Substituting ψ ( x, y ) = 0 into Eqs. (5.17a) and (5.17b), we obtain: τ xy = – G z

dφ dx

τ xz = Gy

dφ dx

(1)

Substituting Eq. (1) into Eq. 5.16, we obtain the following. T =

∫

2 2 2 dφ dφ dφ dφ dφ y ⎛ Gy ⎞ – z ⎛ – G z ⎞ dA = G [ y + z ] dA = G [ ρ ] dA = GJ ⎝ d x⎠ ⎝ dx dx dx d x⎠

A

∫

∫

A

A

(2)

Eq. (2) is the same as Equation 5.7.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.55

Consider the dynamic equilibrium of the differential element shown in Figure P5.55, where T is 2

the internal torque, ρ is the material density, J is the polar area moment of inertia, and ∂ 2φ is the angular ∂t

acceleration. Show 2

∂ φ ∂t

2

= c

2∂

2

φ

∂x

T

G ---ρ

2

where c =

2

5.18 2 ρJ ∂ φ dx 2 ∂t

T+dT

Figure P5.55

dx

dx

Solution

-----------------------------------------------------------By moment equilibrium in Figure P5.55, we obtain the following. 2

T + dT – T = ρJ

From Equation 5.7 we have T = GJ

∂ φ ∂t

2

2

dx

(1)

∂φ . Substituting in Eq. (1) and assuming GJ as constant, we obtain ∂x

2

∂φ ∂ φ ∂ GJ = ρJ 2 ∂x ∂x ∂t

∂ φ ∂T = ρJ 2 ∂x ∂t

or

2

or

GJ

∂ φ ∂x

2

2

= ρJ

∂ φ ∂t

2

2

or

2

G ∂ φ = ⎛ ----⎞ ⎝ ρ⎠ 2 2 ∂x ∂t ∂ φ

(2)

Eq. (2) is same as Eq. 5.18.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.56

Show by substitution that the solution below satisfies Equation 5.18. φ = ⎛ A cos ωx ------- + B sin ωx -------⎞ ( C cos ωt + D sin ωt ) ⎝ c c⎠

5.19

where A, B, C, and D are constants that are determined from the boundary conditions and the initial conditions, ω is the frequency of vibration. Solution

------------------------------------------------------------

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January 2014

The partial derivative of φ with respect to x can be found as shown below. ∂φ = [ C cos ωt + D sin ωt ] ∂ A cos -----ωx- + B sin ωx ------- or ∂x ∂x c c ωx ⎛ ω ωx ∂φ = [ C cos ωt + D sin ωt ] – ⎛ ω ----⎞ A sin ------ + ----⎞ B cos ------⎝ c⎠ ∂x c ⎝ c⎠ c

(1)

The second partial derivative of φ with respect to x can be found as shown below 2

∂ φ ∂x

2

= [ C cos ωt + D sin ωt ]

ω ω ∂ ωx ωx – ⎛ ----⎞ A sin ------- + ⎛ ----⎞ B cos ------- or ∂x ⎝ c ⎠ c ⎝ c⎠ c

2

ω 2 ωx ω 2 ωx ∂ φ = [ C cos ωt + D sin ωt ] – ⎛ ----⎞ A cos ------- – ⎛ ----⎞ B sin ------- or ⎝ c⎠ ⎝ c⎠ 2 c c ∂x 2

2 ω 2 ∂ φ = –⎛ ω ----⎞ [ C cos ωt + D sin ωt ] ⎛ A cos ωx ------- + B sin ωx -------⎞ = – ⎛ ----⎞ φ ⎝ ⎝ ⎠ ⎝ ⎠ 2 c⎠ c c c ∂x

(2)

The partial derivative of φ with respect to can be found as shown below. ∂φ ωx ωx ∂ = A cos ------- + B sin ------- [ C cos ωt + D sin ωt ] or ∂t c c ∂t

∂φ ωx ωx = A cos ------- + B sin ------- [ – ωC sin ωt + ωD cos ωt ] ∂t c c

(3)

The second partial derivative of φ with respect to t can be found as shown below 2

∂ φ ∂t

2

2

∂ φ ∂t

2

ωx ωx ∂ = A cos ------- + B sin ------- [ – ωC sin ωt + ωD cos ωt ] or c c ∂t 2 2 ωx ωx = A cos ------- + B sin ------- [ – ω C cos ωt – ω B sin ωt ] or c c 2

∂ φ ∂t

2

2 2 ωx ωx = – ω A cos ------- + B sin ------- [ C cos ωt + D sin ωt ] = – ω φ c c

(4)

Substituting Eqs. (2) and (4) into Eq. 5.18, we obtain an identity, proving Eq. 5.19 is the solution to Eq. 5.18.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.57 A hollow aluminum shaft of 5 feet length is to carry a torque of 200 in-kips. The inner radius of the shaft is 1 inch. If the maximum torsional shear stress in the shaft is to be limited to 10 ksi, determine the minimum outer radius of the shaft to the nearest 1/8 inch. Solution L = 5 ft T = 200 in-kips Ri = 1 inch τ max ≤ 10 ksi

Ro = ? nearest 1/8 inch ------------------------------------------------------------

The maximum torsional shear stress will exist on the outer radius of the shaft and can be expressed in terms of Ro as shown below. ( 200 ) ( R o ) Tρ τ max = ------- = --------------------------- ≤ 10 4 4 π J --- [ R o – 1 ] 2

or

4

( R o – 12.732R o – 1 ) ≥ 0

(1)

The minimum value of Ro corresponds to the root of the left hand side of Eq. 1, i.e.,

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

4

f ( R o ) = R o – 12.732R o – 1 = 0 . We find the root on the spread sheet as described in Appendix B. The

calculations are shown in the table below. Ro

f(Ro)

Ro

f(Ro)

1.1

-13.542

2.3

-2.300

1.2

-14.205

2.31

-1.938

1.3

-14.696

2.32

-1.569

1.4

-14.984

2.33

-1.194

1.5

-15.036

2.34

-0.812

1.6

-14.818

2.35

-0.423

1.7

-14.293

2.36

-0.028

1.8

-13.421

2.37

0.374

1.9

-12.159

2.38

0.782

2

-10.465

2.39

1.198

2.1

-8.290

2.4

1.620

2.2

-5.586

2.3

-2.300

2.4

1.620

2.5

6.232

From the above table, we see that the root of f(Ro) is between 2.36 and 2.37. The value of Ro to the nearest 3 R o = 2 --- in 8

1/8 that will satisfy Eq. (1) is 2.375. Thus, the minimum value of Ro is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.58 A 4 feet long hollow shaft is to transmit a torque of 100 in-kips. The relative rotation of the two ends of the shaft is limited to 0.06 rads. The shaft can be made of steel or aluminum. The Shear modulus of rigidity G, the allowable shear stress τallow, and the specific weight γ, are given in Table 5.1. The inner radius of the shaft is to be 1 inch. Determine outer radius to the nearest 1/8 inch of the lightest shaft that can be used for transmitting the torque and the corresponding weight. Solution

L = 4 ft. Ri = 1 inch

Δφ ≤ 0.06 rads

T= 100 in-kips Ro = ? to the nearest 1/8 inch

W=?

-----------------------------------------------------------The relative rotation of the ends and maximum shear stress in steel can be found and the following limits on do can be obtained. ( 100 ) ( 4 ) ( 12 ) ( Δφ ) S = ---------------------------------------------------- ≤ 0.06 4 3 π 4 12 ( 10 ) ⎛ ---⎞ [ R o – 1 ] ⎝ 2⎠ ( 100 ) ( R o ) ( τ max ) S = -------------------------------- ≤ 18 4 4 ⎛π ---⎞ [ – 1 ] ⎝ 2⎠ R o

R o ≥ 1.513 in

(1)

R o – 3.5368R o – 1 ≥ 0

(2)

or

or

4

Substituting the value of Ro in Eq. 1 into the left hand side we obtain a negative value, which implies Ro must be greater than 1.513 inch to satisfy Eq. 2. The value of Ro corresponds to the root of the left hand 4

side of Eq. 2, i.e., f ( R o ) = R o – 3.536R o – 1 = 0 . We find the root on the spread sheet as described in

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January 2014

Appendix B. The calculations are shown in the table below. Ro

f(Ro)

Ro

f(Ro)

1.5 1.6

-1.243

1.6

-0.105

-0.105

1.61

0.025

1.7

1.340

1.62

0.158

1.8

3.131

1.63

0.294

1.9

5.312

1.64

0.434

2

7.926

1.65

0.576

From the above table, we see that the root of f(Ro) is between 1.6 and 1.61. Rounding to the nearest 1/ 8 inch, we obtain for steel. R O = 1.625inch

(3)

Using similar calculations for Aluminum shaft we obtain the following. ( 100 ) ( 4 ) ( 12 ) ( Δφ ) Al = ----------------------------------------------- ≤ 0.06 4 π 4 ( 4000 ) ⎛ ---⎞ [ r o – 1 ] ⎝ 2⎠ ( 100 ) ( r o ) ( τ max ) Al = ------------------------------ ≤ 10 4 4 ⎛π ---⎞ [ r – 1 ] ⎝ 2⎠ o

r o ≥ 1.925 in

(4)

r o – 6.366r o – 1 ≥ 0

(5)

or

or

4

Substituting the value of ro in Eq. (4) into the left hand side we obtain a positive value, which implies that ro = 1.925inch. satisfies both Eqs. (4) and (5). Rounding to the nearest 1/8 inch, we obtain for aluminum. r o = 2.0inch

(6)

We can find the weight of each material by taking the product of the material density and the volume of a hollow cylinder as shown below: 2

2

W S = ( 0.285 ) ( π ) ( 1.625 – 1 ) ( 48 ) = 70.51 lbs 2

(7)

2

W Al = ( 0.1 ) ( π ) ( 2 – 1 ) ( 48 ) = 45.24 lbs

(8)

From Eqs 7 and 8 we see that the aluminum shaft is lighter. Thus, the answer is: r o = 2.0 in.

W = 45.24lbs

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.59 Table 5.2 shows the measured radii of a solid tapered shaft at several point along the axis of the shaft. The shaft is made of aluminum (G = 28 GPa) and has a length of 1.5 meters. Determine (a) the rotation of the free end with respect to the wall using numerical integration. (b) the maximum shear stress in

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

the shaft. R(x) T = 35 kN-m B

A

Figure P5.59

x

Table 5.2.. Data for Problem 5.59. x (m)

R(x) (mm)

x (m)

R(x) (mm)

0.0

100.6

0.8

60.1

0.1

92.7

0.9

60.3

0.2

82.6

1.0

59.1

0.3

79.6

1.1

54.0

0.4

75.9

1.2

54.8

0.5

68.8

1.3

54.1

0.6

68.0

1.4

49.4

0.7

65.9

1.5

50.6

Solution

Δφ = φB-φA = ? ------------------------------------------------------------

G = 28 GPa

xA= 0

xB= 1.5 m

τmax= ?

The internal torque is T = 35 kN-m. 3

–6

( 35 ) ( 10 ) 2.5 ( 10 ) dφ T = ------- = ------------------------------------------------------- = ----------------------9 4 4 dx GJ ( 28 ) ( 10 ) ( π ⁄ 2 )R ( x ) πR ( x )

(1) 6

2.5 ( 10 ) In Eq. (1), R(x) is measured in meters. Eq. (1) can be rewritten as dφ = -------------------- , where now R(x) is in dx

4

πR ( x )

millimeters. Integrating from xA= 0 to xB= 1.5, we obtain the following: φB

∫ φA

x B = 1.5

dφ =

∫ xA = 0

1.5

6

2.5 ( 10 ) --------------------- dx 4 πR ( x )

or

Δφ =

∫ 0

6

2.5 ( 10 ) --------------------- dx 4 πR ( x )

(2)

6 ( 10 ) --------------------- and using numerical integration as described in Appendix B, we obtain the Representing f ( x ) = 2.5 4

πR ( x )

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

value of the integral on a spread sheet as shown below. x

R(x)

f(x)

Integral

x

R(x)

f(x)

Integral

0.00

100.60

0.0078

0.0009

0.80

60.10

0.0610

0.0282

0.10

92.70

0.0108

0.0023

0.90

60.30

0.0602

0.0344

0.20

82.60

0.0171

0.0042

1.00

59.10

0.0652

0.0424

0.30

79.60

0.0198

0.0064

1.10

54.00

0.0936

0.0515

0.40

75.90

0.0240

0.0093

1.20

54.80

0.0882

0.0605

0.50

68.80

0.0355

0.0130

1.30

54.10

0.0929

0.0718

0.60

68.00

0.0372

0.0169

1.40

49.40

0.1336

0.0846

0.70

65.90

0.0422

0.0221

1.50

50.60

0.1214

The rotation of the free end is: Δ φ = 0.085 rad; The maximum shear stress will occur at the cross-section where radius is the smallest, i.e., just before point B and can be found as shown below. 3

6 2 35 ( 10 ) ( 0.0506 ) Tρ τ max = ------- = ----------------------------------------- = 171.99 ( 10 ) N ⁄ m or 4 JB ( π ⁄ 2 ) ( 0.0506 )

τ max = 25.8 ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.60 Let the radius of the tapered shaft in problem 5.59 be represented by the equation R ( x ) = a + bx . Using the data in Table 5.2 determine the constant a and b by the least-square method and then find the rotation of section at B by analytical integration. Solution

G = 28 GPa

a=?

b=?

xA= 0

Δφ = φB-φA = ?

xB= 1.5 m

-----------------------------------------------------------We develop the equations for the least square method for a linear representation as described in Appendix B. This difference between the radius value and the value obtained by substituting x = xi in the equation R ( x ) = a + bx is the error ei that can be written as given below. e i = R i – R ( x i ) = R i – ( a + bx i ) N

We define the error E as E =

(1)

∑ ei . The error E is minimized with respect to coefficients a, and b to gen2

i=1

erate a set of linear algebraic equations as shown below. ∂E = 0 ∂a ∂E = 0 ∂b

or

∂e i

∑ 2ei ∂ a ""

or

∑

2e i

∂e i ∂b

= 0

or

∑ 2 [ Ri – ( a + bxi ) ] [ –1 ]

= 0

= 0

or

∑ 2 [ Ri – ( a + bxi ) ] [ –xi ]

= 0

The above equations on the right can be rearranged and written in matrix form as shown below:

∑ xi ⎧⎨ 2 ∑ xi ∑ xi ⎩ N

⎧ a ⎫ = ⎪ ⎬ ⎨ b ⎭ ⎪ ⎩

⎫

∑ Ri ⎪⎬ ∑ xi Ri ⎪⎭

or

⎧ b 11 b 12 ⎧ a ⎫ ⎪ r1 ⎨ ⎬ = ⎨ b 21 b 22 ⎩ b ⎭ ⎪ r2 ⎩

⎫ ⎪ ⎬ ⎪ ⎭

(2)

The coefficients of the b-matrix and the r-vector can be determined by comparison to the matrix form of the equations on the left. The coefficients a and b can be determined by Cramer’s rule. Let D represent the determinant of the b matrix. By Cramer’s rule, we replace the first column in the matrix of b’s by the right hand side and find the determinant of the so constructed matrix and divide by D. Thus, the coefficients a, and b can be written as shown below. D = b 11 b 22 – b 12 b 21

(3)

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

r 1 b 12

a =

r 2 b 22

r 1 b 22 – r 2 b 12 ⁄ D = ------------------------------D

b =

January 2014

r 2 b 11 – r 1 b 21 b 11 r 1 ⁄ D = ------------------------------D b 21 r 2

(4)

The given data and Eq. (2) through (4) can be put in a spread sheet and the coefficients a and b can be found as shown in the table below as: a = 90.226 mm b = – 30.593 xi (m)

Ri (mm)

xi2

xi*Ri

10

0.90

60.30

0.8100

54.270

9.270

11

1.00

59.10

1.0000

59.100

1.10

54.00

1.2100

59.400

xi (m)

Ri (mm)

xi2

xi*Ri

1

0.00

100.60

0.0000

0.000

2

0.10

92.70

0.0100

3

0.20

82.60

0.0400

16.520

12

4

0.30

79.60

0.0900

23.880

13

1.20

54.80

1.4400

65.760

5

0.40

75.90

0.1600

30.360

14

1.30

54.10

1.6900

70.330

6

0.50

68.80

0.2500

34.400

15

1.40

49.40

1.9600

69.160

1.50

50.60

2.2500

75.900

1076.50

12.4000

703.360

7

0.60

68.00

0.3600

40.800

16

8

0.70

65.90

0.4900

46.130

bij & ri

12.0000

9

0.80

60.10

0.6400

48.080

D

54.4

a and b

90.226

-30.593

Thus R(x) = 90.226-30.593x, where x is in meters and R is in millimeters. Substituting this in Eq. (2) of problem 5.59 and integrating, we obtain the rotation as shown below. 1.5

Δφ =

6

6

1.5

2.5 ( 10 ) 2.5 ( 10 ) 1 ----------------------------------------------------- dx = ⎛ ---------------------⎞ ------------------------------------------------------------------------------4 3 ⎝ ⎠ π π ( 90.226 – 30.593x ) ( 3 ) ( 30.593 ) ( 90.226 – 30.593x ) 0 0

∫

= 0.0877 Δφ = 0.088 rad

Thus the rotation of the free end of the shaft is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.61 Table 5.3 shows the value of distributed torque at several point along the axis of a solid steel (G = 12,000 ksi) shaft. The shaft has a length of 36 inches and a diameter of 1 inch. Determine (a) the rotation of the end A with respect to the wall using numerical integration. (b) the maximum shear stress in the shaft.(Hint: Use Equation 5.14 for part (a)) Table 5.3.. Data for Problem 5.61. t(x) in-lb/in

B A x

Figure P5.61

Solution

G = 12,000 ksi

d= 1 inch

x (inches)

t (x) in-lbs./in

x (inches)

t (x) in-lbs./in

0

93.0

21

588.8

3

146.0

24

700.1

6

214.1

27

789.6

9

260.4

30

907.4

12

335.0

33

1040.3

15

424.7

36

1151.4

18

492.6

uA = ?

σmax= ?

-----------------------------------------------------------The equilibrium equation dT + t ( x ) = 0 can be integrated from point A, where T = 0 to any location x, to dx

obtain the internal torque as a function of x as shown below.

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T(x)

T(x) =

∫

January 2014

x

∫

dT = –

TA = 0

t ( x ) dx

(1)

xA = 0 4

4

The polar moment of inertia of the cross-section is J = πd ⁄ 32 = 0.0981in . –9 dφ = T (x) = T( x) -------------------------------------------------------- = 848.8 ( 10 )T ( x ) 6 dx GJ ( 12 ) ( 10 ) ( 0.0981 )

(2)

Integrating from point A i.e., xA =0 to point B, i.e., xB =36. We note that point B is fixed to the wall, hence φB= 0. we obtain the following integral: φB = 0

x B = 36

∫

∫

dφ =

φA

36 –9

[ 848.8 ( 10 )T ( x ) ] dx

or

∫

–9

φ A = – [ 848.8 ( 10 )T ( x ) ] dx

(3)

0

xA = 0

The maximum torsional shear stress τxθ at a section can be found as shown below: Tρ T ( 0.5 ) τ xθ = ------- = ---------------- = 5.0929T J 0.0981

(4)

Eq. (1) can be numerically integrated on a spread sheet to obtain T(xi), the value of internal torque at any xi. Then, Eq (3) can be numerically integrated to obtain the rotation. Eq. (4) can be used to find the torsional shear stress at various xi and the maximum value chosen by inspection. These calculations can be done on a spread sheet as shown below. xi (inches)

t(xi) (in-lbs./in)

T(xi) (in-lbs.)

[φA-φ(xi+1)](10-3) (rads)

τxθ(xi) (ksi)

0

93

0.00

0.456

0.000

3

146.0

-358.50

2.057

-1.826

6

214.1

-898.65

5.251

-4.577

9

260.0

-1609.80

10.487

-8.199

12

335.0

-2502.30

18.310

-12.744

15

424.7

-3641.85

29.334

-18.548

18

492.0

-5016.90

44.174

-25.551

21

588.8

-6638.10

63.539

-33.808

24

700.1

-8571.45

88.211

-43.654

27

789.6

-10806.00

118.970

-55.034

30

907.4

-13351.50

156.689

-67.999

33

1040.3

-16273.05

202.314

-82.878

36

1151.4

-19560.60

-99.621

The rotation of section at A is counter-clockwise with respect to x-axis.

φ A = 0.202 rads

The magnitude of the maximum torsional shear stress is

τ max = 99.6 ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.62 Let the distributed torque t(x) in problem 5.61 be represented by the equation t ( x ) = a + bx + cx 2 . Using the data in Table 5.3 determine the constant a, b, and c by the least-square method and then find the rotation of section at B by analytical integration. Solution

a=?

φA = ? ------------------------------------------------------------

b=?

c=

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

The equilibrium equation dT + t ( x ) = 0 can be integrated from point A, where T = 0 to any location x, to dx

obtain the internal axial force as a function of x as shown below. T(x)

T(x) =

∫

x

∫

dT = –

TA = 0

x

xA = 0

2

3

2 cx bx [ a + bx + cx ] dx = – ax + -------- + -------2 3

∫

t ( x ) dx = –

(1)

xA = 0 4

4

The polar moment of inertia of the cross-section is J = πd ⁄ 32 = 0.0981in . 2

3

–9 bx dφ = T (x) = T(x) cx -------------------------------------------------------- = – 848.82 ( 10 ) ax + -------- + -------6 dx GJ 2 3 ( 12 ) ( 10 ) ( 0.0981 )

(2)

Integrating from point A i.e., xA =0 to point B, i.e., xB =36. We note that point B is fixed to the wall, hence φB = 0

φB= 0. we obtain the following integral:

∫

x B = 36

dφ =

φA 2

3

3

∫ xA = 0

4

bx cx ax φ A = ( 848.82 ) ( 10 ) ⎛⎝ -------- + -------- + --------⎞⎠ 2 6 12 –9

2

–9 bx cx ( – 848.82 ) ( 10 ) ax + -------- + -------- dx or 2 3

36

2

3

4

– 9 a ( 36 ) b ( 36 ) c ( 36 ) = ( 848.82 ) ( 10 ) ---------------- + ---------------- + ---------------2 6 12 0

(3)

Using the Least Square Method described in Appendix B, we obtain the value of the values of constants a, b, and c on a spread sheet as shown in the table below. xi

t(xi)

xi2

xi3

xi4

x*ti

xi2*ti

1

0.0

93

0.0

0.000E+00

0.000E+00

0.000E+00

0.000E+00

2

3.0

146.0

9.0

2.700E+01

8.100E+01

4.380E+02

1.314E+03

3

6.0

214.1

36.0

2.160E+02

1.296E+03

1.285E+03

7.708E+03

4

9.0

260.0

81.0

7.290E+02

6.561E+03

2.340E+03

2.106E+04

5

12.0

335.0

144.0

1.728E+03

2.074E+04

4.020E+03

4.824E+04

6

15.0

424.7

225.0

3.375E+03

5.063E+04

6.371E+03

9.556E+04

7

18.0

492.0

324.0

5.832E+03

1.050E+05

8.856E+03

1.594E+05

8

21.0

588.8

441.0

9.261E+03

1.945E+05

1.236E+04

2.597E+05

9

24.0

700.1

576.0

1.382E+04

3.318E+05

1.680E+04

4.033E+05

10

27.0

789.6

729.0

1.968E+04

5.314E+05

2.132E+04

5.756E+05

11

30.0

907.4

900.0

2.700E+04

8.100E+05

2.722E+04

8.167E+05

12

33.0

1040.3

1089.0

3.594E+04

1.186E+06

3.433E+04

1.133E+06

13

36.0

1151.4

1296.0

4.666E+04

1.680E+06

4.145E+04

1.492E+06

bij &ri

234.0

7142.4

5850.0

1.643E+05

4.918E+06

1.768E+05

5.014E+06

Cij

1.783E+09

1.897E+08

4.216E+06

2.971E+07

7.666E+05

2.129E+04

D

3.453E+09

ai

96.36

15.45

0.39

The values of the constants are: a = 96.36 b = 15.45 Substituting the values of the above constants in Eq. (3) we obtain: 2

3

4

– 9 ( 96.36 ) ( 36 ) ( 15.45 ) ( 36 ) ( 0.39 ) ( 36 ) φ A = ( 848.82 ) ( 10 ) -------------------------------- + -------------------------------- + ----------------------------- = 0.2013 or 2 6 12

c = 0.39

φ A = 0.2013 rads

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.63

A steel (Gs =12,000 ksi) and a bronze (GBr = 5,600 ksi) shaft are securely connected at B as

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

shown. Determine the maximum torsional shear stress in the shaft and rotation of section at B if the applied torque T = 50 in-kips. T A Stee l

B

Bronze

C

2 ft

2 in

4 ft

Figure P5.63

Solution

G AB = 12000 ksi

T = 50in – k ips τ max =? φ B =?

G BC = 5600ksi

-----------------------------------------------------------4

4

The polar moment of inertia for the shaft cross-section can be found as: J = ( π2 ) ⁄ 32 = 1.5708in The following free body diagrams can be drawn with internal torques drawn as per our sign convention. (a)

(b)

TAB TA

A Ste el

T TA A S teel

B

TBC Bronze

By moment equilibrium about the shaft axis we have the following equations. T AB = T A

T BC = T A – T

(1)

The relative rotations of the ends of the each segment can be written as T AB ( x B – x A ) T A ( 2 ) ( 12 ) φ B – φ A = ---------------------------------- = ---------------------------------------- = 1.273 ( 10 –3 )T A G AB J ( 12000 ) ( 1.5708 )

(2)

T BC ( x C – x B ) ( T A – 50 ) ( 4 ) ( 12 ) φ C – φ B = --------------------------------- = ------------------------------------------ = 5.456 ( 10 – 3 ) ( T A – T ) G BC J ( 5600 ) ( 1.5708 )

(3)

Adding Eq’s(2) and (3) and noting that points A and C are built into the wall, hence φ C = 0 and φ A = 0 φ C – φ A = [ 1.273T A + 5.456 ( T A – T ) ] ( 10 –3 ) = 0 or 5.456 T A = -------------------------------------- T = ( 0.8108 ) ( 50 ) = 40.541in – kips ( 1.273 + 5.456 )

From Eq.(2) we obtain φ B = 1.273 ( 10 – 3 ) ( 40.541 ) = 51.608 ( 10 –3 )rads or φ B = 0.0516 rads CCW From Eq.(1), we obtain T AB = 40.541

T BC = – 9.459in – kips

The magnitude of the maximum torsional shear stress will be in segment AB and can be found as T AB ρ max 40.541 ) ( 1 )= (---------------------------τ max = ----------------------1.5708 J

τ max = 25.8 ksi

or

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.64 A steel (Gs =12,000 ksi) and a bronze (GBr = 5,600 ksi) shaft are securely connected at B as shown. Due to the torque T, the section at B rotates by an amount of 0.02 rads. Determine the maximum torsional shear strain and the applied torque T. T A Stee l

B

Bronze

C

2 ft

2 in

4 ft

Figure P5.64

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5-48

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

· φ B = 0.02 rads

Solution

γ max = ?

January 2014

T =?

-----------------------------------------------------------The rate of twist in each segment can be found as shown below: φB – φA – 3 rads 0.02 – 0 ⎛ dφ ------⎞ - = ------------------- = 0.833 ( 10 ) ----------= -----------------⎝ dx⎠ AB in 24 xB – xA φC – φB – 0.02- = – 0.4167 ( 10 – 3 ) rads ⎛ dφ ----------------⎞ = ------------------ = 0 -----------------⎝ dx⎠ BC in 48 xC – xB

(1) (2)

The maximum torsional shear strain will be in AB and can be found as shown –3 dφ γ max = ρ max ⎛ ------⎞ = ( 1 ) ( 0.833 ) ( 10 ) ⎝ dx⎠ AB

γ max = 833 μrads

or

Substituting φ B = 0.02 and φ A = 0 in Eq.(2) in problem 5.53 we obtain 0.02 T A = ----------------------------- = 15.711 in – kips –3 1.273 ( 10 )

Substituting φ B = 0.02 and φ C = 0 in Eq.(3) in problem 5.53 we obtain – 0.02 ( 15.711 – T ) = ----------------------------–3 5.456 ( 10 )

or

T = 19.4 in-kips

T = 19.377 in – kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.65 Two hollow aluminum (G = 10,000 ksi) shafts are securely fastened to a solid aluminum shaft and loaded as shown. Point E is on the inner surface of the shaft. If T= 300 in-kips, Determine (a) the rotation of section at C with respect to rotation the wall at A. (b) the shear strain at point E. T 4 in 2 in

A

B

D

C E 36 in

24 in

24 in

Figure P5.65

Solution

G = 10000 ksi

φC – φA =

T = 300 in – k ips

γE =

?

?

-----------------------------------------------------------The polar moment of cross - sections in various segments can be found as shown below 4 4 π 4 J AB = J CD = ------ ( 4 – 2 ) = 23.562 in 32

(1)

4 π 4 J BC = ------ 4 = 25.133 in 32

(2)

The following free body diagrams can be drawn with internal torques drawn as per our sign convention. (a)

TAB TA A

T

(b) TA A

TBC

T

(c) TA A

B

B

TCD C

By moment equilibrium about the shaft axis we have the following equations. T AB = T A

T BC = T A – T

T CD = T A – T

(3)

The relative rotation of the ends of each segment can be written as shown below T AB ( x B – x A ) ( T A ) ( 24 ) φ B – φ A = ---------------------------------- = ---------------------------------------- = 0.1019 ( 10 –3 )T A GJ AB ( 10000 ) ( 23.562 )

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(4)

5-49

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

T BC ( x C – x B ) ( T A – T ) ( 36 ) φ C – φ B = --------------------------------- = ---------------------------------------- = 0.1432 ( 10 – 3 ) ( T A – T ) GJ BC ( 10000 ) ( 25.133 )

(5)

T CD ( x D – x C ) ( T A – T ) ( 24 ) φ D – φ C = ---------------------------------- = ---------------------------------------- = 0.1019 ( 10 – 3 ) ( T A – T ) GJ CD ( 10000 ) ( 23.562 )

(6)

Adding Eqs. (4),(5) and(6) and noting that points A and C are built into wall, hence φ D = 0 and φ A = 0 –3

φ D – φ A = [ 0.1019T A + 0.1432 ( T A – T ) + 0.1019 ( T A – T ) ] ( 10 ) = 0

or

( 0.1527 + 0.1019 )T T A = ------------------------------------------------------------- = 0.7064T = 211.93 0.1019 + 0.1432 + 0.1019

(7)

Substituting Eq.(7) into Eqs.(4) and (5) and adding, we obtain, –3

–3

φ C – φ A = [ 0.1019 ( 211.93 ) + 0.1432 ( 211.93 – 300 ) ] ( 10 ) = 8.984 ( 10 )

φ C – φ A = 0.00898 rads CCW

From Eq.(3) and (7), we obtain: T CD = – 88.07 in – kips T CD ρ E ( – 88.07 ) ( 1 ) The shear stress at point E is: τ E = ---------------- = ----------------------------- = – 3.738 ksi J CD

23.562

τ –3 3.738 - = – 0.3738 ( 10 ) The shear strain at point E is γ E = ----E- = –--------------G

γ E = – 374 μrads

10000

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.66 Two hollow aluminum (G = 10,000 ksi) shafts are securely fastened to a solid aluminum shaft and loaded as shown. Point E is on the inner surface of the shaft. If the torsional shear strain at point E is -250 μ, determine the rotation of section C and the applied torque T that produced this shear strain. T 4 in 2 in

A

B

D

C E

24 in

36 in

24 in

Figure P5.66

Solution

γ E = – 250 μrads

G = 10000 ksi

φC = ?

T= ?

-----------------------------------------------------------The rotation of the section at C can be found as show below ( φD – φC ) ( 0 – φC ) –6 –6 dφ γ E = ρ E ⎛ ------⎞ or γ E = ( – 250 ) ( 10 ) = ( 1 ) ------------------------ or ( – 250 ) ( 10 ) = ( 1 ) -------------------- or ⎝ dx⎠ CD xD – xC 24 –3

φ C = 0.006 rads CCW;

φ C = 6 ( 10 )rads or

Substituting φ D = 0 and φ C = 0.006 in Eq.(6) of problem 5.55, we obtain 0.006 T A – T = – -------------------------------= – 58.88 –3 0.1019 ( 10 )

(8)

From Eq. (7) in problem 5-55 we have T A = 0.7064 T

Substituting Eq.(3) into Eq.(2) we obtain: ( 0.7064 – 1 )T = – 58.88 or

(9) T = 200.5 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.67 Two solid circular steel (Gs= 80 GPa) shafts and a solid circular bronze (Gbr = 40 GPa) are securely connected by a coupling at C. A torque of T =100 kN-m is applied to the rigid wheel B as shown. If the coupling plates cannot rotate relative to each other, determine the angle of rotation of the wheel B

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

due to the applied torque. T = 10 kN-m A

C

B Steel

Steel 5m

3m

D Bronze

100 mm

4m

Figure P5.67

Solution

G AB = G BC = 80 GPa

G CD = 40 GPa

φ B =?

T = 10 kN – m

-----------------------------------------------------------The polar moment of the cross-section can be found as shown below 4 –6 4 π J = ------ ( 0.1 ) = 9.817 ( 10 ) m 32

(1)

The following free body diagrams can be drawn with internal torques drawn as per our sign convention. T = 10 kN-m T = 10 kN-m TAB (b) (a) TCD TBC (c) A TA

Steel

TAA

TAA

B Steel

Steel

B Steel

Steel

C Bronze

By moment equilibrium about the shaft axis we have the following equations. T AB = – T A

T BC = T – T A

T CD = T – T A

(2)

The relative rotation of the ends of the each segment can be written as shown below. T AB ( x B – x A ) TA ( 5 ) φ B – φ A = ---------------------------------= 6.366 ( 10 – 6 )T A - = ----------------------------------------------------------9 –6 G AB J ( 80 ) ( 10 ) ( 9.817 ) ( 10 )

(3)

T BC ( x C – x B ) ( T –TA ) ( 3 ) - = ----------------------------------------------------------φ C – φ B = --------------------------------= 3.8199 ( 10 – 6 ) ( T – T A ) 9 –6 G BC J ( 80 ) ( 10 ) ( 9.817 ) ( 10 )

(4)

T CD ( x D – x C ) ( T –TA ) ( 4 ) φ D – φ C = ---------------------------------= 10.1864 ( 10 –6 ) ( T – T A ) - = ----------------------------------------------------------9 –6 G CD J ( 40 ) ( 10 ) ( 9.817 ) ( 10 )

(5)

Adding Eqs. (3),(4) and (5) and noting that φ D = 0 and φ A = 0 we obtain the following. –6

φ D – φ A = [ – 6.3661T A + 3.8199 ( T – T A ) + 10.1864 ( T – T A ) ] ( 10 ) = 0

or

( 3.8199 + 10.1864 )T T A = ---------------------------------------------------------------------- = 0.6875T = 6.875 kN – m ( 6.3661 + 3.8199 + 10.1864 ) –6

3

(6)

–3

From Eq.(3) we obtain: φ B = ( – 6.366 ) ( 10 ) ( 6.875 ) ( 10 ) = – 43.77 ( 10 ) φ B = 0.0438 rads CW

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.68 Two solid circular steel (Gs= 80 GPa) shafts and a solid circular bronze (Gbr = 40 GPa) are securely connected by a coupling at C. A torque of T =10 kN-m is applied to the rigid wheel B as shown.If the coupling plates can rotate relative to each other by 0.5o before engaging then what will be the angle of rotation of the wheel B. T = 10 kN-m A

B Steel 5m

Steel

C

3m

D 100 mm Bronze 4m

Figure P5.68

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M. Vable

Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

o

G AB = G BC = 80 GPa G CD = 40 GPa T = 10 kN – m φ C – φ C = 0.5 [ 2 1

φ B =?

-----------------------------------------------------------The coupling plate C1 will rotate clockwise relative to the coupling plate C 2 ,hence the relative rotation of plate C 2 with respect to plate C 1 will be positive.We can write the following –3 0.5π φ C – φ C = ----------- = 8.7267 ( 10 ) 2 1 180

(1)

Eqs (3) and(4) and (5) of problem 5-57 can be rewritten as φ B – φ A = 6.366 ( 10 – 6 )T A

(2)

φ C1 – φ B = 3.8199 ( 10 – 6 ) ( T – T A )

(3)

φ D – φ C2 = 10.1864 ( 10 –6 ) ( T – T A )

(4)

Adding the above four equations and noting that φ D = 0 and φ A = 0 , we obtain –6

3

φ D – φ A = [ – 6.3661T A + 3.8199 ( T – T A ) + 10.1864 ( T – T A ) ] ( 10 ) + 8.7267 ( 10 ) = 0 or 3

( 3.8199 + 10.1864 )T + 8.7267 ( 10 ) T A = ---------------------------------------------------------------------------------------( 6.3661 + 3.8199 + 10.1864 ) 3

(5)

3

3

3

Substituting T = 10 ( 10 ) N – m in Eq.(5) T A = ( 0.6875 ) ( 10 ) ( 10 ) + ( 0.4283 ) ( 10 ) = 7.303 ( 10 ) N – m From Eq. (2) we obtain, –6

3

–3

φ B = ( – 6.366 ) ( 10 ) ( 7.303 ) ( 10 ) = – 46.497 ( 10 ) rads

φ B = – 0.0465 rads CW

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.69 A solid steel (G = 80 GPa) is securely fastened to a solid bronze (G = 40 GPa) shaft that is 2 m long as shown. If T = 10 kN-m in Figure P5.69 , determine (a) the magnitude of maximum torsional shear stress in the shaft. (b) the rotation of section at 1 m from the left wall.

T A

Bronze

B

120 mm Steel 80 mm

2m 1m

Figure P5.69

Solution

G AB = 40GPa

τ max =?

G BC = 80 GPa T = 10 kN – m

φ 1 =?

-----------------------------------------------------------The polar moment of cross - sections in various segments can be found as shown below 4 –6 4 π J AB = ------ ( 0.12 ) = 20.358 ( 10 ) m 32

(1)

4 –6 4 π J BC = ------ ( 0.08 ) = 4.021 ( 10 ) m 32

(2)

The following free body diagrams can be drawn. The internal torques are drawn as per our sign convention. T TAB (b) (a) TA A Bronze

TA A

Bronze

TBC

B

Steel

By moment equilibrium about the shaft axis we have the following equations.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

T AB = T A

January 2014

T BC = T A – T

(3)

The relative rotation of the ends of the each segment can be written as shown below, T AB ( x B – x A ) TA ( 2 ) φ B – φ A = ---------------------------------- = -------------------------------------------------------------= 2.456 ( 10 –6 )T A 9 –6 G AB J AB ( 40 ) ( 10 ) ( 20.358 ) ( 10 )

(4)

T BC ( x C – x B ) ( T –TA ) ( 1 ) φ C – φ B = --------------------------------- = ----------------------------------------------------------= 3.109 ( 10 – 6 ) ( T A – T ) 9 –6 G BC J BC ( 80 ) ( 10 ) ( 4.021 ) ( 10 )

(5)

Adding Equations(4) and (5) and noting that φ C = 0 and φ A = 0 we obtain the following. –6 φ C – φ A = [ 2.456T A + 3.109 ( T A – T ) ] ( 10 ) = 0 or 3 3.109T T A = --------------------------------- = 0.5586T = 5.586 ( 10 )N – m 2.456 + 3.109 3

(6)

3

From Eq.(3) T AB = 5.586 ( 10 )N – m

T BC = ( – 4.413 ( 10 ) )N – m

The maximum torsional shear stress in each segment can be found as shown below 3 T AB ( ρ AB ) max 6 N ( 5.586 ) ( 10 ) ( 0.06 ) ( τ AB ) max = ----------------------------------= ----------------------------------------------- = 16.46 ( 10 ) ------2 – 6 J AB m 20.358 ( 10 ) 3 T BC ( ρ BC ) max 6 N ( – 4.413 ) ( 10 ) ( 0.04 ) ( τ BC ) max = ---------------------------------- = --------------------------------------------------- = – 43.9 ( 10 ) ------2 –6 J BC m 4.021 ( 10 )

τ max = 43.9 MPa

The maximum torsional shear stress is The rotation at 1m from the wall of A can be found as shown below, 3 T AB ( x 1 – x A ) ( 5.586 ) ( 10 ) ( 1 ) φ 1 – φ A = --------------------------------- = -------------------------------------------------------------- = 6.86 ( 10 –3 ) rads 9 –6 G AB J AB ( 40 ) ( 10 ) ( 20.358 ) ( 10 ) Noting φ A = 0 ,we obtain

φ 1 = 0.0069 rads CCW

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.70 A solid steel (G = 80 GPa) is securely fastened to a solid bronze (G = 40 GPa) shaft that is 2 m long as shown. If the section at B rotates by 0.05 rads, determine (a) the maximum torsional shear strain in the shaft. (b) the applied torque T.

T A

Bronze

B

120 mm Steel 80 mm

2m 1m

Figure P5.70

Solution

G AB = 80GPa G BC = 40 GPa φ B = 0.05 rads

τ max = ?

T= ?

-----------------------------------------------------------The rate of twist in the segment AB and BC can be found as φB – φA rads ⎛ dφ ------⎞ = ------------------ = 0.05 ---------- = 0.025 ----------- and ⎝ dx⎠ AB m xB – xA 2

φC – φB – 0.05- = – 0.05 rads ⎛ dφ ------⎞ ----------= ------------------ = 0 -----------------⎝ dx⎠ BC m xC – xB 1

The maximum torsional shear strain in each segment can be found as: dφ ( γ AB ) max = ( ρ AB ) max ⎛ ------⎞ ⎝ dx⎠

–3

= ( 0.06 ) ( 0.025 ) = 1.5 ( 10 ) AB

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

–3 dφ ( γ BC ) max = ( ρ BC ) max ⎛ ------⎞ = ( 0.04 ) ( – 0.05 ) = – 2.0 ( 10 ) ⎝ dx⎠ BC

γ max = – 2000 μ rads

The maximum torsional shear strain is, Substituting φ B = 0.05 and φ A = 0 in Eqs. (4) in problem (5.59) we obtain 3 0.05 T A = ----------------------------- = 20.36 ( 10 )N – m –6 2.456 ( 10 )

From Eq.(6) of problem (5-59),we obtain, 3 TA 3 20.36 ( 10 ) T = --------------- = --------------------------- = 36.44 ( 10 )N – m 0.5586 0.5586

T = 36.44 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.71 Two shafts with shear modulus G1 = G and G2 = 2G are securely fastened at section B. In terms of T, L, G, and d, determine the magnitude maximum torsional shear stress in the shaft and the rotation of section at B. T A G1=G

B G2 = 2G

C

d

L 2.5 L

Figure P5.71

τmax = f(T, L, G, d) =? φB = g(T, L, G, d) =? ------------------------------------------------------------

Solution

The following free body diagrams can be drawn with internal torques drawn as per our sign convention. (a)

(b)

TAB TA

A G 1 =G

T TA

A Ste el

TBC

B G2 = 2 G

By moment equilibrium about the shaft axis we have the following equations T AB = T A

T BC = T A – T

(1)

The relative rotations of each segment can be written as shown below. TA ( L ) T AB ( x B – x A ) - = --------------φ B – φ A = ---------------------------------G AB J GJ

(2)

T BC ( x C – x B ) ( T A – T ) ( 2.5L ) 1.25 - = ------------------------------------φ C – φ B = --------------------------------- = ---------- ( T A – T )L GJ G BC J 2GJ

(3)

Adding Eq’s(2) and (3) and noting that φC=0 and φA=0, we obtain the following L φ C – φ A = ------- [ T A + 1.25 ( T A – T ) ] = 0 GJ

or

5TL From Eq.(2) we obtain φ B = ----------- CCW

1.25 5 T A = ------------------- T = --- T 1.25 + 1 9

(4)

TL φ B = 5.659 ---------- CCW 4 Gd

9GJ

Substituting Eq.(3) into Eq.(1), we obtain 5 T AB = --- T 9

4 T BC = – --- T 9

(5)

The maximum torsional shear stress will be in segment AB and can be found as shown below

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

80 T ( 5T ⁄ 9 ) ( d ⁄ 2 ) τ max = ---------------------------------- = ------ ----3 3 9π d ( πd ⁄ 32 )

January 2014

T τ max = 2.83 ----3 d

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.72 A uniform distributed torque of q in-lb/in is applied to the entire shaft as shown. In addition to the distributed torque a concentrated torque of T=3 q L in-lb is applied at section B. Let the shear modulus be G and radius of the shaft be r. Determine in terms of q, L, G, and r: (a) the rotation of section at B. (b) the magnitude maximum torsional shear stress in the shaft. .

3qL in-lb q in-lb/in

q in-lb/in

C

B

A L

2L

Figure P5.72

Solution

τmax = f(q, L, G, r) =? φB = g(q, L, G, r) =? ------------------------------------------------------------

The following free body diagrams can be drawn with internal torques drawn as per our sign convention. 3qL in-lb (a) q in-lb/in TAB (b) q in-lb/in

TA

q in-lb/in

TBC

TA B

By moment equilibrium about the shaft axis we have the following equations. T AB + qx – T A = 0

or

T BC + qx – T A – 3qL = 0

or

T AB = T A – qx T BC = T A – qx + 3qL

(1) (2)

( T A – qx ) T AB dφ = ---------= ------------------------ dx dx AB GJ GJ

The rate of twist of segment AB can be written as: ⎛⎝ ------⎞⎠ φB

Integrating from xAto xB we obtain

xB = L

∫ dφ

=

φA

∫

( T A – qx ) ------------------------ dx or GJ

xB = 0

1 q 2 φ B – φ A = ------- ⎛ T A x – --- x ⎞ GJ ⎝ 2 ⎠

L 0

q 2 1 = ------- ⎛ T A L – --- L ⎞ 2 ⎠ GJ ⎝

(3)

( T A – qx + 3qL ) T BC dφ - = ----------------------------------------- dx Similarly, the rate of twist of segment BC can be written as ⎛⎝ ------⎞⎠ = --------dx BC GJ GJ φC

Integrating from xB to xC we obtain

∫ φB

x C = 3L

dφ =

∫

( T A – qx + 3qL ) ----------------------------------------- dx or GJ

xB = L

1 q 2 φ c – φ B = ------- ⎛ T A x – --- x + 3qLx⎞ ⎠ GJ ⎝ 2

3L L

2 1 = ------- ( 2T A L + 2qL ) GJ

(4)

Adding Eqs. (3) and (4) and noting that φC=0 and φA=0, we obtain 2 L q 2 φ C – φ A = ------- ⎛ T A L – --- L + 2T A L + 2qL ⎞ = 0 ⎠ GJ ⎝ 2

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or

L T A = – q --2

(5)

5-55

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

4

2

1 q 2 q 2 – qL From Eq. (3) we obtain φ B = ------- ⎛⎝ – --- L – --- L ⎞⎠ = -----------------------or 4 GJ 2 2

2qL φ B = ------------- CW 4 πGr

G ( πr ⁄ 2 )

From Eq.(1)and (2), we obtain q T AB = – --- ⎛ x + L ---⎞ 2⎝ 2⎠

5 T BC = q ⎛ --- L – x⎞ ⎝2 ⎠

(6)

3 ( T BC ) max = --- qL 2

(7)

The maximum value of TAB and TBC is at x = L 3 ( T AB ) max = – --- qL 2

The maximum torsional shear stress will be just after B and can be found as shown below ⎛ 3--- qL⎞ r ( T BC ) max ( ρ BC ) max ⎝2 ⎠ τ max = ------------------------------------------------- = -------------------4 J BC ( πr ⁄ 2 )

3qL τ max = ---------3 πr

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.73 A steel shaft (Gst = 80 GPa) and a bronze shaft (Gbr = 40 GPa) are securely connected at B, as shown in Figure P5.73. The magnitude of maximum torsional shear stresses in steel and bronze are to be limited to 160 MPa and 60 MPa, respectively. Determine the maximum allowable torque T to the nearest N · m that can act on the shaft. T A

Bronze

B

120 mm Steel 80 mm

2m 1m

Figure P5.73

Solution G AB = 40GPa G BC = 80 GPa ( τ AB ) max ≤ 60 MPa

( τ BC ) max ≤ 160 MPa

Text=?

-----------------------------------------------------------The polar moment of cross - sections in various segments can be found as shown below 4 –6 4 π J AB = ------ ( 0.12 ) = 20.358 ( 10 ) m 32

(1)

4 –6 4 π J BC = ------ ( 0.08 ) = 4.021 ( 10 ) m 32

(2)

The following free body diagrams can be drawn. The internal torques are drawn as per our sign convention. T TAB (b) (a) TA A

TA A Bronze

Bronze

TBC

B

Steel

By moment equilibrium about the shaft axis we have the following equations. T AB = T A

T BC = T A – T

(3)

The relative rotation of the ends of the each segment can be written as shown below, T AB ( x B – x A ) TA ( 2 ) φ B – φ A = ---------------------------------- = -------------------------------------------------------------= 2.456 ( 10 –6 )T A 9 –6 G AB J AB ( 40 ) ( 10 ) ( 20.358 ) ( 10 )

(4)

T BC ( x C – x B ) ( T –TA ) ( 1 ) φ C – φ B = --------------------------------- = ----------------------------------------------------------= 3.109 ( 10 – 6 ) ( T A – T ) 9 –6 G BC J BC ( 80 ) ( 10 ) ( 4.021 ) ( 10 )

(5)

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

Adding Equations(4) and (5) and noting that φ C = 0 and φ A = 0 we obtain the following. –6 φ C – φ A = [ 2.456T A + 3.109 ( T A – T ) ] ( 10 ) = 0 or

From Eq.(3) T AB = 0.5586T

3.109T T A = --------------------------------- = 0.5586T 2.456 + 3.109 T BC = – 0.4414T

(6)

The magnitude of maximum torsional shear stress in each material must be less then the allowable value. T AB ( ρ AB ) max 6 ( 0.5586T ) ( 0.06 ) ( τ AB ) max = ----------------------------------= ----------------------------------------- ≤ 60 ( 10 ) –6 J AB 20.358 ( 10 )

T ≤ 36444.6 N-m

or

T BC ( ρ BC ) max 6 ( 0.4414T ) ( 0.04 -) ( τ BC ) max = ---------------------------------≤ 160 ( 10 ) - = ------------------------------------------–6 J BC 4.021 ( 10 )

T ≤ 36438.6 N-m

or

Thus, the maximum torque to the nearest N.m is:

T max = 36438 N-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.74 A steel (Gs = 80 GPa) and a bronze (GBr = 40 GPa) shaft are securely connected at B as shown. The maximum torsional shear stresses in steel and bronze are to be limited to 160 MPa and 60 MPa, respectively and the rotation of section B is limited to 0.05 rads. (a) Determine the maximum allowable torque T to the nearest kN-m that can act on shaft in Figure P5.74 if the diameter of the shaft is d = 100 mm. (b) What is the maximum torsional shear stress and maximum rotation in the shaft corresponding to the answer in part (a). T A Steel

B Bronze

C d

1.5 m 3m

Figure P5.74

Solution

GAB = 80 GPa GBC = 40 GPa

( τ AB ) max ≤ 160 MPa

φ B ≤ 0.05 rads

Tmax =? nearest kN

d = 100 mm

( τ BC ) max ≤ 60 MPa

τmax = ? φB =?

-----------------------------------------------------------The following free body diagrams can be drawn with internal torques drawn as per our sign convention. (a)

TAB TA

(b)

A Ste el

T TA

A Ste el

B

TBC Bronze

By moment equilibrium about the shaft axis we have the following equations. T AB = T A

T BC = T A – T

(1)

The relative rotation of the ends of each segment can be written as shown below. T AB ( x B – x A ) 1.5T A φ B – φ A = ---------------------------------- = --------------------9 G AB J 80 ( 10 )J

(2)

3 ( TA – T ) T BC ( x C – x B ) - = -----------------------φ C – φ B = --------------------------------9 G BC J 40 ( 10 )J

(3)

Adding Eq’s. (2) and (3) and noting that φC=0 and φA=0, we obtain the following

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

1 1.5 φ C – φ A = ---------------------- ------- T + 3 ( T A – T ) = 0 or 9 40 ( 10 )J 2 A 3 T A = ------------------- T = 0.8T 0.75 + 3

(4)

The polar moment of cross - section can be found as 4 –6 4 π J = ------ ( 0.1 ) = 9.817 ( 10 ) m 32

(5)

Substituting Eq.(4) and Eq.(5) into Eq.(2) we obtain –6 ( 1.5 ) ( 0.8T ) φ B = ----------------------------------------------------------- = 1.528T ( 10 ) ≤ 0.05 9 –6 ( 80 ) ( 10 ) ( 9.817 ) ( 10 )

or

(6)

3

T ≤ 32.72 ( 10 ) N – m

(7)

Substituting Eq.(4) into Eq.(1) we obtain T AB = 0.8T

T BC = – 0.2T

(8)

The magnitude of the maximum torsional shear stress in each segment can be written as 3 6 ( 0.8T ) ( 0.05 ) = ( τ AB ) max = ------------------------------4.074 ( 10 )T ≤ 160 ( 10 ) –6 9.817 ( 10 )

or

(9)

3

T ≤ 39.27 ( 10 ) N – m

(10)

3 6 ( 0.2T ) ( 0.05 ) ( τ BC ) max = -------------------------------- = 1.018 ( 10 )T ≤ 60 ( 10 ) –6 9.817 ( 10 )

or

(11)

3

T ≤ 58.90 ( 10 ) N – m

(12)

(103) N-

The maximum value of T that meets all the requirements of Eqs.(7),(10) and (12) is T = 32.72 m.To the nearest kN the maximum value is T max = 32 kN-m –3

Substituting T = 32 kN-m into Eq. (6) we obtain: φ B = 48.896 ( 10 ) rads

φ B = 0.048 rads CCW

Substituting T = 32 kN-m into Eqs. (9) and (11) we obtain the following: 6

2

6

( τ AB ) max = 130.37 ( 10 ) N ⁄ m and ( τ BC ) max = 32.6 ( 10 ) N ⁄ m

2

τ max = 130.4 MPa

The maximum torsional shear stress is in steel and has a value:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.75 A steel (Gs = 80 GPa) and a bronze (GBr = 40 GPa) shaft are securely connected at B as shown in Figure P5.75. The maximum torsional shear stresses in steel and bronze are to be limited to 160 MPa and 60 MPa, respectively and the rotation of section B is limited to 0.05 rads. Determine the minimum diameter ‘d’ of the shaft to the nearest millimeter, if the applied torque T = 20 kN-m. What is the maximum torsional shear stress and maximum rotation in the shaft corresponding to the answer in part (a).

T A Steel

B Bronze

C d

1.5 m 3m

Figure P5.75

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5-58

M. Vable

Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

GAB = 80 GPa

GBC = 40 GPa

( τ AB ) max ≤ 160 MPa

( τ BC ) max ≤ 60 MPa

φ B ≤ 0.05 rads

T = 20 kN-m

τmax = ? ------------------------------------------------------------

dmin =? nearest mm

January 2014

φB =?

T =? nearest kN

From Eq.(4) in problem 5-65 T A = 0.8T = 16 kN – m

(1)

From Eq.(2) in problem 5-65 3

–6 3.056 ( 1.5 ) ( 16 ) ( 10 ) φ B = ------------------------------------- = ------------- ( 10 ) ≤ 0.05 4 9 π 4 d ( 80 ) ( 10 ) ------ d 32

or

(2)

–3

d ≥ 88.417 ( 10 ) m

(3)

Substituting T = 16 k N-m in Eq.(1) of problem 5-65 T AB = 16 kN – m

T BC = 4kN – m

(4)

The magnitude of the maximum torsional shear stress in each segment can be written as 3 d ( 16 ) ( 10 ) ⎛ ---⎞ 3 ⎝ 2⎠ 81.487 ( 10 ) 6 ( τ AB ) max = --------------------------------- = ------------------------------ ≤ 160 ( 10 ) 3 4 d ( πd ⁄ 32 )

or

(5)

–3

d ≥ 79.86 ( 10 ) m

(6)

3 d ( 4 ) ( 10 ) ⎛ ---⎞ 3 ⎝ 2⎠ 6 20.37 ( 10 ) ( τ BC ) max = ------------------------------ = --------------------------- ≤ 60 ( 10 ) 3 4 d ( πd ⁄ 32 )

(7)

or

–3

d ≥ 69.76 ( 10 ) m

(8)

The minimum value of d that meets all the requirements of Eqs (3),(6) and (8) is d = 88.417 mm.To the nearest mm the minimum value is d min = 89 mm –6

–3 ( 10 ) Substituting d = 0.08 m into Eq.(2) we obtain: φ B = 3.056 ----------------------------- = 48.71 ( 10 ) or

( 0.089 )

4

φ B = 0.0487 rads CCW

Substituting d = 0.089 m in Eqs. (5) and (8), we obtain the following. 6

2

6

( τ AB ) max = 115.6 ( 10 ) N ⁄ m and ( τ BC ) max = 39.8 ( 10 ) N ⁄ m

The maximum torsional shear stress is in steel and has a value:

2

τ max = 116 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.76 The solid steel shaft shown in Figure P5.76 has a shear modulus of elasticity of G = 80 GPa and a allowable torsional shear stress of 60 MPa. The allowable rotation of any section is 0.03 rads. The applied torques on shaft are T1 = 10 kN-m and T2 = 25 kN-m. Determine (a) the minimum diameter ‘d’ of the shaft to the nearest millimeter. (b) the maximum torsional shear stress in the shaft and the maximum rotation of any section.

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5-59

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

T1

A

T2

B 1m

January 2014

d

D

C 2.5 m

1.5 m

Figure P5.76

τ max ≤ 60 MPa

SolutionG = 80 GPa

φ max ≤ 0.03 rads

T1 = 10 kN-m

τmax = ? ------------------------------------------------------------

T2 = 25 kN-m

φmax =?

dmin =? nearest mm

The following free body diagrams can be drawn. The internal torques are drawn as per our sign convention. TAB

(a)

T1

(b)

TA

TBC

T2

TCD

TA

TA

A

T1

(c)

A

B

A

B

C

By moment equilibrium about the shaft axis we have the following equations. T AB = T A

T BC = T A + T 1

T CD = T A + T 1 – T 2

(1)

The relative rotation of the ends of each segment can be written as shown below. T AB ( x B – x A ) ( 1 )T φ B – φ A = ---------------------------------- = --------------AGJ GJ

(2)

T BC ( x C – x B ) ( T A + T 1 ) ( 1.5 ) φ C – φ B = --------------------------------- = -----------------------------------GJ GJ

(3)

T CD ( x D – x C ) ( T A + T 1 – T 2 ) ( 2.5 ) φ D – φ C = ---------------------------------- = ------------------------------------------------GJ GJ

(4)

Adding Eq’s(2), (3)and (4) and noting that φA=0 and φD=0, we obtain 1 φ D – φ A = ------- [ T A + 1.5 ( T A + T 1 ) + 2.5 ( T A + T 1 – T 2 ) ] = 0 GJ

or

2.5T 2 – 4T 1 T A = ----------------------------= 0.5T 2 – 0.8T 1 5 (5)

From Eq. (1) we obtain T AB = 0.5T 2 – 0.8T 1 T BC = 0.5T 2 + 0.2T 1

(6)

T CD = – 0.5 T 2 + 0.2T 1

(8)

(7)

Substituting T1 = 10 kN-m and T2 = 25kN-m we obtain T AB = 4.5 kN – m

T BC = 14.5 kN – m

T CD = – 10.5 kN – m

From the values of the internal torques, we conclude that maximum torsional shear stress will be in the signet BC and the maximum rotation will be of the cross-section at C Substituting φD = 0 and φC = φmax in Eq.(4) we obtain 3

–6 –6 – 3.342 3.342 ( – 10.5 ) ( 10 ) ( 2.5 ) – φ C = --------------------------------------------- = ---------------- ( 10 ) or φ max = ------------- ( 10 ) ≤ 0.03 4 4 9 π 4 d d ( 80 ) ( 10 ) ------ d 32 d ≥ 0.1027 m

or

(9) (10)

The magnitude of the maximum torsional shear stress in segment BC can be written as

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

3 d ( 14.5 ) ( 10 ) ⎛ ---⎞ 3 ⎝ 2⎠ 73.847 ( 10 ) 6 τ max = -------------------------------------- = ------------------------------ ≤ 60 ( 10 ) 3 π 4 d ------ d 32 d ≥ 0.1072 m

January 2014

or

(11) (12)

The minimum value of d that meets all the requirements of Eqs (10) and (12) is d = 107.2mm.The nearest mm that meets the requirement is d min = 108 mm; Substituting d = 0.108 m in Eqs (9) and (10), we obtain the following. φ B = 0.025 rads CCW;

φ max = 0.0246 rads 6

τ max = 58.62 ( 10 ) N ⁄ m

2

τ max = 58.62 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.77 The solid steel shaft shown has a shear modulus of elasticity of G = 80 GPa and a allowable torsional shear stress of 60 MPa. The allowable rotation of any section is 0.03 rads.The diameter of the shaft is d = 80 mm. Determine the maximum values of the torques T1 and T2 that can be applied to the shaft. T1

A

T2

B 1m

d

D

C 2.5 m

1.5 m

Figure P5.77

Solution

G = 80 GPa T1 =?

τ max ≤ 60 MPa

T2 =?

φ max ≤ 0.03 rads

d = 80 mm

nearest kN-m

-----------------------------------------------------------The polar moment of cross - section can be found as 4 –6 4 π J = ------ ( 0.08 ) = 4.021 ( 10 ) m 32

(1)

From Eqs. (6),(7) and (8) in problem 5-67, it is clear that for positive values of T1 and T2 the maximum internal torque will be in the segment BC. Hence the maximum torsional shear stress will be in segment BC and can be written as shown below. ( 0.5T 2 + 0.2T 1 ) ( 0.04 ) 6 - = 60 ( 10 ) τ max = -----------------------------------------------------–6 4.021 ( 10 )

or

3

0.5T 2 + 0.2T 1 = 6.032 ( 10 ) N – m

(2)

The rotation of her section at B and C can be written as shown below using Eqs (6) and (8) and noting that φA=0 and φD=0, we obtain T AB ( 1 ) φ B – φ A = ----------------------------------------------------------9 –6 ( 80 ) ( 10 ) ( 4.021 ) ( 10 )

or φ B = ( 0.5T 2 – 0.8T 1 ) ( 3.109 ) ( 10 )

T CD ( 2.5 ) φ D – φ C = ----------------------------------------------------------9 –6 ( 80 ) ( 10 ) ( 4.021 ) ( 10 )

or φ C = – ( 0.2T 1 – 0.5 T 2 ) ( 7.772 ) ( 10 )

–6

–6

(3)

(4)

Assuming φB is greater in magnitude than φC, and φB is clockwise, we obtain –6

( 0.5T 2 – 0.8T 1 ) ( 3.109 ) ( 10 ) = – 0.03

3

or 0.5T 2 – 0.8T 1 = – 9.649 ( 10 )

(5)

Solving Eqs. (2) and (5), we obtain 3

T 1 = 15.68 ( 10 ) N – m

3

and T 2 = 5.792 ( 10 ) N – m

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5-61

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

Substituting Eq.(6) into Eq. (4), we obtain φC = 0.0019, which is less than 0.03, Hence our assumption that φB is greater in magnitude is correct, rounding downwards to the nearest kN, we obtain the maximum values of the torques T 1 = 15 kN-m and T 2 = 5.0 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.78 An aluminum tube and a copper tube are securely fastened to two rigid bars as shown. The bars force the tubes to rotate by equal angles. The tubes are 5 mm thick and 1.5 m long. The mean diameters of the aluminum tube and copper tube are 125 mm and 50 mm, respectively. The shear modulus for aluminum and copper are Gal = 28 GPa and Gcu = 40 GPa, respectively.Under the action of the applied couple the section B of the two tubes rotate by an angle of 0.03 rads. Determine (a) the magnitude maximum torsional shear stress in aluminum and copper. (b) the magnitude of the couple that produced the given rotation. aluminum A

F copper

B

F

Figure P5.78

Solution

t = 5 mm

d al = 125 mm d cu = 50 mm L = 1.5 m G al = 28 GPa G cu = 40 GPa φ 1 = 0.03 rads ( τ al ) max =? ( τ cu ) max =? T = Fd =?

-----------------------------------------------------------The polar moment of cross - section of each material can be found as shown below 4 4 –6 4 π J al = ------ [ ( 0.130 ) – ( 0.120 ) ] = 7.6820 ( 10 ) m 32

(1)

4 4 –6 4 π J cu = ------ [ ( 0.055 ) – ( 0.045 ) ] = 0.4958 ( 10 ) m 32

(2)

Tal and Tcu represents the internal torque on each material cross section. By equilibrium we have T al + T cu = T

(3)

The relative rotation of the ends of the each segment can be written as shown below, T al ( x B – x A ) T al ( 1.5 ) ( φ B ) al – ( φ A ) al = ------------------------------- = ----------------------------------------------------------= 6.973 ( 10 –6 )T al 9 –6 G al J al ( 28 ) ( 10 ) ( 7.682 ) ( 10 )

(4)

T cu ( x C – x B ) T cu ( 1.5 ) ( φ B ) cu – ( φ A ) cu = -------------------------------= -------------------------------------------------------------= 75.63 ( 10 – 6 )T cu 9 –6 G cu J cu ( 40 ) ( 10 ) ( 0.4958 ) ( 10 )

(5)

For both materials, φ A is zero and φ B = 0.03 .From Eqs. (4) and (5) we obtain the following, 3 0.03 T al = ----------------------------- = 4.302 ( 10 ) N – m –6 6.973 ( 10 )

(6)

3 0.03 T cu = ----------------------------- = 0.397 ( 10 ) N – m –6 75.63 ( 10 )

(7)

Substituting Eq (6) and(7) into Eq.(3) we obtain 3

3

T = ( 4.302 + 0.397 ) ( 10 ) = 4.699 ( 10 ) N – m

(8) T = 4.699 kN-m

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

The maximum torsional shear stress in each material can be found as shown below 3 T al ( ρ al ) max 6 2 ( 4.302 ) ( 10 ) ( 0.065 ) ( τ al ) max = ---------------------------- = -------------------------------------------------- = 36.4 ( 10 ) N ⁄ m – 6 J al 7.682 ( 10 )

(9)

( τ al ) max = 36.4 MPa 3 T cu ( ρ cu ) max 6 2 ( 0.397 ) ( 10 ) ( 0.0275 ) ( τ cu ) max = ------------------------------ = ----------------------------------------------------- = 22.0 ( 10 ) N ⁄ m –6 J cu 0.4958 ( 10 )

(10)

( τ cu ) max = 22.0 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.79

Solve Problem 5.78 using Equations (5.18a) and (5.18b).

Solution

t = 5 mm

d al = 125 mm

d cu = 50 mm

L = 1.5 m

φ 1 = 0.03 rads

G al = 28 GPa

G cu = 40 GPa

( τ al ) max =?

( τ cu ) max =?

T = Fd =?

-----------------------------------------------------------The polar moment of cross - section of each material can be found as shown below 4 4 –6 4 π J al = ------ [ ( 0.130 ) – ( 0.120 ) ] = 7.6820 ( 10 ) m 32

(1)

4 4 –6 4 π J cu = ------ [ ( 0.055 ) – ( 0.045 ) ] = 0.4958 ( 10 ) m 32

(2)

The torsional rigidity of the cross section is 9

–6

9

–6

3

ΣG j J j = ( 28 ) ( 10 )7.6820 ( 10 ) + ( 40 ) ( 10 ) ( 0.4958 ) ( 10 ) = 234.93 ( 10 ) N – m

2

(3)

The relative rotation of the ends of the shaft can be written as T AB ( x B – x A ) 3 T ( 1.5 ) φ B – φ A = ---------------------------------- = ------------------------------ = 0.03 or T = 4.669 ( 10 ) N – m 3 ΣG j J j 234.93 ( 10 )

T = 4.7 kN-m

The torsional shear stress at any point on the cross section can be found as shown below 3 T AB ρ 4.669 ( 10 ) τ xθ = G i -------------- = ------------------------------ G i ρ = 0.02G i ρ 3 ΣG j J j 234.93 ( 10 )

(4)

The maximum torsional shear stress in copper will be at ρ = ( d cu ⁄ 2 + t ⁄ 2 ) = 0.0275 m . We obtain: 9

6

( τ cu ) max = ( 0.02 ) ( 40 ) ( 10 ) ( 0.0275 ) = 22 ( 10 ) N ⁄ m

2

( τ cu ) max = 22 MPa

The maximum torsional shear stress in aluminum will be at ρ = ( d al ⁄ 2 + t ⁄ 2 ) = 0.065 m . We obtain: 9

6

( τ al ) max = ( 0.02 ) ( 28 ) ( 10 ) ( 0.065 ) = 36.4 ( 10 ) N ⁄ m

2

( τ al ) max = 36.4 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.80 An aluminum tube and a copper tube are securely fastened to two rigid bars as shown. The bars force the tubes to rotate by equal angles. The tubes are 5 mm thick and 1.5 m long. The mean diameters of the aluminum tube and copper tube are 125 mm and 50 mm, respectively. The shear modulus for aluminum and copper are Gal = 28 GPa and Gcu = 40 GPa, respectively.The applied couple on the tubes shown is 10 kN-m. Determine (a) the magnitude maximum torsional shear stress in aluminum and copper. (b) the rotation of section at B.

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5-63

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

aluminum

January 2014

F

A

copper B

F

Figure P5.80

Solution

t = 5 mm

d al = 125 mm

G cu = 40 GPa

d cu = 50 mm

T = Fd = 10kN – m

L = 1.5 m

G al = 28 GPa

( τ al ) max =?

( τ cu ) max =?

φ B =?

-----------------------------------------------------------The polar moment of cross - section of each material can be found as shown below 4 4 –6 4 π J al = ------ [ ( 0.130 ) – ( 0.120 ) ] = 7.6820 ( 10 ) m 32

(1)

4 4 –6 4 π J cu = ------ [ ( 0.055 ) – ( 0.045 ) ] = 0.4958 ( 10 ) m 32

(2)

The relative rotation of the ends of the each segment can be written as shown below, T al ( x B – x A ) T al ( 1.5 ) ( φ B ) al – ( φ A ) al = ------------------------------- = ----------------------------------------------------------= 6.973 ( 10 –6 )T al 9 –6 G al J al ( 28 ) ( 10 ) ( 7.682 ) ( 10 )

(3)

T cu ( x C – x B ) T cu ( 1.5 ) ( φ B ) cu – ( φ A ) cu = -------------------------------= -------------------------------------------------------------= 75.63 ( 10 – 6 )T cu 9 –6 G cu J cu ( 40 ) ( 10 ) ( 0.4958 ) ( 10 )

(4)

As φAis zero for both material and φB is same for both material, we can equate Eq.(3) and Eq.(4) to obtain –6

–6

6.973 ( 10 )T al = 75.63 ( 10 )T cu or T al = 10.846T cu

(5)

By equilibrium we have 3

T al + T cu = 10 ( 10 ) N – m

(6)

Substituting Eq.(5) into Eq.(6) we obtain 3

( 10.846 + 1 )T cu = 10 ( 10 )

(7)

3

T cu = 0.8442 ( 10 ) N – m

(8)

3

T al = 9.1558 ( 10 ) N – m

(9)

We obtain: –6

–6

–3

φ B = ( 6.973 ) ( 10 ) ( 9.1558 ) ( 10 ) = 63.85 ( 10 )

(10) φ B = 0.064 rads CCW

The maximum torsional shear stress in each material can be found as shown below 3

6 2 ( 9.1558 ) ( 10 ) ( 0.065 ) ( τ al ) max = ----------------------------------------------------- = 77.47 ( 10 ) N ⁄ m –6 7.682 ( 10 )

(11) ( τ al ) max = 77.5 MPa

3

6 2 ( 0.8442 ) ( 10 ) ( 0.065 ) ( τ cu ) max = ----------------------------------------------------- = 46.8 ( 10 ) N ⁄ m –6 0.4958 ( 10 )

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5-64

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

January 2014

( τ cu ) max = 46.8 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.81

Solve Problem 5.80 using Equations (5.18a) and (5.18b).

Solution

t = 5 mm

d al = 125 mm

G cu = 40 GPa

d cu = 50 mm

T = Fd = 10kN – m

L = 1.5 m

G al = 28 GPa

( τ al ) max =?

( τ cu ) max =?

φ B =?

-----------------------------------------------------------The polar moment of cross - section of each material can be found as shown below 4 4 –6 4 π J al = ------ [ ( 0.130 ) – ( 0.120 ) ] = 7.6820 ( 10 ) m 32

(1)

4 4 –6 4 π J cu = ------ [ ( 0.055 ) – ( 0.045 ) ] = 0.4958 ( 10 ) m 32

(2)

The torsional rigidity of the cross section is 9

–6

9

–6

3

ΣG j J j = ( 28 ) ( 10 )7.6820 ( 10 ) + ( 40 ) ( 10 ) ( 0.4958 ) ( 10 ) = 234.93 ( 10 ) N – m

2

(3)

The relative rotation of the ends of the shaft can be written as 3 T AB ( x B – x A ) –3 ( 10 ) ( 10 ) ( 1.5 ) φ B – φ A = ---------------------------------- = ------------------------------------- = 63.85 ( 10 ) 3 ΣG j J j 234.93 ( 10 )

Noting that φA=0, we obtain

φ B = 0.0639 rads CCW;

The torsional shear stress at any point on the cross section can be found as shown below 3 T AB ρ ( 10 ) ( 10 ) τ xθ = G i -------------- = ------------------------------ G i ρ = 0.0426G i ρ 3 ΣG j J j 234.93 ( 10 )

(4)

The maximum torsional shear stress in copper will be at ρ = ( d cu ⁄ 2 + t ) = 0.03 m . We obtain: 9 6 N ( τ cu ) max = ( 0.0426 ) ( 40 ) ( 10 ) ( 0.0275 ) = 46.86 ( 10 ) ------2 m

(5) ( τ cu ) max = 46.9 MPa

The maximum torsional shear stress in aluminum will be at ρ = ( d al ⁄ 2 + t ) = 0.0675 m . We obtain: 9 6 N ( τ al ) max = ( 0.0426 ) ( 28 ) ( 10 ) ( 0.065 ) = 77.53 ( 10 ) ------2 m

(6) ( τ al ) max = 77.5 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.82 Solve Example 5.14 using Equations (5.18a) and (5.18b). Solution

-----------------------------------------------------------We can find the polar moments and torsional rigidities as shown below: 4 –6 4 π J S = ------ ( 0.08 ) = 4.02 ( 10 ) m 32 3

G S J S = 321.6 ( 10 ) N ⋅ m 2

∑ Gj Jj

4 4 –6 4 π J Br = ------ ( 0.12 – 0.08 ) = 16.33 ( 10 ) m 32 2

3

G Br J Br = 653.2 ( 10 ) N ⋅ m 3

= G S J S + G Br J Br = 974.8 ( 10 ) N ⋅ m

2

2

(1) (2) (3)

j=1

The internal torque in AB is

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T AB = 75 kN – m

(4)

We can find the relative rotation of section B with respect to section at A. φ B – φ A = T AB ( x B – x A ) ⁄

2

∑ j=1

3

( 75 ) ( 10 ) ( 2 ) G j J j = -------------------------------- = 0.1538 rad 3 974.8 ( 10 )

(5)

φ B – φ A = 0.1538 rads

We can find the shear stress at any point on the imaginary cross-section in AB as shown below: 2

∑

τ i = ( G i ρT AB ) ⁄

j=1

3

G i ρ ( 75 ) ( 10 ) –3 G j J j = --------------------------------- = ( 76.9 ) ( 10 )G i ρ 3 974.8 ( 10 )

(6)

In AB, shear stress in steel (τs) will be maximum at ρ = 0.04 and the shear stress in bronze (τBr) will be maximum will at ρ = 0.06. Substituting the values of G and ρ, the maximum shear stress in each material in section AB can be found as shown below: –3

9

6

( τ Br )max = ( 76.9 ) ( 10 ) ( 40 ) ( 10 ) ( 0.06 ) = 184.6 ( 10 ) –3

9

6

( τ s ) max = ( 76.9 ) ( 10 ) ( 80 ) ( 10 ) ( 0.04 ) = 246.2 ( 10 ) τ max = 246.2 MPa

We see that the maximum shear stress is:

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.83 The composite shaft shown in Figure P5.83 is constructed from aluminum (Gal = 4000 ksi), bronze (Gbr = 6000 ksi), and steel (Gst = 12,000 ksi). Determine (a) the rotation of the free end with respect to the wall. (b) the maximum torsional shear stress in each material.

30 in-kips

A B 25 in ches

1 in 1 .5 in 2 in

Aluminum Bronze Steel

Figure P5.83

Solution

Gal = 4000 ksi Gbr = 6000 ksi Gst = 12,000 ksi

-----------------------------------------------------------The polar moment of each material cross - section can be found as shown below. 4 4 π 4 J s = ------ ( 2 – 1.5 ) = 1.0738 in 32

4 4 4 π J br = ------ ( 1.5 – 1 ) = 0.3988 in 32

4 π 4 J al = ------ ( 1 ) = 0.0982 in (1) 32

The relative rotation of B with respect to A for each material can be written as: T al ( x B – x A ) T al ( 25 ) ( φ B ) al – ( φ A ) al = ------------------------------- = ------------------------------------- = 0.06365T al G al J al ( 4000 ) ( 0.0982 )

(2)

T br ( x B – x A ) T br ( 25 ) ( φ B ) br – ( φ A ) br = ------------------------------- = ------------------------------------- = 0.01045T br G br J br ( 6000 ) ( 0.3988 )

(3)

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Ts ( xB – xA ) T s ( 25 ) ( φ B ) s – ( φ A ) s = ----------------------------- = ---------------------------------------- = 0.00194T s Gs Js ( 12000 ) ( 1.0738 )

(4)

Equating Equations (2), (3), and (4) we obtain: 0.06365 T br = ------------------- T al = 6.0916T al 0.01045

(5)

0.06365 T s = ------------------- T al = 32.8045T al 0.00194

(6)

T al + T br + T s = 30

(7)

By equilibrium we have: Substituting Equations (5) and (6) into Equation (7) we obtain: 30 T al = --------------------------------------------------- = 0.7519 in – kips 1 + 6.0916 + 32.8045 T br = 6.0916 ( 0.7519 ) = 4.5806 in – kips

(8) (9)

T s = 32.8045 ( 0.7519 ) = 24.6674 in – kips

(10)

From Equation (2) we obtain: ( φ B ) al – ( φ A ) al = 0.06365 ( 0.7519 ) = 0.04786 φ B = 0.0479 rads CCW

The maximum torsional shear stress in each material is: ( 0.7519 ) ( 0.5 ) ( τ al ) max = ---------------------------------- = 3.83 ksi 0.0982

(11)

( 4.5806 ) ( 0.75 -) = 8.61 ksi ( τ br ) max = -----------------------------------0.3988

(12)

( 24.6674 ) ( 1 -) = 22.97 ksi ( τ s ) max = ------------------------------1.0738 ( τ al ) max = 3.83 ksi ( τ br ) max = 8.61 ksi

(13) ( τ s ) max = 22.97 ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.84 Solve Problem 5.83 using Equations (5.18a) and (5.18b). Solution Gal = 4000 ksi Gbr = 6000 ksi Gs = 12000 ksi φB - φA =? γxθ vs. ρ =? τxθ vs. ρ =?

-----------------------------------------------------------The polar moment of each material cross - section can be found as shown below. 4 4 π 4 J s = ------ ( 2 – 1.5 ) = 1.0738 in 32

4 4 4 π J br = ------ ( 1.5 – 1 ) = 0.3988 in 32

4 π 4 J al = ------ ( 1 ) = 0.0982 in (1) 32

The torsional rigidity of the cross-section is 3

ΣG j J j = ( 4000 ) ( 0.0982 ) + ( 6000 ) ( 0.3988 ) + ( 12000 ) ( 1.0738 ) = 15.671 ( 10 ) kips – in

2

The relative rotation of the ends of the shaft can be written as T AB ( x B – x A ) –3 ( 30 ) ( 25 ) = φ B – φ A = ---------------------------------- = ----------------------------47.861 ( 10 ) 3 ΣG j J j 15.671 ( 10 )

Noting that φA=0, we obtain

(2)

φ B = 0.0479 rads CCW

The torsional shear stress and shear strain at any point on the cross section can be found as T AB ρ –3 τ xθ = G i -------------- = 1.914 ( 10 )G i ρ ΣG J

(3)

j j

The maximum torsional shear stress in each material is

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January 2014

–3

( τ Al ) max = 1.914 ( 10 ) ( 4000 ) ( 0.5 ) = 3.828 ksi

(4)

–3

( τ br ) max = 1.914 ( 10 ) ( 6000 ) ( 0.75 ) = 8.613 ksi

(5)

–3

( τ s ) max = 1.914 ( 10 ) ( 12000 ) ( 1 ) = 22.968 ksi ( τ Al ) max = 3.83 ksi

(6)

( τ br ) max = 8.61 ksi

( τ s ) max = 22.97 ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.85 If T = 1500 N · m in Figure P5.85, determine: (a) the magnitude of maximum torsional shear stress in cast iron and copper; (b) the rotation of the section at D with respect to the section at A.

T Cast Iron

T A

Copper

B

C

500 mm

D 400 mm

150 mm

Figure P5.85

Solution

G ir = 70 GPa

G cu = 40 GPa

( d ir ) o = 70 mm ( d ir ) i = 50 mm

( d cu ) o = 50 mm

( d cu ) i = 30 mm

T = 1500 N – m

( τ ir ) max = ?

( τ cu ) max =?

φ D – φ A =?

-----------------------------------------------------------The polar moment of each material cross - section can be found as shown below 4 4 –6 4 π J ir = ------ [ ( 0.07 ) – ( 0.05 ) ] = 1.7436 ( 10 )m 32 4 4 –6 4 π J cu = ------ [ ( 0.05 ) – ( 0.03 ) ] = 0.5341 ( 10 )m 32

1 2

The torsional rigidity of the cross section in segment BC is 9

–6

9

–6

3

ΣG j J j = ( 70 ) ( 10 ) ( 1.7436 ) ( 10 ) + ( 40 ) ( 10 ) ( 0.5341 ) ( 10 ) = 143.41 ( 10 ) N – m

2

The internal torque in all the segments is equal to -T.The relative rotation of the ends of each segment can be written as T AB ( x B – x A ) –3 ( – 1500 ) ( 0.5 ) φ B – φ A = ---------------------------------- = -------------------------------------------------------------- = – 6.145 ( 10 ) 9 –6 G ir J ir ( 70 ) ( 10 ) ( 1.7436 ) ( 10 )

3

T BC ( x C – x B ) –3 ( – 1500 ) ( 0.15 ) φ C – φ B = --------------------------------- = ----------------------------------- = – 1.569 ( 10 ) –3 ΣG j J j ( 143.4 ) ( 10 )

4

T CD ( x D – x C ) –3 ( – 1500 ) ( 0.4 ) φ D – φ C = ---------------------------------- = -------------------------------------------------------------= – 28.08 ( 10 ) 9 –6 G cu J cu ( 40 ) ( 10 ) ( 0.5341 ) ( 10 )

5

Adding Eqs. (3),(4) and (5), we obtain –3

–3

φ D – φ A = – ( 0.145 + 1.569 + 28.08 ) ( 10 ) = 35.798 ( 10 )

φ D – φ A = 0.0358 rads CCW

The magnitude of maximum torsional shear stress in segment AB and CD can be found as shown below T AB ( ρ AB ) max N – 1500 ) ( 0.035 ) = 30.1 ( 10 6 ) -----( τ AB ) max = ----------------------------------= (------------------------------------2 –6 J ir m 1.7436 ( 10 )

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T CD ( ρ CD ) max N – 1500 ) ( 0.025 ) = 70.2 ( 10 6 ) -----( τ CD ) max = ----------------------------------= (------------------------------------2 – 6 J cu m 0.5341 ( 10 )

7

The magnitude of torsional shear stress at any point in segment BC is T BC ρ –3 ( – 1500 ) τ xθ = G i -------------- = ------------------------------ G i ρ = 10.46 ( 10 )G i ρ 3 ΣG j J j 143.41 ( 10 )

8

At ρ=0.035, Gi=Gir, we obtain τ xθ = 25.62 ( 10 ) which is less than in Eqs. (6). Thus, the maximum tor–6

( τ ir ) max = 30.1 MPa

sional shear stress in iron is

At ρ=0.025, Gi=Gcu, we obtain τ xθ = 10.46 ( 10 ) which is less than in Eq.(7). Thus, the maximum tor–6

sional shear stress in copper is

( τ cu ) max = 70.2 MPa

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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5.86 Two steel (G = 80 GPa) shafts AB and CD of diameters 40 mm are connected with gears as shown in Figure P5.86. The radii of gears at B and C are 250 mm and 200 mm, respectively. The bearings at E and F offer no torsional resistance to the shafts. If an input torque of T = 1.5 kN.m is applied at D, determine (a) the maximum torsional shear stress in AB; (b) the rotation of section at D with respect to the fixed section at A. 1.5 m

T D

C E

A

B F

1.2 m

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Figure P5.86

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January 2014

5.87 Two steel (G = 80 GPa) shafts AB and CD of diameters 40 mm are connected with gears as shown in Figure P5.88. The radii of gears at B and C are 250 mm and 200 mm, respectively. The bearings at E and F offer no torsional resistance to the shafts. The allowable shear stress in the shafts is 120 MPa. Determine the maximum torque T that can be applied at section D. 1.5 m

T D

C E

A

B F

Figure P5.87

1.2 m

5.88 Two steel (G = 80 GPa) shafts AB and CDE of 1.5 in. diameters are connected with gears as shown in Figure P5.88. The radii of gears at B and D are 9 in. and 5 in., respectively. The bearings at F, G, and H offer no torsional resistance to the shafts. If an input torque of T = 800 ft.lb is applied at D, determine (a) the maximum torsional shear stress in AB, and (b) the rotation of section at E with respect to the fixed section at C. 5 ft T C

D H

G

A

E

B F

4 ft

Figure P5.88

5.89 Two steel (G = 80 GPa) shafts AB and CD of 60 mm diameters are connected with gears as shown in Figure P5.89. The radii of gears at B and D are 175 mm and 125 mm, respectively. The bearings at E and F offer no torsional resistance to the shafts. If an input torque of T = 2 kN.m is applied, determine (a) the maximum torsional shear stress in AB, and (b) the rotation of section at D with respect to the fixed section

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at C. T

C

D F

A

B E

1.5 m

Figure P5.89

5.90 Two steel (G = 80 GPa) shafts AB and CD of 60 mm diameters are connected with gears as shown in Figure P5.90. The radii of gears at B and D are 175 mm and 125 mm, respectively. The bearings at E and F offer no torsional resistance to the shafts. The allowable shear stress in the shafts is 120 MPa. What is the maximum torque T that can be applied? T

C

D F

A

B E

1.5 m

Figure P5.90

5.91 Two steel (G = 80 GPa) shafts AB and CD of equal diameters d are connected with gears as shown in Figure P5.89. The radii of gears at B and D are 175 mm and 125 mm, respectively. The bearings at E and F offer no torsional resistance to the shafts. The allowable shear stress in the shafts is 120 MPa and the

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input torque is T = 2 kN.m. Determine the minimum diameter d to the nearest millimeter. T

C

D F

A

B E

1.5 m

Figure P5.91

5.92 The allowable shear stress in the stepped shaft shown is 17 ksi. Determine the smallest fillet radius that can be used at section B. Use stress concentration graphs given in Section C.4.3. 2 in 1 in T = 2.5 in-kips

A

B

C

Figure P5.92

Solution

τ max ≤ 17 ksi

r =?

-----------------------------------------------------------The maximum shear stress in section BC can be found as, Tρ max ( 2.5 ) ( 0.5 ) ( τ BC ) max = --------------- = ------------------------- = 12.73 ksi 4 J BC π ( 1 ) ⁄ 32 τ max = K conc ( τ BC ) max = 12.73K conc ≤ 17 or K conc ≤ 1.335

(9)

From Section C.4.3, we obtain the approximate value of r/d corresponding to D/d = 2 and Kconc = 1.335, r --- = 0.135 d

r = ( 0.135 ) ( 1 )

or

r = 0.135 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.93 The fillet radius in the stepped shaft shown is 6 mm. Determine the maximum torque that can act on the rigid wheel if the allowable shear stress is 80 MPa and a Modulus of Rigidity is 28 GPa. Use stress concentration graphs given in Section C.4.3. 60 mm

T

48 mm A

0.9 m

D

C

B

0.75 m

1.0 m

Figure P5.93

Solution

r = 6 mm

τ max ≤ 80 MPa

G = 28 MPa

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Tmax =?

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-----------------------------------------------------------The polar moment of cross-sections in various segments can be found as 4 –6 4 π J AB = ------ ( 0.06 ) = 1.272 ( 10 ) m 32

(1)

4 –6 4 π J BC = J CD = ------ ( 0.048 ) = 0.521 ( 10 ) m 32

(2)

The following free body diagrams can be drawn. The internal torques are drawn as per our sign convention. (a) (b) (c) TAB T T TBC

TA

TA

A

A

CD

TA A

B

C

B

By moment equilibrium about the shaft axis we have the following equations. T AB = T A

T BC = T A

T CD = T A – T

(3)

The relative rotation of the ends of each segment can be written as shown below T AB ( x B – x A ) T A ( 0.9 ) –6 φ B – φ A = ---------------------------------- = ----------------------------------------------------------= 25.26 ( 10 )T A 9 –6 G AB J AB ( 28 ) ( 10 ) ( 1.272 ( 10 ) )

(4)

T BC ( x C – x B ) ( T A ) ( 0.75 ) –6 φ C – φ B = --------------------------------- = ----------------------------------------------------------= 51.40 ( 10 )T A 9 –6 G BC J BC ( 28 ) ( 10 ) ( 0.521 ) ( 10 )

(5)

T CD ( x D – x C ) ( TA – T ) ( 1 ) –6 φ D – φ C = ---------------------------------- = ----------------------------------------------------------= 68.53 ( 10 ) ( T A – T ) 9 –6 G CD J CD ( 28 ) ( 10 ) ( 0.521 ) ( 10 )

(6)

Adding Eq’s(5) and (6) and noting that φA=0 and φD=0, we obtain φ D – φ A = [ 25.26T A + 51.40T A + 68.53 ( T A – T ) ] = 0

From Eq.(3), T AB = 0.472T

T BC = 0.472T

68.53T T A = ---------------------------------------------------- = 0.472T 25.26 + 51.40 + 68.53 T CD = – 0.528T or

The nominal torsional shear stresses in segment BC can be found as shown below T BC ( ρ BC ) max 3 ( 0.472T ) ( 0.04 ) ( τ BC ) max = ---------------------------------- = -------------------------------------- = 21.74T ( 10 ) –6 J BC 0.521 ( 10 ) 6 - = 0.125 Now --r- = ----d

48

and

D 60 ---- = ------ = 1.25 .From Section C.4.3, we obtain the stress concentrad 48

tion factor as Kconc = 1.29.Thus, the maximum shear stress in segment BC is 3

6

( τ max ) BC = 1.29 ( τ BC ) max = ( 1.29 ) ( 21.74 )T ( 10 ) ≤ 80 ( 10 )

or

3

T ≤ 2.853 ( 10 ) N – m

(7)

The magnitude of maximum shear stress in CD is T CD ( ρ CD ) max 3 6 3 ( 0.528T ) ( 0.024 ) ( τ CD ) max = ----------------------------------= ----------------------------------------- = 24.32T ( 10 ) ≤ 80 ( 10 ) or T ≤ 3.29 ( 10 ) N – m – 6 J CD 0.521 ( 10 )

The maximum value of T that meets the requirements in Eq.(7) and (8) is

(8)

T max = 2.853 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.94 Calculate the magnitude of the maximum torsional shear stress in the cross-sections shown that is subjected to a torque of T=100 in-kips

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t = 1/4 in R=2 in

R=2 in

4 in

Figure P5.94

Solution

T=100 in-kips

τmax = ?

-----------------------------------------------------------The enclosed area is: 2 1 2 A E = 2 ⎛ --- π2 ⎞ + ( 4 ) ( 4 ) = 28.566 in ⎝2 ⎠

As thickness is uniform the torsional shear stress is uniform across the cross-section and maximum value of it can be found as shown below. T = --------------------------------------100 τ max = -----------= 7.001 ksi or 2tA E 2 ( 1 ⁄ 4 ) ( 28.566 ) τ max = 7.0 ksi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.95 Calculate the maximum torsional shear stress in the cross-sections shown that is subjected to a torque of T = 900 N-m. t = 3 mm R=50 mm

t = 6 mm

t = 5 mm

100 mm

Figure P5.95

Solution

T = 900 N-m

τmax = ?

-----------------------------------------------------------The enclosed area is: 2 2 1 A E = --- π ( 0.05 ) + ( 0.1 ) ( 0.1 ) = 0.0139 m 2

The maximum torsional shear stress will be where the thickness is minimum, that is t = 3 mm = 0.003 m. Its value can be found as shown below. 6 2 T 900 τ max = ------------ = ------------------------------------------ = 10.77 ( 10 )N ⁄ m or 2tA E 2 ( 0.003 ) ( 0.0139 )

τ max = 10.8 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.96 Calculate the magnitude of the maximum torsional shear stress in the cross-sections shown that is subjected to a torque of T=15 kN-m t = 6 mm 100 mm

100 mm

Figure P5.96

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Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

T = 15 kN-m

January 2014

τmax = ?

-----------------------------------------------------------The enclosed area is: 2 –3 2 1 A E = --- π ( 0.1 ) = 7.854 ( 10 ) m 4

(1)

As thickness is uniform, the torsional shear stress is uniform across the cross-section and maximum value of it can be found as shown below. 3

6 2 15 ( 10 ) T τ max = ------------ = -------------------------------------------------------- = 159.1 ( 10 )N ⁄ m or – 3 2tA E 2 ( 0.006 ) ( 7.854 ) ( 10 )

τ max = 159 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.97 A tube of uniform thickness t has a torque T applied to it. The cross-section of the tube is as shown. Determine the maximum torsional shear stress in terms of t, a, and T.

600

600 a

Figure P5.97

Solution

τmax =f(t, a, T)= ?

-----------------------------------------------------------The enclosed area is: 1 a 3 2 A E = --- ( a ) ⎛ --- tan 60⎞ = ------- a ⎠ 2 ⎝2 4

(1)

As thickness is uniform the torsional shear stress is uniform across the cross-section and maximum value of it can be found as shown below T T τ max = -----------= ---------------------------- or 2tA E 3 2 2 ( t ) ⎛ ------- a ⎞ ⎝ 4 ⎠

(2)

2 T τ max = ------- ------3 a2 t

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.98 A tube of uniform thickness t has a torque T applied to it. The cross-section of the tube is as shown. Determine the maximum torsional shear stress in terms of t, a, and T.

a a Figure P5.98

Solution

τmax =f(t, a, T)= ?

-----------------------------------------------------------The enclosed area is: AE = ( a ) ( a ) = a

2

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(1)

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January 2014

As thickness is uniform the torsional shear stress is uniform across the cross-section and maximum value of it can be found as shown below T T τ max = ------------ = -------------------- or 2 2tA E 2( t)(a )

(2) T τ max = --------2 2a t

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.99 A tube of uniform thickness t has a torque T applied to it. The cross-section of the tube is as shown. Determine the maximum torsional shear stress in terms of t, a, and T.

a

Figure P5.99

Solution

τmax =f(t, a, T)= ?

-----------------------------------------------------------The enclosed area is: A E = πa

2

(3)

As thickness is uniform the torsional shear stress is uniform across the cross-section and maximum value of it can be found as shown below T T τ max = -----------= ------------------------ or 2 2tA E 2 ( t ) ( πa )

(4) T τ max = ------------2 2πa t

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.100 A tube of uniform thickness t has a torque T applied to it. Determine the maximum torsional shear stress in terms of t, a, b, and T.

b a Figure P5.100

Solution

τmax =f(t, a, b, T)= ?

-----------------------------------------------------------The enclosed area is: A E = πab

(1)

As thickness is uniform the torsional shear stress is uniform across the cross-section and maximum value of it can be found as shown below T T τ max = ------------ = ------------------------ or 2tA E 2 ( t ) ( πab )

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T τ max = --------------2πabt

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.101 A hexagonal tube of uniform thickness is loaded as shown. Determine the magnitude of the maximum torsional shear stress in the tube. T4 = 750 N-m T3 = 1500 N-m A T2 = 3250 N-m B T1 = 1000 N-m C D

x

100 mm t = 4 mm

Figure P5.101

Solution

τmax = ?

-----------------------------------------------------------The torque diagram can be drawn using the template and template equation shown. (a) Template

T1

(b) Torque Diagram Text

T N-m

T2

1000

1000 x

A 750

B

C

D

750

T 2 = T 1 + T ext

2250

2250

The maximum internal torque is in segment AB and its value is Tmax = -2250 N-m. The enclosed area is equal to the area of 6 equilateral triangles with sides of 0.1 m. Thus the total enclosed area is: 1 0.1 A E = ( 6 ) --- ( 0.1 ) ⎛ ------- sin 60⎞ ⎝ 2 ⎠ 2

–3

= 12.99 ( 10 ) m

2

(1)

As thickness is uniform the torsional shear stress is uniform across the cross-section and the maximum value of it can be found as shown below. T max 6 2 ( – 2250 ) τ max = -----------= -------------------------------------------------------- = – 21.65 ( 10 )N ⁄ m or –3 2tA E 2 ( 0.004 ) ( 12.99 ) ( 10 )

(2)

τ max = 21.65 MPa

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.102

A rectangular tube is loaded as shown. Determine the maximum shear stress T4 = 2 in-kips

A

B

T3 = 3 in-kips

T1 =2 in-kips

C

T2 =7 in-kips

D

6 in

1/4 in 1/8 in 4 in

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Figure P5.102

Solution

τmax = ?

-----------------------------------------------------------The torque diagram can be drawn using the template and template equation shown. (b) Torque Diagram

Text

(a) Template T1

T2

T 2 in-kips A

5

5

B

C

2 x

T 2 = T 1 + T ext

D 2

2

The maximum internal torque is in segment AB and its value is Tmax = 5 in-kips. The enclosed area is: A E = ( 4 ) ( 6 ) = 24 in

2

(1)

The maximum torsional shear stress will be where the thickness is minimum, that is t = 1/8 in. Its value can be found as shown below. T max ( 5 ) - = 0.833 ksi or τ max = -----------= ---------------------------2 ( 1 ⁄ 8 ) ( 24 ) 2tA E

(2) τ max = 833 psi

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------5.103 The three tubes shown in Problems 5.97 through 5.99 are to be compared for the maximum torque carrying capability, assuming that all tubes have the same thickness t, that the maximum shear stress in each tube can be τ and the amount of material used in the cross-section of each tube is A. Which shape would you use? What is the percentage torque carried by the remaining two shapes in terms of the most efficient structural shape? Solution Shape= Triangle or Square or Circle =? and percentage torque =?

-----------------------------------------------------------Let aT, aS and aC represent the dimension of the triangle, square and circle shown in Figures P5.97, P5.98, P5.99, respectively. The material area in each shape is the product of the perimeter and the thickness. We can find the dimension of each shape in terms of the material area can be found and used in the results of Problems 5.97 through 5.99 as shown below. For triangle: A = ( 3aT )t

or

A- . Substituting this into the result of problem 5.97 and reprea T = ---3t

senting the torque on triangle as TT, we obtain the following. TT 2 TT 2 TT τ max = ------- ------- = ------- --------------- = 10.392 -------2 A-⎞ 2 3 a2 t 3 ⎛ ---A t t T ⎝ 3t⎠

For square: A = ( 4a S )t

or

or

2 1 T T = ---------------- A tτ max 10.392

(1)

A- . Substituting this into the result of problem 5.98 and reprea S = ---4t

senting the torque on square as TS, we obtain the following. TS TS TS - = 8.0 -------τ max = ---------= ----------------2 2 2 A A t 2a S t 2 ⎛ -----⎞ t ⎝ 4t⎠

For Circle: A = ( 2πa C )t

or

or

1 2 T S = ------- A tτ max 8.0

(2)

A - . Substituting this into the result of problem 5.99 and reprea C = ------2πt

senting the torque on circle as TC, we obtain the following

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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 5

TC TC TC - = 6.283 -------τ max = -------------- = ------------------------2 2 2 A A t 2πa C t 2π ⎛ --------⎞ t ⎝ 2πt⎠

or

2 1 T C = ------------- A tτ max 6.283

January 2014

(3)

By inspection of Equations (1), (2), and (3) we conclude that the circular shape will support the maximum Circular shape should be used.

internal torque. Thus our answer is: % torque in triangle relative to circle is:

1 2 ⎛ ------ A tτ max ⎞ TT ⎜ ⎟ 8.0 ⎛ ⎞ = ------- ( 100 ) = ⎜ -----------------------------------⎟ ( 100 ) = 60.46 or ⎝ T C⎠ 2 1 ⎜ -----------⎟ ⎝ 6.283 A tτ max⎠

% torque in triangle relative to circle is 60.46% % torque in square relative to circle is:

2 1 ⎛ --------------- A tτ max⎞ TS ⎜ ⎟ 10.392 = ⎛ -------⎞ ( 100 ) = ⎜ --------------------------------------⎟ ( 100 ) = 78.54 or ⎝ T C⎠ 1 - 2 ⎜ -----------⎟ ⎝ 6.283 A tτ max ⎠

% torque in square relative to circle is 78.54%

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