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M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

6.1 The rigid plate that is welded to the two bars is rotated about the z-axis causing the two bars to bend.The normal strains in bar 1 and 2 were found to be ε 1 = 2000 μ in ⁄ in and ε 2 = – 1500 μ in ⁄ in respectively. Determine the angle of rotation ψ. y x z

D

Bar 2

C

B

Bar 1

A

4 in

ψ 48 in

Solution

ε 1 = 2000 μ in ⁄ in

ψ=?

ε 2 = – 1500 μ in ⁄ in

-----------------------------------------------------------We can draw an exaggerated approximate deformed geometry as shown below. O ψ

R R-4

B1 D1 Ba B a r2 r1

ψ

C A

From the above deformed geometry we obtain the following. AB 1 – AB – 48- = 2000 ( 10 –6 ) ε 1 = ------------------------= Rψ ------------------AB 48

(1)

CD 1 – CD R – 4 )ψ – 48- = – 1500 ( 10 –6 ) ε 2 = ------------------------= (--------------------------------CD 48

(2)

Subtracting Eq. (2) from Eq. (1), we obtain:

–6 4ψ ------- = 3500 ( 10 ) 48

or

ψ = 0.042 rads

or

ψ = 2.41

o

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.2 Determine ‘h’, the location at which a third bar in problem 6.1 must be placed so that there in no normal strain in the third bar. y

D F B

x z

Bar 2 Bar 1

C E 4 in A h

ψ 48 in

Solution

ε 1 = 2000 μ in ⁄ in

ε 2 = – 1500 μ in ⁄ in

h=?

where ε 3 = 0

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

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6-1

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

We draw an exaggerated approximate deformed geometry as shown below.

O

R R-h R-4

ψ

1 FD 1

B1

Ba r2 BB arar 3 1

ψ

C E A

From the deformed geometry we obtain the following. EF 1 – EF ( R – h )ψ – 48- = 0 - = --------------------------------ε 3 = ---------------------EF 48

(1)

Subtracting the above equation from Eq. (1) in problem 6.1, we obtain: hψ ------- = 2000 ( 10 ) or –6

48

, which on substitution yields h = 0.096 ------------- = 2.2857 in or

ψ = 0.042 rads

solution of problem 6.1 we have

h = 0.096 ------------- . From the ψ

0.042 h = 2.29 in.

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.3 The two rigid plates that are welded to six bars are rotated about the z-axis causing the six bars to bend. The normal strains in bar 2 and 5 were found to be zero. What are the strains in the remaining bars? y

z

E C

Bar 3 Bar 2

F D

Bar 6 Bar 5

I H

15 mm

x A

Bar 1

B

Bar 4

G

25 mm

ψ2=2.5o

ψ1=1.25o 2.5 m

3.0 m

Solution

ε2 = 0

ε1 = ?

ε5 = 0

ε3 = ?

ε4 = ?

ε6 = ?

-----------------------------------------------------------The rotation of the rigid plates can be written in radians as shown below. 1.25 ψ 1 = ⎛⎝ ----------⎞⎠ π = 0.02182 rads 180

2.5 ψ 2 = ⎛⎝ ---------⎞⎠ π = 0.04363 rads 180

(1)

We can draw an exaggerated approximate deformed geometry as shown below O1 ψ1

Bar 6 Bar 5 Bar 4

R2+ 0.025 R2

R2- 0.015

E C A

F1 D1B r3 Baar 2 1 B ar B

H G1 1 1ψ 1

ψ1 ψ2 O2

ψ2 I1

5 .01 0 + 5 R 5 R 5 0. 02 5 R

From the deformed geometry, we obtain the following. R2 ψ1 – 3 CD 1 – CD = --------------------- = 0 ε 2 = ------------------------CD 3

or

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R2 ψ1 = 3

(2)

6-2

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

R 2 ψ 1 + ( 0.025 )ψ 1 – 3 AB 1 – AB ( R 2 + 0.025 )ψ 1 – 3 = ---------------------------------------------- = ----------------------------------------------------ε 1 = ------------------------AB 3 3

(3)

0.025 ) ( 0.02182 )- = 181.8 ( 10 –6 ) or Substituting Eq. (2) into Eq. (3) we obtain: ε 1 = (------------------------------------------

ε 1 = 182μ m ⁄ m

3

R 2 ψ 1 – ( 0.015 )ψ 1 – 3 EF 1 – EF ( R 2 – 0.015 )ψ 1 – 3 - = ---------------------------------------------= ----------------------------------------------------ε 3 = ---------------------EF 3 3

(4)

–6 0.015 ) ( 0.02182 ) Substituting Eq. (2) into Eq. (4) we obtain: ε 3 = – (------------------------------------------ = – 109.1 ( 10 ) or

3

ε 3 = – 109.1 μ m ⁄ m R 5 ( ψ 1 + ψ 2 ) – 2.5 D 1 H 1 – DH - = -------------------------------------------- = 0 ε 5 = ---------------------------DH 2.5

R 5 ( ψ 1 + ψ 2 ) = 2.5

or

R 5 ( ψ 1 + ψ 2 ) – ( 0.025 ) ( ψ 1 + ψ 2 ) – 2.5 B 1 G 1 – BG ( R 5 – 0.025 ) ( ψ 1 + ψ 2 ) – 2.5 - = -------------------------------------------------------------------- = --------------------------------------------------------------------------------------------ε 4 = --------------------------BG 2.5 2.5

(5) (6)

0.025 ) ( 0.02182 + 0.04363 )- = – 654.49 ( 10 – 6 ) or Substituting Eq. (5) into Eq. (6) we obtain: ε 4 = – (-------------------------------------------------------------------2.5

ε 4 = – 654 μ m ⁄ m

F 1 I 1 – FI ( R 5 + 0.015 ) ( ψ 1 + ψ 2 ) – 2.5 R 5 ( ψ 1 + ψ 2 ) + ( 0.015 ) ( ψ 1 + ψ 2 ) – 2.5 ε 6 = --------------------- = --------------------------------------------------------------------= --------------------------------------------------------------------------------------------FI 2.5 2.5

(7)

0.015 ) ( 0.02182 + 0.04363 )- = 392.7 ( 10 – 6 ) or ε = 393 μ m ⁄ m Substituting Eq. (5) into Eq. (7) we obtain: ε 6 = (-------------------------------------------------------------------6 2.5

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.4

The strain in bar 1 and 3 were found to be

ε 1 = 800 μ

ε 3 = 500 μ

and

. Determine the strains in the

remaining bars. A

Bar 1

B

Bar 3

E

C

Bar 2

D

Bar 4

F

25 mm 2.5 m

3.0 m

2.0o

3.5o

Figure P6.4

Solution

ε 1 = 800 μ

ε 3 = 500 μ

ε2 = ?

ε4 = ?

-----------------------------------------------------------The rotation of the rigid plates can be written in radians as shown below. 2.0 ψ 1 = ⎛ ---------⎞ π = 0.03491 rads ⎝ 180⎠

3.5 ψ 2 = ⎛⎝ ---------⎞⎠ π = 0.06109 rads 180

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(1)

6-3

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

We can draw an exaggerated approximate deformed geometry as shown below A B ar 1 C Ba r2

B1 Ba Ba r 3 r4

D1 R1

ψ1

R1- 0.025

F1

ψ1 O1

(ψ2 - ψ1)

ψ2

ψ2

R

-0 3

E1

5 .02 R3

O2

From the deformed geometry we obtain the following. AB 1 – AB R1 ψ1 – 3 –6 ε 1 = ------------------------= --------------------- = 800 ( 10 ) AB 3

(2)

CD 1 – CD ( R 1 – 0.025 )ψ 1 – 3 R 1 ψ 1 – 3 0.025ψ 1 ε 2 = ------------------------= ---------------------------------------------= --------------------- – -------------------3 CD 3 3

(3)

Substituting Eqs. (1), (2) and ψ1 into Eq. (3), we obtain the following. –6 ( 0.03491 )- = 509.11 ( 10 – 6 ) or ε 2 = 800 ( 10 ) – 0.025 ------------------------------------3

ε 2 = 509 μm/m

R 3 ( ψ 2 – ψ 1 ) – 2.5 B 1 E 1 – BE –6 - = ------------------------------------------- = 500 ( 10 ) ε 3 = -------------------------BE 2.5

(4)

R 3 ( ψ 2 – ψ 1 ) – 2.5 0.025 ( ψ 2 – ψ 1 ) D 1 F 1 – DF ( R 3 – 0.025 ) ( ψ 2 – ψ 1 ) – 2.5 - = -------------------------------------------------------------------- = ------------------------------------------- – -------------------------------------ε 4 = -------------------------DF 2.5 2.5 2.5

(5)

Substituting Eqs. (1),(4), ψ1 and ψ2 into Eq. (5), we obtain the following. –6 ( 0.06109 – 0.03491 )- = 238.2 ( 10 –6 ) or ε 4 = 500 ( 10 ) – 0.025 -------------------------------------------------------------2.5

ε 4 = 238 μm/m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.5 Due to the action of the external moment Mz and force P, the rigid plate was observed to rotate by 2o and the normal strain in bar 1 was found ε 1 = 2000 μ in ⁄ in . Both bars have an area of cross-section of A = 1/2 in2 and a Modulus of Elasticity of E = 30,000 ksi. Determine the applied moment Mz and force P. y

Bar 2

P x

2 in

Bar 1

4 in

z

Mz

48 in Figure P6.5

Solution

ε 1 = 2000 μ in ⁄ in ψ = 2o

E = 30,000 ksi

A = 1/2 in2

Mz = ?

P= ?

-----------------------------------------------------------1. Strain calculations: The rotation of the rigid plates can be written in radians as shown below. 2.0 ψ = ⎛⎝ ---------⎞⎠ π = 0.03491 rads 180

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(1)

6-4

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

We can draw an exaggerated approximate deformed geometry as shown below in figure (a). (a)

(b)

O

R R-4

ψ

4 in P 2 in

B1 D 1 ψ

13.63 kips o 30 kips

Mz Ba

Ba r1

r2

C A

From the deformed geometry we obtain the following. AB 1 – AB –6 Rψ – 48 ε 1 = ------------------------= -------------------- = 2000 ( 10 ) AB 48

(2)

CD 1 – CD ( R – 4 )ψ – 48- = Rψ – 48- – 4ψ = --------------------------------------------------------ε 2 = ------------------------CD 48 48 48

(3)

Substituting Eqs. (1) and (2) into Eq. (3), we obtain: –6 ( 0.03491 )- = – 908.88 ( 10 – 6 ) ε 2 = 2000 ( 10 ) – 4-------------------------48

(4)

2. Stress calculations: From Hooke’s Law we can find the normal stresses in the bar as shown below. –6

σ 1 = Eε 1 = ( 30, 000 ) ( 2000 ) ( 10 ) = 60 ksi or σ 1 = 60 ksi ( T ) –6

σ 2 = Eε 2 = ( 30, 000 ) ( – 908.88 ) ( 10 ) = – 27.27 ksi or σ 2 = 27.27 ksi ( C )

3. Internal force calculations: The internal normal force in each bar can be found as shown below. N 1 = σ 1 A = ( 60 ) ( 1 ⁄ 2 ) = 30 kips ( T )

N 2 = σ 2 A = ( 27.27 ) ( 1 ⁄ 2 ) = 13.63 kips ( C )

4. External forces and moments calculations: The free body diagram of the rigid plate can be drawn by making imaginary cuts through the rods as shown in Figure (b). By equilibrium of forces we obtain: P + 13.63 – 30 = 0 or P = 16.37 kips By equilibrium of moment about point O we obtain: M z – ( 13.63 ) ( 2 ) – ( 30 ) ( 2 ) = 0 or M z = 87.26 in – kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.6 The rigid plate was observed to rotate 1.25o due to the action of the external moment Mz and Force P. All three bars have an area of cross-section of A= 100 mm2 and a Modulus of Elasticity E=200 GPa. If the strain in bar 2 was measured as zero determine the external moment Mz and Force P. E y

C

Bar 3 Bar 2

x A

Bar 1

z

F 15 mm

D

P

25 mm

B

Mz 3.0 m

Solution ε 2 = 0 ψ = 1.25o

E = 200 GPa

A = 100 mm2

Mz = ?

P= ?

-----------------------------------------------------------1. Strains calculations: The rotation of the rigid plates can be written in radians as shown below. 1.25 ψ = ⎛⎝ ----------⎞⎠ π = 0.02182 rads 180

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(1)

6-5

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

We can draw an exaggerated approximate deformed geometry as shown in figure (a). (b)

(a)

O

R+0.025 R R-0.015

ψ

F

2182N

D B

3636N F1 D1

E C

P

0.025 m

Mz

B1

r3 Ba r 2 Ba a r 1 B

0.015 m

ψ

A

From the deformed geometry, we obtain the following. CD 1 – CD – 3- = 0 = Rψ ---------------ε 2 = ------------------------CD 3

or

(2)

Rψ = 3

AB 1 – AB R + 0.025 )ψ – 3- = --------------------------------------------Rψ + ( 0.025 )ψ – 3= (----------------------------------------ε 1 = ------------------------AB 3 3

(3)

EF 1 – EF R – 0.015 )ψ – 3 = Rψ – ( 0.015 )ψ – 3- = (------------------------------------------------------------------------------------ε 3 = ---------------------EF 3 3

(4)

0.025 ) ( 0.02182 )- = 181.8 ( 10 – 6 ) Substituting Eqs. (1) and (2) into Eq. (3) we obtain: ε 1 = (------------------------------------------

3 (-----------------------------------------0.015 ) ( 0.02182 )–6 Substituting Eqs.(1) and (2) into Eq. (4) we obtain: ε 3 = – = – 109.1 ( 10 ) 3

2. Stress calculations: From Hooke’s Law we can find the normal stresses in the bar as shown below. 9

–6

6

σ 1= Eε 1 = ( 200 ) ( 10 ) ( 181.8 ) ( 10 ) = 36.36 ( 10 ) N ⁄ m 9

–6

6

2

σ 2 = Eε 2 =( 200 ) ( 10 ) ( – 109.1 ) ( 10 ) =– 21.82 ( 10 )N ⁄ m

6

2

or σ 1 = 36.36 ( 10 ) N ⁄ m ( T )

2

6

2

or σ 2 = 21.82 ( 10 ) N ⁄ m ( C )

3. Internal force calculations: The internal normal force in each bar can be found as shown below. 6

–6

6

N 1 = σ 1 A = ( 36.36 ) ( 10 ) ( 100 ) ( 10 ) = 3636N ( T )

–6

N 2 = σ 2 A = ( 21.82 ) ( 10 ) ( 100 ) ( 10 ) = 2182N ( C )

4. External forces and moments calculations: The free body diagram of the rigid plate can be drawn by making imaginary cuts through the rods as shown in figure (b) By equilibrium of forces we obtain: P + 2182 – 3636 = 0 or P = 1454 N By equilibrium of moment about point D we obtain: M z – ( 2182 ) ( 0.015 ) – ( 3636 ) ( 0.025 ) = 0 or M z = 123.6 N – m -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.7 The rigid plates BD and EF were observed to rotate by 2o and 3.5o in the direction of applied moments. All bars have an area of cross-section of A= 125 mm2. Bars 1 and 3 are made of steel ES= 200 GPa. and bars 2 and 4 are made of aluminum Eal = 70 GPa. If the strains in bar 1 and 3 were found to be ε 1 = 800 μ and ε 3 = 500 μ determine the applied moment M1 and M2 and the forces P1 and P2 that act at the center of the rigid plates. A

Bar 1

B

Bar 3

E

C

Bar 2

D P2 Bar 4

F

3.0 m

2.5 m

25 mm

M2

P1

M1

Figure P6.7

Solution

ε 1 = 800 μ

E2 = E4 = 70 GPa

ψ1 = 2o ψ1 = 3.5o A = 125 mm2 M1 = ?

ε 3 = 500 μ

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E1 = E3 =200 GPa P 1= ?

M2 = ?

P2= ?

6-6

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

-----------------------------------------------------------–6 –6 1. Strain calculations: From problem 6.4 we have: ε 4 = 238.2 ( 10 ) and ε 2 = 509.11 ( 10 ) 2. Stress calculations: From Hooke’s Law we can find the normal stresses in the bar as shown below. 9

–6

6

σ 1 = E 1 ε 1 = ( 200 ) ( 10 ) ( 800 ) ( 10 ) = 160 ( 10 ) N ⁄ m 9

–6

2

(T)

6

σ 2 = E 2 ε 2 = ( 70 ) ( 10 ) ( 509.11 ) ( 10 ) = 35.64 ( 10 ) N ⁄ m 9

–6

6

σ 3 = E 3 ε 3 = ( 200 ) ( 10 ) ( 500 ) ( 10 ) = 100 ( 10 ) N ⁄ m 9

–6

2

6

σ 4 = E 4 ε 4 = ( 70 ) ( 10 ) ( 238.2 ) ( 10 ) = 16.67 ( 10 ) N ⁄ m

2

(T)

(T) 2

(T)

3. Internal force calculations: The internal normal force in each bar can be found as shown below. 6

–6

3

N 1 = σ 1 A = ( 160 ) ( 10 ) ( 125 ) ( 10 ) = 20 ( 10 )N = 20kN ( T ) 6

–6

3

N 2 = σ 2 A = ( 35.64 ) ( 10 ) ( 125 ) ( 10 ) = 4.46 ( 10 )N = 4.46kN ( T ) 6

–6

3

N 3 = σ 3 A = ( 100 ) ( 10 ) ( 125 ) ( 10 ) = 12.5 ( 10 )N = 12.5kN ( T ) 6

–6

3

N 4 = σ 4 A = ( 16.67 ) ( 10 ) ( 125 ) ( 10 ) = 2.08 ( 10 )N = 2.08kN ( T )

4. External forces and moments calculations: The free body diagram of the rigid plates can be drawn by making imaginary cuts through the rods as shown below.

(a)

0.0125 m P1 0.0125 m

O1 2.08 kN

(b)

E

12.5 kN

B

20 kN O2

P2

M1

F M1

M2

By equilibrium of forces in Fig. (a) we obtain: P 1 – 12.5 – 2.08 = 0 or

0.0125 m P1 0.0125 m

O1

D

4.46 kN

F

E

P 1 = 14.58kN

By equilibrium of moment about point O1 we obtain: M 1 – ( 12.5 ) ( 0.0125 ) + ( 2.08 ) ( 0.0125 ) = 0 or M 1 = 0.1303 kN – m

M 1 = 130.3N – m

By equilibrium of forces in Fig. (b) we obtain: P 2 + P 1 – 20 – 4.46 = 0 or

P 2 = 9.88kN

By equilibrium of moment about point O1 we obtain: M 2 + M 1 – ( 20 ) ( 0.0125 ) + ( 4.46 ) ( 0.0125 ) = 0 or M 2 = 0.06395 kN – m

M 2 = 64.0N – m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.8 Three wooden beams are glued to form a beam with the cross-section shown in Figure 6.8. The normal strain at due to bending about the z axis is εxx = -0.012y, where y is measured in meters. The modulus of elasticity of wood is 10 GPa. Determine the equivalent internal moment acting at the cross-section. Use tW =20 mm, h =250 mm, tF = 20 mm, and d= 125 mm.

d d y tF h z

tW

h tF

Figure P6.8

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6-7

M. Vable

Solution

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

εxx = - 0.012y

Ewood = 10 GPa tW =20 mm

January 2014

h =250 mm tF = 20 mm df= 125 mm. ------------------------------------------------------------

Mz = ?

From Hooke’s law we can write: 9

6

σ xx = E wood ε xx = ( 10 ) ( 10 ) ( – 0.012y ) = – 120y ( 10 ) N ⁄ m

2

(1)

The differential area dA in the flange is the area of a strip whose width is 2d = 0.25 m and thickness dy i.e., in dA = 0.25dy . The differential area dA in the web is the area of a strip whose width is tW =0.02 m and thickness dy i.e., in wood dA = 0.02dy . The normal stress and dA across the cross-section can be written as shown below.

σ xx

⎧ 6 2 ⎪ – 120y ( 10 ) N ⁄ m ⎪ = ⎨ – 120y ( 10 6 ) N ⁄ m 2 ⎪ ⎪ – 120y ( 10 6 ) N ⁄ m 2 ⎩

dA = 0.25dy

0.250 < y ≤ 0.270

dA = 0.02 dy

– 0.250 < y < 0.250 – 0.270 ≤ y < – 0.250

dA = 0.25dy

(2)

Noting the symmetry with respect to y, the moment equation can be written twice the sum of integrals over the upper half (y>0) of the cross-section. 0.250

M z = – ∫ yσ xx dA = – 2

0.270

∫y ( –120y ) ( 0.02 ) dy

A

+

0 3

y M z = – 2 – 2.4 ----3

∫ y ( –120y ) ( 0.25 ) dy

6

( 10 )

or

0.250

0.250

0.270

3

y – 30 ----3

0

6

3

( 10 ) = – 2 [ – 12.5 – 40.58 ] ( 10 ) or 0.250

M z = 106.2 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.9 Three wooden beams are glued to form a beam with the cross-section shown in Figure 6.9. The normal strain at the cross due to bending about the z axis is εxx = -0.015y, where y is measured in meters. The modulus of elasticity of wood is 10 GPa. Determine the equivalent internal moment acting at the cross-section. Use tW =10 mm, h =50 mm, tF = 10 mm, and d= 25 mm.

d d y tF h

tW

z h tF

Solution

Ewood = 10 GPa tW =10 mm

Figure P6.9

εxx = - 0.15y

h =50 mm tF = 10 mm df= 25 mm. ------------------------------------------------------------

Mz = ?

From Hooke’s law we can write: 9

9

σ xx = E wood ε xx = ( 10 ) ( 10 ) ( – 0.15y ) = – 1.5y ( 10 ) N ⁄ m

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2

(1)

6-8

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

The differential area dA in the flange is the area of a strip whose width is 2d = 0.05 m and thickness dy i.e., dA = 0.05dy . The differential area dA in the web is the area of a strip whose width is tW =0.01 m and thickness dy i.e., dA = 0.01dy . The normal stress and dA across the cross-section can be written as shown below.

σ xx

⎧ 9 2 ⎪ – 1.5y ( 10 ) N ⁄ m ⎪ = ⎨ – 1.5y ( 10 9 ) N ⁄ m 2 ⎪ ⎪ – 1.5y ( 10 9 ) N ⁄ m 2 ⎩

dA = 0.05 dy

0.05 < y ≤ 0.06

dA = 0.01 dy

– 0.05 < y < 0.05 – 0.06 ≤ y < – 0.05

dA = 0.05 dy

(2)

Noting the symmetry with respect to y, the moment equation can be written as twice the sum of integrals in the upper half (y>0) of the cross-section. 0.05

M z = – ∫ yσ xx dA = – 2 A 3

y M z = – 2 – 0.015 ----3

0.06

∫y ( –1.5y ) ( 0.01 ) dy

0 0.05

∫ y ( –1.5y ) ( 0.05 ) dy

9

( 10 ) or

0.05 3

y – 0.075 ----3

0

+

0.06 9

( 10 ) = – 2 [ – 625 – 2275 ] N – m or 0.05

M z = 5.8 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.10 Three wooden beams are glued to form a beam with the cross-section shown in Figure 6.10. The normal strain at the cross due to bending about the z axis is εxx = 0.02y, where y is measured in meters. The modulus of elasticity of wood is 10 GPa. Determine the equivalent internal moment acting at the cross-section. Use tW =15 mm, h =200 mm, tF = 20 mm, and d= 150 mm.

d d y tF h

tW

z h tF

Solution

Ewood = 10 GPa tW =15 mm

Figure P6.10

εxx = 0.02y

h =200 mm tF = 20 mm d= 150 mm. ------------------------------------------------------------

Mz = ?

From Hooke’s law we can write: 9

9

σ xx = E wood ε xx = ( 10 ) ( 10 ) ( 0.02y ) = 0.2y ( 10 ) N ⁄ m

2

(1)

The differential area dA in the flange is the area of a strip whose width is 2d= 0.3 m and thickness dy i.e., dA = 0.3dy . The differential area dA in the web is the area of a strip whose width is tW =0.015 m and thickness dy i.e., dA = 0.015dy . The normal stress and dA across the cross-section can be written as shown below.

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6-9

M. Vable

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

σ xx

⎧ 9 2 ⎪ 0.2y ( 10 ) N ⁄ m ⎪ = ⎨ 0.2y ( 10 9 ) N ⁄ m 2 ⎪ ⎪ 0.2y ( 10 9 ) N ⁄ m 2 ⎩

January 2014

0.20 < y ≤ 0.22

dA = 0.3 dy

2

– 0.20 < y < 0.20 – 0.22 ≤ y < – 0.20

dA = 0.015 dy dA = 0.3 dy

Noting the symmetry with respect to y, the moment equation can be written twice the sum of integrals over the upper half (y>0) of the cross-section. 0.20

M z = – ∫ yσ xx dA = – 2 A 3

y M z = – 2 0.003 ----3

0.22

∫y ( 0.2y ) ( 0.015 ) dy 0 0.20

0

∫ y ( 0.2y ) ( 0.3 ) dy

+

9

( 10 ) or

0.20 3

y + 0.06 ----3

0.22 9

3

( 10 ) = – 2 [ 8 + 52.96 ] ( 10 ) or 0.20

M z = – 121.9 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.11 Steel (Esteel = 30,000 ksi) strips are securely attached to a wooden (Ewood = 2,000 ksi) beam as shown. The

normal strain εxx at the cross-section due to bending about the z-axis, and the dimensions of the cross-section are as given below. Determine the equivalent internal moment Mz.

εxx = - 100y μ. where y is measured in inches. Use d =2 in, hW =4 in and hS= (1/8) in. y d hS

Steel

z

hW Wood

hS x

Steel

Esteel = 30,000 ksi Ewood = 2,000 ksi εxx = - 100y μ. d =2 in hW =4 in hS= (1/8) in. -----------------------------------------------------------From Hooke’s law we can write: Solution

Mz = ?

–6

(1)

–6

(2)

( σ xx ) steel = E steel ε xx = ( 30000 ) ( – 100y ) ( 10 ) = – 3.0y ksi ( σ xx ) wood = E wood ε xx = ( 2000 ) ( – 100y ) ( 10 ) = – 0.2y ksi

The normal stress across the cross-section can be written as shown below. σ xx

⎧ – 3.0y ksi ⎪ = ⎨ – 0.2y ksi ⎪ ⎩ – 3.0y ksi

2.0 < y ≤ 2.125 – 2.0 < y < 2.0 – 2.125 ≤ y < – 2.0

(3)

The differential area dA is the area of a strip whose width is d = 2 inch and thickness dy i.e., dA = 2dy . We can write: – 2.0

M z = – ∫ yσ xx dA = – A 3

y M z = – – 6 ----3

– 2.0

∫

2.0

y ( – 3.0y ) ( 2 ) dy +

– 2.125 2.0 3

y – 0.4 ----3 – 2.125

∫ – 2.0

3

y – 6 ----3 – 2.0

2.125

y ( – 0.2y ) ( 2 ) dy +

∫

y ( – 3.0y ) ( 2 ) dy or

2.0

2.125

= – [ – 3.191 – 2.133 – 3.191 ] or

M z = 8.52 in-kips

2.0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.12 Steel (Esteel = 30,000 ksi) strips are securely attached to a wooden (Ewood = 2,000 ksi) beam as shown. The

normal strain εxx at the cross-section due to bending about the z-axis, and the dimensions of the cross-section are as

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

given below. Determine the equivalent internal moment Mz

εxx = - 50 y μ. where y is measured in inches. Use d =1 in, hW =6 in and hS= (1/4) in. y d hS

Steel

z

hW Wood Steel

hS x

Esteel = 30,000 ksi Ewood = 2,000 ksi εxx = - 50 y μ. d =1 in hW =6 in hS= (1/4) in. -----------------------------------------------------------From Hooke’s law we can write: Solution

Mz = ?

–6

(1)

–6

(2)

( σ xx ) steel = E steel ε xx = ( 30000 ) ( – 50y ) ( 10 ) = – 1.5y ksi ( σ xx ) wood = E wood ε xx = ( 2000 ) ( – 50y ) ( 10 ) = – 0.1y ksi

The normal stress across the cross-section can be written as shown below. ⎧ – 1.5y ksi ⎪ σ xx = ⎨ – 0.1y ksi ⎪ ⎩ – 1.5y ksi

3.0 < y ≤ 3.25 – 3.0 < y < 3.0 – 3.25 ≤ y < – 3.0

(3)

The differential area dA is the area of a strip whose width is d = 1 inch and thickness dy i.e., dA = ( 1 )dy . We can write: – 3.0

M z = – ∫ yσ xx dA = – A

3.0

∫

y ( – 1.5y ) ( 1 ) dy +

– 3.25 3

y M z = – – 1.5 ----3

– 3.0

– 3.25

3

y – 0.1 ----3

∫

3.25

y ( – 0.1y ) ( 1 ) dy +

– 3.0 3.0

– 3.0

3

y – 1.5 ----3

∫

y ( – 1.5y ) ( 1 ) dy or

3.0

3.25

= – [ – 3.664 – 1.800 – 3.664 ] or M z = 9.13 in-kips 3.0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.13 Steel (Esteel = 30,000 ksi) strips are securely attached to a wooden (Ewood = 2,000 ksi) beam as shown. The

normal strain εxx at the cross-section due to bending about the z-axis, and the dimensions of the cross-section are as given below.Determine the equivalent internal moment Mz

εxx = 200 y μ. where y is measured in inches. Use d =1 in, hW =2 in and hS= (1/16) in. y d Steel

z

hS hW

Wood Steel

hS x

Esteel = 30,000 ksi Ewood = 2,000 ksi εxx = 200 y μ. d =1 in hW =2 in hS= (1/16) in. -----------------------------------------------------------From Hooke’s law we can write: Solution

–6

( σ xx ) steel = E steel ε xx = ( 30000 ) ( 200y ) ( 10 ) = 6y ksi –6

( σ xx ) wood = E wood ε xx = ( 2000 ) ( 200y ) ( 10 ) = 0.4y ksi

Mz = ?

(1) (2)

The normal stress across the cross-section can be written as shown below.

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

⎧ 6y ksi ⎪ σ xx = ⎨ 0.4y ksi ⎪ ⎩ 6y ksi

January 2014

1.0 < y ≤ 1.0625 – 1.0 < y < 1.0 – 1.0625 ≤ y < – 1.0

(3)

The differential area dA is the area of a strip whose width is d = 1 inch and thickness dy i.e., dA = ( 1 )dy . We can write: – 1.0

M z = – ∫ yσ xx dA = –

∫

A 3

y M z = – 6 ----3

1.0

y ( 6y ) ( 1 ) dy +

– 1.0625 1.0 3

– 1.0

y + 0.4 ----3 – 1.0625

∫

1.0625

∫

y ( 0.4y ) ( 1 ) dy +

– 1.0 3

y + 6 ----3 – 1.0

y ( 6y ) ( 1 ) dy or

1.0

1.0625

= – [ 0.3989 + 0.2667 + 0.3989 ] or

M z = – 1.06 in.-kips

1.0

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.14 The flanges made from steel (Esteel = 200 GPa) are securely attached to the wooden (Ewood = 10 GPa) beam as shown. The normal strain εxx at the cross-section due to bending about the z-axis and the dimensions of the cross-section are as given below. Determine the equivalent internal moment Mz

εxx = 0.02y, where y is measured in meters. Use tW =15 mm in, hW =200 mm, tf = 20 mm, and df= 150 mm.

tF hW hW

z

df el Stey

Wood

df

tW

el Ste

x

tF

Esteel = 200 GPa Ewood = 10 GPa εxx = 0.02y tW =15 mm hW =200 mm tf = 20 mm -----------------------------------------------------------From Hooke’s law we can write: Solution

9

9

( σ xx ) steel = E steel ε xx = ( 200 ) ( 10 ) ( 0.02y ) = 4y ( 10 ) N ⁄ m 9

9

df= 150 mm.

Mz = ?

2

( σ xx ) wood = E wood ε xx = ( 10 ) ( 10 ) ( 0.02y ) = 0.2y ( 10 ) N ⁄ m

(1)

2

(2)

The differential area dA in the steel is the area of a strip whose width is 2df = 0.3 m and thickness dy i.e., in steel dA = 0.3dy . The differential area dA in the wood is the area of a strip whose width is tW =0.015 m and thickness dy i.e., in wood dA = 0.015dy . The normal stress and dA across the cross-section can be written as shown below.

σ xx

⎧ 9 2 ⎪ 4y ( 10 ) N ⁄ m ⎪ = ⎨ 0.2y ( 10 9 ) N ⁄ m 2 ⎪ ⎪ 4y ( 10 9 ) N ⁄ m 2 ⎩

0.20 < y ≤ 0.22

dA = 0.3 dy dA = 0.015 dy dA = 0.3 dy

– 0.20 < y < 0.20 – 0.22 ≤ y < – 0.20

(3)

Noting the symmetry with respect to y, the integral can be written as twice the sum of integrals over steel and wood in the upper half (y>0) of the cross-section. 0.20

M z = – ∫ yσ xx dA = – 2 A

0.22

∫y ( 0.2y ) ( 0.015 ) dy 0

3

y M z = – 2 0.003 ----3

0.20

0

+

∫ y ( 4y ) ( 0.3 ) dy

9

( 10 ) or

0.20 3

y + 1.2 ----3

0.22 9

3

( 10 ) = – 2 [ 8 + 1059.2 ] ( 10 ) or

M z = – 2134 kN – m

0.20

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.15 A beam of rectangular cross section shown in Figure 6.15 is made from elastic- perfectly plastic material. If

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

the stress distribution across the cross section is as shown determine the equivalent internal bending moment

y

30 ksi

σxx

z 4 in

30 ksi 0.5 in

Solution

Figure P6.15

Mz = ? ------------------------------------------------------------

Fig(a) shows the stress distribution across the cross-section. y = 2.0 y = a = 1.5

30 ksi

(a)

dA = 0.5 dy

y σxx y = a = - 1.5 y = - 2.0

0.5 in

30 ksi

The stress distribution in the linear region is: σ = – 30y --------- = – 20y ksi . We can write the stress expression and value of 1.5

dA for each interval as: σ xx = 30 ksi

dA = 0.5dy

– 2 ≤ y < – 1.5

σ xx = – 20y ksi

dA = 0.5dy

– 1.5 < y < 1.5

σ xx = – 30 ksi

dA = 0.5dy

1.5 < y < 2

1

From the symmetry we conclude that the equivalent internal moment in the lower half ( y < 0 )is same as that in the upper half ( y > 0 ). The integral can be written as twice the integral in the positive y half as shown below. 1.5

M z = – ∫ yσ xx dA = – 2 A

2

∫ y [ σxx ] ( 0.5 dy ) + ∫ y [ σxx ] ( 0.5 dy ) 0

(2)

1.5

Substituting the stress distribution we obtain: 1.5

Mz = –2

∫ 0

2

y [ – 20y ] ( 0.5 dy ) +

∫ 1.5

3

y y [ – 30 ] ( 0.5 dy ) = – 2 – 10 ----3

1.5

0

2

y – 15 ----2

2

= 2 [ 11.25 + 13.125 ]

or

1.5

M z = 48.75 in.-kips

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.16 A rectangular beam cross section has the dimensions shown in Figure 6.16. The normal strain due to bending about the z-axis was found to vary as ε xx = – 0.01y with y measured in meters. Determine the equivalent internal moment that produced the given state of strain. The beam is made from elastic-perfectly plastic material that has a yield stress of σyield= 250 MPa, and a modulus of elasticity E = 200 GPa. Assume material behaves in a similar manner in

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

tension and compression (see problem 3.152)

y 100 mm 100 mm 150mm z 150 mm Figure P6.16

Solution

εxx=-0.01y y in meters

σyield=250MPa

E=200GPa

Mz = ?

-----------------------------------------------------------The strain at yield point is: σ yield ( 10 6 ) = 1.25 ( 10 – 3 ) ---------------------- = 250 ε yield = ------------E 200 ( 10 6 )

(1)

For positive values of y we have contraction, and for negative values of y we have tension. With geometry being symmetric in y, the elastic-plastic boundary will be on both sides ε yield = ± 1.25 ( 10 –3 ) = – 0.01y yield or y yield = ± 0.125m = ± 125mm

(2)

σ xx = Eε xx = ( 200 ) ( 10 9 ) ( – 0.01y ) = – 2000y y in meters.

(3)

In the linear region We have the following stress-distribution for the given state of strain and material. ⎧ – 250 MPa ⎪ σ xx = ⎨ – 2000y MPa ⎪ ⎩ 250 MPa

0.125m < y < 0.150m – 0.125m < y < 0.125m – 0.150m < y < – 0.125m

(4)

Substituting the stress distribution and dA = 0.2 dy in moment equation, we obtain: – 0.125

M z = – ∫ yσ xx dA = – A

∫

0.125

y ( 250 ) ( 0.2 dy ) +

– 0.15 – 0.125 2

y M z = – 50 ----2

– 0.15

3

y – 400 ----3

∫

– 0.125 0.125

– 0.125

0.15

y ( – 2000y ) ( 0.2 dy ) +

∫

6

y ( – 250 ) ( 0.2 dy ) ( 10 ) or

0.125 2

y – 50 ----2

0.15 6

3

( 10 ) = 864.5 ( 10 ) N – m 0.125

M z = 864.5 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.17 A rectangular beam cross section has the dimensions shown in Figure 6.17. The normal strain due to bending about the z-axis was found to vary as ε xx = – 0.01y with y measured in meters. Determine the equivalent internal moment that would produce the given strain. The beam is made from a bi-linear material that has a yield stress of σyield= 200 MPa, Modulus of Elasticity E1 = 250 GPa and E2= 80 GPa. Assume material behaves in a similar manner

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

in tension and compression (see problem 3.153).

y 100 mm 100 mm 150mm z 150 mm Figure P6.17

εxx=-0.01y y in meters σyield=250MPa

Solution

E1 = 200 GPa

E2= 80 GPa

Mz = ?

-----------------------------------------------------------The strain at yield point is: σ yield ( 10 6 ) = 1.25 ( 10 – 3 ) ---------------------- = 250 ε yield = ------------E 200 ( 10 6 )

(1)

For positive values of y we have contraction, and for negative values of y we have tension. With geometry being symmetric in y, the elastic-plastic boundary will be on both sides ε yield = ± 1.25 ( 10 –3 ) = – 0.01y yield or y yield = ± 0.125m = ± 125mm

(2)

In linear region: 9

σ 1 = E 1 ε 1 = ( 200 ) ( 10 ) ( – 0.01y ) = – 2000y MPa

(3)

For y < – 0.125m i.e., on the tensile side 6

9

σ 2 = σ yield + E 2 ( ε xx – ε yield ) = 250 ( 10 ) + ( 80 ) ( 10 ) [ – 0.01y – 1.25 ( 10 –3 ) ] = ( 150 – 800y )MPa

For y > 0.125m i.e., on the compressive side 6

9

σ 2 = – σ yield + E 2 ( ε xx – ( – ε yield ) ) = ( – 250 ) ( 10 ) + ( 80 ) ( 10 ) [ – 0.01y + 1.25 ( 10 – 3 ) ] = ( – 150 – 800y )MPa

We have the following stress-distribution for the given state of strain and material. ⎧ ( – 150 – 800y )MPa ⎪ σ xx = ⎨ – 2000yMPa ⎪ ⎩ ( 150 – 800y )MPa

0.125m < y < 0.150m – 0.125m < y < 0.125m – 0.150m < y < – 0.125m

(4)

Substituting the stress distribution and dA = 0.2 dy in moment equation, we obtain: – 0.125

Mz = –

∫

0.125

∫

y ( 150 – 800y ) ( 0.2 dy ) +

– 0.15

0.15

y ( – 2000y ) ( 0.2 dy ) +

– 0.125 2

3

y y M z = – ⎛⎝ 30 ----- – 40 -----⎞⎠ 2 3

– 0.125

– 0.15

3

y – 400 ----3

0.125

∫

6

y ( – 150 – 800y ) ( 0.2 dy ) ( 10 ) or

0.125 2

3

y y + ⎛⎝ – 30 ----- – 40 -----⎞⎠ 2 3 – 0.125

0.15 6

3

( 10 ) = 765.0 ( 10 ) N – m

or

0.125

M z = 765 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.18 A rectangular beam cross section has the dimensions shown in Figure 6.18. The normal strain due to bending about the z-axis was found to vary as ε xx = – 0.01y with y measured in meters. Determine the equivalent internal moment that would produce the given strain. The beam material has a stress strain relationship given by

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σ = 952ε

0.2

Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

January 2014

MPa . Assume material behaves in a similar manner in tension and compression (see problem 3.154).

y 100 mm 100 mm 150mm z 150 mm Figure P6.18

εxx=-0.01y y in meters

Solution

σ = 952ε

0.2

Mz = ?

MPa

-----------------------------------------------------------As the material behavior in tension and compression is the same, we have: 0.2

⎧ 952ε MPa σ xx = ⎨ ⎩ – 952 ( – ε ) 0.2 MPa

ε>0 ε 0 ). The integral in moment equation can be written as twice the integral in the positive y half as shown below. 0

M z = – ∫ yσ xx dA = – A

∫

0.15

y [ σ xx ] ( 0.2 dy ) +

– 0.15

∫

0.15

y [ σ xx ] ( 0.2 dy ) = – 2

0

∫

y [ σ xx ] ( 0.2 dy )

(3)

0

Substituting the stress distribution we obtain: 0.15

Mz = –2

∫

0.15

y [ – 379 ( y )

0.2

6

] ( 0.2 dy ) ( 10 ) = 151.6

0

∫

y

1.2

2.2

6 y dy ( 10 ) = 151.6 ⎛⎝ ---------⎞⎠ 2.2

0

0.15 6

3

( 10 ) = 1061 ( 10 ) 0

M z = 1061 kN-m

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.19 A solid and a hollow square beam have the same cross-sectional area A. Show that the ratio of second area moment of inertia for the hollow beam (IH) to that of the solid beam (IS) is as given below: 2 IH α +1 ----- = --------------2 IS α –1

aH

aS

z

y α aH

aS

z

y aH

α aH

Solution -----------------------------------------------------------The lengths aS and aH in terms of area can be found as shown below:

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Mechanics of Materials, 2nd edition, Solutions Manual: Chapter 6

2

AS = a S = A 2

or 2

aS = 2

January 2014

A

2

A H = ( αa H ) – a H = ( α – 1 )a H = A

or

A -------------------2 (α – 1)

aH =

Let IS and IH represent the area moment of inertias about the z-axis for the solid cross-section and the hollow cross-section respectively. We can find IS and IH in terms of the area A as shown below using the formula for rectangles in Table A2. 3 1 1 2 I S = ------ a S a S = ------ A 12 12 4

4

3 1 3 1 (α – 1) 4 (α – 1) A I H = ------ ( αa H ) ( αa H ) – ------ a H a H = -------------------- a H = -------------------- ------------------2 12 12 12 12 (α – 1)

2

2 1 ⎛ α + 1-⎞ 2 = ------ ⎜ -------------⎟A 12 ⎝ α 2 – 1⎠ 2

2

The ratio of the two area moment of inertias is: IH ⁄ IS = ( α + 1 ) ⁄ ( α – 1 ) -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.20 Figure 6.20(a) shows four separate wooden strips that bend independently about the neutral axis passing through the centroid of each strip. Figure 6.20(b) shows the four strips are glued together and bend as a unit about the centroid of the glued cross-section. (a) Show that I G = 16I S , where IG is the area moment of inertias for the glued cross-section and IS is the total area moment of inertia of the four separate beams. (b) Also show σ G = σ S ⁄ 4 , where σG and σS are the maximum bending normal stress at any cross-section for the glued and separate beams, respectively. P P (b) Glued Beams

(a)

Separate Beams

lA

xi

s

2b 2b

eu tra

a

a

N

Figure P6.20

N Ne N e u N eu utratral eu t r l A tra al Ax xi l AAx is s xi is s

b

Solution -----------------------------------------------------------The total area moment of inertia for separate beams is four times the area moment of inertia for the individual beam. 3

1 3 ab Using the formula for rectangles in Table A2, Is can be written as: I S = 4 ------ ab = ------12

3

3

1 3 -------------- or The glued beams act as a unit and IG can be written as: I G = ------ a ( 4b ) = 16ab 12

3

I G = 16I S

The maximum bending normal stress σS for separate beams will be at the top and bottom of each beam, i.e. ymax = b/2. From flexure stress formula, the magnitude of the maximum bending normal stress σS for separate beams can be written as: MZ ( b ⁄ 2 ) MZ ( b ⁄ 2 ) 3 ⎛ M Z⎞ - = --- ⎜ -------2-⎟ σ S = ---------------------- = ---------------------3 2 ⎝ ab ⎠ IS ( ab ⁄ 3 )

The maximum bending normal stress σG for glued beams will be at the top and bottom of the glued unit i.e., ymax = 2b. From flexure stress formula, the magnitude of the maximum bending normal stress σG for separate beams can be written as: M Z ( 2b ) M Z ( 2b ) 3 ⎛ M Z⎞ σ G = ------------------= -------------------------= --- ⎜ -------2-⎟ or 3 8 ⎝ ab ⎠ IG ( 16ab ⁄ 3 )

σG = σS ⁄ 4

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------6.21 The cross-section of the beam shown are constructed from thin sheet metal of thickness ‘t’.Assume that the thickness t