• Author / Uploaded
• Shrey Pandey

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B

PROBLEM 2.15

1.2 m

A

A 1.2-m section of aluminum pipe of cross-sectional area 1100 mm2 rests on a fixed support at A. The 15-mm-diameter steel rod BC hangs from a rigid bar that rests on the top of the pipe at B. Knowing that the modulus of elasticity is 200 GPa for steel and 72 GPa for aluminum, determine the deflection of point C when a 60 kN force is applied at C.

0.9 mm C

P

SOLUTION Rod BC:

LBC = 2.1 m, EBC = 200 × 109 Pa ABC =

δ C /B = Pipe AB:

π 4

d2 =

π 4

(0.015)2 = 176.715 × 10−6 m 2

PLBC (60 × 103 )(2.1) = = 3.565 × 10−3 m EBC ABC (200 × 109 )(176.715 × 10−6 )

LAB = 1.2 m, E AB = 72 × 109 Pa, AAB = 1100 mm 2 = 1100 × 10−6 m 2

δ B /A =

PLAB (60 × 103 )(1.2) = = 909.1 × 10−6 m 2 E AB AAB (72 × 109 )(1100 × 10−6 )

δ C = δ B /A + δ C /B = 909.1 × 10−6 + 3.565 × 10−3 = 4.47 × 10−3 = 4.47 mm



PROBLEM 2.17 A 250-mm-long aluminum tube ( E = 70 GPa) of 36-mm outer diameter and 28-mm inner diameter can be closed at both ends by means of single-threaded screw-on covers of 1.5-mm pitch. With one cover screwed on tight, a solid brass rod ( E = 105 GPa) of 25-mm diameter is placed inside the tube and the second cover is screwed on. Since the rod is slightly longer than the tube, it is observed that the cover must be forced against the rod by rotating it one-quarter of a turn before it can be tightly closed. Determine (a) the average normal stress in the tube and in the rod, (b) the deformations of the tube and of the rod.

SOLUTION Atube = A rod =

π 4

π 4

(d o2 − di2 ) = d2 =

π 4

π 4

(362 − 282 ) = 402.12 mm 2 = 402.12 × 10−6 m 2

(25)2 = 490.87 mm 2 = 490.87 × 10−6 m 2

PL P(0.250) = = 8.8815 × 10−9 P 9 −6 Etube Atube (70 × 10 )(402.12 × 10 ) PL P(0.250) = −4.8505 × 10−9 P =− = Erod Arod (105 × 106 )(490.87 × 10−6 )

δ tube = δ rod

1 

 

δ * =  turn  × 1.5 mm = 0.375 mm = 375 × 10−6 m 4 δ tube = δ * + δ rod

or δ tube − δ rod = δ *

8.8815 × 10−9 P + 4.8505 × 10−9 P = 375 × 10−6 P=

(a)

σ tube =

(b)

27.308 × 103 = 67.9 × 106 Pa −6 402.12 × 10

σ tube = 67.9 MPa 

P 27.308 × 103 =− = −55.6 × 106 Pa −6 Arod 490.87 × 10

σ rod = −55.6 MPa 

P Atube

σ rod = −

0.375 × 10−3 = 27.308 × 103 N −9 (8.8815 + 4.8505)(10 )

=

δ tube = (8.8815 × 10−9 )(27.308 × 103 ) = 242.5 × 10−6 m

δ tube = 0.2425 mm 

δ rod = −(4.8505 × 10−9 )(27.308 × 103 ) = −132.5 × 10−6 m

δ rod = −0.1325 mm 

PROBLEM 2.19 Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B.

SOLUTION (a)

AAB = ABC =

π 4

π 4

2 d AB =

2 d BC =

π 4

π 4

(0.020)2 = 314.16 × 10−6 m 2 (0.060)2 = 2.8274 × 10−3 m 2

Force in member AB is P tension. Elongation:

δ AB =

PLAB (4 × 103 )(0.4) = = 72.756 × 10−6 m EAAB (70 × 109 )(314.16 × 10−6 )

Force in member BC is Q − P compression. Shortening:

δ BC =

(Q − P) LBC (Q − P)(0.5) = = 2.5263 × 10−9 (Q − P) EABC (70 × 109 )(2.8274 × 10−3 )

For zero deflection at A, δ BC = δ AB 2.5263 × 10−9 (Q − P) = 72.756 × 10−6 ∴ Q − P = 28.8 × 103 N Q = 28.3 × 103 + 4 × 103 = 32.8 × 103 N

(b)

δ AB = δ BC = δ B = 72.756 × 10−6 m

Q = 32.8 kN 

δ AB = 0.0728 mm ↓ 

PROBLEM 2.25 Each of the links AB and CD is made of aluminum ( E = 75 GPa) and has a cross-sectional area of 258 mm2. Knowing that they support the rigid member BC, determine the deflection of point E.

SOLUTION Use member BC as a free body ΣM C = 0: − (0.64) FAB + (0.44)(5 × 103 ) = 0 FAB = 3.4375 × 103 N ΣM B = 0: (0.64) FCD − (0.20)(5 × 103 ) = 0 FCD = 1.5625 × 103 N

For links AB and CD A = 258 mm 2 = 258 × 10−6 m 2

δ AB =

FAB LAB (3.4375 × 103 )(0.36) = = 63.953 × 10−6 m = δ B EA (75 × 109 )(258 × 10−6 )

δ CD =

FCD LCD (1.5625 × 103 )(0.36) = = 29.07 × 10−6 m = δ C 9 −6 EA (75 × 10 )(258 × 10 ) Slope θ =

δ B − δC lBC

=

34.883 × 10−6 = 54.505 × 10−6 rad 0.64

δ E = δ C + lECθ = 29.07 × 10−6 + (0.44)(54.505 × 10−6 ) = 53.05 × 10−6 m = 0.0531 mm



PROBLEM 2.35 The 1.35 m concrete post is reinforced with six steel bars, each with a 28 mm diameter. Knowing that Es = 200 GPa and Es = 29 GPa, determine the normal stresses in the steel and in the concrete when a 1560 kN axial centric force P is applied to the post.

SOLUTION Let Pc = portion of axial force carried by concrete Ps = portion carried by the six steel rods

δ=

Pc L E Aδ , Pc = c c Ec Ac L

δ=

Ps L E Aδ , Ps = s s Es As L

P = Pc + Ps = ( Ec Ac + Es As )

ε=

δ L

As = 6 Ac =

π

=

π 4

δ L

P Ec Ac + Es As

d s2 =

6π (28) 2 = 3694.5 mm 2 4

dc2 − As =

π

(450) 2 − 3694.5 = 155348.6 mm 2

4 4 L = 1.35 m = 1350 mm

ε=

−1560 × 103 = −297.48 × 10−6  29 × 109   200 × 109    (155348.6)   (3694.5) 6 6  10   10   200 × 109  −6 2  (−297.48 × 10 ) = −59.5 N/mm = −59.5 MPa 6 10  

σ s = Es ε = 



σ c = Ec ε =  29 × 

109 106

 −6 2  (−297.48 × 10 ) = −8.627 N/mm = −8.627 MPa 

 

PROBLEM 2.41 A steel tube ( Es = 200 GPa) with a 30-mm outer diameter and a 3-mm thickness is placed in a vise that is adjusted so that its jaws just touch the ends of the tube without exerting any pressure on them. The two forces shown are then applied to the tube. After these forces are applied, the vise is adjusted to decrease the distance between its jaws by 0.2 mm. Determine (a) the forces exerted by the vise on the tube at A and D, (b) the change in length of the portion BC of the tube.

SOLUTION For the tube

d D = 30 mm

di = do − 2t = 0.03 − 2(0.003) = 0.024 m A=

A to B:

π

(d 7

4

(0.032 − 0.0242 ) = 254.5 × 10−6 m 2

L = 0.075 m

RA (0.075) PL = = 1.473 × 10−9 RA m 9 −6 EA (200 × 10 )(254.5 × 10 )

P = RA + 32 kN L = 0.075 m

δ BC = C to D:

π

)

− di2 =

P = RA kN

δ AB = B to C:

2 o

( RA + 32000)(0.075) PL = = 1.473 × 10−9 RA + 47.15 × 10−6 m 9 −6 EA (200 × 10 )(254.5 × 10 )

P = RA + 8 kN,

δ CD =

L = 0.075 m

( RA + 8000)(0.075) PL = = 1.473 × 10−9 RA + 11.79 × 10−6 9 −6 EA (200 × 10 )(254.5 × 10 )

A to D:

δ AD = δ AB + δ BC + δ CD = 4.419 × 10−9 RA + 58.94 × 10−6 m

Given jaw movement

δ AD = −0.0002 m

(a)

4.419 × 10−9 RA + 58.94 × 10−6 = −0.0002 RA = −58597 N

RA = 58.6 kN

RD = RA + 8000 = −50597 N

(b)

−9

RD = 50.6 kN −6

−6

δ BC = (1.473 × 10 )(−58597) + 47.15 × 10 = 39.2 × 10 m −3

= 39.2 × 10 mm



PROBLEM 2.45 The steel rods BE and CD each have a 16-mm diameter ( E = 200 GPa); the ends of the rods are single-threaded with a pitch of 2.5 mm. Knowing that after being snugly fitted, the nut at C is tightened one full turn, determine (a) the tension in rod CD, (b) the deflection of point C of the rigid member ABC.

SOLUTION Let θ be the rotation of bar ABC as shown. Then

δ B = 0.15θ

But

δ C = δ turn − PCD = =

δ C = 0.25θ PCD LCD ECD ACD

ECD ACD (δ turn − δ C ) LCD (200 × 109 Pa) π4 (0.016 m) 2 2m

(0.0025m − 0.25θ )

= 50.265 × 103 − 5.0265 × 106 θ

δB = PBE =

PBE LBE EBE ABE

or

PBE =

EBE ABE δB LBE

(200 × 109 Pa) π4 (0.016 m) 2 3m

(0.15θ )

= 2.0106 × 106 θ

From free body of member ABC: ΣM A = 0 : 0.15 PBE − 0.25PCD = 0 0.15(2.0106 × 106 θ ) − 0.25(50.265 × 103 − 5.0265 × 106 θ ) = 0

θ = 8.0645 × 10−3 rad (a)

PCD = 50.265 × 103 − 5.0265 × 106 (8.0645 × 10−3 ) = 9.7288 × 103 N

(b)

PCD = 9.73 kN 

δ C = 0.25θ = 0.25(8.0645 × 10−3 ) = 2.0161 × 10−3 m

δ C = 2.02 mm ← 

PROBLEM 2.49 The aluminum shell is fully bonded to the brass core and the assembly is unstressed at a temperature of 15 °C. Considering only axial deformations, determine the stress in the aluminum when the temperature reaches 195 °C.

SOLUTION Brass core: E = 105 GPa

α = 20.9 × 10−6/ °C Aluminum shell: E = 70 GPa

α = 23.6 × 10−6 / °C Let L be the length of the assembly. Free thermal expansion: ΔT = 195 − 15 = 180 °C

Brass core: Aluminum shell:

(δT )b = Lα b ( ΔT ) (δT ) = Lα a (ΔT )

Net expansion of shell with respect to the core:

δ = L(α a − α b )(ΔT )

Let P be the tensile force in the core and the compressive force in the shell. Brass core:

Eb = 105 × 109 Pa Ab =

π

(25) 2 = 490.87 mm 2 4 = 490.87 × 10−6 m 2 PL (δ P )b = Eb Ab

PROBLEM 2.49 (Continued)

Aluminum shell:

Ea = 70 × 109 Pa Aa =

π

(602 − 252 ) 4 = 2.3366 × 103 mm 2

= 2.3366 × 10−3 m 2 δ = (δ P )b + (δ P ) a L(α b − α a )(ΔT ) =

PL PL + = KPL Eb Ab Ea Aa

where K= =

1 1 + Eb Ab Ea Aa 1 1 + −6 9 (105 × 10 )(490.87 × 10 ) (70 × 10 )(2.3366 × 10−3 ) 9

= 25.516 × 10−9 N −1

Then (α b − α a )(ΔT ) K (23.6 × 10−6 − 20.9 × 10−6 )(180) = 25.516 × 10−9 = 19.047 × 103 N

P=

Stress in aluminum:

σa = −

P 19.047 × 103 =− = −8.15 × 106 Pa −3 Aa 2.3366 × 10

σ a = −8.15 MPa 

PROBLEM 2.51 A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of steel ( Es = 200 GPa,α s = 11.7 × 10−6 / °C) and portion BC is made of brass ( Eb = 105 GPa, α b = 20.9 × 10−6 / °C). Knowing that the rod is initially unstressed, determine the compressive force induced in ABC when there is a temperature rise of 50 °C.

SOLUTION AAB = ABC =

π 4

π 4

2 d AB = 2 d BC =

π 4

π 4

(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 (50)2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2

Free thermal expansion:

δ T = LABα s (ΔT ) + LBCα b (ΔT ) = (0.250)(11.7 × 10−6 )(50) + (0.300)(20.9 × 10 −6 )(50) = 459.75 × 10−6 m Shortening due to induced compressive force P:

δP = =

PL PL + Es AAB Eb ABC 0.250 P 0.300 P + −6 9 (200 × 10 )(706.86 × 10 ) (105 × 10 )(1.9635 × 10 −3 ) 9

= 3.2235 × 10−9 P

For zero net deflection, δ P = δT 3.2235 × 10−9 P = 459.75 × 10−6 P = 142.62 × 103 N

P = 142.6 kN 

PROBLEM 2.55 A brass link ( Eb = 105 GPa, α b = 20.9 × 10−6 / °C) and a steel rod ( Es = 200 GPa, α s = 11.7 × 10−6 / °C) have the dimensions shown at a temperature of 20 °C. The steel rod is cooled until it fits freely into the link. The temperature of the whole assembly is then raised to 45 °C. Determine (a) the final stress in the steel rod, (b) the final length of the steel rod.

SOLUTION Initial dimensions at T = 20 °C. Final dimensions at T = 45 °C. ΔT = 45 − 20 = 25 °C

Free thermal expansion of each part: Brass link:

(δ T )b = α b (ΔT ) L = (20.9 × 10−6 )(25)(0.250) = 130.625 × 10−6 m

Steel rod:

(δ T ) s = α s (ΔT ) L = (11.7 × 10−6 )(25)(0.250) = 73.125 × 10−6 m

At the final temperature, the difference between the free length of the steel rod and the brass link is

δ = 120 × 10−6 + 73.125 × 10−6 − 130.625 × 10−6 = 62.5 × 10−6 m Add equal but opposite forces P to elongate the brass link and contract the steel rod. Brass link: E = 105 × 109 Pa Ab = (2)(50)(37.5) = 3750 mm 2 = 3.750 × 10−3 m 2 (δ P ) =

Steel rod:

PL P(0.250) = = 634.92 × 10−12 P EA (105 × 109 )(3.750 × 10−3 )

π

(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 4 PL P(0.250) = = 1.76838 × 10−9 P (δ P ) s = Es As (200 × 109 )(706.86 × 10−6 )

E = 200 × 109 Pa As =

(δ P )b + (δ P ) s = δ : 2.4033 × 10−9 P = 62.5 × 10−6 P = 26.006 × 103 N

P (26.006 × 103 ) =− = −36.8 × 106 Pa −6 As 706.86 × 10

(a)

Stress in steel rod:

σs = −

(b)

Final length of steel rod:

L f = L0 + (δ T ) s − (δ P ) s

σ s = −36.8 MPa 

L f = 0.250 + 120 × 10−6 + 73.125 × 10−6 − (1.76838 × 10−9 )(26.003 × 103 ) = 0.250147 m

L f = 250.147 mm 