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chap 4 68-86


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*4–68. The C83400-red-brass rod AB and 2014-T6- 4–69. The assembly has the diameters and material makeup 2014-T6 Aluminum 304 Stainless
aluminum rod BC are joined at the collar B and fixed A B C indicated. If it fits securely between its fixed supports when steel
C 86100 Bronze
connected at their ends. If there is no load in the members the temperature is T1 = 70°F, determine the average
when T1 = 50°F, determine the average normal stress in normal stress in each material when the temperature A 12 in. 8 in. D
each member when T2 = 120°F. Also, how far will the 3 ft 2 ft reaches T2 = 110°F.
B C 4 in.
collar be displaced? The cross-sectional area of each
member is 1.75 in2.
4 ft 6 ft 3 ft




©Fx = 0; Fbr = Fal = F ©Fx = 0; FA = FB = F

dN>C = 0 F(4)(12)
dA>D = 0; - + 12.8(10 - 6)(110 - 70)(4)(12)
p(6)2(10.6)(106)
Fbr LAB FalLBC
- + aB ¢T LAB - + aal ¢T LBC = 0
AAB Ebr ABCEal F(6)(12)
- + 9.60(10 - 6)(110 - 70)(6)(12)
p(4)2(15)(106)
F(3)(12)
- + 9.80(10 - 6)(120 - 50)(3)(12) F(3)(12)
(1.75)(14.6)(106) - + 9.60(10 - 6)(110 - 70)(3)(12) = 0
p(2)2(28)(106)

F(2)(12) F = 277.69 kip
- + 12.8(10 - 6)(120 - 50)(2)(12) = 0
1.75(10.6)(106)
277.69
F = 17 093.4 lb sal = = 2.46 ksi Ans.
p(6)2

17 093.4 277.69
sbr = sal = = 9.77 ksi Ans. sbr = = 5.52 ksi Ans.
1.75 p(4)2

9.77 ksi < (sg)al and (sg)br OK
277.69
sst = = 22.1 ksi Ans.
p(2)2
17 093.4(3)(12) -6
dB = - + 9.80(10 )(120 - 50)(3)(12)
1.75(14.6)(106)


dB = 0.611(10 - 3) in. : Ans.




Ans:
sal = 2.46 ksi, sbr = 5.52 ksi, sst = 22.1 ksi

251 25 2
4–70. The rod is made of A992 steel and has a diameter of k 1000 lb/ in. k 1000 lb/in. 4–71. If the assembly fits snugly between two rigid L L
0.25 in. If the rod is 4 ft long when the springs are compressed supports A and C when the temperature is at T1 , determine 2 2
0.5 in. and the temperature of the rod is T = 40°F, determine the normal stress developed in both rod segments when the
the force in the rod when its temperature is T = 160°F. 4 ft temperature rises to T2. Both segments are made of the
same material, having a modulus of elasticity of E and C
A B 1d
coefficient of thermal expansion of a. d 2




Compatibility:

+)
(: x = dT - dF

x = 6.60(10 - 6)(160 - 40)(2)(12)
Compatibility Equation: When the assembly is unconstrained, it has a free
1.00(x + 0.5)(2)(12) expansion of d T = a¢TL = a(T2 - T1)L. Using the method of superposition,
- p 2 3 Fig. a,
4 (0.25 )(29.0)(10 )

x = 0.01040 in. +)
(: 0 = d T - dF



0 = a(T2 - T1)L - ≥ ¥
F = 1.00(0.01040 + 0.5) = 0.510 kip Ans.
F(L>2) F(L>2)

a b E a d bE
+
p d 2 p 2
4 2 4

a(T2 - T1)pd2E
F =
10

Normal Stress:

a(T2 - T1)pd2E
F 10 2
sAB = = = a(T2 - T1)E Ans.
AAB p 2 5
d
4

a(T2 - T1)pd2E


a b
F 10 8
sBC = = = a(T2 - T1)E Ans.
ABC p d 2 5
4 2




Ans:
Ans:
2 8
F = 0.510 kip sAB = a (T2 - T1)E, sBC = a (T2 - T1)E
5 5

253 25 4
*4–72. If the assembly fits snugly between the two L L 4–73. The pipe is made of A992 steel and is connected to
supports A and C when the temperature is at T1, determine 2 2 the collars at A and B. When the temperature is 60°F, there is

causes its temperature to rise by ¢T = 140 + 15x2°F, where
the normal stress developed in both segments when the no axial load in the pipe. If hot gas traveling through the pipe
temperature rises to T2. Both segments are made of
the same material having a modulus of elasticity of E and C x is in feet, determine the average normal stress in the pipe.
A B 1d A B
coefficient of the thermal expansion of a. The flexible d 2 The inner diameter is 2 in., the wall thickness is 0.15 in.
8 ft
supports at A and C each have a stiffness k.


Compatibility:
L


L0
0 = dT - dF Where dT = a ¢T dx


0 = 6.60 A 10 - 6 B
Compatibility Equation: When the assembly is unconstrained, it has a free
expansion of dT = a¢TL = a(T2 - T1)L. Using the method of superposition, 8 ft
F(8)
L0
Fig. a, (40 + 15 x) dx -
A(29.0)(103)

0 = 6.60 A 10 - 6 B B 40(8) +
+ ) dC = dT - dF
(: 15(8)2 F(8)
R -
2 A(29.0)(103)
F(L>2) F(L>2)

a b E a d2 bE
F F
= a(T2 - T1)L - D + + T F = 19.14 A
k p d 2 p k
4 2 4 Average Normal Stress:

2
a(T2 - T1)Lpd Ek 19.14 A
F = s = = 19.1 ksi Ans.
10kL + 2pd2E A


Normal Stress:

a(T2 - T1)Lpd2Ek
F 10kL + 2pd2E 4Eka(T2 - T1)L
sAB = = = Ans.
AAB p 2 10kL + 2pd2E
d
4

a(T2 - T1)Lpd2Ek
10kL + 2pd2E 16Eka(T2 - T1)L

a b
F
sBC = = = Ans.
ABC p d 2 10kL + 2pd2E
4 2




Ans:
s = 19.1 ksi

255 25 6
4–74. The bronze C86100 pipe has an inner radius of 4–75. The 40-ft-long A-36 steel rails on a train track are d d
0.5 in. and a wall thickness of 0.2 in. If the gas flowing laid with a small gap between them to allow for thermal
through it changes the temperature of the pipe uniformly expansion. Determine the required gap d so that the rails
from TA = 200°F at A to TB = 60°F at B, determine the just touch one another when the temperature is increased 40 ft
axial fo

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