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CHAPTER 2
PROBLEM 2.1
An 80-m-long wire of 5-mm diameter is made of a steel with E = 200 GPa and an ultimate tensile strength of
400 MPa. If a factor of safety of 3.2 is desired, determine (a) the largest allowable tension in the wire, (b) the
corresponding elongation of the wire.
SOLUTION
(a)
σ U = 400 × 106 Pa
A=
π
4
d2 =
π
4
(5) 2 = 19.635 mm 2 = 19.635 × 10−6 m 2
PU = σ U A = (400 × 106 ) (19.635 × 10−6 ) = 7854 N
Pall =
(b)
δ=
PU
7854
=
= 2454 N
F .S
3.2
PL
(2454) (80)
=
= 50.0 × 10−3 m
AE (19.635 × 10−6 )(200 × 109 )
Pall = 2.45 kN 
δ = 50.0 mm 
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PROBLEM 2.2
A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2-kip tensile load is applied to it.
Knowing that E = 29 × 106 psi, determine (a) the smallest diameter rod that should be used, (b) the
corresponding normal stress caused by the load.
SOLUTION
(a)
δ=
PL
AE
: 0.04 in. =
(2000 lb) (5.5 × 12 in.)
6
A (29 × 10 psi)
1
A = π d 2 = 0.11379 in 2
4
d = 0.38063 in.
(b)
σ=
P
A
=
2000 lb
0.11379 in
2
= 17580 psi
d = 0.381 in. 
σ = 17.58 ksi 
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PROBLEM 2.3
Two gage marks are placed exactly 10 in. apart on a 12 -in.-diameter aluminum rod with E = 10.1 × 106 psi and
an ultimate strength of 16 ksi. Knowing that the distance between the gage marks is 10.009 in. after a load is
applied, determine (a) the stress in the rod, (b) the factor of safety.
SOLUTION
(a)
δ = 10.009 − 10.000 = 0.009 in.
ε=
(b)
F. S . =
σU
16
=
9.09
σ
δ
L
=
σ
E
σ=
Eδ (10.1 × 106 ) (0.009)
=
= 9.09 × 103 psi
L
10
σ = 9.09 ksi 
F. S . = 1.760 
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PROBLEM 2.4
An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam.
It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that E = 200 GPa,
determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.
SOLUTION
(a)
δ=
PL
, or
AE
P=
δ AE
L
1
1
with A = π d 2 = π (0.005) 2 = 19.6350 × 10−6 m 2
4
4
P=
(0.045 m)(19.6350 × 10−6 m 2 )(200 × 109 N/m 2 )
= 9817.5 N
18 m
P = 9.82 kN 
(b)
σ=
P
A
=
9817.5 N
19.6350 × 10
−6
6
m
2
= 500 × 10 Pa
σ = 500 MPa 
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PROBLEM 2.5
A polystyrene rod of length 12 in. and diameter 0.5 in. is subjected to an 800-lb tensile load. Knowing that
E = 0.45 × 106 psi, determine (a) the elongation of the rod, (b) the normal stress in the rod.
SOLUTION
A=
(a)
δ=
(b)
σ=
PL
AE
P
A
=
=
(800) (12)
6
(0.19635) (0.45 ×10 )
800
0.19635
= 4074 psi
π
4
d2 =
π
4
= 0.1086
(0.5) 2 = 0.19635 in 2
δ = 0.1086 in. 
σ = 4.07 ksi 
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PROBLEM 2.6
A nylon thread is subjected to a 8.5-N tension force. Knowing that E = 3.3 GPa and that the length of the
thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.
SOLUTION
(a)
1.1
= 0.011
100
ε=
Stress:
σ = Eε = (3.3 × 109 )(0.011) = 36.3 × 106 Pa
σ =
(b)
δ
Strain:
Area:
A=
Diameter:
d =
Stress:
=
L
P
A
P
σ
=
4A
π
8.5
= 234.16 × 10−9 m 2
36.3 × 106
=
(4)(234.16 × 10−9 )
π
= 546 × 10−6 m
d = 0.546 mm 
σ = 36.3 MPa 
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PROBLEM 2.7
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod. Knowing that, with an
axial load of 6000 N acting on the rod, the distance between the gage marks is 250.18 mm, determine the
modulus of elasticity of the aluminum used in the rod.
SOLUTION
δ = Δ L = L − L0 = 250.18 − 250.00 = 0.18 mm
δ 0.18 mm
ε=
=
= 0.00072
L0
π
250 mm
π
(12)2 = 113.097 mm 2 = 113.097 ×10−6 m 2
4
4
6000
P
σ= =

= 53.052 × 106 Pa
A 113.097 × 10−6
A=
E=
d2 =
σ 53.052 × 106
=
= 73.683 × 109 Pa
ε
0.00072
E = 73.7 GPa 
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PROBLEM 2.8
An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that
E = 10.1 × 106 psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum
allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.
SOLUTION
(a)
δ=
PL
;
AE
L=
Thus,
EAδ
Eδ
(10.1 × 106 ) (0.05)
=
=
σ
P
14 × 103
L = 36.1 in. 
(b)
σ =
Thus,
P
;
A
A=
P
σ
=
127.5 × 103
14 × 103
A = 9.11 in 2 
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PROBLEM 2.9
An aluminum control rod must stretch 0.08 in. when a 500-lb tensile load is applied to it. Knowing that
σ all = 22 ksi and E = 10.1 × 106 psi, determine the smallest diameter and shortest length that can be selected for
the rod.
SOLUTION
σ all = 22 × 103 psi
P = 500 lb, δ = 0.08 in.
σ=
A=
P
< σ all
A
π
4
d2
A>
d=
P
σ all
=
4A
π
500
= 0.022727 in 2
22 × 103
=
(4)(0.022727)
π
Eδ
< σ all
L
Eδ (10.1 × 106 )(0.08)
L>
=
= 36.7 in.
σ all
22 × 103
d min = 0.1701 in. 
σ = Eε =
Lmin = 36.7 in. 
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PROBLEM 2.10
A square yellow-brass bar must not stretch more than 2.5 mm when it is subjected to a tensile load. Knowing
that E = 105 GPa and that the allowable tensile strength is 180 MPa, determine (a) the maximum allowable
length of the bar, (b) the required dimensions of the cross section if the tensile load is 40 kN.
SOLUTION
σ = 180 × 106 Pa P = 40 × 103 N
E = 105 × 109 Pa δ = 2.5 × 10−3 m
(a)
PL σ L
=
AE
E
Eδ (105 × 109 )(2.5 × 10−3 )
=
= 1.45833 m
L=
σ
180 × 106
δ=
L = 1.458 m 
(b)
σ =
A=
P
A
P
σ
A = a2
=
40 × 103
= 222.22 × 10−6 m 2 = 222.22 mm 2
180 × 106
a =
A =
222.22
a = 14.91 mm 
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PROBLEM 2.11
A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when
the rod is subjected to a 10-kN axial load. Knowing that E = 200 GPa, determine the required diameter of
the rod.
SOLUTION
L =4m
δ = 3 × 10−3 m,
σ = 150 × 106 Pa
E = 200 × 109 Pa, P = 10 × 103 N
Stress:
σ =
A=
Deformation:
P
A
P
σ
=
10 × 103
= 66.667 × 10−6 m 2 = 66.667 mm 2
6
150 × 10
δ =
PL
AE
A=
PL
(10 × 103 ) (4)
=
= 66.667 × 10−6 m 2 = 66.667 mm 2
Eδ
(200 × 109 ) (3 × 10−3 )
The larger value of A governs:
A = 66.667 mm 2
A=
π
4
d2
d=
4A
π
=
4(66.667)
π
d = 9.21 mm 
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PROBLEM 2.12
A nylon thread is to be subjected to a 10-N tension. Knowing that E = 3.2 GPa, that the maximum allowable
normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the
required diameter of the thread.
SOLUTION
Stress criterion:
σ = 40 MPa = 40 × 106 Pa P = 10 N
σ =
A=
P
P
10 N
: A=
=
= 250 × 10−9 m 2
A
σ
40 × 106 Pa
π
4
d 2: d = 2
A
π
250 × 10−9
=2
π
= 564.19 × 10−6 m
d = 0.564 mm
Elongation criterion:
δ
L
= 1% = 0.01
δ =
PL
:
AE
A=
P /E 10 N/3.2 × 109 Pa
=
= 312.5 × 10−9 m 2
0.01
δ /L
d =2
A
π
=2
312.5 × 10−9
π
= 630.78 × 10−6 m 2
d = 0.631 mm
The required diameter is the larger value:
d = 0.631 mm 
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PROBLEM 2.13
The 4-mm-diameter cable BC is made of a steel with E = 200 GPa. Knowing
that the maximum stress in the cable must not exceed 190 MPa and that the
elongation of the cable must not exceed 6 mm, find the maximum load P that
can be applied as shown.
SOLUTION
LBC = 62 + 42 = 7.2111 m
Use bar AB as a free body.
 4

3.5 P − (6) 
FBC  = 0
7.2111


P = 0.9509 FBC
Σ M A = 0:
Considering allowable stress: σ = 190 × 106 Pa
A=
π
d2 =
4
FBC
σ=
A
π
4
(0.004) 2 = 12.566 × 10−6 m 2
∴ FBC = σ A = (190 × 106 ) (12.566 × 10−6 ) = 2.388 × 103 N
Considering allowable elongation: δ = 6 × 10−3 m
δ=
FBC LBC
AE
∴ FBC =
AEδ (12.566 × 10−6 )(200 × 109 )(6 × 10−3 )
=
= 2.091 × 103 N
LBC
7.2111
Smaller value governs. FBC = 2.091 × 103 N
P = 0.9509 FBC = (0.9509)(2.091 × 103 ) = 1.988 × 103 N
P = 1.988 kN 
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PROBLEM 2.14
The aluminum rod ABC ( E = 10.1 × 106 psi), which consists of two
cylindrical portions AB and BC, is to be replaced with a cylindrical steel
rod DE ( E = 29 × 106 psi) of the same overall length. Determine the
minimum required diameter d of the steel rod if its vertical deformation is
not to exceed the deformation of the aluminum rod under the same load
and if the allowable stress in the steel rod is not to exceed 24 ksi.
SOLUTION
Deformation of aluminum rod.
δA =
=
PLAB PLBC
+
AAB E ABC E
P  LAB LBC 
+


E  AAB ABC 
 12
+
π
 (1.5)2
4
= 0.031376 in.
=
Steel rod.
28 × 103
10.1 × 106
18
(2.25) 2
4
π




δ = 0.031376 in.
δ=
PL
PL
(28 × 103 )(30)
∴ A=
=
= 0.92317 in 2
6
EA
Eδ (29 × 10 )(0.031376)
σ=
P
A
∴ A=
P
σ
=
28 × 103
= 1.1667 in 2
24 × 103
Required area is the larger value.
A = 1.1667 in 2
Diameter:
d=
4A
π
=
(4)(1.6667)
π
d = 1.219 in. 
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PROBLEM 2.15
A 4-ft section of aluminum pipe of cross-sectional area 1.75 in2 rests on a
fixed support at A. The 58 -in.-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 29 × 106 psi for steel, and 10.4 × 106 psi for aluminum,
determine the deflection of point C when a 15-kip force is applied at C.
SOLUTION
Rod BC:
LBC = 7 ft = 84 in. EBC = 29 × 106 psi
ABC =
δ C/B =
Pipe AB:
π
4
d2 =
π
4
(0.625) 2 = 0.30680 in 2
PLBC
(15 × 103 )(84)
=
= 0.141618 in.
EBC ABC
(29 × 106 )(0.30680)
LAB = 4 ft = 48 in. E AB = 10.4 × 106 psi
AAB = 1.75 in 2
δ B/A =
Total:
PLAB
(15 × 103 ) (48)
=
= 39.560 × 10−3 in.
6
E AB AAB (10.4 × 10 ) (1.75)
δ C = δ B/A + δ C/B = 39.560 × 10−3 + 0.141618 = 0.181178 in.
δ C = 0.1812 in. 
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PROBLEM 2.16
The brass tube AB ( E = 105 GPa) has a cross-sectional area of
140 mm2 and is fitted with a plug at A. The tube is attached at B to a
rigid plate that is itself attached at C to the bottom of an aluminum
cylinder ( E = 72 GPa) with a cross-sectional area of 250 mm2. The
cylinder is then hung from a support at D. In order to close the
cylinder, the plug must move down through 1 mm. Determine the force
P that must be applied to the cylinder.
SOLUTION
Shortening of brass tube AB:
LAB = 375 + 1 = 376 mm = 0.376 m
AAB = 140 mm 2 = 140 × 10−6 m 2
E AB = 105 × 109 Pa
δ AB =
PLAB
P(0.376)
=
= 25.578 × 10−9 P
−6
9
E AB AAB
(105 × 10 )(140 × 10 )
Lengthening of aluminum cylinder CD:
LCD = 0.375 m
δ CD =
Total deflection:
ACD = 250 mm 2 = 250 × 10−6 m 2
ECD = 72 × 109 Pa
PLCD
P(0.375)
=
= 20.833 × 10−9 P
9
−6
ECD ACD (72 × 10 )(250 × 10 )
δ A = δ AB + δ CD where δ A = 0.001 m
0.001 = (25.578 × 10−9 + 20.833 × 10−9 ) P
P = 21.547 × 103 N
P = 21.5 kN 
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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 =
P
27.308 × 103
=
= 67.9 × 106 Pa
Atube 402.12 × 10−6
σ rod = −
(b)
0.375 × 10−3
= 27.308 × 103 N
−9
(8.8815 + 4.8505)(10 )
P
27.308 × 103
=−
= −55.6 × 106 Pa
Arod
490.87 × 10−6
δ tube = (8.8815 × 10−9 )(27.308 × 103 ) = 242.5 × 10−6 m
δ rod = −(4.8505 × 10−9 )(27.308 × 103 ) = −132.5 × 10−6 m
σ tube = 67.9 MPa 
σ rod = −55.6 MPa 
δ tube = 0.2425 mm 
δ rod = −0.1325 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 2.18
The specimen shown is made from a 1-in.-diameter cylindrical steel rod
with two 1.5-in.-outer-diameter sleeves bonded to the rod as shown.
Knowing that E = 29 × 106 psi, determine (a) the load P so that the
total deformation is 0.002 in., (b) the corresponding deformation of the
central portion BC.
SOLUTION
(a)
δ =Σ
Pi Li P Li
= Σ
Ai Ei E Ai
−1
 L 
π
P = Eδ  Σ i  Ai = di2
4
 Ai 
L, in.
d, in.
A, in2
L/A, in–1
AB
2
1.5
1.7671
1.1318
BC
3
1.0
0.7854
3.8197
CD
2
1.5
1.7671
1.1318
6.083
P = (29 × 106 )(0.002)(6.083) −1 = 9.353 × 103 lb
(b)
δ BC =
PLBC P LBC 9.535 × 103
=
=
(3.8197)
ABC E E ABC
29 × 106
← sum
P = 9.53 kips 
δ = 1.254 × 10−3 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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
(0.020)2 = 314.16 × 10−6 m 2
π
(0.060)2 = 2.8274 × 10−3 m 2
4
Force in member AB is P tension.
Elongation:
δ AB =
PLAB
(4 × 103 )(0.4)
=
= 72.756 × 10−6 m
−6
9
EAAB (70 × 10 )(314.16 × 10 )
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 ↓ 
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PROBLEM 2.20
The rod ABC is made of an aluminum for which E = 70 GPa. Knowing that
P = 6 kN and Q = 42 kN, determine the deflection of (a) point A, (b) point B.
SOLUTION
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
PAB = P = 6 × 103 N
PBC = P − Q = 6 × 103 − 42 × 103 = −36 × 103 N
LAB = 0.4 m LBC = 0.5 m
δ AB =
PAB LAB
(6 × 103 )(0.4)
=
= 109.135 × 10−6 m
AAB E A (314.16 × 10−6 )(70 × 109 )
δ BC =
PBC LBC
(−36 × 103 )(0.5)
=
= −90.947 × 10−6 m
ABC E
(2.8274 × 10−3 )(70 × 109 )
(a)
δ A = δ AB + δ BC = 109.135 × 10−6 − 90.947 × 10−6 m = 18.19 × 10−6 m
(b)
δ B = δ BC = −90.9 × 10−6 m = −0.0909 mm
or
δ A = 0.01819 mm ↑ 
δ B = 0.0919 mm ↓ 
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PROBLEM 2.21
Members AB and BC are made of steel ( E = 29 × 106 psi) with crosssectional areas of 0.80 in2 and 0.64 in2, respectively. For the loading
shown, determine the elongation of (a) member AB, (b) member BC.
SOLUTION
(a)
LAB =
62 + 52 = 7.810 ft = 93.72 in.
Use joint A as a free body.
5
FAB − 28 = 0
7.810
= 43.74 kip = 43.74 × 103 lb
ΣFy = 0:
FAB
δ AB =
(b)
FAB LAB
(43.74 × 103 ) (93.72)
=
EAAB
(29 × 106 ) (0.80)
δ AB = 0.1767 in. 
Use joint B as a free body.
ΣFx = 0: FBC −
FBC =
δ BC =
6
FAB = 0
7.810
(6) (43.74)
= 33.60 kip = 33.60 × 103 lb.
7.810
FBC LBC
(33.60 × 103 ) (72)
=
EABC
(29 × 106 ) (0.64)
δ BC = 0.1304 in. 
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PROBLEM 2.22
The steel frame ( E = 200 GPa) shown has a diagonal brace BD with
an area of 1920 mm2. Determine the largest allowable load P if the
change in length of member BD is not to exceed 1.6 mm.
SOLUTION
δ BD = 1.6 × 10−3 m, ABD = 1920 mm 2 = 1920 × 10−6 m 2
LBD =
52 + 62 = 7.810 m, EBD = 200 × 109 Pa
δ BD =
FBD LBD
EBD ABD
FBD =
(200 × 109 ) (1920 × 10−6 )(1.6 × 10−3 )
EBD ABDδ BD
=
7.81
LBD
= 78.67 × 103 N
Use joint B as a free body.
ΣFx = 0:
5
FBD − P = 0
7.810
P=
5
(5)(78.67 × 103 )
FBD =
7.810
7.810
= 50.4 × 103 N
P = 50.4 kN 
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PROBLEM 2.23
For the steel truss ( E = 200 GPa) and loading shown, determine
the deformations of the members AB and AD, knowing that their
cross-sectional areas are 2400 mm2 and 1800 mm2, respectively.
SOLUTION
Statics: Reactions are 114 kN upward at A and C.
Member BD is a zero force member.
LAB = 4.02 + 2.52 = 4.717 m
Use joint A as a free body.
ΣFy = 0 : 114 +
2.5
FAB = 0
4.717
FAB = −215.10 kN
ΣFx = 0 : FAD +
FAD = −
4
FAB = 0
4.717
(4)(−215.10)
= 182.4 kN
4.717
Member AB:
δ AB =
FAB LAB
(−215.10 × 103 )(4.717)
=
EAAB
(200 × 109 )(2400 × 10−6 )
= −2.11 × 10−3 m
Member AD:
δ AD =
δ AB − 2.11 mm 
FAD LAD
(182.4 × 103 )(4.0)
=
EAAD
(200 × 109 )(1800 × 10−6 )
= 2.03 × 10−3 m
δ AD = 2.03 mm 
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PROBLEM 2.24
For the steel truss ( E = 29 × 106 psi) and loading shown, determine the
deformations of the members BD and DE, knowing that their crosssectional areas are 2 in2 and 3 in2, respectively.
SOLUTION
Free body: Portion ABC of truss
ΣM E = 0 : FBD (15 ft) − (30 kips)(8 ft) − (30 kips)(16 ft) = 0
FBD = + 48.0 kips
Free body: Portion ABEC of truss
ΣFx = 0 : 30 kips + 30 kips − FDE = 0 
FDE = + 60.0 kips

δ BD =
PL (+48.0 × 103 lb)(8 × 12 in.)
=
AE
(2 in 2 )(29 × 106 psi)
δ BD = +79.4 × 10−3 in. 
δ DE =
PL (+60.0 × 103 lb)(15 × 12 in.)
=
AE
(3 in 2 )(29 × 106 psi)
δ DE + 124.1 × 10−3 in. 
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PROBLEM 2.25
Each of the links AB and CD is made of aluminum ( E = 10.9 × 106 psi)
and has a cross-sectional area of 0.2 in.2. Knowing that they support the
rigid member BC, determine the deflection of point E.
SOLUTION
Free body BC:
ΣM C = 0 : − (32) FAB + (22) (1 × 103 ) = 0
FAB = 687.5 lb
ΣFy = 0 : 687.5 − 1 × 103 + FCD = 0
FCD = 312.5 lb
FAB LAB
(687.5) (18)
=
= 5.6766 × 10−3 in = δ B
6
EA
(10.9 × 10 ) (0.2)
F L
(312.5) (18)
= CD CD =
= 2.5803 × 10−3 in = δ C
6
EA
(10.9 × 10 ) (0.2)
δ AB =
δ CD
Deformation diagram:
Slope θ =
δ B − δC
LBC
=
3.0963 × 10−3
32
= 96.759 × 10−6 rad
δ E = δ C + LECθ
= 2.5803 × 10−3 + (22) (96.759 × 10−6 )
= 4.7090 × 10−3 in
δ E = 4.71 × 10−3 in ↓ 
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PROBLEM 2.26
3
The length of the 32
-in.-diameter steel wire CD has been adjusted so that
with no load applied, a gap of 161 in. exists between the end B of the rigid
beam ACB and a contact point E. Knowing that E = 29 × 106 psi,
determine where a 50-lb block should be placed on the beam in order to
cause contact between B and E.
SOLUTION
Rigid beam ACB rotates through angle θ to close gap.
θ=
1/16
= 3.125 × 10−3 rad
20
Point C moves downward.
δ C = 4θ = 4(3.125 × 10−3 ) = 12.5 × 10−3 in.
δ CD = δ C = 12.5 × 10−3 in.
ACD =
δ CD
π
d2 =
π 3 
d
F L
= CD CD
EACD
FCD =
2
= 6.9029 × 10−3 in 2


4  32 
EACDδ CD (29 × 106 ) (6.9029 × 10−3 ) (12.5 × 10−3 )
=
12.5
LCD
= 200.18 lb
Free body ACB:
ΣM A = 0: 4 FCD − (50) (20 − x) = 0
(4) (200.18)
= 16.0144
50
x = 3.9856 in.
20 − x =
For contact,
x < 3.99 in. 
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PROBLEM 2.27
Link BD is made of brass ( E = 105 GPa) and has a cross-sectional area of
240 mm2. Link CE is made of aluminum ( E = 72 GPa) and has a crosssectional area of 300 mm2. Knowing that they support rigid member ABC
determine the maximum force P that can be applied vertically at point A if
the deflection of A is not to exceed 0.35 mm.
SOLUTION
Free body member AC:
ΣM C = 0 : 0.350 P − 0.225 FBD = 0
FBD = 1.55556 P
ΣM B = 0 : 0.125 P − 0.225 FCE = 0
FCE = 0.55556 P
FBD LBD
(1.55556 P) (0.225)
=
= 13.8889 × 10−9 P
9
−6
EBD ABD (105 × 10 ) (240 × 10 )
F L
(0.55556 P) (0.150)
= CE CE =
= 3.8581 × 10−9 P
ECE ACE (72 × 109 ) (300 × 10−6 )
δ B = δ BD =
δ C = δ CE
Deformation Diagram:
From the deformation diagram,
Slope,
θ=
δ B + δC
LBC
=
17.7470 × 10−9 P
= 78.876 × 10−9 P
0.225
δ A = δ B + LABθ
= 13.8889 × 10−9 P + (0.125) (78.876 × 10−9 P)
= 23.748 × 10−9 P
Apply displacement limit. δ A = 0.35 × 10−3 m = 23.748 × 10−9 P
P = 14.7381 × 103 N
P = 14.74 kN 
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PROBLEM 2.28
Each of the four vertical links connecting the two rigid
horizontal members is made of aluminum ( E = 70 GPa)
and has a uniform rectangular cross section of 10 × 40 mm.
For the loading shown, determine the deflection of
(a) point E, (b) point F, (c) point G.
SOLUTION
Statics. Free body EFG.
ΣM F = 0 : − (400)(2 FBE ) − (250)(24) = 0
FBE = −7.5 kN = −7.5 × 103 N
ΣM E = 0 : (400)(2 FCF ) − (650)(24) = 0
FCF = 19.5 kN = 19.5 × 103 N
Area of one link:
A = (10)(40) = 400 mm 2
= 400 × 10−6 m 2
Length: L = 300 mm = 0.300 m
Deformations.
δ BE =
FBE L
(−7.5 × 103 )(0.300)
=
= −80.357 × 10−6 m
9
−6
EA
(70 × 10 )(400 × 10 )
δ CF =
FCF L
(19.5 × 103 )(0.300)
=
= 208.93 × 10−6 m
EA
(70 × 109 )(400 × 10−6 )
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PROBLEM 2.28 (Continued)
(a)
Deflection of Point E.
δ E = |δ BF |
δ E = 80.4 μ m ↑ 
(b)
Deflection of Point F.
δ F = δ CF
δ F = 209 μ m ↓ 
Geometry change.
Let θ be the small change in slope angle.
θ=
(c)
δE + δF
LEF
Deflection of Point G.
=
80.357 × 10−6 + 208.93 × 10−6
= 723.22 × 10−6 radians
0.400
δ G = δ F + LFG θ
δ G = δ F + LFG θ = 208.93 × 10−6 + (0.250)(723.22 × 10−6 )
= 389.73 × 10−6 m
δ G = 390 μ m ↓ 
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PROBLEM 2.29
A vertical load P is applied at the center A of the upper section of a
homogeneous frustum of a circular cone of height h, minimum radius a,
and maximum radius b. Denoting by E the modulus of elasticity of the
material and neglecting the effect of its weight, determine the deflection of
point A.
SOLUTION
Extend the slant sides of the cone to meet at a point O and place the origin of the coordinate system there.
tan α =
From geometry,
a1 =
b−a
h
a
b
, b1 =
,
tan α
tan α
r = y tan α
At coordinate point y, A = π r 2
Deformation of element of height dy:
dδ =
Pdy
AE
dδ =
P dy
P
dy
=
2
2
Eπ r
π E tan α y 2
Total deformation.
P
δA =
π E tan 2 α
=

b1
a1
 1
dy
P
=
− 
2
2
π E tan α  y 
y
b1 − a1 P(b1 − a1 )
P
=
2
π Eab
π E tan α a1b1
b1
=
a1
1 1
P
 − 
2
π E tan α  a1 b1 
δA =
Ph
↓ 
π Eab
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PROBLEM 2.30
A homogeneous cable of length L and uniform cross section is suspended from one end. (a) Denoting by ρ the
density (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of the
cable due to its own weight. (b) Show that the same elongation would be obtained if the cable were horizontal
and if a force equal to half of its weight were applied at each end.
SOLUTION
(a)
For element at point identified by coordinate y,
P = weight of portion below the point
= ρ g A(L − y )
Pdy ρ gA( L − y )dy ρ g ( L − y )
dδ =
=
=
dy
EA
EA
E
δ=

=
(b)
Total weight:
L
ρ g (L − y)
0
E
dy =
ρg 
L
1 2
 Ly − y 
E 
2 0
ρg 
2
L 
2
 L −

2 
E 
δ=
1 ρ gL2

2 E
W = ρ gAL
F=
EAδ EA 1 ρ gL2 1
=
⋅
= ρ gAL
L
L 2 E
2
1
F= W 
2
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PROBLEM 2.31
The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initial
diameter of the specimen is d1, show that when the diameter is d, the true strain is ε t = 2 ln(d1 /d ).
SOLUTION
If the volume is constant,
π
4
d 2L =
π
4
d12 L0
L d12  d1 
=
=
L0 d 2  d 
ε t = ln
2
L
d 
= ln  1 
L0
d 
2
ε t = 2ln
d1

d
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PROBLEM 2.32
Denoting by ε the “engineering strain” in a tensile specimen, show that the true strain is ε t = ln (1 + ε ).
SOLUTION
ε t = ln
Thus

L +δ
L
δ 
= ln 0
= ln 1 +  = ln (1 + ε )
L0
L0
L0 

ε t = ln (1 + ε ) 
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PROBLEM 2.33
An axial force of 200 kN is applied to the assembly shown by means of
rigid end plates. Determine (a) the normal stress in the aluminum shell,
(b) the corresponding deformation of the assembly.
SOLUTION
Let Pa = Portion of axial force carried by shell
Pb = Portion of axial force carried by core.
Thus,
with
δ=
Pa L
, or
Ea Aa
Pa =
Ea Aa
δ
L
δ=
Pb L
, or
Eb Ab
Pb =
Eb Ab
δ
L
P = Pa + Pb = ( Ea Aa + Eb Ab )
Aa =
Ab =
π
4
π
4
δ
L
[(0.060)2 − (0.025)2 ] = 2.3366 × 10−3 m 2
(0.025) 2 = 0.49087 × 10−3 m 2
P = [(70 × 109 ) (2.3366 × 10−3 ) + (105 × 109 ) (0.49087 × 10−3 )]
P = 215.10 × 106
Strain:
ε =
δ
L
=
δ
L
δ
L
200 × 103
P
=
= 0.92980 × 10−3
215.10 × 106
215.10 × 106
(a)
σ a = Ea ε = (70 × 109 ) (0.92980 × 10−3 ) = 65.1 × 106 Pa
(b)
δ = ε L = (0.92980 × 10−3 ) (300 mm)
σ a = 65.1 MPa 
δ = 0.279 mm 
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PROBLEM 2.34
The length of the assembly shown decreases by 0.40 mm when an axial
force is applied by means of rigid end plates. Determine (a) the magnitude
of the applied force, (b) the corresponding stress in the brass core.
SOLUTION
Let Pa = Portion of axial force carried by shell and Pb = Portion of axial force carried by core.
Thus,
with
δ=
Pa L
, or
Ea Aa
Pa =
Ea Aa
δ
L
δ=
Pb L
, or
Eb Ab
Pb =
Eb Ab
δ
L
P = Pa + Pb = ( Ea Aa + Eb Ab )
Aa =
Ab =
π
4
π
4
δ
L
[(0.060)2 − (0.025)2 ] = 2.3366 × 10−3 m 2
(0.025)2 = 0.49087 × 10−3 m 2
P = [(70 × 109 ) (2.3366 × 10−3 ) + (105 × 109 ) (0.49087 × 10−3 )]
with
δ
L
= 215.10 × 106
δ
L
δ = 0.40 mm, L = 300 mm
(a)
P = (215.10 × 106 )
(b)
σb =
0.40
= 286.8 × 103 N
300
Pb
Eδ
(105 × 109 )(0.40 × 10−3 )
= b =
= 140 × 106 Pa
−3
Ab
L
300 × 10
P = 287 kN 
σ b = 140.0 MPa 
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PROBLEM 2.35
A 4-ft concrete post is reinforced with four steel bars, each with a 34 -in. diameter.
Knowing that Es = 29 × 106 psi and Ec = 3.6 × 106 psi, determine the normal
stresses in the steel and in the concrete when a 150-kip axial centric force P is
applied to the post.
SOLUTION
 π  3 2 
As = 4     = 1.76715 in 2
 4  4  
Ac = 82 − As = 62.233 in 2
δs =
Ps L
Ps (48)
=
= 0.93663 × 10−6 Ps
As Es (1.76715)(29 × 106 )
δc =
Pc L
Pc (48)
=
= 0.21425 × 10−6 Pc
Ac Ec (62.233)(3.6 × 106 )
But δ s = δ c : 0.93663 × 10−6 Ps = 0.21425 × 10−6 Pc
Ps = 0.22875 Pc
Also:
Substituting (1) into (2):
Ps + Pc = P = 150 kips
(1)
(2)
1.22875Pc = 150 kips
Pc = 122.075 kips
From (1):
Ps = 0.22875(122.075) = 27.925 kips
σs = −
Ps
27.925
=−
As
1.76715
σ s = −15.80 ksi 
σc = −
Pc
122.075
=−
Ac
62.233
σ c = −1.962 ksi 
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PROBLEM 2.36
A 250-mm bar of 15 × 30-mm rectangular cross
section consists of two aluminum layers, 5-mm thick,
brazed to a center brass layer of the same thickness.
If it is subjected to centric forces of magnitude
P = 30 kN, and knowing that Ea = 70 GPa and
Eb = 105 GPa, determine the normal stress (a) in the
aluminum layers, (b) in the brass layer.
SOLUTION
For each layer,
A = (30)(5) = 150 mm 2 = 150 × 10−6 m 2
Let Pa = load on each aluminum layer
Pb = load on brass layer
Pa L Pb L
=
Ea A Eb A
Deformation.
δ=
Total force.
P = 2 Pa + Pb = 3.5 Pa
Solving for Pa and Pb,
Pa =
2
P
7
Pb =
Pb =
Eb
105
Pa =
Pa = 1.5 Pa
70
Ea
3
P
7
(a)
σa = −
Pa
2P
2 30 × 103
=−
=−
= −57.1 × 106 Pa
−6
A
7 A
7 150 × 10
σ a = −57.1 MPa 
(b)
σb = −
Pb
3P
3 30 × 103
=−
=−
= −85.7 × 106 Pa
A
7 A
7 150 × 10−6
σ b = −85.7 MPa 
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PROBLEM 2.37
Determine the deformation of the composite bar of
Prob. 2.36 if it is subjected to centric forces of
magnitude P = 45 kN.
PROBLEM 2.36 A 250-mm bar of 15 × 30-mm
rectangular cross section consists of two aluminum
layers, 5-mm thick, brazed to a center brass layer of
the same thickness. If it is subjected to centric forces
of magnitude P = 30 kN, and knowing that Ea = 70
GPa and Eb = 105 GPa, determine the normal stress
(a) in the aluminum layers, (b) in the brass layer.
SOLUTION
For each layer,
A = (30)(5) = 150 mm 2 = 150 × 10−6 m 2
Let Pa = load on each aluminum layer
Pb = load on brass layer
Pa L
PL
=− b
Ea A
Eb A
Deformation.
δ =−
Total force.
P = 2 Pa + Pb = 3.5 Pa
δ =−
=−
Pb =
Eb
105
Pa =
Pa = 1.5 Pa
70
Ea
Pa =
2
P
7
Pa L
2 PL
=−
Ea A
7 Ea A
2 (45 × 103 )(250 × 10−3 )
7 (70 × 109 )(150 × 10−6 )
= −306 × 10−6 m
δ = −0.306 mm 
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PROBLEM 2.38
Compressive centric forces of 40 kips are applied at both ends
of the assembly shown by means of rigid plates. Knowing
that Es = 29 × 106 psi and Ea = 10.1 × 106 psi, determine (a) the
normal stresses in the steel core and the aluminum shell,
(b) the deformation of the assembly.
SOLUTION
Let Pa = portion of axial force carried by shell
Ps = portion of axial force carried by core
Total force.
δ=
Pa L
Ea Aa
Pa =
Ea Aa
δ
L
δ=
Ps L
Es As
Ps =
Es As
δ
L
P = Pa + Ps = ( Ea Aa + Es As )
δ
L
Data:
δ
=ε =
L
P
Ea Aa + Es As
P = 40 × 103 lb
Aa =
As =
π
4
π
4
(d 02 − di2 ) =
2
d =
π
4
π
4
(2.52 − 1.0)2 = 4.1233 in 2
2
(1) = 0.7854 in 2
ε=
− 40 × 103
= −620.91 × 10−6
6
6
(10.1 × 10 )(4.1233) + (29 × 10 )(0.7854)
(a)
σ s = Es ε = (29 × 106 )(−620.91 × 10−6 ) = −18.01 × 103 psi

σ a = Ea ε = (10.1 × 106 )(620.91 × 10−6 ) = −6.27 × 103 psi 
(b)
δ = Lε = (10)(620.91 × 10−6 ) = −6.21 × 10−3
−18.01 ksi 
−6.27 ksi 
δ = −6.21 × 10−3 in. 
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PROBLEM 2.39
Three wires are used to suspend the plate shown. Aluminum wires of 18 -in.
diameter are used at A and B while a steel wire of 121 -in. diameter is used at
C. Knowing that the allowable stress for aluminum ( Ea = 10.4 × 106 psi) is
14 ksi and that the allowable stress for steel ( Es = 29 × 106 psi) is 18 ksi,
determine the maximum load P that can be applied.
SOLUTION
By symmetry,
PA = PB , and δ A = δ B
Also,
δC = δ A = δ B = δ
Strain in each wire:
δ
ε A = εB =
2L
, εC =
δ
L
= 2ε A
Determine allowable strain.
Wires A&B:
εA =
σA
EA
=
14 × 103
= 1.3462 × 10−3
10.4 × 106
ε C = 2 ε A = 2.6924 × 10−4
Wire C:
18 × 103
= 0.6207 × 10−3
EC 29 × 106
1
ε A = ε B = ε C = 0.3103 × 10−6
2
εC =
σC
=
σ A = E Aε A
σ C = 18 × 103 psi
∴
Allowable strain for wire C governs,
PA = AA E Aε A
π 1
2
(10.4 × 106 )(0.3103 × 10−6 ) = 39.61 lb
4  8 
PB = 39.61 lb
=
σ C = EC ε C
PC = ACσ C =
π1
2
(18 × 103 ) = 98.17 lb
4  12 
For equilibrium of the plate,
P = PA + PB + PC = 177.4 lb
P = 177.4 lb 
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PROBLEM 2.40
A polystyrene rod consisting of two cylindrical portions AB and BC is
restrained at both ends and supports two 6-kip loads as shown. Knowing
that E = 0.45 × 106 psi, determine (a) the reactions at A and C, (b) the
normal stress in each portion of the rod.
SOLUTION
(a)
We express that the elongation of the rod is zero:
δ=
But
PAB LAB
d2 E
4 AB
π
+
PAB = + RA
PBC LBC
π
4
2
d BC
E
=0
PBC = − RC
Substituting and simplifying:
RA LAB RC LBC
− 2
=0
2
d AB
d BC
RC =
LAB
LBC
2
2
 d BC 
25  2 

 RA = 
 RA
15
d
 1.25 
 AB 
RC = 4.2667 RA
From the free body diagram:
Substituting (1) into (2):
RA + RC = 12 kips
(2)
5.2667 RA = 12
RA = 2.2785 kips
From (1):
(1)
RA = 2.28 kips ↑ 
RC = 4.2667 (2.2785) = 9.7217 kips
RC = 9.72 kips ↑ 
(b)
σ AB =
PAB + RA
2.2785
=
=
AAB AAB π4 (1.25)2
σ AB = +1.857 ksi 
σ BC =
PBC − RC −9.7217
=
=
π
ABC
ABC
(2)2
4
σ BC = −3.09 ksi 
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PROBLEM 2.41
Two cylindrical rods, one of steel and the other of brass, are joined at
C and restrained by rigid supports at A and E. For the loading shown
and knowing that Es = 200 GPa and Eb = 105 GPa, determine
(a) the reactions at A and E, (b) the deflection of point C.
SOLUTION
A to C:
E = 200 × 109 Pa
π
(40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2
4
EA = 251.327 × 106 N
A=
C to E:
E = 105 × 109 Pa
π
(30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2
4
EA = 74.220 × 106 N
A=
A to B:
P = RA
L = 180 mm = 0.180 m
RA (0.180)
PL
=
δ AB =
EA 251.327 × 106
= 716.20 × 10−12 RA
B to C:
P = RA − 60 × 103
L = 120 mm = 0.120 m
δ BC =
PL ( RA − 60 × 103 )(0.120)
=
EA
251.327 × 106
= 447.47 × 10−12 RA − 26.848 × 10−6
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PROBLEM 2.41 (Continued)
C to D:
P = RA − 60 × 103
L = 100 mm = 0.100 m
PL ( RA − 60 × 103 )(0.100)
=
EA
74.220 × 106
= 1.34735 × 10−9 RA − 80.841 × 10−6
δ BC =
D to E:
P = RA − 100 × 103
L = 100 mm = 0.100 m
PL ( RA − 100 × 103 )(0.100)
=
EA
74.220 × 106
= 1.34735 × 10−9 RA − 134.735 × 10−6
δ DE =
A to E:
δ AE = δ AB + δ BC + δ CD + δ DE
= 3.85837 × 10−9 RA − 242.424 × 10−6
Since point E cannot move relative to A,
(a)
(b)
δ AE = 0
3.85837 × 10−9 RA − 242.424 × 10−6 = 0 RA = 62.831 × 103 N
RA = 62.8 kN ← 
RE = RA − 100 × 103 = 62.8 × 103 − 100 × 103 = −37.2 × 103 N
RE = 37.2 kN ← 
δ C = δ AB + δ BC = 1.16367 × 10−9 RA − 26.848 × 10−6
= (1.16369 × 10−9 )(62.831 × 103 ) − 26.848 × 10−6
= 46.3 × 10−6 m
δ C = 46.3 μ m → 
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PROBLEM 2.42
Solve Prob. 2.41, assuming that rod AC is made of brass and
rod CE is made of steel.
PROBLEM 2.41 Two cylindrical rods, one of steel and the
other of brass, are joined at C and restrained by rigid supports at A
and E. For the loading shown and knowing that Es = 200 GPa
and Eb = 105 GPa, determine (a) the reactions at A and E, (b) the
deflection of point C.
SOLUTION
A to C:
E = 105 × 109 Pa
π
(40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2
4
EA = 131.947 × 106 N
A=
C to E:
E = 200 × 109 Pa
π
(30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2
4
EA = 141.372 × 106 N
A=
A to B:
P = RA
L = 180 mm = 0.180 m
RA (0.180)
PL
=
δ AB =
EA 131.947 × 106
= 1.36418 × 10−9 RA
B to C:
P = RA − 60 × 103
L = 120 mm = 0.120 m
δ BC =
PL ( RA − 60 × 103 )(0.120)
=
EA
131.947 × 106
= 909.456 × 10−12 RA − 54.567 × 10−6
C to D:
P = RA − 60 × 103
L = 100 mm = 0.100 m
δ CD =
PL ( RA − 60 × 103 )(0.100)
=
EA
141.372 × 106
= 707.354 × 10−12 RA − 42.441 × 10−6
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PROBLEM 2.42 (Continued)
D to E:
P = RA − 100 × 103
L = 100 mm = 0.100 m
PL ( RA − 100 × 103 )(0.100)
=
EA
141.372 × 106
= 707.354 × 10−12 RA − 70.735 × 10−6
δ DE =
A to E:
δ AE = δ AB + δ BC + δ CD + δ DE
= 3.68834 × 10−9 RA − 167.743 × 10−6
Since point E cannot move relative to A,
(a)
(b)
δ AE = 0
3.68834 × 10−9 RA − 167.743 × 10−6 = 0 RA = 45.479 × 103 N
R A = 45.5 kN ← 
RE = RA − 100 × 103 = 45.479 × 103 − 100 × 103 = −54.521 × 103
RE = 54.5 kN ← 
δ C = δ AB + δ BC = 2.27364 × 10−9 RA − 54.567 × 10−6
= (2.27364 × 10−9 )(45.479 × 103 ) − 54.567 × 10−6
= 48.8 × 10−6 m
δ C = 48.8 μ m → 
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PROBLEM 2.43
The rigid bar ABCD is suspended from four identical wires. Determine the
tension in each wire caused by the load P shown.
SOLUTION
Deformations Let θ be the rotation of bar ABCD and δ A , δ B , δ C and δ D be the deformations of wires A, B,
C, and D.
From geometry,
θ=
δB − δ A
L
δ B = δ A + Lθ
δ C = δ A + 2 Lθ = 2δ B − δ A
(1)
δ D = δ A + 3Lθ = 3δ B − 2δ A
(2)
Since all wires are identical, the forces in the wires are proportional to the deformations.
TC = 2TB − TA
(1′)
TD = 3TB − 2TA
(2′)
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PROBLEM 2.43 (Continued)
Use bar ABCD as a free body.
ΣM C = 0 : − 2 LTA − LTB + LTD = 0
(3)
ΣFy = 0 : TA + TB + TC + TD − P = 0
(4)
Substituting (2′) into (3) and dividing by L,
−4TA + 2TB = 0
TB = 2TA
(3′)
Substituting (1′), (2′), and (3′) into (4),
TA + 2TA + 3TA + 4TA − P = 0
10TA = P
TA =
 1 
TB = 2TA = (2)   P
 10 
1
P 
10
TB =
1   1 
TC = (2)  P  −  P 
 5   10 
TC =
1 
 1 
TD = (3)  P  − (2)  P 
5


 10 
TD =
1
P 
5
3
P 
10
2
P 
5
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PROBLEM 2.44
The rigid bar AD is supported by two steel wires of 161 -in. diameter
( E = 29 × 106 psi) and a pin and bracket at D. Knowing that the
wires were initially taut, determine (a) the additional tension in each
wire when a 120-lb load P is applied at B, (b) the corresponding
deflection of point B.
SOLUTION
Let θ be the rotation of bar ABCD.
Then δ A = 24θ
δ C = 8θ 
δA =
PAE LAE
AE
PAE =
6
2
EAδ A (29 × 10 ) π4 ( 161 ) (24θ )
=
LAE
15
= 142.353 × 103θ
δC =
PCF LCF
AE
6 π 1
EAδ C (29 × 10 ) 4 ( 16 ) (8θ )
=
=
8
LCF
2
PCF
= 88.971 × 103θ
Using free body ABCD,
ΣM D = 0 :
−24PAE + 16P − 8PCF = 0
−24(142.353 × 103θ ) + 16(120) − 8(88.971 × 103θ ) = 0
θ = 0.46510 × 10−3 rad哷
(a)
(b)
PAE = (142.353 × 103 ) (0.46510 × 10−3 )
PAE = 66.2 lb 
PCF = (88.971 × 103 ) (0.46510 × 10−3 )
PCF = 41.4 lb 
δ B = 16θ = 16(0.46510 × 10−3 )
δ B = 7.44 × 10−3 in. ↓ 
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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 ← 
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PROBLEM 2.46
Links BC and DE are both made of steel ( E = 29 × 106 psi) and are
1
in. wide and 14 in. thick. Determine (a) the force in each link
2
when a 600-lb force P is applied to the rigid member AF shown,
(b) the corresponding deflection of point A.
SOLUTION
Let the rigid member ACDF rotate through small angle θ clockwise about point F.
δ C = δ BC = 4θ in. →
δ D = −δ DE = 2θ in. →
Then
δ=
For links:
FL
EA
or
F=
EAδ
L
 1  1 
A =    = 0.125 in 2
 2  4 
LBC = 4 in.
LDE = 5 in.
FBC =
EA δ BC (29 × 106 )(0.125)(4θ )
=
= 3.625 × 106 θ
4
LBC
FDE =
EAδ DE (29 × 106 )(0.125)( −2θ )
=
= −1.45 × 106 θ
5
LDE
Use member ACDF as a free body.
ΣM F = 0 : 8 P − 4 FBC + 2 FDE = 0
1
1
FBC − FDE
2
4
1
1
600 = (3.625 × 106 )θ − ( −1.45 × 106 )θ = 2.175 × 106 θ
2
4
−3
+
θ = 0.27586 × 10 rad 哶
P=
(a)
(b)
FBC = (3.625 × 106 )(0.27586 × 10−3 )
FBC = 1000 lb 
FDE = −(1.45 × 106 )(0.27586 × 10−3 )
FDE = −400 lb 
Deflection at Point A.
δ A = 8θ = (8)(0.27586 × 10−3 )
δ A = 2.21 × 10−3 in → 
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PROBLEM 2.47
The concrete post ( Ec = 3.6 × 106 psi and α c = 5.5 × 10−6 / °F) is reinforced
with six steel bars, each of 78 -in. diameter ( Es = 29 × 106 psi and
α s = 6.5 × 10−6 / °F). Determine the normal stresses induced in the steel and
in the concrete by a temperature rise of 65°F.
SOLUTION
As = 6
π
4
d2 = 6
π 7
2
= 3.6079 in 2
4  8 
Ac = 102 − As = 102 − 3.6079 = 96.392 in 2
Let Pc = tensile force developed in the concrete.
For equilibrium with zero total force, the compressive force in the six steel rods equals Pc .
Strains:
εs = −
Matching: ε c = ε s
Pc
+ α s (ΔT )
Es As
εc =
Pc
+ α c ( ΔT )
Ec Ac
Pc
P
+ α c (ΔT ) = − c + α s (ΔT )
Ec Ac
Es As
 1
1 
+

 Pc = (α s − α c )(ΔT )
 Ec Ac Es As 


1
1
−6
+

 Pc = (1.0 × 10 )(65)
6
6
 (3.6 × 10 )(96.392) (29 × 10 )(3.6079) 
Pc = 5.2254 × 103 lb
σc =

Pc 5.2254 × 103
=
= 54.210 psi
Ac
96.392
σs = −
Pc
5.2254 × 103
=−
= −1448.32 psi 
As
3.6079
σ c = 54.2 psi
σ s = −1.448 ksi 
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PROBLEM 2.48
The assembly shown consists of an aluminum shell ( Ea = 10.6 × 106 psi,
αa = 12.9 × 10–6/°F) fully bonded to a steel core ( Es = 29 × 106 psi,
αs = 6.5 × 10–6/°F) and is unstressed. Determine (a) the largest allowable
change in temperature if the stress in the aluminum shell is not to exceed
6 ksi, (b) the corresponding change in length of the assembly.
SOLUTION
Since α a > α s , the shell is in compression for a positive temperature rise.
σ a = −6 ksi = −6 × 103 psi
Let
Aa =
As =
π
4
π
4
(d
)
2
o
− di2 =
d2 =
π
4
π
4
(1.252 − 0.752 ) = 0.78540 in 2
(0.75) 2 = 0.44179 in 2
P = −σ a Aa = σ s As
where P is the tensile force in the steel core.
σs = −
ε=
(α a − α s )(ΔT ) =
(6.4 × 10−6 )(ΔT ) =
(a)
ΔT = 145.91°F
(b)
ε=
σ a Aa
σs
Es
σs
Es
As
=
(6 × 103 )(0.78540)
= 10.667 × 103 psi
0.44179
+ α s ( ΔT ) =
−
σa
Ea
+ α a ( ΔT )
σa
Ea
10.667 × 103
6 × 103
+
= 0.93385 × 10−3
29 × 106
10.6 × 106
ΔT = 145.9°F 
10.667 × 103
+ (6.5 × 10−6 )(145.91) = 1.3163 × 10−3
29 × 106
or
ε=

−6 × 103
+ (12.9 × 10−6 )(145.91) = 1.3163 × 10−3
10.6 × 106
δ = Lε = (8.0)(1.3163 × 10−3 ) = 0.01053 in. 
δ = 0.01053 in. 
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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
π
(25) 2 = 490.87 mm 2
4
= 490.87 × 10−6 m 2
PL
(δ P )b =
Eb Ab
Ab =
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PROBLEM 2.49 (Continued)
Aluminum shell:
Ea = 70 × 109 Pa
π
(602 − 252 )
4
= 2.3366 × 103 mm 2
Aa =
= 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 
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PROBLEM 2.50
Solve Prob. 2.49, assuming that the core is made of steel ( Es = 200 GPa,
α s = 11.7 × 10−6 / °C) instead of brass.
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
E = 70 GPa α = 23.6 × 10−6 / °C
Aluminum shell:
Let L be the length of the assembly.
ΔT = 195 − 15 = 180 °C
Free thermal expansion:
Steel core:
(δT ) s = Lα s (ΔT )
Aluminum shell:
(δT ) a = Lα a (ΔT )
δ = L(α a − α s )( ΔT )
Net expansion of shell with respect to the core:
Let P be the tensile force in the core and the compressive force in the shell.
Es = 200 × 109 Pa, As =
Steel core:
(δ P ) s =
Aluminum shell:
π
4
(25) 2 = 490.87 mm 2 = 490.87 × 10−6 m 2
PL
Es As
Ea = 70 × 109 Pa
(δ P ) a =
PL
Ea Aa
π
(602 − 25) 2 = 2.3366 × 103 mm 2 = 2.3366 × 10−3 m 2
4
δ = (δ P ) s + (δ P )a
Aa =
L(α a − α s )(ΔT ) =
PL
PL
+
= KPL
Es As Ea Aa
where
K=
=
1
1
+
Es As Ea Aa
1
1
+
−6
9
(200 × 10 )(490.87 × 10 ) (70 × 10 )(2.3366 × 10−3 )
9
= 16.2999 × 10−9 N −1
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PROBLEM 2.50 (Continued)
Then
P=
(α a − α s )(ΔT ) (23.6 × 10−6 − 11.7 × 10−6 )(180)
=
= 131.41 × 103 N
K
16.2999 × 10−9
Stress in aluminum: σ a = −
P
131.19 × 103
=−
= −56.2 × 106 Pa
−3
Aa
2.3366 × 10
σ a = −56.2 MPa 
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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:
δT = LABα s (ΔT ) + LBC α 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 = δ T
3.2235 × 10−9 P = 459.75 × 10−6
P = 142.62 × 103 N
P = 142.6 kN 
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PROBLEM 2.52
A steel railroad track ( Es = 200 GPa, α s = 11.7 × 10−6 /°C) was laid out at a temperature of 6°C. Determine
the normal stress in the rails when the temperature reaches 48°C, assuming that the rails (a) are welded to
form a continuous track, (b) are 10 m long with 3-mm gaps between them.
SOLUTION
(a)
δT = α ( ΔT ) L = (11.7 × 10−6 )(48 − 6)(10) = 4.914 × 10−3 m
δP =
PL Lσ
(10)σ
=
=
= 50 × 10−12 σ
AE
E
200 × 109
δ = δT + δ P = 4.914 × 10−3 + 50 × 10−12 σ = 0
σ = −98.3 × 106 Pa
(b)
σ = −98.3 MPa
δ = δT + δ P = 4.914 × 10−3 + 50 × 10−12 σ = 3 × 10−3
3 × 10−3 − 4.914 × 10−3
50 × 10−12
= −38.3 × 106 Pa
σ=
σ = −38.3 MPa 
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PROBLEM 2.53
A rod consisting of two cylindrical portions AB and BC is restrained at
both ends. Portion AB is made of steel ( Es = 29 × 106 psi,
α s = 6.5 × 10−6 /°F) and portion BC is made of aluminum
( Ea = 10.4 × 106 psi, α a = 13.3 × 10−6 /°F). Knowing that the rod is
initially unstressed, determine (a) the normal stresses induced in portions
AB and BC by a temperature rise of 70°F, (b) the corresponding
deflection of point B.
SOLUTION
AAB =
Free thermal expansion.
π
4
(2.25) 2 = 3.9761 in 2 ABC =
π
4
(1.5) 2 = 1.76715 in 2
ΔT = 70°F
(δ T ) AB = LABα s (ΔT ) + (24)(6.5 × 10−6 )(70) = 10.92 × 10−3 in
(δT ) BC = LBC α a (ΔT ) = (32)(13.3 × 10−6 (70) = 29.792 × 10−3 in.
δT = (δT ) AB + (δT ) BC = 40.712 × 10−3 in.
Total:
Shortening due to induced compressive force P.
PLAB
24 P
=
= 208.14 × 10−9 P
Es AAB (29 × 106 )(3.9761)
PLBC
32 P
=
=
= 1741.18 × 10−9 P
Ea ABC (10.4 × 106 )(1.76715)
(δ P ) AB =
(δ P ) BC
δ P = (δ P ) AB + (δ P ) BC = 1949.32 × 10−9 P
Total:
For zero net deflection, δ P = δ T
(a)
(b)
1949.32 × 10−9 P = 40.712 × 10−3
P = 20.885 × 103 lb
σ AB = −
P
20.885 × 103
=−
= −5.25 × 103 psi
3.9761
AAB
σ AB = −5.25 ksi 
σ BC = −
P
20.885 × 103
=−
= −11.82 × 103 psi
1.76715
ABC
σ BC = −11.82 ksi 
(δ P ) AB = (208.14 × 10−9 )(20.885 × 103 ) = 4.3470 × 10−3 in.
δ B = (δT ) AB → + (δ P ) AB ← = 10.92 × 10−3 → + 4.3470 × 10−3 ←
δ B = 6.57 × 10−3 in. → 
or
(δ P ) BC = (1741.18 × 10−9 )(20.885 × 103 ) = 36.365 × 10−3 in.
δ B = (δT ) BC ← + (δ P ) BC → = 29.792 × 10−3 ← + 36.365 × 10−3 → = 6.57 × 10−3 in. →
(checks)
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PROBLEM 2.54
Solve Prob. 2.53, assuming that portion AB of the composite rod is made
of aluminum and portion BC is made of steel.
PROBLEM 2.53 A rod consisting of two cylindrical portions AB and BC
is restrained at both ends. Portion AB is made of steel ( Es = 29 × 106 psi,
α s = 6.5 × 10−6 /°F) and portion BC is made of aluminum
( Ea = 10.4 × 106 psi, α a = 13.3 × 10−6 /°F). Knowing that the rod is
initially unstressed, determine (a) the normal stresses induced in portions
AB and BC by a temperature rise of 70°F, (b) the corresponding
deflection of point B.
SOLUTION
AAB =
π
4
(2.25) 2 = 3.9761 in 2
Free thermal expansion.
ABC =
π
4
(1.5) 2 = 1.76715 in 2
ΔT = 70°F
(δ T ) AB = LABα a ( ΔT ) = (24)(13.3 × 10−6 )(70) = 22.344 × 10−3 in.
(δT ) BC = LBC α s (ΔT ) = (32)(6.5 × 10−6 )(70) = 14.56 × 10−3 in.
δT = (δT ) AB + (δT ) BC = 36.904 × 10−3 in.
Total:
Shortening due to induced compressive force P.
PLAB
24 P
=
= 580.39 × 10−9 P
Ea AAB (10.4 × 106 )(3.9761)
PLBC
32 P
=
=
= 624.42 × 10−9 P
6
Es ABC (29 × 10 )(1.76715)
(δ P ) AB =
(δ P ) BC
δ P = (δ P ) AB + (δ P ) BC = 1204.81 × 10−9 P
Total:
For zero net deflection, δ P = δ T
(a)
1204.81 × 10−9 P = 36.904 × 10−3
P = 30.631 × 103 lb
σ AB = −
P
30.631 × 103
=−
= −7.70 × 103 psi
3.9761
AAB
σ AB = −7.70 ksi 
σ BC = −
P
30.631 × 103
=−
= −17.33 × 103 psi
1.76715
ABC
σ BC = −17.33 ksi 
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PROBLEM 2.54 (Continued)
(b)
(δ P ) AB = (580.39 × 10−9 ) (30.631 × 103 ) = 17.7779 × 10−3 in.
δ B = (δT ) AB → + (δ P ) AB ← = 22.344 × 10−3 → + 17.7779 × 10−3 ←

or
δ B = 4.57 × 10−3 in. → 
(δ P ) BC = (624.42 × 10−9 )(30.631 × 103 ) = 19.1266 × 10−3 in.
δ B = (δT ) BC ← + (δ P ) BC → = 14.56 × 10−3 ← + 19.1266 × 10−3 → = 4.57 × 10−3 in. →
(checks)
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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:
(δT )b = α b (ΔT ) L = (20.9 × 10−6 )(25)(0.250) = 130.625 × 10−6 m
Steel rod:
(δT ) 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 )
E = 200 × 109 Pa
(δ P ) s =
As =
π
4
(30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2
PL
P(0.250)
=
= 1.76838 × 10−9 P
Es As (200 × 109 )(706.86 × 10−6 )
(δ P )b + (δ P ) s = δ : 2.4033 × 10−9 P = 62.5 × 10−6 P = 26.006 × 103 N
σs = −
P
(26.006 × 103 )
=−
= −36.8 × 106 Pa
−6
As
706.86 × 10
(a)
Stress in steel rod:
(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 
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PROBLEM 2.56
Two steel bars ( Es = 200 GPa and α s = 11.7 × 10−6 /°C) are used to
reinforce a brass bar ( Eb = 105 GPa, α b = 20.9 × 10−6 / °C) that is
subjected to a load P = 25 kN. When the steel bars were fabricated, the
distance between the centers of the holes that were to fit on the pins
was made 0.5 mm smaller than the 2 m needed. The steel bars
were then placed in an oven to increase their length so that they
would just fit on the pins. Following fabrication, the temperature
in the steel bars dropped back to room temperature. Determine
(a) the increase in temperature that was required to fit the steel
bars on the pins, (b) the stress in the brass bar after the load is
applied to it.
SOLUTION
(a)
Required temperature change for fabrication:
δT = 0.5 mm = 0.5 × 10−3 m
Temperature change required to expand steel bar by this amount:
δT = Lα s ΔT , 0.5 × 10−3 = (2.00)(11.7 × 10−6 )(ΔT ),
ΔT = 0.5 × 10−3 = (2)(11.7 × 10−6 )(ΔT )
ΔT = 21.368 °C
(b)
21.4 °C 
Once assembled, a tensile force P* develops in the steel, and a compressive force P* develops in the
brass, in order to elongate the steel and contract the brass.
As = (2)(5)(40) = 400 mm2 = 400 × 10−6 m 2
Elongation of steel:
(δ P ) s =
F *L
P* (2.00)
=
= 25 × 10−9 P*
As Es (400 × 10−6 )(200 × 109 )
Contraction of brass: Ab = (40)(15) = 600 mm 2 = 600 × 10−6 m 2
(δ P )b =
P* L
P* (2.00)
=
= 31.746 × 10−9 P*
Ab Eb (600 × 10−6 )(105 × 109 )
But (δ P ) s + (δ P )b is equal to the initial amount of misfit:
(δ P ) s + (δ P )b = 0.5 × 10−3 , 56.746 × 10−9 P* = 0.5 × 10−3
P* = 8.811 × 103 N
Stresses due to fabrication:
Steel:
σ s* =
P* 8.811 × 103
=
= 22.03 × 106 Pa = 22.03 MPa
As 400 × 10−6
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PROBLEM 2.56 (Continued)
Brass:
σ b* = −
P*
8.811 × 103
=−
= −14.68 × 106 Pa = −14.68 MPa
−6
Ab
600 × 10
To these stresses must be added the stresses due to the 25 kN load.
For the added load, the additional deformation is the same for both the steel and the brass. Let δ ′ be the
additional displacement. Also, let Ps and Pb be the additional forces developed in the steel and brass,
respectively.
δ′ =
Ps L
PL
= b
As Es Ab Eb
As Es
(400 × 10−6 )(200 × 109 )
δ′ =
δ ′ = 40 × 106 δ ′
2.00
L
Ab Eb
(600 × 10−6 )(105 × 109 )
δ′ =
δ ′ = 31.5 × 106 δ ′
Pb =
2.00
L
Ps =
P = Ps + Pb = 25 × 103 N
Total
40 × 106 δ ′ + 31.5 × 106 δ ′ = 25 × 103
δ ′ = 349.65 × 10−6 m
Ps = (40 × 106 )(349.65 × 10−6 ) = 13.986 × 103 N
Pb = (31.5 × 106 )(349.65 × 10−6 ) = 11.140 × 103 N
σs =
Ps 13.986 × 103
=
= 34.97 × 106 Pa
As
400 × 10−6
σb =
Pb 11.140 × 103
=
= 18.36 × 106 Pa
Ab
600 × 10−6
Add stress due to fabrication.
Total stresses:
σ s = 34.97 × 106 + 22.03 × 106 = 57.0 × 106 Pa
σ s = 57.0 MPa
σ b = 18.36 × 106 − 14.68 × 106 = 3.68 × 106 Pa
σ b = 3.68 MPa 
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PROBLEM 2.57
Determine the maximum load P that may be applied to the brass bar of
Prob. 2.56 if the allowable stress in the steel bars is 30 MPa and the
allowable stress in the brass bar is 25 MPa.
PROBLEM 2.56 Two steel bars ( Es = 200 GPa and α s = 11.7 × 10–6/°C)
are used to reinforce a brass bar ( Eb = 105 GPa, α b = 20.9 × 10–6/°C)
that is subjected to a load P = 25 kN. When the steel bars were fabricated,
the distance between the centers of the holes that were to fit on the pins was
made 0.5 mm smaller than the 2 m needed. The steel bars were then placed
in an oven to increase their length so that they would just fit on the pins.
Following fabrication, the temperature in the steel bars dropped back to
room temperature. Determine (a) the increase in temperature that was
required to fit the steel bars on the pins, (b) the stress in the brass bar after
the load is applied to it.
SOLUTION
See solution to Problem 2.56 to obtain the fabrication stresses.
σ s* = 22.03 MPa
σ b* = 14.68 MPa
Allowable stresses:
σ s ,all = 30 MPa, σ b ,all = 25 MPa
Available stress increase from load.
σ s = 30 − 22.03 = 7.97 MPa
σ b = 25 + 14.68 = 39.68 MPa
Corresponding available strains.
εs =
εb =
σs
Es
σb
Eb
=
7.97 × 106
= 39.85 × 10−6
200 × 109
=
39.68 × 106
= 377.9 × 10−6
105 × 109
Smaller value governs ∴ ε = 39.85 × 10−6
Areas: As = (2)(5)(40) = 400 mm2 = 400 × 10−6 m 2
Ab = (15)(40) = 600 mm 2 = 600 × 10−6 m 2
Forces Ps = Es As ε = (200 × 109 )(400 × 10−6 )(39.85 × 10−6 ) = 3.188 × 103 N
Pb = Eb Abε = (105 × 109 )(600 × 10−6 )(39.85 × 10−6 ) = 2.511 × 10−3 N
Total allowable additional force:
P = Ps + Pb = 3.188 × 103 + 2.511 × 103 = 5.70 × 103 N
P = 5.70 kN 
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PROBLEM 2.58
Knowing that a 0.02-in. gap exists when the temperature is 75 °F,
determine (a) the temperature at which the normal stress in the
aluminum bar will be equal to −11 ksi, (b) the corresponding exact
length of the aluminum bar.
SOLUTION
σ a = −11 ksi = −11 × 103 psi
P = −σ a Aa = (11 × 103 )(2.8) = 30.8 × 103 lb
Shortening due to P:
δP =
=
PLb
PLa
+
Eb Ab Ea Aa
(30.8 × 103 )(14) (30.8 × 103 )(18)
+
(15 × 106 )(2.4) (10.6 × 106 )(2.8)
= 30.657 × 10−3 in.
Available elongation for thermal expansion:
δT = 0.02 + 30.657 × 10−3 = 50.657 × 10−3 in.
But δ T = Lbα b (ΔT ) + Laα a (ΔT )
= (14)(12 × 10−6 )( ΔT ) + (18)(12.9 × 10−6 )(ΔT ) = 400.2 × 10−6 ) ΔT
Equating, (400.2 × 10−6 )ΔT = 50.657 × 10−3
(a)
(b)
ΔT = 126.6 °F
Thot = Tcold + ΔT = 75 + 126.6 = 201.6 °F
δ a = Laα a (ΔT ) −
Thot = 201.6°F 
PLa
Ea Aa
= (18)(12.9 × 10−6 )(26.6) −
(30.8 × 103 )(18)
= 10.712 × 10−3 in.
6
(10.6 × 10 )(2.8)
Lexact = 18 + 10.712 × 10−3 = 18.0107 in.
L = 18.0107 in. 
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PROBLEM 2.59
Determine (a) the compressive force in the bars shown after a
temperature rise of 180 °F, (b) the corresponding change in length of
the bronze bar.
SOLUTION
Thermal expansion if free of constraint:
δT = Lbα b (ΔT ) + Laα a (ΔT )
= (14)(12 × 10−6 )(180) + (18)(12.9 × 10−6 )(180)
= 72.036 × 10−3 in.
Constrained expansion: δ = 0.02in.
Shortening due to induced compressive force P:
δ P = 72.036 × 10−3 − 0.02 = 52.036 × 10−3 in.
δP =
But
PLb
PLa  Lb
L
+
=
+ a
Eb Ab Ea Aa  Eb Ab Ea Aa

P



14
18
−9
=
+
 P = 995.36 × 10 P
6
6
 (15 × 10 )(2.4) (10.6 × 10 )(2.8) 
995.36 × 10−9 P = 52.036 × 10−3
Equating,
P = 52.279 × 103 lb
P = 52.3 kips 
(a)
(b)
δ b = Lbα b (ΔT ) −
PLb
Eb Ab
= (14)(12 × 10−6 )(180) −
(52.279 × 103 )(14)
= 9.91 × 10−3 in.
(15 × 106 )(2.4)
δ b = 9.91 × 10−3 in. 
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PROBLEM 2.60
At room temperature (20 °C) a 0.5-mm gap exists between the ends
of the rods shown. At a later time when the temperature has reached
140°C, determine (a) the normal stress in the aluminum rod, (b) the
change in length of the aluminum rod.
SOLUTION
ΔT = 140 − 20 = 120 °C
Free thermal expansion:
δT = Laα a (ΔT ) + Lsα s ( ΔT )
= (0.300)(23 × 10−6 )(120) + (0.250)(17.3 × 10−6 )(120)
= 1.347 × 10−3 m
Shortening due to P to meet constraint:
δ P = 1.347 × 10−3 − 0.5 × 10−3 = 0.847 × 10−3 m
PLa
PLs  La
L 
+
=
+ s P
Ea Aa Es As  Ea Aa Es As 


0.300
0.250
=
+
P
9
−6
9
−6 
 (75 × 10 )(2000 × 10 ) (190 × 10 )(800 × 10 ) 
= 3.6447 × 10−9 P
δP =
3.6447 × 10−9 P = 0.847 × 10−3
Equating,
P = 232.39 × 103 N
P
232.39 × 103
=−
= −116.2 × 106 Pa
Aa
2000 × 10−6
(a)
σa = −
(b)
δ a = Laα a (ΔT ) −
σ a = −116.2 MPa 
PLa
Ea Aa
= (0.300)(23 × 10−6 )(120) −
(232.39 × 103 )(0.300)
= 363 × 10−6 m
9
−6
(75 × 10 )(2000 × 10 )
δ a = 0.363 mm 
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PROBLEM 2.61
A 600-lb tensile load is applied to a test
coupon made from 161 -in. flat steel plate
(E = 29 × 106 psi and v = 0.30). Determine
the resulting change (a) in the 2-in. gage
length, (b) in the width of portion AB of the
test coupon, (c) in the thickness of
portion AB, (d) in the cross-sectional
area of portion AB.
SOLUTION
 1  1 
A =    = 0.03125 in 2
 2  16 
600
P
σ= =
= 19.2 × 103 psi
A 0.03125
σ 19.2 × 103
εx = =
= 662.07 × 10−6
E
29 × 106
(a)
δ x = L0ε x = (2.0)(662.07 × 10−6 )
δ l = 1.324 × 10−3 in. 
ε y = ε z = −vε x = −(0.30)(662.07 × 10−6 ) = −198.62 × 10−6
1
(b)
δ width = w0ε y =   ( −198.62 × 10−6 )
2
(c)
δ thickness = t0ε z =   (−198.62 × 10−6 )
 16 
(d)
 1 
δ w = −99.3 × 10−6 in. 
δ t = −12.41 × 10−6 in. 
A = wt = w0 (1 + ε y )t0 (1 + ε z )
= w0t0 (1 + ε y + ε z + ε y ε z )
ΔA = A − A0 = w0t0 (ε y + ε z + ε y ε z )
 1  1 
=    ( −198.62 × 10−6 − 198.62 × 10−6 + negligible term)
 2  16 
= −12.41 × 10−6 in 2
ΔA = −12.41 × 10−6 in 2 
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PROBLEM 2.62
In a standard tensile test, a steel rod of 22-mm diameter is subjected
to a tension force of 75 kN. Knowing that v = 0.3 and
E = 200 GPa, determine (a) the elongation of the rod in a 200-mm
gage length, (b) the change in diameter of the rod.
SOLUTION
P = 75 kN = 75 × 103 N
A=
σ =
εx =
π
4
d2 =
π
4
(0.022)2 = 380.13 × 10−6 m 2
P
75 × 103
=
= 197.301 × 106 Pa
A 380.13 × 10−6
σ
E
=
197.301 × 106
= 986.51 × 10−6
200 × 109
δ x = Lε x = (200 mm)(986.51 × 10−6 )
(a)
δ x = 0.1973 mm 

ε y = −vε x = −(0.3)(986.51 × 10−6 ) = −295.95 × 10−6 

δ y = d ε y = (22 mm)(−295.95 × 10−6 ) 

(b) δ y = −0.00651 mm 
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PROBLEM 2.63
A 20-mm-diameter rod made of an experimental plastic is subjected to a tensile force of magnitude P = 6 kN.
Knowing that an elongation of 14 mm and a decrease in diameter of 0.85 mm are observed in a 150-mm length,
determine the modulus of elasticity, the modulus of rigidity, and Poisson’s ratio for the material.
SOLUTION
Let the y-axis be along the length of the rod and the x-axis be transverse.
π
(20)2 = 314.16 mm 2 = 314.16 × 10−6 m 2
4
P
6 × 103
σy = =
= 19.0985 × 106 Pa
−6
A 314.16 × 10
δ y 14 mm
εy =
=
= 0.093333
L 150 mm
A=
Modulus of elasticity: E =
εx =
σy
εy
δx
d
Poisson’s ratio:
v=−
Modulus of rigidity:
G=
=
19.0985 × 106
= 204.63 × 106 Pa
0.093333
=−
P = 6 × 103 N
E = 205 MPa 
0.85
= −0.0425
20
εx
−0.0425
=−
0.093333
εy
E
204.63 × 106
=
= 70.31 × 106 Pa
2(1 + v)
(2)(1.455)
v = 0.455 
G = 70.3 MPa 
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PROBLEM 2.64
The change in diameter of a large steel bolt is carefully measured as the
nut is tightened. Knowing that E = 29 × 106 psi and v = 0.30,
determine the internal force in the bolt, if the diameter is observed to
decrease by 0.5 × 10−3 in.
SOLUTION
δ y = −0.5 × 10−3 in.
εy =
εy
d
v=−
=−
d = 2.5 in.
0.5 × 10−3
= −0.2 × 10−3
2.5
εy
:
εx
εx =
−ε y
v
=
0.2 × 10−3
= 0.66667 × 10−3
0.3
σ x = Eε x = (29 × 106 )(0.66667 × 10−3 ) = 19.3334 × 103 psi
A=
π
4
d2 =
π
4
(2.5) 2 = 4.9087 in 2
F = σ x A = (19.3334 × 103 )(4.9087) = 94.902 × 103 lb
F = 94.9 kips 
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PROBLEM 2.65
A 2.5-m length of a steel pipe of 300-mm outer diameter and 15-mm
wall thickness is used as a column to carry a 700-kN centric axial load.
Knowing that E = 200 GPa and v = 0.30, determine (a) the change in
length of the pipe, (b) the change in its outer diameter, (c) the change in
its wall thickness.
SOLUTION
d o = 0.3 m
t = 0.015 m
L = 2.5 m
di = do − 2t = 0.3 − 2(0.015) = 0.27 m
A=
(a)
δ =−
π
4
( do2 − di2 ) =
δ
L
4
(0.32 − 0.272 ) = 13.4303 × 10−3 m 2
PL
(700 × 103 )(2.5)
=−
EA
(200 × 109 )(13.4303 × 10−3 )
= −651.51 × 10−6 m
ε=
π
P = 700 × 103 N
=
δ = −0.652 mm 
−651.51 × 10−6
= −260.60 × 10−6
2.5
ε LAT = −vε = −(0.30)(−260.60 × 10−6 )
= 78.180 × 10−6
(b)
Δdo = do ε LAT = (300 mm)(78.180 × 10−6 )
Δd o = 0.0235 mm 
(c)

Δt = tε LAT = (15 mm)(78.180 × 10−6 )
Δt = 0.001173 mm 
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PROBLEM 2.66
An aluminum plate ( E = 74 GPa and v = 0.33) is subjected to a
centric axial load that causes a normal stress σ. Knowing that, before
loading, a line of slope 2:1 is scribed on the plate, determine the slope
of the line when σ = 125 MPa.
SOLUTION
The slope after deformation is
tan θ =
2(1 + ε y )
1+ εx
125 × 106
= 1.6892 × 10−3
E
74 × 109
ε y = −vε x = −(0.33)(1.6892 × 10−3 ) = −0.5574 × 10−3
εx =
tan θ =
σx
=
2(1 − 0.0005574)
= 1.99551
1 + 0.0016892
tan θ = 1.99551 
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PROBLEM 2.67
The block shown is made of a magnesium alloy, for which
E = 45 GPa and v = 0.35. Knowing that σ x = −180 MPa,
determine (a) the magnitude of σ y for which the change in the
height of the block will be zero, (b) the corresponding change in
the area of the face ABCD, (c) the corresponding change in the
volume of the block.
SOLUTION
(a)
δy = 0
εy = 0
σz = 0
1
(σ x − vσ y − vσ z )
E
σ y = vσ x = (0.35)(−180 × 106 )
εy =
= −63 × 106 Pa
σ y = −63 MPa 
1
v
(0.35)( −243 × 106 )
(σ z − vσ x − vσ y ) = − (σ x + σ y ) =
= −1.89 × 10−3
E
E
45 × 109
σ x − vσ y
1
157.95 × 106
ε x = (σ x − vσ y − vσ Z ) =
=−
= −3.51 × 10−3
E
E
45 × 109
εz =
(b)
A0 = Lx Lz
A = Lx (1 + ε x ) Lz (1 + ε z ) = Lx Lz (1 + ε x + ε z + ε xε z )
Δ A = A − A0 = Lx Lz (ε x + ε z + ε x ε z ) ≈ Lx Lz (ε x + ε z )
Δ A = (100 mm)(25 mm)(−3.51 × 10−3 − 1.89 × 10−3 )
(c)
Δ A = −13.50 mm 2 
V0 = Lx Ly Lz
V = Lx (1 + ε x ) Ly (1 + ε y ) Lz (1 + ε z )
= Lx Ly Lz (1 + ε x + ε y + ε z + ε xε y + ε y ε z + ε z ε x + ε x ε y ε z )
ΔV = V − V0 = Lx Ly Lz (ε x + ε y + ε z + small terms)
ΔV = (100)(40)(25)(−3.51 × 10−3 + 0 − 1.89 × 10−3 )
ΔV = −540 mm3 
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PROBLEM 2.68
A 30-mm square was scribed on the side of a large steel pressure
vessel. After pressurization, the biaxial stress condition at the square
is as shown. For E = 200 GPa and v = 0.30, determine the change
in length of (a) side AB, (b) side BC, (c) diagnonal AC.
SOLUTION
Given:
σ x = 80 MPa
σ y = 40 MPa
Using Eq’s (2.28):
εx =
1
80 − 0.3(40)
(σ x − vσ y ) =
= 340 × 10−6
200 × 103
E
εy =
1
40 − 0.3(80)
(σ y − vσ x ) =
= 80 × 10−6
200 × 103
E
(a) Change in length of AB.
δ AB = ( AB)ε x = (30 mm)(340 × 10−6 ) = 10.20 × 10−3 mm
δ AB = 10.20 μ m 
(b) Change in length of BC.
δ BC = ( BC )ε y = (30 mm)(80 × 10−6 ) = 2.40 × 10−3 mm
δ BC = 2.40 μ m 
(c) Change in length of diagonal AC.
From geometry,
Differentiate:
But
Thus,
( AC )2 = ( AB)2 + ( BC ) 2
2( AC ) Δ( AC ) = 2( AB)Δ( AB) + 2( BC )Δ( BC )
Δ( AC ) = δ AC
Δ( AB) = δ AB
Δ( BC ) = δ BC
2( AC )δ AC = 2( AB)δ AB + 2( BC )δ BC
δ AC =
AB
BC
1
1
(10.20 μ m)+
(2.40 μ m)
δ AB +
δ BC =
AC
AC
2
2
δ AC = 8.91 μ m 
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PROBLEM 2.69
The aluminum rod AD is fitted with a jacket that is used to apply a
hydrostatic pressure of 6000 psi to the 12-in. portion BC of the rod.
Knowing that E = 10.1 × 106 psi and v = 0.36, determine (a) the change in
the total length AD, (b) the change in diameter at the middle of the rod.
SOLUTION
σ x = σ z = − P = −6000 psi
σy = 0
1
(σ x − vσ y − vσ z )
E
1
=
[−6000 − (0.36)(0) − (0.36)( −6000)]
10.1 × 106
= −380.198 × 10−6
1
ε y = (−vσ x + σ y − vσ z )
E
1
[−(0.36)(−6000) + 0 − (0.36)( −6000)]
=
10.1 × 106
= 427.72 × 10−6
εx =
Length subjected to strain ε x: L = 12 in.
(a)
δ y = Lε y = (12)(427.72 × 10−6 )
(b)
δ x = d ε x = (1.5)(−380.198 × 10−6 )
δ l = 5.13 × 10−3 in. 
δ d = −0.570 × 10−3 in. 
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PROBLEM 2.70
For the rod of Prob. 2.69, determine the forces that should be applied to
the ends A and D of the rod (a) if the axial strain in portion BC of the rod
is to remain zero as the hydrostatic pressure is applied, (b) if the total
length AD of the rod is to remain unchanged.
PROBLEM 2.69 The aluminum rod AD is fitted with a jacket that is
used to apply a hydrostatic pressure of 6000 psi to the 12-in. portion BC
of the rod. Knowing that E = 10.1 × 106 psi and v = 0.36, determine
(a) the change in the total length AD, (b) the change in diameter at the
middle of the rod.
SOLUTION
Over the pressurized portion BC,
σx =σz = −p
σy =σy
1
(−vσ x + σ y − vσ z )
E
1
= (2vp + σ y )
E
(ε y ) BC =
(a)
(ε y ) BC = 0
2vp + σ y = 0
σ y = −2vp = −(2)(0.36)(6000)
= −4320 psi
A=
π
d2 =
π
(1.5) 2 = 1.76715 in 2
4
4
F = Aσ y = (1.76715)(−4320) = −7630 lb
i.e.,
(b)
Over unpressurized portions AB and CD,
(ε y ) AB
7630 lb compression 
σx =σz = 0
σy
= (ε y )CD =
E
For no change in length,
δ = LAB (ε y ) AB + LBC (ε y ) BC + LCD (ε y )CD = 0
( LAB + LCD )(ε y ) AB + LBC (ε y ) BC = 0
σy
12
(2vp + σ y ) = 0
E
E
24vp
(24)(0.36)(6000)
σy = −
=−
= −2592 psi
20
20
P = Aσ y = (1.76715)(−2592) = −4580 lb
P = 4580 lb compression 
(20 − 12)
+
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PROBLEM 2.71
In many situations, physical constraints prevent
strain from occurring in a given direction.
For example, ε z = 0 in the case shown, where
longitudinal movement of the long prism is
prevented at every point. Plane sections
perpendicular to the longitudinal axis remain
plane and the same distance apart. Show that for
this situation, which is known as plane strain,
we can express σ z , ε x , and ε y as follows:
σ z = v(σ x + σ y )
1
[(1 − v 2 )σ x − v(1 + v)σ y ]
E
1
ε y = [(1 − v 2 )σ y − v(1 + v)σ x ]
E
εx =
SOLUTION
εz = 0 =
1
(−vσ x − vσ y + σ z ) or σ z = v(σ x + σ y )
E
1
(σ x − vσ y − vσ z )
E
1
= [σ x − vσ y − v 2 (σ x + σ y )]
E
1
= [(1 − v 2 )σ x − v(1 + v)σ y ]
E

εx =
1
( −vσ x + σ y − vσ z )
E
1
= [−vσ x + σ y − v 2 (σ x + σ y )]
E
1
= [(1 − v 2 )σ y − v(1 + v)σ x ]
E

εy =

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PROBLEM 2.72
In many situations, it is known that the normal stress in a given direction is zero,
for example, σ z = 0 in the case of the thin plate shown. For this case, which is
known as plane stress, show that if the strains εx and εy have been determined
experimentally, we can express σ x , σ y , and ε z as follows:
σx = E
ε x + vε y
1− v
2
σy = E
ε y + vε x
1− v
2
εz = −
v
(ε x + ε y )
1− v
SOLUTION
σz = 0
εx =
1
(σ x − vσ y )
E
(1)
εy =
1
( −vσ x + σ y )
E
(2)
Multiplying (2) by v and adding to (1),
ε x + vε y =
1 − v2
σx
E
or
σx =
E
(ε x + vε y )
1 − v2

or
σy =
E
(ε y + vε x )
1 − v2

Multiplying (1) by v and adding to (2),
ε y + vε x =
1 − v2
σy
E
1
v
E
(−vσ x − vσ y ) = −
⋅
(ε x + vε y + ε y + vε x )
E
E 1 − v2
v(1 + v)
v
=−
(ε x + ε y ) = −
(ε x + ε y )
2
1− v
1− v
εz =

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PROBLEM 2.73
For a member under axial loading, express the normal strain ε′ in a
direction forming an angle of 45° with the axis of the load in terms of the
axial strain εx by (a) comparing the hypotenuses of the triangles shown in
Fig. 2.49, which represent, respectively, an element before and after
deformation, (b) using the values of the corresponding stresses of σ ′ and
σx shown in Fig. 1.38, and the generalized Hooke’s law.
SOLUTION
Figure 2.49
(a)
[ 2(1 + ε ′)]2 = (1 + ε x ) 2 + (1 − vε x ) 2
2(1 + 2ε ′ + ε ′2 ) = 1 + 2ε x + ε x2 + 1 − 2vε x + v 2ε x2
4ε ′ + 2ε ′2 = 2ε x + ε x2 − 2vε x + v 2ε x2
4ε ′ = 2ε x − 2vε x
Neglect squares as small
(A)
ε′ =
1− v
εx 
2
(B)
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PROBLEM 2.73 (Continued)
(b)
vσ ′
E
E
1− v P
=
⋅
E 2A
1− v
σx
=
2E
ε′ =
=
σ′
−
1− v
εx
2

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PROBLEM 2.74
The homogeneous plate ABCD is subjected to a biaxial loading as
shown. It is known that σ z = σ 0 and that the change in length of
the plate in the x direction must be zero, that is, ε x = 0. Denoting
by E the modulus of elasticity and by v Poisson’s ratio, determine
(a) the required magnitude of σ x , (b) the ratio σ 0 / ε z .
SOLUTION
σ z = σ 0 , σ y = 0, ε x = 0
εx =
1
1
(σ x − vσ y − vσ z ) = (σ x − vσ 0 )
E
E
σ x = vσ 0 
(a)
(b)
εz =
1
1
1 − v2
(−vσ x − vσ y + σ z ) = (−v 2σ 0 − 0 + σ 0 ) =
σ0
E
E
E
σ0
E
=

ε z 1 − v2
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PROBLEM 2.75
A vibration isolation unit consists of two blocks of hard rubber
bonded to a plate AB and to rigid supports as shown. Knowing
that a force of magnitude P = 25 kN causes a deflection
δ = 1.5 mm of plate AB, determine the modulus of rigidity of the
rubber used.
SOLUTION
F =
1
1
P = (25 × 103 N) = 12.5 × 103 N
2
2
τ =
F
12.5 × 103 N)
=
= 833.33 × 103 Pa
A (0.15 m)(0.1 m)
δ = 1.5 × 10−3 m
h = 0.03 m
1.5 × 10−3
= 0.05
h
0.03
τ
833.33 × 103
G = =
= 16.67 × 106 Pa
γ
0.05
γ=
δ
=
G = 16.67 MPa 
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PROBLEM 2.76
A vibration isolation unit consists of two blocks of hard rubber with a
modulus of rigidity G = 19 MPa bonded to a plate AB and to rigid
supports as shown. Denoting by P the magnitude of the force applied
to the plate and by δ the corresponding deflection, determine the
effective spring constant, k = P/δ , of the system.
SOLUTION
δ
h
Shearing strain:
γ =
Shearing stress:
τ = Gγ =
Force:
Gδ
h
1
GAδ
P = Aτ =
2
h
P
P=
k =
with
A = (0.15)(0.1) = 0.015 m 2
k =
2GAδ
h
2GA
h
Effective spring constant:
δ
=
or
h = 0.03 m
2(19 × 106 Pa)(0.015 m 2 )
= 19.00 × 106 N/m
0.03 m
k = 19.00 × 103 kN/m 
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PROBLEM 2.77
The plastic block shown is bonded to a fixed base and to a horizontal rigid
plate to which a force P is applied. Knowing that for the plastic used
G = 55 ksi, determine the deflection of the plate when P = 9 kips.
SOLUTION
Consider the plastic block. The shearing force carried is P = 9 × 103 lb
The area is A = (3.5)(5.5) = 19.25 in 2
Shearing stress:
τ =
Shearing strain:
γ =
But
γ =
P 9 × 103
=
= 467.52 psi
A
19.25
τ
G
=
467.52
= 0.0085006
55 × 103
δ
∴ δ = hγ = (2.2)(0.0085006)
h
δ = 0.0187 in. 
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PROBLEM 2.78
A vibration isolation unit consists of two blocks of hard rubber bonded to
plate AB and to rigid supports as shown. For the type and grade of rubber
used τ all = 220 psi and G = 1800 psi. Knowing that a centric vertical force
of magnitude P = 3.2 kips must cause a 0.1-in. vertical deflection of
the plate AB, determine the smallest allowable dimensions a and b of the
block.
SOLUTION
Consider the rubber block on the right. It carries a shearing force equal to
1
2
1
2
P.
P
The shearing stress is
τ=
or required area
A=
But
A = (3.0)b
Hence,
b=
Use
b = 2.42 in
Shearing strain.
γ=
But
γ=
δ
a
Hence,
a=
δ
0.1
=
= 0.818 in.
γ 0.12222
A
P 3.2 × 103
=
= 7.2727 in 2
2τ (2)(220)
A
= 2.42 in.
3.0
τ
G
=
and
bmin = 2.42 in. 
τ = 220 psi
220
= 0.12222
1800
amin = 0.818 in. 
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PROBLEM 2.79
The plastic block shown is bonded to a rigid support and to a vertical plate to
which a 55-kip load P is applied. Knowing that for the plastic used G = 150 ksi,
determine the deflection of the plate.

SOLUTION
A = (3.2)(4.8) = 15.36 in 2
P = 55 × 103 lb
τ =
P 55 × 103
=
= 3580.7 psi
A
15.36
G = 150 × 103 psi
γ =

τ
=
G
h = 2 in.
3580.7
= 23.871 × 10−3
150 × 103
δ = hγ = (2)(23.871 × 10−3 )
= 47.7 × 10−3 in.
δ = 0.0477 in. ↓ 
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PROBLEM 2.80
What load P should be applied to the plate of Prob. 2.79 to produce a
1
16
-in. deflection?
PROBLEM 2.79 The plastic block shown is bonded to a rigid support and to a vertical plate to which a
55-kip load P is applied. Knowing that for the plastic used G = 150 ksi, determine the deflection of the plate.
SOLUTION
1
in. = 0.0625 in.
16
h = 2 in.
δ =
γ =
δ
h
=
0.0625
= 0.03125
2
G = 150 × 103 psi
τ = Gγ = (150 × 103 )(0.03125)
= 4687.5 psi
A = (3.2)(4.8) = 15.36 in 2
P = τ A = (4687.5)(15.36)
= 72 × 103 lb
72 kips 
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PROBLEM 2.81
Two blocks of rubber with a modulus of rigidity G = 12 MPa are
bonded to rigid supports and to a plate AB. Knowing that c = 100 mm
and P = 45 kN, determine the smallest allowable dimensions a and b of
the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa
and the deflection of the plate is to be at least 5 mm.
SOLUTION
Shearing strain:
γ=
a=
Shearing stress:
τ=
b=
δ
=
a
Gδ
τ
1
2
P
A
τ
G
=
(12 × 106 Pa)(0.005 m)
= 0.0429 m
1.4 × 106 Pa
=
P
2bc
P
45 × 103 N
=
= 0.1607 m
2cτ 2(0.1 m) (1.4 × 106 Pa)
a = 42.9 mm 
b = 160.7 mm 
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PROBLEM 2.82
Two blocks of rubber with a modulus of rigidity G = 10 MPa are bonded
to rigid supports and to a plate AB. Knowing that b = 200 mm and
c = 125 mm, determine the largest allowable load P and the smallest
allowable thickness a of the blocks if the shearing stress in the rubber
is not to exceed 1.5 MPa and the deflection of the plate is to be at least
6 mm.
SOLUTION
Shearing stress:
τ=
1
2
P
A
=
P
2bc
P = 2bcτ = 2(0.2 m)(0.125 m)(1.5 × 103 kPa)
Shearing strain:
γ=
a=
δ
a
=
Gδ
τ
P = 75.0 kN 
τ
G
=
(10 × 106 Pa)(0.006 m)
= 0.04 m
1.5 × 106 Pa
a = 40.0 mm 
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PROBLEM 2.83∗
Determine the dilatation e and the change in volume of the 200-mm
length of the rod shown if (a) the rod is made of steel with E = 200 GPa
and v = 0.30, (b) the rod is made of aluminum with E = 70 GPa and
v = 0.35.
SOLUTION
π
π
d 2 = (22) 2 = 380.13 mm 2 = 380.13 × 10−6 m 2
4
4
3
P = 46 × 10 N
P
σ x = = 121.01 × 106 Pa
A
σy =σz = 0
A=
εx =
σ
1
(σ x − vσ y − vσ z ) = x
E
E
ε y = ε z = −vε x = −v
σx
E
(1 − 2v)σ x
1
e = ε x + ε y + ε z = (σ x − vσ x − vσ x ) =
E
E
Volume:
(a)
V = AL = (380.13 mm 2 )(200 mm) = 76.026 × 103 mm3
ΔV = Ve
Steel:
e=
(1 − 0.60)(121.01 × 106 )
= 242 × 10−6
9
200 × 10
ΔV = (76.026 × 103 )(242 × 10−6 ) = 18.40 mm3
(b)
Aluminum:
e=
(1 − 0.70)(121.01 × 106 )
= 519 × 10−6
9
70 × 10
ΔV = (76.026 × 103 )(519 × 10−6 ) = 39.4 mm3
e = 242 × 10−6 
ΔV = 18.40 mm3 
e = 519 × 10−6 
ΔV = 39.4 mm3 
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PROBLEM 2.84∗
Determine the change in volume of
the 2-in. gage length segment AB in
Prob. 2.61 (a) by computing the
dilatation of the material, (b) by
subtracting the original volume of
portion AB from its final volume.
SOLUTION
From Problem 2.61, thickness =
(a)
1
in.,
16
E = 29 × 106 psi, v = 0.30.
 1  1 
A =    = 0.03125 in 2
 2  16 
Volume: V0 = AL 0 = (0.03125)(2.00) = 0.0625 in 3
P
600
=
= 19.2 × 103 psi
σy =σz = 0
A 0.03125
σ
1
19.2 × 103
= 662.07 × 10−6
ε x = (σ x − vσ y − vσ z ) = x =
E
E
29 × 106
σx =
ε y = ε z = −v ε x = −(0.30)(662.07 × 10−6 ) = −198.62 × 10−6
e = ε x + ε y + ε z = 264.83 × 10−6
ΔV = V0 e = (0.0625) (264.83 × 10−6 ) = 16.55 × 10−6 in 3
(b)

From the solution to Problem 2.61,
δ x = 1.324 × 10−3 in.,
δ y = −99.3 × 10−6 in.,
δ z = −12.41 × 10−6 in.
The dimensions when under a 600-lb tensile load are:
Length:
L = L0 + δ x = 2 + 1.324 × 10−3 = 2.001324 in.
Width:
w = w0 + δ y =
Thickness:
Volume:
1
− 99.3 × 10−6 = 0.4999007 in.
2
1
t = t0 + δ z = − 12.41 × 10−6 = 0.06248759 in.
16
V = Lwt = 0.062516539 in 3
ΔV = V − V0 = 0.062516539 − 0.0625 = 16.54 × 10−6 in 3

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PROBLEM 2.85∗
A 6-in.-diameter solid steel sphere is lowered into the ocean to a point where the pressure is 7.1 ksi (about
3 miles below the surface). Knowing that E = 29 × 106 psi and v = 0.30, determine (a) the decrease in
diameter of the sphere, (b) the decrease in volume of the sphere, (c) the percent increase in the density of the
sphere.
SOLUTION
For a solid sphere,
V0 =
π
6
d 03
π
(6.00)3
6
= 113.097 in.3
=
σ x = σ y = σ z = −p
= −7.1 × 103 psi
1
(σ x − vσ y − vσ z )
E
(1 − 2v) p
(0.4)(7.1 × 103 )
=−
=−
E
29 × 106
εx =
= −97.93 × 10−6
Likewise,
ε y = ε z = −97.93 × 10−6
e = ε x + ε y + ε z = −293.79 × 10−6
(a)
−Δd = −d 0ε x = −(6.00)(−97.93 × 10−6 ) = 588 × 10−6 in.
(b)
−ΔV = −V0 e = −(113.097)(−293.79 × 10−6 ) = 33.2 × 10−3 in 3
(c)
Let m = mass of sphere.
−Δd = 588 × 10−6 in. 
− ΔV = 33.2 × 10−3 in 3 
m = constant.
m = ρ0V0 = ρV = ρV0 (1 + e)
ρ − ρ0 ρ
V
m
1
=
−1 =
× 0 −1 =
−1
V0 (1 + e) m
1+ e
ρ0
ρ0
= (1 − e + e 2 − e3 + ) − 1 = −e + e 2 − e3 + 
≈ −e = 293.79 × 10−6
ρ − ρ0
× 100% = (293.79 × 10−6 )(100%)
ρ0
0.0294% 
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PROBLEM 2.86∗
(a) For the axial loading shown, determine the change in height and the
change in volume of the brass cylinder shown. (b) Solve part a,
assuming that the loading is hydrostatic with σ x = σ y = σ z = −70 MPa.
SOLUTION
h0 = 135 mm = 0.135 m
π
π
(85) 2 = 5.6745 × 103 mm 2 = 5.6745 × 10−3 m 2
4
4
V0 = A0 h0 = 766.06 × 103 mm3 = 766.06 × 10−6 m3
A 0=
(a)
d 02 =
σ x = 0, σ y = −58 × 106 Pa, σ z = 0
σy
1
58 × 106
(−vσ x + σ y − vσ z ) =
=−
= −552.38 × 10−6
E
E
105 × 109
εy =
Δh = h 0ε y = (135 mm)( − 552.38 × 10−6 )
e=
(1 − 2v)σ y (0.34)(−58 × 106 )
1 − 2v
(σ x + σ y + σ z ) =
=
= −187.81 × 10−6
9
E
E
105 × 10
ΔV = V0 e = (766.06 × 103 mm3 )(−187.81 × 10−6 )
(b)
Δh = −0.0746 mm 
σ x = σ y = σ z = −70 × 106 Pa
εy =
σ x + σ y + σ z = −210 × 106 Pa
1
1 − 2v
(0.34)(−70 × 106 )
(−vσ x + σ y − vσ z ) =
= −226.67 × 10−6
σy =
E
E
105 × 109
Δh = h 0ε y = (135 mm)( −226.67 × 10−6 )
e=
ΔV = −143.9 mm3 
Δh = −0.0306 mm 
1 − 2v
(0.34)( −210 × 106 )
(σ x + σ y + σ z ) =
= −680 × 10−6
9
E
105 × 10
ΔV = V0 e = (766.06 × 103 mm3 )(−680 × 10−6 )
ΔV = −521 mm3 
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PROBLEM 2.87∗
A vibration isolation support consists of a rod A of radius R1 = 10 mm and
a tube B of inner radius R 2 = 25 mm bonded to an 80-mm-long hollow
rubber cylinder with a modulus of rigidity G = 12 MPa. Determine the
largest allowable force P that can be applied to rod A if its deflection is not
to exceed 2.50 mm.
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, R1 ≤ r ≤ R2 .
Shearing stress τ acting on a cylindrical surface of radius r is
τ=
P
P
=
A 2π rh
The shearing strain is
γ=
τ
=
G
P
2π Ghr
Shearing deformation over radial length dr,
dδ
=γ
dr
d δ = γ dr =
P dr
2π Gh r
Total deformation.
δ=

R2
R1
dδ =
P
2π Gh
R2
P
ln r
R1
2π Gh
R
P
ln 2
=
R1
2π Gh
=
Data:
R1 = 10 mm = 0.010 m,
R1
R2 = 25 mm = 0.025 m, h = 80 mm = 0.080 m
G = 12 × 106 Pa
P=
dr
r
P
(ln R2 − ln R1 )
=
2π Gh
2π Ghδ
or P =
ln( R2 / R1 )

R2
δ = 2.50 × 10−3 m
(2π )(12 × 106 ) (0.080) (2.50 × 10−3 )
= 16.46 × 103 N
ln (0.025/0.010)
16.46 kN 
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PROBLEM 2.88
A vibration isolation support consists of a rod A of radius R1 and a tube B of
inner radius R2 bonded to a 80-mm-long hollow rubber cylinder with a
modulus of rigidity G = 10.93 MPa. Determine the required value of the ratio
R2/R1 if a 10-kN force P is to cause a 2-mm deflection of rod A.
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, R1 ≤ r ≤ R2 .
Shearing stress τ acting on a cylindrical surface of radius r is
τ=
P
P
=
A 2π rh
The shearing strain is
γ=
τ
G
=
P
2π Ghr
Shearing deformation over radial length dr,
dδ
=γ
dr
d δ = γ dr
P dr
drδ =
2π Gh r
Total deformation.
δ=

R2
R1
dδ =
P
2π Gh
R2
P
ln r
R1
2π Gh
R
P
=
ln 2
R1
2π Gh
=
ln
dr
R1 r
P
(ln R2 − ln R1 )
=
2π Gh

R2
R2 2π Ghδ (2π ) (10.93 × 106 ) (0.080) (0.002)
=
=
= 1.0988
R1
P
10.103
R2
= exp (1.0988) = 3.00
R1
R2 /R1 = 3.00 
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PROBLEM 2.89∗
The material constants E, G, k, and v are related by Eqs. (2.33) and (2.43). Show that any one of these
constants may be expressed in terms of any other two constants. For example, show that
(a) k = GE/(9G − 3E) and (b) v = (3k – 2G)/(6k + 2G).
SOLUTION
k=
(a)
1+ v =
and
G=
E
2(1 + v)
E
E
or v =
−1
2G
2G
k=
(b)
E
3(1 − 2v)
E
2 EG
2 EG
=
=

 E
  3[2G − 2 E + 4G ] 18G − 6 E
− 1 
3 1 − 2 
 2G  

k=
EG

9G − 6 E
v=
3k − 2G

6k + 2G
k
2(1 + v)
=
G 3(1 − 2v)
3k − 6kv = 2G + 2Gv
3k − 2G = 2G + 6k
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PROBLEM 2.90∗
Show that for any given material, the ratio G/E of the modulus of rigidity over the modulus of elasticity is
always less than 12 but more than 13 . [Hint: Refer to Eq. (2.43) and to Sec. 2.13.]
SOLUTION
G=
E
2(1 + v)
Assume v > 0 for almost all materials, and v <
or
1
2
E
= 2(1 + v)
G
for a positive bulk modulus.
E
 1
< 2 1 +  = 3
G
2

Applying the bounds,
2≤
Taking the reciprocals,
1
G 1
>
>
2
E 3
or
1 G 1
< < 
3
E 2
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PROBLEM 2.91∗
A composite cube with 40-mm sides and the properties shown is
made with glass polymer fibers aligned in the x direction. The cube
is constrained against deformations in the y and z directions and is
subjected to a tensile load of 65 kN in the x direction. Determine
(a) the change in the length of the cube in the x direction, (b) the
stresses σ x , σ y , and σ z .
Ex = 50 GPa
vxz = 0.254
E y = 15.2 GPa vxy = 0.254
Ez = 15.2 GPa vzy = 0.428
SOLUTION
Stress-to-strain equations are
εx =
σx
Ex
εy = −
εz = −
−
v yxσ y
Ey
vxyσ x
Ex
+
−
σy
Ey
vzxσ z
Ez
−
vzyσ z
Ez
vxzσ x v yzσ y σ z
−
+
Ex
Ey
Ez
vxy
Ex
v yz
Ey
=
=
v yx
Ey
vzy
Ez
vzx vxz
=
Ez Ex
(1)
(2)
(3)
(4)
(5)
(6)
ε y = 0 and ε z = 0.
The constraint conditions are
Using (2) and (3) with the constraint conditions gives
vzy
vxy
1
σy −
σz =
σx
Ey
Ez
Ex
−
v yz
Ey
σy +
V
1
σ z = xz σ x
Ez
Ex
(7)
(8)
1
0.428
0.254
σy −
σz =
σ x or σ y − 0.428σ z = 0.077216 σ x
15.2
15.2
50
0.428
1
0.254
−
σy +
σz =
σ x or − 0.428 σ y + σ z = 0.077216 σ x
15.2
15.2
50
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PROBLEM 2.91∗ (Continued)
Solving simultaneously,
σ y = σ z = 0.134993 σ x
Using (4) and (5) in (1),
εx =
Ex =
=
vxy
v
1
σx −
σ y − xz σ z
Ex
Ex
E
1
[1 − (0.254) (0.134993) − (0.254)(0.134993)]σ x
Ex
0.93142 σ x
Ex
A = (40)(40) = 1600 mm 2 = 1600 × 10−6 m 2
P
65 × 103
=
= 40.625 × 106 Pa
A 1600 × 10−6
(0.93142) (40.625 × 103 )
εx =
= 756.78 × 10−6
9
50 × 10
σx =
(a)
δ x = Lx ε x = (40 mm) (756.78 × 10−6 )
δ x = 0.0303 mm 
(b)
σ x = 40.625 × 106 Pa
σ x = 40.6 MPa 

σ y = σ z = (0.134993) (40.625 × 106 ) = 5.48 × 106 Pa 
σ y = σ z = 5.48 MPa 
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PROBLEM 2.92∗
The composite cube of Prob. 2.91 is constrained against
deformation in the z direction and elongated in the x direction by
0.035 mm due to a tensile load in the x direction. Determine (a) the
stresses σ x , σ y , and σ z , (b) the change in the dimension in the
y direction.
Ex = 50 GPa
vxz = 0.254
E y = 15.2 GPa vxy = 0.254
Ez = 15.2 GPa vzy = 0.428
SOLUTION
εx =
σx
Ex
εy = −
εz = −
−
v yxσ y
vxyσ x
Ex
Ey
+
−
σy
Ey
vzxσ z
Ez
−
vzyσ z
Ez
vxzσ x v yzσ y σ z
−
+
Ex
Ey
Ez
vxy
Ex
v yz
Ey
=
=
v yx
Ey
vzy
Ez
vzx vxz
=
Ez Ex
Constraint condition:
Load condition :
(1)
(2)
(3)
(4)
(5)
(6)
εz = 0
σy = 0
0=−
From Equation (3),
σz =
vxz
1
σx + σz
Ex
Ez
vxz Ez
(0.254) (15.2)
σx =
= 0.077216 σ x
Ex
50
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PROBLEM 2.92∗ (Continued)
From Equation (1) with σ y = 0,
εx =
=
1
1
[σ x − 0.254 σ z ] =
[1 − (0.254) (0.077216)]σ x
Ex
Ex
=
0.98039
σx
Ex
σx =
But,
εx =
(a)
σx =
δx
Lx
=
v
v
1
1
σ x − zx σ z = σ x − xz σ z
Ex
Ez
Ex
Ex
Exε x
0.98039
0.035 mm
= 875 × 10−6
40 mm
(50 × 109 ) (875 × 10−6 )
= 44.625 × 103 Pa
0.98039
σ x = 44.6 MPa 
σy = 0
σ z = (0.077216) (44.625 × 106 ) = 3.446 × 106 Pa
From (2),
εy =
vxy
Ex
σx +
σ z = 3.45 MPa 
vzy
1
σy −
σz
Ey
Ez
(0.254) (44.625 × 106 )
(0.428) (3.446 × 106 )
0
+
−
50 × 109
15.2 × 109
−6
= −323.73 × 10
=−
(b)
δ y = Ly ε y = (40 mm) ( − 323.73 × 10−6 )
δ y = −0.0129 mm 
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PROBLEM 2.93
Two holes have been drilled through a long steel bar that is subjected to a centric
axial load as shown. For P = 6.5 kips, determine the maximum value of the
stress (a) at A, (b) at B.
SOLUTION
(a)
At hole A:
r =
1 1 1
⋅ = in.
2 2 4
d =3−
1
= 2.50 in.
2
1
Anet = dt = (2.50)   = 1.25 in 2
2
σ non =
P
6.5
=
= 5.2 ksi
Anet 1.25
2 ( 14 )
2r
=
= 0.1667
D
3
From Fig. 2.60a,
K = 2.56
σ max = Kσ non = (2.56)(5.2)
(b)
r =
At hole B:
1
(1.5) = 0.75,
2
σ max = 13.31 ksi 
d = 3 − 1.5 = 1.5 in.
1
Anet = dt = (1.5)   = 0.75 in 2 ,
2
σ non =
P
6.5
=
= 8.667 ksi
Anet
0.75
2r
2(0.75)
=
= 0.5
D
3
From Fig. 2.60a,
K = 2.16
σ max = Kσ non = (2.16)(8.667)
σ max = 18.72 ksi 
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PROBLEM 2.94
Knowing that σ all = 16 ksi, determine the maximum allowable value of the
centric axial load P.
SOLUTION
At hole A:
r =
1 1 1
⋅ = in.
2 2 4
d =3−
1
= 2.50 in.
2
1
Anet = dt = (2.50)   = 1.25 in 2
2
2 ( 14 )
2r
=
= 0.1667
D
3
From Fig. 2.60a,
K = 2.56
σ max =
At hole B:
r =
KP
Anet
∴ P=
Anetσ max
(1.25)(16)
=
= 7.81 kips
K
2.56
1
(1.5) = 0.75 in,
2
d = 3 − 1.5 = 1.5 in.
1
Anet = dt = (1.5)   = 0.75 in 2 ,
2
2r
2(0.75)
=
= 0.5
D
3
From Fig. 2.60a,
K = 2.16
P=
Smaller value for P controls.
Anetσ max
(0.75)(16)
=
= 5.56 kips
K
2.16
P = 5.56 kips 
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PROBLEM 2.95
Knowing that the hole has a diameter of 9 mm, determine (a) the radius
rf of the fillets for which the same maximum stress occurs at the hole A
and at the fillets, (b) the corresponding maximum allowable load P if
the allowable stress is 100 MPa.
SOLUTION
1
r =   (9) = 4.5 mm
2
For the circular hole,
d = 96 − 9 = 87 mm
2r
2(4.5)
=
= 0.09375
D
96
Anet = dt = (0.087 m)(0.009 m) = 783 × 10−6 m 2
K hole = 2.72
From Fig. 2.60a,
σ max =
P=
(a)
For fillet,
K hole P
Anet
Anetσ max
(783 × 10−6 )(100 × 106 )
=
= 28.787 × 103 N
K hole
2.72
D = 96 mm, d = 60 mm
D 96
=
= 1.60
d
60
Amin = dt = (0.060 m)(0.009 m) = 540 × 10−6 m 2
σ max =
(5.40 × 10−6 )(100 × 106 )
K fillet P
A σ
∴ K fillet = min max =
Amin
P
28.787 × 103
= 1.876
From Fig. 2.60b,
rf
d
≈ 0.19 ∴ rf ≈ 0.19d = 0.19(60)
r f = 11.4 mm 
(b) P = 28.8 kN 
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PROBLEM 2.96
For P = 100 kN, determine the minimum plate thickness t required if
the allowable stress is 125 MPa.
SOLUTION
At the hole:
rA = 20 mm
d A = 88 − 40 = 48 mm
2rA
2(20)
=
= 0.455
DA
88
From Fig. 2.60a,
K = 2.20
σ max =
t =
At the fillet:
From Fig. 2.60b,
KP
KP
KP
=
∴ t =
Anet
d At
d Aσ max
(2.20)(100 × 103 N)
= 36.7 × 10−3 m = 36.7 mm
6
(0.048 m)(125 × 10 Pa)
D
88
=
= 1.375
dB
64
D = 88 mm,
d B = 64 mm
rB = 15 mm
rB
15
=
= 0.2344
dB
64
K = 1.70
σ max =
t =
KP
KP
=
Amin
d Bt
KP
d Bσ max
=
(1.70)(100 × 103 N)
= 21.25 × 10−3 m = 21.25 mm
(0.064 m)(125 × 106 Pa)
The larger value is the required minimum plate thickness.

t = 36.7 mm 
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PROBLEM 2.97
The aluminum test specimen shown is subjected to two equal and opposite
centric axial forces of magnitude P. (a) Knowing that E = 70 GPa
and σ all = 200 MPa, determine the maximum allowable value of P and the
corresponding total elongation of the specimen. (b) Solve part a, assuming that
the specimen has been replaced by an aluminum bar of the same length and a
uniform 60 × 15-mm rectangular cross section.
SOLUTION
σ all = 200 × 106 Pa E = 70 × 109 Pa
Amin = (60 mm) (15 mm) = 900 mm 2 = 900 × 10−6 m 2
(a)
Test specimen.
D = 75 mm, d = 60 mm, r = 6 mm
D 75
=
= 1.25
d 60
From Fig. 2.60b
K = 1.95
P=
r
6
=
= 0.10
d 60
σ max = K
P
A
Aσ max (900 × 10−6 ) (200 × 106 )
=
= 92.308 × 103 N
K
1.95
P = 92.3 kN 
Wide area A* = (75 mm) (15 mm) = 1125 mm 2 = 1.125 × 10−3 m 2
δ =Σ
Pi Li
P L 92.308 × 103  0.150
0.300
0.150 
= Σ i =
+
+

−3
−6
9
Ai Ei E Ai
70 × 10
900 × 10
1.125 × 10−3 
1.125 × 10
= 7.91 × 10−6 m
(b)
Uniform bar.
P = Aσ all = (900 × 10−6 )(200 × 106 ) = 180 × 103 N

δ = 0.791 mm 
δ=
PL
(180 × 103 )(0.600)
=
= 1.714 × 10−3 m 
AE (900 × 10−6 )(70 × 109 )
P = 180.0 kN 
δ = 1.714 mm 
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PROBLEM 2.98
For the test specimen of Prob. 2.97, determine the maximum value of the
normal stress corresponding to a total elongation of 0.75 mm.
PROBLEM 2.97 The aluminum test specimen shown is subjected to two equal
and opposite centric axial forces of magnitude P. (a) Knowing that E = 70 GPa
and σ all = 200 MPa, determine the maximum allowable value of P and the
corresponding total elongation of the specimen. (b) Solve part a, assuming that
the specimen has been replaced by an aluminum bar of the same length and a
uniform 60 × 15-mm rectangular cross section.
SOLUTION
δ =Σ
Pi Li P Li
= Σ
Ei Ai E Ai
δ = 0.75 × 10−3 m
L1 = L3 = 150 mm = 0.150 m,
L2 = 300 mm = 0.300 m
A1 = A3 = (75 mm) (15 mm) = 1125 mm 2 = 1.125 × 10−3 m 2
A2 = (60 mm) (15 mm) = 900 mm2 = 900 × 10−6 m 2
Σ
Li
0.150
0.300
0.150
=
+
+
= 600 m −1
−3
−6
−3
Ai 1.125 × 10
900 × 10
1.125 × 10
P=
Stress concentration.
(70 × 109 ) (0.75 × 10−3 )
Eδ
=
= 87.5 × 103 N
Li
600
Σ
Ai
D = 75 mm, d = 60 mm, r = 6 mm
D 75
=
= 1.25
d 60
From Fig. 2.60b
6
r
=
= 0.10
d 60
K = 1.95
σ max = K
P
(1.95) (87.5 × 103 )
=
= 189.6 × 106 Pa
−6
Amin
900 × 10
σ max = 189.6 MPa 
Note that σ max < σ all . 
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PROBLEM 2.99
A hole is to be drilled in the plate at A. The diameters of the bits
available to drill the hole range from 12 to 11/2 in. in 14 -in.
increments. If the allowable stress in the plate is 21 ksi,
determine (a) the diameter d of the largest bit that can be used if
the allowable load P at the hole is to exceed that at the fillets, (b)
the corresponding allowable load P.
SOLUTION
At the fillets:
D
4.6875
=
= 1.5
d
3.125
From Fig. 2.60b,
K = 2.10
r
0.375
=
= 0.12
d
3.125
Amin = (3.125)(0.5) = 1.5625 in 2
σ max = K
Pall =
Aminσ all
(1.5625)(21)
=
= 15.625 kips
K
2.10
Anet = ( D − 2r )t , K from Fig. 2.60a
At the hole:
σ max = K
with
Hole diam.
0.5 in.
Pall
= σ all
Amin
r
0.25 in.
P
= σ all
Anet
∴
Pall =
D = 4.6875 in.
t = 0.5 in.
d = D − 2r
2r/D
4.1875 in.
0.107
Anetσ all
K
σ all = 21 ksi
K
2.68
Anet
Pall
2.0938 in2
16.41 kips
2
0.75 in.
0.375 in.
3.9375 in.
0.16
2.58
1.96875 in
16.02 kips
1 in.
0.5 in.
3.6875 in.
0.213
2.49
1.84375 in2
15.55 kips
2
1.25 in.
0.625 in.
3.4375 in.
0.267
2.41
1.71875 in
14.98 kips
1.5 in.
0.75 in.
3.1875 in.
0.32
2.34
1.59375 in2
14.30 kips
(a)
Largest hole with Pall > 15.625 kips is the
(b)
Allowable load Pall = 15.63 kips
3
4
-in. diameter hole.


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PROBLEM 2.100
(a) For P = 13 kips and d = 12 in., determine the maximum stress
in the plate shown. (b) Solve part a, assuming that the hole at A is
not drilled.
SOLUTION
Maximum stress at hole:
Use Fig. 2.60a for values of K.
2r
0.5
=
= 0.017,
D
4.6875
K = 2.68
Anet = (0.5)(4.6875 − 0.5) = 2.0938 in 2
σ max = K
P
(2.68)(13)
=
= 16.64 ksi,
Anet
2.0938
Maximum stress at fillets:
Use Fig. 2.60b for values of K.
r
0.375
=
= 0.12
d
3.125
D
4.6875
=
= 1.5
d
3.125
K = 2.10
Amin = (0.5)(3.125) = 1.5625 in 2
σ max = K
(2.10)(13)
P
=
= 17.47 ksi
Amin
1.5625
(a)
With hole and fillets:
17.47 ksi 
(b)
Without hole:
17.47 ksi 
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PROBLEM 2.101
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa.
A force P is applied to the rod and then removed to give it a permanent set
δ p = 2 mm. Determine the maximum value of the force P and the maximum
amount δ m by which the rod should be stretched to give it the desired permanent
set.
SOLUTION
AAB =
ABC =
π
4
π
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
(40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2
4
Pmax = Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N
Pmax = 176.7 kN 
δ′ =
P′LAB P ′LBC
(176.715 × 103 )(0.8)
(176.715 × 103 )(1.2)
+
=
+
EAAB
EABC (200 × 109 )(706.86 × 10−6 ) (200 × 109 )(1.25664 × 10−3 )
= 1.84375 × 10−3 m = 1.84375 mm
δ p = δ m − δ ′ or δ m = δ p + δ ′ = 2 + 1.84375
δ m = 3.84 mm 
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PROBLEM 2.102
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa.
A force P is applied to the rod until its end A has moved down by an amount
δ m = 5 mm. Determine the maximum value of the force P and the permanent
set of the rod after the force has been removed.
SOLUTION
AAB =
4
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
π
(40) 2 = 1.25664 × 103 mm 2 = 1.25644 × 10−3 m 2
4
= Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N
ABC =
Pmax
π
Pmax = 176.7 kN 
δ′ =
P′LAB P ′LBC
(176.715 × 103 )(0.8)
(176.715 × 103 )(1.2)
+
=
+
EAAB
EABC (200 × 109 )(706.68 × 10−6 ) (200 × 109 )(1.25664 × 10−3 )
= 1.84375 × 10−3 m = 1.84375 mm
δ p = δ m − δ ′ = 5 − 1.84375 = 3.16 mm
δ p = 3.16 mm 
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without permission.
PROBLEM 2.103
The 30-mm square bar AB has a length L = 2.2 m; it is made of a mild steel that is
assumed to be elastoplastic with E = 200 GPa and σ Y = 345 MPa. A force P is applied
to the bar until end A has moved down by an amount δ m . Determine the maximum
value of the force P and the permanent set of the bar after the force has been removed,
knowing that (a) δ m = 4.5 mm, (b) δ m = 8 mm.
SOLUTION
A = (30)(30) = 900 mm2 = 900 × 10−6 m 2
δY = Lε Y =
Lσ Y (2.2)(345 × 106 )
=
= 3.795 × 10−3 = 3.795 mm
E
200 × 109
If δ m ≥ δ Y , Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.5 × 103 N
Unloading: δ ′ =
Pm L σ Y L
=
= δY = 3.795 mm
AE
E
δ P = δm − δ ′
(a)
δ m = 4.5 mm > δY Pm = 310.5 × 103 N
δ perm = 4.5 mm − 3.795 mm
(b)
δ m = 8 mm > δY Pm = 310.5 × 103 N

δ perm = 8.0 mm − 3.795 mm
δ m = 310.5 kN 
δ perm = 0.705 mm 
δ m = 310.5 kN 
δ perm = 4.205 mm 
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PROBLEM 2.104
The 30-mm square bar AB has a length L = 2.5 m; it is made of mild steel that is assumed
to be elastoplastic with E = 200 GPa and σ Y = 345 MPa. A force P is applied to the bar
and then removed to give it a permanent set δ p . Determine the maximum value of the
force P and the maximum amount δ m by which the bar should be stretched if the desired
value of δ p is (a) 3.5 mm, (b) 6.5 mm.
SOLUTION
A = (30)(30) = 900 mm 2 = 900 × 10−6 m 2
δY = Lε Y =
Lσ Y (2.5)(345 × 106 )
=
= 4.3125 × 103 m = 4.3125 mm
9
E
200 × 10
When δ m exceeds δ Y , thus producing a permanent stretch of δ p , the maximum force is
Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.5 × 103 N
= 310.5 kN 
δ p = δ m − δ ′ = δ m − δY ∴ δ m = δ p + δY
(a)
δ p = 3.5 mm δ m = 3.5 mm + 4.3125 mm
= 7.81 mm 
(b)
δ p = 6.5 mm δ m = 6.5 mm + 4.3125 mm
= 10.81 mm 
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PROBLEM 2.105
Rod AB is made of a mild steel that is assumed to be elastoplastic
with E = 29 × 106 psi and σ Y = 36 ksi. After the rod has been attached
to the rigid lever CD, it is found that end C is 83 in. too high. A vertical
force Q is then applied at C until this point has moved to position C ′.
Determine the required magnitude of Q and the deflection δ1 if the
lever is to snap back to a horizontal position after Q is removed.
SOLUTION
Since the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where
PY = Aσ Y =
π 3
2
(36) = 3.976 kips
4  8 
Referring to the free body diagram of lever CD,
ΣM D = 0: 33 Q − 22 P = 0
22
(22)(3.976)
Q=
P=
= 2.65 kips
33
33
Q = 2.65 kips 
During unloading, the spring back at B is
δ B = LABε Y =
LABσ Y (60)(36 × 103 )
=
= 0.0745 in.
E
29 × 106
From the deformation diagram,
Slope:
θ=
δB
22
=
δC
33
∴ δC =
33
δ B = 0.1117 in.
−22
δ C = 0.1117 in. 
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PROBLEM 2.106
Solve Prob. 2.105, assuming that the yield point of the mild steel is
50 ksi.
PROBLEM 2.105 Rod AB is made of a mild steel that is assumed to
be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi. After the rod
has been attached to the rigid lever CD, it is found that end C is 83 in.
too high. A vertical force Q is then applied at C until this point has
moved to position C′. Determine the required magnitude of Q and the
deflection δ1 if the lever is to snap back to a horizontal position after
Q is removed.
SOLUTION
Since the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where
PY = Aσ Y =
π 3
2
(50) = 5.522 kips
4  8 
Referring to the free body diagram of lever CD,
ΣM D = 0: 33Q − 22 P = 0
Q=
22
(22)(5.522)
P=
= 3.68 kips
33
33
Q = 3.68 kips 
During unloading, the spring back at B is
δ B = LAB ε Y =
LABσ Y (60)(50 × 103 )
=
= 0.1034 in.
E
29 × 106
From the deformation diagram,
Slope:
θ=
δB
22
=
δC
33
∴
δC =
33
δB
22
δ C = 0.1552 in. 
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PROBLEM 2.107
Each cable has a cross-sectional area of 100 mm2 and is made of an
elastoplastic material for which σ Y = 345 MPa and E = 200 GPa. A force
Q is applied at C to the rigid bar ABC and is gradually increased from 0 to
50 kN and then reduced to zero. Knowing that the cables were initially taut,
determine (a) the maximum stress that occurs in cable BD, (b) the
maximum deflection of point C, (c) the final displacement of point C.
(Hint: In Part c, cable CE is not taut.)
SOLUTION
Elongation constraints for taut cables.
Let θ = rotation angle of rigid bar ABC.
θ=
δ BD
LAB
δ BD =
=
δ CE
LAC
LAB
1
δ CE = δ CE
LAC
2
(1)
Equilibrium of bar ABC.
M A = 0 : LAB FBD + LAC FCE − LAC Q = 0
Q = FCE +
LAB
1
FBD = FCE + FBD
2
LAC
Assume cable CE is yielded.
FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 N
From (2),
FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 N
(2)
Since FBD < Aσ Y = 34.5 × 103 N, cable BD is elastic when Q = 50 kN.
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PROBLEM 2.107 (Continued)
(a)
σ CE = σ Y = 345 MPa
Maximum stresses.
σ BD =
(b)
FBD 31.0 × 103
=
= 310 × 106 Pa
−6
A
100 × 10
σ BD = 310 MPa 
Maximum of deflection of point C.
δ BD =
From (1),
FBD LBD
(31.0 × 103 )(2)
=
= 3.1 × 10−3 m
EA
(200 × 109 )(100 × 10−6 )
δ C = δ CE = 2δ BD = 6.2 × 10−3 m
Permanent elongation of cable CE: (δ CE ) p = (δ CE ) −
6.20 mm ↓ 
σ Y LCE
E
FCE LCE
σ L
= (δ CE ) max − Y CE
EA
E
6
(345 × 10 )(2)
= 6.20 × 10−3 −
= 2.75 × 10−3 m
9
200 × 10
(δ CE ) P = (δ CE ) max −
(c)
Unloading. Cable CE is slack ( FCE = 0) at Q = 0.
From (2),
FBD = 2(Q − FCE ) = 2(0 − 0) = 0
Since cable BD remained elastic, δ BD =
FBD LBD
= 0.
EA

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PROBLEM 2.108
Solve Prob. 2.107, assuming that the cables are replaced by rods of the
same cross-sectional area and material. Further assume that the rods are
braced so that they can carry compressive forces.
PROBLEM 2.107 Each cable has a cross-sectional area of 100 mm2
and is made of an elastoplastic material for which σ Y = 345 MPa and
E = 200 GPa. A force Q is applied at C to the rigid bar ABC and is
gradually increased from 0 to 50 kN and then reduced to zero. Knowing
that the cables were initially taut, determine (a) the maximum stress
that occurs in cable BD, (b) the maximum deflection of point C, (c) the
final displacement of point C. (Hint: In Part c, cable CE is not taut.)
SOLUTION
Elongation constraints.
Let θ = rotation angle of rigid bar ABC.
θ=
δ BD =
δ BC
LAB
=
δ CE
LAC
LAB
1
δ CE = δ CE
LAC
2
(1)
Equilibrium of bar ABC.
M A = 0: LAB FBD + LAC FCE − LAC Q = 0
Q = FCE +
LAB
1
FBD = FCE + FBD
LAC
2
(2)
Assume cable CE is yielded. FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 N
From (2),
FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 N
Since FBD < Aσ Y = 34.5 × 103 N, cable BD is elastic when Q = 50 kN.
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PROBLEM 2.108 (Continued)
(a)
σ CE = σ Y = 345 MPa
Maximum stresses.
σ BD =
(b)
FBD 31.0 × 103
=
= 310 × 106 Pa
−6
A
100 × 10
σ BD = 310 MPa 
Maximum of deflection of point C.
δ BD =
FBD LBD
(31.0 × 103 )(2)
=
= 3.1 × 10−3 m
EA
(200 × 109 )(100 × 10−6 )
δ C = δ CE = 2δ BD = 6.2 × 10−3 m
From (1),
6.20 mm ↓ 
′ = δ C′
Unloading. Q′ = 50 × 103 N, δ CE
′ = 12 δ C′
δ BD
From (1),
′′ =
Elastic FBD
′ =
FCE
(200 × 109 )(100 × 10−6 )( 12 δ C′ )
′
EAδ BD
=
= 5 × 106 δ C′
LBD
2
′
EAδ CE
(200 × 109 )(100 × 10−6 )(δ C′ )
=
= 10 × 106 δ C′
LCE
2
From (2),
′ + 12 FBD
′ = 12.5 × 106 δ C′
Q′ = FCE
Equating expressions for Q′,
12.5 × 106 δ C′ = 50 × 103
δ C′ = 4 × 10−3 m
(c)
Final displacement.
δ C = (δ C ) m − δ C′ = 6.2 × 10−3 − 4 × 10−3 = 2.2 × 10−3 m
2.20 mm ↓ 
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PROBLEM 2.109
Rod AB consists of two cylindrical portions AC and BC, each with a cross-sectional
area of 1750 mm2. Portion AC is made of a mild steel with E = 200 GPa and
σ Y = 250 MPa, and portion CB is made of a high-strength steel with E = 200 GPa
and σ Y = 345 MPa. A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
SOLUTION
Displacement at C to cause yielding of AC.
L σ
(0.190)(250 × 106 )
δ C ,Y = LAC ε Y , AC = AC Y , AC =
= 0.2375 × 10−3 m
9
E
200 × 10
FAC = Aσ Y , AC = (1750 × 10−6 )(250 × 106 ) = 437.5 × 103 N
Corresponding force.
FCB = −
EAδ C
(200 × 109 )(1750 × 10−6 )(0.2375 × 10−3 )
=−
= −437.5 × 103 N
0.190
LCB
For equilibrium of element at C,
FAC − ( FCB + PY ) = 0
PY = FAC − FCB = 875 × 103 N
Since applied load P = 975 × 103 N > 875 × 103 N, portion AC yields.
FCB = FAC − P = 437.5 × 103 − 975 × 103 N = −537.5 × 103 N
(a)
δC = −
FCB LCD
(537.5 × 103 )(0.190)
=
= 0.29179 × 10−3 m
EA
(200 × 109 )(1750 × 10−6 )
0.292 mm 
(b)
Maximum stresses: σ AC = σ Y , AC = 250 MPa
(c)
FBC
537.5 × 103
=−
= −307.14 × 106 Pa = −307 MPa
−6
A
1750 × 10
Deflection and forces for unloading.
P′ L
P′ L
L
′ = − PAC
′ AC = − PAC
′
δ ′ = AC AC = − CB CB ∴ PCB
EA
EA
LAB
σ BC =

250 MPa 
−307 MPa 
′ − PCB
′ = 2 PAC
′ PAC
′ = 487.5 × 10−3 N
P′ = 975 × 103 = PAC
δ′ =
(487.5 × 103 )(0.190)
= 0.26464 × 103 m
−6
9
(200 × 10 )(1750 × 10 )
δ p = δ m − δ ′ = 0.29179 × 10−3 − 0.26464 × 10−3
= 0.02715 × 10−3 m
0.0272 mm 
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PROBLEM 2.110
For the composite rod of Prob. 2.109, if P is gradually increased from zero until the
deflection of point C reaches a maximum value of δ m = 0.3 mm and then decreased
back to zero, determine (a) the maximum value of P, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C after the load is removed.
PROBLEM 2.109 Rod AB consists of two cylindrical portions AC and BC, each with a
cross-sectional area of 1750 mm2. Portion AC is made of a mild steel with E = 200 GPa
and σ Y = 250 MPa, and portion CB is made of a high-strength steel with E = 200 GPa
and σ Y = 345 MPa. A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
SOLUTION
Displacement at C is δ m = 0.30 mm. The corresponding strains are
ε AC =
δm
LAC
ε CB = −
=
δm
LCB
0.30 mm
= 1.5789 × 10−3
190 mm
=−
0.30 mm
= −1.5789 × 10−3
190 mm
Strains at initial yielding:
ε Y, AC =
ε Y, CB =
(a)
σ Y, AC
E
σ Y, BC
E
250 × 106
= 1.25 × 10−3
(yielding)
200 × 109
345 × 106
=−
= −1.725 × 10−3
(elastic)
200 × 109
=
Forces: FAC = Aσ Y = (1750 × 10−6 )(250 × 106 ) = 437.5 × 10−3 N
FCB = EAε CB = (200 × 109 )(1750 × 10−6 )(−1.5789 × 10−3 ) = −552.6 × 10−3 N
For equilibrium of element at C, FAC − FCB − P = 0
P = FAC − FCD = 437.5 × 103 + 552.6 × 103 = 990.1 × 103 N
(b)
Stresses: AC : σ AC = σ Y, AC
CB : σ CB =
FCB
552.6 × 103
=−
= −316 × 106 Pa
−6
A
1750 × 10
= 990 kN 
= 250 MPa 
−316 MPa 
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PROBLEM 2.110 (Continued)
(c) Deflection and forces for unloading.
δ′ =
′ LAC
PAC
P′ L
L
′ = − PAC
′ AC = − PAC
= − CB CB ∴ PCB
EA
EA
LAB
′ − PCB
′ = 2 PAC
′ = 990.1 × 103 N ∴ PAC
′ = 495.05 × 103 N
P′ = PAC
δ′ =
(495.05 × 103 )(0.190)
= 0.26874 × 10−3 m = 0.26874 mm
(200 × 109 )(1750 × 10−6 )
δ p = δ m − δ ′ = 0.30 mm − 0.26874 mm
0.031mm 
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PROBLEM 2.111
Two tempered-steel bars, each 163 -in. thick, are bonded to a 12 -in. mild-steel
bar. This composite bar is subjected as shown to a centric axial load of
magnitude P. Both steels are elastoplastic with E = 29 × 106 and with yield
strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild
steel. The load P is gradually increased from zero until the deformation of
the bar reaches a maximum value δ m = 0.04 in. and then decreased back to
zero. Determine (a) the maximum value of P, (b) the maximum stress in the
tempered-steel bars, (c) the permanent set after the load is removed.
SOLUTION
1
For the mild steel, A1 =   (2) = 1.00 in 2
2
δY1 =
Lσ Y 1 (14)(50 × 103 )
=
= 0.024138 in.
E
29 × 106
 3
For the tempered steel, A2 = 2   (2) = 0.75 in 2
 16 
δY 2 =
Lσ Y 2 (14)(100 × 103 )
=
= 0.048276 in.
E
29 × 103
Total area: A = A1 + A2 = 1.75 in 2
δ Y 1 < δ m < δY 2 . The mild steel yields. Tempered steel is elastic.
(a)
Forces: P1 = A1σ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb
P2 =
EA2δ m (29 × 103 )(0.75)(0.04)
=
= 62.14 × 103 lb
14
L
P = P1 + P2 = 112.14 × 103 lb = 112.1kips
(b)
Stresses: σ1 =
σ2 =
Unloading:
(c)
P = 112.1 kips 
P1
= σ Y 1 = 50 × 103 psi = 50 ksi
A1
P2 62.14 × 103
=
= 82.86 × 103 psi = 82.86 ksi
A2
0.75
δ′ =
82.86 ksi 
PL (112.14 × 103 )(14)
=
= 0.03094 in.
EA
(29 × 106 )(1.75)
Permanent set: δ p = δ m − δ ′ = 0.04 − 0.03094 = 0.00906 in.
0.00906in. 
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PROBLEM 2.112
For the composite bar of Prob. 2.111, if P is gradually increased from zero to
98 kips and then decreased back to zero, determine (a) the maximum
deformation of the bar, (b) the maximum stress in the tempered-steel bars,
(c) the permanent set after the load is removed.
PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded to
a 12 -in. mild-steel bar. This composite bar is subjected as shown to a centric
axial load of magnitude P. Both steels are elastoplastic with E = 29 × 106 psi
and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the
tempered and mild steel.
SOLUTION
Areas:
Mild steel:
1
A1=   (2) = 1.00 in 2
2
Tempered steel:
 3
A2 = 2   (2) = 0.75 in 2
 16 
A = A1+ A2 = 1.75 in 2
Total:
Total force to yield the mild steel:
σY1 =
PY
∴ PY = Aσ Y 1 = (1.75)(50 × 103 ) = 87.50 × 103 lb
A
P > PY , therefore, mild steel yields.
Let P1 = force carried by mild steel.
P2 = force carried by tempered steel.
P1 = A1σ1 = (1.00)(50 × 103 ) = 50 × 103 lb
P1 + P2 = P, P2 = P − P1 = 98 × 103 − 50 × 103 = 48 × 103 lb
(a)
δm =
P2 L
(48 × 103 )(14)
=
EA2 (29 × 106 )(0.75)
(b)
σ2 =
P2 48 × 103
=
= 64 × 103 psi
A2
0.75
Unloading: δ ′ =
(c)
= 0.03090in. 
= 64 ksi 
PL
(98 × 103 )(14)
=
= 0.02703 in.
EA (29 × 106 )(1.75)
δ P = δ m − δ ′ = 0.03090 − 0.02703
= 0.00387 in. 
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PROBLEM 2.113
The rigid bar ABC is supported by two links, AD and BE, of uniform
37.5 × 6-mm rectangular cross section and made of a mild steel that is
assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa.
The magnitude of the force Q applied at B is gradually increased from
zero to 260 kN. Knowing that a = 0.640 m, determine (a) the value of
the normal stress in each link, (b) the maximum deflection of point B.
SOLUTION
ΣM C = 0 : 0.640(Q − PBE ) − 2.64 PAD = 0
Statics:
δ A = 2.64θ , δ B = aθ = 0.640θ
Deformation:
Elastic analysis:
A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2
PAD =
σ AD
PBE
σ BE
EA
(200 × 109 )(225 × 10−6 )
δA =
δ A = 26.47 × 106 δ A
LAD
1.7
= (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ
P
= AD = 310.6 × 109 θ
A
(200 × 109 )(225 × 10−6 )
EA
=
δB =
δ B = 45 × 106 δ B
1.0
LBE
= (45 × 106 )(0.640θ ) = 28.80 × 106 θ
P
= BE = 128 × 109 θ
A
From Statics, Q = PBE +
2.64
PAD = PBE + 4.125PAD
0.640
= [28.80 × 106 + (4.125)(69.88 × 106 ] θ = 317.06 × 106 θ
θY at yielding of link AD:
σ AD = σ Y = 250 × 106 = 310.6 × 109 θ
θY = 804.89 × 10−6
QY = (317.06 × 106 )(804.89 × 10−6 ) = 255.2 × 103 N
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PROBLEM 2.113 (Continued)
(a)
Since Q = 260 × 103 > QY , link AD yields.
σ AD = 250 MPa 
PAD = Aσ Y = (225 × 10−6 )(250 × 10−6 ) = 56.25 × 103 N
From Statics, PBE = Q − 4.125PAD = 260 × 103 − (4.125)(56.25 × 103 )
PBE = 27.97 × 103 N
σ BE =
(b)
δB =
PBE 27.97 × 103
=
= 124.3 × 106 Pa
−6
A
225 × 10
PBE LBE
(27.97 × 103 )(1.0)
=
= 621.53 × 10−6 m
EA
(200 × 109 )(225 × 10−6 )
σ BE = 124.3 MPa 
δ B = 0.622 mm ↓ 
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PROBLEM 2.114
Solve Prob. 2.113, knowing that a = 1.76 m and that the magnitude of
the force Q applied at B is gradually increased from zero to 135 kN.
PROBLEM 2.113 The rigid bar ABC is supported by two links, AD
and BE, of uniform 37.5 × 6-mm rectangular cross section and made
of a mild steel that is assumed to be elastoplastic with E = 200 GPa
and σ Y = 250 MPa. The magnitude of the force Q applied at B is
gradually increased from zero to 260 kN. Knowing that a = 0.640 m,
determine (a) the value of the normal stress in each link, (b) the
maximum deflection of point B.
SOLUTION
ΣM C = 0 : 1.76(Q − PBE ) − 2.64 PAD = 0
Statics:
δ A = 2.64θ , δ B = 1.76θ
Deformation:
Elastic Analysis:
A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2
PAD =
σ AD
PBE
σ BE
EA
(200 × 109 )(225 × 10−6 )
δA =
δ A = 26.47 × 106 δ A
LAD
1.7
= (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ
P
= AD = 310.6 × 109 θ
A
(200 × 109 )(225 × 10−6 )
EA
=
δB =
δ B = 45 × 106 δ B
1.0
LBE
= (45 × 106 )(1.76θ ) = 79.2 × 106 θ
P
= BE = 352 × 109 θ
A
From Statics, Q = PBE +
2.64
PAD = PBE + 1.500 PAD
1.76
= [73.8 × 106 + (1.500)(69.88 × 106 ] θ = 178.62 × 106 θ
θY at yielding of link BE:
σ BE = σ Y = 250 × 106 = 352 × 109 θY
θY = 710.23 × 10−6
QY = (178.62 × 106 )(710.23 × 10−6 ) = 126.86 × 103 N
Since Q = 135 × 103 N > QY , link BE yields.
σ BE = σ Y = 250 MPa 
PBE = Aσ Y = (225 × 10−6 )(250 × 106 ) = 56.25 × 103 N

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PROBLEM 2.114 (Continued)
From Statics, PAD =
1
(Q − PBE ) = 52.5 × 103 N
1.500
(a)
From elastic analysis of AD,
(b)
σ AD =
PAD 52.5 × 103
=
= 233.3 × 106
−6
A
225 × 10
θ=
PAD
= 751.29 × 10−3 rad
69.88 × 106
δ B = 1.76θ = 1.322 × 10−3 m
σ AD = 233 MPa 
δ B = 1.322 mm ↓ 
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PROBLEM 2.115∗
Solve Prob. 2.113, assuming that the magnitude of the force Q
applied at B is gradually increased from zero to 260 kN and then
decreased back to zero. Knowing that a = 0.640 m, determine (a)
the residual stress in each link, (b) the final deflection of point B.
Assume that the links are braced so that they can carry compressive
forces without buckling.
PROBLEM 2.113 The rigid bar ABC is supported by two links,
AD and BE, of uniform 37.5 × 6-mm rectangular cross section and
made of a mild steel that is assumed to be elastoplastic with
E = 200 GPa and σ Y = 250 MPa. The magnitude of the force Q
applied at B is gradually increased from zero to 260 kN. Knowing
that a = 0.640 m, determine (a) the value of the normal stress in
each link, (b) the maximum deflection of point B.
SOLUTION
See solution to Problem 2.113 for the normal stresses in each link and the deflection of Point B after loading.
σ AD = 250 × 106 Pa
σ BE = 124.3 × 106 Pa
δ B = 621.53 × 10−6 m
The elastic analysis given in the solution to Problem 2.113 applies to the unloading.
Q′ = 317.06 × 106 θ ′
Q′ =
Q
260 × 103
=
= 820.03 × 10−6
317.06 × 106 317.06 × 106
σ ′AD = 310.6 × 109 θ = (310.6 × 109 )(820.03 × 10−6 ) = 254.70 × 106 Pa
′ = 128 × 109 θ = (128 × 109 )(820.03 × 10−6 ) = 104.96 × 106 Pa
σ BE
δ B′ = 0.640θ ′ = 524.82 × 10−6 m
(a)
(b)
Residual stresses.
′ = 250 × 106 − 254.70 × 10−6 = −4.70 × 106 Pa
σ AD , res = σ AD − σ AD
= −4.70 MPa 
′ = 124.3 × 106 − 104.96 × 106 = 19.34 × 106 Pa
σ BE , res = σ BE − σ BE
= 19.34 MPa 
δ B , P = δ B − δ B′ = 621.53 × 10−6 − 524.82 × 10−6 = 96.71 × 10−6 m
= 0.0967 mm ↓ 
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PROBLEM 2.116
A uniform steel rod of cross-sectional area A is attached to rigid supports
and is unstressed at a temperature of 45 °F. The steel is assumed to be
elastoplastic with σ Y = 36 ksi and E = 29 × 106 psi. Knowing that
α = 6.5 × 10−6 /°F, determine the stress in the bar (a) when the temperature
is raised to 320 °F, (b) after the temperature has returned to 45 °F .
SOLUTION
Let P be the compressive force in the rod.
Determine temperature change to cause yielding.
σ L
PL
+ L α (ΔT ) = − Y + Lα ( ΔT )Y = 0
AE
E
3
σ
36 × 10
(ΔT )Y = Y =
= 190.98 °F
Eα (29 × 106 )(6.5 × 10−6 )
δ =−
But ΔT = 320 − 45 = 275 °F > (ΔTY )
(a)
σ = −σ Y = −36 ksi 
Yielding occurs.
Cooling:
(ΔT)′ = 275 °F
δ ′ = δ P′ = δ T′ = −
P′L
+ Lα (ΔT )′ = 0
AE
P′
= − Eα (ΔT )′
A
= −(29 × 106 )(6.5 × 10−6 )(275) = −51.8375 × 103 psi
σ′=
(b) Residual stress:
σ res = −σ Y − σ ′ = −36 × 103 + 51.8375 × 103 = 15.84 × 10 psi
15.84 ksi 
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PROBLEM 2.117
The steel rod ABC is attached to rigid supports and is unstressed at a
temperature of 25 °C. The steel is assumed elastoplastic, with E = 200
GPa and σ y = 250 MPa. The temperature of both portions of the rod is
then raised to 150 °C . Knowing that α = 11.7 × 10−6 / °C, determine
(a) the stress in both portions of the rod, (b) the deflection of point C.
SOLUTION
AAC = 500 × 10−6 m 2
LAC = 0.150 m
ACB = 300 × 10−6 m 2
LCB = 0.250 m
δ P + δT = 0
Constraint:
Determine ΔT to cause yielding in portion CB.
−
PLAC PLCB
−
= LABα ( ΔT )
EAAC EACB
ΔT =
P  LAC LCB 
+


LAB Eα  AAC ACB 
At yielding, P = PY = ACBσ Y = (300 × 10−6 )(2.50 × 106 ) = 75 × 103 N
(ΔT )Y =
=
PY
LAB Eα
 LAC LCB 
+


 AAC ACB 
75 × 103
0.250
 0.150
+
9
−6 
−6
(0.400)(200 × 10 )(11.7 × 10 )  500 × 10
300 × 10−6
Actual ΔT:
150 °C − 25 °C = 125 °C > (ΔT )Y
Yielding occurs. For ΔT > (ΔT )Y ,
(a)
(b)

 = 90.812 °C

P = PY = 75 × 103 N
σ AC = −
PY
75 × 103
=−
= −150 × 10−6 Pa
AAC
500 × 10−6
σ AC = −150 MPa 
σ CB = −
PY
= −σ Y
ACB
σ CB = −250 MPa 
For ΔT > (ΔT )Y , portion AC remains elastic.
δ C /A = −
=−
PY LAC
+ LACα (ΔT )
EAAC
(75 × 103 )(0.150)
+ (0.150)(11.7 × 10−6 )(125) = 106.9 × 10−6 m
(200 × 109 )(500 × 10−6 )
Since Point A is stationary, δ C = δ C /A = 106.9 × 10−6 m
δ C = 0.1069 mm → 
\
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PROBLEM 2.118∗
Solve Prob. 2.117, assuming that the temperature of the rod is raised
to 150°C and then returned to 25 °C.
PROBLEM 2.117 The steel rod ABC is attached to rigid supports
and is unstressed at a temperature of 25 °C. The steel is assumed
elastoplastic, with E = 200 GPa and σ Y = 250 MPa. The
temperature of both portions of the rod is then raised to 150 °C.
Knowing that α = 11.7 × 10−6 / °C, determine (a) the stress in both
portions of the rod, (b) the deflection of point C.
SOLUTION
AAC = 500 × 10−6 m 2
Constraint:
LAC = 0.150 m
ACB = 300 × 10−6 m 2
LCB = 0.250 m
δ P + δT = 0
Determine ΔT to cause yielding in portion CB.
−
PLAC PLCB
−
= LABα ( ΔT )
EAAC EACB
ΔT =
P
LAB Eα
 LAC LCB 
+


 AAC ACB 
At yielding, P = PY = ACBσ Y = (300 × 10−6 )(250 × 106 ) = 75 × 103 N
PY  LAC LCB 
75 × 103
0.250 
 0.150
+
+

=

9
−6 
−6
LAB Eα  AAC ACB  (0.400)(200 × 10 )(11.7 × 10 )  500 × 10
300 × 10−6 
= 90.812 °C
(ΔT )Y =
Actual
ΔT : 150 °C − 25 °C = 125 °C > (ΔT )Y
Yielding occurs. For
ΔT > (ΔT )Y
Cooling: (ΔT )′ = 125 °C P′ =
P = PY = 75 × 103 N
ELABα (ΔT )′
(
LAC
AAC
+
LCB
ACB
)
=
(200 × 109 )(0.400)(11.7 × 10−6 )(125)
= 103.235 × 103 N
0.150
0.250
+
−6
−6
500 × 10
300 × 10
Residual force: Pres = P′ − PY = 103.235 × 103 − 75 × 103 = 28.235 × 103 N (tension)
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PROBLEM 2.118∗ (Continued)
(a)
(b)
Residual stresses.
σ AC =
Pres 28.235 × 103
=
AAC
500 × 10−6
σ AC = 56.5 MPa 
σ CB =
Pres 28.235 × 103
=
ACB
300 × 10−6
σ CB = 9.41 MPa 
Permanent deflection of point C. δ C =
Pres LAC
EAAC
δ C = 0.0424 mm → 
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PROBLEM 2.119∗
For the composite bar of Prob. 2.111, determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero to
98 kips and then decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are
bonded to a 12 -in. mild-steel bar. This composite bar is subjected as
shown to a centric axial load of magnitude P. Both steels are
elastoplastic with E = 29 × 106 psi and with yield strengths equal to
100 ksi and 50 ksi, respectively, for the tempered and mild steel. The
load P is gradually increased from zero until the deformation of the bar
reaches a maximum value δ m = 0.04 in. and then decreased back to
zero. Determine (a) the maximum value of P, (b) the maximum stress
in the tempered-steel bars, (c) the permanent set after the load is
removed.
SOLUTION
Areas:
Mild steel:
1
A1 =   (2) = 1.00 in 2
2
 3
Tempered steel: A2 = (2)   (2) = 0.75 in 2
 16 
Total: A = A1+ A2 = 1.75 in 2
Total force to yield the mild steel: σ Y 1 =
PY
A
∴ PY = Aσ Y 1 = (1.75)(50 × 103 ) = 87.50 × 103 lb
P > PY; therefore, mild steel yields.
P1 = force carried by mild steel.
Let
P2 = force carried by tempered steel.
P1 = A1σ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb
P1 + P2 = P, P2 = P − P1 = 98 × 103 − 50 × 103 = 48 × 103 lb
σ2 =
Unloading: σ ′ =
P2 48 × 103
=
= 64 × 103 psi
A2
0.75
P 98 × 103
=
= 56 × 103 psi
A
1.75
Residual stresses.
Mild steel:
σ1,res = σ 1 − σ ′ = 50 × 103 − 56 × 103 = −6 × 10−3 psi = −6 ksi
Tempered steel:
σ 2,res = σ 2 − σ 1 = 64 × 103 − 56 × 103 = 8 × 103 psi
8.00 ksi 
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PROBLEM 2.120∗
For the composite bar in Prob. 2.111, determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero until the
deformation of the bar reaches a maximum value δ m = 0.04 in. and is
then decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded
to a 12 -in. mild-steel bar. This composite bar is subjected as shown to a
centric axial load of magnitude P. Both steels are elastoplastic
with E = 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi,
respectively, for the tempered and mild steel. The load P is gradually
increased from zero until the deformation of the bar reaches a maximum
value δ m = 0.04 in. and then decreased back to zero. Determine (a) the
maximum value of P, (b) the maximum stress in the tempered-steel bars,
(c) the permanent set after the load is removed.
SOLUTION
Lδ
(14)(50 × 103 )
1
= 0.024138 in
A1 =   (2) = 1.00 in 2 δ Y 1 = Y 1 =
E
29 × 106
2
For the mild steel,
Lδ
(14)(100 × 103 )
 3
For the tempered steel, A2 = 2   (2) = 0.75 in 2 δ Y 2 = Y 2 =
= 0.048276 in.
E
29 × 106
 16 
A = A1 + A2 = 1.75 in 2
Total area:
δY 1 < δ m < δ Y 2
The mild steel yields. Tempered steel is elastic.
P1 = A1δY 1 = (1.00)(50 × 103 ) = 50 × 103 lb
Forces:
P2 =
Stresses:
Unloading:
σ1 =
EA2δ m (29 × 106 )(0.75)(0.04)
=
= 62.14 × 103 lb
14
L
P1
P
62.14 × 103
= δY 1 = 50 × 103 psi σ 2 = 2 =
= 82.86 × 103 psi
0.75
A1
A2
σ′ =
P 112.14
=
= 64.08 × 103 psi
A
1.75
Residual stresses. σ1,res = σ 1 − σ ′ = 50 × 103 − 64.08 × 103 = −14.08 × 103 psi = −14.08 ksi
σ 2,res = σ 2 − σ ′ = 82.86 × 103 − 64.08 × 103 = 18.78 × 103 psi = 18.78 ksi

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PROBLEM 2.121∗
Narrow bars of aluminum are bonded to the two sides of a thick steel plate as
shown. Initially, at T1 = 70°F, all stresses are zero. Knowing that the
temperature will be slowly raised to T2 and then reduced to T1, determine
(a) the highest temperature T2 that does not result in residual stresses, (b) the
temperature T2 that will result in a residual stress in the aluminum equal to
58 ksi. Assume α a = 12.8 × 10−6 / °F for the aluminum and α s = 6.5 × 10−6 / °F
for the steel. Further assume that the aluminum is elastoplastic, with
E = 10.9 × 106 psi and σ Y = 58 ksi. (Hint: Neglect the small stresses in the
plate.)
SOLUTION
Determine temperature change to cause yielding.
δ=
PL
+ Lα a ( ΔT )Y = Lα s ( ΔT )Y
EA
P
= σ = − E (α a − α s )(ΔT )Y = −σ Y
A
σY
58 × 103
(ΔT )Y =
=
= 844.62 °F
6
E (α a − α s ) (10.9 × 10 )(12.8 − 6.5)(10−6 )
(a)
T2Y = T1 + (ΔT )Y = 70 + 844.62 = 915 °F
915 °F 
After yielding,
δ=
σY L
E
+ Lα a (ΔT ) = Lα s (ΔT )
Cooling:
δ′ =
P′L
+ Lα a (ΔT )′ = Lα s (ΔT )′
AE
The residual stress is
σ res = σ Y −
Set
σ res = −σ Y
−σ Y = σ Y − E (α a − α s )(ΔT )
ΔT =
(b)
P′
= σ Y − E (α a − α s )(ΔT )
A
2σ Y
(2)(58 × 103 )
=
= 1689 °F
E (α a − α s ) (10.9 × 106 )(12.8 − 6.5)(10−6 )
T2 = T1 + ΔT = 70 + 1689 = 1759 °F
1759 °F 
If T2 > 1759 °F, the aluminum bar will most likely yield in compression.
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PROBLEM 2.122∗
Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel
that is assumed to be elastoplastic with E = 200 GPa and
σ Y = 250 MPa. Knowing that the force F increases from 0 to 520 kN
and then decreases to zero, determine (a) the permanent deflection of
point C, (b) the residual stress in the bar.
SOLUTION
A = 1200 mm 2 = 1200 × 10−6 m 2
Force to yield portion AC:
PAC = Aσ Y = (1200 × 10−6 )(250 × 106 )
= 300 × 103 N
For equilibrium, F + PCB − PAC = 0.
PCB = PAC − F = 300 × 103 − 520 × 103
= −220 × 103 N
δC = −
PCB LCB (220 × 103 )(0.440 − 0.120)
=
EA
(200 × 109 )(1200 × 10−6 )
= 0.293333 × 10−3 m
PCB
220 × 103
=
A 1200 × 10−6
= −183.333 × 106 Pa
σ CB =
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PROBLEM 2.122∗ (Continued)
Unloading:
δ C′ =
′ =
PAC
′ LAC
′ ) LCB
PAC
P′ L
( F − PAC
= − CB CB =
EA
EA
EA
L  FL
L
′  AC + BC  = CB
PAC
EA 
EA
 EA
FLCB
(520 × 103 )(0.440 − 0.120)
=
= 378.182 × 103 N
LAC + LCB
0.440
′ = PAC
′ − F = 378.182 × 103 − 520 × 103 = −141.818 × 103 N
PCB
′
PAC
378.182 × 103
=
= 315.152 × 106 Pa
−6
A
1200 × 10
P′
141.818 × 103
′ = BC = −
σ BC
= −118.182 × 106 Pa
−6
A
1200 × 10
(378.182)(0.120)
= 0.189091 × 10−3 m
δ C′ =
9
6
(200 × 10 )(1200 × 10 )
′ =
σ AC
(a)
δ C , p = δ C − δ C′ = 0.293333 × 10−3 − 0.189091 × 10−3 = 0.1042 × 10−3 m
= 0.1042 mm 
(b)
σ AC , res = σ Y − σ ′AC = 250 × 106 − 315.152 × 106 = −65.2 × 106 Pa
= −65.2 MPa 
′ = −183.333 × 106 + 118.182 × 106 = −65.2 × 106 Pa
σ CB , res = σ CB − σ CB
= −65.2 MPa 
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PROBLEM 2.123∗
Solve Prob. 2.122, assuming that a = 180 mm.
PROBLEM 2.122 Bar AB has a cross-sectional area of 1200 mm2 and
is made of a steel that is assumed to be elastoplastic with E = 200 GPa
and σ Y = 250 MPa. Knowing that the force F increases from 0 to
520 kN and then decreases to zero, determine (a) the permanent
deflection of point C, (b) the residual stress in the bar.
SOLUTION
A = 1200 mm 2 = 1200 × 10−6 m 2
Force to yield portion AC:
PAC = Aσ Y = (1200 × 10−6 )(250 × 106 )
= 300 × 103 N
For equilibrium, F + PCB − PAC = 0.
PCB = PAC − F = 300 × 103 − 520 × 103
= −220 × 103 N
δC = −
PCB LCB (220 × 103 )(0.440 − 0.180)
=
EA
(200 × 109 )(1200 × 10−6 )
= 0.238333 × 10−3 m
PCB
220 × 103
=−
A
1200 × 10−6
= −183.333 × 106 Pa
σ CB =
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PROBLEM 2.123∗ (Continued)
Unloading:
′ LAC
′ ) LCB
PAC
P′ L
( F − PAC
= − CB CB =
EA
EA
EA
L  FL
L
′  AC + BC  = CB
= PAC
EA 
EA
 EA
δ C′ =
′ =
PAC
FLCB
(520 × 103 )(0.440 − 0.180)
=
= 307.273 × 103 N
0.440
LAC + LCB
′ = PAC
′ − F = 307.273 × 103 − 520 × 103 = −212.727 × 103 N
PCB
δ C′ =
(307.273 × 103 )(0.180)
= 0.230455 × 10−3 m
(200 × 109 )(1200 × 10−6 )
′
PAC
307.273 × 103
=
= 256.061 × 106 Pa
−6
A
1200 × 10
′
PCB
−212.727 × 103
′ =
σ CB
=
= −177.273 × 106 P
A
1200 × 10−6
′ =
σ AC
(a)
δ C , p = δ C − δ C′ = 0.238333 × 10−3 − 0.230455 × 10−3 = 0.00788 × 10−3 m
(b)
σ AC ,res = σ AC − σ ′AC = 250 × 106 − 256.061 × 106 = −6.06 × 106 Pa
= −6.06 MPa 
′ = −183.333 × 106 + 177.273 × 106 = −6.06 × 106 Pa
σ CB ,res = σ CB − σ CB
= −6.06 MPa 
= 0.00788 mm 

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PROBLEM 2.124
Rod BD is made of steel ( E = 29 × 106 psi) and is used to brace the axially
compressed member ABC. The maximum force that can be developed in member
BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in
length of BD must not exceed 0.001 times the length of ABC, determine the
smallest-diameter rod that can be used for member BD.
SOLUTION
FBD = 0.02 P = (0.02) (130) = 2.6 kips = 2.6 × 103lb
Considering stress: σ = 18 ksi = 18 × 103 psi
σ=
FBD
A
∴ A=
FBD
σ
=
2.6
= 0.14444 in 2
18
Considering deformation: δ = (0.001)(144) = 0.144 in.
δ=
FBD LBD
AE
∴
A=
FBD LBD
(2.6 × 103 ) (54)
=
= 0.03362 in 2
6
Eδ
(29 × 10 ) (0.144)
Larger area governs. A = 0.14444 in 2
A=
π
4
d2 ∴
d=
4A
π
=
(4) (0.14444)
π
d = 0.429in. 
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PROBLEM 2.125
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel ( E = 200 GPa) and rod BC of brass ( E = 105 GPa). Determine
(a) the total deformation of the composite rod ABC, (b) the deflection of
point B.
SOLUTION
Rod AB:
FAB = − P = −30 × 103 N
LAB = 0.250 m
E AB = 200 × 109 GPa
δ AB
Rod BC:
π
(30)2 = 706.85 mm 2 = 706.85 × 10−6 m 2
4
F L
(30 × 103 )(0.250)
= AB AB = −
= −53.052 × 10−6 m
−6
9
E AB AAB
(200 × 10 )(706.85 × 10 )
AAB =
FBC = 30 + 40 = 70 kN = 70 × 103 N
LBC = 0.300 m
EBC = 105 × 109 Pa
δ BC
π
(50) 2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2
4
F L
(70 × 103 )(0.300)
= BC BC = −
= −101.859 × 10−6 m
9
−3
EBC ABC
(105 × 10 )(1.9635 × 10 )
ABC =
δ tot = δ AB + δ BC = −154.9 × 10−6 m
(a)
Total deformation:
(b)
Deflection of Point B. δ B = δ BC
= −0.1549 mm 
δ B = 0.1019 mm ↓ 
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PROBLEM 2.126
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel ( E = 29 × 106 psi), and rod BC of brass ( E = 15 × 106 psi).
Determine (a) the total deformation of the composite rod ABC, (b) the
deflection of point B.
SOLUTION
Portion AB:
PAB = 40 × 103 lb
LAB = 40 in.
d = 2 in.
E AB
δ AB =
Portion BC:
π
d2 =
π
(2) 2 = 3.1416 in 2
4
4
= 29 × 106 psi
AAB =
PAB LAB
(40 × 103 )(40)
=
= 17.5619 × 10−3 in.
E AB AAB (29 × 106 )(3.1416)
PBC = −20 × 103 lb
LBC = 30 in.
d = 3 in.
ABC =
π
d2 =
π
(3) 2 = 7.0686 in 2
4
4
EBC = 15 × 106 psi
δ BC =
PBC LBC
(−20 × 103 )(30)
=
= −5.6588 × 10−3 in.
6
EBC ABC (15 × 10 )(7.0686)
(a)
δ = δ AB + δ BC = 17.5619 × 10−6 − 5.6588 × 10−6
δ = 11.90 × 10−3 in. ↓ 
(b)
δ B = −δ BC
δ B = 5.66 × 10−3 in. ↑ 
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PROBLEM 2.127
The uniform wire ABC, of unstretched length 2l, is attached to the
supports shown and a vertical load P is applied at the midpoint B.
Denoting by A the cross-sectional area of the wire and by E the
modulus of elasticity, show that, for δ  l , the deflection at the
midpoint B is
δ =l3
P
AE
SOLUTION
Use approximation.
sin θ ≈ tan θ ≈
Statics:
l
ΣFY = 0 : 2 PAB sin θ − P = 0
PAB =
Elongation:
δ
P
Pl
≈
2sin θ 2δ
δ AB =
PAB l
Pl 2
=
AE 2 AEδ
Deflection:
From the right triangle,
(l + δ AB ) 2 = l 2 + δ 2
2
δ 2 = l 2 + 2l δ AB + δ AB
− l2
 1 δ AB
= 2l δ AB 1 +
2 l

≈
δ3 ≈

 ≈ 2l δ AB

Pl 3
AEδ
Pl 3
P
∴ δ ≈ l3
AE
AE

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PROBLEM 2.128
The brass strip AB has been attached to a fixed support
at A and rests on a rough support at B. Knowing that
the coefficient of friction is 0.60 between the strip and
the support at B, determine the decrease in temperature
for which slipping will impend.
SOLUTION
Brass strip:
E = 105 GPa
α = 20 × 10−6 / °C
ΣFy = 0 : N − W = 0
N =W
ΣFx = 0 : P − μ N = 0
δ =−
Data:
PL
+ Lα ( ΔT ) = 0
EA
P = μW = μ mg
ΔT =
μ mg
P
=
EAα EAα
μ = 0.60
A = (20)(3) = 60 mm 2 = 60 × 10−6 m 2
m = 100 kg
g = 9.81 m/s 2
E = 105 × 109 Pa
ΔT =
(0.60)(100)(9.81)
(105 × 109 )(60 × 10−6 )(20 × 106 )
ΔT = 4.67 °C 
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PROBLEM 2.129
Members AB and CD are 1 18 -in.-diameter steel rods, and members BC and AD
are 78 -in.-diameter steel rods. When the turnbuckle is tightened, the diagonal
member AC is put in tension. Knowing that E = 29 × 106 psi and h = 4 ft,
determine the largest allowable tension in AC so that the deformations in
members AB and CD do not exceed 0.04 in.
SOLUTION
δ AB = δ CD = 0.04 in.
h = 4ft = 48 in. = LCD
ACD =
δ CD
π
d2 =
π
4
4
FCD LCD
=
EACD
FCD =
(1.125) 2 = 0.99402 in 2
EACDδ CD (29 × 106 )(0.99402)(0.04)
=
= 24.022 × 103 lb
48
LCD
Use joint C as a free body.
ΣFy = 0 : FCD −
FAC =
4
5
FAC = 0 ∴ FAC = FCD
5
4
5
(24.022 × 103 ) = 30.0 × 103 lb
4
FAC = 30.0 kips 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 2.130
The 1.5-m concrete post is reinforced with six steel bars, each with a 28-mm
diameter. Knowing that Es = 200 GPa and Ec = 25 GPa, determine the
normal stresses in the steel and in the concrete when a 1550-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
Ec Ac
Pc =
Ec Acδ
L
δ=
Ps L
Es As
Ps =
Es Asδ
L
P = Pc + Ps = ( Ec Ac + Es As )
δ
L
δ
−P
ε= =
L Ec Ac + Es As
6π
(28) 2 = 3.6945 × 103 mm 2
4
4
= 3.6945 × 10−3 m 2
As = 6
Ac =
π
π
d s2 =
d c2 − As =
π
(450) 2 − 3.6945 × 103
4
4
= 155.349 × 103 mm 2
= 153.349 × 10−3 m 2
L = 1.5 m
ε=
1550 × 103
= 335.31 × 10−6
(25 × 109 )(155.349 × 10−3 ) + (200 × 109 )(3.6945 × 10−3 )
σ s = Es ε = (200 × 109 )(335.31 × 10−6 ) = 67.1 × 106 Pa
σ s = 67.1 MPa 
σ c = Ecε = (25 × 109 )( −335.31 × 10−6 ) = 8.38 × 106 Pa
σ c = 8.38 MPa 
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PROBLEM 2.131
The brass shell (α b = 11.6 × 10−6 /°F) is fully bonded to the steel core
(α s = 6.5 × 10−6 /°F). Determine the largest allowable increase in
temperature if the stress in the steel core is not to exceed 8 ksi.
SOLUTION
Let Ps = axial force developed in the steel core.
For equilibrium with zero total force, the compressive force in the brass shell is Ps .
Strains:
εs =
Ps
+ α s (ΔT )
Es As
εb = −
Matching:
Ps
+ α b (ΔT )
Eb Ab
ε s = εb
Ps
P
+ α s (ΔT ) = − s + α b (ΔT )
Es As
Eb Ab
 1
1 
+

 Ps = (α b − α s )(ΔT )
 Es As Eb Ab 
(1)
Ab = (1.5) (1.5) − (1.0) (1.0) = 1.25 in 2
As = (1.0) (1.0) = 1.0 in 2
αb − α s = 5.1 × 10−6 /°F
Ps = σ s As = (8 × 103 ) (1.0) = 8 × 103 lb
1
1
1
1
+
=
+
= 87.816 × 10−9 lb −1
Es As Eb Ab (29 × 106 ) (1.0) (15 × 106 ) (1.25)
From (1),
(87.816 × 10−9 ) (8 × 103 ) = (5.1 × 10−6 ) (ΔT )
ΔT = 137.8 °F 
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PROBLEM 2.132
A fabric used in air-inflated structures is subjected to a biaxial loading
that results in normal stresses σ x = 120 MPa and σ z = 160 MPa.
Knowing that the properties of the fabric can be approximated as
E = 87 GPa and v = 0.34, determine the change in length of (a) side AB,
(b) side BC, (c) diagonal AC.
SOLUTION
σ x = 120 × 106 Pa,
σ y = 0,
σ z = 160 × 106 Pa
1
1
(σ x − vσ y − vσ z ) =
[120 × 106 − (0.34)(160 × 106 )] = 754.02 × 10−6
E
87 × 109
1
1
ε z = ( −vσ x − vσ y + σ z ) =
[−(0.34)(120 × 106 ) + 160 × 106 ] = 1.3701 × 10−3
E
87 × 109
εx =
(a)
δ AB = ( AB)ε x = (100 mm)(754.02 × 10−6 )
= 0.0754 mm 
(b)
δ BC = ( BC )ε z = (75 mm)(1.3701 × 10−6 )
= 0.1028 mm 
Label sides of right triangle ABC as a, b, and c.
c2 = a 2 + b2
Obtain differentials by calculus.
2c dc = 2a da + 2b db
dc =
a
b
da + db
c
c
But a = 100 mm,
b = 75 mm,
c = (1002 + 752 ) = 125 mm
da = δ AB = 0.0754 mm
db = δ BC = 0.1370 mm
(c)
δ AC = dc =
100
75
(0.0754) +
(0.1028)
125
125
= 0.1220 mm 
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PROBLEM 2.133
An elastomeric bearing (G = 0.9 MPa) is used to support a bridge
girder as shown to provide flexibility during earthquakes. The beam
must not displace more than 10 mm when a 22-kN lateral load is
applied as shown. Knowing that the maximum allowable shearing
stress is 420 kPa, determine (a) the smallest allowable dimension b,
(b) the smallest required thickness a.
SOLUTION
Shearing force:
P = 22 × 103 N
Shearing stress:
τ = 420 × 103 Pa
P
P
∴ A=
τ
A
22 × 103
=
= 52.381 × 10−3 m 2
420 × 103
= 52.381 × 103 mm 2
A = (200 mm)(b)
τ=
(a)
b=
γ=
(b)
A
52.381 × 103
=
= 262 mm
200
200
τ
G
=
But γ =
b = 262 mm 
420 × 103
= 466.67 × 10−3
6
0.9 × 10
δ
a
∴ a=
δ
10 mm
=
= 21.4 mm
γ 466.67 × 10−3
a = 21.4 mm 
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PROBLEM 2.134
Knowing that P = 10 kips, determine the maximum stress when (a) r = 0.50 in.,
(b) r = 0.625 in.
SOLUTION
P = 10 × 103 lb D = 5.0 in. d = 2.50 in.
5.0
D
=
= 2.00
2.50
d
3
Amin = dt = (2.50)   = 1.875 in 2
4
(a)
r = 0.50 in.
From Fig. 2.60b,
σ max =
0.50
r
=
= 0.20
2.50
d
K = 1.94
KP
(1.94) (10 × 103 )
=
= 10.35 × 103 psi
Amin
1.875
10.35 ksi 
(b)
r = 0.625 in.
σ max =
r
0.625
=
= 0.25 K = 1.82
d
2.50
KP
(1.82) (10 × 103 )
=
= 9.71 × 103 psi
Amin
1.875
9.71 ksi 
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PROBLEM 2.135
The uniform rod BC has a cross-sectional area A and is made of a
mild steel that can be assumed to be elastoplastic with a modulus of
elasticity E and a yield strength σ y . Using the block-and-spring
system shown, it is desired to simulate the deflection of end C of the
rod as the axial force P is gradually applied and removed, that is, the
deflection of points C and C′ should be the same for all values of P.
Denoting by μ the coefficient of friction between the block and the
horizontal surface, derive an expression for (a) the required mass m
of the block, (b) the required constant k of the spring.
SOLUTION
Force-deflection diagram for Point C or rod BC.
P < PY = Aσ Y
For
PL
EA
= PY = Aσ Y
δC =
Pmax
P=
EA
δC
L
Force-deflection diagram for Point C′ of block-and-spring system.
ΣFy = 0 : N − mg = 0
N = mg
ΣFx = 0 : P − F f = 0
P = Ff
If block does not move, i.e., F f < μ N = μ mg or P < μ mg ,
δ c′ =
then
P
K
or
P = kδ c′
If P = μ mg, then slip at P = Fm = μ mg occurs.
If the force P is the removed, the spring returns to its initial length.
(a)
Equating PY and Fmax,
(b)
Equating slopes,
Aσ Y = μ mg
k=
EA
L
m=
Aσ Y
μg


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