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 control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN.
Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the
smallest diameter that can be selected for the rod, (b) the corresponding maximum length of the rod.
SOLUTION
(a)
σ=
A=
P
A
π
4
A=
d2 d =
P
σ
=
4A
π
4 × 103
= 22.222 × 10−6 m 2
180 × 106
=
(4)(22.222 × 10−6 )
π
= 5.32 × 10−3 m
d = 5.32 mm 
(b)
δ=
PL
AE
L=
−6
−3
AEδ (22.222 × 10 )(105 × 10 )(3 × 10 )
=
P
4 × 103
9
L = 1.750 m 
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PROBLEM 2.3
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E = 73 GPa and
an ultimate strength of 140 MPa. Knowing that the distance between the gage marks is 250.28 mm after a load
is applied, determine (a) the stress in the rod, (b) the factor of safety.
SOLUTION
E = 73 × 109 Pa
δ = 250.28 − 250.00 = 0.28 mm = 0.28 × 10−3 m
L = 250 mm = 250 × 10−3 m
(a)
σ = Eε =
(b)
F. S . =
Eδ (73 × 109 )(0.28 × 10−3 )
=
= 81.76 × 106 Pa
E
250 × 10−3
σU
140 × 106
=
= 1.712
σ
81.76 × 106
81.8 MPa 

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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 300 mm and diameter 12 mm is subjected to a 3-kN tensile load. Knowing that
E = 3.1 GPa, determine (a) the elongation of the rod, (b) the normal stress in the rod.
SOLUTION
A=
(a)
δ=
(b)
σ =
PL
AE
P
A
=
=
π
4
d2 =
(3000)(0.3)
−6
9
(113.1 × 10 )(3.1×10 )
3000
113.1 × 10
−6
= 26.525 Pa
π
4
(0.012) 2 = 113.1 × 10−6 m 2
= 0.002567 m
δ = 2.6 mm 
σ = 26.5 MPa 
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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)
Strain:
Stress:
(b)
ε=
δ
=
L
σ = Eε = (3.3 × 109 )(0.011) = 36.3 × 106 Pa
σ =
P
A
Area:
A=
P
Diameter:
d =
Stress:
1.1
= 0.011
100
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
A=
π
250 mm
d2 =
π
(12)2 = 113.097 mm 2 = 113.097 ×10−6 m 2
4
4
P
6000
σ= =
= 53.052 × 106 Pa
−6
A 113.097 × 10
E=
σ 53.052 × 106
=
= 73.683 × 109 Pa
ε
0.00072

E = 73.7 GPa 
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PROBLEM 2.8
A cast-iron tube is used to support a compressive load. Knowing that E = 69 GPa and that the maximum
allowable change in length is 0.025%, determine (a) the maximum normal stress in the tube, (b) the minimum
wall thickness for a load of 7.2 kN if the outside diameter of the tube is 50 mm.
SOLUTION
E = 69 GPa = 69 × 109 Pa
ε=
(a)
(b)
δ
L
=
0.00025L
= 0.00025
L
σ = Eε = (69 × 10 )(0.00025) = 17.25 × 106 Pa
9
σ = 17.25 MPa 
P
P
7.2 × 103
A= =
= 417.39 × 10−6 m 2 = 417.39 mm 2
6
A
σ 17.25 × 10
4A
(4)(417.39)
π 2
= 502 −
= 1968.56 mm 2
A=
do − di2 di2 = d o2 −
4
π
π
1
1
di = 44.368 mm t = (d o − di ) = (50 − 44.368)
2
2
σ=
(
)
t = 2.82 mm 
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PROBLEM 2.9
An aluminum control rod must stretch 2 mm when a 2-kN tensile load is applied to it. Knowing that
σ all = 154 MPa and E = 70 GPa, determine the smallest diameter and shortest length which may be selected for
the rod.
SOLUTION
P = 2 kN, δ = 2 mm
σ=
A=
P
≤ σ all
A
π
4
d2
A≥
d=
P
σ all
σ all = 154 MPa
=
4A
π
2000
= 12.987 mm 2
154
=
(4)(12.987)
π
Eδ
≤ σ all
L
Eδ (70 × 109 )(0.002)
=
= 0.909 m
L≥
σ all
154 × 106
d min = 4.07 mm 
σ = Eε =
Lmin = 0.91 m 
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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 =
P
A
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:
Deformation:
σ =
P
A
A=
P
σ
=
10 × 103
= 66.667 × 10−6 m 2 = 66.667 mm 2
6
150 × 10
δ =
PL
AE
A=
(10 × 103 ) (4)
PL
=
= 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
Rod BD is made of steel ( E = 200 GPa) 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 126 MPa 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)(520) = 10.4 kN
Considering stress: σ = 126 MPa
σ=
FBD
A
∴ A=
FBD
σ
=
10400
= 82.54 mm 2
126
Considering deformation: δ = (0.001)(3600) = 3.6 mm
δ=
FBD LBD
AE
∴
A=
FBD LBD (10.4 × 103 )(13.50)
=
= 19.5 mm 2
3
Eδ
(200 × 10 )(3.6)
Larger area governs. A = 19.5 mm 2
A=
π
4
d2 ∴
d=
4A
π
=
(4)(19.5)
π
d = 5 mm 
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B
PROBLEM 2.15
1.2 m
A
A 1.2-m section of aluminum pipe of cross-sectional area 1100 mm2 rests on a
fixed support at A. The 15-mm-diameter steel rod BC hangs from a rigid bar
that rests on the top of the pipe at B. Knowing that the modulus of elasticity is
200 GPa for steel and 72 GPa for aluminum, determine the deflection of point
C when a 60 kN force is applied at C.
0.9 mm
C
P
SOLUTION
Rod BC:
LBC = 2.1 m, EBC = 200 × 109 Pa
ABC =
δ C /B =
Pipe AB:
π
4
d2 =
π
4
(0.015)2 = 176.715 × 10−6 m 2
PLBC
(60 × 103 )(2.1)
=
= 3.565 × 10−3 m
EBC ABC (200 × 109 )(176.715 × 10−6 )
LAB = 1.2 m, E AB = 72 × 109 Pa, AAB = 1100 mm 2 = 1100 × 10−6 m 2
δ B /A =
PLAB
(60 × 103 )(1.2)
=
= 909.1 × 10−6 m 2
E AB AAB (72 × 109 )(1100 × 10−6 )
δ C = δ B /A + δ C /B = 909.1 × 10−6 + 3.565 × 10−3 = 4.47 × 10−3
= 4.47 mm

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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
9
−6
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
−6
9
Etube Atube (70 × 10 )(402.12 × 10 )
PL
P(0.250)
=−
=
= −4.8505 × 10−9 P
Erod Arod (105 × 106 )(490.87 × 10−6 )
δ tube =
δ rod
1



δ * =  turn  × 1.5 mm = 0.375 mm = 375 × 10−6 m
4
δ tube = δ * + δ rod
or δ tube − δ rod = δ *
8.8815 × 10−9 P + 4.8505 × 10−9 P = 375 × 10−6
P=
(a)
σ tube =
(b)
27.308 × 103
= 67.9 × 106 Pa
−6
402.12 × 10
σ tube = 67.9 MPa 
P
27.308 × 103
=−
= −55.6 × 106 Pa
−6
Arod
490.87 × 10
σ rod = −55.6 MPa 
P
Atube
σ rod = −
0.375 × 10−3
= 27.308 × 103 N
−9
(8.8815 + 4.8505)(10 )
=
δ tube = (8.8815 × 10−9 )(27.308 × 103 ) = 242.5 × 10−6 m
δ tube = 0.2425 mm 
δ rod = −(4.8505 × 10−9 )(27.308 × 103 ) = −132.5 × 10−6 m
δ rod = −0.1325 mm 
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PROBLEM 2.18
A single axial load of magnitude P = 58 kN is applied at end C of the
brass rod ABC. Knowing that E = 105 GPa, determine the diameter d of
portion BC for which the deflection of point C will be 3 mm.
SOLUTION
δC =
P  LAB
 A E = E  A
Pi Li
i
AB
+
LBC 

ABC 
LBc Eδ C LAB (105 × 109 )(3 × 10−3 )
=
=
=
−
ABC
P
AAB
58 × 103
ABC =
ABC =
LBC
3.7334 × 10
π
4
3
=
12
= 3.7334 × 103 m −1
2
(0.030)
4
π
0.8
= 214.28 × 10−6 m 2
3.7334 × 103
2
d BC
∴ d BC =
4 ABC
π
=
(4)(214.28 × 10−6 )
π
= 16.52 × 10−3 m
= 16.52 mm

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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
π
4
(0.020)2 = 314.16 × 10−6 m 2
(0.060)2 = 2.8274 × 10−3 m 2
Force in member AB is P tension.
Elongation:
δ AB =
PLAB
(4 × 103 )(0.4)
=
= 72.756 × 10−6 m
EAAB (70 × 109 )(314.16 × 10−6 )
Force in member BC is Q − P compression.
Shortening:
δ BC =
(Q − P) LBC
(Q − P)(0.5)
=
= 2.5263 × 10−9 (Q − P)
EABC
(70 × 109 )(2.8274 × 10−3 )
For zero deflection at A, δ BC = δ AB
2.5263 × 10−9 (Q − P) = 72.756 × 10−6 ∴ Q − P = 28.8 × 103 N
Q = 28.3 × 103 + 4 × 103 = 32.8 × 103 N
(b)
δ AB = δ BC = δ B = 72.756 × 10−6 m
Q = 32.8 kN 
δ AB = 0.0728 mm ↓ 
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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 = 200 GPa) with crosssectional areas of 516 mm2 and 412 mm2, respectively. For the loading
shown, determine the elongation of (a) member AB, (b) member BC.
SOLUTION
(a)
LAB = 1.82 + 1.52 = 2.34 m = 2340 mm
Use joint A as a free body.
1.5
FAB − 125 = 0
2.34
= 195 kN
ΣFy = 0
FAB
δ AB =
(b)
FAB LAB (125 × 103 )(2340)
=
= 2.83 mm
EAAB
(200000)(516)

Use joint B as a free body.
ΣFx = 0: FBC −
FBC =
1.8
FAB = 0
2.34
(1.8)(195)
= 150 kN
2.34
δ BC =
FBC LBC
(150 × 103 )(1800)
=
= 3.27 mm
EABC
(200 000)(412)

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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 =
EBD ABDδ BD
(200 × 109 )(1920 × 10−6 )(1.6 × 10−3 )
=
LBD
7.81
= 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 crosssectional 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
Members AB and CD are 30-mm-diameter steel rods, and members BC
and AD are 22-mm-diameter steel rods. When the turnbuckle is tightened,
the diagonal member AC is put in tension. Knowing that E = 200 GPa,
determine the largest allowable tension in AC so that the deformations in
members AB and CD do not exceed 1.0 mm.
SOLUTION
δ AB = δ CD = 1 mm
= LCD
π
d 2 = (0.03) 2 = 706.9 × 10−6 m 2
4
4
F L
= CD CD
EACD
ACD =
δ CD
π
h = 1.2 m
FCD =
E ACD δ CD (200 × 109 )(706.9 × 10−6 )(0.001)
=
LCD
1.2
= 117.8 kN
Use joint C as a free body.
ΣFy = 0 : FCD −
FAC =
1.2
1.5
FAC = O ∴ FAC =
FCD
1.5
1.2
1.5
(117.8) = 147.3 kN
1.2

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PROBLEM 2.25
Each of the links AB and CD is made of aluminum ( E = 75 GPa) and has
a cross-sectional area of 258 mm2. Knowing that they support the rigid
member BC, determine the deflection of point E.
SOLUTION
Use member BC as a free body
ΣM C = 0: − (0.64) FAB + (0.44)(5 × 103 ) = 0
FAB = 3.4375 × 103 N
ΣM B = 0: (0.64) FCD − (0.20)(5 × 103 ) = 0
FCD = 1.5625 × 103 N
For links AB and CD
A = 258 mm 2 = 258 × 10−6 m 2
δ AB =
FAB LAB
(3.4375 × 103 )(0.36)
=
= 63.953 × 10−6 m = δ B
EA
(75 × 109 )(258 × 10−6 )
δ CD =
FCD LCD
(1.5625 × 103 )(0.36)
=
= 29.07 × 10−6 m = δ C
9
−6
EA
(75 × 10 )(258 × 10 )
Slope θ =
δ B − δC
lBC
=
34.883 × 10−6
= 54.505 × 10−6 rad
0.64
δ E = δ C + lECθ
= 29.07 × 10−6 + (0.44)(54.505 × 10−6 )
= 53.05 × 10−6 m = 0.0531 mm

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PROBLEM 2.26
Members ABC and DEF are joined with steel links ( E = 200 GPa).
Each of the links is made of a pair of 25 × 35-mm plates. Determine
the change in length of (a) member BE, (b) member CF.
SOLUTION
Use member ABC as a free-body
ΣM B = 0
(0.260)(18 × 103 ) − (0.180) FCF = 0
FCF =
(0.260)(18 × 103 )
= 26 × 103 N
0.180
Σ MC = 0
FBE = −
(0.440)(18 × 103 ) + (0.180) FBE = 0
(0.440)(18 × 103 )
= −44 × 103 N
0.180
Area for link made of two plates
A = (2)(0.025)(0.035) = 1.75 × 10−3 m 2
(a)
δ BE =
FBE LBE
(−44 × 103 )(0.240)
=
= −30.2 × 10−6 m = −0.0302 mm
EA
(200 × 109 )(1.75 × 10−3 )

(b)
δ CF =
FCF LCF
(26 × 103 )(0.240)
=
= 17.83 × 10−6 m = 0.01783 mm
EA
(200 × 109 )(1.75 × 10−3 )

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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
δ B = δ BD =
FBD LBD
(1.55556 P)(0.225)
=
= 13.8889 × 10−9 P
EBD ABD (105 × 109 )(240 × 10−6 )
δ C = δ CE =
FCE LCE
(0.55556 P) (0.150)
=
= 3.8581 × 10−9 P
ECE ACE (72 × 109 )(300 × 10−6 )
Deformation Diagram:
From the deformation diagram,
Slope,
θ=
δ B + δC
=
LBC
δ A = δ B + LABθ
17.7470 × 10−9 P
= 78.876 × 10−9 P
0.225
= 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
Determine the deflection of the apex A of a homogenous circular cone of height h, density ρ, and modulus of
elasticity E, due to its own weight.
SOLUTION
Let b = radius of the base and
r = radius at section with coordinate y.
r=
b
y
h
Volume of portion above element
1
1 b2
V = π r 2 y = π 2 y3
3
3 h
πρ gb 2 y 3
P = ρ gV =
3h 2
δ=
=

h
PΔy
P dy
=
=
EA 0 EA
ρ g y2
3E 2

h
=
0
ρ gh 2
6E
A = π r2 =
h

0
π ρ gb 2 y 3
3h
2
π b2
h2
⋅
y2
h
h2
Eπ b y
2
2
dy =
ρ gy
 3E
dy
0

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PROBLEM 2.30
A homogenous 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 gA ( L − y )
dδ =
Pdy ρ gA( L − y )dy ρ g ( L − y )
=
=
dy
EA
EA
E
δ=
=
(b)
Total weight:

L
ρ g ( L − y)
0
E
ρg 
L
1 2
dy =
 Ly − y 
E 
2 0
ρg 
L2 
2
 L − 
2 
E 
δ=
1 ρ gL2

2 E
W = ρ gAL
F=
EAδ EA 1 ρ gL2 1
=
⋅
= ρ gAL
2
L
L 2 E
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.
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 )
Thus,
Aa =
with
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 × 10
Strain:
ε =
δ
L
=
δ
L
6 δ
L
P
200 × 103
=
= 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
The 1.35 m concrete post is reinforced with six steel bars, each with a 28 mm
diameter. Knowing that Es = 200 GPa and Es = 29 GPa, determine the normal
stresses in the steel and in the concrete when a 1560 kN axial centric force P
is applied to the post.
SOLUTION
Let Pc = portion of axial force carried by concrete
Ps = portion carried by the six steel rods
δ=
Pc L
E Aδ
, Pc = c c
Ec Ac
L
δ=
Ps L
E Aδ
, Ps = s s
Es As
L
P = Pc + Ps = ( Ec Ac + Es As )
ε=
δ
L
As = 6
Ac =
π
=
π
4
δ
L
P
Ec Ac + Es As
d s2 =
6π
(28) 2 = 3694.5 mm 2
4
dc2 − As =
π
(450) 2 − 3694.5 = 155348.6 mm 2
4
4
L = 1.35 m = 1350 mm
ε=
−1560 × 103
= −297.48 × 10−6
 29 × 109 
 200 × 109 

 (155348.6) 
 (3694.5)
6
6
 10

 10

 200 × 109 
−6
2
 (−297.48 × 10 ) = −59.5 N/mm = −59.5 MPa
6
10


σ s = Es ε = 

σ c = Ec ε =  29 ×

109
106

−6
2
 (−297.48 × 10 ) = −8.627 N/mm = −8.627 MPa



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PROBLEM 2.36
An axial centric force of magnitude P = 450 kN is applied to the
composite block shown by means of a right end plate. Knowing that
h = 10 mm, determine the normal stress in (a) the brass core, (b) the
aluminum plates.
SOLUTION
Let Pb = portion of axial force carried by brass core
Pa = portion carried by two aluminum plates
δ=
Pb L
E Aδ
, Pb = b b
Eb Ab
L
δ=
Pa L
E Aδ
, Pa = a a
Ea Aa
L
P = Pb + Pa = ( Eb Ab + Eb Aa )
δ
L
δ
P
ε= =
L Eb Ab + Ea Aa
Ab = (60)(40) = 2400 mm 2 = 2400 × 10−6 m 2
Aa = (2)(60)(10) = 2400 mm 2 = 2400 × 10−6 m 2
ε=
450 × 103
= 1.3393 × 10−3
(105 × 109 )(2400 × 10−6 ) + (70 × 109 )(1200 × 10−6 )
(a)
σ b = Ebε = (105 × 109 )(1.3393 × 10−3 ) = 140.6 × 106 Pa = 140.6 MPa

(b)
σ a = Ea ε = (70 × 109 )(1.393 × 10−3 ) = 93.75 × 106 Pa = 93.75 MPa

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PROBLEM 2.37
For the composite block shown in Prob. 2.36, determine (a) the value of h
if the portion of the load carried by the aluminum plates is half the portion
of the load carried by the brass core, (b) the total load if the stress in the
brass is 80 MPa.
SOLUTION
Let Pb = portion of axial force carried by brass core
Pa = portion carried by the two aluminum plates
(a)
Given
δ=
Pb L
E Aδ
, Pb = b b
Eb Ab
L
δ=
Pa L
E Aδ
, Pa = a a
Ea Aa
L
1
Pb
2
Ea Aaδ 1 Eb Abδ
=
L
2 L
1 Eb
Aa =
Ab
2 Ea
Pa =
Ab = (40)(60) = 2400 mm 2 = 2400 × 10−6 m 2
Aa =
h=
(b)
σb =
1 105 × 109
2400 = 1800 mm 2 = (2)(60)h
2 70 × 109
1800
= 15 mm
(2)(60)
Pb
Ab
Pb = Abσ b = (2400 × 10−6 )(80 × 106 ) = 192 × 103 N
1
Pb = 96 × 103 N
2
P = Pb + Pa = 288 × 103 N = 288 kN
Pa =
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PROBLEM 2.38
Compressive centric forces of 160 kN are applied at both ends of
the assembly shown by means of rigid plates. Knowing that
Es = 200 GPa and Ea = 70 GPa, 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 = 160 kN
Aa =
As =
π
4
π
4
(d o2 − di2 ) =
d2 =
π
4
π
4
(0.0632 − 0.025) 2 = 0.002528 m
(0.025) 2 = 0.000491 m 2
ε=
−160000
= −581.5 × 10−6
9
(70 × 10 )(0.002528) + (200 × 10 )(0.000491)
9
(a)
σ s = Es ε = (200 × 109 )(−581.5 × 10−6 ) = −116.3 MPa


σ a = Ea ε = (70 × 109 )(−581.5 × 10−6 ) = −40.7 MPa

(b)
−6
δ = Lε = (0.25)(−581.5 × 10 ) = −14.5 × 10
−6
m = −0.145 mm

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PROBLEM 2.39
Three wires are used to suspend the plate shown. Aluminum wires are used at
A and B with a diameter of 3 mm and a steel wire is used at C with a diameter
of 2 mm. Knowing that the allowable stress for aluminum ( E = 70 GPa) is
98 MPa and that the allowable stress for steel ( E = 200 GPa) is 126 MPa,
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:
ε A = εB =
δ
2L
, εC =
δ
L
= 2ε A
Determine allowable strain.
Wires A&B:
εA =
σA
EA
=
98
= 1.4 × 10−3
70 × 103
ε C = 2 ε A = 4.8 × 10−3
σC
126
= 0.63 × 10−3
EC 200 × 103
1
ε A = ε B = ε C = 0.315 × 10−3
2
σ C = 126 MPa
Allowable strain for wire C governs,
∴
Wire C:
εC =
=
σ A = E Aε A
PA = AA E Aε A
π
(0.003) 2 (70 × 109 )(0.315 × 10−3 )
4
PB = 155.87 N
=
σ C = EC ε C
PC = AC σ C =
π
4
(0.002)2 (126 × 106 ) = 395.84 N
For equilibrium of the plate,
P = PA + PB + PC = 707.6 N

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PROBLEM 2.40
Three steel rods ( E = 200 GPa) support a 36-kN load P. Each of the rods
AB and CD has a 200-mm2 cross-sectional area and rod EF has a 625-mm2
cross-sectional area. Neglecting the deformation of rod BED, determine
(a) the change in length of rod EF, (b) the stress in each rod.
SOLUTION
Use member BED as a free body.
By symmetry, or by ΣM E = 0
PCD = PAB
ΣFy = 0:
PAB + PCD + PEF − P = 0
P = 2 PAB + PEF
δ AB =
PAB LAB
,
E AAB
LAB = LCD
Since
δ CD =
PCD LCD
,
E ACD
AAB = ACD ,
and
Since points A, C, and E are fixed,
δ B = δ AB , δ D = δ CD , δ E = δ EF
Since member BED is rigid,
δ E = δ B , = δC
PAB LAB PEF LEF
=
E AAB
E AEF
∴ PAB =
δ EF =
PEF LDF
E AEF
δ AB = δ CD
AAB LEF
200 400
PEF =
PEF
⋅
⋅
625 500
AEF LAB
= 0.256 PEF
P = 2 PAB + PEF = (2)(0.256) PEF + PEF = 1.512 PEF
36 × 103
P
=
= 23.810 × 103 N
1.512
1.512
= PCD = (0.256)(23.810 × 103 ) = 6.095 × 103 N
PEF =
PAB
(a)
(b)
δ = δ EF =
(23.810 × 103 )(400 × 10−3 )
= 76.2 × 10−6 m
9
−6
(200 × 10 )(625 × 10 )
or δ = δ AB =
(6.095 × 103 )(500 × 10−3 )
= 76.2 × 10−6 m
9
−6
(200 × 10 )(200 × 10 )
σ AB = σ CD =
PAB 6.095 × 103
=
= 30.5 × 106 Pa = 30.5 MPa
AAB
200 × 10−6
σ EF = −
PEF
23.810 × 103
=−
= −38.1 × 106 Pa = 38.1 MPa
−6
AEF
625 × 10
= 0.0762 mm 


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PROBLEM 2.41
A steel tube ( Es = 200 GPa) with a 30-mm outer diameter and a 3-mm thickness is placed in a vise that is
adjusted so that its jaws just touch the ends of the tube without exerting any pressure on them. The two forces
shown are then applied to the tube. After these forces are applied, the vise is adjusted to decrease the distance
between its jaws by 0.2 mm. Determine (a) the forces exerted by the vise on the tube at A and D, (b) the
change in length of the portion BC of the tube.
SOLUTION
For the tube
d D = 30 mm
di = do − 2t = 0.03 − 2(0.003) = 0.024 m
A=
A to B:
π
(d
7
4
(0.032 − 0.0242 ) = 254.5 × 10−6 m 2
L = 0.075 m
RA (0.075)
PL
=
= 1.473 × 10−9 RA m
9
−6
EA (200 × 10 )(254.5 × 10 )
P = RA + 32 kN L = 0.075 m
δ BC =
C to D:
π
)
− di2 =
P = RA kN
δ AB =
B to C:
2
o
( RA + 32000)(0.075)
PL
=
= 1.473 × 10−9 RA + 47.15 × 10−6 m
9
−6
EA (200 × 10 )(254.5 × 10 )
P = RA + 8 kN,
δ CD =
L = 0.075 m
( RA + 8000)(0.075)
PL
=
= 1.473 × 10−9 RA + 11.79 × 10−6
9
−6
EA (200 × 10 )(254.5 × 10 )
A to D:
δ AD = δ AB + δ BC + δ CD = 4.419 × 10−9 RA + 58.94 × 10−6 m
Given jaw movement
δ AD = −0.0002 m
(a)
4.419 × 10−9 RA + 58.94 × 10−6 = −0.0002
RA = −58597 N
RA = 58.6 kN
RD = RA + 8000 = −50597 N
(b)
−9
δ BC = (1.473 × 10 )(−58597) + 47.15 × 10
RD = 50.6 kN
−6
−6
= 39.2 × 10 m

−3
= 39.2 × 10 mm
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PROBLEM 2.42
Solve Problem 2.41, assuming that after the forces have been applied, the vise is adjusted to increase the
distance between its jaws by 0.1 mm.
PROBLEM 2.41 A steel tube ( Es = 200 GPa) with a 30-mm outer diameter and a 3-mm thickness is placed in
a vise that is adjusted so that its jaws just touch the ends of the tube without exerting any pressure on them.
The two forces shown are then applied to the tube. After these forces are applied, the vise is adjusted to
decrease the distance between its jaws by 0.2 mm. Determine (a) the forces exerted by the vise on the tube at
A and D, (b) the change in length of the portion BC of the tube.
SOLUTION
For the tube
do = 30 mm
di = do − 2t = 0.03 − 2(0.003) = 0.024 m
A=
A to B:
π
(d
7
RA (0.075)
PL
=
= 1.473 × 10−9 RA m
EA (200 × 109 )(254.5 × 10−6 )
L = 0.075 m
( RA + 32000)(0.075)
PL
=
= 1.473 × 10−9 RA + 47.15 × 10−6 m
EA (200 × 109 )(254.5 × 10−6 )
P = RA + 8000 N
δ CD =
π
(0.032 − 0.0242 ) = 254.5 × 10−6 m 2
4
L = 0.075 m
P = RA + 32 kN
δ BC =
C to D:
)
− di2 =
P = RA
δ AB =
B to C:
2
o
L = 0.075 m
( RA + 8000)(0.075)
PL
=
= 1.473 × 10−9 RA + 11.79 × 10−6 m
EA (200 × 109 )(254.5 × 10−6 )
A to D:
δ AD = δ AB + δ BC + δ CD = 4.419 × 10−9 RA + 58.94 × 10−6
Given jaw movement
δ AD = −0.0001 m
(a)
4.419 × 10−9 RA + 58.94 × 10−6 = −0.0001
RA = −35967 N
RD = RA + 8000 = −27967 N
(b)
−9
δ BC = (1.473 × 10 )( −35967) + 47.15 × 10
−6
RA = 35.97 kN

RD = 27.97 kN

−6
= −5.83 × 10 m

−3
= −5.83 × 10 mm
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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 1.5 mm diameter
( E = 200 GPa) and a pin and bracket at D. Knowing that the wires
were initially taught, determine (a) the additional tension in each wire
when a 1.0 kN load P is applied at D, (b) the corresponding deflection
of point D.
SOLUTION
Let θ be the rotation of bar ABCD
δ B = 300θ
δ C = 600θ
Then
δB =
PBE =
PBE LBE
AE
EA δ BE
LBE

109
 200 × 6
10
=
= 424.1 × 103θ
P L
δ C = CF CF
EA

109
 200 × 6
10
EAδ CF 
PCF =
=
LCF
π
 (1.5)(300θ )
4
250
π
 (1.5)(600θ )
4
450
= 471.25 × 103θ
Using free body ABCD
ΣM A = 0
300 PBE + 600 PCF − 900 P = 0
(300)(421.1 × 103θ ) + (600)(471.25 × 103θ ) − (900)(1000) = 0
409.08 × 106 θ = (9 × 105 ) ∴ θ = 0.00220 rad
(a)
(b)
PBE = (424.1 × 103 )(0.00220) = 933 N
PC = (471.25 × 103 )(0.00220) = 1036.7 N

δ D = 900θ = (900)(0.00220) = 1.98 mm

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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 = 200 GPa) and are
12 mm wide and 6 mm thick. Determine (a) the force in each
link when a 2.4-kN 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 = 0.1θ m →
δ D = −δ DE = 0.05θ m →
Then
δ=
For links:
FL
EA
or
F=
EAδ
L
A = (0.012)(0.006) = 72 × 10−6 m 2 , LBC = 0.1 m, LDE = 0.125 m
FBC =
EAδ BC (200 × 109 )(72 × 10−6 )(0.1θ )
=
= 14.4 × 106 θ
0.1
LBC
FDE =
EAδ DE (200 × 109 )(72 × 10−6 )(−0.05θ )
=
= −5.76 × 106 θ
0.125
LDE
Use member ACDF as a free body.
ΣM F = 0 : 0.2 P − 0.1FBC + 0.05 FDE = 0
1
1
FBC − FDE
2
4
1
1
2400 = (14.4 × 106 )θ − (−5.76 × 106 )θ = 8.64 × 106 θ
2
4
θ = 0.27778 × 10−3 rad
P=
(a)
FBC = (14.4 × 106 )(0.27778 × 10−3 )
FDE = −(5.76 × 106 )(0.27778 × 10−3 )
(b)
FBC = 4 kN 
FDE = −1.6 kN 
Deflection at Point A.
δ A = 0.2θ = (0.2)(0.27778 × 10−3 )
δ A = 0.056 mm → 
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PROBLEM 2.47
A 1.2-m concrete post is reinforced by four steel bars, each of 18 mm
diameter. Knowing that Es = 200 GPa, α s = 11.7 × 10−6 /°C and Ec = 25 GPa
and α c = 9.9 × 10−6 /°C, determine the normal stresses induced in the steel
and in the concrete by a temperature rise of 27°C.
SOLUTION
As = 4
π
π 
d 2 = (4)   (0.018) 2 = 0.0010179 m 2
4
4
Ac = A − As = 0.22 − 0.0010179 = 0.039 m 2 .
Let Pc be tensile in the concrete. For equilibrium with zero total force, the compressive force in the
four steel rods is − Pc .
Strains:
εs =
Pc
P
+ α s ( ΔT ) = − c + α s (ΔT )
Es As
Es As
εc =
Pc
+ α c (ΔT )
Ec Ac
Matching: ε c = ε s
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
+

 Pc
9
9
 (2.5 × 10 )(0.039) (200 × 10 )(0.0010179) 
Pc = 8185 N
= (11.7 − 9.9)(10−6 )(27)
Ps = −8185 N
Stress in steel
σs =
Ps −8.185 × 103
=
= −8.04 MPa
0.0010179
As

Stress in concrete
σc =
Pc 8.185 × 103
=
= 0.21 MPa
0.039
Ac

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\
PROBLEM 2.48
The assembly shown consists of an aluminum shell ( Ea = 73 GPa,
α a = 23.2 × 10−6 /°C) fully bonded to a steel core ( Es = 200 GPa,
α s = 11.7 × 106 /°C) and is unstressed. Determine (a) the largest
allowable change in temperature if the stress in the aluminum shell
is not to exceed 41 MPa, (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 = −41 MPa = −41 N/mm 2
π 2
π
2
2
Let
Aa =
As =
(d
4
o
)
− di =
π
(30 − 182 ) = 452.4 mm 2
4
π
d 2 = (18)2 = 254.5 mm 2
4
4
P = −σ a Aa = σ s As where P is the tensile force in the steel core.
σs = −
α a Aa
As
=
(41)(452.4)
= 72.88 N/mm 2
254.5
= 72.88 MPa
ε=
(α a − α )( ΔT ) =
(11.5 × 10−6 )( ΔT ) =
(a)
σs
Es
σs
Es
+ σ s (ΔT ) =
−
σa
Ea
+ α a (ΔT )
σa
Ea
72.88
9
200 × 106
10
+
41
9
73 × 106
= 0.926 × 10−3
10
ΔT = 80.53°C
72.88
ε=
+ (11.7 × 10−6 )(80.53) = 1.3066 × 10−3
109
200 × 6
10
or ε =
−41
9
73 × 106
10
+ (23.2 × 10−6 )(80.53) = 1.3066 × 10−3
δ = Lε = (200)(1.3066 × 10−3 ) = 0.261 mm
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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
Ab =
π
(25) 2 = 490.87 mm 2
4
= 490.87 × 10−6 m 2
PL
(δ P )b =
Eb Ab
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PROBLEM 2.49 (Continued)
Aluminum shell:
Ea = 70 × 109 Pa
Aa =
π
(602 − 252 )
4
= 2.3366 × 103 mm 2
= 2.3366 × 10−3 m 2
δ = (δ P )b + (δ P ) a
L(α b − α a )(ΔT ) =
PL
PL
+
= KPL
Eb Ab Ea Aa
where
K=
=
1
1
+
Eb Ab Ea Aa
1
1
+
−6
9
(105 × 10 )(490.87 × 10 ) (70 × 10 )(2.3366 × 10−3 )
9
= 25.516 × 10−9 N −1
Then
(α b − α a )(ΔT )
K
(23.6 × 10−6 − 20.9 × 10−6 )(180)
=
25.516 × 10−9
= 19.047 × 103 N
P=
Stress in aluminum:
σa = −
P
19.047 × 103
=−
= −8.15 × 106 Pa
−3
Aa
2.3366 × 10
σ a = −8.15 MPa 
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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
Aluminum shell:
E = 70 GPa α = 23.6 × 10−6 / °C
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:
δ T = LABα s (ΔT ) + LBCα b (ΔT )
= (0.250)(11.7 × 10−6 )(50) + (0.300)(20.9 × 10 −6 )(50)
= 459.75 × 10−6 m
Shortening due to induced compressive force P:
δP =
=
PL
PL
+
Es AAB Eb ABC
0.250 P
0.300 P
+
−6
9
(200 × 10 )(706.86 × 10 ) (105 × 10 )(1.9635 × 10−3 )
9
= 3.2235 × 10−9 P
For zero net deflection, δ P = δT
3.2235 × 10−9 P = 459.75 × 10−6
P = 142.62 × 103 N
P = 142.6 kN 
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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)
δ T = 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 = δ T + δ P = 4.914 × 10−3 + 50 × 10−12 σ = 0
σ = −98.3 × 106 Pa
(b)
σ = −98.3 MPa
−3
δ = δ T + δ P = 4.914 × 10 + 50 × 10
3 × 10−3 − 4.914 × 10−3
50 × 10−12
= −38.3 × 106 Pa
−12
σ = 3 × 10
−3
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 brass ( Eb = 105 GPa, α b = 20.9 × 10−6 / °C) and
portion BC is made of aluminum ( Es = 72 GPa, α s = 23.9 × 10−6 /°C).
Knowing that the rod is initially unstressed, determine (a) the normal
stresses induced in portions AB and BC by a temperature rise of 42°C, (b)
the corresponding deflection of point B.
SOLUTION
AAB =
π
4
π
2
d AB
=
π
4
(60)2 = 2.8274 × 103 mm 2 = 2.8274 × 10−3 m 2
π
(40) 2 = 1.2566 × 103 mm 2 = 1.2566 × 10−3 m 2
4
δ T = LABα b (ΔT ) + LBCα a ( ΔT )
ABC =
Free thermal expansion.
4
2
d BC
=
= (1.1)(20.9 × 10−6 )(42) + (1.3)(23.9 × 10−6 )(42)
= 2.2705 × 10−3 m
Shortening due to induced compressive force
δP =
=
PLBC
PLAB
=
Eb AAB Ea ABC
1.1 P
1.3P
+
(105 × 109 )(2.8274 × 10−3 ) (72 × 109 )(1.2566 × 10−3 )
= 18.074 × 10−9 P
For zero net deflection, δ P = δT
18.074 × 10−9 P = 2.2705 × 10−3
P = 125.62 × 103 N
(a)
(b)
σ AB = −
P
125.62 × 103
=−
= −44.4 × 106 Pa = −44.4 MPa
AAB
2.8274 × 10−3

σ BC = −
P
125.62 × 103
=−
= −100.0 × 106 Pa = −100.0 MPa
−3
ABC
1.2566 × 10

δB = +
=
PLAB
− LABα b (ΔT )
Eb AAB
(125.62 × 103 )(1.1)
− (1.1)(20.9 × 10−6 )(42)
(105 × 109 )(2.3274 × 10−3 )
= −500 × 10−6 m = −0.500 mm
i.e. 0.500 mm 
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PROBLEM 2.54
Solve Problem 2.53, assuming that portion AB of the composite rod is
made of steel and portion BC is made of brass.
PROBLEM 2.53 A rod consisting of two cylindrical portions AB and BC
is restrained at both ends. Portion AB is made of brass ( Eb = 105 GPa,
α b = 20.9 × 10−6 / °C) and portion BC is made of aluminum ( Ea = 72 GPa,
α a = 23.9 × 10−6 /°C). Knowing that the rod is initially unstressed,
determine (a) the normal stresses induced in portions AB and BC by a
temperature rise of 42°C, (b) the corresponding deflection of point B.
SOLUTION
AAB =
ABC =
Free thermal expansion.
π
4
π
2
d AB
=
2
d BC
=
4
ΔT = 42°
π
4
π
4
(0.06) 2 = 2.8274 × 10−6 m 2
(0.04) 2 = 1.2566 × 10−6 m 2
(δT ) AB = LABα s (ΔT ) = (1.1)(11.7 × 10−6 )(42) = 0.541 × 10−3 m
(δ T ) BC = LBCα s (ΔT ) = (1.3)(23.9 × 10−6 )(42) = 1.305 × 10−3
δ T = (δ T ) AB + (δ T ) BC = 1.846 × 10−3 m
Total
Shortening due to induced compressive force P.
PLAB
1.1 P
=
= 1945.25 × 10−9 P
9
Es AAB (200 × 10 )(2.8274 × 10−6 )
P LBC
1.3P
=
=
= 14368.58 × 10−9 P
9
−6
Eb ABC (72 × 10 )(1.2566 × 10 )
(δ P ) AB =
(δ P ) BC
δ P = (δ P ) AB + (δ P ) BC = 16313.8 × 10−9 P
Total
For zero net deflection, δ P = δT
(a)
16313.8 × 10−9 P = 1.846 × 10−3
P = 113.16 N
σ AB = −
P
113.16
=−
= −40.02 MPa
AAB
2.8274 × 10−6
σ AB = −40 MPa 
σ BC = −
P
113.16
=−
= −90.05 MPa
ABC
1.2566 × 10−6
σ BC = −90.1 MPa 
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PROBLEM 2.54 (Continued)
(b)
(δ P ) AB = (1945.25 × 10−9 )(113.16) = 0.22 × 10−3 m
δ B = (δT ) AB + (δ P ) AB = 0.541 × 10−3 + 0.22 × 10−3
or
δ B = 0.32 mm 
(δ P ) BC = (14368.58 × 10−9 )(113.16) = 1.626 × 10−3 m
δ B = (δT ) BC + (δ P ) BC = 1.305 × 10−3 1.626 × 10−3
δ B = 0.32 mm
(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:
(δ T )b = α b (ΔT ) L = (20.9 × 10−6 )(25)(0.250) = 130.625 × 10−6 m
Steel rod:
(δ T ) s = α s (ΔT ) L = (11.7 × 10−6 )(25)(0.250) = 73.125 × 10−6 m
At the final temperature, the difference between the free length of the steel rod and the brass link is
δ = 120 × 10−6 + 73.125 × 10−6 − 130.625 × 10−6 = 62.5 × 10−6 m
Add equal but opposite forces P to elongate the brass link and contract the steel rod.
Brass link: E = 105 × 109 Pa
Ab = (2)(50)(37.5) = 3750 mm 2 = 3.750 × 10−3 m 2
(δ P ) =
Steel rod:
PL
P(0.250)
=
= 634.92 × 10−12 P
EA (105 × 109 )(3.750 × 10−3 )
π
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
4
PL
P(0.250)
(δ P ) s =
=
= 1.76838 × 10−9 P
Es As (200 × 109 )(706.86 × 10−6 )
E = 200 × 109 Pa As =
(δ P )b + (δ P ) s = δ : 2.4033 × 10−9 P = 62.5 × 10−6 P = 26.006 × 103 N
P
(26.006 × 103 )
=−
= −36.8 × 106 Pa
−6
As
706.86 × 10
(a)
Stress in steel rod:
σs = −
(b)
Final length of steel rod:
L f = L0 + (δ T ) s − (δ P ) s
σ s = −36.8 MPa 
L f = 0.250 + 120 × 10−6 + 73.125 × 10−6 − (1.76838 × 10−9 )(26.003 × 103 )
= 0.250147 m
L f = 250.147 mm 
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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.
Elongation of steel: As = (2)(5)(40) = 400 mm 2 = 400 × 10−6 m 2
(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
−6
As 400 × 10
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PROBLEM 2.56 (Continued)
Brass:
σ b* = −
P*
8.811 × 103
=−
= −14.68 × 106 Pa = −14.68 MPa
Ab
600 × 10−6
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
AE
(600 × 10−6 )(105 × 109 )
δ ′ = 31.5 × 106 δ ′
Pb = b b δ ′ =
L
2.00
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
Allowable stresses:
σ b* = 14.68 MPa
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 =
Smaller value governs ∴ ε = 39.85 × 10
σs
Es
σb
Eb
=
7.97 × 106
= 39.85 × 10−6
9
200 × 10
=
39.68 × 106
= 377.9 × 10−6
9
105 × 10
−6
Areas: As = (2)(5)(40) = 400 mm 2 = 400 × 10−6 m 2
Ab = (15)(40) = 600 mm 2 = 600 × 10−6 m 2
Forces Ps = Es Asε = (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 0.5-mm gap exists when the temperature is 24°C,
determine (a) the temperature at which the normal stress in the
aluminum bar will be equal to −75 MPa, (b) the corresponding exact
length of the aluminum bar.
SOLUTION
σ a = −75 MPa
P = −σ a Aa = −(75 × 106 )(1800 × 10−6 ) = −135 kN
Shortening due to P:
δP =
=
PLb
PLa
+
Eb Ab Ea Aa
(135000)(0.35)
(135000 × 0.45)
+
9
−6
(105 × 10 )(1500 × 10 ) (73 × 109 )(1800 × 10−6 )
= 762.3 × 10−6 m
Available elongation for thermal expansion:
δ T = (0.0005 + 762.3 × 10−6 ) = 1.2623 × 10−3 m
But δ T = Lbα b (ΔT ) + Laα a (ΔT )
= (0.35)(21.6 × 10−6 )(ΔT ) + (0.45)(23.2 × 10−6 )( ΔT ) = (18 × 10−6 )ΔT
Equating, (18 × 10−6 )ΔT = 1.2623 × 10−3
(a)
(b)
ΔT = 70.1°C
Thot = Tcold + ΔT = 24 + 70.1 = 94.1°C
δ a = Laα a (ΔT ) −

PLa
Ea Aa
= (0.45)(23.2 × 10−6 )(70.1) −
(135000)(0.45)
= 0.27 × 10−3 m
(73 × 109 )(1800 × 10−6 )
Lexact = 0.45 + 0.27 × 10−3 = 0.45027 m

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PROBLEM 2.59
Determine (a) the compressive force in the bars shown after a
temperature rise of 82 °C, (b) the corresponding change in length of
the bronze bar.
SOLUTION
Thermal expansion if free of constraint:
δ T = Lbα b (ΔT ) + Laα a (ΔT )
= (0.35)(21.6 × 10−6 )(82) + (10.45)(23.2 × 10−6 )(82)
= 1.476 × 10−3 m
Constrained expansion.
δ = 0.5 mm
Shortening due to induced compressive force P:
δ P = 1.476 × 10−3 − 0.0005 = 0.976 × 10−3 m
δP =
But
PLb
PLa  Lb
L 
+
=
+ a P
Eb Ab Ea Aa  Eb Ab Ea Aa 


0.35
0.45
P = 5.647 × 10−9
=
+
9
−6
9
−6 
(105
10
)(1500
10
)
(73
10
)(1800
10
)
×
×
×
×


(a)
5.647 × 10−9 P = 0.976 × 10−3
P = 172835 N
Equating,
172.8 kN 
(b)
δ b = Lbα b (ΔT ) −
PLb
Eb Ab
= (0.35)(21.6 × 10−3 )(82) −
(172835)(0.35)
= 0.2358 × 10−3 m = 0.236 mm
(105 × 109 )(1500 × 10−6 )

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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:
δ T = 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
δP =
PLa
PLs  La
L
+
=
+ s
Ea Aa Es As  Ea Aa Es As

P



0.300
0.250
=
+
P
9
−6
9
−6 
 (75 × 10 )(2000 × 10 ) (190 × 10 )(800 × 10 ) 
= 3.6447 × 10−9 P
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
(75 × 109 )(2000 × 10−6 )
δ a = 0.363 mm 
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PROBLEM 2.61
A 2.75 kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate ( E = 200 GPa, v = 0.30).
Determine the resulting change (a) in the 50-mm 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
A = (1.6)(12) = 19.2 mm 2
= 19.2 × 10−6 m 2
P = 2.75 × 103 N
P 2.75 × 103
=
= 143.229 × 106 Pa
A 19.2 × 10−6
σ
143.229 × 106
εx = x =
= 716.15 × 10−6
E
200 × 109
ε y = ε z = −vEx = −(0.30)(716.15 × 10−6 ) = −214.84 × 10−6
σx =
(a)
L = 0.050 m
δ x = Lε x = (0.050)(716.15 × 10−6 ) = 35.81 × 10−6 m
−6
0.0358 mm 
−6
(b)
W = 0.012 m
δ y = W ε y = (0.012)(−214.84 × 10 ) = −2.578 × 10 m
(c)
t = 0.0016 m
δ z = tε z = (0.0016)( −214.84 × 10−6 ) = −3437 × 10−9 m
(d)
A = w0 (1 + ε y )t0 (1 + ε z ) = w0t0 (1 + ε y + ε z + ε y ε z )
−0.00258 mm 
−0.0003437 mm 
A0 = w0 t0
ΔA = A − A0 = w0 t0 (ε y + ε z + negligibly form) = 2 w0 t0ε y
= (2)(0.012)(0.0016)(−214.84 × 10−6 ) = −8.25 × 10−9 m 2 = −0.00825 mm 2

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PROBLEM 2.62
In a standard tensile test, a steel an aluminum rod of 20-mm diameter is
subjected to a tension force of P = 30 kN . Knowing that v = 0.35 and
E = 70 GPa, determine (a) the elongation of the rod in a 150-mm gage
length, (b) the change in diameter of the rod.
SOLUTION
P = 30 × 106 N
A=
π
d2 =
π
(20) 2 = 314.159 mm 2 = 314.159 × 10−6 m
4
4
P
30 × 103
σy = =
= 95.493 × 106 Pa
A 314.159 × 10−6
σ y 95.493 × 106
=
= 1.36418 × 10−3
εy =
E
70 × 106
(a)
δ y = Lε y = (150 × 10−3 )(1.36418 × 10−3 ) = 205 × 10−6 m

ε x = −vε y = −(0.35)(1.36418 × 10−3 ) = −477.46 × 10−6
(b)
δ x = d ε x = (20 × 10−3 )(−477.46 × 10−6 ) = −9.55 × 10−6 m 
0.205 mm 
−0.00955 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
= 19.0985 × 106 Pa
σy = =
A 314.16 × 10−6
δ y 14 mm
=
= 0.093333
εy =
L 150 mm
A=
Modulus of elasticity: E =
δx
0.85
= −0.0425
d
20
ε
−0.0425
v=− x =−
εy
0.093333
εx =
Poisson’s ratio:
σ y 19.0985 × 106
=
= 204.63 × 106 Pa
0.093333
εy
Modulus of rigidity: G =
P = 6 × 103 N
E = 205 MPa 
=−
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 = 200 GPa and v = 0.30,
determine the internal force in the bolt, if the diameter is observed to
decrease by 13 μm.
SOLUTION
δ y = −13 × 10−6 m
εy =
εy
d
v=−
=
εy
εx
d = 60 × 10−3 m
13 × 10−6
= −216.67 × 10−6
60 × 10−3
∴ εx = −
εy
v
=
216.67 × 10−6
= 722.22 × 10−6
0.30
σ x = Eε x = (200 × 109 )(722.22 × 10−6 ) = 144.44 × 106 Pa
A=
π
4
d2 =
π
4
(60) 2 = 2.827 × 103 m 2 = 2.827 × 10−3 m 2
F = σ x A = (144.44 × 106 )(2.827 × 10−3 ) = 408 × 103 N
= 408 kN

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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
do = 0.3 m
t = 0.015 m
L = 2.5 m
di = d o − 2t = 0.3 − 2(0.015) = 0.27 m
A=
(a)
δ =−
π
(d
4
δ
L
)
− di2 =
π
4
(0.32 − 0.27 2 ) = 13.4303 × 10−3 m 2
PL
(700 × 10 )(2.5)
=−
EA
(200 × 109 )(13.4303 × 10−3 )
3
= −651.51 × 10−6 m
ε=
2
o
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 = doε LAT = (300 mm)(78.180 × 10−6 )
Δdo = 0.0235 mm 
(c)

Δt = tε LAT = (15 mm)(78.180 × 10−6 )
Δt = 0.001173 mm 
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PROBLEM 2.66
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
E
200 × 103
εy =
1
40 − 0.3(80)
(σ y − vσ x ) =
= 80 × 10−6
3
E
200 × 10
(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 μm 
(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 μm 
(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 μm)
δ AB +
δ BC =
AC
AC
2
2
δ AC = 8.91 μm 
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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
=−
= −3.51 × 10−3
ε x = (σ x − vσ y − vσ Z ) =
9
E
E
45 × 10
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 xε 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 xε 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
An aluminum plate ( E = 74GPa 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
σx
125 × 106
= 1.6892 × 10−3
9
E
74 × 10
ε y = −vε x = −(0.33)(1.6892 × 10−3 ) = −0.5574 × 10−3
εx =
tan θ =
=
2(1 − 0.0005574)
= 1.99551
1 + 0.0016892
tan θ = 1.99551 
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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 42 MPa to the 300-mm portion BC of the rod.
Knowing that E = 70 GPa 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 = −42 MPa
σy =0
1
(σ x − vσ y − vσ z )
E
1
=
[−42 × 106 − (0.36)(0) − (0.36)(−42 × 106 )]
9
70 × 10
= −384 × 10−6
1
ε y = (−vσ x + σ y − vσ z )
E
1
=
[−(0.36)(−42 × 106 ) + 0 − (0.36)(−42 × 10−6 )]
70 × 109
= 432 × 10−6
εx =
Length subjected to strain ε x: L = 0.3 m
(a)
δ y = Lε y = (0.3)(432 × 10−6 ) = 0.1296 × 10−3 m = 0.13 mm

(b)
δ x = d ε x = (0.038)(−384 × 10−6 ) = −0.01459 × 10−3 m = −0.0146 mm

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PROBLEM 2.70
For the rod of Problem 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 42 MPa to the 300-mm portion
BC of the rod. Knowing that E = 70 GPa 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)(42 × 106 )
= −30.24 MPa
π
π
(0.0038) 2 = 1.134 × 10−3 m 2
4
4
F = Aσ y = (1.134 × 10−3 )(−30.24 × 106 ) = −34292 N
A=
d2 =
i.e.,
(b)
Over unpressurized portions AB and CD,
(ε y ) AB
34.3 kN 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
0.3
+
(2vp + σ y ) = 0
E
E
0.6vp
(0.6)(0.36)(42 × 106 )
σy = −
=−
= −90.72 MPa
0.1
0.1
P = Aσ y = (1.134 × 10−3 )( −90.72 × 106 ) = −102876 N
(0.5 − 0.3)
i.e.,
102.9 kN compression 
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PROBLEM 2.71
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
1
(σ x − vσ y )
E
1
ε y = (−vσ x + σ y )
E
εx =
(1)
(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 2.72
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
1
ε y = (−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

εx =


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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. 2.50, 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 ′)] = (1 + ε x ) + (1 − vε x )
2
2
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
ε′ =
4ε ′ = 2ε x − 2vε x
Neglect squares as small
(A)
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
1− v
=
εx
2
ε′ =
σ′
−

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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
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=
A
τ
22 × 103
=
= 52.381 × 10−3 m 2
420 × 103
= 52.381 × 103 mm 2
A = (200 mm)(b)
τ=
(a)
b=
τ
420 × 103
= 466.67 × 10−3
G 0.9 × 106
δ
δ
10 mm
∴ a= =
= 21.4 mm
But γ =
a
γ 466.67 × 10−3
γ=
(b)
A
52.381 × 103
=
= 262 mm
200
200
b = 262 mm 
=
a = 21.4 mm 
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PROBLEM 2.76
Two blocks of rubber, each of width w = 60 mm, are bonded to rigid
supports and to the movable plate AB. Knowing that a force of
magnitude P = 19 kN causes a deflection δ = 3 mm, determine the
modulus of rigidity of the rubber used.
SOLUTION
Consider upper block of rubber. The force carried is
1
2
1
P.
2
P
The shearing stress is
τ=
where
A = (180 mm)(60 mm) = 10.8 × 103 mm 2 = 10.8 × 10−3 m 2
A
P = 19 × 103 N
τ=
( 12 ) (19 × 103 ) = 0.87963 × 106 Pa
10.8 × 10−3
3 mm
δ
= 0.085714
γ= =
h 35 mm
τ 0.87963 × 106
= 10.26 × 106 Pa
G= =
0.085714
γ
= 10.26 MPa

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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 = 385 MPa, determine the deflection of the plate
when P = 36 kN.
SOLUTION
Consider the plastic block. The shearing force carried is P = 36 kN
The area is A = (0.084)(0.132) = 11.088 × 10−3 m 2
Shearing stress:
Shearing strain:
But
P
36 × 103
=
= 3.24675 MPa
A 11.088 × 10−3
τ 3.24675
γ= =
= 0.008433
385
G
τ=
γ=
δ
∴ δ = hγ = (53)(0.008433) = 0.45 mm
h

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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 = 1.54 MPa and G = 12.6 MPa. Knowing that a centric vertical
force of magnitude P = 13 kN must cause a 2.5-mm 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 12 P.
1
2
P
The shearing stress is
τ=
or required area
A=
But
A = (3.0)b
Hence,
b=
Use
b = 56.3 mm
Shearing strain.
γ=
But,
Hence,
A
P
13 × 103
=
= 4.221 × 103 m 2
2τ (2)(1.54 × 106 )
A
= 0.05628 m
0.075
and
bmin = 56.3 mm 
τ = 1.54 MPa
τ 1.54
=
= 0.12222
G 12.6
δ
a
δ
2.5
a= =
= 20.45 mm
γ 0.12222
γ=
amin = 20.5 mm 
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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 240-kN load P is applied. Knowing that for the plastic used
G = 1050 MPa, determine the deflection of the plate.

A = (80)(120) = 9.6 × 103 mm 2 = 9.6 × 10−3 m 2
SOLUTION
P = 240 × 103 N
P 240 × 103
=
= 25 × 106 Pa
A 9.6 × 103
G = 1050 × 106 Pa
τ=
τ
25 × 106
= 23.810 × 10−3
6
G 1050 × 10
h = 50 mm = 0.050 m
γ=

=
δ = hγ = (0.050)(23.310 × 10−3 )
= 1.190 × 10−3 m
1.091 mm 
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PROBLEM 2.80
What load P should be applied to the plate of Problem 2.79 to produce a 1.5-mm
deflection?
PROBLEM 2.79 The plastic block shown is bonded to a rigid support and to a
vertical plate to which a 240-kN load P is applied. Knowing that for the plastic
used G = 1050 MPa, determine the deflection of the plate.

SOLUTION
δ = 1.5 mm = 1.5 × 10−3 m
h = 50 mm = 50 × 10−3 m
1.5 × 10−3
= 30 × 10−3
h 50 × 10−3
G = 1050 × 106 Pa
γ=
δ
=
τ = Gγ = (1050 × 106 )(30 × 10−3 )
= 31.5 × 106 Pa

A = (120)(80) = 9.6 × 103 mm 2 = 9.6 × 10−3 m 2
P = τ A = (31.5 × 106 )(9.6 × 10−3 ) = 302 × 103 N
30.2 kN 
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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
A=
π
d2 =
π
4
4
3
P = 46 × 10 N
(22) 2 = 380.13 mm 2 = 380.13 × 10−6 m 2
P
= 121.01 × 106 Pa
A
σy =σz = 0
σx =
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: V = AL = (380.13 mm 2 )(200 mm) = 76.026 × 103 mm3
ΔV = Ve
(a)
Steel:
e=
(1 − 0.60)(121.01 × 106 )
= 242 × 10−6
200 × 109
ΔV = (76.026 × 103 )(242 × 10−6 ) = 18.40 mm3
(b)
Aluminum:
e=
(1 − 0.70)(121.01 × 106 )
= 519 × 10−6
70 × 109
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 50-mm gage length segment AB in Problem 2.61 (a) by computing the
dilatation of the material, (b) by subtracting the original volume of portion AB from its final volume.
PROBLEM 2.61 A 2.75 kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate
( E = 200 GPa, v = 0.30), Determine the resulting change (a) in the 50-mm 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
(a)
A0 = (12)(1.6) = 19.2 mm 2 = 19.2 × 10−6 m 2
Volume V0 = L0 A0 = (50)(19.2) = 960 mm3
σx =
P
2.75 × 103
=
= 143.229 × 106 Pa
A0 19.2 × 10−6
εx =
σ
1
143.229 × 106
(σ x − vσ y − vσ z ) = x =
= 716.15 × 10−6
9
E
E
200 × 10
σy =σz = 0
ε y = ε z = −v ε x = −(0.30)(716.15 × 10−3 ) = −214.84 × 10−6
e = ε x + ε y + ε z = 286.46 × 10−6
ΔV = V0 e = (960)(286.46 × 10−6 ) = 0.275 mm3
(b)

From the solution to Problem 2.62
δ x = 0.03581 mm
δ y = −0.002578 mm
δ z = −0.0003437 mm
The dimensions when under 2.75 kN load are:
Length
L = L0 + δ x = 50 + 0.03581 = 50.03581 mm
Width
w = w0 + δ y = 12 − 0.002578 = 11.997422 mm
Thickness
t = t0 + δ z = 1.6 − 0.0003437 = 1.5996563 mm
Volume
V = Lwt = (50.03581)(11.997422)(1.5996563) = 960.275 mm3
ΔV = V − V0 = 960.275 − 960 = 0.275mm3

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PROBLEM 2.85∗
A 150-mm-diameter solid steel sphere is lowered into the ocean to a point where the pressure is 50 MPa
(about 5 km below the surface). Knowing that E = 200 GPa 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,
εx =
Likewise,
π
π
(0.15)3 = 1.767 × 10−3 m3
6
σ x = σ y = σ z = − p = 50 MPa
V0 =
6
d03 =
1
(1 − 2v) p
(0.4)(50 × 106 )
(σ x − vσ y − vσ z ) = −
=−
= −100 × 10−6
E
E −6
29 × 109
ε y = ε z = −100 × 10
e = ε x + ε y + ε z = −300 × 10−6
(a)
−Δd = −d 0ε x = −(0.15)(100 × 10−6 ) = 15 × 10−6 m

(b)
−Δv = −v0 e = −(1.767 × 10−3 )(−300 × 10−6 ) = 530 × 10−9 m3

(c)
Let m = mass of sphere.
m = constant.
m = ρ0V0 = ρV = ρV0 (1 + e)
ρ − ρ0 ρ
V
m
1
=
−1 =
× 0 −1 =
−1
1+ e
ρ0
ρ0
V0 (1 + e) m
= (1 − e + e 2 − e3 + ) − 1 = −e + e 2 − e3 + 
≈ −e = 300 × 10−6
ρ − ρ0
× 100% = (300 × 10−6 )(100%) = 0.03%
ρ0

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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
A0=
π
d 02 =
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)
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
9
E
E
105 × 10
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
E
E
105 × 109
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 =
9
E
E
105 × 10
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
E
105 × 109
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=
The shearing strain is
γ=
P
P
=
A 2π rh
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,
R2
R2 = 25 mm = 0.025 m, h = 80 mm = 0.080 m
G = 12 × 106 Pa
P=
dr
R1 r
P
=
(ln R2 − ln R1 )
2π Gh
2π Ghδ
or P =
ln( R2 / R1 )

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
=
=
−
+
− 6E
3[2
2
4
]
18
G
E
G
G


 E

3 1 − 2 
− 1 
 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 <
Applying the bounds,
Taking the reciprocals,
or
or
1
2
E
= 2(1 + v)
G
for a positive bulk modulus.
E
 1
< 2 1 +  = 3
G
 2
1
G 1
>
>
2
E 3
2≤
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
The constraint conditions are
σx
v yxσ y
vzxσ z
Ex
Ey
Ez
vxyσ x σ y vzyσ z
εy = −
+
−
Ex
Ey
Ez
v yzσ y σ z
v σ
ε z = − xz x −
+
Ex
Ey
Ez
vxy v yx
=
Ex E y
v yz vzy
=
E y Ez
vzx vxz
=
Ez Ex
ε y = 0 and ε z = 0.
εx =
−
−
Using (2) and (3) with the constraint conditions gives
vzy
vxy
1
σy − σz =
σx
Ey
Ez
Ex
v yz
V
1
−
σ y + σ z = xz σ x
Ey
Ez
Ex
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
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
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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
σx =
65 × 103
P
=
= 40.625 × 106 Pa
−6
A 1600 × 10
εx =
(0.93142)(40.625 × 103 )
= 756.78 × 10−6
50 × 109
(a)
δ x = Lxε 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
50
Ex
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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, ε x =
(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 
σ y = 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−
9
50 × 10
15.2 × 109
= −323.73 × 10−6
=−
(b)
δ y = Ly ε y = (40 mm)(−323.73 × 10−6 )
δ y = −0.0129 mm 
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PROBLEM 2.93
Knowing that P = 40 kN, determine the maximum stress when (a) r = 12 mm,
(b) r = 15 mm.
SOLUTION
P = 40 kN
D = 125 mm
d = 62 mm
D 125
=
= 2.016
62
d
Amin = dt = (0.062)(0.018) = 1.116 × 10−3 m 2
(a)
r 12
=
= 0.1935
d 62
r = 12 mm
K = 1.94
From Fig. 2.64b,
σ max =
(b)
r = 15 mm
r 15
=
= 0.24
d 62
KP (1.94)(40 × 103 )
=
= 69.5 MPa
Amin
1.116 × 10−3

K = 1.86
σ max =
KP (1.86)(40 × 103 )
=
= 66.7 MPa
Amin
1.116 × 10−3

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PROBLEM 2.94
Knowing that, for the plate shown, the allowable stress is 110 MPa, determine
the maximum allowable value of P when (a) r = 10 mm, (b) r = 18 mm.
SOLUTION
σ max = 110 MPa
D
= 2.016
d
= dt = (0.062)(0.018) = 1.116 × 10−3 m 2
D = 125 mm d = 62 mm
Amin
(a)
r 10
=
= 0.161
d 62
r = 10 mm
K = 2.09
From Fig. 2.64b
σ max =
P=
(b)
r = 18 mm
r
18
=
= 0.29
d 0.62
KP
Amin
Aminσ max (1.116 × 10−3 )(110 × 106 )
=
= 58.7 kN
2.09
K

K = 1.76
P=
Aminσ max (1.116 × 10−3 )(110 × 106 )
=
= 69.75 kN
K
1.76

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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
96
D
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 =
From Fig. 2.60b,
rf
d
K fillet P
A σ
(5.40 × 10−6 )(100 × 106 )
∴ K fillet = min max =
Amin
P
28.787 × 103
= 1.876
≈ 0.19 ∴ rf ≈ 0.19d = 0.19(60)
rf = 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 mm2 = 1.125 × 10−3 m 2
δ =Σ
Pi Li
P L 92.308 × 103  0.150
0.300
0.150 
= Σ i =
+
+

9
−3
−6
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 
−6
9
AE (900 × 10 )(70 × 10 )
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 L
= Σ i
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 mm 2 = 900 × 10−6 m 2
Σ
Li
0.150
0.300
0.150
=
+
+
= 600 m −1
Ai 1.125 × 10−3 900 × 10−6 1.125 × 10−3
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
r
6
=
= 0.10
d 60
K = 1.95
σ max = K
P
(1.95)(87.5 × 103 )
=
= 189.6 × 106 Pa
Amin
900 × 10−6
σ max = 189.6 MPa 
Note that σ max < σ all . 
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PROBLEM 2.99
Two holes have been drilled through a long steel bar that is subjected
to a centric axial load as shown. For P = 30 kN, determine the
maximum value of the stress (a) at A, (b) at B.
SOLUTION
(a)
At hole A:
r=
1
⋅ 12 = 6 mm
2
d = 76 − 6 = 70 mm
Anet = dt = (70)(12) = 840 mm 2
σ nom =
P
30 × 103
=
= 35.7 N/mm 2
Anet
840
r
6
=
= 0.0857
d 70
From Fig. 2.60a,
K = 2.70
σ max = Kσ nom = (2.70)(35.7) = 96.4 N/mm2
(b)
r=
At hole B:
1
(38) = 19,
2

d = 76 − 38 = 38 mm
Anet = dt = (38)(12) = 456 mm 2 ,
σ nom =
P
30 × 103
=
= 65.8 N/mm 2
Anet
456
r 19
=
= 0.5
d 38
From Fig. 2.64a,
K = 2.10
σ max = Kσ nor = (2.10)(65.8) = 138.2 N/mm 2

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PROBLEM 2.100
Knowing that σ all = 100 MPa, determine the maximum allowable value
of the centric axial load P.
SOLUTION
At hole A:
r=
1
(12) = 6 mm
2
d = 76 − 6 = 70 mm
Anet = dt = (70)(12) = 840 mm 2
r
6
=
= 0.0857
d 70
From Fig. 2.60a,
K = 2.70
σ max =
At hole B:
r=
KP
Anet
∴ P=
Anetσ max (840 × 10−6 )(110 × 106 )
=
= 34.2 kN
K
2.70
1
(38) = 19 mm,
2
d = 76 − 38 = 38 mm
Anet = dt = (38)(12) = 456 mm 2
r 19
=
= 0.5
d 38
From Fig. 2.64a,
K = 2.10
P=
Smaller value for P controls
Anetσ max (4.56 × 10−6 )(10 × 106 )
=
= 23.9 kN
2.10
K
P = 23.9 kN 
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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
π
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
4
π
= (40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2
4
= Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N
AAB =
ABC
Pmax
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
π
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
4
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
AAB =
ABC
Pmax
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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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
E
200 × 109
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 = 200 GPa and σ γ = 345 MPa. After the rod has been attached to the rigid
lever CD, it is found that end C is 6 mm 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
AAB =
π
4
(9)2 = 53.617 mm 2 = 63.617 × 10−6 m 2
Since rod AB is to be stretched permanently,
( FAB )max = AABσ γ = (63.617 × 10−6 )(345 × 106 )
= 21.948 × 103 N
ΣM D = 0: 1.1Q − 0.7 FAB = 0
Qmax =
0.7
(21.948 × 103 ) = 13.967 × 103 N
1.1
13.97 kN 
( FAB ) max LAB
(21.948 × 103 )(1.25)
=
= 2.15625 × 10−3 m
EAAB
(200 × 109 )(63.617 × 10−6 )
δ′
θ ′ = AB = 3.0804 × 10−3 rad
0.7
′ =
δ AB
δ1 = 1.1θ ′ = 3.39 × 10−3 m
3.39 mm 
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PROBLEM 2.106
Solve Problem 2.105, assuming that the yield point of the mild steel is
250 MPa.
PROBLEM 2.105 Rod AB is made of a mild steel that is assumed to
be elastoplastic with E = 200 GPa and σ Y = 345 MPa. After the rod has
been attached to the rigid lever CD, it is found that rod C is 6 mm 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
AAB =
π
4
(9)2 = 63.617 mm 2 = 63.617 × 10−6 m 2
Since rod AB is to be stretched permanently,
( FAB )max = AABσ Y = (63.617 × 10−6 )(250 × 106 )
= 15.9043 × 103 N
ΣM D = 0: 1.1Q − 0.7 FAB = 0
Qmax =
0.7
(15.9043 × 103 ) = 10.12 × 103 N
1.1
10.12 kN 
( FAB ) max LAB
(15.9043 × 103 )(1.25)
=
= 1.5625 × 10−3 m
EAAB
(200 × 109 )(63.617 × 10−6 )
δ′
θ ′ = AB = 2.2321 × 10−3 rad
0.7
′ =
δ AB
δ1 = 1.1θ ′ = 2.46 × 10−3 m
2.46 mm 
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PROBLEM 2.107
Each of the three 6-mm-diameter steel cables is made of an
elastoplastic material for which σ Y = 345 MPa and E = 200 GPa. A
force P is applied to the rigid bar ABC until the bar has moved
downward a distance d = 2 mm. Knowing that the cables were initially
taut, determine (a) the maximum value of P, (b) the maximum stress
that occurs in cable AD, (c) the final displacement of the bar after the
load is removed (Hint: In part c, cable BE is not taut.)
SOLUTION
For each cable
A=
Strain at initial yielding
εY =
π
4
(0.006) 2 = 28.274 × 10−6 m 2
σY
345 × 106
= 1.725 × 10−3
E 200 × 109
δ
2 mm
= ε CF =
=
= 1.25 × 10−3
LAD 1600 mm
Strain in cables AD and CF:
ε AD
Strain in cable BE:
ε BE =
δ
LBE
=
=
2 mm
= 2.50 × 10−3
800 mm
Since
ε AD < ε Y , σ AD = Eε AD = (200 × 109 )(1.25 × 10−3 ) = 250 × 106 Pa
Since
ε BE > ε Y , σ BE = σ Y = 345 × 106 Pa
Forces:
PAD = CCF = Aσ AD = (28.274 × 10−6 )(250 × 106 ) = 7.0685 × 103 N)
PBE = Aσ BE = (28.274 × 10−6 )(345 × 106 ) = 9.7545 × 10−3 N
For equilibrium of bar ABC
(a)
PAD + PBE + PCF − P = 0
P = PAD + PBE + PCF = (7.0685 + 9.7545 + 7.0685) × 103 N
= 23.9 × 103 N = 23.9 kN
(b)

σ AD = 2.50 × 10 Pa = 250 MPa

6
After unloading P = 0
Cable BE is not taut PBE = 0
By symmetry PAD = PCF
For equilibrium PAD = PCF = 0
(c)
Final displacement δ is controlled by the final lengths of cables AD and CF. Since these cables were
never permanently deformed, the final displacement is
δ = δ AD = δ CF = 0
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PROBLEM 2.108
Solve Problem 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 of the three 6-mm-diameter steel cables is
made of an elastoplastic material for which σ γ = 345 MPa and
E = 200 GPa. A force P is applied to the rigid bar ABC until the bar has
moved downward a distance d = 2 mm. Knowing that the cables were
initially taut, determine (a) the maximum value of P, (b) the maximum
stress that occurs in cable AD, (c) the final displacement of the bar after
the load is removed (Hint: In part c, cable BE is not taut.)
SOLUTION
For each cable
A=
Strain at initial yielding
εY =
π
4
(0.006) 2 = 28.274 × 10−6 m 2
σY
345 × 106
= 1.725 × 10−3
E 200 × 109
δ
2 mm
= ε CF =
=
= 1.25 × 10−3
LAD 1600 mm
Strain in rods AD and CF:
ε AD
Strain in rods BE:
ε BE =
δ
LBE
=
=
2 mm
= 2.50 × 10−3
800 mm
Since ε AD < ε Y ,
σ AD = Eε AD = (200 × 109 )(1.25 × 10−3 ) = 250 × 106 Pa
Since ε BE > ε Y ,
σ BE = σ Y = 345 × 106 Pa
Forces:
PAD = CCF = Aσ AD = (28.274 × 10−6 )(250 × 106 ) = 7.0685 × 103 N
PBE = Aσ BE = (28.274 × 10−6 ) (345 × 106 ) = 9.7545 × 10−3 N
For equilibrium of bar ABC
PAD + PBE + PCF − P = 0
(a)
P = PAD + PBE + PCF = (7.0685 + 9.7545 + 7.0685) × 103 N = 23.9 kN
(b)
σ AD = 250 × 106 Pa = 250 MPa

Let δ ′ = change in displacement during unloading
′ =
PAD
EA
(200 × 109 )(28.274 × 10−6 )
δ′ =
δ ′ = 3.534 × 106 δ ′ = PCF
LAD
1600 × 10−3
′ =
PBE
(200 × 109 )(28.274 × 10−6 )
EA
δ′ =
δ ′ = 7.0685 × 106 δ ′
−3
LBE
800 × 10
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PROBLEM 2.108 (Continued)
′ + PBE
′ + PCF
′ = 14.137 × 106 δ ′
P′ = PAD
For equilibrium
But
P − P′ = 0
P′ = P = 23.89 × 103 N
δ′ =
23.89 × 103
= 1.690 × 10−3 m
6
14.137 × 10
Permanent displacement of bar
δ final = δ max − δ ′ = 2 × 10−3 − 1.690 × 10−3 = 0.310 × 10−3 m
= 0.310 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
E
200 × 109
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
LCB
0.190
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
(200 × 109 )(1750 × 10−6 )
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)
9
200 × 10
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 5 mm thick, are bonded to a 12 mm mildsteel bar. This composite bar is subjected as shown to a centric axial
load of magnitude P. Both steels are elastoplastic with E = 200 GPa and
with yield strengths equal to 690 MPa and 345 MPa, 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 = 1 mm
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
A1 = (12)(50) = 600 mm2
For the mild steel
δY 1 =
For the tempered steel
Lσ Y 1 (0.350)(345 × 106 )
=
= 0.06 × 10−3 m
9
E
200 × 10
A2 = 2(5)(50) = 500 mm 2
δY 2 =
Lσ Y 2 (0.350)(690 × 106 )
=
= 1.2 × 10−3 m
E
200 × 109
A = A1 + A2 = 1100 mm2
Total area:
δY 1 < δ m < δY 2 . The mild steel yields. Tempered steel is elastic.
(a)
Forces
P1 = A1σ Y 1 = (600 × 10−6 )(345 × 106 ) = 207 × 103 N = 207 kN
P2 =
EA2δ m (29 × 109 )(500 × 10−6 )(1.0 × 10−3 )
=
= 285.71 kN
0.350
L
P = P1 + P2 = 492.71 kN
(b)
Stresses
σ1 =
P1
= σ Y 1 = 345 MPa
A1
σ2 =
P2 285.71 × 103
=
= 571.4 MPa
500
A2
Unloading δ ′ =
(c)


PL (492.7 × 103 )(350)
=
= 0.783 mm
EA
(1100)(2 × 105 )
Permanent set δ p = δ m − δ ′ = 1.0 − 0.783
= 0.217 mm

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PROBLEM 2.112
For the composite bar of Problem 2.111, if P is gradually increased from
zero to 440 kN 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 5 mm thick, are bonded
to a 12 mm mild-steel bar. This composite bar is subjected as shown to a
centric axial load of magnitude P. Both steels are elastoplastic with
E = 200 GPa and with yield strengths equal to 690 MPa and 345 MPa,
respectively, for the tempered 200 GPa and mild steel.
SOLUTION
Areas:
Mild steel
A1= (12)(50) = 600 mm2
Tempered steel
A2 = 2(5)(50) = 500 mm 2
A = A1+ A2 = 1100 mm2
Total:
Total force to yield the mild steel:
P
σ Y 1 = Y ∴ PY = Aσ Y 1 = (1100 × 10−6 )(345 × 106 ) = 379.5 kN
A
P > PY , therefore, mild steel yields.
P1 = force carried by mild steel.
Let
P2 = force carried by tempered steel.
P1 = A1σ 1 = (600 × 10−6 )(345 × 106 ) = 207 kN
P1 + P2 = P, P2 = P − P1 = 440 − 207 = 233 kN
(a)
δm =
P2 L
(233 × 103 )(0.350)
=
= 0.816 × 10−3 m = 0.816 mm
EA2 (200 × 109 )(500 × 10−6 )
(b)
σ2 =
P2
233 × 103
=
= 466 × 106 Pa = 466 MPa
A2 500 × 10−6
Unloading δ ′ =
(c)
PL
(440 × 103 )(0.350)
=
= 0.7 × 10−3 m = 0.70 mm
EA (200 × 109 )(1100 × 10−6 )
δ p = δ m − δ ′ = 0.816 − 0.700 = 0.116 mm



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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
EA
(200 × 109 )(225 × 10−6 )
=
δB =
δ B = 45 × 106 δ B
LBE
1.0
= (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′ =
260 × 103
Q
=
= 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 7 °C. The steel is assumed to be elastoplastic
with σ Y = 250 MPa and E = 200 GPa. Knowing that α = 11.7 × 10−6 /°C,
determine the stress in the bar (a) when the temperature is raised to 160 °C, (b)
after the temperature has returned to 7 °C.
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
6
σ
250 × 10
( ΔT )Y = Y =
= 106.8 °C
Eα (200 × 109 )(11.7 × 10−6 )
δ =−
But ΔT = 160 − 7 = 153 °C > ( ΔTY )
(a)
σ = −σ Y = 250 MPa 
Yielding occurs.
Cooling:
( ΔT )′ = 153 °C.
δ ′ = δ P′ = δT′ = −
P′L
+ Lα ( ΔT )′ = 0
AE
P′
= − Eα ( ΔT )′
A
= −(200 × 109 )(11.7 × 10−6 )(153) = −358 MPa
σ′=
(b) Residual stress:
σ res = −σ Y − σ ′ = −250 × 106 + 358 × 106 = 108 MPa
108 MPa 
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PROBLEM 2.117
The steel rod ABC is attached to rigid supports and is unstressed at a
temperature of 3 °C. The steel is assumed elastoplastic, with σ Y = 248 MPa
and E = 200 GPa. The temperature of both portions of the rod is then
raised to 120 °C. Knowing that α = 11.7 × 10−6 / °C, determine (a) the stress
in portion AC, (b) the deflection of point C.
SOLUTION
δ B /A = δ B /A, P = 0 (constraint)
Determine ΔT to cause yielding in AC.
PL
PL
− AC − CB = LABα (ΔT ) = 0
EAAC EACB
( ΔT ) =
P  LAC LCB 
+


LAB Eα  AAC ACB 
At yielding, P = AAC αY
( ΔT )Y =
=
Actual
AAC σ Y  LAC LCB 
+


LAB Eα  AAC ACB 
(440 × 10−6 )(248 × 106 )
0.350
 0.175
+

(525 × 10−3 )(200 × 109 )(11.7 × 10−6 )  440 × 10−6 625 × 10−6

 = 85.06 °C

ΔT = 120 − 3 = 117°C > (ΔT )Y ∴ yielding occurs.
σ AC = −σ Y = −248 MPa

−6
P = σ Y AAC = (248 × 10 )(440 × 10 ) = 109.1 kN
6
δ C = −δ C/B =
=
PLCB
− LCBα (ΔT )
EACB
(109.1 × 103 )(0.350)
− (0.350)(11.7 × 10−6 )(117)
(200 × 109 )(625 × 10−6 )
= 0.305 × 10−3 m − 0.479 × 10−3 m = −0.174 × 10−3 m
δ C = 0.174 mm

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PROBLEM 2.118
Solve Problem 2.117, assuming that the temperature of the rod is raised to
120°C and then returned to 3°C.
PROBLEM 2.117 The steel rod ABC is attached to rigid supports and is
unstressed at a temperature of 3 °C. The steel is assumed elastoplastic, with
σ Y = 248 MPa and E = 200 GPa. The temperature of both portions of the
rod is then raised to 120 °C. Knowing that α = 11.7 × 10−6 / °C, determine
(a) the stress in portions AC, (b) the deflection of point C.
SOLUTION
δ B /A = δ B /A, P + δ B /A,T = 0 (constraint)
Determine ΔT to cause yielding in AC.
PL
PL
− AC − CB = LABα (ΔT ) = 0
EAAC EACB
ΔT =
P  LAC LCB
+

LAB Eα  AAC ACB

P
0.350 
 0.175
+
=

9
−6 
−6
625 × 10−6 
 (0.525)(200 × 10 )(11.7 × 10 )  440 × 10
= 0.78 × 10−3 P
PY = σ Y AAC = (248 × 106 )(440 × 10−6 ) = 109 kN
At yielding,
(ΔT )Y = (0.78 × 10−3 )(109 × 103 ) = 85.2°C
Actual
ΔT = 120 − 3 = 117°C > (ΔT )Y ∴ yielding occurs.
σ AC = −σ Y = −248 MPa
δ C = −δ B / C =
PLCB
(109 × 103 )(0.350)
− LBCα (ΔT ) −
EACB
(200 × 109 )(625 × 10−6 )
− (0.350)(11.7 × 10−6 )(117)
= 0.3052 × 10−3 − 0.479 × 10−3
Cooling
= −0.1738 × 10−3 m = −0.1738 mm
ΔT
117
ΔT ′ = 117°C
P′ =
=
= 150 kN
−3
0.78 × 10
0.78 × 10−3
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PROBLEM 2.118 (Continued)
(a)
Residual stress in AC
σ AC , res = −σ Y +

δ C′ = −δ B′ / C = −
P′
150 × 103
= −248 × 106 +
= 92.9 × 106 Pa
−6
AAC
440 × 10
= 92.9 MPa

P ′LCB
+ LCBα (ΔT )
EACB
(150 × 103 )(0.350)
+ (0.350)(11.7 × 10−6 )(117)
(200 × 109 )(625 × 10−6 )
= −0.00042 + 0.00048 = 0.00006 m = 0.06 m
= δ C + δ C′ = −0.1738 + 0.06 = −0.1138 mm
= −0.1138 mm
=−
δC,P

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PROBLEM 2.119
For the composite bar of Problem 2.111, determine the residual stresses
in the tempered-steel bars if P is gradually increased from zero to
440 kN and then decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 5 mm thick, are
bonded to a 12 mm mild-steel bar. This composite bar is subjected as
shown to a centric axial load of magnitude P. Both steels are elastoplastic
with E = 200 GPa and with yield strengths equal to 690 MPa and
345 MPa, respectively, for the tempered and mild steel.
SOLUTION
Areas:
Mild steel:
A1 = (12)(50) = 600 mm 2
Tempered steel:
A2 = (50)(5)(2) = 500 mm 2
A = A 1+ A2 = 1100 mm 2
Total:
Total force to yield the mild steel: σ Y 1 =
PY
A
∴ PY = Aσ Y 1 = (1100 × 10−6 )(345 × 106 ) = 379.5 kN
P > Pr; therefore, mild steel yields.
P1 = force carried by mild steel.
Let
P2 = force carried by tempered steel.
P1 = A1σ Y 1 = (600 × 10−6 )(345 × 10−6 ) = 207 kN
P1 + P2 = P, P2 = P − P1 = 440 − 207 = 233 kN
Unloading:
σ2 =
P2
233 × 103
=
= 466 × 106 Pa = 466 MPa
−6
A2 500 × 10
σ′=
P
440 × 103
=
= 400 × 106 Pa = 400 MPa
A 1100 × 10−6
Residual stresses
Mild steel
σ 1,res = σ 1 − σ ′ = 345 − 400 = −45 MPa
Tempered steel
σ 2,res = σ 2 − σ1 = 466 MPa − 400 MPa
= 66 MPa

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PROBLEM 2.120
For the composite bar in Problem 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 = 1 mm and is then
decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 5 mm thick, are
bonded to a 12 mm mild-steel bar. This composite bar is subjected as
shown to a centric axial load of magnitude P. Both steels are elastoplastic
with E = 200 GPa and with yield strengths equal to 690 MPa and
345 MPa, respectively, for the tempered and mild steel.
SOLUTION
A1 = (12)(50) = 600 mm 2
For the mild steel,
Lσ Y 1 (0.350)(345 × 106 )
=
E
200 × 109
= 0.603 × 10−3 m = 0.603 mm
A2 = 2(5)(50) = 500 mm 2
δY 1 =
For the tempered steel
Lδ Y 2 (0.350)(690 × 106 )
=
= 1.207 mm
E
200 × 109
A = A1 + A2 = 1100 mm 2
δY 2 =
Total area:
δ Y 1 < δ m < δY 2
Forces
The mild steel yields. Tempered steel is elastic
P1 = A1δ Y 1 = (600 × 10−6 )(345 × 106 ) = 207 kN
EA2δ m (200 × 109 )(500 × 10−6 )(1.0 × 10−3 )
=
= 285.7 kN
L
0.350
P
σ 1 = 1 = σ Y 1 = 345 MPa
A1
P2 =
Stresses:
Unloading
Residual stresses.
σ2 =
P2 285.7 × 103
=
= 571.4 MPa
A2 500 × 10−6
σ′=
P 492.7 × 103
=
= 447.9 MPa
A 1100 × 10−6
σ 1,res = σ 1 − σ ′ = 345 − 447.9 = −102.9 MPa
σ 2,res = σ 2 − σ ′ = 571.4 − 447.9 = 123.5 MPa

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PROBLEM 2.121∗
A narrow bar of aluminum is bonded to the side of a thick steel plate as shown.
Initially, at T1 = 21°C, 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 400 MPa. Assume α a = 23 × 10−6 /°C for
the aluminum and α s = 11.7 × 10−6 /°C for the steel. Further assume that the
aluminum is elastoplastic, with E = 75 GPa and σ Y = 400 MPa (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
400 × 106
( ΔT )Y =
=
= 472°C
E (α a − α s ) (75 × 109 )(23 − 11.7)(10−6 )
(a)
T2Y = T1 + (ΔT )Y = 21 + 472 = 493 °C

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
(400 × 106 )
=
= 944 °C
E (α a − α s ) (75 × 109 )(23 − 11.7)(10−6 )
T2 = T1 + ΔT = 21 + 944 = 965 °C

If T2 > 965 °C, 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
0.440
LAC + LCB
′ = 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 = −
= −118.182 × 106 Pa
σ BC
−6
A
1200 × 10
(378.182)(0.120)
δ C′ =
= 0.189091 × 10−3 m
(200 × 109 )(1200 × 106 )
′ =
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
′
P
−212.727 × 103
′ = CB =
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
The aluminum rod ( E = 70 GPa), which consists of two cylindrical
portions, AB and BC, is to be replaced with a cylindrical steel rod
DE ( E = 200 GPa) of the same overall length. Determine the minimum
required diameter of 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 165 MPa.
SOLUTION
Deformation of aluminum rod
δA =
=
PLAB PLBC P  LAB LBC 
+
= 
+

AAB E ABC E E  AAB ABC 
125 × 103  0.300 0.450 
−3
+

 = 0.798 × 10 m = 0.798 mm
70 × 109  0.038 0.056 
δ = 0.7989 mm
Steel rod
δ=
P
A
∴
A=
PL
(125 × 103 )(0.750)
=
= 0.5874 × 10−3 m 2
Eδ (200 × 109 )(0.798 × 10−3 )
= 587.4 mm 2
δ=
P
A
∴
A=
125 × 103
= 757.5 × 10−6 m 2
6
σ 165 × 10
= 757.5 mm 2
P
=
Required area is the larger value A = 757.5 mm 2
d=
4A
π
=
(4)(757.5)
π
= 31 mm

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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
The specimen shown is made from a 24-mm-diameter cylindrical
steel rod with two 36-mm-outer-diameter sleeves bonded to the rod as
shown. Knowing that E = 200 GPa, determine (a) the load P so that
the total deformation is 0.004 mm, (b) the corresponding deformation
of the central portion BC.
SOLUTION
(a)
δ=
 AE
Pi Li
i

P = Eδ 

=
i
P
E
Li 
 A 
A
Li
i
−1
Ai =
i
π
4
di2
L, mm
d, mm
A, mm2
L/A, mm–1
AB
4B
36
1017.9
0.04716
BC
72
24
452.4
0.15915
CD
48
36
1017.9
0.04716
0.25347
P = (200 × 103 )(0.004)(0.25347) −1 = 31.562 × 103 N
(b)
δ BC =
PLBC P LBC 31.562 × 10
=
=
(0.15915)
ABC E E ABC
200 × 103
3
sum
P = 31.6 kN 
δ = 0.025 mm 
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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
− l2
δ 2 = l 2 + 2l δ AB + δ AB
 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
For the steel truss ( E = 200 GPa) and loading shown, determine the
deformations of the members BD and DE, knowing that their crosssectional areas are 1250 mm2 and 1875 mm2, respectively.
SOLUTION
Free body: Portion ABC of truss
ΣM E = 0 : FBD (4.5 m) − (120 kN)(2.4 m) − (120 kN)(4.8 m) = 0
FBD = + 192 kN
Free body: Portion ABEC of truss
ΣFx = 0 : 120 kN + 120 kN − FDE = 0
FDE = + 240 kN

δ BD =
PL
(+192 × 103 N)(2400 mm)
=
AE (1250 mm 2 )(200000 MPa)
δ BD = +1.84 mm 
δ DE =
PL
(+240 × 103 N)(4500 mm)
=
AE (1875 mm 2 )(200000 MPa)
δ DE = +2.88 mm 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 2.130
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
Ea
70
3
P
7
(a)
σa = −
Pa
2P
2 30 × 103
=−
=−
= −57.1 × 106 Pa
−6
7 A
7 150 × 10
A
σ a = −57.1 MPa 
(b)
σb = −
Pb
3P
3 30 × 103
=−
=−
= −85.7 × 106 Pa
7 A
7 150 × 10−6
A
σ b = −85.7 MPa 
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PROBLEM 2.131
The brass shell (α b = 20.9 × 106 / °C) is fully bonded to the steel core
(α s = 11.7 × 10−6 /°F). Determine the largest allowable increase in
temperature if the stress in the steel core is not to exceed 55 MPa.
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 .
εs =
Strains:
Ps
+ α s ( ΔT )
Es As
εb = −
Ps
+ α b ( ΔT )
Eb Ab
ε s = εb
Matching:
Ps
P
+ α s (ΔT ) = − s + α b (ΔT )
Es As
Eb Ab
 1
1
+

 Es As Eb Ab

 Ps = (α b − α s )(ΔT )

As = (0.020)(0.020) = 400 × 10−6 m 2
Ab = (0.030)(0.030) − (0.020)(0.020) = 500 × 10−6 m 2
α b − α s = 9.2 × 10−6 / °C
Ps = σ s As = (55 × 106 )(400 × 10−6 ) = 22 × 103 N
1
1
1
1
+
=
+
= 31.55 × 10−9 N −1
9
−6
9
Es As Eb Ab (200 × 10 )(400 × 10 ) (105 × 10 )(500 × 10−6 )
(31.55 × 10−9 )(22 × 102 ) = (9.2 × 10−6 )( ΔT )
ΔT = 75.4 °C 
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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
9
E
87 × 10
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
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 =
12.5 × 103 N)
F
=
= 833.33 × 103 Pa
A (0.15 m)(0.1 m)
δ = 1.5 × 10−3 m
δ
h = 0.03 m
−3
1.5 × 10
= 0.05
0.03
τ
833.33 × 103
G = =
= 16.67 × 106 Pa
0.05
γ
γ=
h
=
G = 16.67 MPa 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 2.134
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
Shearing strain:
γ =
Shearing stress:
τ = Gγ =
Force:
h
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 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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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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