Solution 1.15
PROBLEM 1.15
Knowing that a force P of magnitude 50 kN is required to punch a hole
of diameter d = 20 mm in an aluminum sheet of thickness t = 5 mm,
determine the average shearing stress in the aluminum at failure.
SOLUTION
Area of failure in plate:
A π=
dt π ( 0.020 m )( 0.005 m )
=
= 3.1416 × 10−4 m 2
Average shearing stress:
τ avg =
=
P
A
50 × 103 N
3.1416 × 10−4 m 2
τ avg = 159.2 MPa 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 1.18
PROBLEM 1.18
A load P is applied to a steel rod supported as shown by an aluminum plate
into which a 12-mm-diameter hole has been drilled. Knowing that the
shearing stress must not exceed 180 MPa in the steel rod and 70 MPa in the
aluminum plate, determine the largest load P that can be applied to the rod.
SOLUTION
For steel:
=
A1 π=
dt π (0.012 m)(0.010 m)
= 376.99 × 10−6 m 2
τ1 =
P
∴ P = A1τ1 = (376.99 × 10−6 m 2 )(180 × 106 Pa)
A
= 67.858 × 103 N
=
A2 π=
dt π (0.040 m)(0.008 =
m) 1.00531 × 10−3 m 2
For aluminum:
τ2 =
P
∴ P = A2τ 2 = (1.00531 × 10−3 m 2 )(70 × 106 Pa) = 70.372 × 103 N
A2
Limiting value of P is the smaller value, so
P = 67.9 kN 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 1.41
PROBLEM 1.41
In the truss shown, members AC and AD consist of rods made of the
same metal alloy. Knowing that AC is of 25-mm diameter and that the
ultimate load for that rod is 345 kN, determine (a) the factor of safety
for AC, (b) the required diameter of AD if it is desired that both rods
have the same factor of safety.
SOLUTION
Forces in AC and AD.
Joint C:
=
ΣFy 0:
Joint D:
1.5
0
F=
AC − 48 kN
11.25
FAC = 107.331 kN T
=
ΣFy 0:
PU
FAC
345 kN
107.331 kN
(a)
Factor of safety for AC.
(b)
For the same factor of safety in AC and AD, σ AD = σ AC .
F.S. =
F.S. =
1.5
F=
0
AD − 48 kN
38.25
FAD = 197.909 kN T
F.S. = 3.21 
FAD
F
= AC
AAD
AAC
AAD
=
Required diameter:
FAD
197.909 kN  π 
m) 2 9.0513 × 10−4 m 2
AAC
=
=
  (0.025
107.331 kN  4 
FAC
=
d AD
4 AAD
=
π
(4)(9.0513 × 10−4 m 2 )
π
d AD = 33.9 mm 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 1.43
PROBLEM 1.43
Two wooden members are joined by plywood splice plates that are fully glued on
the contact surfaces. Knowing that the clearance between the ends of the members
is 6 mm and that the ultimate shearing stress in the glued joint is 2.5 MPa, determine
the length L for which the factor of safety is 2.75 for the loading shown.
SOLUTION
=
τ all
2.5 MPa
= 0.90909 MPa
2.75
On one face of the upper contact surface,
A=
L − 0.006 m
(0.125 m)
2
Since there are 2 contact surfaces,
τ all =
P
2A
16 × 103
0.90909 × 106 =
( L − 0.006)(0.125)
L = 0.14680 m
146.8 mm 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 1.48
PROBLEM 1.48
For the support of Prob. 1.47, knowing that the diameter of the pin is
d = 16 mm and that the magnitude of the load is P = 20 kN, determine
(a) the factor of safety for the pin, (b) the required values of b and c if the
factor of safety for the wooden members is the same as that found in part a
for the pin.
PROBLEM 1.47 A load P is supported as shown by a steel pin that has been
inserted in a short wooden member hanging from the ceiling. The ultimate
strength of the wood used is 60 MPa in tension and 7.5 MPa in shear, while
the ultimate strength of the steel is 145 MPa in shear. Knowing that
b = 40 mm, c = 55 mm, and d = 12 mm, determine the load P if an
overall factor of safety of 3.2 is desired.
SOLUTION
=
P 20 kN
= 20 × 103 N
(a)
Pin:
=
A
Double shear:
=
τ
π
=
d2
4
π
4
P
=
τU
2A
2
(0.016)
2.01.06 × 10−6 m 2
=
PU
2A
PU = 2 Aτ U = (2)(201.16 × 10−6 )(145 × 106 ) = 58.336 × 103 N
F=
.S.
(b)
Tension in wood:
PU
=
A
PU
w(b − d )
where =
w 40 mm
= 0.040 m
PU
58.336 × 103
=
0.016 +
=
40.3 × 10−3 m
wσ U
(0.040)(60 × 106 )
Shear in wood:
b = 40.3 mm 
=
PU 58.336 × 103 N for same F.S.
Double shear: each area is A = wc
=
c
F .S . = 2.92 
=
PU 58.336 × 103 N for same F.S.
σ=
U
b=
d +
PU
58.336 × 103
=
P
20 × 103
=
τU
PU
=
2A
PU
58.336 × 103
=
=
97.2 × 10−3 m
6
2wτ U
(2)(0.040)(7.5 × 10 )
PU
2wc
c = 97.2 mm 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.9
PROBLEM 2.9
A 9-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine
the smallest diameter wire that can be used, knowing that the normal stress must not exceed
150 MPa and that the increase in length of the wire must not exceed 25 mm.
SOLUTION
Stress:
Deformation:
σ =
P
A
=
A
P
=
σ
δ =
PL
AE
=
A
PL
=
Eδ
The larger value of A governs:
9 × 103 N
= 60 × 10−6 m 2
6
150 × 10 Pa
(9 × 103 )(50)
= 90 × 10−6 m 2
−3
9
(200 × 10 )(25 × 10 )
A = 90 mm 2
=
A
π
4
d 2=
d
4A
=
π
4(90 mm 2 )
π
d = 10.70 mm 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.14
PROBLEM 2.14
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

F=
3.5P − (6) 
BC  0
7.2111


P = 0.9509 FBC
M A 0:
Σ=
Considering allowable stress, =
σ 190 × 106 Pa
π
π
2
d2
=
=
(0.004)
12.566 × 10−6 m 2
4
4
FBC
∴ FBC =σ A =(190 × 106 )(12.566 × 10−6 ) =2.388 × 103 N
σ=
A
A
=
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 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.17
PROBLEM 2.17
The specimen shown has been cut from a 5-mm-thick
sheet of vinyl (E = 3.10 GPa) and is subjected to a
1.5-kN tensile load. Determine (a) the total
deformation of the specimen, (b) the deformation of its
central portion BC.
SOLUTION
=
δ AB
PLAB
(1.5 × 103 N)(40 × 10−3 m)
=
= 154.839 × 10−6 m
EAAB (3.1 × 109 N/m 2 )(25 × 10−3 m)(5 × 10−3 m)
=
δ BC
PLBC
(1.5 × 103 N)(50 × 10−3 m)
=
= 483.87 × 10−6 m
−3
−3
9
2
EABC (3.1 × 10 N/m )(10 × 10 m)(5 × 10 m)
δ=
δ=
154.839 × 10−6 m
CD
AB
(a)
Total deformation:
δ = δ AB + δ BC + δ CD
=
δ 793.55 × 10−6 m
δ = 0.794 mm 
(b)
Deformation of portion BC :
δ BC = 0.484 mm 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.24
PROBLEM 2.24
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
52 + 62 = 7.810 m, EBD = 200 × 109 Pa
LBD =
δ BD =
FBD
=
FBD LBD
EBD ABD
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
=
FBD
7.810
= 50.4 × 103 N
(5)(78.67 × 103 )
7.810
P = 50.4 kN 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.33
PROBLEM 2.33
An axial centric force of magnitude P = 450 kN is applied
to the composite block shown by means of a rigid end
plate. Knowing that h = 10 mm, determine the normal
stress in (a) the brass core, (b) the aluminum plates.
SOLUTION
δ=
δ=
δ;
P PA + PB
=
A
B
PA L
PB L
=
δ =
and δ
E A AA
EB AB
δ 
δ 
Therefore, =
PA (=
E A AA )   ;
PB ( EB AB )  
L
 
L
δ
Substituting,
=
PA ( E A AA + E B AB )  
 L
δ
P
=
∈ =
L ( E A AA + E B AB )
∈=
(450 × 103 N)
(70 × 10 Pa)(2)(0.06 m)(0.01 m) + (105 × 109 Pa)(0.06 m)(0.04 m)
9
=
∈ 1.33929 × 10−3
σ = E∈
Now,
(a)
Brass-core:
σB =
(105 × 109 Pa)(1.33929 × 10−3 )
= 1.40625 × 108 Pa
σ B = 140.6 MPa 
(b)
Aluminum:
σA =
(70 × 109 Pa)(1.33929 × 10 −3 )
= 9.3750 × 107 Pa
σ A = 93.8 MPa 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.48
PROBLEM 2.48
The brass shell (αb = 20.9 × 10–6/°C) is fully bonded to the steel
core (αs = 11.7 × 10–6/°C). 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 .
Ps
ε s=
+ α s (∆T )
Strains:
Es As
P
εb =
− s + α b (∆T )
Eb Ab
Matching:
ε s = εb
Ps
P
+ α s (∆T ) =
− s + α b (∆T )
Es As
Eb Ab
 1
1 
+

 Ps = (α b − α s )(∆T )
 Es As Eb Ab 
(1)
Ab =
(0.030)(0.030) − (0.020)(0.020) =
500 × 10−6 m 2
=
As (0.020)(0.020)
= 400 × 10−6 m 2
α b − α s = 9.2 × 10−6 / °C
σ s As =
(55 × 106 )(400 × 10−6 ) =
22 × 103 N
Ps =
1
1
1
1
+
=
+
= 31.548 × 10−9 N −1
−6
−6
9
9
Es As Eb Ab (200 × 10 )(400 × 10 ) (105 × 10 )(500 × 10 )
From (1),
(31.548 × 10−9 )(22 × 103 ) = (9.2 × 10−6 )(∆T )
∆T= 75.4 °C 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.51
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
π
π
π
π
2
=
=
d AB
(30)2 706.86 =
mm 2 706.86 × 10−6 m 2
4
4
2
= (50)2 =
ABC = d BC
1.9635 × 103 mm 2 =
1.9635 × 10−3 m 2
4
4
Free thermal expansion:
=
δT LABα s (∆T ) + LBCα b (∆T )
= (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
+
(200 × 109 )(706.86 × 10−6 ) (105 × 109 )(1.9635 × 10−3 )
= 3.2235 × 10−9 P
For zero net deflection, δ P = δT
3.2235 × 10−9 P = 459.75 × 10−6
=
P 142.624 × 103 N
P = 142.6 kN 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.60
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
∆T = 140 − 20 = 120°C
Free thermal expansion:
Laα a (∆T ) + Lsα s (∆T )
δ=
T
= (0.300)(23 × 10−6 )(120) + (0.250)(17.3 × 10−6 )(120)
= 1.347 × 10−3 m
Shortening due to P to meet constraint:
δ P= 1.347 × 10−3 − 0.5 × 10−3= 0.847 × 10−3 m
PLa
PLs  La
L 
+
=
+ s P
Ea Aa Es As  Ea Aa Es As 


0.300
0.250
= 
+
P
9
9
−6
−6 
 (75 × 10 )(2000 × 10 ) (190 × 10 )(800 × 10 ) 
δP =
= 3.6447 × 10−9 P
3.6447 × 10−9 P = 0.847 × 10−3
Equating,
=
P 232.39 × 103 N
P
Aa
232.39 × 103
=
−116.2 × 106 Pa
2000 × 10−6
(a)
σa =
−
=
−
(b)
δ=
Laα a (∆T ) −
a
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 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.70
PROBLEM 2.70
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)=
δ y 0=
ε y 0=
σz 0
1
(σ x − vσ y − vσ z )
E
σ y = vσ x = (0.35)(−180 × 106 )
ε y=
=
−63 × 106 Pa
σ y = −63.0 MPa 
(0.35)(−243 × 106 )
=
+1.890 × 10−3
45 × 109
σ x − vσ y
1
157.95 × 106
ε x =(σ x − vσ y − vσ Z ) =
=
−
=
−3.510 × 10−3
E
E
45 × 109
1
E
v
E
ε z =(σ z − vσ x − vσ y ) =
− (σ x + σ y ) =
−
(b)
A0 = Lx Lz
A= Lx (1 + ε x ) Lz (1 + ε z )= Lx Lz (1 + ε x + ε z + ε xε z )
∆ A = A − A0 = Lx Lz (ε x + ε z + ε xε z ) ≈ Lx Lz (ε x + ε z )
=
∆ A (100 mm)(25 mm)(−3.510 × 10−3 + 1.890 × 10−3 )
(c)
∆A=
−4.05 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 )
∆V = V − V0 = Lx Ly Lz (ε x + ε y + ε z + small terms)
=
∆V (100)(40)(25)(−3.510 × 10−3 + 0 + 1.890 × 10−3 )
∆V =
−162.0 mm3 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
Solution 2.81
PROBLEM 2.81
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 of plate
AB, determine the modulus of rigidity of the rubber used.
SOLUTION
Each block of rubber carries ½ of the applied force P. Thus the shearing stress is
τ=
1 / 2P
where A =
10.8 103 mm 2 =×
10.8 10−3 m 2
(180 mm )( 60 mm ) =×
A
1
(19 × 103 N)
2
τ
=
= 0.87963 × 106 Pa
10.8 × 10−3
δ
3 mm
=
= 0.085714
γ =
h
35 mm
0.87963 × 103
τ
G
= =
= 10.26 × 106 Pa
0.085714
γ
G = 10.26 MPa 
Copyright ©2020 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.