PROBLEM 9.46
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
140 20 120 C
∆= − = °
T
Free thermal expansion:
66
3
() ()
(0.300)(23 10 )(120) (0.250)(17.3 10 )(120)
1.347 10 m
T aa ss
L TL T
δa a
−−
−
= ∆+ ∆
=×+ ×
= ×
Shortening due to P to meet constraint:
33 3
1.347 10 0.5 10 0.847 10 m
P
δ
−− −
= × −× = ×
9 6 96
9
0.300 0.250
(75 10 )(2000 10 ) (190 10 )(800 10 )
3.6447 10
as as
Paa ss aa ss
PL PL L L P
EA EA EA EA
P
P
δ
−−
−
=+= +
= +
× × ××
= ×
Equating,
93
3
3.6447 10 0.847 10
232.39 10 N
−−
×=×
= ×
P
P
(a)
36
6
232.39 10 116.2 10 Pa
2000 10
aa
P
A
σ
−
×
=−=− =− ×
×
116.2 MPa
σ
= −
a
(b)
()
a
a aa aa
PL
LT
EA
δa
= ∆−
3
66
96
(232.39 10 )(0.300)
(0.300)(23 10 )(120) 363 10 m
(75 10 )(2000 10 )
−−
−
×
=×− =×
××
0.363 mm
δ
=
a
consent of McGraw–Hill Education.
PROBLEM 9.47
A brass link
( 105 GPa,
b
E=
6
20.9 10 / C)
b
α
−
=×°
and a steel rod
(Es
200 GPa,=
6
11.7 10 / C)
s
α
−
=×°
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
20 C.T= °
Final dimensions at
45 C.T= °
45 20 25 CT∆= − = °
Free thermal expansion of each part:
Brass link:
66
( ) ( ) (20.9 10 )(25)(0.250) 130.625 10 m
Tb b
TL
δα
−−
=∆= × = ×
Steel rod:
66
( ) ( ) (11.7 10 )(25)(0.250) 73.125 10 m
Ts s
TL
δα
−−
=∆= × = ×
At the final temperature, the difference between the free length of the steel rod and the brass link is
6 6 66
120 10 73.125 10 130.625 10 62.5 10 m
δ
− − −−
=×+ ×− ×=×
Add equal but opposite forces P to elongate the brass link and contract the steel rod.
Brass link:
9
105 10 PaE= ×
2 32
12
93
(2)(50)(37.5) 3750 mm 3.750 10 m
(0.250)
( ) 634.92 10
(105 10 )(3.750 10 )
b
P
A
PL P P
EA
δ
−
−
−
= = = ×
= = = ×
××
Steel rod:
9
200 10 PaE= ×
2 2 62
(30) 706.86 mm 706.86 10 m
4
s
A
π
−
= = = ×
9
96
(0.250)
( ) 1.76838 10
(200 10 )(706.86 10 )
Ps ss
PL P P
EA
δ
−
−
= = = ×
××
96 3
( ) ( ) : 2.4033 10 62.5 10 26.006 10 N
δ δδ
−−
+= × =× = ×
Pb Ps
PP
(a) Stress in steel rod:
36
6
(26.006 10 ) 36.8 10 Pa
706.86 10
ss
P
A
s
−
×
=−=− =− ×
×
36.8 MPa
s
s
= −
(b) Final length of steel rod:
0() ()
f Ts Ps
LL
δδ
=+−
6 6 93
0.250 120 10 73.125 10 (1.76838 10 )(26.003 10 )
f
L
−− −
= +×+ ×− × ×
0.250147 m=
250.147 mm
f
L=
consent of McGraw–Hill Education.
PROBLEM 9.48
Two steel bars
6
( 200 GPa and 11.7 10 / C)
ss
E
a
−
= =×°
are used to
reinforce a brass bar
6
( 105 GPa, 20.9 10 / C)
a
−
= =×°
bb
E
that is subjected
to a load
25 kN.P=
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:
3
0.5 mm 0.5 10 m
T
δ
−
= = ×
36
36
, 0.5 10 (2.00)(11.7 10 )( ),
0.5 10 (2)(11.7 10 )( )
21.368 C
Ts
LT T
TT
T
δa
−−
−−
=∆ ×= × ∆
∆= × = × ∆
∆= °
21.4 C°
Elongation of steel:
2 62
(2)(5)(40) 400 mm 400 10 m
s
A−
= = = ×
** 9*
69
(2.00)
( ) 25 10
(400 10 )(200 10 )
δ
−
−
= = = ×
××
Ps ss
FL P P
AE
Contraction of brass:
2 62
(40)(15) 600 mm 600 10 m
b
A
−
= = = ×
** 9*
69
(2.00)
( ) 31.746 10
(600 10 )(105 10 )
δ
−
−
= = = ×
××
Pb bb
PL P P
AE
Ps Pb
3 9* 3
*3
( ) ( ) 0.5 10 , 56.746 10 0.5 10
8.8112 10 N
δδ
− −−
+=× × =×
= ×
Ps Pb P
P
Stresses due to fabrication:
Steel:
*3
*6
6
8.8112 10 22.028 10 Pa 22.028 MPa
400 10
ss
P
A
s
−
×
== = ×=
×
PROBLEM 9.48 (Continued)
Brass:
*3
*6
6
8.8112 10 14.6853 10 Pa 14.685 MPa
600 10
bb
P
A
s
−
×
=−=− =− × =−
×
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.
69 6
(400 10 )(200 10 ) 40 10
2.00
sb
ss bb
ss
s
PL PL
AE AE
AE
PL
δ
δ δδ
−
′= =
××
′ ′′
= = = ×
PROBLEM 9.49
In a standard tensile test, a steel rod of 22-mm diameter is subjected to a
tension force of 75 kN. Knowing that
0.3=v
and
200 GPa,=
E
determine (a) the elongation of the rod in a 200-mm gage length, (b) the
change in diameter of the rod.
SOLUTION
3 2 2 62
75 kN 75 10 N (0.022) 380.13 10 m
44
P Ad
ππ
−
==×===×
36
6
66
9
6
75 10 197.301 10 Pa
380.13 10
197.301 10 986.51 10
200 10
(200 mm)(986.51 10 )
x
xx
P
A
E
L
σ
σ
ε
δε
−
−
−
×
= = = ×
×
×
= = = ×
×
= = ×
(a)
0.1973 mm
x
δ
=
66
(0.3)(986.51 10 ) 295.95 10
yx
v
εε
−−
=−=− × =− ×
6
(22 mm)( 295.95 10 )
δε
−
== −×
yy
d
(b)
0.00651 mm
δ
= −
y
consent of McGraw–Hill Education.
PROBLEM 9.50
A standard tension test is used to determine the properties of an experimental plastic.
The test specimen is a
5
8
–in.–diameter rod and it is subjected to an 800–lb tensile
force. Knowing that an elongation of 0.45 in. and a decrease in diameter of 0.025 in.
are observed in a 5–in. gage length, determine the modulus of elasticity, the modulus
of rigidity, and Poisson’s ratio for the material.
SOLUTION
2
22
50.306796 in
4 48
ππ
= = =
Ad
800 lbP=
3
800 2.6076 10 psi
0.306796
yP
A
s
= = = ×
0.45 0.090
5.0
= = =
y
yL
δ
ε
0.025 0.040
0.625
x
x
d
δ
ε
−
= = = −
33
2.6076 10 28.973 10 psi
0.090
×
= = = ×
y
y
E
δ
ε
3
29.0 10 psiE= ×
0.040 0.44444
0.090
−
=−= =
x
y
v
ε
ε
0.444
=v
33
28.973 10 10.0291 10 psi
2(1 ) (2)(1 0.44444)
E
v
s
×
= = = ×
++
3
10.03 10 psi
s
= ×
PROBLEM 9.51
A 2–m length of an aluminum pipe of 240–mm outer diameter and 10–mm
wall thickness is used as a short column to carry a 640–kN centric axial load.
Knowing that E = 73 GPa and v = 0.33, determine (a) the change in length of
the pipe, (b) the change in its outer diameter, (c) the change in its wall
thickness.
SOLUTION
0.240 0.010 2.0= = =
o
dtL
3
2 0.240 2(0.010) 0.220 m 640 10 N= −= − = = ×
io
dd t P
( )
2 2 32
(0.240 0.220) 7.2257 10 m
44
ππ
−
= −= − = ×
oi
A dd
(a)
3
93
(640 10 )(2.0)
(73 10 )(7.2257 10 )
δ
−
×
=−=−××
PL
EA
3
2.4267 10 m
−
=−×
2.43 mm= −
δ
3
2.4267 1.21335 10
2.0
−
−
== =−×
L
δ
ε
3
(0.33)( 1.21335 10 )
εε
−
=−=− − ×
LAT v
4
4.0041 10−
= ×
(b)
42
(240 mm)(4.0041 10 ) 9.6098 10 mm
o o LAT
dd
ε
−−
∆= = × = ×
0.0961 mm
o
d∆=
43
(10 mm)(4.0041 10 ) 4.0041 10 mm
−−
∆= = × = ×
LAT
tt
ε
0.00400 mmt∆=
consent of McGraw–Hill Education.
PROBLEM 9.52
The change in diameter of a large steel bolt is carefully measured as the
nut is tightened. Knowing that
6
29 10 psiE= ×
and
0.30,=
v
determine the internal force in the bolt if the diameter is observed to
decrease by
3
0.5 10 in.
−
×
SOLUTION
3
33
0.5 10 in. 2.5 in.
0.5 10 0.2 10
2.5
y
y
y
d
d
d
ε
ε
−
−−
=−× =
×
= =− =−×
33
0.2 10
: 0.66667 10
0.3
yy
x
x
vv
εε
ε
ε
−−
−×
=−===×
63 3
(29 10 )(0.66667 10 ) 19.3334 10 psi
xx
E
sε
−
==× ×= ×
22 2
(2.5) 4.9087 in
44
ππ
= = =Ad
33
(19.3334 10 )(4.9087) 94.902 10 lb
x
FA
s
==×=×
94.9 kipsF=
consent of McGraw–Hill Education.
PROBLEM 9.53
An aluminum plate
( 74E=
GPa and
0.33)v=
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
125
σ
=
MPa.
SOLUTION
2(1 )
y
ε
+
PROBLEM 9.46
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
140 20 120 C
∆= − = °
T
Free thermal expansion:
66
3
() ()
(0.300)(23 10 )(120) (0.250)(17.3 10 )(120)
1.347 10 m
T aa ss
L TL T
δa a
−−
−
= ∆+ ∆
=×+ ×
= ×
Shortening due to P to meet constraint:
33 3
1.347 10 0.5 10 0.847 10 m
P
δ
−− −
= × −× = ×
9 6 96
9
0.300 0.250
(75 10 )(2000 10 ) (190 10 )(800 10 )
3.6447 10
as as
Paa ss aa ss
PL PL L L P
EA EA EA EA
P
P
δ
−−
−
=+= +
= +
× × ××
= ×
Equating,
93
3
3.6447 10 0.847 10
232.39 10 N
−−
×=×
= ×
P
P
(a)
36
6
232.39 10 116.2 10 Pa
2000 10
aa
P
A
σ
−
×
=−=− =− ×
×
116.2 MPa
σ
= −
a
(b)
()
a
a aa aa
PL
LT
EA
δa
= ∆−
3
66
96
(232.39 10 )(0.300)
(0.300)(23 10 )(120) 363 10 m
(75 10 )(2000 10 )
−−
−
×
=×− =×
××
0.363 mm
δ
=
a
consent of McGraw–Hill Education.
PROBLEM 9.47
A brass link
( 105 GPa,
b
E=
6
20.9 10 / C)
b
α
−
=×°
and a steel rod
(Es
200 GPa,=
6
11.7 10 / C)
s
α
−
=×°
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
20 C.T= °
Final dimensions at
45 C.T= °
45 20 25 CT∆= − = °
Free thermal expansion of each part:
Brass link:
66
( ) ( ) (20.9 10 )(25)(0.250) 130.625 10 m
Tb b
TL
δα
−−
=∆= × = ×
Steel rod:
66
( ) ( ) (11.7 10 )(25)(0.250) 73.125 10 m
Ts s
TL
δα
−−
=∆= × = ×
At the final temperature, the difference between the free length of the steel rod and the brass link is
6 6 66
120 10 73.125 10 130.625 10 62.5 10 m
δ
− − −−
=×+ ×− ×=×
Add equal but opposite forces P to elongate the brass link and contract the steel rod.
Brass link:
9
105 10 PaE= ×
2 32
12
93
(2)(50)(37.5) 3750 mm 3.750 10 m
(0.250)
( ) 634.92 10
(105 10 )(3.750 10 )
b
P
A
PL P P
EA
δ
−
−
−
= = = ×
= = = ×
××
Steel rod:
9
200 10 PaE= ×
2 2 62
(30) 706.86 mm 706.86 10 m
4
s
A
π
−
= = = ×
9
96
(0.250)
( ) 1.76838 10
(200 10 )(706.86 10 )
Ps ss
PL P P
EA
δ
−
−
= = = ×
××
96 3
( ) ( ) : 2.4033 10 62.5 10 26.006 10 N
δ δδ
−−
+= × =× = ×
Pb Ps
PP
(a) Stress in steel rod:
36
6
(26.006 10 ) 36.8 10 Pa
706.86 10
ss
P
A
s
−
×
=−=− =− ×
×
36.8 MPa
s
s
= −
(b) Final length of steel rod:
0() ()
f Ts Ps
LL
δδ
=+−
6 6 93
0.250 120 10 73.125 10 (1.76838 10 )(26.003 10 )
f
L
−− −
= +×+ ×− × ×
0.250147 m=
250.147 mm
f
L=
consent of McGraw–Hill Education.
PROBLEM 9.48
Two steel bars
6
( 200 GPa and 11.7 10 / C)
ss
E
a
−
= =×°
are used to
reinforce a brass bar
6
( 105 GPa, 20.9 10 / C)
a
−
= =×°
bb
E
that is subjected
to a load
25 kN.P=
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:
3
0.5 mm 0.5 10 m
T
δ
−
= = ×
36
36
, 0.5 10 (2.00)(11.7 10 )( ),
0.5 10 (2)(11.7 10 )( )
21.368 C
Ts
LT T
TT
T
δa
−−
−−
=∆ ×= × ∆
∆= × = × ∆
∆= °
21.4 C°
Elongation of steel:
2 62
(2)(5)(40) 400 mm 400 10 m
s
A−
= = = ×
** 9*
69
(2.00)
( ) 25 10
(400 10 )(200 10 )
δ
−
−
= = = ×
××
Ps ss
FL P P
AE
Contraction of brass:
2 62
(40)(15) 600 mm 600 10 m
b
A
−
= = = ×
** 9*
69
(2.00)
( ) 31.746 10
(600 10 )(105 10 )
δ
−
−
= = = ×
××
Pb bb
PL P P
AE
Ps Pb
3 9* 3
*3
( ) ( ) 0.5 10 , 56.746 10 0.5 10
8.8112 10 N
δδ
− −−
+=× × =×
= ×
Ps Pb P
P
Stresses due to fabrication:
Steel:
*3
*6
6
8.8112 10 22.028 10 Pa 22.028 MPa
400 10
ss
P
A
s
−
×
== = ×=
×
PROBLEM 9.48 (Continued)
Brass:
*3
*6
6
8.8112 10 14.6853 10 Pa 14.685 MPa
600 10
bb
P
A
s
−
×
=−=− =− × =−
×
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.
69 6
(400 10 )(200 10 ) 40 10
2.00
sb
ss bb
ss
s
PL PL
AE AE
AE
PL
δ
δ δδ
−
′= =
××
′ ′′
= = = ×
PROBLEM 9.49
In a standard tensile test, a steel rod of 22-mm diameter is subjected to a
tension force of 75 kN. Knowing that
0.3=v
and
200 GPa,=
E
determine (a) the elongation of the rod in a 200-mm gage length, (b) the
change in diameter of the rod.
SOLUTION
3 2 2 62
75 kN 75 10 N (0.022) 380.13 10 m
44
P Ad
ππ
−
==×===×
36
6
66
9
6
75 10 197.301 10 Pa
380.13 10
197.301 10 986.51 10
200 10
(200 mm)(986.51 10 )
x
xx
P
A
E
L
σ
σ
ε
δε
−
−
−
×
= = = ×
×
×
= = = ×
×
= = ×
(a)
0.1973 mm
x
δ
=
66
(0.3)(986.51 10 ) 295.95 10
yx
v
εε
−−
=−=− × =− ×
6
(22 mm)( 295.95 10 )
δε
−
== −×
yy
d
(b)
0.00651 mm
δ
= −
y
consent of McGraw–Hill Education.
PROBLEM 9.50
A standard tension test is used to determine the properties of an experimental plastic.
The test specimen is a
5
8
–in.–diameter rod and it is subjected to an 800–lb tensile
force. Knowing that an elongation of 0.45 in. and a decrease in diameter of 0.025 in.
are observed in a 5–in. gage length, determine the modulus of elasticity, the modulus
of rigidity, and Poisson’s ratio for the material.
SOLUTION
2
22
50.306796 in
4 48
ππ
= = =
Ad
800 lbP=
3
800 2.6076 10 psi
0.306796
yP
A
s
= = = ×
0.45 0.090
5.0
= = =
y
yL
δ
ε
0.025 0.040
0.625
x
x
d
δ
ε
−
= = = −
33
2.6076 10 28.973 10 psi
0.090
×
= = = ×
y
y
E
δ
ε
3
29.0 10 psiE= ×
0.040 0.44444
0.090
−
=−= =
x
y
v
ε
ε
0.444
=v
33
28.973 10 10.0291 10 psi
2(1 ) (2)(1 0.44444)
E
v
s
×
= = = ×
++
3
10.03 10 psi
s
= ×
PROBLEM 9.51
A 2–m length of an aluminum pipe of 240–mm outer diameter and 10–mm
wall thickness is used as a short column to carry a 640–kN centric axial load.
Knowing that E = 73 GPa and v = 0.33, determine (a) the change in length of
the pipe, (b) the change in its outer diameter, (c) the change in its wall
thickness.
SOLUTION
0.240 0.010 2.0= = =
o
dtL
3
2 0.240 2(0.010) 0.220 m 640 10 N= −= − = = ×
io
dd t P
( )
2 2 32
(0.240 0.220) 7.2257 10 m
44
ππ
−
= −= − = ×
oi
A dd
(a)
3
93
(640 10 )(2.0)
(73 10 )(7.2257 10 )
δ
−
×
=−=−××
PL
EA
3
2.4267 10 m
−
=−×
2.43 mm= −
δ
3
2.4267 1.21335 10
2.0
−
−
== =−×
L
δ
ε
3
(0.33)( 1.21335 10 )
εε
−
=−=− − ×
LAT v
4
4.0041 10−
= ×
(b)
42
(240 mm)(4.0041 10 ) 9.6098 10 mm
o o LAT
dd
ε
−−
∆= = × = ×
0.0961 mm
o
d∆=
43
(10 mm)(4.0041 10 ) 4.0041 10 mm
−−
∆= = × = ×
LAT
tt
ε
0.00400 mmt∆=
consent of McGraw–Hill Education.
PROBLEM 9.52
The change in diameter of a large steel bolt is carefully measured as the
nut is tightened. Knowing that
6
29 10 psiE= ×
and
0.30,=
v
determine the internal force in the bolt if the diameter is observed to
decrease by
3
0.5 10 in.
−
×
SOLUTION
3
33
0.5 10 in. 2.5 in.
0.5 10 0.2 10
2.5
y
y
y
d
d
d
ε
ε
−
−−
=−× =
×
= =− =−×
33
0.2 10
: 0.66667 10
0.3
yy
x
x
vv
εε
ε
ε
−−
−×
=−===×
63 3
(29 10 )(0.66667 10 ) 19.3334 10 psi
xx
E
sε
−
==× ×= ×
22 2
(2.5) 4.9087 in
44
ππ
= = =Ad
33
(19.3334 10 )(4.9087) 94.902 10 lb
x
FA
s
==×=×
94.9 kipsF=
consent of McGraw–Hill Education.
PROBLEM 9.53
An aluminum plate
( 74E=
GPa and
0.33)v=
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
125
σ
=
MPa.
SOLUTION
2(1 )
y
ε
+