PROBLEM 9.38
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:
6
105 GPa
20.9 10 / C
E
a
−
=
=×°
Aluminum shell:
6
70 GPa
23.6 10 / C
E
a
−
=
=×°
Free thermal expansion:
195 15 180 CT∆= − = °
Brass core:
() ( )
Tb b
LT
δa
= ∆
Aluminum shell:
() ( )
= ∆
Ta a
LT
δa
ab
Brass core:
9
22
62
105 10 Pa
(25) 490.87 mm
4
490.87 10 m
()
b
b
Pb bb
E
A
PL
EA
π
δ
−
= ×
= =
= ×
=
consent of McGraw–Hill Education.
PROBLEM 9.39
The concrete post
6
( 3.6 10
c
E= ×
psi and
6
5.5 10 / F)
α
−
=×°
c
is reinforced with six steel
bars, each of
7
8-in.
diameter
6
( 29 10
s
E= ×
psi and
6
6.5 10 / F).
s
α
−
=×°
Determine the
normal stresses induced in the steel and in the concrete by a temperature rise of 65°F.
SOLUTION
2
22
7
6 6 3.6079 in
4 48
ππ
= = =
s
Ad
22 2
10 10 3.6079 96.392 in= −= − =
cs
AA
Let
c
P=
tensile force developed in the concrete.
For equilibrium with zero total force, the compressive force in the six steel rods equals
.
c
P
Strains:
() ()
cc
s sc c
ss cc
PP
TT
EA EA
ε αε α
=− +∆ = +∆
Matching:
cs
εε
=
() ()
cc
cs
cc ss
PP
TT
EA EA
αα
+ ∆=− + ∆
6
66
3
11 ( )( )
11
(1.0 10 )(65)
(3.6 10 )(96.392) (29 10 )(3.6079)
5.2254 10 lb
αα
−
+ =−∆
+=×
××
= ×
c sc
cc ss
c
c
PT
EA EA
P
P
3
5.2254 10 54.210 psi
96.392
c
cc
P
A
s
×
= = =
54.2 psi
c
s
=
3
5.2254 10 1448.32 psi
3.6079
c
ss
P
A
s
×
=−=− =−
1.448 ksi
s
s
= −
consent of McGraw–Hill Education.
PROBLEM 9.40
The steel rails of a railroad track (Es = 200 GPa, αs = 11.7 × 102–6/°C) were laid 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)
63
( ) (11.7 10 )(48 6)(10) 4.914 10 m
TTL
δα
−−
=∆= × − = ×
12
9
(10) 50 10
200 10
P
PL L
AE E
σσ
δσ
−
= = = = ×
×
3 12
4.914 10 50 10 0
TP
δδ δ σ
−−
= + = × +× =
6
98.3 10 Pa
σ
=−×
98.3 MPa
σ
= −
(b)
3 12 3
4.914 10 50 10 3 10
TP
δδ δ σ
−− −
= + = × +× =×
33
12
6
3 10 4.914 10
50 10
38.3 10 Pa
σ
−−
−
×− ×
=×
=−×
38.3 MPa
σ
= −
consent of McGraw–Hill Education.
PROBLEM 9.41
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel
( 200 GPa,
s
E=
6
11.7 10 / C)
s
α
−
=×°
and
portion BC is made of brass
( 105 GPa,
b
E=
6
20.9 10 / C).
b
α
−
=×°
Knowing
that the rod is initially unstressed, determine the compressive force induced in
ABC when there is a temperature rise of
50 C.°
SOLUTION
2 2 2 62
2 2 3 2 32
(30) 706.86 mm 706.86 10 m
44
(50) 1.9635 10 mm 1.9635 10 m
44
AB AB
BC BC
Ad
Ad
ππ
ππ
−
−
= = = = ×
===×=×
Free thermal expansion:
66
6
() ()
(0.250)(11.7 10 )(50) (0.300)(20.9 10 )(50)
459.75 10 m
T AB s BC b
L TL T
δα α
−−
−
= ∆+ ∆
= ×+ ×
= ×
Shortening due to induced compressive force P:
9 69 3
9
0.250 0.300
(200 10 )(706.86 10 ) (105 10 )(1.9635 10 )
3.2235 10
δ
−−
−
= +
= +
×× ××
= ×
Ps AB b BC
PL PL
EA EA
PP
P
For zero net deflection,
PT
δδ
=
96
3
3.2235 10 459.75 10
142.624 10 N
P
P
−−
×= ×
= ×
142.6 kNP=
consent of McGraw–Hill Education.
PROBLEM 9.42
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel
6
( 29 10 psi,= ×
s
E
6
6.5 10 / F)
α
−
=×°
s
and
portion BC is made of aluminum
6
( 10.4 10 psi,= ×
a
E
6
13.3 10 /°F).
α
−
= ×
a
Knowing that the rod is initially unstressed, determine (a) the normal stresses
induced in portions AB and BC by a temperature rise of 70°F, (b) the
corresponding deflection of point B.
SOLUTION
22 2 2
(2.25) 3.9761 in (1.5) 1.76715 in
44
ππ
= = = =
AB BC
AA
Free thermal expansion.
70 FT∆= °
Total:
63
63
3
( ) ( ) (24)(6.5 10 )(70) 10.92 10 in.
( ) ( ) (32)(13.3 10 )(70) 29.792 10 in.
( ) ( ) 40.712 10 in.
δα
δα
δδ δ
−−
−−
−
= ∆= × = ×
= ∆= × = ×
=+=×
T AB AB s
T BC BC a
T T AB T BC
LT
LT
Shortening due to induced compressive force P.
9
6
9
6
24
( ) 208.14 10
(29 10 )(3.9761)
32
( ) 1741.18 10
(10.4 10 )(1.76715)
δ
δ
−
−
= = = ×
×
= = = ×
×
AB
P AB s AB
BC
P BC a BC
PL PP
EA
PL PP
EA
Total:
9
( ) ( ) 1949.32 10
δδ δ
−
=+= ×
P P AB P BC
P
For zero net deflection,
PT
δδ
=
93
1949.32 10 40.712 10
−−
×= ×P
3
20.885 10 lb= ×P
(a)
33
20.885 10 5.25 10 psi
3.9761
AB AB
P
A
s
×
=− =− =−×
5.25 ksi
AB
s
= −
33
20.885 10 11.82 10 psi
1.76715
BC BC
P
A
s
×
=−=− =− ×
11.82 ksi
BC
s
= −
(b)
93 3
( ) (208.14 10 )(20.885 10 ) 4.3470 10 in.
P AB
δ
−−
= × ×= ×
33
( ) ( ) 10.92 10 4.3470 10
B T AB P AB
δδ δ
−−
= →+ ←= × →+ × ←
3
6.57 10 in.
B
δ
−
=×→
or
93 3
( ) (1741.18 10 )(20.885 10 ) 36.365 10 in.
P BC
δ
−−
= × ×= ×
3 33
( ) ( ) 29.792 10 36.365 10 6.57 10 in.
B T BC P BC
δδ δ
− −−
= ←+ →= × ←+ × →= × →
(checks)
consent of McGraw–Hill Education.
PROBLEM 9.43
Solve Prob. 9.42, assuming that portion AB of the composite rod is made of
aluminum and portion BC is made of steel.
PROBLEM 9.42 A rod consisting of two cylindrical portions AB and BC is
restrained at both ends. Portion AB is made of steel
6
( 29 10 psi,
= ×
s
E
6
6.5 10 / F)
α
−
=×°
s
and portion BC is made of aluminum
6
( 10.4 10 psi,
a
E= ×
6
13.3 10 /°F).
α
−
= ×
a
Knowing that the rod is initially unstressed, determine
(a) the normal stresses induced in portions AB and BC by a temperature rise of
70°F, (b) the corresponding deflection of point B.
SOLUTION
22
(2.25) 3.9761 in
4
π
= =
AB
A
22
(1.5) 1.76715 in
4
π
= =
BC
A
Free thermal expansion.
70 FT∆= °
63
63
( ) ( ) (24)(13.3 10 )(70) 22.344 10 in.
( ) ( ) (32)(6.5 10 )(70) 14.56 10 in.
T AB AB a
T BC BC s
LT
LT
δα
δα
−−
−−
= ∆= × = ×
= ∆= × = ×
Total:
3
( ) ( ) 36.904 10 in.
T T AB T BC
δδ δ
−
=+=×
Shortening due to induced compressive force P.
9
6
9
6
24
( ) 580.39 10
(10.4 10 )(3.9761)
32
( ) 624.42 10
(29 10 )(1.76715)
δ
δ
−
−
= = = ×
×
= = = ×
×
AB
P AB a AB
BC
P BC s BC
PL PP
EA
PL PP
EA
Total:
9
( ) ( ) 1204.81 10
δδ δ
−
=+= ×
P P AB P BC
P
For zero net deflection,
PT
δδ
=
93
1204.81 10 36.904 10P
−−
×= ×
3
30.631 10 lbP= ×
(a)
33
30.631 10 7.70 10 psi
3.9761
AB AB
P
A
s
×
=− =− =−×
7.70 ksi
AB
s
= −
33
30.631 10 17.33 10 psi
1.76715
BC BC
P
A
s
×
=−=− =− ×
17.33 ksi
BC
s
= −
consent of McGraw–Hill Education.
PROBLEM 9.43 (Continued)
(b)
93 3
( ) (580.39 10 )(30.631 10 ) 17.7779 10 in.
δ
−−
= × ×= ×
P AB
33
( ) ( ) 22.344 10 17.7779 10
B T AB P AB
δδ δ
−−
= →+ ←= × →+ × ←
3
4.57 10 in.
δ
−
=×→
B
or
93 3
( ) (624.42 10 )(30.631 10 ) 19.1266 10 in.
δ
−−
= × ×= ×
P BC
3 33
( ) ( ) 14.56 10 19.1266 10 4.57 10 in. (checks)
B T BC P BC
δδ δ
− −−
= ←+ →= × ←+ × →= × →
consent of McGraw–Hill Education.
PROBLEM 9.44
Determine (a) the compressive force in the bars shown after a
temperature rise of
180 F,°
(b) the corresponding change in length of
the bronze bar.
SOLUTION
Thermal expansion if free of constraint:
66
3
() ()
(14)(12 10 )(180) (18)(12.9 10 )(180)
72.036 10 in.
δα α
−−
−
= ∆+ ∆
=× +×
= ×
T bb aa
L TL T
Constrained expansion:
0.02 in.
δ
=
Shortening due to induced compressive force P:
33
72.036 10 0.02 52.036 10 in.
P
δ
−−
= ×−= ×
But
ba ba
Pbb aa bb aa
PL PL L L P
EA EA EA EA
δ
=+= +
9
66
14 18 995.36 10
(15 10 )(2.4) (10.6 10 )(2.8)
−
=+=×
××
PP
Equating,
93
3
995.36 10 52.036 10
52.279 10 lb
−−
×= ×
= ×
P
P
(a)
52.3 kipsP=
(b)
()
b
b bb bb
PL
LT
EA
δα
= ∆−
3
63
6
(52.279 10 )(14)
(14)(12 10 )(180) 9.91 10 in.
(15 10 )(2.4)
−−
×
=×− =×
×
3
9.91 10 in.
δ
−
= ×
b
consent of McGraw–Hill Education.
PROBLEM 9.38
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:
6
105 GPa
20.9 10 / C
E
a
−
=
=×°
Aluminum shell:
6
70 GPa
23.6 10 / C
E
a
−
=
=×°
Free thermal expansion:
195 15 180 CT∆= − = °
Brass core:
() ( )
Tb b
LT
δa
= ∆
Aluminum shell:
() ( )
= ∆
Ta a
LT
δa
ab
Brass core:
9
22
62
105 10 Pa
(25) 490.87 mm
4
490.87 10 m
()
b
b
Pb bb
E
A
PL
EA
π
δ
−
= ×
= =
= ×
=
consent of McGraw–Hill Education.
PROBLEM 9.39
The concrete post
6
( 3.6 10
c
E= ×
psi and
6
5.5 10 / F)
α
−
=×°
c
is reinforced with six steel
bars, each of
7
8-in.
diameter
6
( 29 10
s
E= ×
psi and
6
6.5 10 / F).
s
α
−
=×°
Determine the
normal stresses induced in the steel and in the concrete by a temperature rise of 65°F.
SOLUTION
2
22
7
6 6 3.6079 in
4 48
ππ
= = =
s
Ad
22 2
10 10 3.6079 96.392 in= −= − =
cs
AA
Let
c
P=
tensile force developed in the concrete.
For equilibrium with zero total force, the compressive force in the six steel rods equals
.
c
P
Strains:
() ()
cc
s sc c
ss cc
PP
TT
EA EA
ε αε α
=− +∆ = +∆
Matching:
cs
εε
=
() ()
cc
cs
cc ss
PP
TT
EA EA
αα
+ ∆=− + ∆
6
66
3
11 ( )( )
11
(1.0 10 )(65)
(3.6 10 )(96.392) (29 10 )(3.6079)
5.2254 10 lb
αα
−
+ =−∆
+=×
××
= ×
c sc
cc ss
c
c
PT
EA EA
P
P
3
5.2254 10 54.210 psi
96.392
c
cc
P
A
s
×
= = =
54.2 psi
c
s
=
3
5.2254 10 1448.32 psi
3.6079
c
ss
P
A
s
×
=−=− =−
1.448 ksi
s
s
= −
consent of McGraw–Hill Education.
PROBLEM 9.40
The steel rails of a railroad track (Es = 200 GPa, αs = 11.7 × 102–6/°C) were laid 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)
63
( ) (11.7 10 )(48 6)(10) 4.914 10 m
TTL
δα
−−
=∆= × − = ×
12
9
(10) 50 10
200 10
P
PL L
AE E
σσ
δσ
−
= = = = ×
×
3 12
4.914 10 50 10 0
TP
δδ δ σ
−−
= + = × +× =
6
98.3 10 Pa
σ
=−×
98.3 MPa
σ
= −
(b)
3 12 3
4.914 10 50 10 3 10
TP
δδ δ σ
−− −
= + = × +× =×
33
12
6
3 10 4.914 10
50 10
38.3 10 Pa
σ
−−
−
×− ×
=×
=−×
38.3 MPa
σ
= −
consent of McGraw–Hill Education.
PROBLEM 9.41
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel
( 200 GPa,
s
E=
6
11.7 10 / C)
s
α
−
=×°
and
portion BC is made of brass
( 105 GPa,
b
E=
6
20.9 10 / C).
b
α
−
=×°
Knowing
that the rod is initially unstressed, determine the compressive force induced in
ABC when there is a temperature rise of
50 C.°
SOLUTION
2 2 2 62
2 2 3 2 32
(30) 706.86 mm 706.86 10 m
44
(50) 1.9635 10 mm 1.9635 10 m
44
AB AB
BC BC
Ad
Ad
ππ
ππ
−
−
= = = = ×
===×=×
Free thermal expansion:
66
6
() ()
(0.250)(11.7 10 )(50) (0.300)(20.9 10 )(50)
459.75 10 m
T AB s BC b
L TL T
δα α
−−
−
= ∆+ ∆
= ×+ ×
= ×
Shortening due to induced compressive force P:
9 69 3
9
0.250 0.300
(200 10 )(706.86 10 ) (105 10 )(1.9635 10 )
3.2235 10
δ
−−
−
= +
= +
×× ××
= ×
Ps AB b BC
PL PL
EA EA
PP
P
For zero net deflection,
PT
δδ
=
96
3
3.2235 10 459.75 10
142.624 10 N
P
P
−−
×= ×
= ×
142.6 kNP=
consent of McGraw–Hill Education.
PROBLEM 9.42
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel
6
( 29 10 psi,= ×
s
E
6
6.5 10 / F)
α
−
=×°
s
and
portion BC is made of aluminum
6
( 10.4 10 psi,= ×
a
E
6
13.3 10 /°F).
α
−
= ×
a
Knowing that the rod is initially unstressed, determine (a) the normal stresses
induced in portions AB and BC by a temperature rise of 70°F, (b) the
corresponding deflection of point B.
SOLUTION
22 2 2
(2.25) 3.9761 in (1.5) 1.76715 in
44
ππ
= = = =
AB BC
AA
Free thermal expansion.
70 FT∆= °
Total:
63
63
3
( ) ( ) (24)(6.5 10 )(70) 10.92 10 in.
( ) ( ) (32)(13.3 10 )(70) 29.792 10 in.
( ) ( ) 40.712 10 in.
δα
δα
δδ δ
−−
−−
−
= ∆= × = ×
= ∆= × = ×
=+=×
T AB AB s
T BC BC a
T T AB T BC
LT
LT
Shortening due to induced compressive force P.
9
6
9
6
24
( ) 208.14 10
(29 10 )(3.9761)
32
( ) 1741.18 10
(10.4 10 )(1.76715)
δ
δ
−
−
= = = ×
×
= = = ×
×
AB
P AB s AB
BC
P BC a BC
PL PP
EA
PL PP
EA
Total:
9
( ) ( ) 1949.32 10
δδ δ
−
=+= ×
P P AB P BC
P
For zero net deflection,
PT
δδ
=
93
1949.32 10 40.712 10
−−
×= ×P
3
20.885 10 lb= ×P
(a)
33
20.885 10 5.25 10 psi
3.9761
AB AB
P
A
s
×
=− =− =−×
5.25 ksi
AB
s
= −
33
20.885 10 11.82 10 psi
1.76715
BC BC
P
A
s
×
=−=− =− ×
11.82 ksi
BC
s
= −
(b)
93 3
( ) (208.14 10 )(20.885 10 ) 4.3470 10 in.
P AB
δ
−−
= × ×= ×
33
( ) ( ) 10.92 10 4.3470 10
B T AB P AB
δδ δ
−−
= →+ ←= × →+ × ←
3
6.57 10 in.
B
δ
−
=×→
or
93 3
( ) (1741.18 10 )(20.885 10 ) 36.365 10 in.
P BC
δ
−−
= × ×= ×
3 33
( ) ( ) 29.792 10 36.365 10 6.57 10 in.
B T BC P BC
δδ δ
− −−
= ←+ →= × ←+ × →= × →
(checks)
consent of McGraw–Hill Education.
PROBLEM 9.43
Solve Prob. 9.42, assuming that portion AB of the composite rod is made of
aluminum and portion BC is made of steel.
PROBLEM 9.42 A rod consisting of two cylindrical portions AB and BC is
restrained at both ends. Portion AB is made of steel
6
( 29 10 psi,
= ×
s
E
6
6.5 10 / F)
α
−
=×°
s
and portion BC is made of aluminum
6
( 10.4 10 psi,
a
E= ×
6
13.3 10 /°F).
α
−
= ×
a
Knowing that the rod is initially unstressed, determine
(a) the normal stresses induced in portions AB and BC by a temperature rise of
70°F, (b) the corresponding deflection of point B.
SOLUTION
22
(2.25) 3.9761 in
4
π
= =
AB
A
22
(1.5) 1.76715 in
4
π
= =
BC
A
Free thermal expansion.
70 FT∆= °
63
63
( ) ( ) (24)(13.3 10 )(70) 22.344 10 in.
( ) ( ) (32)(6.5 10 )(70) 14.56 10 in.
T AB AB a
T BC BC s
LT
LT
δα
δα
−−
−−
= ∆= × = ×
= ∆= × = ×
Total:
3
( ) ( ) 36.904 10 in.
T T AB T BC
δδ δ
−
=+=×
Shortening due to induced compressive force P.
9
6
9
6
24
( ) 580.39 10
(10.4 10 )(3.9761)
32
( ) 624.42 10
(29 10 )(1.76715)
δ
δ
−
−
= = = ×
×
= = = ×
×
AB
P AB a AB
BC
P BC s BC
PL PP
EA
PL PP
EA
Total:
9
( ) ( ) 1204.81 10
δδ δ
−
=+= ×
P P AB P BC
P
For zero net deflection,
PT
δδ
=
93
1204.81 10 36.904 10P
−−
×= ×
3
30.631 10 lbP= ×
(a)
33
30.631 10 7.70 10 psi
3.9761
AB AB
P
A
s
×
=− =− =−×
7.70 ksi
AB
s
= −
33
30.631 10 17.33 10 psi
1.76715
BC BC
P
A
s
×
=−=− =− ×
17.33 ksi
BC
s
= −
consent of McGraw–Hill Education.
PROBLEM 9.43 (Continued)
(b)
93 3
( ) (580.39 10 )(30.631 10 ) 17.7779 10 in.
δ
−−
= × ×= ×
P AB
33
( ) ( ) 22.344 10 17.7779 10
B T AB P AB
δδ δ
−−
= →+ ←= × →+ × ←
3
4.57 10 in.
δ
−
=×→
B
or
93 3
( ) (624.42 10 )(30.631 10 ) 19.1266 10 in.
δ
−−
= × ×= ×
P BC
3 33
( ) ( ) 14.56 10 19.1266 10 4.57 10 in. (checks)
B T BC P BC
δδ δ
− −−
= ←+ →= × ←+ × →= × →
consent of McGraw–Hill Education.
PROBLEM 9.44
Determine (a) the compressive force in the bars shown after a
temperature rise of
180 F,°
(b) the corresponding change in length of
the bronze bar.
SOLUTION
Thermal expansion if free of constraint:
66
3
() ()
(14)(12 10 )(180) (18)(12.9 10 )(180)
72.036 10 in.
δα α
−−
−
= ∆+ ∆
=× +×
= ×
T bb aa
L TL T
Constrained expansion:
0.02 in.
δ
=
Shortening due to induced compressive force P:
33
72.036 10 0.02 52.036 10 in.
P
δ
−−
= ×−= ×
But
ba ba
Pbb aa bb aa
PL PL L L P
EA EA EA EA
δ
=+= +
9
66
14 18 995.36 10
(15 10 )(2.4) (10.6 10 )(2.8)
−
=+=×
××
PP
Equating,
93
3
995.36 10 52.036 10
52.279 10 lb
−−
×= ×
= ×
P
P
(a)
52.3 kipsP=
(b)
()
b
b bb bb
PL
LT
EA
δα
= ∆−
3
63
6
(52.279 10 )(14)
(14)(12 10 )(180) 9.91 10 in.
(15 10 )(2.4)
−−
×
=×− =×
×
3
9.91 10 in.
δ
−
= ×
b
consent of McGraw–Hill Education.