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PROBLEM 9.80
The assembly shown consists of an aluminum shell
6
( 10.6 10 psi,= ×
a
E
a
a = 12.9 × 10−6/°F) fully bonded to a steel core
6
( 29 10 psi,= ×
s
E
a
s = 6.5 × 10−6/°F) and is unstressed. Determine (a) the largest allowable
change in temperature if the stress in the aluminum shell is not to exceed
6 ksi, (b) the corresponding change in length of the assembly.
SOLUTION
Since
,
as
aa
>
the shell is in compression for a positive temperature rise.
Let
3
6 ksi 6 10 psi
a
s
=− =−×
( )
22 2 2 2
(1.25 0.75 ) 0.78540 in
44
ππ
= −= − =
a oi
A dd
22 2
(0.75) 0.44179 in
44
ππ
= = =
s
Ad
aa ss
P AA
ss
=−=
where P is the tensile force in the steel core.
33
(6 10 )(0.78540) 10.667 10 psi
0.44179
aa
ss
A
A
s
s
×
=−= = ×
33
63
66
() ()
( )( )
10.667 10 6 10
(6.4 10 )( ) 0.93385 10
29 10 10.6 10
sa
sa
sa
sa
as sa
TT
EE
TEE
T
ss
εa a
ss
aa
−−
= + ∆= + ∆
− ∆= −
××
× ∆= + = ×
××
(a)
145.91 FT∆= °
145.9 F∆= °T
(b)
363
6
10.667 10 (6.5 10 )(145.91) 1.3163 10
29 10
ε
−−
×
= +× = ×
×
or
363
6
6 10 (12.9 10 )(145.91) 1.3163 10
10.6 10
ε
−−
−×
= +× = ×
×
3
(8.0)(1.3163 10 ) 0.01053 in.
δε
−
== ×=L
0.01053 in.
δ
=
consent of McGraw-Hill Education.
PROBLEM 9.81
The block shown is made of a magnesium alloy, for which
45E=
GPa and
0.35.v=
Knowing that
180
x
σ
= −
MPa,
determine (a) the magnitude of
y
σ
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)
6
000
1()
(0.35)( 180 10 )
yy z
y xyz
yx
vv
E
v
δεσ
ε σσσ
σσ
= = =
= −−
= = −×
6
63 10 Pa=−×
63.0 MPa
y
σ
= −
63
9
63
9
1 (0.35)( 243 10 )
( ) ( ) 1.890 10
45 10
1 157.95 10
( ) 3.510 10
45 10
ε σ σ σ σσ
σσ
ε σσσ
−
−
−×
= −− =− +=− =+ ×
×
−×
= −− = =− =− ×
×
z z x y xy
xy
x xyZ
v
vv
EE
v
vv
EE
(b)
0
0
(1 ) (1 ) (1 )
( )()
xz
xxzzxzxzxz
xz x z xz xz x z
A LL
A L L LL
A A A LL LL
ε ε ε ε εε
εεεε εε
=
= + + = +++
∆=− = + + ≈ +
33
(100 mm)(25 mm)( 3.510 10 1.890 10 )A
−−
∆= − × + ×
2
4.05 mm∆=−A
(c)
0
0
(1 ) (1 ) (1 )
(1 )
( small terms)
εεε
ε ε ε εε εε εε εεε
εεε
=
=+++
= ++++ + + +
∆=− = + + +
xyz
x xy yz z
xyz x y z xy yz zx xyz
xyz x y z
V LLL
VL L L
LLL
V V V LLL
33
(100)(40)(25)( 3.510 10 0 1.890 10 )
−−
∆= − × ++ ×
V
3
162.0 mm∆=−
V
PROBLEM 9.82
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
all
220
τ
=
psi and
1800G=
psi. Knowing that a centric vertical force
of magnitude
3.2P=
kips must cause a 0.1-in. vertical deflection of
the plate AB, determine the smallest allowable dimensions a and b of the
consent of McGraw-Hill Education.
PROBLEM 9.83
A hole is to be drilled in the plate at A. The diameters of the
bits available to drill the hole range from
1
2
to 11/2 in. in
1
4
-in.
increments. If the allowable stress in the plate is 21 ksi,
determine (a) the diameter d of the largest bit that can be
used if the allowable load P at the hole is to exceed that at
consent of McGraw-Hill Education.
PROBLEM 9.84
(a) For
13 kipsP=
and
1
2
in.,d=
determine the maximum
stress in the plate shown. (b) Solve part a, assuming that the
hole at A is not drilled.
SOLUTION
consent of McGraw-Hill Education.
PROBLEM 9.80
The assembly shown consists of an aluminum shell
6
( 10.6 10 psi,= ×
a
E
a
a = 12.9 × 10−6/°F) fully bonded to a steel core
6
( 29 10 psi,= ×
s
E
a
s = 6.5 × 10−6/°F) and is unstressed. Determine (a) the largest allowable
change in temperature if the stress in the aluminum shell is not to exceed
6 ksi, (b) the corresponding change in length of the assembly.
SOLUTION
Since
,
as
aa
>
the shell is in compression for a positive temperature rise.
Let
3
6 ksi 6 10 psi
a
s
=− =−×
( )
22 2 2 2
(1.25 0.75 ) 0.78540 in
44
ππ
= −= − =
a oi
A dd
22 2
(0.75) 0.44179 in
44
ππ
= = =
s
Ad
aa ss
P AA
ss
=−=
where P is the tensile force in the steel core.
33
(6 10 )(0.78540) 10.667 10 psi
0.44179
aa
ss
A
A
s
s
×
=−= = ×
33
63
66
() ()
( )( )
10.667 10 6 10
(6.4 10 )( ) 0.93385 10
29 10 10.6 10
sa
sa
sa
sa
as sa
TT
EE
TEE
T
ss
εa a
ss
aa
−−
= + ∆= + ∆
− ∆= −
××
× ∆= + = ×
××
(a)
145.91 FT∆= °
145.9 F∆= °T
(b)
363
6
10.667 10 (6.5 10 )(145.91) 1.3163 10
29 10
ε
−−
×
= +× = ×
×
or
363
6
6 10 (12.9 10 )(145.91) 1.3163 10
10.6 10
ε
−−
−×
= +× = ×
×
3
(8.0)(1.3163 10 ) 0.01053 in.
δε
−
== ×=L
0.01053 in.
δ
=
consent of McGraw-Hill Education.
PROBLEM 9.81
The block shown is made of a magnesium alloy, for which
45E=
GPa and
0.35.v=
Knowing that
180
x
σ
= −
MPa,
determine (a) the magnitude of
y
σ
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)
6
000
1()
(0.35)( 180 10 )
yy z
y xyz
yx
vv
E
v
δεσ
ε σσσ
σσ
= = =
= −−
= = −×
6
63 10 Pa=−×
63.0 MPa
y
σ
= −
63
9
63
9
1 (0.35)( 243 10 )
( ) ( ) 1.890 10
45 10
1 157.95 10
( ) 3.510 10
45 10
ε σ σ σ σσ
σσ
ε σσσ
−
−
−×
= −− =− +=− =+ ×
×
−×
= −− = =− =− ×
×
z z x y xy
xy
x xyZ
v
vv
EE
v
vv
EE
(b)
0
0
(1 ) (1 ) (1 )
( )()
xz
xxzzxzxzxz
xz x z xz xz x z
A LL
A L L LL
A A A LL LL
ε ε ε ε εε
εεεε εε
=
= + + = +++
∆=− = + + ≈ +
33
(100 mm)(25 mm)( 3.510 10 1.890 10 )A
−−
∆= − × + ×
2
4.05 mm∆=−A
(c)
0
0
(1 ) (1 ) (1 )
(1 )
( small terms)
εεε
ε ε ε εε εε εε εεε
εεε
=
=+++
= ++++ + + +
∆=− = + + +
xyz
x xy yz z
xyz x y z xy yz zx xyz
xyz x y z
V LLL
VL L L
LLL
V V V LLL
33
(100)(40)(25)( 3.510 10 0 1.890 10 )
−−
∆= − × ++ ×
V
3
162.0 mm∆=−
V
PROBLEM 9.82
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
all
220
τ
=
psi and
1800G=
psi. Knowing that a centric vertical force
of magnitude
3.2P=
kips must cause a 0.1-in. vertical deflection of
the plate AB, determine the smallest allowable dimensions a and b of the
consent of McGraw-Hill Education.
PROBLEM 9.83
A hole is to be drilled in the plate at A. The diameters of the
bits available to drill the hole range from
1
2
to 11/2 in. in
1
4
-in.
increments. If the allowable stress in the plate is 21 ksi,
determine (a) the diameter d of the largest bit that can be
used if the allowable load P at the hole is to exceed that at
consent of McGraw-Hill Education.
PROBLEM 9.84
(a) For
13 kipsP=
and
1
2
in.,d=
determine the maximum
stress in the plate shown. (b) Solve part a, assuming that the
hole at A is not drilled.
SOLUTION
consent of McGraw-Hill Education.
