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106
CHAPTER 2
Axially Loaded Numbers
Problem 2.53
A rigid bar of weight
W
5
750 lb hangs from three
equally spaced wires, two of steel and one of aluminum (see figure).
The diameter of the wires is
1
/
8
in. Before they were loaded, all three
wires had the same length.
What temperature increase
D
T
in all three wires will result in the
entire load being carried by the steel wires? (Assume
E
s
5
30
3
10
6
psi,
a
s
5
6.5
3
10
2
6
±°F, and
a
a
5
12
3
10
2
6
±°F.)
Solution 2.53
Bar supported by three wires
W
= 750 lb
SAS
S
5
steel
A
5
aluminum
W
5
750 lb
E
s
5
30
3
10
6
psi
E
s
A
s
5
368,155 lb
a
s
5
6.5
3
10
2
6
±
8
F
a
a
5
12
3
10
2
6
±
8
F
L
5
Initial length of wires
d
1
5
increase in length of a steel wire due to
temperature increase
D
T
5
a
s
(
D
T
)
L
A
s
5
p
d
2
4
5
0.012272
in.
2
d
5
1
8
in.
d
2
5
increase in length of a steel wire due to load
W
±2
d
3
5
increase in length of aluminum wire due to
temperature increase
D
T
5
a
a
(
D
T
)
L
For no load in the aluminum wire:
d
1
1
d
2
5
d
3
or
Substitute numerical values:
N
OTE
: If the temperature increase is larger than
D
T
,
the aluminum wire would be in compression, which
is not possible. Therefore, the steel wires continue to
carry all of the load. If the temperature increase is
less than
D
T
, the aluminum wire will be in tension
and carry part of the load.
5
185
8
F
Ê
—
¢
T
5
750
lb
(2)(368,155
lb)(5.5
3
10
2
6
/
8
F)
¢
T
5
W
2
E
s
A
s
(
a
a
2
a
s
)
Ê
—
a
s
(
¢
T
)
L
1
WL
2
E
s
A
s
5
a
a
(
¢
T
)
L
5
WL
2
E
s
A
s
W
Rigid
Bar
W
2
W
2
d
3
d
1
d
2
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View Full DocumentSECTION 2.5
Thermal Effects
107
Problem 2.54
A steel rod of diameter 15 mm is held snugly (but
without any initial stresses) between rigid walls by the arrangement
shown in the figure.
Calculate the temperature drop
D
T
(degrees Celsius) at which
the average shear stress in the 12mm diameter bolt becomes 45 MPa.
(For the steel rod, use
a
5
12
3
10
2
6
/°C and
E
5
200 GPa.)
Solution 2.54
Steel rod with bolted connection
15 mm
12 mm diameter bolt
R
5
rod
B
5
bolt
P
5
tensile force in steel rod due to temperature drop
D
T
A
R
5
crosssectional area of steel rod
From Eq. (217) of Example 27:
P
5
EA
R
a
(
D
T
)
Bolt is in double shear.
V
5
shear force acting over one cross section of the
bolt
t
5
average shear stress on cross section of the bolt
A
B
5
crosssectional area of bolt
t
5
V
A
B
5
EA
R
a
(
¢
T
)
2
A
B
V
5
P
/
2
5
1
2
EA
R
a
(
¢
T
)
S
UBSTITUTE NUMERICAL VALUES
:
t
5
45 MPa
d
B
5
12 mm
d
R
5
15 mm
a
5
12
3
10
2
6
/
8
C
E
5
200 GPa
¢
T
5
24
8
C
Ê
—
¢
T
5
2(45
MPa)(12
mm)
2
(200
GPa)(12
3
10
2
6
/
8
C)(15
mm)
2
¢
T
5
2
t
d
B
2
E
a
d
R
2
—
A
R
5
p
d
R
2
4
Ê
where
d
R
5
diameter
of
steel
rod
A
B
5
p
d
B
2
4
Ê
where
d
B
5
diameter
of
bolt
Solve
for
¢
T
:
¢
T
5
2
t
A
B
EA
R
a
15 mm
12 mm diameter bolt
B
R
Problem 2.55
A bar
AB
of length
L
is held between rigid supports and
heated nonuniformly in such a manner that the temperature increase
D
T
at
distance
x
from end
A
is given by the expression
D
T
5 D
T
B
x
3
/
L
3
, where
D
T
B
is the increase in temperature at end
B
of the bar (see figure).
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This note was uploaded on 03/19/2008 for the course E M 316 taught by Professor Korkolis during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Korkolis

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