106
CHAPTER 2
Axially Loaded Members
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
This preview has intentionally blurred sections. Sign up to view the full version.
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).
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Sharobeam
 mechanics, Trigraph, Tensile strength, Compressive stress, Tensile stress, Eqs.

Click to edit the document details