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Shear Stresses in Circular Beams
Problem 5.91
A wood pole of solid circular cross section (
d
5
diameter) is subjected to a horizontal force
P
5
450 lb (see figure). The
length of the pole is
L
5
6 ft, and the allowable stresses in the wood are
1900 psi in bending and 120 psi in shear.
Determine the minimum required diameter of the pole based upon (a)
the allowable bending stress, and (b) the allowable shear stress.
Solution 5.91
Wood pole of circular cross section
338
CHAPTER 5
Stresses in Beams (Basic Topics)
L
P
d
d
P
5
450 lb
L
5
6 ft
5
72 in.
s
allow
5
1900 psi
t
allow
5
120 psi
Find diameter
d
(a) B
ASED UPON BENDING STRESS
M
max
5
PL
5
(450 lb)(72 in.)
5
32,400 lbin.
d
min
5
5.58 in.
(b) B
ASED UPON SHEAR STRESS
V
max
5
450 lb
d
min
5
2.52 in.
(Bending stress governs.)
t
5
4
V
3
A
5
16
V
3
p
d
2
Ê
d
2
5
16
V
max
3
p
t
allow
5
6.366 in.
2
s
5
M
S
5
32
M
p
d
3
Ê
d
3
5
32
M
max
p
s
allow
5
173.7 in.
3
L
P
d
d
Problem 5.92
A simple log bridge in a remote area consists of two
parallel logs with planks across them (see figure). The logs are Douglas
fir with average diameter 300 mm. A truck moves slowly across the
bridge, which spans 2.5 m. Assume that the weight of the truck is equally
distributed between the two logs.
Because the wheelbase of the truck is greater than 2.5 m, only one set
of wheels is on the bridge at a time. Thus, the wheel load on one log is
equivalent to a concentrated load
W
acting at any position along the span.
In addition, the weight of one log and the planks it supports is equivalent
to a uniform load of 850 N/m acting on the log.
Determine the maximum permissible wheel load
W
based upon (a) an
allowable bending stress of 7.0 MPa, and (b) an allowable shear stress of
0.75 MPa.
W
2.5 m
x
850 N/m
300 m
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View Full DocumentSolution 5.92
Log bridge
SECTION 5.9
Shear Stresses in Circular Beams
339
Diameter
d
5
300 mm
s
allow
5
7.0 MPa
t
allow
5
0.75 MPa
Find allowable load
W
(a) B
ASED UPON BENDING STRESS
Maximum moment occurs when wheel is at midspan
(
x
5
L
/2).
5
0.625
W
1
664.1 (N
?
m)
(
W
5
newtons)
M
max
5
S
s
allow
5
(2.651
3
10
2
3
m
3
)(7.0 MPa)
5
18,560 N
?
m
[
0.625
W
1
664.1
5
18,560
W
5
28,600 N
5
28.6 kN
S
5
p
d
3
32
5
2.651
3
10
2
3
m
3
M
max
5
WL
4
1
qL
2
8
5
W
4
(2.5 m)
1
1
8
(850 N
/
m)(2.5 m)
2
(b) B
ASED UPON SHEAR STRESS
Maximum shear force occurs when wheel is adjacent
to support (
x
5
0).
5
W
1
1062.5 N
(
W
5
newtons)
5
39,760 N
[
W
1
1062.5 N
5
39,760 N
W
5
38,700 N
5
38.7 kN
V
max
5
3
A
t
allow
4
5
3
4
(0.070686 m
2
)(0.75 MPa)
t
max
5
4
V
max
3
A
A
5
p
d
2
4
5
0.070686 m
2
V
max
5
W
1
qL
2
5
W
1
1
2
(850 N
/
W
L
= 2.5 m
x
q
= 850 N/m
Problem 5.93
A sign for an automobile service station is supported by
two aluminum poles of hollow circular cross section, as shown in the
figure. The poles are being designed to resist a wind pressure of 75 lb/ft
2
against the full area of the sign. The dimensions of the poles and sign are
h
1
5
20 ft,
h
2
5
5 ft, and
b
5
10 ft. To prevent buckling of the walls of
the poles, the thickness
t
is specified as onetenth the outside diameter
d
.
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 Spring '08
 Muliana

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