sm7 43

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PROBLEM (7.43) A simple beam with two overhangs is loaded as shown in Fig. P7.43, with w = 1.2
kips/ft. Determine:
(a) The shearing stress at a point 1.5 ft to the left of B and ¾ in. below the top of the beam.
(b) The maximum shear stress in the beam.
SOLUTION
From solution of Prob. 7.31: .763.1.63.13 4inyinI ==
(a) 3
2.7 10 [0.75 1(2.237 0.375)] 277
13.63(1)
VQ
p
si
Ib
τ
××−
== =
(b) 3
4.5 10 [3 1(3 2.237)] 756
13.63(1)
EVQ
p
si
Ib
τ
××−
== =
We have, at the neutral axis:
.503.2)2237.2(237.21 inQz
=
×=
3
4.5 10 (2.503) 826
13.63(1)
zVQ
p
si
Ib
τ
×
== =
Thus,
psi826
max =
τ
1 in. 1 in.
1 in.
1.5
ft
C
B
V, kips
x
3 in.
A
-3.6
3.6
7.5 f
t
1.2 ki
p
s/f
t
1.5 in.
y
3 ft
z
2.237 in.
C
4.5
-4.5
8.1 kips
8.1 ki
p
s 1.763 in.
E
¾in
V=-2.7 kips
(from geometry)
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