sm9 63

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PROBLEM (9.63) Reconsider the cantilever beam illustrated in Fig. P9.61and described in
Prob. 9.61 to calculate he maximum shear stress max
τ
at point B.
SOLUTION
The internal forces and moments and area properties at the section containing
points A and B are as found in the solution of Prob. 9.61.
At point B (Fig. a):
(60 25) 35
B
y
mm=− − =−
40
B
zmm=−
yB
zB
xzy
M
z
MyP
AI I
σ =− +
33 3
66 6
100(10 ) 3.3(10 )( 0.035) 3(10 )( 0.04)
9600(10 ) 11.52(10 ) 5.12(10 )
−− −
−−
=+ +
10.417 10.026 23.438 23.05
M
Pa=−−=−
See Figure (b):
y
xy z
VQ
I
b
τ =
Where
2
15 11.52
y
VkNI kNm==⋅ 80bmm
=
25
* (80 25)(35 )
2
QAy==× +
33
95(10 ) mm=
and
36
6
15(10 )(95 10 ) 1.546
11.52(10 )(0.08)
xy
M
Pa τ
−
−
×
==
Therefore
22 2 2
max ( ) (11.525) (1.546)
2xxy
σ
ττ
=+= +
11.63
M
Pa=
x
σ
B
x
y
τ
Figure (a)
35 mm
C
y
z
80 mm
60 mm
15 kN
Figure (b)
B
y
A
*