________________________________________________________________________
PROBLEMS (7.64 through 7.66) The maximum shearing stress in a beam with circular, thin-
walled circular, or triangular crosss section (Fig. P7.64 through P7.66) acted upon by a vertical force
V may be expressed as follows:
max V
kA
τ
= (P7.64)
Here A is the cross-sectional area of the beam and k represents a numerical factor, or shear
coefficient. Determine:
(a) The location of the point at which max
τ
occurs.
(b) The value of k.
(c) The maximum vertical shear force V the section may carry for c = 2 in., b = h = 4 in., r = 1
in., t = 0.1 in., and max
τ
= 9 ksi.
SOLUTION (7.64)
2c
4
3c
π
(a) is maximum at the center. Q
Thus max
τ
occurs at the center.
(b) 24 4cAcI
ππ
==
23
max 1142
()
2233
c
QAyc
ππ
== =
c
3
max
max 42
(2 3) 4
(4)(2)3
VQ Vc V
I
bcc
τ
c
π
π
== =
44
33
Vk
A
==
(c) 32
max 910()(2) 84.82
43
A
Vk
k
τπ
×
== = ips
SOLUTION (7.65)
(a) Q is maximum at the center. So w
τ
occurs at the center.
(b) Refer to Table B.6.
z
y
y
t
C
r
For a thin tube:
32
I
rt A rt
π
π
==
For a semicircular thin tube.
2r
Art y
π
π
== 2bt
=
2
2QAy rt==
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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Continued on next slide
Therefore, for a thin tube:
2
max 3
(2 ) 2
()(2)
VQ V r t V
I
brtt
τπ
== =
A
where
2k=
(c) 3
max 9(10 ) (2 1 0.1)
2
A
VK
τπ
== ××
= 2.83 kips
SOLUTION (7.66)
(a)
3
1
36
I
bh=
y
tb
h
=
VQ
I
t
τ
=
2
1
()1
2()
3
yt y
Q
y
hy
tt
==−
In order
τ
be maximum:
2
1
() [( )]
3
ddVQVdyh y
dy dy It I dy 0
τ
== −=
or
1
2
y
h=
(b) For :2hy =
22
max 13
()
32 4 12
Vh h Vh V
h
I
Ibh
τ
=−==
33
22
Vk
A
==
y
z
22
33
hy
y=−
b
C
h/3
2y/3
t
h
y
z’
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.
Continued on next slide
(c) 3
max 91044 48
32 2
A
Vk
k
τ
××
== =
ips
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.
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