sm7 135 136

________________________________________________________________________
PROBLEMS (*7.135 and 7.136) Determine the distance e that locates the shear center S of a
thin-walled beam of cross section shown in Figs. P7.135 and P7.136.
SOLUTION (*7.135)
y
From A to B:
2
2
10
22
AB AB y A
z
s
Qts V
I
ττ == =
23
1
23
ByABB
zz
bb
VF bt
6
y
t
V
I
I
ττ ===
From B to D:
2
5
() 2()
222
Dbb
Qtb tb bt=+ =
22
52 5
(2 ) 4
D
yy
zz
bt b
VV
It I
τ ==
22 3
115
( ) (2 ) ( )2
2224
y
BD B D y
zz
V
bb bt
Fbt tb
II
ττ =+ =+ =
7
4
V
Then
3
0: 2 , (1 )
3
yDFAByDF z
bt
FFFVF I =−= =+
y
V
∑
0: 2 ( )
SDFBDAB
M
FeFbFbe=⋅=++
∑
or
(2
2
BD AB
DF AB
bF F )
FF
+
=−
e
Substituting the forces & simplifying:
4
25
12 z
bt
e
I
=
Vy
z
S
C
2
t
e
A
F
BD
F
D
F
FAB
F
F
G
t
B
D
A
E
s
b
b/2 b/2
b
D
τ
F
G
B
D
A
E
D
τ
B
τ
E
τ
1
13B
Ab
τ
=
(See : Table B.6)
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Continued on next slide

where 32
11
2[ ( )(2 ) (2 )( ) ]
12 3 3
0
z
I
tb btb bt==+
Thus, 5
8
eb =
SOLUTION (7.136)
y
3
I
rt
π
= (by Table B.6)
2
0(sin)( ) (1 cos)Qr trdrt
θ
φ
φθ =⋅=−
∫
Hence
(1 cos )VQ V dA t rd
Ib rt
θ
τ
θ
π
−
== =⋅
Thus
2
0
(1 cos )
() 2
cVr
TVe dAr
π
Vr
θ
τπ
−
== =
∫∫
from which
2er=
z
θ
d
φ
φ
V
r
S
C
e
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.