sm7 128

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PROBLEM (7.128) A channel section of uniform thickness is loaded as shown in Fig. P7.128.
Calculate:
(a) The distance e to the shear center.
(b) The shearing stress at D.
(c) The maximum shearing stress.
Given: b = 120 mm, h = 180 mm, t = 5 mm, Vy = 2 kN
SOLUTION
Equation (h) of Example 7.18:
y
32
11
12 2
z
I
th bth=+
3
1(5)(180)
12
=
+2
1(120 5)(180)
2×
46
1015.12 mm×=
(a) Equation (7.52):
120 43.64
2 ( 3 ) 2 (180 240)
b
em
hb
== =
++ m
(b) 3
6
2 10 (0.12)(0.09) 1.78
12.15 10
Dy
bc
VM
I
τ
−
×
== =
×Pa
(c) The first moment of the shaded section about the z-axis:
11
(2) (4) (4 )
28
z
Qbth hth htbh=+ =+ (1)
Shear stress formula is thus:
max 2
(8)(4 )
1(6 )
12
yz y
z
VQ V ht b h
It th b h t
τ
+
==
+
3(4 )
2(6 )
y
Vbh
th b h
+
=+ (7.51)
Substitute the given data
3
max 3(2 10 )(4 0.12 0.18)
2(0.05 0.18)(6 0.12 0.18)
τ
××+
=××+ 244 kPa
=
z
S C
D
t
e h/2
h/2
Vy
b
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