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Problem 5.5
Given: By means of Eq. 5.6 and 5.7 and the formula for shear stress in a beam of rectangular cross
section given in Problem 5.4, show that the correction coefficient k is 1.20.
Solution:
UV
U
0
⌠
⎮
⎮
⌡
d
=
Eq. 5.6
U
0
1
2E
⋅
σ
xx
2
σ
yy
2
+
σ
zz
2
+
⎛
⎝
⎞
⎠
⋅
ν
E
σ
xx
σ
yy
⋅
σ
yy
σ
zz
⋅
+
σ
xx
σ
zz
⋅
+
()
⋅
−
1
2G
⋅
σ
xy
2
σ
zy
2
+
σ
xz
2
+
⎛
⎝
⎞
⎠
⋅
+
=
Eq. 5.7
From Problem 5.4
σ
zy
τ
max
1
4y
2
⋅
h
2
−
⎛
⎜
⎝
⎞
⎟
⎠
⋅
=
3V
y
⋅
2A
⋅
1
2
⋅
h
2
−
⎛
⎜
⎝
⎞
⎟
⎠
⋅
=
h is the hieght of the beam
P
Q
z
y
V
V
M
M
dz
x
y
dA
b
h
U
s
V
σ
zy
2
⋅
⌠
⎮
⎮
⎮
⌡
d
=
z
A
9V
y
2
⋅
8A
2
⋅
G
⋅
1
2
⋅
h
2
−
⎛
⎜
⎝
⎞
⎟
⎠
2
⋅
⌠
⎮
⎮
⎮
⎮
⌡
d
⌠
⎮
⎮
⎮
⎮
⌡
d
=
Let
dA
b dy
⋅
=
where b is the thickness of the beam.
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 Spring '10
 MechanicsofSolids
 Shear, Stress

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