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Problem 6.15
Derive the relation for the shear stress
distribution on the x axis for the equilateral
triangle in Fig. 6.10.
Solution:
From Eq. 6.48, page 215:
ϕ
G
θ
⋅
2h
⋅
x3
y
⋅
−
⋅
3
−
⎛
⎝
⎞
⎠
⋅
y
⋅
+
⋅
3
−
⎛
⎝
⎞
⎠
⋅
x
h
3
+
⎛
⎝
⎞
⎠
⋅
=
From the definition of the Prandtl Stress Funcion
σ
zy
x
ϕ
d
d
−
=
σ
zy
x
G
θ
⋅
⋅
y
⋅
−
⋅
3
−
⎛
⎝
⎞
⎠
⋅
y
⋅
+
⋅
3
−
⎛
⎝
⎞
⎠
⋅
x
h
3
+
⎛
⎝
⎞
⎠
⋅
⎡
⎣
⎤
⎦
d
d
−
=
simplify
σ
zy
G
θ
⋅
⋅
x
⋅
3x
2
⋅
−
3y
2
⋅
+
()
⋅
⋅
=
→
Along the x axis, y=0 and thus:
σ
zy
G
θ
⋅
⋅
2x
⋅
h
⋅
2
−
⋅
=
σ
zx
y
ϕ
d
d
=
σ
zx
y
G
θ
⋅
⋅
y
⋅
−
⋅
3
−
⎛
⎝
⎞
⎠
⋅
y
⋅
+
⋅
3
−
⎛
⎝
⎞
⎠
⋅
x
h
3
+
⎛
⎝
⎞
⎠
⋅
⎡
⎣
⎤
⎦
d
d
=
simplify
σ
zx
G
θ
⋅
y
⋅
h3
x
⋅
+
⋅
h
−
=
→
Along the x axis, y=0 and thus:
σ
zx
0
=
Here is some extra work:
Note that at
x
h
3
−
=
σ
zy
G
θ
⋅
⋅
2
h
3
−
⎛
⎝
⎞
⎠
⋅
h
⋅
3
h
3
−
⎛
⎝
⎞
⎠
2
−
⎡
⎢
⎣
⎤
⎥
⎦
⋅
=
G
θ
⋅
h
⋅
2
2
3
−
1
3
−
⎛
⎝
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 Spring '10
 MechanicsofSolids
 Shear, Stress

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