7.6 Torsion of Elliptical Cross-Section
237
θ =
M t
G I P
=
M t
G J
Therefore,
M t = G J θ
= Gθ
πa
3 b
3
a 2 + b 2
or
θ =
M t
G
a
2 +b
2
πa 3 b 3
The shearing stresses are given by
τ yz = Gθ
∂ψ
∂ y
+ x
= M t
a
2
+ b
2
πa 3 b 3
b
2
− a
2
b 2 + a 2 + 1
x
or
τ yz =
2M t x
πa 3 b
Similarly,
τ xz =
2M t y
πab 3
Therefore, the resultant shearing stress at any point (x, y) is
τ =
τ 2
yz + τ 2
xz =
2M t
πa 3 b 3
b
4 x
2
+ a
4 y
2
1
2
(7.20)
Determination of Maximum Shear Stress
To determine where the maximum shear stress occurs, substitute for x
2 from
x
2
a 2 +
y
2
b 2 = 1,
or
x
2
= a
2
1 − y
2
/b
2
and