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where a and b are the semi-axes. The stress function
ψ
must be constant along the boundary
of the cross-section. We choose to let
ψ =
0 on the boundary. A stress function that would
satisfy this condition would be
C y 2
1
z 2
b 2
ψ =
a 2 +
(2)
where C is a constant to be determined. Substitution of this stress function into Poisson's
equation of Eq. (1.159) gives
a 2 b 2
a 2
φ
C
=−
b 2 G
(3)
+
From (2) and (3), the stress function is now
φ y 2
1
a 2 b 2
a 2
z 2
b 2
ψ =−
b 2 G
a 2 +
(4)
+
The stress components for this elliptical cross-section are given by Eq. (1.155) with (4)
2 a 2 z
a 2
2 b 2 y
a 2
= ∂ψ
=− ∂ψ
φ
φ
τ
z =−
b 2 G
τ
y =
b 2 G
(5)
xy
xz
+
+
The maximum shear stress occurs at the boundary closest to the centroid, i.e., at z
=
b
.
2 a 2 b
a 2
φ
τ
= τ
|
=−
b 2 G
(6)
max
xy
z
=
b
+
From Eq. (1.163) the torque becomes
φ 1
2
a 2
A
b 2
A
a 2 b 2
a 2
1
y 2 dy dz
z 2 dy dz
M t =
A ψ
dy dz
=−
b 2 G
+
+
dy dz
A
φ I yy
A
a 2 b 2
a 2
I zz
b 2
=−
b 2 G
a 2 +
(7)
+
where
y 2 dy dz,
z 2 dy dz,
I yy
=
I zz
=
(8)
A
A
= A dy dz
a 3 b
ab 3
and A
.
For an ellipse, I yy
= π
/
4 ,I zz
= π
/
4 and A
= π
ab
.
If these values
are placed in (7), we find
a 3 b 3
π
φ
M t
=
G
(9)
(
a 2
+
b 2
)
Then the stresses of (5) become
2
ab 3
2
a 3 b
τ
=−
M t z
τ
=
M t y
(10)
xy
xz
π
π
φ =
From
M t
/
GJ and (9),
a 3 b 3
= π
J
(11)
a 2
+
b 2
These results are the same obtained using the displacement formulation with a warping
function (Problem 1.63).
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