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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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