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FIGURE 7.3
Bar of elliptical cross-section.
Introduction of the strains and stresses of Eqs. (1.142) and (1.143) leads to
G
∂ω
∂
z
∂
∂
φ
∂ω
∂
z
x
δφ
+
φ
∂
−
y
+
y
δω
y
+
∂
V
∂ω
∂
y
∂
∂
φ
∂ω
∂
y
dV
x
δφ
+
φ
∂
+
z
−
z
δω
z
−
+
m
x
δφ
dx
=
0
(2)
∂
L
This can be separated into two terms that are equal to zero, one containing
δφ
and the other
with
δω
. The
δω
terms appear as
φ
2
∂
∂
y
dV
y
δω
∂ω
y
+
∂
z
δω
∂ω
z
+
∂
−
∂
∂
G
y
δω
z
z
δω
∂
∂
∂
∂
V
∂
∂
∂
∂
y
dA
G
L
φ
2
dx
y
δω
∂ω
y
+
∂
z
δω
∂ω
−
∂
∂
=
+
y
δω
z
z
δω
=
0
(3)
∂
∂
∂
z
A
Thus, the equation that can be solved for
ω
, and that will permit the cross-sectional char-
acteristics to be computed, takes the form
∂
∂
∂
∂
y
dA
y
δω
∂ω
y
+
∂
z
δω
∂ω
−
∂
∂
+
y
δω
z
z
δω
=
0
(4)
∂
∂
∂
z
A
or
A
δω
[
(
y
∂∂
y
+
z
∂∂
z
)ω
+
(
y
∂
z
−
z
∂
y
)
]
dA
=
0
(5)
Suppose the trial solution is
ω
=
ω
yz
(6)
This volume integral should now be a (cross-sectional) surface integral. Substitution of (6)
in (5) gives
[
z
2
y
2
δω
(ω
+
1
)
+
(ω
−
1
)
]
dA
=
0
(7)
A
or
[
z
2
y
2
(ω
+
1
)
+
(ω
−
1
)
]
dA
=
0
(8)
A
For an ellipse,
z
2
dA
ab
3
y
2
dA
a
3
b
=
π/
4
,
=
π/
4
A
A
and
ab
3
a
3
b
a
2
b
2
π
π
−
(ω
+
1
)
+
(ω
−
1
)
=
0or
ω
=
(9)
4
4
a
2
+
b
2
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