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that is,
x
δφ
x
δφ
∂τ
dA d x
y
+
∂τ
xy
xz
−
ω(τ
xz
a
z
+
τ
xy
a
y
)
ds dx
+
A
ω
∂
∂
z
x
φ
∂τ
dA
y
+
∂τ
xy
xz
+
δ
A
ω
−
δ
ω(τ
xz
a
z
+
τ
xy
a
y
)
ds
∂
∂
z
xz
dA
dA
dx
1
G
δτ
1
G
δτ
−
τ
+
τ
+
A
(
−
δτ
xy
z
+
δτ
xz
y
)
xy
xy
xz
A
+
[
(
M
−
M
t
)δφ
]
L
−
[
(φ
−
φ)δ
M
t
]
0
=
0
(15)
From Eq. (4), and Chapter 1, Eqs. (1.142) and (1.143),
δφ
ω(τ
xz
a
z
+
τ
xy
a
y
)
−
ds dx
x
x
δφ
dA d x
xy
∂ω
∂
xz
∂ω
∂
=−
τ
y
+
τ
z
A
x
δφ
∂τ
dA d x
y
+
∂τ
xy
xz
−
A
ω
∂
∂
z
xy
∂ω
∂
z
xz
∂ω
∂
y
dA d x
x
δφ
=−
τ
y
+
+
τ
z
−
A
∂τ
xy
dA d x
x
δφ
x
δφ
y
+
∂τ
xz
+
A
(τ
xy
z
−
τ
xz
y
)
dA d x
−
A
ω
∂
∂
z
∂ω
∂
z
2
∂ω
∂
y
2
dA d x
G
x
δφ
M
t
dx
x
δφ
=
y
+
+
z
−
−
A
∂τ
xy
∂
dA d x
x
δφ
y
+
∂τ
xz
+
A
ω
∂
z
∂τ
xy
∂
dA d x
x
δφ
y
+
∂τ
xz
G J
x
δφ
φ
dx
L
0
=
−
δφ
M
t
|
−
A
ω
(16)
∂
z
A
z
2
y
2
dA
with
J
=
∂ω/∂
y
+
+
∂ω/∂
z
−
.
Substitution of (16) into (15) gives
x
δφ
Applicable to the
direction of
the bar axis.
φ
dx
GJ
−
[
M
t
δφ
]
L
−
[
φδ
M
t
]
0
+
[
δ(
M
t
φ)
]
0
x
φ
∂τ
dA
y
+
∂τ
xy
xz
+
δ
A
ω
−
δ
ω(τ
xz
a
z
+
τ
xy
a
y
)
ds
Applicable to
the bar's
cross-section.
∂
∂
z
xz
dA
dA
dx
1
G
δτ
1
G
δτ
−
τ
+
τ
+
A
(
−
δτ
xy
z
+
δτ
xz
y
)
=
0
xy
xy
xz
A
(17)
The axial and cross-sectional parts must be equal to zero separately. For the cross-sectional
terms, using
τ
τ
xy
xz
τ
xy
=
τ
xz
=
and
φ
)
φ
)
(
G
(
G
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