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one obtains
∂τ
xy
∂
dA
y
+
∂τ
xz
ω(τ
xz
a
z
+
τ
xy
a
y
)
δ
A
ω
−
δ
ds
∂
z
A
(τ
xy
δτ
xy
+
τ
xz
δτ
xz
)
A
(
−
δτ
xy
z
+
δτ
xz
−
dA
+
y
)
dA
=
0
(18)
The symmetry of the relations can be observed in the matrix form of (18)
0
∂
∂
ω
τ
xz
τ
xy
z
y
dA
ωτ
xz
τ
xy
]
A
δ
∂
−
10
[
z
∂
0
−
1
y
ω
τ
xz
τ
xy
0
a
z
a
y
ωτ
xz
τ
xy
]
ds
−
δ
[
a
z
00
a
y
00
0
−
dA
ωτ
xz
τ
xy
]
−
A
δ
[
y
=
0
(19)
z
Note that the relationships
AB
and
AD
are equivalent in the sense that one can be
transformed into the other. To observe this, rewrite
AB
by multiplying
B
by
−
1 to obtain
∂τ
dA d x
y
+
∂τ
xy
xz
A
δ(φ
ω)
−
A
(γ
δτ
+
γ
δτ
)
dA d x
xy
xy
xz
xz
∂
∂
z
x
x
δ(φ
ω)(τ
−
xy
a
y
+
τ
xz
a
z
)
ds dx
+
[
(
M
−
M
t
)δφ
]
L
−
[
(φ
−
φ)δ
M
t
]
0
x
∂ω
∂
z
∂ω
∂
y
δτ
xz
dA d x
x
φ
−
y
+
δτ
xy
+
z
−
=
0
(20)
A
The final term of (20) can be written as
∂ω
∂
δτ
xz
dA d x
x
φ
y
δτ
xy
+
∂ω
z
δτ
xz
+
z
δτ
xy
−
y
∂
A
x
φ
∂
∂
dA
∂δτ
xy
∂
dA
y
(ω δτ
xy
)
+
∂
+
∂δτ
xz
∂
=
z
(ω δτ
xz
)
−
A
ω
∂
y
z
A
dA
dx
+
A
(
z
δτ
xy
−
y
δτ
xz
)
x
φ
∂δτ
dA
+
∂δτ
xy
xz
=
ω(δτ
xy
a
y
+
δτ
xz
a
z
)
ds
−
A
ω
∂
y
∂
z
dA
dx
+
A
(
z
δτ
−
y
δτ
)
(21)
xy
xz
Substitution of (21) into (20) results in (14). This indicates the equivalence of
AB
and
AD
.
Since (20) is essentially the same as (12), the two matrix expressions of (13) and (19) are
equivalent.
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