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area
A
, bounded by the lines
x
=
0
,x
=
a, y
=
0, and
y
=
b
. Begin with the final integral
on the right-hand side of Eq. (1.117). Introduce
D
T
of Eq. (1.50a) and set
D
T
D
T
σ
=
σ
x
σ
y
τ
xy
dA
[
u
x
u
y
]
∂
x
0
∂
y
u
T
D
T
σ
σ
dA
=
0
∂
y
∂
x
A
A
b
a
a
b
a
b
=
u
x
σ
x,x
dx dy
+
u
x
τ
xy, y
dy dx
+
u
y
σ
y, y
dy dx
0
0
0
0
0
0
b
a
+
u
y
τ
xy,x
dx dy
0
0
b
u
x
σ
x
a
u
x,x
σ
x
dx
dy
a
u
x
τ
xy
b
u
x, y
τ
xy
dy
dx
a
0
−
b
0
−
=
+
0
0
0
0
a
u
y
σ
y
b
u
y, y
σ
y
dy
dx
b
u
y
τ
xy
a
u
y, x
τ
xy
dx
dy
b
0
−
a
0
−
+
+
0
0
0
0
σ
x
σ
y
τ
xy
σ
x
σ
y
τ
xy
[
u
x
u
y
]
100
001
[
u
x
u
y
]
001
010
b
a
a
b
=
dy
+
dx
0
0
0
0
b
a
b
a
−
u
x,x
σ
x
dx dy
−
u
y, x
τ
xy
dx dy
0
0
0
0
a
b
a
b
−
u
x, y
τ
xy
dy dx
−
u
y, y
σ
y
dy dx
0
0
0
0
σ
x
σ
y
τ
xy
ds
σ
x
σ
y
τ
xy
dA
[
u
x
u
y
]
a
x
[
u
x
u
y
]
x
∂
0
a
y
0
∂
y
=
−
0
a
y
a
x
0
∂
∂
y
x
S
A
u
T
A
T
σ
ds
u
T
u
D
T
σ
dA
=
−
(1.118)
S
A
where
a
x
and
a
y
represent the direction cosines of the unit normal to
s
and
x
∂
are
partial derivatives of the preceding quantities (
u
x
and
u
y
as appropriate). Thus,
D
u
and
D
T
σ
and
y
∂
are adjoints in the sense of Eq. (1.117) for the case of two-dimensional elasticity.
The situation is similar for the case of beams with shear deformation. For a beam
element from
x
=
a
to
x
=
b
,
N
V
M
dx
b
b
d
x
00
u
T
D
s
s
dx
=
[
u
0
wθ
]
0
d
x
0
a
a
0
−
1
d
x
b
[
u
0
N
+
w
V
+
θ(
M
−
=
V
)
]
dx
a
b
a
(
b
a
u
0
N
+
w
V
+
θ
M
=
(
u
0
N
+
w
V
+
θ
M
)
|
−
+
θ
V
)
dx
N
V
M
b
+
N
V
M
a
100
010
001
−
100
0
=
[
u
0
wθ
]
[
u
0
wθ
]
10
00
−
−
1
N
V
M
dx
b
x
d
00
0
x
d
0
01
x
d
−
[
u
0
wθ
]
a
b
u
T
A
T
s
b
a
u
T
u
D
T
s
dx
=
|
−
a
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