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Then
V
(δ
σ
T
E
−
1
σ
+
σ
T
E
−
1
1
2
δ
∗
=
p
T
u
dS
δ
σ
)
dV
−
S
u
δ
V
δ
σ
T
dV
p
T
u
dS
=
−
S
u
δ
=
0
(2)
−
δ
W
i
−
δ
W
e
=
This expression is the same as
0 of Eq. (2.78). For the beam, using the
notation of Example 2.6,
x
δ
0
x
γ
κ
u
0
w
θ
dx
δ
∗
=
[
NVM
]
−
δ
[
NVM
]
(3)
S
u
[
NVM
]
T
. In order to
where
σ
and
p
of (2) are replaced by the stress resultants
s
, with
s
=
derive the proper relations, it is necessary to add the identity
∂
00
x
s
T
D
u
u
dV
s
T
D
u
u
dV
V
δ
−
V
δ
=
0
,
D
u
=
0
∂
1
(4)
x
00
∂
x
to the right-hand side of (3). This leads to
δ
∗
=
u
0
)
+
δ
(γ
−
w
−
θ)
+
δ
(κ
−
θ
)
[
δ
N
(
−
V
M
]
dx
0
x
x
δ
θ
L
Nu
0
+
δ
(w
+
θ)
+
δ
θ
]
dx
+
[
δ
V
M
−
N u
0
+
δ
V
w
+
δ
M
0
=
0
(5)
x
S
u
where the terms due to (4) are indicated by an underline. Integration by parts of the second
integral gives
Nu
0
+
δ
(w
+
θ)
+
δ
θ
]
dx
[
δ
V
M
x
δ
θ
L
N
u
0
+
δ
V
w
+
(δ
M
−
δ
=
Nu
0
+
δ
V
w
+
δ
M
0
−
[
δ
V
)θ
]
dx
(6)
x
S
u
N
=
V
=
M
−
Since for equilibrium (Chapter 1, Section 1.8.3)
δ
0
,
δ
0
,
δ(
V
)
=
0 along
x
, (6)
becomes
δ
θ
L
0
Nu
0
+
δ
(w
+
θ)
+
δ
θ
]
dx
[
δ
V
M
=
Nu
0
+
δ
V
w
+
δ
M
(7)
x
S
u
δ
∗
in the form
Substitution of (7) into (5) gives
δ
∗
=
u
0
)
+
δ
(γ
−
w
−
θ)
+
δ
(κ
−
θ
)
[
δ
N
(
−
V
M
]
dx
0
x
x
L
0
−
δ
N
(
u
0
−
u
0
)
+
δ
V
(w
−
w)
+
δ
M
(θ
−
θ)
(8)
S
u
or
δ
∗
=
s
T
p
T
V
δ
(
−
)
−
S
u
δ
(
−
)
=
D
u
u
u
u
dS
0
(9)
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