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Substitute
V
σ
δ
u
i
dV
=
p
i
δ
u
i
dS
−
V
σ
δ
u
i, j
dV
ij, j
ij
S
of Eq. (2.40) into Eq. (2.44). Also, introduce Eq. (2.45) to find
V
σ
δ
−
p
Vi
δ
−
p
i
δ
−
δ
=
u
i, j
dV
u
i
dV
u
i
dS
p
i
u
i
dS
0
(2.46)
ij
V
S
p
S
u
To introduce strain into Eq. (2.46), we recall that in Appendix I, the virtual displacement
δ
u
i
is defined by introducing a family of neighboring functions,
u
i
=
u
i
+
δ
u
i
The corresponding strain-displacement relationship of Chapter 1, Eq. (1.21) is
=
D
u
. Then
2
(δ
,i
1
2
(
1
2
(
1
=
u
i, j
+
u
j,i
)
=
u
i, j
+
u
j,i
)
+
u
i
)
+
(δ
u
j
)
ij
,j
1
2
(
1
2
δ(
=
u
i, j
+
u
j,i
)
+
u
i, j
+
u
j,i
)
=
ij
+
δ
ij
(2.47)
This uses Eq. (I.6) in Appendix I,
(δ
u
i
)
,j
=
δ(
u
i, j
).
It follows that if the displacements
u
and
u
obey their respective strain-displacement relationships, then
1
2
δ(
δ
=
u
i, j
+
u
j,i
)
in
V
(2.48)
ij
By using the summation convention and Eq. (2.48) it can be verified that (Problem 2.13)
σ
ij
δ
u
i, j
=
σ
ij
δ
ij
and Eq. (2.46) can be written as
V
σ
δ
ij
dV
−
p
Vi
δ
u
i
dV
−
p
i
δ
u
i
dS
−
p
i
δ
u
i
dS
=
0
ij
V
S
p
S
u
or
T
σ
dV
u
T
p
V
dV
u
T
p
dS
u
T
p
dS
V
δ
−
V
δ
−
S
p
δ
−
S
u
δ
=
0
(2.49)
If
u
i
and
u
i
both satisfy the geometric boundary conditions
(
u
i
=
u
i
on
S
u
)
,
δ
u
i
=
0on
S
u
(2.50)
Virtual displacements that satisfy both Eqs. (2.48) and (2.50) are said to be kinematically
admissible. Applying Eq. (2.50) to Eq. (2.49) causes the integral over
S
u
[in Eq. (2.49)] to be
equal to zero. Thus,
V
σ
δ
ij
dV
−
p
Vi
δ
u
i
dV
−
p
i
δ
u
i
dS
=
0
ij
V
S
p
or
(2.51)
V
δ
T
σ
dV
u
T
p
V
dV
u
T
p
dS
−
V
δ
−
S
p
δ
=
0
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