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In general, the external work done by the prescribed forces
p
i
, p
V
i
can be expressed
as
u
i
u
i
W
e
=
p
i
du
i
dS
+
p
V
i
du
i
dV
(2.26)
S
p
0
V
0
In the instance where the so
li
d is
li
near in its response (zero to
u
i
) to a load applied from a
zero value to its final value
p
i
or
p
Vi
,
1
2
1
2
1
2
1
2
u
T
p
dS
u
T
p
V
dV
W
e
=
p
i
u
i
dS
+
p
Vi
u
i
dV
=
+
(2.27)
S
p
V
S
p
V
To e
st
ablish the quantity
complementary external work
performed by prescribe
d
displace-
ments
u
i
, a similar pattern can be followed. For the bar with prescribed elongation
u
,
N
F
W
e
=
udN
(2.28)
0
where
N
F
should be interpreted as the reactive force developed in the bar. In general,
p
i
W
e
=
u
i
dp
i
dS
(2.29)
S
u
0
If the relationship between displacements and forces is linear, then
1
2
1
2
W
e
=
p
T
u
dS
u
i
p
i
dS
=
(2.30)
S
u
S
u
which is referred to as the complementary work of the prescribed displacements. The
change in the work from the initial to the final configuration is given by Eq. (2.26), where
the integration from zero to
u
i
is to be made from the initial state to the final state. Similar
to the case of internal energy with Eq. (2.22), the conditions for the existence of potential
functions for applied loads are
∂
T
p
T
∂
T
U
0
e
(
1
)
∂
U
0
e
(
2
)
∂
p
V
=−
=−
(2.31)
u
u
where the
U
0
e
(
k
)
,k
1
,
2 are density functions w
it
h the subscript
k
distingui
sh
ing between
those portions of the potential related to forces
p
and those resulting from
p
V
. In order to
understand better the relations of Eq. (2.31), consider the related energy expression
=
u
i
u
i
U
e
=−
p
i
du
i
dS
−
p
Vi
du
i
dV
S
p
0
V
0
=
U
0
e
(
1
)
dS
+
U
0
e
(
2
)
dV
(2.32a)
S
p
V
with the energy density of the prescribed forces
u
i
u
i
U
0
e
(
1
)
=−
p
i
du
i
and
U
0
e
(
2
)
=−
p
Vi
du
i
0
0
Equation (2.32) satisfies Eq. (2.31), i.e.,
p
i
du
i
u
i
∂
U
0
e
(
1
)
∂
∂
u
i
=
−
=−
p
i
∂
u
i
0
u
i
p
Vi
du
i
∂
U
0
e
(
2
)
∂
∂
u
i
=
−
=−
p
Vi
∂
u
i
0
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