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The first term in (4) corresponds to the complementary strain energy for the stretching
displacement
u
0
, and the second term corresponds to that for the lateral displacement
.
The complementary strain energy expression of (4) can also be obtained by simply substi-
tuting the expression for the normal stress [(Chapter 1, Eq. (1.137)]
w
N
A
+
M
I
σ
x
=
z
(5)
into
V
σ
x
E
1
2
U
i
=
dV
(6)
Thus,
L
N
2
A
2
+
I
2
dA dx
L
N
2
EA
+
dx
M
2
z
2
M
2
EI
1
2
1
E
2
NMz
AI
1
2
U
i
=
+
=
(7)
0
A
0
where
A
zdA
0 and
A
z
2
dA
=
=
I
.
EXAMPLE 2.2 Beam Theory Including Shear Deformation
In the previous example, it was assumed that shear deformation effects can be ignored. If
this is not the case,
γ
=
γ
=
0, so that from Eq. (2.7)
xz
1
2
U
i
=
V
(σ
+
τ
γ
)
dV
(1)
x
x
xz
xz
Use
=
σ
N
A
+
M
I
V
A
=
V
k
s
GA
x
E
,
σ
=
z,
τ
=
τ
=
k
s
G
γ
,
γ
=
γ
=
(2)
x
x
xz
xz
The expressions for
τ
and
γ
xz
are taken from Chapter 1, Eq. (1.109). Substitute (2) into (1)
to obtain
N
2
EA
+
dx
L
M
2
EI
+
V
2
k
s
GA
1
2
U
i
=
(3)
0
2.1.3
Work and Potential Energy of the Applied Loading
In order to
ob
tain the total work, it
i
s necessary to take into account the work done by the
bo
dy forces
p
V
, the external forces
p
applied to the surface area
S
p
, and the displacements
u
applied on the surface area
S
u
S
p
. These are referred
t
o as the
applied loadings.
Consider again the simple extension bar wi
th
axial force
N
and axial extension. As the
axial force increases from zero to its final value
N
F
, the displacement
u
in the axial direction
increases from zero to
u
F
. “External” work
W
e
is performed until the final configuration is
reached. From the definition of work, we have
=
S
−
u
F
W
e
=
Ndu
(2.24)
0
If the displacement is proportional to the load, i.e.,
u
=
k N
, where
k
is a constant of
proportionality, then
u
F
u
F
u
F
k
u
k
1
2
1
2
N
F
u
F
W
e
=
Ndu
=
du
=
=
(2.25)
0
0
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