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N
N
x, u
(a)
N
N
N
N
(b)
FIGURE 2.1
Element of an extension bar.
2.1.2 Work and Potential Energy of the Internal Forces
Consider an elastic extension bar, a one-dimensional syst
em
, of length
L
and cross-sectional
area
A
Suppose that it is loaded with an axial force
N
which induces the axial stress
σ
x
. Applied quantities are identified with overbars. What is the work done during the
deformation of an element of length
dx
of uniform cross-section (Fig. 2.1a)?
The internal force
N
.
=
A
σ
x
dA
is by equilibrium equal to the applied force
N
but
opposite in sign. Refer to the fictitious cut in the bar of Fig. 2.1b. Then, the internal work
done in a differential element
dx
is
−
σ
x
d
x
Adx
=−
σ
x
d
x
dV
with the strain
d
=
d
(
du
/
dx
)
x
and the volume
dV
Adx
of the element. The negative sign is due to the expression being
formed on the ficticious cut of Fig. 2.1b where the stresses and stress resultant
N
are defined
as internal variables. Thus, the total work of the bar as the strains (defined as being positive
in the direction of the x-axis) increase from zero to their final values has a negative sign and
the value
=
L
x
0
σ
W
i
=−
x
d
x
Adx
(2.4)
0
Upon substitution of Hooke's law,
σ
x
=
E
x
, which is the relationship between stress
σ
x
and strain
x
L
A
x
0
L
L
1
2
1
2
2
W
i
=−
E
x
d
x
dx
=−
EA
x
dx
=−
N
x
dx
(2.5)
0
0
0
The expression for the work of the internal forces is similar in the case of a three-
dimensional body with six independent components of the stress tensor
σ
ij
and the strain
tensor
ij
. For an infinitesimal rectangular parallelepiped of volume
(
dx dydz
)
, the work
done during the strain increments
d
x
,d
y
,
...
,d
γ
yz
by the stresses would be
(σ
x
d
x
+
σ
y
d
y
+···+
τ
yz
d
γ
yz
)
dx dydz
where
d
∂
,
d
∂v
∂
,
d
∂w
∂
u
y
+
∂v
d
x
=
d
y
=
...
,d
γ
yz
=
∂
x
y
∂
z
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