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and with
G
taken from (4), the stiffness matrix of (8) becomes
1
EA
−
1
k
i
=
(10)
−
11
EXAMPLE 4.2 Loading Vector Due to the Weight of the Bar
Find the loading vector for the axial deformation of a vertical bar loaded by its own weight.
The weight per unit length along the bar is
is the weight density of the
material. Assume the bar is in a vertical position and
x
is positive upwards. The external
virtual work would be
γ
A,
where
γ
0
δ
u
T
0
N
u
(
−
γ
δ
W
e
=
u
(
−
γ
A
)
dx
=
δ
A
)
dx
v
T
G
T
0
N
u
(
−
γ
v
T
p
i
0
=
δ
A
)
dx
=
δ
(1)
where
u
,
G
,
N
u
,
and
v
are given in Example 4.1. The same expression, i.e.,
0
δ
δ
W
e
=
u p
x
dx
(2)
A,
follows from Chapter 2, Example 2.6. Also, it is analogous to
b
a
with
p
x
=−
γ
δw
p
z
dx
of Eq. (4.52), which is intended for transverse motion. Upon integration, (1) leads to
1
1
=−
γ
A
p
i
0
(3)
2
EXAMPLE 4.3 Loading Vector Due to Thermal Loading
Find the loading vector
p
i
0
for axial displacement due to a temperature change along the
x
bar of
T
x
=
T
=
T
a
+
(
T
−
T
a
)
where
T
and
T
a
are reference temperature changes and
α
is the thermal expansion coefficient.
The imposed strain due to
0
Equation (1.43) can be inserted in
Eq. (2.35) to obtain an expression for virtual work. This “external” virtual work takes the
form
T
would be
=
α
T
.
0
δ
0
δ
W
e
=−
EA
dx
(1)
From Example 4.1, Eqs. (1) and (4)
G
T
0
1
N
u
G
v
T
δ
=
δ
=
δ
δ
=
δ
=
δ
=
δ
u
N
u
u
N
u
G
v
and
v
[0
1]
G
v
(2)
Then (1) becomes
v
T
G
T
0
0
1
δ
W
e
=
δ
(
EA
α
T
)
dx
(3)
v
T
p
i
0
. Finally,
Also,
δ
W
e
=
δ
0
1
EA
T
a
+
(
dx
G
T
0
x
p
i
0
=−
α
T
−
T
a
)
T
a
+
−
1
2
(
1
1
=
EA
α
T
−
T
a
)
(4)
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