Civil Engineering Reference
In-Depth Information
And, since the domain of interest is the volume of the element, the Galerkin
residual equations become
N
i
(
x
)
E
d
V
N
i
E
A
d
x
L
d
2
u
d
x
2
d
2
u
d
x
2
=
=
0
i
=
1, 2
(5.33)
V
0
where
d
V
A
d
x
and
A
is the constant cross-sectional area of the element.
Integrating by parts and rearranging, we obtain
=
N
i
AE
d
u
d
x
L
L
d
N
i
d
x
d
u
d
x
AE
d
x
=
(5.34)
0
0
which, utilizing Equation 5.32, becomes
x
=
0
=−
L
d
N
1
d
x
d
d
x
(
u
1
N
1
+
d
u
d
x
AE
u
2
N
2
)d
x
=−
AE
AE
ε
|
x
=
0
=−
A
|
x
=
0
(5.35a)
0
x
=
L
L
d
N
2
d
x
d
x
(
u
1
N
1
+
u
2
N
2
)d
x
=
AE
d
u
d
AE
=
AE
ε
|
x
=
L
=
A
x
=
L
d
x
(5.35b)
0
From the right sides of Equation 5.35, we observe that, for the bar element, the
gradient boundary condition simply represents the applied nodal force since
A
=
.
Equation 5.35 is readily combined into matrix form as
F
d
N
1
d
x
d
N
1
d
x
d
N
1
d
x
d
N
2
d
x
d
x
u
1
u
2
F
1
F
2
L
AE
=
(5.36)
d
N
1
d
x
d
N
2
d
x
d
N
2
d
x
d
N
2
d
x
0
where the individual terms of the matrix are integrated independently.
Carrying out the indicated differentiations and integrations, we obtain
AE
L
1
u
1
u
2
F
1
F
2
−
1
=
(5.37)
−
11
which is the same result as obtained in Chapter 2 for the bar element. This sim-
ply illustrates the equivalence of Galerkin's method and the methods of equilib-
rium and energy (Castigliano) used earlier for the bar element.
5.4.2 Beam Element
Application of the Galerkin method to the beam element begins with consid-
eration of the equilibrium conditions of a differential section taken along the
