Civil Engineering Reference
In-Depth Information
If the true variation in
u
is higher order, as will often be the case, greater accuracy
could be achieved by introducing higher order shape functions or by including more linear
subdivisions.
When (2.4) is substituted in (2.3), we have
N
2
]
u
1
u
2
EA
d
2
d
x
2
[
N
1
+
F
=
R
(2.6)
where
is a measure of the error in the approximation and is called the
residual
.The
differential equation has thus been replaced by an equation in terms of the nodal values
u
1
and
u
2
. The problem now reduces to one of finding “good” values for
u
1
and
u
2
in order
to minimise the residual
R
.
Many methods could be used to achieve this. For example Griffiths and Smith (1991)
discuss collocation, subdomain, Galerkin, and least squares techniques. Of these, Galerkin's
method, for example Finlayson (1972), is the most widely used in finite element work. The
method consists of multiplying or “weighting” the residual in (2.6) by each shape function
in turn, integrating over the element and equating to zero. Thus
L
R
N
1
N
2
EA
d
2
d
x
u
1
u
2
L
N
1
N
2
0
0
2
[
N
1
N
2
] d
x
+
F
d
x
=
(2.7)
0
0
Note that in the present example, in which the shape functions are linear, double dif-
ferentiation of these functions would cause them to vanish. This difficulty is resolved by
applying Green's theorem (integration by parts) to yield typically
2
N
i
∂
N
j
∂x
∂N
i
∂x
∂N
j
∂x
d
x
=−
d
x
+
boundary terms, which we usually ignore
(2.8)
2
Hence, assuming
EA
and
F
are not functions of
x
, (2.7) becomes
∂N
1
∂x
∂N
1
∂x
∂N
1
∂x
∂N
2
∂x
u
1
u
2
N
1
N
2
d
x
=
0
0
EA
L
0
L
−
d
x
+
F
(2.9)
∂N
2
∂x
∂N
1
∂x
∂N
2
∂x
∂N
2
∂x
0
On evaluation of the integrals,
=
1
L
1
L
2
2
u
1
u
2
0
0
−
−
EA
+
F
(2.10)
1
L
1
L
−
The above case is for a uniformly distributed force
F
acting along the element, and it
should be noted that the Galerkin procedure has resulted in the total force
FL
being shared
equally between the two nodes. If in Figure 2.1(a) the loading is applied only at the nodes
we have
u
1
u
2
f
x
1
f
x
2
EA
L
1
−
1
=
(2.11)
−
11
