Civil Engineering Reference
In-Depth Information
qL
2
qL
2
q
qL
2
12
qL
2
12
1
2
x
L
(a)
(b)
Figure 4.8
Work-equivalent nodal forces and moments for a uniform
distributed load.
Comparison of Equations 4.52 and 4.53 shows that
L
F
1
q
=
(4.54)
q
(
x
)
N
1
(
x
)d
x
0
L
M
1
q
=
q
(
x
)
N
2
(
x
)d
x
(4.55)
0
L
F
2
q
=
(4.56)
q
(
x
)
N
3
(
x
)d
x
0
L
M
2
q
=
q
(
x
)
N
4
(
x
)d
x
(4.57)
0
Hence, the nodal force vector representing a distributed load on the basis of work
equivalence is given by Equations 4.54-4.57. For example, for a uniform load
q
(
x
)
=
q
=
constant, integration of these equations yields
qL
2
qL
2
12
qL
2
F
1
q
M
1
q
F
2
q
M
2
q
=
(4.58)
qL
2
12
−
The equivalence of a uniformly distributed load to the corresponding nodal loads
on an element is shown in Figure 4.8.
EXAMPLE 4.2
The simply supported beam shown in Figure 4.9a is subjected to a uniform transverse
load, as shown. Using two equal-length elements and work-equivalent nodal loads, ob-
tain a finite element solution for the deflection at midspan and compare it to the solution
given by elementary beam theory.
