Civil Engineering Reference
In-Depth Information
0420
2121
1212
0 000
−
y
y
y
y
V
V
V
V
0
0
−
108
EI
1
1
−
=
L
3
−
−
2
2
3
3
−+
+− +
−+
y y
yyyy
y
42
V
V
V
V
1
2
0
2
2
2
yyy y
108
EI
0
1
2
3
1
−
=
L
3
+−
2
0
1
2
3
2
0
3
35 27
34 92
19 90
0
.
.
.
V
V
V
V
0
1
kips
=
2
3
The exact solution at the end is
V
0
=
50
.
00
k
, which is a 16.67% error,
while the value in the center is exact. The large error at the end is due to
the fact that the shear drops there, which creates a discontinuity in the
equation. Using more segments would reduce the error, but it would still
be more inaccurate than the other values.
The next example is similar to the previous one, but is included to
show differences in modeling and accuracy. It also uses the higher order
(smaller error) of error equations for more accuracy.
Example 3.11
Fixed beam with difference operator
Calculate the shear, moment, rotation, and deflection for a 30 ft long fixed
end beam with a uniformly distributed load of 5 k/ft using central differ-
ence operator of order of error
h
4
at 1/6th points. The beam has
E
= 29,000
ksi and
I
= 1000 in
4
.
The primary difference in the simply supported beam in Example 3.10
and the fixed end beam in this example is the model of the deflected curve
beyond the boundary as shown in Figure 3.7.
The solution to this example is very similar to Example 3.10 and only
the setup and solutions are presented. The central difference expressions
with error of order
h
4
will be used to solve for the values. Since the load
is known, we will use the fourth derivative relationship between load and
deflection.
