Civil Engineering Reference
In-Depth Information
144
144
216
−+
−− +
Q
Q
−
2
Q
2
2
2
L
L
L
252
144
288
−+
2
Q
Q
−+
−
−
2
Q
L
2
L
2
L
2
252
54432
936
Q
41472
576
Q
2
2
0
=
−
2
Q
−
+
4
Q
−
−
−
2
Q
2
4
2
4
2
L
L
L
L
L
Multiplying the values for like terms yields the following:
144
504
Q
10386
72
Q
−
2
0
=− −
Q
31104
+ −+ −
2
Q
L
2
L
2
L
4
L
2
36
41472
576
Q
18144 144
Q
+
−
+
2
Q
2
−
−
2
4
2
4
2
L
L
L
L
L
Multiplying all the values and combining like terms:
1119744
49248
576
0
=
−
Q
+
L
QQ
2
−
2
3
L
6
L
4
2
The solution to these cubic equations yields the following general value
of
Q
:
P
EI
Q
c
==0 0004
.
Substituting in the particular values for the column of
E
and
I
gives
P
cr
=
11,600
kips. The exact values using the Euler buckling equation with an
effective length factor of
k=
0
.
5 for a fixed end column is
P
cr
=
12,721
kips.
This central difference solution has an error of 8.8%.
3.10
PARtiAL DiffEREntiAL EQuAtiOn
DiffEREncE OPERAtORS
More advanced structural analysis problems may require the solution of
partial differential equations. One example is the bending of a plate under
uniform lateral load. “Theory of Plates and Shells” by Timoshenko and
Woinowsky-Krieger (1959) contains an exact solution to general plates.
The differential relationship for plate bending uses partial differential
equations. Their difference operators can be derived from the basic dif-
ference operators using two basic principles. If you add two differential
