Civil Engineering Reference
In-Depth Information
Example 5.17
Geometric stiffness
Determine the deformations at the free end of the beam by including both
the elastic and geometric stiffness contributions to the stiffness solution.
The beam is shown in Figure 5.20.
Z
A=10 in
2
I=100 in
4
100kips
X
1kip
200in
E=10,000 ksi
Figure 5.20.
Example 5.17 Geometric stiffness.
This is the same as Example 5.16, except the axial force has been
increased to 100 kips.
500
0
0
00 0
0610
0 0
∆
∆
100
1
0
ix
0 5
.
−
150
+−
−
.
=
iz
i
q
0
−
150 20 000
,
2666 7
.
500
0
0
∆
∆
100
1
0
ix
009
.
−
140
=
iz
i
q
0
−
140 17 333 3
,
.
∆
∆
0 2000
4 33333
0 00350
.
.
.
ix
=−
−
iz
i
q
Observe that the
i-
i-end of the beam moved in the negative Z direction. This
does not make logical sense. The reason of the backward motion is that the
member has buckled elastically. The actual elastic buckling load of this
column is 61.685 kips.
5.10
gEOMEtRic StiffnESS, 3-D SYStEM
By combining Equations 5.29 and 5.30 and adding the torsional stiffness
terms, the geometric stiffness matrix in the 3-D Cartesian coordinate sys-
tem can be found.
