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We could take the column moments, compute the lateral deflection, increase the mo-
ment by P
, recalculate the lateral deflection and the increased moment, and so on. Although
about two cycles would be sufficient, this would still be a tedious and impractical procedure.
It can be shown 5 that the increased moment can be estimated very well by multiply-
ing the primary moment by 1/(1
P / P c ), where P is the axial load and P c is the Euler
u ) 2 .
In Example 11.2 this expression is used to estimate the magnified moment in a later-
ally loaded column. It will be noted that in this problem the primary moment of 75 ft-k is
estimated to increase by 7.4 ft-k. If we computed the deflection due to the lateral load, we
would get 0.445 in. For this value P
2 EI /( k
buckling load
(150)(0.445)
66.75 in.-k
5.6 ft-k. This mo-
ment causes more deflection, which causes more moment, and so on.
EXAMPLE 11.2
(a) Compute the primary moment in the column shown in Figure 11.6 due to the lateral 20-k load.
(b) Determine the estimated total moment, including the secondary moment due to lateral de-
flection using the appropriate magnification factor just presented. E 3.16 10 3 ksi. As-
sume k 1.0 and u 15 ft.
SOLUTION
(a) Primary moment due to lateral load:
M u (20)(15)
4
75 ft- k
Figure 11.6
5 Timoshenko, S. P., and Gere, J. M., 1961, Theory of Elastic Stability , 2nd ed. (New York: McGraw-Hill),
pp. 319-356.
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