Stability Analysis - Mechanics of Structures: Variational and Computational Methods - page 694

0 becomes EIA /

The first three columns and rows vanish and the condition det K = L = 0 .

L is not zero, the quantity A can be zero for particular values (eigenvalues) of

Since EI / ε.

ε( sin ε − ε cos ε)

A =

ε =

0 (2)

2 ( 1 − cos ε) − ε sin

Thus,

2 cos

ε ε − ε ε = ε = ε

sin 0 or tan (3)

Equation (3) is plotted in Fig. 11.28b and the first two eigenvalues identified. Normally, the
structure will buckle for the first eigenvalue

ε 1 = 4 . 493 . This gives

2 EI
L 2

2 EI 2 EI

L 2

P cr = ε

π

05 π =

2 =

2 . (4)

( 0 . 699 L )

where is introduced so that P cr appears in the same form as the Euler load of the previous

example. The quantity 0.699 L , which is referred to as the equivalent length L e , is shown in
Fig. 11.28b.
The equivalent length is often used in the analysis of frameworks when an approximate
calculation involving the buckling load of an equivalent column replaces a more rigorous,
complex analysis.

π

EXAMPLE 11.8 Column with Moment Release
Find the critical axial force of the column with a moment release shown in Fig. 11.29.
A simple approach to solving this problem involves establishing as one of the boundary
conditions the relationship between the deflection and shear force at the moment release.
Consider the fre e- body diagram of the link between the hinges of Fig. 11.30. Summation of
moments gives P w( L ) = VL 1 or

P w( L )

V = (1)

L 1

FIGURE 11.29
Column with moment release.

FIGURE 11.30
Free-body diagram.