Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

672
COLuMNS
If we let
2
P
k =
(16.2)
EI
then Equation (16.1) can be rewritten as
2
d v
2
+ k v = 0
2
dx
The general solution of this homogeneous equation is
v = C1sinkx + C2 cos kx
(16.3)
where C 1 and C 2 are constants that must be evaluated with the use of the boundary conditions.
From the boundary conditions v(0) = 0, we obtain
0 = C sin(0) + C cos(0) = C (0) + C (1)
1 2 1 2
∴ C = 0
From the boundary conditions v(L) = 0, we obtain
2
(c)
0 = C1 sin( kL)
(d)
One solution of Equation (d) is C 1 = 0; however, this is a trivial solution, since it would
imply that v = 0, and hence the column would remain perfectly straight. The other solution
of Equation (d) is sin(kL) = 0, so
kL = nπ
n = 1, 2,3,…
(e)
because the sine function equals zero for integer multiples of π.
Now, taking the square root of both sides of Equation (16.2) gives
k =
P
EI
This expression for k can be substituted into Equation (e) to give
which may be solved for the load P:
P
= π
EI L n
2 2
n π EI
P = n = 1, 2,3, …
2
L
(16.4)
Euler Buckling Load and Buckling modes
The purpose of this analysis is to determine the minimum load P at which lateral deflections
occur in the column; therefore, the smallest load P that causes buckling occurs for
n = 1 in Equation (e), since that value of n gives the minimum value of P for a nontrivial
solution. As in the elementary column-buckling model (see Figure 16.1), this load is called
the critical buckling load; P cr an ideal column, where
P
cr
2
π EI
=
2
L
(16.5)