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

682
COLuMNS
where the first two terms are the homogeneous solution (which is identical to the homogeneous
solution for the pinned–pinned column) and the third term is the particular solution.
The constants C 1 and C 2 must be evaluated with the use of the boundary conditions. From
the boundary condition v(0) = 0, we obtain
By
= C + C + = +
P L C BL y
0 sin(0) cos(0) ( )
P
From the boundary condition v(L) = 0, we obtain
which can be simplified to
1 2 2 (c)
B
0 = C sin( kL) + C cos( kL) + y
1 2
( −
P L L )
The derivative of Equation (16.11) with respect to x is
dv
dx
0 = C1tan( kL)
+ C2 (d)
By
= Ck 1 coskx −Ck 2 sin kx −
P
From the boundary condition v′(0) = 0, the following expression is obtained:
By
By
0 = Ckcos(0) −Cksin(0)
− = Ck−
P P
1 2 1 (e)
To obtain a nontrivial solution, B y is eliminated from Equations (c) and (e), yielding an
expression for C 2 . From Equation (e), B y = C 1 kP, and this expression can be substituted into
Equation (c) to obtain
C
2
BL y CkPL 1
=− =− =− CkL 1
P P
(f)
Upon substitution of this result into Equation (d), the following equation is obtained:
This equation can be simplified to
0 = C tan( kL) + C = C tan( kL)
−C kL
1 2 1 1
tan( kL) = kL
(16.12)
The solution of Equation (16.12) gives the critical buckling load for a fixed–pinned
column. Since Equation (16.12) is a transcendental equation, it has no explicit solution.
However, the following solution may be obtained by numerical methods:
kL = 4.4934 (g)
Note that only the smallest value of kL that satisfies Equation (16.12) is of interest here.
Since k 2 = P/EI, Equation (g) can then be expressed as
and solved for the critical buckling load P cr :
P
=
EI L 4.4934
P
20.1907 EI
2
2.0457π
EI
= =
2
L
L
cr 2
(16.13)