Buckling length of column

[rc2008]

Homework Statement

I was given this question to find the buckling length of the non-sway frame of column AB.

Relevant Equations

The general equation is v" +$\omega\$^2 v = MB/EI + Vx/EI,

Taking B as origin, the boundary condition provided was v(x=0)= 0 , v(x=L)=0 , v'(x=0)= $\ theta B\$
v'(x=L)=0 , ans also v'' (x=0) = MB/ EI

However, I had problem of expressing the equation in terms of $\omega\$ L
Can anyone help ?

 

[rc2008]

$BC1: At x=0x = 0x=0, v=0v = 0v=0: c2+θBL=0c_2 + \frac{\theta_B}{L} = 0c2+LθB=0• BC2: At x=Lx = Lx=L, v=0v = 0v=0: c1sin⁡(ωL)+c2cos⁡(ωL)+VEIL+θBL=0c_1 \sin(\omega L) + c_2 \cos(\omega L) + \frac{V}{EI} L + \frac{\theta_B}{L} = 0c1sin(ωL)+c2cos(ωL)+EIVL+LθB=0• BC3: At x=0x = 0x=0, v′=θBv' = \theta_Bv′=θB: c1ω+VEI=θBc_1 \omega + \frac{V}{EI} = \theta_Bc1ω+EIV=θB• BC4: At x=Lx = Lx=L, v′=0v' = 0v′=0: c1ωcos⁡(ωL)−c2ωsin⁡(ωL)+VEI=0c_1 \omega \cos(\omega L) - c_2 \omega \sin(\omega L) + \frac{V}{EI} = 0c1ωcos(ωL)−c2ωsin(ωL)+EIV=0• BC5: At x=0x = 0x=0, Moment MB=EIv′′M_B = EI v''MB=EIv′′: −c2ω2=2EILθB-c_2 \omega^2 = \frac{2EI}{L} \theta_B−c2ω2=L2EθB$