Question 8.11: A rectangular portal frame ABCD is rigidly fixed to a founda......
A rectangular portal frame ABCD is rigidly fixed to a foundation at A and D and is subjected to a compression load P applied at each end of the horizontal member BC (see Fig. P.8.11). If all the members have the same bending stiffness EI, show that the buckling loads for modes that are symmetrical about the vertical center line are given by the transcendental equation,
{\frac{\lambda a}{2}}=-{\frac{1}{2}}\left({\frac{a}{b}}\right)\tan\left({\frac{\lambda a}{2}}\right)where
λ² = P/EI
The deflected shape of each of the members AB and BC is shown in Fig. S.8.11. For the member AB and from Eq. (8.1),
E I{\cfrac{\mathrm{d}^{2}\nu}{\mathrm{d}z^{2}}}=-P_{\mathrm{CR}}\nu (8.1)
E I{\frac{\mathrm{d}^{2}v_{1}}{\mathrm{d}z_{1}^{2}}}=-M_{\mathrm{B}}
so that
E I{\cfrac{\mathrm{d}{v}_{1}}{\mathrm{d}z_{1}}}=-M_{\mathrm{{B}}}z_{1}+AWhen z_{1}=b,\mathrm{d}v_{1}/\mathrm{d}z_{1}=0. Thus A=M_{\mathrm{{B}}}b and
E I{\cfrac{\mathrm{d}{v}_{1}}{\mathrm{d}z_{1}}}=-M_{\mathrm{{B}}}(z_{1}-b) (i)
At B, when z_{1}=0, Eq. (i) gives
{\frac{\mathrm{d}v_{1}}{\mathrm{d}z_{1}}}={\frac{M_{\mathrm{B}}b}{E I}} (ii)
In BC, Eq. (8.1) gives
E I{\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}=-P v+M_{\mathrm{B}}or
E I{\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}+P v=M_{\mathrm{B}} (iii)
The solution of Eq. (iii) is
v=B\cos{\lambda z}+C\sin{\lambda{z}}+M_{\mathrm{{B}}}/P (iv)
When z = 0, υ = 0 so that B=-M_{\mathrm{B}}/P.
When z=a/2, dυ/dz = 0 so that
C=B\tan{\frac{\lambda a}{2}}=-{\frac{M_{\mathrm{B}}}{P}}\tan{\frac{\lambda a}{2}}Eq. (iv) then becomes
\upsilon=-\frac{M_{\mathrm{B}}}{P}{\Big(}\cos{\lambda z}+\tan\frac{\lambda a}{2}\sin{\lambda z}-1{\Big)}so that
\frac{\mathrm{d}\upsilon}{\mathrm{d}z}=-\frac{M_{\mathrm{B}}}{P}\left(-\lambda\sin\lambda z+\lambda\tan\frac{\lambda a}{2}\cos\lambda z\right)At B, when z = 0,
{\frac{\mathrm{d}v}{\mathrm{d}z}}=-{\frac{M_{\mathrm{B}}}{P}}\lambda\mathrm{tan}{\frac{\lambda a}{2}} (v)
Since \mathrm{d}v_{1}/\mathrm{d}z_{1}=\mathrm{d}v/\mathrm{d}z at B, then, from Eqs (ii) and (v),
{\frac{b}{E I}}=-{\frac{\lambda}{P}}\tan{\frac{\lambda a}{2}}whence,
\frac{\lambda a}{2}=-\frac{1}{2}\Big(\frac{a}{b}\Big)\tan\frac{\lambda a}{2}