Question 21.P.7: A rectangular portal frame ABCD is rigidly fixed to foundati...

The deflected shape of each of the members AB and BC is shown in Fig. S.21.7.

For the member AB and from Eq. (13.3) \frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}=\frac{M}{E I}

E I \frac{\mathrm{d}^{2} v_{1}}{\mathrm{~d} x_{1}^{2}}=-M_{B}

Then

E I \frac{\mathrm{d} v_{1}}{\mathrm{~d} x_{1}}=-M_{\mathrm{B}} x_{1}+C_{1}

When x_{1} = b, dv_{1}/dx_{1} = 0 so that C_{1} = M_{B}b and

E I \frac{\mathrm{d} v_{1}}{\mathrm{~d} x_{1}}=-M_{\mathrm{B}}\left(x_{1}-b\right)   (i)

At B when x_{1} = 0 Eq. (i) gives

\frac{\mathrm{d} v_{1}}{\mathrm{~d} x_{1}}=\frac{M_{\mathrm{B}} b}{E I}   (ii)

In the member BC Eq. (13.3) gives

E I \frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}=-P v+M_{\mathrm{B}}

or

\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}+\mu^{2} v=\frac{M_{\mathrm{B}}}{E I}  (iii)

where μ²=P/EI. The solution of Eq. (iii) is

v=C_{2} \cos \mu x+C_{3} \sin \mu x+\frac{M_{\mathrm{B}}}{P}    (iv)

When x = 0, v = 0 so that C_{2} = -M_{B}/P. Also, when x = a/2, dv/dx = 0 which gives

C_{3}=C_{2} \tan \frac{\mu a}{2}=-\frac{M_{\mathrm{B}}}{P} \tan \frac{\mu a}{2}

Eq. (iv) then becomes

v=-\frac{M_{\mathrm{B}}}{P}\left(\cos \mu x+\tan \frac{\mu a}{2} \sin \mu x-1\right)

Then

\frac{\mathrm{d} v}{\mathrm{~d} x}=-\frac{M_{\mathrm{B}}}{P}\left(-\mu \sin \mu x+\mu \tan \frac{\mu a}{2} \cos \mu x\right)

At B when x = 0,

\frac{\mathrm{d} v}{\mathrm{~d} x}=-\frac{M_{\mathrm{B}}}{P} \mu \tan \frac{\mu a}{2}  (v)

Since d_{v1}/d_{x1} = dv/dx at B then, from Eqs (ii) and (v)

\frac{b}{E I}=-\frac{\mu}{P} \tan \frac{\mu \mathrm{a}}{2}

Rearranging

\frac{\mu a}{2}=\frac{-1}{2}\left(\frac{a}{b}\right) \tan \frac{\mu a}{2}