Question 8.5: A truss supports load W as shown in Figure 8.12 (a). Length ...

From the free-body diagram of joint B shown in Figure 8.12(b), we can write for equilibrium that

\sum F_y=0=F_{ BC } \sin \theta-W

or        F_{ BC }=\frac{W}{\sin \theta}=W \operatorname{cosec} \theta          (1)

Now considering critical buckling load of BC,

\left(P_{ Cr }\right)_{ BC }=\left[\frac{\pi^2 E(I)}{\left(L_{ e }\right)^2}\right]_{ BC }

\left(P_{ Cr }\right)_{ BC }=\frac{\lambda}{L_1^2} \cos ^2 \theta \quad\left[\text { as } \lambda=\pi^2 E(I)_{ BC }\right]           (2)

Combining Eqs. (1) and (2)

W \operatorname{cosec} \theta=\frac{\lambda}{L_1^2} \cos ^2 \theta

or        W=\frac{\lambda}{L_1^2} \cos ^2 \theta \sin \theta=\frac{\lambda}{L_1^2}\left(\sin \theta-\sin ^3 \theta\right)

Obviously, for W_{\max }, d (W) / d \theta=0 \text { and } d ^2(W) / d \theta^2<0 . \text { Now, } d (W) / d \theta=0 gives

\cos \theta-3 \sin ^2 \theta \cdot \cos \theta=0

or        \cos \theta\left(1-3 \sin ^2 \theta\right)=0

which gives either \cos \theta=0 \text { or } \sin ^2 \theta=1 / 3 . Now,

\frac{ d ^2(W)}{ d \theta^2}=\frac{\lambda}{L_1^2}\left(-\sin \theta-6 \sin \theta \cos ^2 \theta+3 \sin ^3 \theta\right)

and        \left.\frac{ d ^2 W}{ d \theta^2}\right|_{\cos \theta=0 ; \sin \theta=1}=\frac{\lambda}{L_1^2}(-1+3)>0

and        \left.\frac{ d ^2 W}{ d \theta^2}\right|_{\sin ^2 \theta=1 / 3}=\frac{\lambda}{L_1^2}\left[-\frac{1}{\sqrt{3}}-\frac{6}{\sqrt{3}} \times \frac{2}{3}+\frac{1}{\sqrt{3}}\right]<0

Therefore, for w=w_{\max }

\sin ^2 \theta=\frac{1}{3} \quad \text { or } \quad \sin \theta=\frac{1}{\sqrt{3}} \quad \text { or } \quad \theta=35.3^{\circ}

Therefore,

w_{\max }=\frac{\lambda}{L_1^2} \times \frac{2}{3} \times \frac{1}{\sqrt{3}}=\frac{2 \pi^2 E I}{3 \sqrt{3} L_1^2}