Question 8.3: A column with rectangular cross-section is pin-supported at ...

A rectangular section possesses two principal axes of buckling as shown in Figure 8.10. From the figure

I_{x x}=\frac{1}{12} b h^3 \quad \text { and } \quad I_{y y}=\frac{1}{12} \lambda b^3

Now radius of gyration,

K_{x x}=\sqrt{\frac{1}{12} b h^3 \times \frac{1}{b h}}=\frac{h}{2 \sqrt{3}}

K_{y y}=\sqrt{\frac{1}{12} h b^3 \times \frac{1}{b h}}=\frac{b}{2 \sqrt{3}}

From the cross-section configuration, h>b \text {, so } K_{y y}>K_{x x} \text {. Thus, } K_{y y}=K_{\min } . As the column is free to buckle in any direction, it will buckle in a plane whose normal is the y-axis. It means, in this case, the column with buckle in the ‘in-plane’ or in the plane of the figure. Now, for in-plane buckling in the present configuration,

\left(P_{ Cr }\right)_I=\frac{\pi^2 E(I)_{\min }}{L_{ e }^2}

where L_{ e } is the effective length; here it is L/2. Therefore, substituting, we get

\left(P_{ Cr }\right)_I=\frac{\pi^2 E(I)_{\min }}{(L / 2)^2}=\frac{\pi^2 E\left(h b^3 / 12\right)}{(L / 2)^2}

\left.\left(P_{ Cr }\right)_I=\frac{h b^3}{3} \lambda \quad \text { (as } \lambda=\pi^2 E / L^2\right)           (1)

For ‘out-of-plane’ buckling, L_{ e }=L as there is no additional constraint at B in the plane perpendicular to the plane of figure. So

\left(P_{ Cr }\right)_0=\frac{\pi^2 E}{L^2}\left\lgroup \frac{b h^3}{12} \right\rgroup

Thus,

\left(P_{ Cr }\right)_0=\lambda \cdot \frac{b h^3}{12}            (2)

According to the problem, \left(P_{ Cr }\right)_I=\left(P_{ Cr }\right)_0 , therefore

\frac{h b^3}{3}=\frac{b h^3}{12}

or        \frac{h}{b}=2

So, h : b = 2 : 1.