Question 26.3.48: With one fixed end and other free end, a column of length L ...

(d)

EXPLANATION

For a column of one end fixed and other free, the bucking load is

P_1=\frac{\pi^2 E I}{4 L^2}

For a second column of same length and same cross-sectional area when both ends are fixed, the buckling load is

P_2=\frac{4 \pi^2 E I}{L^2}

∴              \frac{P_2}{P_1}=\frac{\left(\frac{4 \pi^2 E I}{L^2}\right)}{\left(\frac{\pi^2 E I}{4 L^2}\right)}=4 \times 4= \pmb{1 6.}