Question (contd.) diameter may be taken as 0.8 times the external diameter. Take $\sigma_c=550N/mm^2\ and\ \alpha=\frac{1}{1600}$ in Rankin's formula.

Given data:

$L=4m=4000mm\\ P=250\ kN=250\times10^3\ N\\ \sigma_c=550\ N/mm^2\\ \alpha=\frac{1}{1600}\\ FOS=5\\ d=0.80D $

Let D = Extrenal dia.

d = internal diameter

crippling load $(P_c)$ = Axial force x F.O.S. = $250 \times 10^3 \times 5$

$P_c = 1.25 \times 10^6 N$

To find: Design the column.

Since both the ends …