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 are fixed

Effective length $L_e=\frac{L}{2}=\frac{4000}{2}=2000mm $

Moment of inertia $I=\frac{\pi}{64}(D^4-d^4)=\frac{\pi}{64}[D^4=(0.8D)^4]\\ I=26.98\times10^{-3}D^4$

Area, $A=\frac{\pi}{4}(D^2-d^2)=\frac{\pi}{4}(D^2-(0.8D)^2)=0.2827D^2 $

Radius of Gyration, $k=\sqrt{\frac{I}{A}}=\sqrt{\frac{26.98\times10^{-3}D^4}{0.2827D^2}}=0.389D$

Crippling load by Rankine formula,

$P_c=\frac{\sigma_cA}{1+\alpha\left( \frac{1}{k}\right)^2}\\ \therefore 1.25\times10^6 =\frac{550\times0.22827D^2}{1+\frac{1}{1600}\left( \frac{2000}{0.3089}\right)^2}\\ \therefore 1.25\times10^6=\frac{155.485D^2}{1+\frac{26.2\times10^3}{D^2}}\\ 1.25\times10^6=\frac{155.485D^4}{D^2+26.2\times10^3}\\ 8.039\times10^3=\frac{D^4}{D^2+26.2\times10^3}\\ \therefore D^4-8.039\times10^3D^2-210.63\times10^6=0\\ \therefore D^2=19078.93\\ D=138.12mm\\ \therefore d=0.8\times138.12=110.5mm$

$\therefore$ The external diameter is $\underline{138.12mm}$

& Internal diameter is $\underline{110.5mm}$