Question

The safe stress , for a hollow steel column which carries an axial load of 2200 kN is 120 MN/m^2. If the external diameter of the column is 25cm, determine the internal diameter.

Answer 1

Given axial load, $P= 2.2\times10^3kN$

Stress $\delta= 120MN/m^2=120N/mm^2=12N/cm$

External diameter = 25cm

Area, A =$\frac{Load}{Stress}=\frac{2.2\times10^3}{12}=183.333$

But the area of hollow A =$\frac{\pi}{4}(D^2-d^2)=183.333\implies \frac{\pi}{4}(25^2-d^2)=183.333$

$0.785714285\times(625-d^2)=183.333\implies d^2=391.6667089$

$d=\sqrt(391.6667089)=19.79 cm$