Given:

For short circular (hollow) column

D = 40 cm = 400 mm

d = 20 cm = 200 mm

Criteria - no tensional base

Solution:

C/S Area = $A = \frac{\pi}{4}(D^{2} - d^{2}) = \frac{\pi}{4}(400^{2} - 200^{2}) = 94.25 \times 10^{3} mm^{2}$

M I = $I = \frac{\pi}{64}(D^{4} - d^{4}) = \frac{\pi}{64}(400^{4} - 200^{4}) = 11.78 \times 10^{8} mm^{4}$

$y_{max} = \frac{D}{2} = \frac{400}{2} = 200 mm$

For no tension Condition,

$\sigma_{o} = \sigma_{b}$

$\therefore \frac{P}{A} = \frac{P.e.y_{max}}{I}$

$\therefore e = \frac{I}{A \times y_{max}} = \frac{11.78 \times 10^{8}}{94.25 \times 10^{3} \times 200}$

$e = 62.49 mm$

A hollow circular column of 25 cm external and 20 cm internal diameter respectively carries an axial load of 200 kN . It also carries a load of 100 kN on a bracket whose line of action is 20 cm from the axis of the column. Determine the maximum and minimum stress at the base action.