Given: For rectangular s=column section -

b = 200 mm , d = 100 mm , $P = 180 \times 10^{3}$,

e = 100mm in the plane bisecting thickness.

Solution:

Eccentricity about YY-axis

$\therefore I_{YY} = \frac{100 \times 200^{3}}{12} = 66.67 \times 10^{6} mm^{4}$

$Y_{max} = 200/2 = 100 mm$

A = 200 x 100 = $2000mm^{2}$

1) Direct stress = $\sigma_{o} = \frac{P}{A} = \frac{180 \times 10^{3}}{20000} = 9N/mm^{2}$

2) bending stress = $\sigma_{b} = \pm \frac{P.e. y_{max}}{I}$

$ = \pm \frac{180 \times 10^{3} 100 \times 100}{66.67 \times 10^{6}}$

$\sigma_{b} = \pm 27 N/mm^{2}$

3) $\sigma_{max} = \sigma_{o} + \sigma_{b} = 9 + 27 = 36 N / mm^{2} (comp)$

4) $\sigma_{min} = \sigma_{o} - \sigma_{b} = 9 - 27 = 18 N/mm^{2} (tensile)$