Note: In Resultant stress calculation at point A,B,C,D the tensile stress of the point is taken as negative and compressive stress is taken as positive

X = 0.25 m, Y = 0.5 and W = 800 KN

[1] Area of the section = $A = (2.5 + 2.5) \times (1.5 + 1.5) = 5 \times 3$

= $15 \ m^2$

[2] Section modulus about the XX – axis is,

$Z_{xx} = \frac{I_{xx}}{y} = (\frac{3 \times 5^3}{12}) / (\frac{5}{2}) = 12.5 \ m^3$

[3] Section modulus about the YY – axis is,

$Z_{yy} = \frac{I_{yy}}{y} = (\frac{5 \times 3^3}{12}) / (\frac{3}{2}) = 7.5 \ m^3$

[4] Uniform direct stress due to load:

$P_d = \frac{W}{A} = \frac{800}{15} = 53.33 \ KN/m^2$ [c]

(W = 800 KN given)

[5] Max bending stress due to eccentricity of the load about the XX is, (compressive at D & C, and tensile at A & B).

$= \pm \frac{M_{xx}}{Z_{xx}} = \frac{W \times y}{Zxx} = \frac{800 \times 0.25}{12.5} = 32 \ KN/m^2$

[6] Max bending stress due to eccentricity of the load about the YY is, (compressive at B & C, and tensile at A & D).

$= \pm \frac{M_{yy}}{Z_{yy}} = \frac{Wxx}{Zyy} = \frac{800 \times 0.25}{7.5} = 26.67 \ KN/m^2$

[7] Hence, the resultant stresses at four corners are as follows:

Stress at A = 53.33 – 32 – 26.67 = -5.34 $KN/m^2$ (tensile)

Stress at B = 53.33 – 32 + 26.67 = 48 $KN/m^2$ (compressive)

Stress at C = 53.33 + 32 + 26.67 = 112 $KN/m^2$ (compressive)

Stress at D = 53.33 + 32 - 26.67 = 58.66 $KN/m^2$ (compressive)

Note: In Resultant stress calculation at point A,B,C,D the tensile stress of the point is taken as negative and compressive stress is taken as positive