It carries a load 80 kN at a point 5 cm away from the center of the section along one of the diagonals. Calculate the intensity of the stress at the corners of the column in the plan.

tan $\theta = \frac{300}{400}$

$sin \ \theta = \frac{300}{500} = 0.6$

$Area \ A = 400 \times 300 = 12 \times 10^4 \ mm^2$

$I_{xx} = \frac{400 \times 300^3}{12} = 900 \times 10^6 \ mm^4$

$I_{yy} = \frac{300 \times 400^3}{12} = 1.6 \times 10^9 \ mm^4 $

$p_d = \frac{P}{A}$

$P_d = + [ \frac{80 \times 10^3}{12 \times 10^4}] = 0.667 \ Mpa$ - - - [c]

(Direct Stress)

Extreme stress due to eccentricity about XX – axis.

(Compressive at A, B and tensile at C and D)

Extreme stress due to eccentricity about axis YY.

(Compressive on B, C and tensile on AD)

$Pr]_A = 0.667 + 0.4 – 0.4 = 0.677 \ Mpa$ [C]

$Pr]_B = 0.667 + 0.4 + 0.4 = 1.467 \ Mpa$ [C]

$Pr]_C = 0.667 - 0.4 + 0.4 = 0.667 \ Mpa$ [C]

$Pr]_D = 0.667 - 0.4 – 0.4 = 0.133 = 0.133 \ Mpa$ [T]

NOTE: When Pr value comes negative it means Tension.

Additional: Axial Load for no tension =

$0.133 \times A = 0.133 \times 400 \times 300$

$= 15.96 \times 10^3 \ N$