Question

The vertical stress under the corner of a uniformly loaded rectangular area of size 2m*4m at depth 5m and load of 80 kN/m^2 is given by ___________

(a) 6.22 kN/m^2

(b) 7.45 kN/m^2

(c) 8.12 kN/m^2

(d) 9.23 kN/m^2

This question was addressed to me in an online interview.

The origin of the question is Stress Distribution topic in division Stress Distribution of Geotechnical Engineering I

Answer 1

Right option is (b) 7.45 kN/m^2

The best explanation: Given,

Load q=80 kN/m^2

Size of rectangle=2m*4m

Depth z=5m

Therefore m=2/5=0.4

n=4/5=0.8

∴ the vertical stress at the corner of rectangle is given by,

\(σ_z=\frac{q}{4π}\left[\frac{2mn\sqrt{(m^2+n^2+1)}}{m^2+n^2+m^2 n^2+1}* \frac{m^2+n^2+2}{m^2+n^2+1}+tan^{-1}⁡\frac{2mn\sqrt{(m^2+n^2+1)}}{m^2+n^2+m^2 n^2+1} \right] \)

∴ \(σ_z=\frac{80}{4π}\left[\frac{2*0.4*0.8\sqrt{(0.4^2+0.8^2+1)}}{0.4^2+0.8^2+0.4^2 0.8^2+1}* \frac{0.4^2+0.8^2+2}{0.4^2+0.8^2+1}+tan^{-1}\frac{2*0.4*0.8\sqrt{(0.4^2+0.8^2+1)}}{0.4^2+0.8^2-0.4^2 0.8^2+1} \right] \)

∴ σz=7.45 kN/m^2.