Question

The influence factor for the vertical stress under the corner of a uniformly loaded rectangular area of size 1m*2m at depth 5m and load of 80 kN/m^2 is given by ___________

(a) 0.6212

(b) 0.7465

(c) 0.0328

(d) 0.0624

This question was addressed to me during a job interview.

I'd like to ask this question from Stress Distribution topic in portion Stress Distribution of Geotechnical Engineering I

Answer 1

Correct answer is (c) 0.0328

For explanation I would say: Given,

Load q=80 kN/m^2

Size of rectangle=1m*2m

Depth z=5m

Therefore m=1/5=0.2

n=2/5=0.4

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

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

∴ σz=kq

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

\(k=\frac{1}{4π} \left[\frac{2*0.2*0.\sqrt{(0.2^2+0.4^2+1)}}{0.2^2+0.4^2+0.2^2 0.4^2+1}*\frac{0.2^2+0.4^2+2}{0.2^2+0.4^2+1}+tan^{-1}\frac{20.20.4\sqrt{(0.2^2+0.4^2+1)}}{0.2^2+0.4^2-0.2^2 0.4^2+1}\right] \)

∴k=0.0328.