Question

The elastic strain ∆ of the subgrade is __________

(a) \(∆=\frac{p}{E}\left[(2-2μ^2)(a^2+z^2)^\frac{1}{2}-\frac{(1+μ) z^2}{(a^2+z^2)^\frac{1}{2}}+(μ+2μ^2-1)z\right]\)

(b) \(∆=\frac{p}{2}\left[\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+ \frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right]\)

(c) \(∆=\frac{p}{2}\left[1-2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+ \frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right]\)

(d) \(∆=\frac{p}{2}\left[1+2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+ \frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right]\)

The question was asked during an interview.

My question is taken from Stresses in Flexible Pavements topic in division Design of Flexible & Rigid Pavement of Geotechnical Engineering II

Answer 1

Correct answer is (a) \(∆=\frac{p}{E}\left[(2-2μ^2)(a^2+z^2)^\frac{1}{2}-\frac{(1+μ) z^2}{(a^2+z^2)^\frac{1}{2}}+(μ+2μ^2-1)z\right]\)

Easiest explanation: The vertical strain is given by,

\(∆=\frac{1}{E}[σ_z-2μ(σ_r)],\) also the vertical and horizontal stresses are given by,

\(σ_z=p\left[1-\left[\frac{1}{1+(\frac{a}{z})^2}\right]^\frac{3}{2}\right]\, and\, σ_r=\frac{p}{2}\left[1+2μ-\frac{2(1+μ)z}{(a^2+z^2)^\frac{1}{2}}+ \frac{z^3}{(a^2+z^2)^\frac{3}{2}}\right],\)

therefore from the three equations, we get,

\(∆=\frac{p}{E}\left[(2-2μ^2)(a^2+z^2)^\frac{1}{2}-\frac{(1+μ) z^2)}{(a^2+z^2)^\frac{1}{2}}+(μ+2μ^2-1)z\right]\)