Question

For a Poisson’s ratio of 0.5, the elastic strain ∆ is ___________

(a) \(∆=\frac{3pa^2}{2E} \)

(b) \(∆=\frac{pa^2}{2E(a^2+z^2)^\frac{1}{2}} \)

(c) \(∆=\frac{3pa^2}{2E(a^2+z^2)^\frac{1}{2}} \)

(d) \(∆=\frac{a^2}{2E(a^2+z^2)^\frac{1}{2}} \)

This question was addressed to me in an interview.

Origin of the question is Stresses in Flexible Pavements in chapter Design of Flexible & Rigid Pavement of Geotechnical Engineering II

Answer 1

Right answer is (c) \(∆=\frac{3pa^2}{2E(a^2+z^2)^\frac{1}{2}} \)

The explanation is: The elastic strain is given by,

\(∆=\frac{p}{E}\left[(2-2μ^2)(a^2+z^2)^\frac{1}{2}-\frac{(1+μ) z^2)}{(a^2+z^2)^\frac{1}{2}}+(μ+2μ^2-1)z\right]\)substituting μ=0.5,

∴ \(∆=\frac{3pa^2}{2E(a^2+z^2)^\frac{1}{2}}. \)