Question

The compatibility equation in terms of plane stress case is given by ________

(a) \((\frac{∂^2}{∂x^2} +\frac{∂^2}{∂z^2})=0\)

(b) \((\frac{∂^2}{∂y^2} +\frac{∂^2}{∂z^2})(σ_y+σ_z )=0\)

(c) \((\frac{∂^2}{∂x^2} +\frac{∂^2}{∂y^2})(σ_x+σ_y )=0\)

(d) \((\frac{∂^2}{∂x^2} +\frac{∂^2}{∂z^2})(σ_x+σ_z )=0\)

The question was posed to me in an internship interview.

Query is from Elasticity Elements topic in chapter Elements of Elasticity of Geotechnical Engineering I

Answer 1

The correct answer is (d) \((\frac{∂^2}{∂x^2} +\frac{∂^2}{∂z^2})(σ_x+σ_z )=0\)

Explanation: For two dimensional case, the six compatibility equations are evidently reduced to one single equation;

\(\frac{∂^2 ε_x}{∂z^2} +\frac{∂^2 ε_z}{∂x^2} = \frac{∂^2 Γ_{xz}}{∂x∂z} ———————(1)\)

From the Hooke’s law equation,

\(ε_x=\frac{1}{E} (σ_x-μσ_z ),\)

\(ε_z=\frac{1}{E} (σ_z-μσ_x ) \) and

\(Γ_{xz}=\frac{2(1+μ)}{E} τ_{zx}\)

Substituting these values in (1) and simplifying further we get,

\((\frac{∂^2}{∂x^2} + \frac{∂^2}{∂z^2})(σ_x+σ_z)=0.\)