Question

For both plane stress as well as plain strain case the equilibrium equation in z-direction is _______

(a) \(\frac{∂τ_{xz}}{∂x}+\frac{∂σ_z}{∂z}+γ=0\)

(b) \(\frac{∂σ_x}{∂x}+\frac{∂τ_{zx}}{∂z}+γ=1\)

(c) \(\frac{∂σ_x}{∂x}+\frac{∂τ_{yx}}{∂y}+γ=0\)

(d) \(\frac{∂σ_x}{∂x}+\frac{∂τ_{zx}}{∂z}=0\)

This question was addressed to me during an online exam.

The doubt is from Elasticity Elements in portion Elements of Elasticity of Geotechnical Engineering I

Answer 1

The correct answer is (a) \(\frac{∂τ_{xz}}{∂x}+\frac{∂σ_z}{∂z}+γ=0\)

The best explanation: The equilibrium equation in x-direction is \(\frac{∂τ_{xz}}{∂x} + \frac{∂τ_{yz}}{∂y} +\frac{∂σ_z}{∂z}+γ=0.\)

In the plain strain case, one dimension (y) is very large in comparison to the other two directions. So, the strain components in this direction are zero. Also in plain stress condition, the stresses in y-direction are considered as zero.

∴ The equation reduces to \(\frac{∂τ_{xz}}{∂x}+\frac{∂σ_z}{∂z}+γ=0\)