Question

The equilibrium equation obtained by summing all forces on x-direction is ________

(a) \(\frac{∂σ_x}{∂x} + \frac{∂τ_{yx}}{∂y} + \frac{∂τ_{zx}}{∂z} +X=0\)

(b) \(\frac{∂τ_{xy}}{∂x} + \frac{∂σ_y}{∂y} +\frac{∂τ_{zy}}{∂z}+Y=0\)

(c) \(\frac{∂τ_{xz}}{∂x} +\frac{∂τ_{yz}}{∂y} +\frac{∂σ_z}{∂z} +Z=0\)

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

I had been asked this question during a job interview.

I want to ask this question from Elasticity in section Elements of Elasticity of Geotechnical Engineering I

Answer 1

Correct answer is (a) \(\frac{∂σ_x}{∂x} + \frac{∂τ_{yx}}{∂y} + \frac{∂τ_{zx}}{∂z} +X=0\)

Explanation: The sum of forces acting on the element in the x-direction is given by,

\(\{(σ_x+\frac{∂σ_x}{∂x} \frac{dx}{2})dy.dz-(σ_x-\frac{∂σ_x}{∂x}\frac{dx}{2})dy.dz\} +

\{(τ_{yx}+\frac{∂τ_{yx}}{∂y} \frac{dy}{2})dx.dz-(τ_{yx}-\frac{∂τ_{yx}}{∂y}\frac{dy}{2})dx.dz\} + \)

\(\{(τ_{zx}+\frac{∂τ_{zx}}{∂z}\frac{dz}{2})dx.dy-(τ_{zx}-\frac{∂τ_{zx}}{∂z}\frac{dz}{2})dx.dy\} + Xdx.dy.dz = 0\)

Now dividing all the terms by dx.dy.dz we get,

\(\frac{∂σ_x}{∂x} + \frac{∂τ_{yx}}{∂y} + \frac{∂τ_{zx}}{∂z} +X=0.\)