Question

The equilibrium equation obtained by summing all forces on z-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 got this question in exam.

The origin of the question is Elasticity in chapter Elements of Elasticity of Geotechnical Engineering I

Answer 1

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

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

\(\{(σ_z+\frac{∂σ_z}{∂z}\frac{dz}{2})dx.dy-(σ_z-\frac{∂σ_z}{∂z} \frac{dz}{2})dx.dy\} +

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

\( \{(τ_{yz}+\frac{∂τ_{yz}}{∂y} \frac{dy}{2})dz.dx-(τ_{yz}-\frac{∂τ_{yz}}{∂y}\frac{dy}{2})dz.dx\} + Zdx.dy.dz = 0\)

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

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