Question

The equilibrium equation obtained by summing all forces on y-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 by my college professor while I was bunking the class.

My doubt stems from Elasticity in chapter Elements of Elasticity of Geotechnical Engineering I

Answer 1

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

Best explanation: The sum of forces acting on the element in the y-direction is given by,

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

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

\(\{(τ_{xy}+\frac{∂τ_{xy}}{∂x}\frac{dx}{2})dy.dz-(τ_{xy}-\frac{∂τ_{xy}}{∂x}\frac{dx}{2})dy.dz\} + Ydx.dy.dz = 0\)

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

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