# Soil Mechanics Questions and Answers – Elasticity Elements – Stress Functions

This set of Soil Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Elasticity Elements – Stress Functions”.

1. The stress function was introduced by __________
a) G.B Airy
b) Terzaghi
c) Darcy
d) Meyerhof

Explanation: The stress function was introduced by G.B Airy in 1862. The stress function is denoted by Φ and is called Airy stress function. Darcy gave the law of flow of water through soils.

2. The σx in terms of stress function is given by __________
a) $$\frac{∂Φ}{∂z}$$
b) $$\frac{∂^2 Φ}{∂z^2}$$
c) $$\frac{∂^2 Φ}{∂x2}$$
d) $$\frac{∂Φ}{∂x}$$

Explanation: It is convenient to reduce the compatibility and equilibrium equations into a single equation in terms of stress function Φ.
The stress function for stress in the x-direction is given by,
$$\frac{∂^2 Φ}{∂z^2}.$$

3. The σz in terms of stress function is given by __________
a) $$\frac{∂Φ}{∂z}$$
b) $$\frac{∂^2 Φ}{∂z^2}$$
c) $$\frac{∂^2 Φ}{∂x^2}$$
d) $$\frac{∂Φ}{∂x}$$

Explanation: It is convenient to reduce the compatibility and equilibrium equations into a single equation in terms of stress function Φ.
The stress function for stress in the z-direction is given by,
$$σ_z=\frac{∂^2 Φ}{∂x^2}.$$

4. The shear stress τxz in terms of stress function is given by __________
a) $$\frac{∂Φ}{∂z}-γx$$
b) $$-\frac{∂^2Φ}{∂x∂z}-γx$$
c) $$\frac{∂^2Φ}{∂x^2} -γx$$
d) $$\frac{∂Φ}{∂x}-γx$$

Explanation: It is convenient to reduce the compatibility and equilibrium equations into a single equation in terms of stress function Φ.
The shear stress function τxz is given by,
$$τ_{xz}=-\frac{∂^2Φ}{∂x∂z}-γx$$

5. For both plane stress as well as plain strain case the equilibrium equation in x-direction is _______
a) $$\frac{∂σ_x}{∂x}+\frac{∂τ_{yx}}{∂y}=0$$
b) $$\frac{∂σ_x}{∂x}+\frac{∂τ_{yx}}{∂y}+\frac{∂τ_{zx}}{∂z}+X=1$$
c) $$\frac{∂σ_x}{∂x}+\frac{∂τ_{yx}}{∂y}+X=0$$
d) $$\frac{∂σ_x}{∂x}+\frac{∂τ_{zx}}{∂z}=0$$

Explanation: The equilibrium equation in x-direction is $$\frac{∂σ_x}{∂x}+\frac{∂τ_{yx}}{∂y}+\frac{∂τ_{zx}}{∂z}+X=0$$
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 $$\frac{∂σ_x}{∂x}+\frac{∂τ_{zx}}{∂z}=0$$
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6. 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) $$\frac{∂σ_x}{∂x}+\frac{∂τ_{zx}}{∂z}=0$$

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 $$\frac{∂τ_{xz}}{∂x}+\frac{∂σ_z}{∂z}+γ=0$$

7. For two dimensional case, for both plane stress as well as plain strain case the compatibility equation is _______
a) $$\frac{∂^2 ε_x}{∂z^2} +\frac{∂^2 ε_z}{∂x^2} =\frac{∂^2 Γ_{xz}}{∂x∂z}$$
b) $$\frac{∂^2 ε_z}{∂z^2} +\frac{∂^2 ε_y}{∂x^2} =\frac{∂^2 Γ_{zy}}{∂z∂y}$$
c) $$\frac{∂^2 ε_x}{∂y^2} +\frac{∂^2 ε_y}{∂x^2} =\frac{∂^2 Γ_{xy}}{∂x∂y}$$
d) $$\frac{∂^2 ε_z}{∂z^2} +\frac{∂^2 ε_y}{∂x^2} =0$$

Explanation: For two dimensional case, the six compatibility equations are evidently reduced to one single equation;
$$\frac{∂^2 ε_x}{∂z^2} +\frac{∂^2 ε_z}{∂x^2} =\frac{∂^2 Γ_{xz}}{∂x∂z}.$$

This is because, 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.

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

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

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