# Soil Mechanics Questions and Answers – Stress Distribution – Vertical and Horizontal Pressure

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This set of Soil Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Stress Distribution – Vertical and Horizontal Pressure”.

1. In simple radial distribution, the three stress components σr, σθ and τ are given by ___________

a) $$σ_r=K \frac{Q cos⁡θ}{r}, σ_θ=0 \,and\, τ_{rθ}=0$$
b) σr=KQ, σθ=0 and τ=0
c) $$σ_r=\frac{Q cos⁡θ}{r}, σ_θ=0 \,and\, τ_{rθ}=0$$
d) σr=0, σθ=0 and τ= 0

Explanation: At any radial distance r and polar angle θ, Mitchell found that the three stress components σr, σθ and τ are given by,
$$σ_r=K \frac{Q cos⁡θ}{r}, σ_θ=0 \,and\, τ_{rθ}=0.$$
Where K is a constant to be found by boundary conditions. The above solution is valid only if it satisfies the equilibrium equations and the compatibility equation.

2. The equilibrium equation in polar coordinates is given by _____________
a) $$\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}+\frac{σ_r-σ_θ}{r}=0$$
b) $$\frac{∂σ_r}{∂r}+\frac{∂τ_{rθ}}{∂θ}+\frac{σ_r-σ_θ}{r}=0$$
c) $$\frac{∂σ_r}{∂r}+\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}+\frac{σ_r-σ_θ}{r}=0$$
d) $$\frac{∂σ_r}{∂r}+\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}=0$$

Explanation: The equilibrium equations in polar coordinates are given by,
1. $$\frac{∂σ_r}{∂r}+\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}+\frac{σ_r-σ_θ}{r}=0$$
2. $$\frac{1}{r}\frac{∂σ_θ}{∂θ}+\frac{∂τ_{rθ}}{∂r}+\frac{2τ_{rθ}}{r}=0.$$

3. The equilibrium equation in polar coordinates is given by _____________
a) $$\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}+\frac{σ_r-σ_θ}{r}=0$$
b) $$\frac{∂σ_r}{∂r}+\frac{∂τ_{rθ}}{∂θ}+\frac{σ_r-σ_θ}{r}=0$$
c) $$\frac{∂σ_r}{∂r}+\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}+\frac{σ_r-σ_θ}{r}=0$$
d) $$\frac{∂σ_r}{∂r}+\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}=0$$

Explanation: The equilibrium equations in polar coordinates are given by,
1. $$\frac{∂σ_r}{∂r}+\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}+\frac{σ_r-σ_θ}{r}=0$$
2. $$\frac{1}{r}\frac{∂σ_θ}{∂θ}+\frac{∂τ_{rθ}}{∂r}+\frac{2τ_{rθ}}{r}=0.$$

4. The compatibility equation in terms of stress components in polar coordinates are given by ____________
a) $$(\frac{∂^2}{∂r^2} +\frac{1}{r} \frac{∂}{∂r}+\frac{1}{r^2} \frac{∂^2}{∂θ^2} )(σ_r+σ_θ )=0$$
b) $$(\frac{∂^2}{∂r^2} +\frac{1}{r} \frac{∂}{∂r}+\frac{1}{r^2} \frac{∂^2}{∂θ^2} )(σ_θ )=0$$
c) $$(\frac{∂^2}{∂r^2} +\frac{1}{r} \frac{∂}{∂r}+\frac{1}{r^2} \frac{∂^2}{∂θ^2} )(σ_r )=0$$
d) $$(\frac{∂^2}{∂r^2} +\frac{1}{r} \frac{∂}{∂r}+\frac{1}{r^2} \frac{∂^2}{∂θ^2} )(σ_r+σ_θ )=1$$

Explanation: The compatibility equation is the additional equation to solve the stress problem. The compatibility equation in terms of stress components in polar coordinates are given by,
$$(\frac{∂^2}{∂r^2} +\frac{1}{r} \frac{∂}{∂r}+\frac{1}{r^2} \frac{∂^2}{∂θ^2} )(σ_r+σ_θ )=0.$$

5. In simple radial distribution, if $$σ_r=K \frac{Q cos⁡θ}{r},$$ then the value of K is ________

a) K=$$\frac{2}{2α+sin2α}$$
b) K=2α+sinα
c) K=2α-sinα
d) K=sinα

Explanation: Considering the equilibrium of the wedge aob,
We have $$KQ(α+\frac{1}{2} sin2α)=Q$$
∴ K=$$\frac{2}{2α+sin2α}.$$

6. When the ground is horizontal, $$α=\frac{π}{2}$$ in constant K. What will be the radial stress σr due to vertical line load?

a) $$σ_r=\frac{Q cos⁡θ}{r}$$
b) $$σ_r=\frac{2Q cos⁡θ}{πr}$$
c) $$σ_r=\frac{Q sin⁡θ}{r}$$
d) $$σ_r=\frac{2Q sin⁡θ}{r}$$

Explanation: At any radial distance r and polar angle θ, Mitchell found that the radial stress component σr is given by,
$$σ_r=K\frac{Q cos⁡θ}{r} \,where\, K=\frac{2}{2α+sin2α}$$
When the ground is horizontal =$$\frac{π}{2},$$
∴ $$σ_r=\frac{2Q cos⁡θ}{πr}.$$

7. The relation between the stress component in x-direction on a horizontal plane in Cartesian coordinates and polar coordinates for vertical line load is ___________
a) σxr tan2⁡θ
b) σxr cosec2⁡θ
c) σxr cos⁡θ
d) σxr sin2⁡θ

Explanation: From the figure,

On a horizontal plane, the relation between the stress component in x-direction in Cartesian coordinates and polar coordinates is,
σxr sin2⁡θ.

8. The relation between the stress component in z-direction on a horizontal plane in Cartesian coordinates and polar coordinates for vertical line load is ___________
a) σzr cos2⁡θ
b) σzr cosec2⁡θ
c) σzr cos⁡θ
d) σzr sin2⁡θ

Explanation: From the figure,

On a horizontal plane, the relation between the stress component in z-direction in Cartesian coordinates and polar coordinates is,
σzr cos2⁡θ.

9. The relation between the shear stress component in xz-plane in Cartesian coordinates and polar coordinates for vertical line load is ___________
a) τxzr tan2⁡θ
b) τxzr cosec2⁡θ
c) τxzr sinθcos⁡θ
d) τxzr sin2⁡θ

Explanation: From the figure,

On xz-plane , the relation between the shear stress component in Cartesian coordinates and polar coordinates is,
τxzr sinθcos⁡θ.

10. The stress component in x-direction on a horizontal plane in Cartesian coordinates for horizontal line load is ___________
a) $$σ_x=\frac{2Q}{xzsinθcos⁡θ}$$
b) $$σ_x=\frac{2Qxz^2}{π(x^2+z^2)^2}$$
c) $$σ_x=\frac{2Qx^3}{π(x^2+z^2)^2}$$
d) $$σ_x=\frac{2Qx^2 z}{π(x^2+z^2)^2}$$

Explanation: On a horizontal plane, the relation between the stress component in x-direction in Cartesian coordinates and polar coordinates is,
$$σ_x=σ_r sin^2⁡θ \,where\, σ_r=K \frac{Q cos⁡θ}{r}$$
And $$K=\frac{2}{2α+sin2α}$$
∴ $$σ_x=\frac{2Qx^3}{π(x^2+z^2)^2}.$$

11. The stress component in x-direction on a horizontal plane in Cartesian coordinates for horizontal line load is ___________
a) $$σ_x=\frac{2Q}{xzsinθcos⁡θ}$$
b) $$σ_x=\frac{2Qxz^2}{π(x^2+z^2)^2}$$
c) $$σ_x=\frac{2Qx^3}{π(x^2+z^2)^2}$$
d) $$σ_x=\frac{2Qx^2 z}{π(x^2+z^2)^2}$$

Explanation: On a horizontal plane, the relation between the stress component in z-direction in Cartesian coordinates and polar coordinates is,
$$σ_z=σ_r cos^2⁡θ \,where\, σ_r=K \frac{Q cos⁡θ}{r}$$
And $$K=\frac{2}{2α+sin2α}$$
∴ $$σ_x=\frac{2Qx^3}{π(x^2+z^2 )^2}.$$

12. The shear stress component in xz-plane in Cartesian coordinates for horizontal line load is ___________
a) $$τ_{xz}=\frac{2Q}{xzsinθcos⁡θ}$$
b) $$τ_{xz}=\frac{2Qxz^2}{π(x^2+z^2)^2}$$
c) $$τ_{xz}=\frac{2Qx^3}{π(x^2+z^2)^2}$$
d) $$τ_{xz}=\frac{2Qx^2 z}{π(x^2+z^2)^2}$$

Explanation: On a xz-plane, the relation between the shear stress component in Cartesian coordinates and polar coordinates is,
$$τ_{xz}=σ_r sinθcos⁡θ \,where\, σ_r=K \frac{Q cos⁡θ}{r}$$
And $$K=\frac{2}{2α+sin2α}$$
∴$$τ_{xz}=\frac{2Qx^2 z}{π(x^2+z^2)^2}.$$

13. The radial stress component σr due to inclined line load of intensity Q per unit length is given by ___________

a) $$σ_r=\frac{2Q}{r}(\frac{cosβcosθ}{2α+sin2α})$$
b) $$σ_r=\frac{2Q}{r} (\frac{cosβcosθ}{2α+sin2α}+\frac{sinβsinθ}{2α-sin2α})$$
c) $$σ_r=\frac{Q}{r} (\frac{cosβcosθ}{2α+sin2α}+\frac{sinβsinθ}{2α-sin2α})$$
d) $$σ_r=\frac{2Q}{r}(\frac{sinβsinθ}{2α-sin2α})$$

Explanation: The stresses due to inclined load of intensity Q per unit length can be found by resolving the inclined load into horizontal and vertical components, we get,
$$σ_r=\frac{2Q}{r} (\frac{cosβcosθ}{2α+sin2α}+\frac{sinβsinθ}{2α-sin2α}).$$

14. When the ground is horizontal, $$α=\frac{π}{2}$$ in constant K. What will be the radial stress σr due to inclined line load at the horizontal ground surface?

a) $$σ_r=\frac{Q cos⁡θ}{r}$$
b) $$σ_r=\frac{2Q cos(θ-β)}{πr}$$
c) $$σ_r=\frac{Q sin⁡θ}{r}$$
d) $$σ_r=\frac{2Q sin⁡θ}{r}$$

Explanation: At any radial distance r and polar angle θ, the radial stress component σr due to inclined line load is given by,
$$σ_r=\frac{2Q}{r} (\frac{cosβcosθ}{2α+sin2α}+\frac{sinβsinθ}{2α-sin2α}).$$
When the ground is horizontal =$$\frac{π}{2}$$,
∴ $$σ_r=\frac{2Q cos(θ-β)}{πr}.$$

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