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

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 ___________
Find the three stress components σr, σθ & τrθ in simple radial distribution
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
View Answer

Answer: a
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) \(\frac{∂σ_r}{∂r}+\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}=0\)
View Answer

Answer: c
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) \(\frac{∂σ_r}{∂r}+\frac{1}{r} \frac{∂τ_{rθ}}{∂θ}=0\)
View Answer

Answer: c
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. \(\frac{1}{r}\frac{∂σ_θ}{∂θ}+\frac{∂τ_{rθ}}{∂r}+\frac{2τ_{rθ}}{r}=0.\)
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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) \((\frac{∂^2}{∂r^2} +\frac{1}{r} \frac{∂}{∂r}+\frac{1}{r^2} \frac{∂^2}{∂θ^2} )(σ_r+σ_θ )=1\)
View Answer

Answer: a
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 ________
Find the three stress components σr, σθ & τrθ in simple radial distribution
a) K=\(\frac{2}{2α+sin2α}\)
b) K=2α+sinα
c) K=2α-sinα
d) K=sinα
View Answer

Answer: a
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?
Find the three stress components σr, σθ & τrθ in simple radial distribution
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) \(σ_r=\frac{2Q sin⁡θ}{r}\)
View Answer

Answer: b
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
View Answer

Answer: d
Explanation: From the figure,
Find the three stress components σr, σθ & τrθ in simple radial distribution
On a horizontal plane, the relation between the stress component in x-direction in Cartesian coordinates and polar coordinates is,
σxr sin2⁡θ.
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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
View Answer

Answer: a
Explanation: From the figure,
Find the three stress components σr, σθ & τrθ in simple radial distribution
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
View Answer

Answer: c
Explanation: From the figure,
Find the three stress components σr, σθ & τrθ in simple radial distribution
On xz-plane , the relation between the shear stress component in Cartesian coordinates and polar coordinates is,
τxzr sinθcos⁡θ.
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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) \(σ_x=\frac{2Qx^2 z}{π(x^2+z^2)^2} \)
View Answer

Answer: c
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) \(σ_x=\frac{2Qx^2 z}{π(x^2+z^2)^2} \)
View Answer

Answer: b
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) \(τ_{xz}=\frac{2Qx^2 z}{π(x^2+z^2)^2} \)
View Answer

Answer: d
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 ___________
Find the three stress components σr, σθ & τrθ in simple radial distribution
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) \(σ_r=\frac{2Q}{r}(\frac{sinβsinθ}{2α-sin2α})\)
View Answer

Answer: b
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?
Find the three stress components σr, σθ & τrθ in simple radial distribution
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) \(σ_r=\frac{2Q sin⁡θ}{r}\)
View Answer

Answer: b
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 =\(\frac{π}{2}\),
∴ \(σ_r=\frac{2Q cos(θ-β)}{πr}.\)

Sanfoundry Global Education & Learning Series – Soil Mechanics.

To practice all areas of Soil Mechanics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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