Soil Mechanics Questions and Answers – Consolidation Equation in Polar Coordinates

This set of Soil Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Consolidation Equation in Polar Coordinates”.

1. The transformation from Cartesian to plane coordinates in x-direction is given by ______
a) x=rsinθ
b) x=rcosθ
c) x=rcos2θ
d) x=rsin2θ
View Answer

Answer: b
Explanation: The transformation from Cartesian to plane coordinates in x-direction is given by,
x=rcosθ. Where, x=coordinate in Cartesian form
r=coordinate in polar form.
r2=x2+y2 and θ=tan-1(y/x).

2. The transformation from Cartesian to plane coordinates in y-direction is given by ______
a) y=rsinθ
b) y=rcosθ
c) y=rcos2θ
d) y=rsin2θ
View Answer

Answer: a
Explanation: The transformation from Cartesian to plane coordinates in y-direction is given by,
y=rsinθ. Where, y=coordinate in Cartesian form
r=coordinate in polar form.
r2=x2+y2 and θ=tan-1(y/x).

3. The transformation from Cartesian to plane coordinates in x-direction is given by ______
a) z=rsinθ
b) z=rcosθ
c) z=z
d) z=r2sinθcosθ
View Answer

Answer: b
Explanation: In the z-direction, since it does not make any angle with the axis, the z-direction in both the Cartesian as well as the polar coordinates is the same.
∴ z=z.
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4. In polar form the term, \(\frac{∂r}{∂x}\) is given by______
a) \(\frac{∂r}{∂x}=sinθ\)
b) \(\frac{∂r}{∂x}=cosθsinθ\)
c) \(\frac{∂r}{∂x}=cosθ \)
d) \(\frac{∂r}{∂x}=sin2θ\)
View Answer

Answer: c
Explanation: Since we know, r2=x2+y2 —————-(1)
∴ differentiating (1) with respect to x, we get,
\(\frac{∂r}{∂x}=\frac{x}{r}=\frac{rcosθ}{r}=cosθ.\)

5. In polar form the term, \(\frac{∂r}{∂y}\) is given by______
a) \(\frac{∂r}{∂y}=sinθ\)
b) \(\frac{∂r}{∂y}=cosθsinθ\)
c) \(\frac{∂r}{∂y}=cosθ\)
d) \(\frac{∂r}{∂y}=sin2θ\)
View Answer

Answer: a
Explanation: Since we know, r2=x2+y2 —————-(1)
∴ differentiating (1) with respect to y, we get,
\(\frac{∂r}{∂y}=\frac{y}{r}=\frac{rsinθ}{r}=sinθ.\)
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6. In polar form the term, \(\frac{∂θ}{∂x}\) is given by______
a) \(\frac{∂θ}{∂x}=\frac{sinθ}{r} \)
b) \(\frac{∂θ}{∂x}=-cosθsinθ \)
c) \(\frac{∂θ}{∂x}=-\frac{cosθ}{r}\)
d) \(\frac{∂θ}{∂x}=\frac{-sinθ}{r}\)
View Answer

Answer: d
Explanation: Since we know,
θ=tan-1(y/x) —————-(1)
∴ differentiating (1) with respect to x, we get,
\(\frac{∂θ}{∂x}=\frac{-sinθ}{r}.\)

7. In polar form the term, \(\frac{∂θ}{∂y}\) is given by______
a) \(\frac{∂θ}{∂y}=\frac{sinθ}{r} \)
b) \(\frac{∂θ}{∂y}=cosθsinθ\)
c) \(\frac{∂θ}{∂y}=\frac{cosθ}{r}\)
d) \(\frac{∂θ}{∂y}=\frac{sin2θ}{r}\)
View Answer

Answer: c
Explanation: Since we know,
θ=tan-1(y/x) —————-(1)
∴ differentiating (1) with respect to y, we get,
\(\frac{∂θ}{∂y}=\frac{cosθ}{r}.\)
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8. The partial differentiation of excess hydrostatic pressure \overline{u} as a function of r and θ with respect to x is given by _______
a) \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ\)
b) \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r} \frac{∂\overline{u}}{∂θ} cosθ\)
c) \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} sinθ-\frac{1}{r} \frac{∂\overline{u}}{∂θ} sinθ\)
d) \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} sinθ-\frac{1}{r} \frac{∂\overline{u}}{∂θ} cosθ\)
View Answer

Answer: a
Explanation: Partially differentiating the excess hydrostatic pressure \overline{u} with respect to x,
\(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} \frac{∂r}{∂x}+\frac{∂\overline{u}}{∂θ}\frac{∂θ}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ.\)
∴ \(\frac{∂\overline{u}}{∂x}=\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} sinθ.\)

9. The term \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}\) in terms of r and θ is given by _______
a) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}-\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}\)
b) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}+\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}\)
c) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}-\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}\)
d) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}+\frac{1}{c^2}\frac{∂^2 \overline{u}}{∂θ^2}\)
View Answer

Answer: b
Explanation: The second order of the partial differentiation of excess hydrostatic pressure u with respect to x is,
\(\frac{∂^2 \overline{u}}{∂x^2}=(\frac{∂}{∂r} cosθ-\frac{1}{r} sinθ \frac{∂}{∂θ})(\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} θsinθ) ————-(1)\)
also, the second order of the partial differentiation of excess hydrostatic pressure \overline{u} with respect to y is,
\(\frac{∂^2 \overline{u}}{∂y^2}=(\frac{∂}{∂r} cosθ-\frac{1}{r} sinθ \frac{∂}{∂θ})(\frac{∂\overline{u}}{∂r} cosθ-\frac{1}{r}\frac{∂\overline{u}}{∂θ} θsinθ)\) ———-(2)
∴ adding (1) and (2),
\(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}+\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}.\)
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10. In case of radial symmetry, \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}\) is_________
a) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}\)
b) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}\)
c) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=-\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}\)
d) \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=-\frac{∂^2 \overline{u}}{∂r^2}-\frac{1}{r} \frac{∂\overline{u}}{∂r}\)
View Answer

Answer: a
Explanation: The term \(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}\) in terms of r and θ is given by,
\(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}+\frac{1}{r^2}\frac{∂^2 \overline{u}}{∂θ^2}.\)
In case of radial symmetry, \overline{u} is independent of θ and hence we get,
\(\frac{∂^2 \overline{u}}{∂x^2}+\frac{∂^2 \overline{u}}{∂y^2}=\frac{∂^2 \overline{u}}{∂r^2}+\frac{1}{r} \frac{∂\overline{u}}{∂r}\)

11. In three dimensional consolidation of sand drain, having radial symmetry, the governing consolidation equation is _______
a) \(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})-C_{vz}\frac{∂^2 \overline{u}}{∂z^2}\)
b) \(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}\)
c) \(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}-\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}\)
d) \(\frac{∂\overline{u}}{∂t}=C_{vz} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}\)
View Answer

Answer: b
Explanation: In three dimensional consolidation of sand drain, for the case of radial symmetry,
Cvx=Cvy =Cvr
∴ the governing consolidation equation is,
\(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}.\)

12. The radial flow part of governing consolidation equation of three dimensional consolidation having radial symmetry is _______
a) \(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})\)
b) \(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}\)
c) \(\frac{∂\overline{u}}{∂t}=C_{vz} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}\)
d) \(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})\)
View Answer

Answer: a
Explanation: In three dimensional consolidation of sand drain, having radial symmetry, the governing consolidation equation is,
\(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2},\) in this the radial flow part of equation is,
\(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r}).\)

13. The one dimensional flow part of governing consolidation equation of three dimensional consolidation having radial symmetry is _______
a) \(\frac{∂\overline{u}}{∂t}=C_{vr} \frac{∂\overline{u}}{∂r^2}\)
b) \(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{∂\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}\)
c) \(\frac{∂\overline{u}}{∂t}=C_{vz} (\frac{∂\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}\)
d) \(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{∂\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})\)
View Answer

Answer: a
Explanation: In three dimensional consolidation of sand drain, having radial symmetry, the governing consolidation equation is,
\(\frac{∂\overline{u}}{∂t}=C_{vr} (\frac{∂\overline{u}}{∂r^2}+\frac{1}{r}\frac{∂\overline{u}}{∂r})+C_{vz}\frac{∂^2 \overline{u}}{∂z^2}\), in this the one dimensional flow part of equation is,
\(\frac{∂\overline{u}}{∂t}=C_{vr} \frac{∂\overline{u}}{∂r^2}\).

14. The equation given by Carillo in 1942 relating the degree of consolidation in one dimensional flow (Uz) and radial flow (Ur) is _______
a) (1-U)=(1-Uz)(1+Ur)
b) (1-U)=(1-Uz)(1-Ur)
c) (1-U)=(1+Uz)(1-Ur)
d) (1-U)=(1+Uz)(1+Ur)
View Answer

Answer: b
Explanation: The equation given by Carillo in 1942 relating the degree of consolidation in one dimensional flow (Uz) and radial flow (Ur) is,
(1-U)=(1-Uz)(1-Ur). This is the combination of the one dimensional flow and radial flow parts from the governing consolidation equation.

15. The time factor Tv for the vertical flow is given by _______
a) \(T_v=\frac{C_{vz} t}{H^2} \)
b) \(T_v=\frac{-C_{rz} t}{H^2} \)
c) \(T_v=\frac{C_{vz}}{H^2} \)
d) \(T_v=\frac{C_{vz} t}{H}\)
View Answer

Answer: a
Explanation: The time factor Tv for the vertical flow is given by,
\(T_v=\frac{C_{vz} t}{H^2}, \)
Where Cvz is the coefficient of consolidation in z-direction
t=elapsed time
H=average drainage path.

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