Foundation Engineering Questions and Answers – Passive Earth Pressure: Rankine’s Theory

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This set of Foundation Engineering Multiple Choice Questions & Answers focuses on “Passive Earth Pressure: Rankine’s Theory”.

1. In case of passive state of plastic state of equilibrium, the lateral pressure is the _________
a) minor principal stress
b) major principal stress
c) major principal plane
d) minor principal plane
View Answer

Answer: b
Explanation: In case of passive state of plastic state of equilibrium, the lateral pressure is the major principal stress while the vertical pressure is the minor principal stress.
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2. The passive earth pressure intensity is given by _______
a) pp=γz cot2 α
b) pp=γz tan2 α
c) pp=γz sin2 α
d) pp=γz cos2 α
View Answer

Answer: b
Explanation: In case of passive state of plastic state of equilibrium,
pph1, and σ3v=γz,
∴ for non-cohesive soils, σ13 tan2 α,
∴ pp=γz tan2 α.

3. The passive earth pressure intensity in terms of coefficient of passive earth pressure is _________
a) pp=Kp γH
b) pp=γH
c) pp=Kp H
d) pp=Kp γ
View Answer

Answer: a
Explanation: The passive earth pressure intensity is given by,
pp=γz tan2 α,
since, Kp= tan2 α,
∴ pp=Kp γH.
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4. The ratio between coefficients of passive and active earth pressure is __________
a) cot2 \((45°+\frac{φ}{2})\)
b) cos2 \((45°+\frac{φ}{2})\)
c) tan4 \((45°+\frac{φ}{2})\)
d) sin2 \((45°+\frac{φ}{2})\)
View Answer

Answer: c
Explanation: The coefficient of passive earth pressure is Kp= tan2 α and that of the coefficient of active earth pressure is \(K_a=\frac{1}{tan^2(45°+\frac{φ}{2})},\)
∴ \(\frac{K_p}{K_a}\) = tan2 α* tan2 α
\(\frac{K_p}{K_a} =tan^4 (45°+\frac{φ}{2}).\)

5. For angle of internal friction of 250, the ratio of coefficient of passive and active earth pressure is __________
a) 2
b) 4
c) 6
d) 8
View Answer

Answer: c
Explanation: Given,
angle of internal friction φ=250,
∴ \(\frac{K_p}{K_a} =tan^4 (45°+\frac{φ}{2}) = tan^4 (45°+\frac{25}{2}),\)
Kp/Ka =6.
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6. The relation between Kp and Ka is _____________
a) Kp=Ka
b) Kp=1+Ka
c) Kp=1-Ka
d) Kp=1/Ka
View Answer

Answer: d
Explanation: The coefficient of passive earth pressure is,
Kp= tan2 α,
the coefficient of active earth pressure is,
\(K_a=\frac{1}{tan^2(45°+\frac{φ}{2})},\)
∴ Kp=1/Ka.

7. For a cohesion less soil, the total passive pressure Pp at a depth H is ___________
a) Pp=\(\frac{1}{2}\) Kp γH2
b) Pp=γH2
c) Pp=Kp γH2
d) Pp=Kp H2
View Answer

Answer: a
Explanation: The pressure intensity pp is given by,
pp=Kp γz,
The total earth pressure P0 at rest per unit length is,
Pp=\(\int_0^H\)Kp γz.dz
∴ Pp=\(\frac{1}{2}\) Kp γH2.
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8. For a cohesion less soil with surcharge Q, the passive pressure intensity is ___________
a) pp=Kp γz+Kp q
b) pp=Kp γz-Kp q
c) pp=Kp γz*Kp q
d) pp=Kp γz/Kp q
View Answer

Answer: a
Explanation: When the backfill is horizontal and carries a surcharge q, then the vertical pressure increment will be by q. Due to this, the lateral pressure will increase by Kp q. Hence, lateral pressure at the depth of wall H is pp=Kp γz+Kp q.

9. For a surcharge angle β, the passive earth pressure is __________
a) \(p_p=γz \frac{\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}} \)
b) \(p_p=γz \frac{cosβ+\sqrt{cos^2 β-cos^2 φ}}{cosβ-\sqrt{cos^2 β-cos^2 φ}}\)
c) \(p_p=γz \frac{cosβ-\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}}\)
d) \(p_p=γzcosβ \frac{cosβ+\sqrt{cos^2 β-cos^2 φ}}{cosβ-\sqrt{cos^2 β-cos^2 φ}}\)
View Answer

Answer: d
Explanation: If the backfill is having its top surface inclined to the horizontal at an angle β, then the passive earth pressure is given by,
\(p_p=γzcosβ \frac{cosβ+\sqrt{cos^2 β-cos^2 φ}}{cosβ-\sqrt{cos^2 β-cos^2 φ}},\)
where, β= surcharge angle
γ=unit weight of the soil
φ= angle of internal friction.
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10. For a surcharge angle β the coefficient of passive earth pressure is _______
a) Kp=\(\frac{\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}} \)
b) Kp=\(cosβ \frac{cosβ+\sqrt{cos^2 β-cos^2 φ}}{cosβ-\sqrt{cos^2 β-cos^2 φ}}\)
c) Kp=\(\frac{cosβ-\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}}\)
d) Kp=\(cosβ \frac{cosβ-\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}}\)
View Answer

Answer: b
Explanation: The vertical pressure in case of backfill with surcharge,
σ= γzcosβ,
Therefore, from the assumption of stresses being conjugate,
Kp=\(cosβ \frac{cosβ+\sqrt{cos^2 β-cos^2 φ}}{cosβ-\sqrt{cos^2 β-cos^2 φ}}\)

11. When the surcharge angle β is reduced to zero, then the coefficient of passive earth pressure is ______
a) 1
b) \(\frac{1-sinφ}{1+sinφ} \)
c) \(\frac{1+sinφ}{1-sinφ} \)
d) 0
View Answer

Answer: c
Explanation: For backfill with sloping surface, the coefficient of passive earth pressure is given by,
Kp=\(cosβ \frac{cosβ+\sqrt{cos^2 β-cos^2 φ}}{cosβ-\sqrt{cos^2 β-cos^2 φ}},\)
When the surcharge angle reduces to zero, β=0,
substituting this in the equation, we get,
Ka=\(\frac{1+sinφ}{1-sinφ}.\)

12. In case of cohesive backfill, the passive pressure intensity is given by _______
a) pp=γz tan2 α-2c tan⁡α
b) pp=γz tan2 α+2c tan⁡α
c) pp=-2c tan⁡α
d) pp=γz tan2 α/2c tan⁡α
View Answer

Answer: b
Explanation: Since the principal stress relationship on a failure plane is given by,
σ1=2c tan⁡α+σ3 tan2 α,
σ1=pp and σ3=γz,
∴ pp=2c tan⁡α+γztan2 α
∴ pp=γz tan2 α+2c tan⁡α.

13. In case of cohesive backfill, the total passive pressure is _________
a) Pp=\(\frac{1}{2}\) Kp γH2+2cH tanα
b) Pp=γH2+2c tanα
c) Pp=Kp γH2-2c tanα
d) Pp=Kp H2-2c tanα
View Answer

Answer: a
Explanation: The pressure intensity pp for a cohesive soil is given by,
pp=Kp γz+2c tanα,
The total earth pressure P0 at rest per unit length is,
Pp=\(\int_0^H\)Kp γz+2c tanα.dz
∴ Pp=\(\frac{1}{2}\) Kp γH2+2cH tanα.

14. With respect to the flow value, the total passive pressure for cohesive backfill is _______
a) Pp=\(\frac{1}{2}\) Nφ γH2+2cH √Nφ
b) Pp=γH2+2c √Nφ
c) Pp=Nφ γH2-2c √Nφ
d) Pp=Nφ H2-2c √Nφ
View Answer

Answer: a
Explanation: The flow value is given by,
Nφ= tan2 α=Kp,
since the the total passive pressure for cohesive backfill is given by,
Pp=\(\frac{1}{2}\) Kp γH2+2cH tanα,
therefore on substituting the value of Nφ in the equation, we get,
Pp=\(\frac{1}{2}\) Nφ γH2+2cH √Nφ.

15. At the ground surface, there will be no passive pressure in the case of cohesive backfill.
a) True
b) False
View Answer

Answer: b
Explanation: The passive pressure intensity is given by,
pp=Kp γz+2c tanα,
At the ground surface, z=0,
∴ pp=2c tanα.

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