Geotechnical Engineering Questions and Answers – Active Earth Pressure: Rankine’s Theory – 2

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This set of Foundation Engineering Questions and Answers for Freshers focuses on “Active Earth Pressure: Rankine’s Theory – 2”.

1. For a dry backfill with no surcharge, the active earth pressure intensity is _________
a) pa=Ka γH
b) pa=γH
c) pa=Ka H
d) pa=Ka γ
View Answer

Answer: a
Explanation: The active earth pressure intensity is given by,
pa=Ka γH,
where, \(\frac{σ_1}{σ_3} = \frac{σ_v}{σ_h} = K_a=\frac{1}{tan^2(45°+\frac{φ}{2})},\)
γ=unit weight of the back fill
H= height of the retaining wall.
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2. The resultant active pressure per unit length of wall for dry backfill with no surcharge is _______
a) \(P_a=\frac{1}{2}K_aγH^2\)
b) Pa=γH2
c) Pa=Ka γH2
d) Pa=Ka H2
View Answer

Answer: a
Explanation: The pressure intensity p0 is given by,
pa=Ka γz,
The total earth pressure P0 at rest per unit length is,
\(P_a=\int_0^HK_a γz.dz\)
∴ \(P_a=\frac{1}{2}K_aγH^2\)

3. The resultant active pressure per unit length of wall for dry backfill with no surcharge acting at _________ above the base of wall.
a) H/2
b) H
c) H/6
d) H/3
View Answer

Answer: d
Explanation: The pressure distribution of the stresses in the retaining wall due to the backfill is triangular one. Since the pressure distribution is triangular, the resultant active pressure per unit length of wall will act at the centroid of the wall, which is at a distance of H/3 from the base of the wall.
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4. For a submerged backfill, the active earth pressure is given by _________
a) pa=Kaγ’z
b) pa=Kaγ’z-γwz
c) pa=Kaγ’z+γwz
d) pa=Kaγ’z*γwz
View Answer

Answer: c
Explanation: For a submerged backfill, the lateral pressure is made up of two components,

  • lateral pressure due to submerged weight of backfill
  • lateral pressure due to water

∴ pa=Kaγ’z+γwz.

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5. If free water stands on both side of a retaining wall, the lateral earth pressure is given by ____
a) pa=Ka γ’ z
b) pa=Ka γ’ z-γw z
c) pa=Ka γ’ z+γw z
d) pa=Ka γ’ z*γw z
View Answer

Answer: a
Explanation: When the free water stands on both side of a retaining wall, the water pressure is not considered and the net pressure is given by,
pa=Ka γ’ z.

6. If the angle of internal friction decreases, then Ka ___________
a) decreases
b) increases
c) equal to zero
d) does not change
View Answer

Answer: b
Explanation: Since the coefficient of earth pressure for active state of plastic equilibrium is given by,
\(K_a=\frac{1-sinφ}{1+sinφ}.\) Therefore, from the equation it is clear that as the angle of internal friction decreases, then Ka increases.
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7. For the same value of φ, the backfill is partly submerged to height H2 and the backfill is moist to a depth H1. Find the lateral pressure intensity at the base of wall.
a) pa=Ka γH1-Ka γ’ H2w H2
b) pa=Ka γH1+Ka γ’ H2w H2
c) pa=Ka γH1w H2
d) pa=Ka γH1+Ka γ’ H2
View Answer

Answer: b
Explanation: When the backfill is partly submerged to height H2 and the backfill is moist to a depth H1, lateral pressure intensity is due to:

  • lateral pressure due to moist weight of backfill
  • lateral pressure due to saturated weight of backfill
  • lateral pressure due to water

∴ pa=Ka γH1+Ka γ’ H2w H2.

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8. For the different value of φ, the backfill is partly submerged to height H2 and the backfill is moist to a depth H1. Find the lateral pressure intensity at the base of wall for φ12.
a) pa=Ka2 γH1-Ka2 γ’ H2w H2
b) pa=Ka2 γH1+Ka2 γ’ H2w H2
c) pa=Ka2 γH1w H2
d) pa=Ka2 γH1+Ka2 γ’ H2
View Answer

Answer: b
Explanation: For the different value of φ, the coefficient of active earth pressure is different. Since, the lateral pressure intensity is due to:

  • lateral pressure due to moist weight of backfill
  • lateral pressure due to saturated weight of backfill
  • lateral pressure due to water

∴ pa=Ka2 γH1+Ka2 γ’ H2w H2. We have to consider Ka2 over Ka1 as the φ12.

9. If the backfill carries a uniform surcharge q, then the lateral pressure at the depth of wall H is ____________
a) pa=Ka γz+Ka q
b) pa=Ka γz-Ka q
c) pa=Ka γz*Ka q
d) pa=Ka γz/Ka 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 Ka q. Hence, lateral pressure at the depth of wall H is pa=Ka γz+Ka q.

10. The height of fill Ze, equivalent to uniform surcharge intensity is __________
a) q/γ
b) q-γ
c) q+γ
d) q*γ
View Answer

Answer: a
Explanation: The height of fill Ze, equivalent to uniform surcharge intensity is given by,
Ka γze=Ka q,
∴ \(z_e=\frac{q}{γ}.\)

11. For finding out the active earth pressure for a backfill with sloping surface, the Rankine’s theory makes as additional assumption of ________
a) vertical and lateral stresses are normal to surcharge
b) vertical and lateral stresses are tangential to surcharge
c) vertical and lateral stresses are conjugate
d) vertical and lateral stresses are negligible
View Answer

Answer: c
Explanation: The additional assumption of vertical and lateral stresses are conjugate is made as, it can be shown that stresses on a given plane at a given point is parallel to another plane, the stresses on the latter plane at the same point must be parallel to the first plane.

12. The vertical and the lateral pressures have the same angle of obliquity β.
a) True
b) False
View Answer

Answer: a
Explanation: In Rankine’s theory, the additional assumption of the vertical and lateral stresses are conjugate is made. Being conjugate, both the vertical and the lateral pressures have the same angle of obliquity β.

13. The Rankine’s lateral pressure ratio is given by ________
a) \(K=\frac{\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}} \)
b) \(K=\frac{cosβ+\sqrt{cos^2 β-cos^2 φ}}{cosβ-\sqrt{cos^2 β-cos^2 φ}} \)
c) \(K=\frac{cosβ-\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}} \)
d) \(K=cosβ\frac{cosβ-\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}} \)
View Answer

Answer: c
Explanation: The ratio K is the conjugate ratio or the Rankine’s lateral pressure ratio, which is given by,
\(K=\frac{cosβ-\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}}, \)
where, β=surcharge angle
φ=angle of internal friction.

14. For backfill with sloping surface, the coefficient of active earth pressure is given by ______
a) \(K_a=\frac{\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}} \)
b) \(K_a=\frac{cosβ+\sqrt{cos^2 β-cos^2 φ}}{cosβ-\sqrt{cos^2 β-cos^2 φ}} \)
c) \(K_a=\frac{cosβ-\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}} \)
d) \(K_a=cosβ\frac{cosβ-\sqrt{cos^2 β-cos^2 φ}}{cosβ+\sqrt{cos^2 β-cos^2 φ}} \)
View Answer

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

15. When the surcharge angle reduces to zero, the coefficient of active earth pressure is given by
a) Ka=1
b) \(K_a=\frac{1-sinφ}{1+sinφ}\)
c) \(K_a=\frac{1+sinφ}{1-sinφ}\)
d) Ka=0
View Answer

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter