Foundation Engineering Questions and Answers – Earth Pressure Problems

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This set of Foundation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Earth Pressure Problems”.

1. For an angle of internal friction of 300, the values of coefficients of active and passive earth pressures are given by a _____ and ____ respectively.
a) \(\frac{1}{3}\), 3
b) \(\frac{1}{5}\), 5
c) 3, \(\frac{1}{3}\)
d) 5, \(\frac{1}{5}\)
View Answer

Answer: a
Explanation: Given,
Φ=300,
Therefore \(K_a=\frac{1-sin⁡φ}{1+sin⁡φ}=\frac{1+sin⁡30}{1+sin⁡30}=\frac{1}{3}\)
\(K_p=\frac{1}{K_a} =3.\)

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2. Compute the active pressure intensity when the backfill has unit weight of 18 KN/m3 and height of wall is 6 m. The angle of internal friction is 30°.
a) 45 KN/m2
b) 36 KN/m3
c) 87 KN/m2
d) 27 KN/m2
View Answer

Answer: b
Explanation: Given,
γ= 18 kN/m3
H=6m
Φ=30°
Therefore \(K_a=\frac{1-sin⁡φ}{1+sin⁡φ}=\frac{1+sin⁡30}{1+sin⁡30}=\frac{1}{3}\)
Pp= KaγH = \((\frac{1}{3})\)×18×6 = 36 kN/m2.

3. Compute the passive pressure intensity when the backfill has unit weight of Ka=\(\frac{1}{3}\) and height of wall is 6 m. The coefficient of active earth pressure Ka=\(\frac{1}{3}.\)
a) 176 KN/m3
b) 154 KN/m3
c) 324 KN/m3
d) 476 KN/m3
View Answer

Answer: c
Explanation: Given,
γ= 18 kN/m3
H=6m
Ka=\(\frac{1}{3}\)
Kp=\(\frac{1}{K_a}=3 \)
Therefore Pp = KpγH =3×18×6=324 KN/m3.
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4. The water table is at the ground level and backfill has saturated unit weight as 22KN/m3. The height of wall is 6m and Ka=\(\frac{1}{3}.\) Find the pressure intensity.
a) 45.22 KN/m2
b) 18.45 KN/m2
c) 83.24 KN/m2
d) 67.83 KN/m2
View Answer

Answer: c
Explanation: Given,
γsat=22 KN/m3
γ’=γsatω=22-9.81= 12.19
Ka = \(\frac{1}{3}\) H = 6m
We know that, Pa=Ka γ’ H+ γω H
Pa = \((\frac{1}{3})\)×12.19×6 + 9.81×6 = 83.24 kN/m2.

5. The water table is at the ground level and backfill has saturated unit weight as 22KN/m3. The height of wall is 6m and Ka=1/3. Find the pressure intensity, if free water stands on other face of wall.
a) 45.82 KN/m2
b) 78.28 KN/m2
c) 24.38 KN/m2
d) 56.27 KN/m2
View Answer

Answer: c
Explanation: Given,
Free water stands on both sides of wall, so the effect of water pressure is not considered
Pa=Ka γ’ H=\((\frac{1}{3})\) ×12.19×6
Pa= 24.38 kN/m2.
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6. Compute the active pressure intensity when the backfill has unit weight of 18 KN/m3 and height of wall is 6 m. The angle of internal friction is 300. The backfill has surcharge of 36 KN/m2.
a) 45 KN/m2
b) 75 KN/m2
c) 48 KN/m2
d) 78 KN/m2
View Answer

Answer: c
Explanation: Given,
q = 36 KN/m2
Pa=Ka γH+ Ka q= \((\frac{1}{3})\) ×18×6 + \((\frac{1}{3})\) ×36
Pa= 48 KN/m2.

7. Compute the passive pressure intensity when the backfill has unit weight of 18 KN/m3 and height of wall is 6 m. The angle of internal friction is 300. The backfill has surcharge of 36 KN/m2.
a) 378 KN/m2
b) 432 KN/m2
c) 875 KN/m2
d) 972 KN/m2
View Answer

Answer: b
Explanation: Given,
q = 36 KN/m2
Pa=Ka γH+ Ka q= 3×18×6 + 3×36
Pp= 432 kN/m2.
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8. If the backfill has unit weight of 18 KN/m2 and surcharge of 36 KN/m2, then find the equivalent height of surcharge.
a) 4m
b) 56m
c) 2m
d) 86m
View Answer

Answer: c
Explanation: Given,
q= 36 kN/m3
γ= 18 kN/m3
\(Z_e = \frac{q}{γ} = \frac{36}{18}=2m.\)

9. If a backfill has unit weight of 12 KN/m3 and the equivalent height of surcharge as 3m, then find the surcharge.
a) 12 KN/m2
b) 25 KN/m2
c) 36 KN/m2
d) 75 KN/m2
View Answer

Answer: c
Explanation: Given,
γ = 12 kN/m3
Ze = 3m
Therefore q = Ze γ= 3×12= 36 kN/m2.
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10. When the height of retaining wall is 6m, then the distance of point of application of the resultant pressure from the base is ________
a) 10m
b) 2m
c) 3m
d) 7m
View Answer

Answer: b
Explanation: Given,
H = 6m
The point of application is \(\frac{H}{3}= \frac{6}{3} = 2m.\)

11. Find the critical height of an unsupported vertical cut in cohesive soil of C= 10KN/m2 and γ=20KN/m2.
a) 5m
b) 2m
c) 8m
d) 9m
View Answer

Answer: b
Explanation: Given,
c= 10 KN/m2
γ=20KN/m2
Hc= 4c/γ tan⁡∝
Since φ=0, tan⁡∝=1
\(H_c= 4×\frac{10}{20}×1= 2m.\)

12. For a soil deposit, C=20 KN/m2 and is found be to a purely cohesive soil. It has a critical height of an unsupported vertical cut as 4m. Find the unit weight of soil.
a) 13 kN/m3
b) 20 kN/m3
c) 56 kN/m3
d) 77 kN/m3
View Answer

Answer: b
Explanation: Given,
C= 20 kN/m2
Φ=0,this makes tan⁡∝=1
Hc= 4 m
Therefore \(H_c=\frac{4c}{γ} tan⁡∝\)
γ = 4×\(\frac{20}{4}×1\)= 20 kN/m3.

13. For a saturated clay of 20KN/m3 and cu=24 kN/m2, find the height of tension cracks.
a) 2.4 m
b) 6.9 m
c) 3.5 m
d) 1.7 m
View Answer

Answer: a
Explanation: Given,
cu=24 KN/m2
γsat=20 KN/m3
Therefore height of the tension cracks is,
\(Z_0=\frac{2c_u}{γ_{sat}} = \frac{2×24}{20}=2.4 m. \)

14. For a saturated clay cu=18KN/m2, find the height of tension cracks is 7m. Find the saturated unit weight.
a) 43 KN/m3
b) 98 KN/m3
c) 18 KN/m3
d) 86 KN/m3
View Answer

Answer: c
Explanation: Given,
cu=18 KN/m2
zo= 2m
we know that, \(Z_0=\frac{2c_u}{γ_{sat}} \)
\(γ_{sat}=\frac{2c_u}{Z_0}=\frac{2×18}{2}=18\)KN/m3.

15. For an angle of internal friction of 30°, find the flow value Nφ
a) 4
b) 10
c) 3
d) 0
View Answer

Answer: c
Explanation: Given,
φ=30°
Therefore flow value \(N_φ=tan^2⁡ (45+\frac{φ}{2})\)
\(N_φ=tan^2⁡(45+\frac{30}{2})\)
Nφ=3.

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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