Soil Mechanics Questions and Answers – Earth Pressure Problems

This set of Soil Mechanics 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 \(K_a=\frac{1-sin⁡φ}{1+sin⁡φ}=\frac{1+sin⁡30}{1+sin⁡30}=\frac{1}{3}\)
\(K_p=\frac{1}{K_a} =3.\)

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.
advertisement

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.
Free 30-Day Python Certification Bootcamp is Live. Join Now!

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.

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.
advertisement

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.

Sanfoundry Global Education & Learning Series – Foundation Engineering.

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

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
I’m Manish - Founder and CTO at Sanfoundry. I’ve been working in tech for over 25 years, with deep focus on Linux kernel, SAN technologies, Advanced C, Full Stack and Scalable website designs.

You can connect with me on LinkedIn, watch my Youtube Masterclasses, or join my Telegram tech discussions.

If you’re in your 40s–60s and exploring new directions in your career, I also offer mentoring. Learn more here.