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
Explanation: Given,
Φ=300,
Therefore \(K_a=\frac{1-sinφ}{1+sinφ}=\frac{1+sin30}{1+sin30}=\frac{1}{3}\)
\(K_p=\frac{1}{K_a} =3.\)
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
Explanation: Given,
γ= 18 kN/m3
H=6m
Φ=30°
Therefore \(K_a=\frac{1-sinφ}{1+sinφ}=\frac{1+sin30}{1+sin30}=\frac{1}{3}\)
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
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.
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
Explanation: Given,
γsat=22 KN/m3
γ’=γsat-γω=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
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.
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
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
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
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
Explanation: Given,
γ = 12 kN/m3
Ze = 3m
Therefore q = Ze γ= 3×12= 36 kN/m2.
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
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
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
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
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
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
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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