This set of Soil Mechanics Objective Questions & Answers focuses on “Consolidation Problems – 3”.
1. For a mass of 3.192g of soil with voids ratio 0.532, find its saturated unit weight.
a) 20.44 KN/m3
b) 76.22
c) 23.55
d) 50.43
View Answer
Explanation:
\(ρ_{sat}=\frac{M}{1+e}=\frac{3.192}{1+0.532}=2.084\) g/cm3
γsat = 9.81ρsat = 9.81×2.084 = 20.44 kN/m3.
2. For a remolded sample of clay with liquid limit of 45%, find its compression index.
a) 0.753
b) 0.561
c) 0.493
d) 0.245
View Answer
Explanation: Given,
WL= 45%, for remolded sample of clay
Cc=0.007(WL-10%)
=0.245.
3. For an ordinary clay of medium sensitivity, the value of compression index is _____ if liquid limit is 50%.
a) 0.36
b) 0.45
c) 0.12
d) 0.64
View Answer
Explanation: WL=50%
For ordinary sample of clay
Cc=0.009(WL -10)
=0.36.
4. If change in voids ratio is 0.18 and change in effective pressure is 100KN/m2, then the coefficient of compressibility is ______
a) 2.5×10-3 m2/kN
b) 2×10-3 m2/kN
c) 9×10-3 m2/kN
d) 9.5×10-3 m2/kN
View Answer
Explanation: Given,
∆e=0.18
∆σ’=100 kN/m3
\(a_w = \frac{∆e}{∆σ’} = \frac{0.18}{100}\)=2×10-3 m2/kN.
5. Two clay layer of A and B of thickness 2cm and 3 cm is load with pressure of 200KN/m2. If the time taken by soil A to reach 50% consolidation is 1/4th of that required by soil B to reach 50% consolidation, then find the ratio of coefficients of consolidation.
a) 2.4432
b) 1.7778
c) 4.3312
d) 5.3489
View Answer
Explanation: Given,
dA=2/2 = 1cm
dB=3/2= 1.5 cm
\(\frac{t_B}{t_A} = \frac{4}{1}\)
The ratio of coefficient of consolidation is, \(\frac{(C_v)_A}{(C_v)_B} \)
\(\frac{(C_v)_A}{(C_v)_B} = \frac{(d_A)^2}{(d_B)^2} × \frac{t_B}{t_A} = \frac{(2/2)^2}{(3/2)^2} × \frac{4}{1}=1.7778.\)
6. Two clay layer of A and B of thickness 2cm and 3 cm is load with pressure of 200KN/m2. If the time taken by soil A to reach 50% consolidation is 1/4th of that required by soil B to reach 50% consolidation, then find the ratio of coefficients of Permeability. Given that ratio of coefficient of volume change is 1.751.
a) 5.234
b) 2.456
c) 3.114
d) 4.233
View Answer
Explanation: Given,
\(\frac{(m_v)_A}{(m_v)_B} = 1.751 \)
\(\frac{K_B}{K_A} = \frac{(C_v)_A}{(C_v)_B} × \frac{(m_v)_A}{(m_v)_B} = 1.7778×1.751 \)
\(\frac{K_B}{K_A} = 3.114. \)
7. A clay sample 24mm thick takes 20 minutes to consolidate 50% with double drainage. The field clay layer from which sample was obtained takes 386 days to consolidate 50% with double drainage. Find the depth of the clay layer in the field.
a) 4m
b) 7m
c) 3m
d) 2m
View Answer
Explanation: Given,
t1= 20 min
t2= 386 days
d1= 2.4/2 = 1.2 cm
since, \(\frac{t_2}{t_1} =(\frac{d_2}{d_1})^2\)
therefore \(d_2 = d_1 \sqrt{\frac{t_2}{t_1}}=1.2\sqrt{\frac{386×60×24}{20}}=200cm=2m\)
since double drainage is allowed, the depth is 4m.
8. A clay sample 24mm thick takes 20 minutes to consolidate 50% with double drainage. The field clay layer from which sample was obtained takes 386 days to consolidate 50% with single drainage. Find the depth of the clay layer in the field.
a) 5m
b) 2m
c) 22m
d) 16m
View Answer
Explanation: Given, the sample has double drainage while the field layer has single drainage.
t1= 20 min
t2= 386 days
d1= 2.4/2 = 1.2 cm
since, \(\frac{t_2}{t_1} =(\frac{d_2}{d_1})^2\)
therefore \(d_2 = d_1 \sqrt{\frac{t_2}{t_1}}=1.2\sqrt{\frac{386×60×24}{20}}=200cm=2m.\)
9. A clay sample 24mm thick takes 20 minutes to consolidate 50% with single drainage. The field clay layer from which sample was obtained takes 386 days to consolidate 50% with single drainage. Find the depth of the clay layer in the field.
a) 200cm
b) 367cm
c) 400cm
d) 569cm
View Answer
Explanation: Given, both the sample and the field layer has single drainage,
t1= 20 min
t2= 386 days
d1= 2.4cm
since, \(\frac{t_2}{t_1} =(\frac{d_2}{d_1})^2\)
therefore \(d_2 = d_1 \sqrt{\frac{t_2}{t_1}}=2.4\sqrt{\frac{386×60×24}{20}}=400cm=4m.\)
10. A clay sample 24mm thick takes 20 minutes to consolidate 50% with single drainage. The field clay layer from which sample was obtained takes 386 days to consolidate 50% with double drainage. Find the depth of the clay layer in the field.
a) 800cm
b) 500cm
c) 600cm
d) 480cm
View Answer
Explanation: Given, the sample has single drainage while the field layer has double drainage.
t1= 20 min
t2= 386 days
d1= 2.4 cm
since, \(\frac{t_2}{t_1} =(\frac{d_2}{d_1})^2\)
therefore \(d_2 = d_1 \sqrt{\frac{t_2}{t_1}}=2.4\sqrt{\frac{386×60×24}{20}}=400cm=4m.\)
Since double drainage is allowed in field, the depth is 400×2= 800 cm.
11. For a foundation, find the elapsed time in which 10% of ultimate settlement will occur. The coefficient of consolidation is 4*10-4cm2/s. The average drainage path is 50cm.
a) 19.9 hours
b) 15.5 hours
c) 13.9 hours
d) 17.5 hours
View Answer
Explanation: Given,
d=50 cm
Cv=4×10-4 cm2/s
For consolidation of 10%, U<60%
Tv=\(\frac{π}{4} (\frac{v}{100})^2=\frac{π}{4} (\frac{10}{100})^2=0.008\)
Therefore time elapsed \(t_{10}=\frac{(T_v)_{10} d^2}{C_v} = \frac{0.008×(50)^2}{4×10^{-4}} s = 13.9 hours.\)
12. For a foundation, find the elapsed time in which 90% of ultimate settlement will occur. The coefficient of consolidation is 4*10-4cm2/s. The average drainage path is 50cm.
a) 56.38 days
b) 43.78 days
c) 61.34 days
d) 78.23 days
View Answer
Explanation: Given,
d=50 cm
Cv=4×10-4 cm2/s
For consolidation of 90%, U>60%
Therefore Tv= 1.7813 – 0.9332 log10 (100-U)
Tv= 1.7813 – 0.9332 log10 (100-90)
Tv= 0.848
\(G_0 = \frac{(T_v)_{90}d^2}{C_v} = \frac{0.848×(50)^2}{4×10^{-4}}s\)
G0= 61.34 days.
13. If initial voids ratio is 1.08, then find the compression index.
a) 0.524
b) 0.243
c) 0.728
d) 0.871
View Answer
Explanation: Given,
From Hough equation,
Cc=0.3(e0-0.27)
Cc=0.3(1.08-0.27)
Cc= 0.243.
14. If excess hydraulic pressure is 19.62 KN/m2 then find the hydraulic head.
a) 2m
b) 4m
c) 8m
d) 16m
View Answer
Explanation: Given,
u=19.62 kN/m2
Therefore \(h=\frac{\overline{u}}{γ_ω} = \frac{19.62}{9.81}=2m\)
h = 2m.
15. Find the hydrostatic pressure of water at depth 10m.
a) 98.1KN/m2
b) 54. 1KN/m2
c) 32. 1KN/m2
d) 43. 1KN/m2
View Answer
Explanation: Given,
h=10m
\(h=\frac{\overline{u}}{γ_ω} \)
u=hγω=10×9.81=98.1 kN/m2.
Sanfoundry Global Education & Learning Series – Soil Mechanics.
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