Consolidation Problems – 3

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

Answer: a
Explanation:
ρsat=M1+e=3.1921+0.532=2.084 g/cm³
γ_sat = 9.81ρ_sat = 9.81×2.084 = 20.44 kN/m³.

 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

Answer: d
Explanation: Given,
W_L= 45%, for remolded sample of clay
C_c=0.007(W_L-10%)
=0.245.

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

Answer: a
Explanation: W_L=50%
For ordinary sample of clay
C_c=0.009(W_L -10)
=0.36.

 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

Answer: b
Explanation: Given,
∆e=0.18
∆σ’=100 kN/m³
aw=ΔeΔσ′=0.18100=2×10^-3 m²/kN.

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 1/4^th 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

Answer: b
Explanation: Given,
d_A=2/2 = 1cm
d_B=3/2= 1.5 cm
tBtA=41
The ratio of coefficient of consolidation is, (Cv)A(Cv)B
(Cv)A(Cv)B=(dA)2(dB)2×tBtA=(2/2)2(3/2)2×41=1.7778.

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 1/4^th 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

Answer: c
Explanation: Given,
(mv)A(mv)B=1.751
KBKA=(Cv)A(Cv)B×(mv)A(mv)B=1.7778×1.751
KBKA=3.114.

 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

Answer: a
Explanation: Given,
t₁= 20 min
t₂= 386 days
d₁= 2.4/2 = 1.2 cm
since, t2t1=(d2d1)2
therefore d2=d1t2t1−−√=1.2386×60×2420−−−−−−−√=200cm=2m
since double drainage is allowed, the depth is 4m.

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

Answer: b
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, t2t1=(d2d1)2
therefore d2=d1t2t1−−√=1.2386×60×2420−−−−−−−√=200cm=2m.

 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

Answer: c
Explanation: Given, both the sample and the field layer has single drainage,
t₁= 20 min
t₂= 386 days
d₁= 2.4cm
since, t2t1=(d2d1)2
therefore d2=d1t2t1−−√=2.4386×60×2420−−−−−−−√=400cm=4m.

 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

Answer: a
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, t2t1=(d2d1)2
therefore d2=d1t2t1−−√=2.4386×60×2420−−−−−−−√=400cm=4m.
Since double drainage is allowed in field, the depth is 400×2= 800 cm.

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

Answer: c
Explanation: Given,
d=50 cm
C_v=4×10^-4 cm²/s
For consolidation of 10%, U<60%
T_v=π4(v100)2=π4(10100)2=0.008
Therefore time elapsed t10=(Tv)10d2Cv=0.008×(50)24×10−4s=13.9hours.

 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

Answer: c
Explanation: Given,
d=50 cm
C_v=4×10-4 cm²/s
For consolidation of 90%, U>60%
Therefore T_v= 1.7813 – 0.9332 log₁₀ (100-U)
T_v= 1.7813 – 0.9332 log₁₀ (100-90)
T_v= 0.848
G0=(Tv)90d2Cv=0.848×(50)24×10−4s
G₀= 61.34 days.

If initial voids ratio is 1.08, then find the compression index.
a) 0.524
b) 0.243
c) 0.728
d) 0.871

Answer: b
Explanation: Given,
From Hough equation,
C_c=0.3(e₀-0.27)
C_c=0.3(1.08-0.27)
C_c= 0.243.

 If excess hydraulic pressure is 19.62 KN/m² then find the hydraulic head.
a) 2m
b) 4m
c) 8m
d) 16m

Answer: a
Explanation: Given,
u=19.62 kN/m²
Therefore h=u¯¯¯γω=19.629.81=2m
h = 2m.

 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

Answer: a
Explanation: Given,
h=10m
h=u¯¯¯γω
u=hγ_ω=10×9.81=98.1 kN/m².