Engineering Questions with Answers - Multiple Choice Questions

Home » MCQs » Engineering MCQs » Consolidation Problems – 3

# 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/m^{3}

b) 76.22

c) 23.55

d) 50.43

**
View Answer**

Answer: a

Explanation:

ρsat=M1+e=3.1921+0.532=2.084 g/cm^{3}

γ_{sat} = 9.81ρ_{sat} = 9.81×2.084 = 20.44 kN/m^{3}.

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**

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

**
View Answer**

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/m^{2}, then the coefficient of compressibility is ______

a) 2.5×10^{-3} m^{2}/kN

b) 2×10^{-3} m^{2}/kN

c) 9×10^{-3} m^{2}/kN

d) 9.5×10^{-3} m^{2}/kN

**
View Answer**

Answer: b

Explanation: Given,

∆e=0.18

∆σ’=100 kN/m^{3}

aw=ΔeΔσ′=0.18100=2×10^{-3} m^{2}/kN.

Two clay layer of A and B of thickness 2cm and 3 cm is load with pressure of 200KN/m^{2}. 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

**
View Answer**

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/m^{2}. 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

**
View Answer**

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

**
View Answer**

Answer: a

Explanation: Given,

t_{1}= 20 min

t_{2}= 386 days

d_{1}= 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

**
View Answer**

Answer: b

Explanation: Given, the sample has double drainage while the field layer has single drainage.

t_{1}= 20 min

t_{2}= 386 days

d_{1}= 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

**
View Answer**

Answer: c

Explanation: Given, both the sample and the field layer has single drainage,

t_{1}= 20 min

t_{2}= 386 days

d_{1}= 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

**
View Answer**

Answer: a

Explanation: Given, the sample has single drainage while the field layer has double drainage.

t_{1}= 20 min

t_{2}= 386 days

d_{1}= 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^{-4}cm^{2}/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**

Answer: c

Explanation: Given,

d=50 cm

C_{v}=4×10^{-4} cm^{2}/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^{-4}cm^{2}/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**

Answer: c

Explanation: Given,

d=50 cm

C_{v}=4×10-4 cm^{2}/s

For consolidation of 90%, U>60%

Therefore T_{v}= 1.7813 – 0.9332 log_{10} (100-U)

T_{v}= 1.7813 – 0.9332 log_{10} (100-90)

T_{v}= 0.848

G0=(Tv)90d2Cv=0.848×(50)24×10−4s

G_{0}= 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

**
View Answer**

Answer: b

Explanation: Given,

From Hough equation,

C_{c}=0.3(e_{0}-0.27)

C_{c}=0.3(1.08-0.27)

C_{c}= 0.243.

If excess hydraulic pressure is 19.62 KN/m^{2} then find the hydraulic head.

a) 2m

b) 4m

c) 8m

d) 16m

**
View Answer**

Answer: a

Explanation: Given,

u=19.62 kN/m^{2}

Therefore h=u¯¯¯γω=19.629.81=2m

h = 2m.

Find the hydrostatic pressure of water at depth 10m.

a) 98.1KN/m^{2}

b) 54. 1KN/m^{2}

c) 32. 1KN/m^{2}

d) 43. 1KN/m^{2}

**
View Answer**

Answer: a

Explanation: Given,

h=10m

h=u¯¯¯γω

u=hγ_{ω}=10×9.81=98.1 kN/m^{2}.