Engineering Questions with Answers - Multiple Choice Questions

# Consolidation Problems – 1

1 - Question

A clay sample in laboratory test has 24mm thickness. It is tested with double drainage and consolidated 50%. The clay layer from which sample was obtained is 4m thick. Find the time to consolidate 50% with double drainage.
a) 365 days
b) 386 days
c) 750 days
d) 150 days

Explanation: Given, double drainage for both specimen and field,
t1=20 min
d1=24 mm/2 = 2.4 cm/2 = 1.2 cm
d2 =4/2 m 200cm
Since, for same degree of consolidation of 50%, Tv is same,
td2Cr
Therefore, t2t1=(d2d1)2
So, t2=t1(d2d1)2=20(2001.2)2=365days.

2 - Question

A clay sample in laboratory test has 24mm thickness. It is tested with double drainage and consolidated 50%. The clay layer from which sample was obtained is 4m thick. Find the time to consolidate 50% with single drainage.
a) 1544 days
b) 1455 days
c) 1322 days
d) 1799 days

Explanation: Given, double drainage for sample and single drainage for field layer,
t1=20 min
d1=24 mm/2 = 2.4 cm/2 = 1.2 cm
d2 = 4 m = 400cm
Since, for same degree of consolidation of 50%, Tv is same,
td2Cr
Therefore, t2t1=(d2d1)2
So, t2=t1(d2d1)2=20(4001.2)2=1544days.

3 - Question

A clay sample in laboratory test has 24mm thickness. It is tested with single drainage and consolidated 50%. The clay layer from which sample was obtained is 4m thick. Find the time to consolidate 50% with double drainage.
a) 44 days
b) 55 days
c) 97 days
d) 107 days

Explanation: Given, single drainage for sample and double drainage for field layer,
t1 = 2 0 min
d1 = 24 mm = 2.4 cm
d2 = 2 m = 200cm
Since, for same degree of consolidation of 50%, Tv is same,
td2Cr
Therefore, t2t1=(d2d1)2
So, t2=t1(d2d1)2=20(2002.4)2=97days.

4 - Question

A clay sample in laboratory test has 24mm thickness. It is tested with single drainage and consolidated 50%. The clay layer from which sample was obtained is 4m thick. Find the time to consolidate 50% with single drainage.
a) 386 days
b) 390 days
c) 404 days
d) 175 days

Explanation: Given, single drainage for both the sample and for the field layer,
t1=20 min
d1=24 mm= 2.4 cm
d2 =4 m = 400cm
Since, for same degree of consolidation of 50%, Tv is same,
td2Cr
Therefore, t2t1=(d2d1)2
So, t2=t1(d2d1)2=20(4002.4)2=386days.

5 - Question

For a dry soil mass of 180.4 g, specific gravity 2.68 and cross-sectional area of specimen as 50 cm2, find the height of solids Hs.
a) 13.45 mm
b) 14 mm
c) 17 mm
d) 19.5 mm

Explanation: Given,
Md=180.4 g
G=2.68
A=50cm2
Since Hs=MdGAρw
Hs=180.4102.68501=13.45mm.

6 - Question

If the specimen height H is 24mm and height of solids Hs is 13.45mm, then find the voids ratio.
a) 0.786
b) 0.432
c) 2
d) 34

Explanation: Given,
H=24mm
Hs =13.45mm
e=HHsHs
e=2413.4513.45
e=0.786.

7 - Question

If the final water content is 30% and specific gravity is 2.65 for a soil sample, then its final void ratio is _________
a) 0.900
b) 0.795
c) 0.667
d) 0.549

Explanation: Wf=30%=0.30
G=2.65
ef=WfG
ef=0.3*2.65=0.795.

8 - Question

If the final voids ratio is 0.864 for a soil sample of specific gravity 2.7, then its final water content will be _______________
a) 32%
b) 48%
c) 56%
d) 62%

Explanation: Given,
ef=0.864
G=2.7
ef=WfG
efG=Wf
Wf=32%.

9 - Question

If height of solids Hs is 6.725mm, mass is 180.4g with specific gravity G=2.68, then final the Cross-sectional area of specimen.
a) 220 cm2
b) 342 cm2
c) 100 cm2
d) 659 cm2

Explanation: Given,
Hs=6.725mm
G=2.68
Md=180.4g
Since Hs=MdGHsρw
A=MdGHsρw
A=180.4×102.68×6.725×1=100cm2.

10 - Question

If the final height of specimen Hf is 13.45mm and final voids is 0.864, then the change of voids ratio ∆e with respect to ∆H is ____________
a) ∆e=0.139∆H
b) ∆e=0.123∆H
c) ∆e=0.178∆H
d) ∆e=0.148∆H

Explanation: Given, Hf=13.45
ef=0.864
Δe=1+efHfΔH
Δe=1+0.86413.45ΔH
∆e=0.139∆H.

11 - Question

The change in voids ratio is 0.18 and initial void ratio is 0.68 with the increase of pressure as 200KN/m2. Find the coefficient of volume change.
a) 0.536 m2/MN
b) 0.111 m2/MN
c) 0.832 m2/MN
d) 0.356 m2/MN

Explanation: Given,
∆ef=0.18
eo=0.68
∆σ’=200kN/m2
mv=Δe1+eo1Δσ
mv=0.181+0.68+1200=0.536m2/MN.

12 - Question

The change in voids ratio is 0.18 and initial void ratio is 0.68 with the increase of pressure as 200KN/m2. Find the coefficient of Compressibility.
a) 0.9 m2/MN
b) 0.7 m2/MN
c) 0.4 m2/MN
d) 0.5 m2/MN

Explanation: Given,
∆ef=0.18
eo=0.68
∆σ’=200kN/m2
mv=Δe1+eo1Δσ
since, mv=av1+e0
av=mv(1+e0)=0.536(1+0.68)
av=0.9 m2/MN.

13 - Question

If coefficient of volume change is 0.291 m2/MN, ∆e=0.1 and the initial voids ratio is 0.72, then find the change in effective pressure.
a) 200 KN/m2
b) 50 KN/m2
c) 300 KN/m2
d) 00 KN/m2

Explanation: Given,
mv=0.291 m2/MN
∆e=0.10
eo=0.72
Δσ=Δe1+eo1mv
Δσ=0.101+0.72×1×1030.291=200KN/m2.

14 - Question

If a soil A has coefficient of volume change as 0.536m2/MN and the ratio of coefficient of volume change of A and B soils is 1.845, then find mv for soil B.
a) 0.255 m2/MN
b) 0.291 m2/MN
c) 0.234 m2/MN
d) 0.345 m2/MN

Explanation: Given,
Mva=0.536 m2/MN
mvamvb=1.845
Mvb=0.5361.845=0.291m2/MN.

15 - Question

A soil sample is having voids ratio 0.72 under a pressure of 200KN/m2. If the void ratio is reduced to 0.5 under 450kn/m2 load, then find the change in effective pressure.
a) 4000 kN/m2
b) 250 kN/m2
c) 350 kN/m2
d) 2000 kN/m2