Soil Mechanics Questions and Answers – Consolidation Problems – 1

This set of Soil Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Consolidation Problems – 1”.

1. 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,
$$t∝\frac{d^2}{C_r}$$
Therefore, $$\frac{t_2}{t_1} =(\frac{d_2}{d_1})^2$$
So, $$t_2=t_1 (\frac{d_2}{d_1} )^2=20(\frac{200}{1.2})^2=365 \,days.$$

2. 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,
$$t∝\frac{d^2}{C_r}$$
Therefore, $$\frac{t_2}{t_1} =(\frac{d_2}{d_1} )^2$$
So, $$t_2=t_1 (\frac{d_2}{d_1} )^2=20(\frac{400}{1.2})^2=1544 \,days.$$

3. 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,
$$t∝\frac{d^2}{C_r}$$
Therefore, $$\frac{t_2}{t_1} =(\frac{d_2}{d_1})^2$$
So, $$t_2=t_1 (\frac{d_2}{d_1} )^2=20(\frac{200}{2.4})^2=97 \,days.$$

4. 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,
$$t∝\frac{d^2}{C_r}$$
Therefore, $$\frac{t_2}{t_1} =(\frac{d_2}{d_1} )^2$$
So, $$t_2=t_1 (\frac{d_2}{d_1})^2=20(\frac{400}{2.4})^2=386 \,days.$$

5. 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 $$H_s=\frac{M_d}{GAρ_w}$$
$$H_s=\frac{180.4*10}{2.68*50*1} =13.45mm.$$

6. 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
$$e=\frac{H-H_s}{H_s}$$
$$e=\frac{24-13.45}{13.45}$$
e=0.786.

7. 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. 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
$$\frac{e_f}{G}=W_f$$
Wf=32%.

9. 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 $$H_s= \frac{M_d}{GH_s ρ_w}$$
$$A=\frac{M_d}{GH_s ρ_w}$$
$$A=\frac{180.4×10}{2.68×6.725×1}=100cm^2.$$

10. 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
$$∆e=\frac{1+e_f}{H_f} ∆H$$
$$∆e=\frac{1+0.864}{13.45} ∆H$$
∆e=0.139∆H.

11. 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
$$m_v=\frac{-∆e}{1+e_o}\frac{1}{∆σ’}$$
$$m_v=\frac{0.18}{1+0.68}+\frac{1}{200}=0.536m^2/MN.$$

12. 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
$$m_v=\frac{-∆e}{1+e_o}\frac{1}{∆σ’}$$
since, $$m_v=\frac{a_v}{1+e_0}$$
av=mv(1+e0)=0.536(1+0.68)
av=0.9 m2/MN.

13. 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
$$∆σ’=\frac{-∆e}{1+e_o}\frac{1}{m_v}$$
$$∆σ’=\frac{0.10}{1+0.72}×\frac{1×10^3}{0.291}=200KN/m^2.$$

14. 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
$$\frac{m_{va}}{m_{vb}} =1.845$$
$$M_{vb}=\frac{0.536}{1.845}=0.291 m^2/MN.$$

15. 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

Explanation: Given,
σ0’=200kN/m2
σ’=450kN/m2
∆σ’=σ’-σ0’=450-200
∆σ’=250kN/m2 .

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