# Soil Mechanics Questions and Answers – Consolidation Problems – 2

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This set of Soil Mechanics written test Questions & Answers focuses on “Consolidation Problems – 2”.

1. If the height of soil specimen is reduced from 0.28 cm to 1.926cm due to an increment of load of 100KN/ m2. Find the coefficient of volume change.
a) 0.74m2/MN
b) 0.98m2/MN
c) 0.84m2/MN
d) 0.67m2/MN

Explanation: Given,
H0= 2.08 cm
H=1.92 cm
∆σ’=100kN/m2
$$m_v=-\frac{∆H}{H_0}\frac{1}{∆σ’}$$
$$m_v=-\frac{0.154}{2.08} \frac{1}{100}=0.74 m^2/MN.$$

2. For n soil of height 2.08cm, coefficient of volume change mv=0.74 m2/MN. Find the settlement.
a) 0.154cm
b) 0.342cm
c) 0.432cm
d) 0.455cm

Explanation: Given,
H0= 2.08cm
H=1.926cm
∆H=-0.154cm
∆σ’=100kN/m2
ρf=m∆σ’H
ρf =0.74×100×2.08×10-3
ρf=0.154 cm.

3. The mv for pressure range 100 to 200 KN/m2 is 0.74 m2 /MN. The change in void ratio is 0.157. Find the initial void ratio.
a) 2.456
b) 5.645
c) 1.121
d) 8.553

Explanation: Given,
mv= 0.74 m2/MN= 0.74×10-3 m2/kN
∆σ’=200-100= 100kN/m2
∆e= 0.157
$$1+e_0=\frac{∆e_0}{m_v}\frac{1}{∆σ’}$$
e0= 1.121.

4. If a clay layer is expected to settle 12cm under loading settles 3cm at the end of one month, then find the degree of consolidation.
a) 25%
b) 50%
c) 60%
d) 56%

Explanation: Given,
ρ=3 cm
ρf=12 cm
degree of consolidation $$U=\frac{ρ}{ρ_f}×100=\frac{3}{12}×100=25%.$$

5. If the initial voids ratio is 0.964 for a layer of height 100cm. It is located with 200KN/m2 and the void ratio becomes 0.822. Find the settlement.
a) 8.54 cm
b) 7.43 cm
c) 7.22 cm
d) 8.44 cm

Explanation: Given,
e0=0.964
E=0.822
H=100cm
$$ρ_f=\frac{e_0-e}{1+e_0}×H$$
$$ρ_f=\frac{0.964-0.822}{1+0.964}×100=7.22cm.$$
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6. For the consolidation of 50%, find the time factor.
a) 0.197
b) 0.564
c) 0.253
d) 0.826

Explanation: Given, U=50%
For U < 60%,
$$T_v=\frac{π}{4}(\frac{V}{100})^2$$
$$(T_v)_{50}=\frac{π}{4}(\frac{50}{100})^2=0.197.$$

7. For the consolidation of 90%, find the time factor.
a) 0.898
b) 0.848
c) 0.976
d) 0.934

Explanation: Given, U=90%
For U > 60 %,
Tv=1.7813-0.9322log10⁡(100-U%)= 0.848.

8. For a time factor Tv of 0.159, Find the degree of consolidation.
a) 45%
b) 98%
c) 34%
d) 18%

Explanation: As we know,
The time factor for 50% consolidation is 0.287
since, 0.159<0.287,
U > 60%,
Therefore $$T_v=\frac{π}{4}$$(U%/100)2
$$0.159=\frac{π}{4}$$(U%/100)2
U=45%.

9. For a time factor Tv of 0.567, find the degree of consolidation.
a) 37%
b) 60%
c) 80%
d) 95%

Explanation: As we know,
The time factor for 60% consolidation is 0.287
Since 0.287 < 0.567
U>60%
Therefore, Tv=1.7813-0.9322log10⁡(100-U%)
0.567=1.7813-0.9322log10⁡(100-U%)
U=80%.

10. For a dry mass of soil of 180.4g and height of solids as 13.45mm with cross-sectional area as 50cm2, find the specific gravity of soil.
a) 2.68
b) 8.33
c) 1.64
d) 5.32

Explanation: Given,
Md=180.4 g
A= 5- cm2
Hs=13.45 mm
We know that, $$H_s=\frac{M_d}{GAρ_w}$$
$$G=\frac{M_d}{H_s Aρ_w}$$
$$G=\frac{180.4×10}{13.45×50×1}=2.68.$$

11. For final void ratio 0.886 and water content of 30%, find the specific gravity.
a) 4.68
b) 2.68
c) 1.45
d) 2.45

Explanation: Given,
ef=0.886
Wf=33%=0.33
ef=WfG
$$G=\frac{e_f}{W_f} = \frac{0.886}{0.33}=2.68.$$

12. The diagram shows a soil profile with following properties: Sand layer: γsat=20.86 kN/m3
Clay layer: W=38%, Cc=0.26, G=2.72
Find the effective pressure in sand layer.
a) 44.16 KN/m3
b) 35.33 kN/m3
c) 56.77 kN/m3
d) 23.73 kN/m3

Explanation: Given,
Sand layer: γsat=20.86 kN/m3
Z=4m
γsat’= γsat – γw=20.86-9.81=11.05 kN/m3.
σ’=11.05×4=44.16 kN/m3.

13. The diagram shows a soil profile with following properties
Sand layer: γsat=20.86 kN/m3
Clay layer: W=38%, Cc=0.26, G=2.72, Z=2m
Find the effective pressure in clay layer. a) 34.5 KN/m3
b) 45.0 KN/m3
c) 16.6 KN/m3
d) 81.5 KN/m3

Explanation: Given,
Clay layer: W=38%, Cc=0.26, G=2.72, Z=2m
e=WsatG=0.38×2.72=1.0336
$$γ’=\frac{(G-1) γ_ω}{1+e}=8.3 kN/m^3$$
σ’=8.3×2=16.6 kN/m3.

14. The diagram shows a soil profile with following properties
Sand layer:
Clay layer: W=38%, Cc=0.26, G=2.72
Find the settlement of the clay layer. a) 0.571
b) 0.321 m
c) 0.937 m
d) 0.534 m

Explanation: Given,
Sand layer:
γsat=20.86 kN/m3
Z=4m
γsat’= γsat –γw=20.86-9.81=11.05 kN/m3.
W=38%, Cc=0.26, G=2.72, Z=2m
e=WsatG=0.38×2.72=1.0336
$$γ’=\frac{(G-1)γ_ω}{1+e}=8.3 kN/m^3$$
σ0‘=11.05×4+8.3×1 = 52.5 kN/m2
$$ρ_f=\frac{C_c H}{1+e}log_10⁡\frac{σ_0’+∆σ}{σ_0′}=\frac{0.26×2}{1+1.0336}log_{10}\frac{52.5+120}{52.5} = 0.1321 m.$$

15. For a mass of 3.192g of soil with voids ratio 0.532, Find its saturated mass density.
a) 2.084 g/cm3
b) 1.097 g/cm3
c) 3.593 g/cm3
d) 5.237 g/cm3

Explanation: Given,
M=3.192g
e=0.532
$$ρ_{sat}=\frac{M}{1+e}=\frac{3.192}{1+0.532}=2.084 g/cm^3.$$

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