Engineering Questions with Answers - Multiple Choice Questions

Home » MCQs » Engineering MCQs » Consolidation Problems – 1

# Consolidation Problems – 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

**
View Answer**

Answer: b

Explanation: Given, double drainage for both specimen and field,

t_{1}=20 min

d_{1}=24 mm/2 = 2.4 cm/2 = 1.2 cm

d_{2} =4/2 m 200cm

Since, for same degree of consolidation of 50%, T_{v} is same,

t∝d2Cr

Therefore, t2t1=(d2d1)2

So, t2=t1(d2d1)2=20(2001.2)2=365days.

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

**
View Answer**

Answer: a

Explanation: Given, double drainage for sample and single drainage for field layer,

t_{1}=20 min

d_{1}=24 mm/2 = 2.4 cm/2 = 1.2 cm

d_{2} = 4 m = 400cm

Since, for same degree of consolidation of 50%, T_{v} is same,

t∝d2Cr

Therefore, t2t1=(d2d1)2

So, t2=t1(d2d1)2=20(4001.2)2=1544days.

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

**
View Answer**

Answer: c

Explanation: Given, single drainage for sample and double drainage for field layer,

t_{1} = 2 0 min

d_{1} = 24 mm = 2.4 cm

d_{2} = 2 m = 200cm

Since, for same degree of consolidation of 50%, T_{v} is same,

t∝d2Cr

Therefore, t2t1=(d2d1)2

So, t2=t1(d2d1)2=20(2002.4)2=97days.

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

**
View Answer**

Answer: a

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

t_{1}=20 min

d_{1}=24 mm= 2.4 cm

d_{2} =4 m = 400cm

Since, for same degree of consolidation of 50%, T_{v} is same,

t∝d2Cr

Therefore, t2t1=(d2d1)2

So, t2=t1(d2d1)2=20(4002.4)2=386days.

For a dry soil mass of 180.4 g, specific gravity 2.68 and cross-sectional area of specimen as 50 cm^{2}, find the height of solids H_{s}.

a) 13.45 mm

b) 14 mm

c) 17 mm

d) 19.5 mm

**
View Answer**

Answer: a

Explanation: Given,

M_{d}=180.4 g

G=2.68

A=50cm^{2}

Since Hs=MdGAρw

Hs=180.4∗102.68∗50∗1=13.45mm.

If the specimen height H is 24mm and height of solids H_{s} is 13.45mm, then find the voids ratio.

a) 0.786

b) 0.432

c) 2

d) 34

**
View Answer**

Answer: a

Explanation: Given,

H=24mm

H_{s} =13.45mm

e=H−HsHs

e=24−13.4513.45

e=0.786.

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

**
View Answer**

Answer: b

Explanation: W_{f}=30%=0.30

G=2.65

e_{f}=W_{f}G

e_{f}=0.3*2.65=0.795.

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%

**
View Answer**

Answer: a

Explanation: Given,

e_{f}=0.864

G=2.7

e_{f}=W_{f}G

efG=Wf

W_{f}=32%.

If height of solids H_{s} is 6.725mm, mass is 180.4g with specific gravity G=2.68, then final the Cross-sectional area of specimen.

a) 220 cm^{2}

b) 342 cm^{2}

c) 100 cm^{2}

d) 659 cm^{2}

**
View Answer**

Answer: c

Explanation: Given,

H_{s}=6.725mm

G=2.68

M_{d}=180.4g

Since Hs=MdGHsρw

A=MdGHsρw

A=180.4×102.68×6.725×1=100cm2.

If the final height of specimen H_{f} 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

**
View Answer**

Answer: a

Explanation: Given, H_{f}=13.45

e_{f}=0.864

Δe=1+efHfΔH

Δe=1+0.86413.45ΔH

∆e=0.139∆H.

The change in voids ratio is 0.18 and initial void ratio is 0.68 with the increase of pressure as 200KN/m^{2}. Find the coefficient of volume change.

a) 0.536 m^{2}/MN

b) 0.111 m^{2}/MN

c) 0.832 m^{2}/MN

d) 0.356 m^{2}/MN

**
View Answer**

Answer: a

Explanation: Given,

∆e_{f}=0.18

e_{o}=0.68

∆σ’=200kN/m^{2}

mv=−Δe1+eo1Δσ′

mv=0.181+0.68+1200=0.536m2/MN.

The change in voids ratio is 0.18 and initial void ratio is 0.68 with the increase of pressure as 200KN/m^{2}. Find the coefficient of Compressibility.

a) 0.9 m^{2}/MN

b) 0.7 m^{2}/MN

c) 0.4 m^{2}/MN

d) 0.5 m^{2}/MN

**
View Answer**

Answer: a

Explanation: Given,

∆e_{f}=0.18

e_{o}=0.68

∆σ’=200kN/m^{2}

mv=−Δe1+eo1Δσ′

since, mv=av1+e0

a_{v}=m_{v}(1+e_{0})=0.536(1+0.68)

a_{v}=0.9 m^{2}/MN.

If coefficient of volume change is 0.291 m^{2}/MN, ∆e=0.1 and the initial voids ratio is 0.72, then find the change in effective pressure.

a) 200 KN/m^{2}

b) 50 KN/m^{2}

c) 300 KN/m^{2}

d) 00 KN/m^{2}

**
View Answer**

Answer: a

Explanation: Given,

m_{v}=0.291 m^{2}/MN

∆e=0.10

e_{o}=0.72

Δσ′=−Δe1+eo1mv

Δσ′=0.101+0.72×1×1030.291=200KN/m2.

If a soil A has coefficient of volume change as 0.536m^{2}/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 m^{2}/MN

b) 0.291 m^{2}/MN

c) 0.234 m^{2}/MN

d) 0.345 m^{2}/MN

**
View Answer**

Answer: b

Explanation: Given,

M_{va}=0.536 m^{2}/MN

mvamvb=1.845

Mvb=0.5361.845=0.291m2/MN.

A soil sample is having voids ratio 0.72 under a pressure of 200KN/m^{2}. If the void ratio is reduced to 0.5 under 450kn/m^{2} load, then find the change in effective pressure.

a) 4000 kN/m^{2}

b) 250 kN/m^{2}

c) 350 kN/m^{2}

d) 2000 kN/m^{2}

**
View Answer**

Answer: b

Explanation: Given,

σ_{0}’=200kN/m^{2}

σ’=450kN/m^{2}

∆σ’=σ’-σ_{0}’=450-200

∆σ’=250kN/m^{2} .