This set of Soil Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Consolidation Process – 4”.

1. Who conducted consolidation test on a number of clays from different parts of the world?

a) Terzaghi

b) Taylor

c) Skempton

d) Darcy

View Answer

Explanation: Skempton conducted consolidation tests on a number of clays from different parts of the world and gave the following expression for the compression index for remoulded sample of clay,

C

_{c}=0.007(w

_{L}-10%).

2. In practical cases, the final consolidation settlement is calculated by the equation ___________

a) \(ρ_f=∫_0^H m_v Hdz\)

b) \(ρ_f=∫_0^HH∆σ’dz\)

c) \(ρ_f=∫_0^Hm_v H∆σ’ dz\)

d) \(ρ_f=∫_0^Hm_v ∆σ’ dz\)

View Answer

Explanation: The consolidation settlement ∆ρ

_{f}of element of thickness dz is calculated under average effective pressure increment ∆σ’. Therefore, the final settlement is the integration of the ∆ρ

_{f}.

∴ \(ρ_f=∫_0^Hm_v ∆σ’ dz.\)

3. The integration of the equation \(ρ_f=∫_0^Hm_v ∆σ’dz\) can be performed by _____________

a) numerical method only

b) graphical method only

c) both numerical method and graphical method

d) can not be performed

View Answer

Explanation: The integration of the equation \(ρ_f=∫_0^Hm_v ∆σ’dz\) can be performed by:

- numerical method
- graphical method.

4. The numerical integration may be performed by _______________

a) dividing the total thickness H into a number of thin layers

b) dividing the total thickness H into only two thin layers

c) dividing the total thickness H by a factor

d) excluding the thickness of the soil layer

View Answer

Explanation: The numerical integration may be performed by dividing the total thickness H into a number of thin layers. The settlement of each thin layer is found and is summed up to give the final settlement.

5. In numerical integration of the equation, the total settlement of the layer is equal to _____________

a) product of individual settlements of the various thin layers

b) sum of individual settlements of the various thin layers

c) difference of individual settlements of the various thin layers

d) division of individual settlements of the various thin layers

View Answer

Explanation: The numerical integration may be performed by dividing the total thickness H into a number of thin layers. The settlement of each thin layer is found and is summed up to give the final settlement. Therefore, the total settlement of the layer is equal to sum of individual settlements of the various thin layers.

6. The relation between difference in thickness and voids ratio is given by ___________

a) \(∆H = \frac{e_o-e}{1+e_o}\)

b) \(\frac{∆H}{H_0} =\frac{e_o-e}{1+e_o}\)

c) \(H_0 = \frac{e_o-e}{1+e_o}\)

d) \(\frac{∆H}{H_0} = \frac{1}{1+e_o}\)

View Answer

Explanation: When the soil is laterally confined, there is decrease in the volume. Since the soil is laterally confined, there will be change of volume only due to the decrease in the thickness which implies to the changes in volume is proportional to change in thickness ∆H. This is also true for the initial conditions.

∴ \(\frac{∆H}{H_0} = \frac{e_o-e}{1+e_o}\)

Where e

_{o}-e=∆e.

7. The final settlement in terms of voids ratio is given by ____________

a) \(ρ_f=\frac{e_o-e}{e_o}H\)

b) \(ρ_f=\frac{e}{1+e_o}H\)

c) \(ρ_f=\frac{e_o-e}{1+e_o}\)

d) \(ρ_f=\frac{e_o-e}{1+e_o}H\)

View Answer

Explanation: Since the final settlement is equal to the change in thickness, and the change in thickness is given by,

\(\frac{∆H}{H_0} = \frac{e_o-e}{1+e_o},\)

∴ \(ρ_f = ∆H = \frac{e_o-e}{1+e_o}H.\)

8. The compression index for normally consolidated soil is ____________

a) constant

b) variable

c) zero

d) unity

View Answer

Explanation: The compression index for normally consolidated soil is constant. To achieve normal consolidation, the total load applied on the sample is greater than that which is experienced in the field layer of soil from which the sample is collected.

9. In terms of compression index and voids ratio for normally consolidated soil, the final settlement is _________

a) \(ρ_f=Hlog_{10} \frac{σ’}{σ’_0}\)

b) \(ρ_f=H \frac{C_c}{1+e_o} log_{10} \frac{σ’}{σ’_0}\)

c) \(ρ_f=\frac{C_c}{1+e_o} log_{10} \frac{σ’}{σ’_0}\)

d) \(ρ_f=H \frac{1}{1+e_o} log_{10} \frac{σ’}{σ’_0}\)

View Answer

Explanation: The final settlement is given by the equation,

\(ρ_f=\frac{e_o-e}{1+e_o}H,\) the relationship between the compression index and voids ratio is \(C_c=\frac{e_0-e}{log_{10}\frac{σ’}{σ’_0}},\)

Therefore, substituting for compression index,

We get, \(ρ_f=H \frac{C_c}{1+e_o} log_{10} \frac{σ’}{σ’_0}.\)

10. In case of pre-consolidated soil, the final settlement is ___________

a) small

b) negligible

c) large

d) very large

View Answer

Explanation: In case of pre-consolidated soil, the final settlement is small. This is because, the recompression index or the swelling index C

_{s}is very small in comparison to the compression index C

_{c}. The pre-consolidated soil has been subjected to large overburden loads in the past.

11. The recompression index C_{s} is________ compared to compression index C_{c} for pre-consolidated soils.

a) small

b) very small

c) large

d) very large

View Answer

Explanation: In the case of pre-consolidated soils, the settlement of it is very small as the recompression index or the swelling index C

_{s}is very small when compared to the compression index C

_{c}.

12. When the effective pressure is smaller than pre-consolidation pressure, the final settlement is ____________

a) \(ρ_f=Hlog_{10} \frac{σ’}{σ’_0}\)

b) \(ρ_f=H \frac{C_c}{1+e_o} log_{10} \frac{σ’}{σ’_0}\)

c) \(ρ_f=H \frac{C_S}{1+e_o} log_{10} \frac{σ’}{σ’_0}\)

d) \(ρ_f=H \frac{1}{1+e_o} log_{10} \frac{σ’}{σ’_0}\)

View Answer

Explanation: When the effective pressure σ’ is smaller than pre-consolidation pressure σ

_{p}‘, the final settlement is given in terms of swelling index,

\(ρ_f=H \frac{C_S}{1+e_o} log_{10} \frac{σ’}{σ’_0}.\)

13. For a finite surface loading, the intensity of change in effective pressure decreases with depth of layer in linear manner.

a) True

b) False

View Answer

Explanation: For a finite surface loading as in the case of practical loads, the intensity of change in effective pressure decreases with depth of layer in non-linear manner. In such cases, the consolidation settlement ∆ρ

_{f}of element of thickness dz is calculated under average effective pressure increment ∆σ’.

14. When the pre-consolidation pressure σ_{p}‘ is greater than initial effective pressure σ’_{0} but smaller than effective pressure σ’, then the final settlement is calculated on the basis of ___________

a) settlement due to σ’_{0} to σ_{p}‘ using C_{S} only

b) settlement due to σ_{p}‘ to σ’ using C_{c} only

c) settlement due to σ’_{0} to σ_{p}‘ using C_{S} and settlement due to σ_{p}‘ to σ’ using C_{c}

d) settlement due to σ’ to σ_{p}‘ using C_{c} and settlement due to σ_{p}‘ to σ’ using C_{c}

View Answer

Explanation: In the determination of the final settlement by voids ratio, when the pre-consolidation pressure σ

_{p}‘ is greater than initial effective pressure σ’

_{0}but smaller than effective pressure σ’, then the final settlement is calculated using the two parts:

- settlement due to σ’
_{0}to σ_{p}‘ using C_{S} - settlement due to σ
_{p}‘ to σ’ using C_{c}.

15. When the pre-consolidation pressure σ_{p}‘ is greater than initial effective pressure σ’_{0} but smaller than effective pressure σ’, then the final settlement is calculated on the basis of equation _________

a) \(ρ_f=H \frac{C_c}{1+e_o}log_{10}\frac{σ_p’}{σ’_0}+H \frac{C_c}{1+e_o}log_{10}\frac{σ’}{σ_p’}\)

b) \(ρ_f=H \frac{C_S}{1+e_o}log_{10}\frac{σ_p’}{σ’_0}+H \frac{C_S}{1+e_o}log_{10}\frac{σ’}{σ_p’}\)

c) \(ρ_f=\frac{C_S}{1+e_o}log_{10}\frac{σ_p’}{σ’_0}+\frac{C_c}{1+e_o}log_{10} \frac{σ’}{σ_p’}\)

d) \(ρ_f=H\frac{C_S}{1+e_o}log_{10}\frac{σ_p’}{σ’_0}+H \frac{C_c}{1+e_o}log_{10} \frac{σ’}{σ_p’}\)

View Answer

Explanation: In the determination of the final settlement by voids ratio, when the pre-consolidation pressure σ

_{p}‘ is greater than initial effective pressure σ’

_{0}but smaller than effective pressure σ’, then the final settlement is calculated using the two parts:

settlement due to σ’

_{0}to σ

_{p}‘ using C

_{S}

settlement due to σ

_{p}‘ to σ’ using C

_{c}.

Therefore, the final settlement is given by the equation,

\(ρ_f=H\frac{C_S}{1+e_o}log_{10}\frac{σ_p’}{σ’_0}+H\frac{C_c}{1+e_o}log_{10}\frac{σ’}{σ_p’}.\)

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