# Soil Mechanics Questions and Answers – Consolidation Process – 4

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This set of Soil Mechanics Questions and Answers for Entrance exams 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

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,
Cc=0.007(wL-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) $$ρ_f=∫_0^Hm_v ∆σ’ dz$$

Explanation: The consolidation settlement ∆ρf of element of thickness dz is calculated under average effective pressure increment ∆σ’. Therefore, the final settlement is the integration of the ∆ρf.
∴ $$ρ_f=∫_0^Hm_v ∆σ’ dz.$$

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

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

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

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.
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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) $$\frac{∆H}{H_0} = \frac{1}{1+e_o}$$

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 eo-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) $$ρ_f=\frac{e_o-e}{1+e_o}H$$

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

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

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

Explanation: In case of pre-consolidated soil, the final settlement is small. This is because, the recompression index or the swelling index Cs is very small in comparison to the compression index Cc. The pre-consolidated soil has been subjected to large overburden loads in the past.

11. The recompression index Cs is________ compared to compression index Cc for pre-consolidated soils.
a) small
b) very small
c) large
d) very large

Explanation: In the case of pre-consolidated soils, the settlement of it is very small as the recompression index or the swelling index Cs is very small when compared to the compression index Cc.

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

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

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 ∆ρf of element of thickness dz is calculated under average effective pressure increment ∆σ’.

14. When the pre-consolidation pressure σp‘ 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 σp‘ using CS only
b) settlement due to σp‘ to σ’ using Cc only
c) settlement due to σ’0 to σp‘ using CS and settlement due to σp‘ to σ’ using Cc
d) settlement due to σ’ to σp‘ using Cc and settlement due to σp‘ to σ’ using Cc

Explanation: In the determination of the final settlement by voids ratio, when the pre-consolidation pressure σp‘ 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 σp‘ using CS
• settlement due to σp‘ to σ’ using Cc.

15. When the pre-consolidation pressure σp‘ 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) $$ρ_f=H\frac{C_S}{1+e_o}log_{10}\frac{σ_p’}{σ’_0}+H \frac{C_c}{1+e_o}log_{10} \frac{σ’}{σ_p’}$$

Explanation: In the determination of the final settlement by voids ratio, when the pre-consolidation pressure σp‘ 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 σp‘ using CS
settlement due to σp‘ to σ’ using Cc.
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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