Consolidation MCQ Quiz - Objective Question with Answer for Consolidation - Download Free PDF

Explanation:

The coefficient of volume change is given as:

\(m_v = \frac{Δ V}{V}\frac{1}{Δ\sigma}\)

ΔV/V= change in volume per unit volume which can be expressed as

\(\frac{\Delta V}{V} = \frac{\Delta e}{1 + e}\)

ΔV = Change in Volume, V = Original Volume

Δσ = change in effec­tive stress or pressure

Δe = change in void ratio and e is the initial void ratio

Calculation:

Given that

Initial void ratio, e = 1.121 and final void ratio, e_f = 0.964

Initial effective stress, σ_o = 100 kN/m²

Final stress or pressure, σ_f = 200 kN/m²

Substituting these values in the above formula, 

The coefficient of volume change

\(m_v = \frac{(1.121 - 0.964)}{1 + 1.121}\times \frac{1}{(200-100)}\)

∴ m_v = 7.4 × 10^-5 m²/kN

Explanation

Degree of consolidation

The degree of consolidation is defined as the ratio of settlement at a particular time to the ultimate settlement, expressed in percentage. Also, the degree of consolidation is the function of the time factor (T_v).

So \({T_V} = {C_V} \times \frac{t}{{{d^2}}}\)      ----(1)

where TV = Time Tractor, CV = Efficient & Conciliation

t = time in sec, d = length or thickness of drainage path

Calculations:

Now U < 60% 

So \({T_V} = \frac{\pi }{4} \times {(U)^2}\)

Put the values in equation 1, we get

\(\frac{\pi }{4} \times {(U)^2} = {C_V} \times \frac{t}{{{d^2}}}\)

\({(U)^2} = \frac{{{C_V} \times 4}}{\pi } \times \frac{t}{{{d^2}}}\)

\(U = \sqrt {\frac{{4x{C_V} \times t}}{{\pi \times {d^2}}}} \)

\(U = \frac{2}{d} \times {\left( {\frac{{{C_V} \times t}}{\pi }} \right)^{{1\over2}}}\)

Additional Information

If U > 60%

Then

\({T_v} = 1.781 - 0.933{\log _{10}}\left( {100 - U\% } \right)\)

Explanation:

1. Consolidation refers to the gradual reduction in volume and settlement of a soil due to the expulsion of water from its pores under an applied load.

2. In the case of fine-grained soils (like clays), lowering the water level can cause the soil particles to settle further, leading to consolidation settlement.

3. Fine-grained soils have a high water retention capacity, and when the water level is lowered, the water pressure in the pores decreases, resulting in further compression of the soil over time.

 Additional InformationElastic Compression

• Elastic compression refers to the temporary deformation of a material when subjected to stress, where the material returns to its original shape once the stress is removed.

• In soils, this refers to the immediate deformation caused by external loads, such as the weight of a structure.

• Unlike consolidation, which is a time-dependent process involving the expulsion of water, elastic compression is instantaneous and reversible when the applied load is removed.

Erosion

• Erosion is the process by which soil particles are worn away and removed from their original location due to the action of wind, water, or other natural forces.

• It generally occurs in coarse-grained soils or near water bodies where the flow can transport particles.

• Erosion doesn’t directly relate to settlement caused by the lowering of water levels in fine-grained soils. However, it can occur if water levels fluctuate rapidly or if the soil is disturbed by water movement.

Plastic Compression

• Plastic compression refers to the permanent deformation of a material when subjected to a load beyond its elastic limit, causing it to change shape irreversibly.

• In soil, this might occur in coarse-grained soils under extreme pressure but is not a time-dependent process like consolidation. It’s more typical in materials that do not return to their original shape after load removal.

• Plastic compression doesn’t directly describe the process that occurs when water levels are lowered in fine-grained soils. It’s more relevant in soils subjected to extreme load conditions.

Explanation:

Assumptions made in Terzaghi’s one-dimensional consolidation theory are:

1. Flow is one-dimensional (vertical)
2. Darcy’s law is valid
3. The soil is homogeneous
4. The soil is completely saturated
5. The soil grains and water are both incompressible.
6. Strains are small i.e with the applied load increment there is no change in the thickness.

Explanation:

The total permissible settlement is given by I.S. code:

Concept:

Over Consolidation Ratio:

Over Consolidation Ratio ( OCR ) is defined as the ratio of maximum pressure experienced in the past

( Preconsoildation pressure ) to the present overburden pressure.

\({\bf{OCR}} = \frac{{{\bf{Maximum}}\;{\bf{pressure}}\;{\bf{in}}\;{\bf{past}}}}{{{\bf{Present}}\;{\bf{overburden}}\;{\bf{pressure}}}}\)

OCR > 1 ⇒ It is Over consolidated soil

OCR = 1 It is Normally consolidated soil

OCR < 1It is Under consolidated soil

Calculation:

OCR of clay = 4

Since the OCR of the given soil is greater than 1, it is overconsolidated clay.

A-factor:

A – factor is a function of the over-consolidation ratio in the case of overconsolidated clay.

For OC clays, A = f (OCR)

Concept:

The coefficient of compression index (Cc)

\({C_c} = \frac{{{e_1} - {e_2}}}{{{{\log }_{10}}\left( {\frac{{{{\bar \sigma }_1}}}{{{{\bar \sigma }_o}}}} \right)}}\)

The coefficient of compressibility (av)

\({a_v} = \left| {\frac{{\Delta e}}{{{{\bar \sigma }_o} - {{\bar \sigma }_1}}}} \right|\)

Calculation:

Given,

Δe = 0.2

\({{\bar \sigma }_o} = 50\) kN/m2

\({{\bar \sigma }_1} = 100\) kN/m2

Cc = 0.664 m2/kN

\({a_v} = \left| {\frac{{\Delta e}}{{{{\bar \sigma }_o} - {{\bar \sigma }_1}}}} \right| = \frac{{0.2}}{{50}} = 0.004\)

∴ av = 0.004 m2/kN

Concept:

Degree of consolidation (U) \(= \frac{{{\rm{\Delta }}h}}{{{\rm{\Delta }}H}} \times 100\) 

Δh = settlement of soil after time ‘t’.

ΔH = Ultimate consolidation settlement when U = 100%

Δh will depend on two way or one-way drainage condition

Time factor [T_V]

\({T_V} = \frac{\pi }{4}{\left( U \right)^2},\;when\;U \le 60\%\)

T_V = (-0.9332 log₁₀ (1 - U) – 0.0851) when U > 60%

\({T_V} = \frac{{{C_v}t}}{{{d^2}}}\)

Where,

d = length of drainage path

\(d = \frac{{{H_0}}}{2}\left[ {Two\;way\;drainage} \right]\)

d = H₀ [one way drainage]

where,

H₀ = thickness of clay layer

t = time required for degree of consolidation equal to (U%)

C_V = coefficient of consolidation​

Calculation:

For U₁ = 30% then t₁ = 180 days

It is said further 20% consolidation So U2 = 50% then t₂ = ?

\(\frac{{{t_2}}}{{{t_1}}} = {\left( {\frac{{{U_2}}}{{{U_1}}}} \right)^2}\)

\({t_2} = {\left( {\frac{{50}}{{30}}} \right)^2} \times 180 = 500\;days\)

∴ The additional time required = 500 – 180 = 320 days

Explanation:

The final consolidation settlement is given by,

\({\rm{Δ H}} = \left( {\frac{{{C_c}\; × \;{H_0}}}{{1\; + \;{e_0}}}} \right) \;× \;{\log _{10}}\frac{{{σ _0}' \;+\; {\rm{Δ }}σ }}{{{σ _0}'}}\)

ΔH - ultimate consolidation settlement

Cc - Compression index, H0 - Initial thickness of soil layer

σ0' - Initial effective stress, Δσ'  - Increase in effective stress

Settlement proportional to log (\({\left( {\frac{{{{\bar \sigma }_f}}}{{{\sigma _0}}}} \right)} \))

For 2 conditions

\(\begin{array}{l} \frac{{{S_{c1}}}}{{{S_{c2}}}} = \frac{{\log {{\left( {\frac{{{{\bar \sigma }_f}}}{{{\sigma _0}}}} \right)}_1}}}{{\log {{\left( {\frac{{{{\bar \sigma }_f}}}{{{\sigma _0}}}} \right)}_2}}} \end{array}\)

Calculation:

Given

Case 1: effective stress was increased from 80 to 160 kN/m2

The ratio of stresses is 160/80 = 2

Case 2: effective stress is increased from 160 to 320 kN/m2

The ratio of stresses is 320/160 = 2

Since \({\left( {\frac{{{{\bar \sigma }_f}}}{{{\sigma _0}}}} \right)_1} = {\left( {\frac{{{{\bar \sigma }_f}}}{{{\sigma _0}}}} \right)_2}\)

\(\begin{array}{l} \frac{{{S_{c1}}}}{{{S_{c2}}}} = \frac{{\log {{\left( {\frac{{{{\bar \sigma }_f}}}{{{\sigma _0}}}} \right)}_1}}}{{\log {{\left( {\frac{{{{\bar \sigma }_f}}}{{{\sigma _0}}}} \right)}_2}}} \end{array}=1\)

Hence settlement will be the same. So Settlement = 2 cm

When the pressure is applied to saturated soil, the soil undergoes immediate small compression due to the expulsion of excess pore water because of seepage of water from the voids. This immediate settlement is determined by using elastic theory and the process is called initial consolidation.

 After the initial consolidation, primary consolidation starts which is a time-dependent phenomenon. This process continues until excess pore water becomes zero.

After the reduction of all excess pore pressure to zero and primary consolidation is completed. Secondary consolidation occurs under constant effective stress at a very slow rate where plastic readjustment of solid particles compression is due to the expulsion of highly viscous water. Also called Secondary Compression.

In all these stages soil is undersaturation, only excess pore water is squeezed out. Thus results in a reduction of volume of soil.

Explanation:

Assumptions made in Terzaghi’s one-dimensional consolidation theory are:

1. Flow is one-dimensional and no change in the area cross-sectional occurs.

2. Darcy’s law is valid.

3. The soil is homogeneous

4. Volume of the soil solids is constant.

5. The soil is completely saturated

6. The soil grains and water are both incompressible.

7. Strains are small i.e with the applied load increment there is no change in the thickness.

8. Load is applied only in vertical direction and consolidation occurs only in the direction of the load applied, i.e., the soil is laterally confined and restrained against lateral consolidation.

9. Drainage of pore water occurs only in the vertical direction.

10. The boundary is a free surface offering no resistance to the flow of pore water from the soil.

11. The time lag in consolidation is due entirely to the low permeability of the soil, and thus, the secondary consolidation is disregarded.

12. There is a unique relationship between the void ratio and the effective stress, and this relationship remains constant during the load increment. 

13. The consolidation occurs due to expulsion of water from the voids and consequent decrease in the void ratio.

14. The coefficient of permeability of the soil has the same value at all points, and it remains constant during the entire period of consolidation.

Concept:

Computation of settlement when change in void ratio is given:

Therefore, \(\frac{{Δ H}}{{{H_o}}} = \frac{{Δ e}}{{1 + {e_o}}}\) 

Where

H_o = thickness of clay deposit

ΔH = settlement layer thickness

Δe = e₁ - e_o

e_o = initial void ratio

e₁ = final void ratio

Calculation:

Given,

H₁ = 10 cm

e_o = 0.5

e₂ = 0.2

\(\frac{{Δ H}}{{{H_o}}} = \frac{{Δ e}}{{1 + {e_o}}}\)

\(\Delta H = \frac{{\Delta e}}{{1 + {e_o}}} \times {H_o} = \frac{{(0.5 - 0.2)}}{{1 + 0.5}} \times 10 = 2cm\)

Concept:

As per the theory of consolidation, Time factor is calculated as,

\({T_v} = {C_v}.\frac{t}{{{d^2}}}\)

Where T_v = time factor which depends upon the degree of consolidation

C_v = Coefficient of consolidation

d = Length of the drainage path

t = time

For one way draiange, d = H

For two way drainage, d = H/2

Calculation:

Given, For case 1:

degree of consolidation = 50%, d₁ = d/2 (double drainage), t₁ = t

For Case 2:

Degree of consolidation = 50%, d₂ = 3d (Single drainage), t₂ = ?

∵ \({T_v} = {C_v}.\frac{t}{{{d^2}}}\)

\({T_{v1}} = {C_{v1}}.\frac{{{t_1}}}{{d_1^2}} = {C_v}.\frac{t}{{{{\left( {\frac{d}{2}} \right)}^2}}}\) .......(i)

\( {T_{v2}} = {C_{v2}}.\frac{{{t_2}}}{{d_2^2}} = {C_v}.\frac{t_2}{{{{\left( {3d} \right)}^2}}}\).........(ii)

From (i) and (ii)

t₂ = 36t

Given,

Degree of consolidation = 50%

U = 50%

Time factor

\(T_v = \frac \pi 4(U^2)\) when U ≤ 0.6

T_v = 1.781 - 0.933 log₁₀ (100 - U%)

U > 0.6

Than \(T_v = \frac \pi 4 \left(\frac {50}{100}\right)^2\)

\(= \frac \pi 4 \left(\frac 1 2 \right)^2\)

T_v = 0.196

So, T_v ≃ 0.2

Concept:

Determination of Coefficient of Consolidation(C_v):

The coefficient of consolidation, C_v can readily be estimated from the time-settlement curve using graphical methods. There are two most commonly used methods are as follows:

1. Casagrande Logarithm of Time Fitting Method (Casagrande and Fadum, 1940):

The coefficient of consolidation, C_V, is determined by estimating the time at 50% consolidation (t₅₀), Then, C_V can be estimated as:

\(C_V = 0.197 \times {H_{dr}^2 \over t_{50}}\)

where H_dr is the drainage path. Given the initial height of the specimen (H_i) and the compression of the soil sample at 50% consolidation (ΔΗ), the drainage path (for double drainage), H_dr, is computed as:

H_dr = (H_i - ΔΗ) / 2

2. Taylor Square Root of Time Fitting Method (Taylor, 1948):

In this method, the daily readings are plotted against the square root of time. The coefficient of consolidation, C_V, is determined by estimating the time at 90% consolidation (t₉0), Then, C_V can be estimated as:

\(C_V = 0.848 \times {H_{dr}^2 \over t_{90}}\)

Where  Hdr is the average drainage path.