Consolidation

 Consolidation

Q: Explain Terzaghi’s spring-mass analogy

Q: Explain Terzaghi’s consolidation theory

  • Consolidation is process of reduction in volume due to expulsion of water particle with the application of static load for long duration

Terzaghi’s Spring Mass Analogy

Assumptions:

  • The soil medium is completely saturated 

  • The soil medium is isotropic and homogeneous 

  • Darcy’s law is valid for flow of water 

  • Flow is one dimensional in the vertical direction 

  • The coefficient of permeability is constant 

  • The coefficient of volume compressibility is constant 

  • The increase in stress on the compressible soil deposit is constant (∆σ´ = constant) 

  • Soil particles and water are incompressible 

One dimensional theory is based on the following hypothesis 

1. The change in volume of soil is equal to volume of pore water expelled. 

2. The volume of pore water expelled is equal to change in volume of voids. 

3. Since compression is in one direction the change in volume is equal to change in height.

  • T=0

  • σ = u  and  σ´=0

  • Soil=Spring

  • Water= Water Void

  • After application of load - primary consolidation Start

  • No reduction in volume

  • 0<T<∞

  • σ = u  +  σ´

  • All load is resisted by water

  • flow is upward

  • reduction in volume 

  • spring & water share load

  • T = ∞

  • All the load is resisted by spring

  • Water doesn’t carry any load

  • Soil solids carry external pressure

  • Excess pore water pressure is zero

  • End of primary consolidation

Theory 

  • Terzaghi’s model consists of a vessel with a piston attached to spring as shown.

  • The space below piston is filled with water. The piston has outlet to allow for passage of water. 

  • Piezometers can be inserted to measure the excess pore water pressure.

  • Terzaghi has correlated the spring mass compression process with the consolidation of saturated clay subjected to external load σ.

  • The springs and the surrounding water represent the saturated soil. 

  • The springs represent the soil skeleton networks of soil grains and water in the vessels represents the water in the voids. 

  • In this arrangement the compression is one dimensional and flow will be in the vertical direction.

  • When pressure σ is applied this will be borne by water surrounding the spring

σ = u  and  σ´=0 at time t =0

where, u is called excess hydrostatic pressure and there will be no volume change.

  • After sometime ‘t’ there will be flow of water through outlet.

  • Since the flow is in upward direction/segment there will be reduction in volume.

  • Due to this spring get compressed and being to carry a portion of the applied load. 

  • This signifies a reduction in excess pore water pressure and increase in effective stress.

σ = u  + σ´🡪 load shared by both water and spring

  • At time t =  ∞  when no more pore water flows out the excess hydrostatic pressure will be and the entire load is carried by spring.

σ = σ’  and  u=0 at time t =∞

Q: Explain laboratory consolidation test

The test is conducted to determine the settlement due to primary consolidation. To determine :

  • Rate of consolidation under normal load.

  • Degree of consolidation at any time.

  • Pressure-void ratio relationship.

  • Coefficient of consolidation at various pressures.

  • Compression index.

From the above information it will be possible for us to predict the time rate and extent of settlement of structures founded on fine-grained soils. It is also helpful in analyzing the stress history of soil. Since the settlement analysis of the foundation depends mainly on the values determined by the test, this test is very important for foundation design. 

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Principal Involved

When a compressive load is applied to soil mass, a decrease in its volume takes place, the decrease in volume of soil mass under stress is known as compression and the property of soil mass pertaining to its tendency to decrease in volume under pressure is known as compressibility. In a saturated soil mass having its void filled with incompressible water, decrease in volume or compression can take place when water is expelled out of the voids. Such a compression resulting from a long time static load and the consequent escape of pore water is termed as consolidation.

Procedure

  1. Saturate two porous stones by keeping them submerged in the distilled water for 4 to 8 hrs. Wipe away excess water. 

  2. Assemble the consolidometer, with the soil specimen and porous stones at top and bottom of specimen, providing a filter paper between the soil specimen and porous stone. Position the pressure pad centrally on the top porous stone.

  3. Mount the mould assembly on the loading frame, and center it such that the load applied is axial.

  4. Position the dial gauge to measure the vertical compression of the specimen. The dial gauge holder should be set so that the dial gauge is in the begging of its releases run, allowing sufficient margin for the swelling of the soil, if any.

  5. Connect the mould assembly to the water reservoir and the sample is allowed to saturate. The level of the water in the reservoir should be at about the same level as the soil specimen.

  6. Apply an initial load to the assembly. The magnitude of this load should be chosen by trial, such that there is no swelling. 

  7. The load should be allowed to stand until there is no change in dial gauge readings for two consecutive hours or for a maximum of 24 hours.

  8. Note the final dial reading under the initial load. Apply first load of intensity 0.1 kg/cm2 start the stop watch simultaneously. Record the dial gauge readings at various time intervals. The dial gauge readings are taken until 90% consolidation is reached. Primary consolidation is gradually reached within 24 hrs.

  9. At the end of the period, specified above take the dial reading and time reading. Double the load intensity and take the dial readings at various time intervals. Repeat this procedure for successive load increments. The usual loading intensity are as follows :

    1. 0.1, 0.2, 0.5, 1, 2, 4 and 8 kg/cm2.

  10. After the last loading is completed, reduce the load to half of the value of the last load and allow it to stand for 24 hrs. Reduce the load further in steps of half the previous intensity till an intensity of 0.1 kg/cm2 is reached. Take the final reading of the dial gauge.

  11. Reduce the load to the initial load, keep it for 24 hrs and note the final readings of the dial gauge.

  12. Quickly dismantle the specimen assembly and remove the excess water on the soil specimen in oven, note the dry weight of it.

Analysis:

  1. Calculate the initial water content and specific gravity of the soil.

  2. For each pressure increment, construct a semi-log plot of the consolidation dial readings versus the log time (in minutes). Determine the coefficient of consolidation (cv). 

  3. Calculate the void ratio at the end of primary consolidation for each pressure increment. Plot log pressure versus void ratio. Based on this plot, calculate compression index, recompression index and preconsolidation pressure (maximum past pressure).

Q: Explain Log – time curve fitting method to find coefficient of consolidation (Cv)

The basis for this method is the theoretical (Uz ) versus log Tv curve and experimental dial gauge reading and log(t) curves are similar.

Steps

  1. Plot the dial reading of compression for a given pressure increment versus time to log scale as shown in figure.

  2. Plot two points P and Q on the upper portion of the consolidation curve (say compression line) corresponding to time t1 and t2 such that t 2 =4t 1

  3. Let x be the difference in dial reading between P and Q. locate R at a vertical distance x above point P 

  4. Draw a horizontal line RS the dial reading corresponding to this line is d 0 which corresponds with 0% consolidation.

  5. Project the straight line portion of primary and secondary consolidation to intersect at point T. The dial reading corresponding to T is d 100  and this corresponds to 100% consolidation.

  6. Determine the point V on the consolidation curve which corresponds to the dial reading of

d 50 = (d 0 +d 100)/2 . The time corresponding to point V is t 50 i.e time for 50% consolidation.

Log – time curve fitting method

Q: Explain Square root of time method to find coefficient of consolidation (Cv)

  1. Plot the dial reading and the corresponding square-root-of-time 𝑡 as shown in Figure 

  2. Draw the tangent PQ to the early portion of the plot. 

  3. Draw a line PR such that OR = (1.15)(OQ). 

  4. The abscissa of the point S (i.e., the intersection of PR and the consolidation curve) will give √𝑡90 ( i.e., the square-root-of-time for 90% consolidation). 

  5. The value of 𝑇𝑣 for Uav = 90% 𝑖𝑠 0.848. so,

Square root of time method

Q: Explain soil compressibility characteristics

Q: Explain normally consolidated, over and under consolidated soil

  • A laboratory soil specimen of dia 60mm and height 20mm is extracted from the undisturbed soil sample obtained from the field. 

  • This sample is subjected to 1D consolidation in the lad under various pressure increments. 

  • Each pressure increment is maintained for 24 hrs and equilibrium void ratio is recorded before the application of the next pressure increment. 

  • Then a plot of void ratio versus effective stress is made as shown in Fig. 

  • When the sample is recompressed from point D it follows DE and beyond C it merges along BCF and it compresses as it moves along BCF

  • During the initial stages (at low effective stress) sample follows recompression path (portion AB) and undergoes less compression. 

  • Beyond this is the virgin compression line (portion BC) also called the normal compression line and the sample undergoes large compression. 

1. BC – Virgin compression curve also called normal consolidation line

2. From ‘C’ when the sample is unloaded, sample expands and traces path CD (expansion curve unloading) 

3. Sample undergoes Permanent strain due to irreversible soil structure and there is a small elastic recovery. 

4. The deformation recovered is due to elastic rebound

5. When the sample is reloaded-reloading curve lies above the rebound curve and makes an hysteresis loop between expansion and reloading curves. 

6. The reloaded soils shows less compression. 

7. Loading beyond ‘C’ makes the curve to merge smoothly into portion EF as if the soil is not unloaded.

Soil Compressibility


It is the maximum effective stress experienced by a soil in its stress history (past existence)

  • For the soil loaded along the recompression curve AB the effective stress close to point B will be the preconsolidation pressure. 

  • If the soil is compressed along BC and unloaded along CD and then reloaded along DC the effective stress close to point C will be the new preconsolidation pressure.

Effect of Stress History

It is based on the stress history (preconsolidation pressure) soils are classified as 

1. Normally Consolidated Soils 

2. Over Consolidated Soils 

3. Under Consolidated Soils

Normally Consolidated Soils It is a soil deposit that has never subjected to a vertical effective stress greater than the present vertical stress.

Under Consolidated Soils A soil deposit that has not consolidated under the present overburden pressure (effective stress) is called Under Consolidated Soil. These soils are susceptible to larger deformation and cause distress in buildings built on these deposits.

Over Consolidated Soils It is a soil deposit that has been subjected to vertical effective stress greater than the present vertical effective stress.

Q: how to determine pre-consolidation pressure ?

Determination of Preconsolidation Pressure (Yield Stress)

Step 1. Conduct an oedometer test on the undisturbed soil sample obtained from the field.

Step 2. Plot e - log σ´ plot as shown. The equilibrium void ratio at the end of each of the pressure increments are used in obtaining e - log σ´ plot.

Step 3. Select the point of maximum curvature (Point A) on the e - log σ´ curve 

Step 4. Draw a tangent at the point of maximum curvature (Point A) 

Step 5. Draw a horizontal line AC 

Step 6. Draw the bisector line AD between the tangent and horizontal line 

Step 7. Extend the normally consolidated line to intersect the bisector line at ‘O’ 

Step 8. The vertical effective stress corresponding to point of intersection (O) is the preconsolidation pressure (σ´pc)

Q: Explain compressibility characteristics mv, av,Cc

Compressibility Characteristics

The compressibility of soils under one-dimensional compression can be described from the decrease in the volume of voids with the increase of effective stress. This relation of void ratio and effective stress can be depicted either as an arithmetic plot or a semi-log plot.

It can be said that the compressibility of a soil decreases as the effective stress increases. This can be represented by the slope of the void ratio – effective stress relation, which is called the coefficient of compressibility, av.

av= e1- e22'- 1'

Co-efficient of volume compressibility (mv) It is the ratio of change in volume of a soil per unit initial volume due to unit increase in effective stress and is given by mv= av1+ eo

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Coefficient of compression/compression index (Cc):It is the slope of the normal consolidation line in a plot of void ratio-logarithm of effective stress (e - logσ´). It is given by

Empirical correlations 

Cc = 0.009 (LL-10) Undisturbed clays 

Cc = 0.007 (LL-10) Remoulded soil sample 

Cc = 1.15 (e0-0.30) Upper bound values 

Cc = 0.30 (e0-0.27) Lower bound values 

The value of Cc is constant for a given soil. The compression index is used to determine primary consolidation settlement of normally consolidated soils. A high value of Cc indicates high compressibility and higher consolidation settlement.

Formulae in Consolidation chapter:

  • Compression index 

  • It is a slope of pressure-voids ratio curve when plotted on logarithmic graph

  • cc= e0- e'0'

  • Coefficient of compressibility

  • It is a slope of pressure-voids ratio curve when plotted on natural graph

  •  av= e0- e'-  0'

  • Coefficient of volume change

  • mv= av1  +  e0

  • Where, e0 is initial voids ratios when pressure = 0', e is voids ratio when pressure change to '

  • Consolidation settlement

  • f= mv H ∆'

  • H = soil thickness, ∆'= pressure increment

  • Coefficient of Consolidation 

  • cv=kmv w 

  • Where, k is coefficient of permeability

  • From consolidation test data

    • cv=Tv t d2 

    • Tv= 4 U1002….when degree of consolidation =U < 60%

    • Tv= -0.9332 1-U100- 0.0851 ….when degree of consolidation =U > 60%

    • Where, Tv is time factor,  t is time required for consolidation, d is drainage path

    • d = soil thickness when soil has single drainage

    • d = (soil thickness/2) when soil has double drainage

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