Settlement: Consolidation Of Soil

COMPRESSIBILITY OF SOIL

TERMS AND DEFINITIONS  Settlement  Total vertical deformation at soil surface resulting from the load  Consolidation (volume change velocity)  Rate of decrease in volume with respect to time  Compressibility (volume change flexibility)  Volume decrease due to a unit load

TERMS AND DEFINITIONS  Contraction (temperature expansion)  Change in volume of soil due to a change in temperature

 Swelling  Volume expansion of soil due to increase in water content  Shrinkage  Volume contraction of soil due to reduction in water content

Causes of compression:  Deformation of soil particles  Relocations of soil particles  Expulsion of water or air from the void spaces   

When stressed, soil deforms Stressed released, deformation remains Soil deformation:  Distortion (change in shape)  Compression (change in volume)  Both

Components of Settlement:  Immediate Settlement, Si  caused by the elastic deformation of dry, moist, and saturated soils, without any change in moisture content  Primary Consolidation Settlement, Sc  caused by a volume change in saturated cohesive soils due to expulsion of water that occupies the void spaces  Secondary Consolidation Settlement, Ss  Caused by plastic adjustment of soil fabrics. It is an additional form of compression that occurs at constant effective stress

Total Settlement, St

 St = Si + Sc + Ss

 Soil Movement:  Downward: load increase or lowering water table  Upward: temporary or permanent excavation

 Points of Interest:  HOW MUCH SETTLEMENT OCCURS? o depends on the rigidity of soil skeleton o compression of sand occurs instantly o consolidation of cohesive soil is time dependent  HOW FAST SETTLEMENT OCCURS? o depends on permeability of soil

Compressibility of Soil  Assumption in settlement:  100 % saturated and 1D (vertical) soil deformation  When soil is loaded it will compress because of:  Deformation of soil grains (small ~ negligible)  Compression of air and water in the voids  Squeezing out of water and air from the voids  Compressible soil mostly found below the water table ~ considered fully saturated  As pore fluid squeezed out:  Soil grain rearrange themselves ~ stable/denser configuration  Decrease in volume ~ surface settlement resulted

CONSOLIDATION OF CLAY  At equilibrium: time, t=0 

     

System is analog to soil layer at equilibrium with weight of all soil layer (overburden) above it. Valve is closed. Piston is loaded, compresses a spring in chamber. Hydrostatic pressure = uo

Spring ~ soil skeleton Water ~ water in pores Valve ~ pore sizes in soil/permeability

 Under load Δp (0
   

uo+Δu



   

Spring ~ soil skeleton Water ~ water in pores Valve ~ pore sizes in soil/ permeability



Soil is loaded by increment Δp. Valve initially closed. Pressure(Δp) is transferred to the water. As water is incompressible and valve still closed, no water is out, no deformation of piston. Pressure gauge read: Δu = Δp where Δu is excess hydrostatic pressure. To simulate a fine-grained cohesive soil, where permeability is low, valve can be opened. Water slowly leave chamber.

 At equilibrium (t=∞)



po+Δp

  

  

Spring ~ soil skeleton Water ~ water in pores Valve ~ pore sizes in soil/ permeability

 

To simulate a fine-grained cohesive soil, where permeability is low, valve can be opened. Water slowly leave chamber. As water flows out, load (Δp) is transferred to the spring. At equilibrium, no further water squeezed out, pore water pressure back to its hydrostatic condition. Spring is in equilibrium with load po + Δp. Settlement “s” exist.

SETTLEMENT PROCESS:  Initially all external load is transferred into excess pore water (excess hydrostatic pressure)  No change in the effective stress in the soil  Gradually, as water squeezed out under pressure gradient, the soil skeleton compress, take up the load, and the effective stress increase.  Eventually, excess hydrostatic pressure becomes zero and the pore water pressure is the same as hydrostatics pressure prior to loading.

NORMALLY CONSOLIDATED AND OVERCONSOLIDATED CLAYS  When soil is loaded to a stress level greater than it ever experienced in the past, the soil structure is no longer able to sustain the increased load, and start to breakdown.

 Preconsolidation Pressure, Pc:  Maximum pressure experienced by soil in the past  Normally Consolidated: OCR = 1  When the preconsolidation pressure is equal to the existing effective vertical overburden pressure, Pc = P’o  Present effective overburden pressure is the maximum pressure that soil has been subjected in the past

 Preconsolidation Pressure, Pc:  Maximum pressure experienced by soil in the past  Overconsolidated: OCR > 1  When the preconsolidation pressure is greater than the existing effective vertical overburden pressure Pc > P’o  Present effective overburden pressure is less than that which the soil has been subjected in the past

 Preconsolidation Pressure, Pc:  Maximum pressure experienced by soil in the past  Underconsolidated: OCR < 1  When the preconsolidation pressure is less than the existing effective vertical overburden pressure Pc < P’o  e.g. recently deposited soil geologically or manually.

Mechanism causing preconsolidation:  Change in total stress due to:  Removal of overburden  Past structures  Glaciation  Change in pore water pressure:  Change in water table elevation  Artesian pressure  Deep pumping, flow into tunnel  Desiccation due to surface drying and plant life  Environmental changes such as pH, temperature, and salt concentration  Chemical alteration due to weathering, precipitation, cementing agents, ion exchange

Consolidation Test Data Plots

How to determine Pc?

SETTLEMENT CALCULATION NORMALLY CONSOLIDATED CLAY

SETTLEMENT CALCULATION

Example  From the given soil profile shown, the ground surface is subjected to a uniform increase in vertical pressure of 12 N/cm2.  Compute the buoyant unit weight of clay.  Compute the overburden pressure Po of mid-height of the compressible clay layer.  Compute the total settlement due to primary consolidation. ΔP = 12 N/cm2 4.6 m

Sand

5.86 m

y’ = 10.4 kN/m3 Clay

7.6 m

ydry = 17.6 kN/m3

LL = 45 w = 40 % ys = 27.8 kN/m3

Example  From the given soil profile shown, given B = 1.5 m, and L = 2.5 m. The footing carries a load of 120 kN.  Compute the average effective pressure at mid-height of clay layer.  Compute the average increase of effective pressure in the clay layer using 2:1 method.  Compute the primary consolidation settlement of the foundation. 1.5 m

Sand

1.5 m

ysat = 18 kN/m3 Clay

2.5 m

ysat = 15 kN/m3

LL = 38 w = 35 % Gs = 2.7

Example  Prior to placement of a fill covering a large area at a site, the thickness of a compressible soil layer was 10 m. Its original in situ void ratio was 1.0. Some time after the fill was constructed, measurements indicated that the average void ratio was 0.8. Estimate the settlement of the soil layer.

Example

Example  The laboratory consolidation data for an undisturbed clay specimen are as follows:  e1 = 1.12 P1 = 90 kPa  e2 = 0.90 P2 = 460 kPa  Compute the compression index.  Find the void ratio for a pressure of 600 kPa.  Determine the coefficient of compressibility.

Example  In a laboratory consolidation test on a clay sample the following results were obtained:  e1 = 0.92 P1 = 50 kPa  e2 = 0.78 P2 = 120 kPa  Thickness of the sample of clay = 25 mm  Time for 50 % consolidation = 2.5 min  Tv = 0.197  Find the coefficient of volume compressibility  Determine the coefficient of consolidation if sample of clay was drained on both sides.  Compute the hydraulic conductivity of the clay.