Compressibility & Consolidation

CHAPTER Compressibility & Consolidation Settlement If a structure is placed on soil surface, then the soil will undergo an elastic and plastic deformation. In engineering practice, the deformation or reduction in the soil volume is seen as settlement or heave depending on either the load is increased or decreased. Components of Total Settlement S = S i + S c + S s S i = Immediate Settlement S c = Consolidation Settlement S s = Secondary compression

Components of Total Settlement Immediate settlement (Elastic settlement) This compression is usually taken as occurring immediately after application of the load. Primary consolidation The process of compression due to extrusion of water from the voids in a soil as a result of increased loading. Secondary consolidation Soil compression and additional associated settlement continue at a very slow rate, the result of plastic readjustment of soil grains. Compression of soil The compression of soil is due to Elastic Deformation Deformation of soil grains (small and negligible) Deformation of solid particles secondary compression Compression of air and water in the voids (water is incompressible) Squeezing out of water and air from the voids Expulsion of air from pores Compaction Dissipation of pore-water Consolidation Immediate Settlement Cohesive Soil If saturated clay is loaded rapidly, the soil will deform with virtually no volume change due to without any dissipation of water from the soil..

Immediate/Elastic Settlement General S = i q o B ( ν )α E Immediate/Elastic Settlement A rigid foundation of size 3m x 3m carrying a uniform load of 800 kn is lying on a compressible (medium soft) clay layer of infinite thickness. The undrained elastic modulus of clay (Eu) is 40Mpa. Poisson ratio of the clay is 0.5.Determine the average immediate settlement under this foundation. Immediate/Elastic Settlement q o B Si = ( ν )Cs E (800/(3x3) (3) Si = ( 0.5 )(0.8) 40000 S i =0.009m 3

Immediate Settlement of Footing on Sand S = C C (q q) i q C = 0.5 qo q o B Iz Δz E 0 s Consolidation Settlement Non-linear & irreversible Time dependent CONSOLIDATION SETTLEMENT When saturated soil is loaded Load is taken by the water water dissipate Load is taken by the soil skeleton Deformation 4

Oedometer Test Oedometer Test Load Porous stone Soil Specimen Porous stone Water Confining ring Ref: BS 377 (Par 5) Oedometer Test Soil sample is placed in a metal ring (Figure 4). Purpose of the disks are to allow water to flow vertically into and out of the soil sample. Pressure is applied to the soil sample and dial readings (deformation) and corresponding time observation are made. Normally, this is done over a 4-hour period. Then graph a graph is prepared using these data with tie on logarithmic scale (Figure 5). The procedure is repeated after the sample reached 00% of consolidation. The applied load then will be double. For each graph, the void ratio (e) and coefficient of consolidation (c v ) that correspond to the specific applied pressure (p) are determined. 5

Assumptions Saturated clay layer Laterally confined One dimensional flow Constant permeability Can be measured by oedometer Test DATA ANALYSIS Determination of C v Determination of C v Coefficient of consolidation (cv) indicates how rapidly (or slowly) the process of consolidation takes place. Log time method Cassagrande Square root time method- Taylor 6

Determination of Cv, Log time method Most of the time it is difficult to get this straight line. Plot a graph relating dial reading (mm) versus log time. Produce a straight line for primary consolidation and secondary consolidation part of the graph. The two lines will meet at point C. 3. The ordinate of point C is D 00 = the deformation corresponds to U = 00% 4. Choose time t (point A), t = 4t (point B), t 3 = 4t etc. The difference in the dial reading is equal to x. 5. An equal distance x set off above point A fixes the point D 0 = the deformation corresponds to U =0%. Notes that D 0 is not essentially equal to the initial reading may be due to small compression of air within the sample. 6. The compression between D 0 and D 00 is called the primary consolidation. 7. A point corresponding to U = 50% can be located midway between D 0 and D 00. The value of T corresponds to U = 50% is 0.96. 8. Thus where H d = half the thickness of the specimen 0.96 Hd Cv = t 50 Determination of c v, Square root of time method 7

. Extent the straight line part of the curve to intersect the ordinate (t = 0) at point D. The point shows the initial reading (D o ). The intersection of this line with the abscissa is P.. Take point Q such that OQ =.5 OP. 3. The intersection of line DQ and the curve is called point G 4. Draw horizontal line from G to the ordinate (D 90 ). The point shows the value of t 90. The value of T corresponds to U = 90% is 0.848. 5. Thus 0.848 H d Cv = H d is half the thickness of specimen t90 for a particular pressure increment. How do we get the factor.5 d 90 = F 0.848 d 90 9 d 5 50 9 5 d = 9 5 50 F 0.97 F 0.848 = =.55.5 9 F 0.97 5 SQUARE-ROOT OF TIME VS LOG TIME METHOD Generally square root of time method is better Square root of time methods usually gives lower C v Square root of time method is easier to program in computer k computed from C v almost always less then measured value; slower compression To decide which is correct compare the k value 8

SQUARE-ROOT OF TIME VS LOG TIME METHOD The square root of time method works well based on the assumption of NO secondary consolidation Use strictly for vertical drainage Cannot be used for some soils such as peat, k changes very much when subjects to change in effective stress. Square root of time method is OK because Terzaghi theory does not account for secondary compression anyway Data Analysis Stress Strain curve Results of consolidation testing are presented by plotting: the void ratio (e) at the end of each increment period against the corresponding effective stress (linear scale) or e-p' curve 0.75 0.70 a v e e e = = σ σ ' σ ' void ratio (e) 0.65 0.60 0.55 0.50 0.45 0.40 0 00 400 600 800 000 00 400 600 800 pressure (p) av m v = + e o = + e o eo e σ ' σ ' Problems: The value of m v is not constant, and only true for a certain range of stress. 9

the void ratio (e) at the end of each increment period against the corresponding effective stress (log scale) or e-log p' curve 0.75 0.70 0.65 Cc e e e = = logσ σ log σ void ratio (e) 0.60 0.55 0.50 C r C c 0.45 0.40 0 00 000 0000 pressure (p) Empirical formula & Correlation with other properties Empirical Formulas, correlation of Cc with liquid limit, void ratio, and water content Empirical formula Cc = 0.009 (LL 0) Cc = 0.007 (LL 0) Cc =.5 (e o 0.35) Cc = 0.30 (e o 0.7) Cc =.5 x 0 - ωn Cc = 0.75 (e o 0.50) Soil type All clay (undisturbed) Remolded soil All clay (undisturbed) An organic soil Organic soil Low plasticity soil Based on Critical State model (Wroth & Wood (978) C c = Gs PI 00 Variation of C c (normalized by +e o ) with natural water content (Lambe&Whitman, 979) Correlation between C c and C r Recompression index (C r ) can be estimated as 0.05 0.0 Cc Critical State concept C r = C c (-A) for G s =.7, and A = 0.8 C r = PI/370 0

OVERCONSOLIDATED SOIL Over-consolidated soil is the soil that has undergone recompression in oedometer test. Over-consolidation is judged if pre consolidation pressure σ c obtained from e-log p curve based on oedometer test is greater than the currently existing overburden pressure (σ o ). OR C c obtained from test < C c = 0.009 (LL-0) OR LI is between 0 to 0.6 c σ ' OCR = σ ' The pre-consolidation pressure should be estimated based on e-log p curve o. Choose by eye the point of maximum curvature on the consolidation curve (point A). Draw a horizontal line from point A 3. Draw a line tangent to the curve at point A 4. Bisect the angle made by step and 3 5. Extent the straight line portion of the virgin compression curve up to where it meets the bisector line obtained in step 4. The point of intersection of these two lines is the approximate value of the pre-consolidation pressure (point B)

Over-consolidation is judged if pre consolidation pressure σ c obtained from e-log p curve based on oedometer test is greater than the currently existing overburden pressure (σ o ). σ ' OCR = σ ' c o Insitu e-log p curve Due to effect of sampling and preparation, the specimen in the Oedometer test will be slightly disturbed decrease in the slope of the compression line C c actual > C c measured The field virgin compression line should be estimated based on the e-log p curve Estimation of field consolidation curve In-situ curve Slightly disturbed

Obtaining the field consolidation line Obtaining the field consolidation line. Produce two horizontal lines at e o and 0.4 e o. Produce a vertical line at pressure equal to σ o or σ c. This line should intersect the horizontal line at e o at point. 3. Plot compression line (C c ) based on the oedometer test. The line will intersect the horizontal line at 0.4 e o at point. 4. Draw a line from point to point. The slope of this line is the field compression index (C cf ). Settlement Calculation 3

CONSOLIDATION SETTLEMENT: ONE-DIMENSION H z dz Sc σ eo e σ σ o σ Consider a layer of saturated clay of thickness H, due to construction, the total vertical stress in an elemental layer of thickness dz at depth z is increased by σ Initial stress + stress increment Effective overburden pressure σ o Applied pressure calculated using stress distribution (elastic) σ Settlement of Normally Consolidated Soil H σ' o + Δσ Sc = Cc log + eo σ' o If we use m v, then S c = mv σ H or S c = av/(+e o ) σ H 4

Settlement of Over-consolidated Soil H σ' o + Δσ Sc = Cr log + eo σ' o H σ' c H σ' o + Δσ Sc = Cr log + Cc log + eo σ' o + eo σ' c DATA ANALYSIS Settlement-Time Relationship Rate of Consolidation Terzaghi -D consolidation theory Assumptions:. Compression and flow are -dimensional. The compressible layer is homogenous and saturated 3. Soil particles and water is incompressible 4. Darcy s law is valid 5. Strain due to external load is small and within the range of elasticity 6. The coefficient m v and k remain constant throughout the process 7. There is a unique relationship, independent of time, between void ratio and effective stress 5

uniform load dh h sand water table a u o = σ b d= H/ z dz u z σ' z clay H dx d= H/ c excess pore pressure distribution d Distribution of excess pore water pressure in a clay layer subjected to uniform load PRIMARY CONSOLIDATION The period of consolidation where the volume change of soil is due to the drainage of water driven by excess pore water pressure Basic Equation: where u u Cv = z t Cv = k mvγw General solution to basic consolidation equation is given by Taylor (948) n ( ) = u = σ ' σ' f( Z ) f ( T ) n= 0 where Z and T are non-dimensional parameters. Z is geometry factor, which is equal to z/h, T is Time factor, in which T = Cv t H d = k mv t γ H w d General solution to basic consolidation equation is given by Taylor (948) n ( ) = u = σ ' σ' f( Z ) f( T ) n = 0 where Z and T are non-dimensional parameters. Z is geometry factor, which is equal to z/h, T is Time factor, in which T = C v H t d k t = mvγ H w d 6

Degree of Consolidation (U) The degree of consolidation can be expressed in the form of void ratio or change in effective stress orthe change in p.w.p. e e U z = e e U σ' σ' σ' σ' u u = i z = = = σ' σ' Δσ ui Initially U = 0, U =00% when void ratio is equal to e f. U z = n= 0 u u f (Z) f (T) Relationship between U and Tv is presented in the form of Table, Graph or formula i f Relationship among U, T, and depth Isochrones Single drainage vs Double drainage Pervious Soil layer Drainage, H d = H H Pervious Impervious H H d = H/ drainage drainage Soil layer Pervious 7

Relationship between T v and U Consolidation Percentage 0% 0% 30% 40% 50% 60% 70% 80% 90% 95% 00% U average 0. 0. 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95.0 T v Double drainage 0.008 0.03 0.07 0.6 0.96 0.87 0.403 0.567 0.848.36 α Cassagrande: U = 0% to 60% T v = π/4 (U%/00) U > 60% T v =.78 0.933 log(00%-u%) Relationship between T v and U Time Factor, Tv..0 0.8 0.6 0.4 0. 0.0 0 0 0 30 40 50 60 70 80 90 00 Percent consolidation (%) 8