Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay

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2 Module 3: Lecture - 4 on Compressibility and Consolidation

3 Contents Stresses in soil from surface loads; Terzaghi s 1-D consolidation theory; pplication in different boundary conditions; Ramp loading; Determination of Coefficient of consolidation; Normally and Over-consolidated soils; Compression curves; Secondary consolidation; Radial consolidation; Settlement of compressible soil layers and Methods for accelerating consolidation settlements.

4 Terzaghi s 1-D consolidation theory Where: Where: Basic DE of Terzaghi s 1D consolidation theory and can be solved with proper boundary conditions.

5 Terzaghi s 1-D consolidation theory To solve the equation, we assume u to be the product of two functions, i.e., the product of a function of z and a function of t, or --(3) The constants in Eq. (3) can be evaluated from the boundary conditions, which are as follows:

6 Terzaghi s 1-D consolidation theory Note that H is the length of the longest drainage path. In this case, which is a two-way drainage condition (top and bottom of the clay layer), H is equal to half the total thickness of the clay layer, H t

7 Terzaghi s 1-D consolidation theory From the above, a general solution can be given in the form ---(4) To satisfy the first boundary condition, we must have the coefficients of n such that ---(5)

8 Terzaghi s 1-D consolidation theory Equation (4) is a Fourier sine series, and n can be given by ---(6) Combining Eqs. (4) and (6), So far, no assumptions have been made regarding the variation of u i with the depth of the clay layer.

9 Terzaghi s 1-D consolidation theory If u i is constant with depth i.e., if u i = u 0 z u 0 H t = 2H So, u 0 H t = H Note that the term 1 cosnπ in the above equation is zero for cases when n is even; therefore u is also zero. For the nonzero terms it is convenient to substitute n = 2m+1, where m is an integer.

10 Terzaghi s 1-D consolidation theory t a given time the degree of consolidation at any depth z is defined as u = u i (1 - U z ) σ = Increase of effective stress at a depth z due to consolidation

11 Terzaghi s 1-D consolidation theory In most cases, however, we need to obtain the average degree of consolidation for the entire layer. This is given by:

12 Variation of U z with z/h and T v The variation of U z with depth for various values of the nondimensional time factor T V ; these curves are called isochrones. The diagrams which represent the successive stages of a process of consolidation by means of isochrones, will briefly be called piezographs.

13 Example problem Consider the case of an initial excess hydrostatic pore water that is constant with depth, i.e., u i = u 0. For T v = 0.3, determine the degree of consolidation at a depth H/3 measured from the top of the layer. Here z = H/3, or z/h = 1/3, and M = (2m+1)π/2

14 Example problem For m = 0, 1 and 2 Here z = H/3, or z/h = 1/3, and M = (2m+1)π/2 m > 3 is negligible; Hence not required. Using the value of calculated in step 9(above), the degree of consolidation at depth H/3 is

15 Variation of U z with z/h and T v For z/h = 1/3 and T V = 0.3 U z = 0.69 u = u i (1 - U z ) = u i ( ) =0.31 u i

16 Terzaghi s 1-D consolidation theory The average degree of consolidation is also the ratio of consolidation settlement at any time to maximum consolidation settlement. Note, in this case, that H t = 2H and u i = u 0. Combining the following equations:

17 Terzaghi s 1-D consolidation theory Terzaghi suggested the following equations for U av approximate the values from to

18 Log T v vs U or U av. log T V Tangent U (%) Equation of a parabola For open layer top and bottom Infinitely long consolidation symptote

19 For values of U% greater than 52.6 the Log T v vs U curve is almost identical with a curve with the equation It should be noted that the radius of curvature of the curve U (%) vs log (T V ) increases steadily until U% becomes approximately equal to 50, then decreases once more and assumes a second minimum at about U% = 85. The curve thus obtained has a point of inflection at about U% = 75. In the vicinity of U% = 95 it flattens rapidly and approaches horizontal asymptote corresponding to U% = 100. On overall, the curve represents an equation a

20 Substituting for T v = tc v /H 2 If the degree of consolidation of two beds of clay with different values of cv/h 2 is plotted against the logarithm of time, the time-consolidation curves thus obtained have the same shape but they are separated from each other by a horizontal distance log lo (c v /H 2 ). For T v /t = 1 the time-consolidation curve becomes identical with the time factor-consolidation curve. This important property of the semi-logarithmic time-consolidation graph facilitates comparison of empirical consolidation curves with the theoretical standard curve for the purpose of detecting deviations of the real process from the theoretical one. Therefore in many cases the semilogarithmic plot is preferable to the arithmetic plot.

21 Example γ = 19.5 kn/m 3 γ = 18.7 kn/m 3 γ = 19 kn/m 3 γ = 16.5 kn/m 3 5 m 1 m Silty sand 2 m 4.8 m 12 m Clay Proposed fill Before Fill: σ = ( 18.7)( 1) + ( 19)( 2) + ( 16.5)( 4.8) u = ( 9.81)( 6.8) = 67kPa σ ' = σ u = 69kPa Short Term fter Fill: σ = ( 19.5)( 5) = 234kPa u = u + u = u + σ = 67 + ( 19.5)( 5) σ ' 0 = σ u 0 = 69kPa Long Term fter Fill: σ = ( 19.5)( 5) = 234 u σ ' = u 0 = σ = 67kPa u = 167kPa kpa = 136kPa = 165kPa

22 Before Fill: Example σ = ( 18.7)( 1) + ( 19)( 2) + ( 16.5)( 4.8) = u = ( 9.81)( 6.8) = 67kPa σ ' = σ u = 69kPa Short Term fter Fill: σ = ( 19.5)( 5) = 234kPa u = u + u = u + σ = 67 + ( 19.5)( 5) σ ' 0 = σ u 0 = 69kPa Long Term fter Fill: σ = ( 19.5)( 5) = 234 u σ ' = u 0 = σ = 67kPa u = 167kPa kpa 136kPa = 165kPa

23 Proposed fill σ = 100 kn/m 2 Example PWP in clay after 5 years γ = 17.2 kn/m m u (kn/m 2 ) γ = 18 kn/m 3 γ = 19 kn/m 3 Clay 12 m 3 m Sand e 0 = 0.62 w 0 = 23.2% Soft clay G s = 2.66 C v = 8x 10-8 m 2 /s 0 m t = 0 years Dense Sand 12 m

24 Example Using T v = t c v /(H dr2 ) For t = 5 years C v = 8x 10-8 m 2 /s = 8 x10-8 x m 2 /year H dr = 2H = 12/2 m (Double drainage) T v = z = 0 m z/h = 0 U z = 100 % z = 3 m z/h = 3/6 = 0.5 U z = 61 % z = 6 m z/h = 6/6 = 1.0 z = 9 m z/h = 9/6 = 1.5 U z = 46 % U z = 61 % z = 12 m z/h = 12/6 = 2 U z = 100 %

25 Using Variation of U z with z/h and T v u = u i (1 - U z z = 0m u = 0 kpa z z = 0 m for top of z = 3 m u = 100(1-0.61) = 39 z = 6 m u = 100 (1-0.46) = 54 z = 9 m u = 39 z = 12 m u = 0 kpa z = H t =2H m for bottom of clay U z = 0.46

26 Proposed fill σ = 100 kn/m 2 Example PWP in clay after 5 years γ = 17.2 kn/m m u (kn/m 2 ) γ = 18 kn/m 3 γ = 19 kn/m 3 Clay 12 m 3 m Sand e 0 = 0.62 w 0 = 23.2% Soft clay G s = 2.66 C v = 8x 10-8 m 2 /s 0 m t = 0 years Dense Sand 12 m 150 EPWP yet to be dissipated after 5 years 250

27 Degree of consolidation under time-dependent Loading (Ramp loading) Olson (1977) presented a mathematical solution for onedimensional consolidation due to a single ramp load.

28 Ramp loading The expression for the excess pore water pressure for the case where u i = u 0 is given by: s stated above, the applied load is a function of time: q = f (t a ) where t a is the time of application of any load. For a differential load dq applied at time t a, the instantaneous pore pressure increase will be du i = dq. t time t the remaining excess pore water pressure du at a depth z can be given by the expression: --- (a)

29 Ramp loading The average degree of consolidation can be defined as: --- (b) where αq c is the total load per unit area applied at the time of the analysis. The settlement at time t = is, of course, the ultimate settlement. Note that the term q c in the denominator = instantaneous EPWP u i = q c that might have been generated throughout the clay layer had the stress q c been applied instantaneously.

30 Ramp loading Proper integration of Eqs. (a) and (b) gives the following:

31 Ramp loading Plot of against time factor for single ramp load (fter Olson, 1977)

32 Example Based on one-dimensional consolidation test results on a clay, the coefficient of consolidation for a given pressure range was obtained as mm2/s. In the field there is a 2-m-thick layer of the same clay with two-way drainage. Based on the assumption that a uniform surcharge of 70kN/m 2 was to be applied instantaneously, the total consolidation settlement was estimated to be 150 mm. However, during the construction, the loading was gradual; the resulting surcharge can be approximated as: Estimate the settlement at t = 30 and 120 days after beginning of construction (Given t c = 60 days)

33 Solution:

34 Ramp loading U av ~27% Plot of against time factor for single ramp load (fter Olson, 1977)

35 Solution:

36 Limitations of 1D consolidation In the derviation of 1D equation the permeability (K z ) and coefficient of volume compressibility (m v ) are assumed constant, but as consolidation progresses void spaces decrease and this results in decrease of permeability and therefore permeability is not constant. The coefficient of volume compressibility also changes with stress level. Therefore C v is not constant. The flow is assumed to be 1D but in reality flow is three dimensional The application of external load is assumed to produce excess pore water pressure over the entire soil stratum but in some cases the excess pore water pressure does not develop over the entire clay stratum.

Time Rate of Consolidation Settlement

Time Rate of Consolidation Settlement We know how to evaluate total settlement of primary consolidation S c which will take place in a certain clay layer. However this settlement usually takes place over