Lecture-05 Consolidation-Rate of Settlement-II.pptx

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1. 1 CE-313 (2 Credit Hours) Geotechnical Engineering-II Time Rate of Consolidation Instructor: Dr Irshad Ahmad Lecture-05 Department of Civil Engineering University of Engineering and Technology, Peshawar

2. 2 Theme of this lecture Terzaghi's theory of 1D consolidation Derivation of differential equation relating u, z, t Solution of differential equation Determination of c v Log time method (Casagrande Method) Root-time method (Taylor method) Isochrones U and T relationship Examples Contents

3. 3 Terzaghi’s Theory of 1D consolidation u e =f( z, t)

4. 4 Terzaghi’s Theory of 1D consolidation

5. Consider an element having dimensions dx , dy & dz within a clay layer of thickness 2d, as shown in below figure. An increment of total vertical stress ∆  is applied to the element. The flow velocity through the element is given by Darcy’s law as v z = k  i z = -k ( h/ z) -ve sign shows decrease in total head in the direction of v z Since any change in total head (h) is due only to a change in pore pressure: u e =  w h  h = u e /  w v z = (k/ w ) (u e / z) 5 Terzaghi’s Theory of 1D consolidation

6. 6 [ v z +(  v z /z)dz] dy dx = q in dx dz q out – q in = dV/dt [v z +( v z /z)dz] dx dy - (v z ) dy dx = dV/dt ( v z /z)dz dx dy = dV/dt We know v z = (k/ w ) (u e / z) Terzaghi’s Theory of 1D consolidation dx dy = q out (v z ) The volume of water entering the element per unit time is = v z dx dy = q in The volume of water leaving the element per unit time is = [ v z + (  v z / z)dz]dx dy = q out If, however, the volume of the element is undergoing change, the equation of continuity becomes (  v z / z) dx dy dz = dV / dz Here dV /dt is the volume change per unit time. A= dy dx

7. 7 The rate of volume change can be expressed in term m v ; m v = V/(V o ) V = m v V o  = m v (dx dy dz )  V /t = m v (dx dy dz )  /t dV /dt = m v (dx dy dz ) (/t) When t 0, The total stress increment is gradually transferred to the soil skeleton, increasing effective stress, as the excess pore water pressure decreases. Hence the rate of volume change can be expressed as; (/t) = - ( u e /t) dV /dt = - mv (u e /t) dx dy dz As we know from previous slide Or Terzaghi’s Theory of 1D consolidation c v = k/(  w m v ) c v is being defined as the coefficient of consolidation, suitable unit being m 2 /year. Since k and m v are assumed as constants, c v is constant during consolidation.

8. 8 Solution of 1D Consolidation Equation The differential of the Terzaghi consolidation theory is a parabolic equation and can be solved using several different methods. Incidentally, this differential equation is identical in form to the equation governing the diffusion process, flow of electricity and dissipation of heat. u e =f( z,t ) u e is differentiated twice w.r.t. z and once w.r.t. t. therefore for a solution of the differential equation, two conditions wrt to z (boundary), and one condition wrt t (initial).

9. 9 The initial Condition Sand (Free draining boundary) Sand ( Free draining boundary) Clay 2d z z u e @ t=0 u e = u i u i

10. 10 Two boundary conditions at z=0, u e = 0 when t>0 Sand (Free draining boundary) Sand ( Free draining boundary) Clay 2d z at z=2d, u e = 0 when t>0

11. 11 Solution of Consolidation Equation for constant u i throughout the depth

12. 12 Isochrones

13. 13 Isochrones

14. 14 U and T relationship Approximate solution Solution The Relationship between U and T v is represented by curve 1 as shown in the next slide

15. 15 U and T relationship

16. 16 Example-01 A 12m thick layer of Chicago clay is doubly drained. (This means that a very pervious layer compared to the clay exists on top of and under the 12 m clay layer.) The coefficient of consolidation c v = 8.0  10 -8 m 2 /s. Required: Find the degree or percent consolidation for the clay 5 yr after loading at depths of 3, 6, 9, and 12 m. Free draining boundary 2d 0 m Clay c v = 8.0 x 10 -8 m 2 /s z 3 m 6 m 9 m 12 m Free draining boundary

17. 17 Example-01 sol. Free draining boundary 2d 0 m Clay c v = 8.0 x 10 -8 m 2 /s z 3 m 6 m 9 m 12 m Free draining boundary T v  

18. 18 Example-01 sol. cont... At z=3m, z/d =3/6= 0.50, Uz = ? At z=6m, z/d =6/6= 1.00, Uz = ? U z =0.61 U z = 0.46 61% 46% T=0.35 T=0.35

19. 19 Example-01 sol. cont... At z=3m, z/d = 3/6 = 0.50, U z = 61% At z=6m, z/d = 6/6 = 1.00, U z = 46% At z=9m, z/d = 9/6 = 1.50, U z = 61% At z=12m, z/d = 12/6 =2.00, U z = 100%

20. 20 m= 0,1,2,3,4,…….,infinity For m = 0 M = /2 u e = u e-1 = 53.68557139 Uz = 100(1-ue/ui)= 46.31442861 For m=1  M=3/2 u e-2 = 53.68557139+ u e-2 = 53.66769351 Uz =100(1-ue/ui)= 46.33230649 And so on… Using series 0.35 6 6

21. 21 Example 02 Free draining boundary 2d 0 m Clay z 3 m 6 m 9 m 12 m Free draining boundary  = 100

22. 22 Example 02 sol… Solution:   at z= 3m, u e = ? u e = 100 (1 – 0.61) = 39 kPa u e = u i (1 - U z ) From the solution in Example 9.1 we obtain: At z = 3 m U z = 61% u e = 39 kPa At z = 6 m U z = 46% u e = 54 kPa At z = 9 m U z = 61% u e = 39 kPa At z = 12 m U z = 100% u e = 0 kPa Note that they are excess pore pressures, that is, they are above the hydrostatic water pressure u i =  = 100 kPa U z = 61% u e = 100 (1 - U z )

23. 23 Example 03 Use data of Example -01, what will be the average degree of consolidation after 5 years. Solution: Using Approximate solution U = 0.667 > 0.6, so our assumption is not true for T v = 0.35, U = ? Assuming, U < 0.6, 0.35 U 2   Assuming, U > 0.6 T v = -0.933 log(1-U) – 0.085 0.35 = -0.933 log(1-U) – 0.085 U = 0.66 > 0.6 Ok Using another solution

24. 24 Series Method m= 0,1,2,3,4,…….,infinity For m = 0 M = /2 U = 65.8564188 For m=1 M = 3 /2 U = 65.8526585 For m=2 M = 5/2 U = 65.8526585 0.35