Universitat Politècnica de Catalunya BARCELONATECH Escola Tècnica Superior d Enginyers de Camins, Canals i Ports. Soil Mechanics.

Transcription

1 Universitat Politècnica de Catalunya BARCELONATECH Escola Tècnica Superior d Enginyers de Camins, Canals i Ports Soil Mechanics Chapter 8 Consolidation

2 Chapter 6 1. One-dimensional consolidation theory. Consolidation with radial flow Exercises

3 6.1 One-dimensional consolidation theory

4 Introduction: coupling between the hydraulic and mechanical problems

5 Introduction We have seen that to study failure in soils: if it is short term work with total stress because we don t know how to calculate p w if it is long term work with effective stress because we know how to calculate p w : either it is the hydrostatic pressure or it can be obtained from the flow conditions steady state in both cases (no time-dependent) that is why we have not dealt directly with calculating p w in the failure analysis

6 Equations so far: (a) Equilibrium equations total stress: effective stress: p b b x x x ij ij w i i j j i b (0,0, ) nat (b) Compatibility equations (c) Constitutive law (d) Boundary conditions water pressure: hydrostatic flow

7 Porewater pressure To understand what happens with the porewater pressure, we need more equations Add the continuity (mass conservation) equation and Darcy s law (chapter ): Fn Conservation: div q 0 Darcy: t q K grad h Assume saturated soil; h = piezometric level F

8 Equations: (a) Equilibrium (b) Compatibility (c) Constitutive law (d) Boundary conditions + (e) Continuity of mass of water (f) Darcy s law Let s study in detail equations (e) and (f), combining them, but assuming now that the porosity n is not constant (as was done in chapter )

9 Continuity + Darcy In general, the mass conservation equation, with one phase only, would be: r + div( rv) = 0 ( 1) t Þ r t + v gradr + r div v = 0 Þ dr dt + r div v = 0 Mass conservation equation in spatial form, where v is the absolute velocity (taking a control volume fixed in space)

10 Continuity + Darcy But in soils there are actually two phases It is easy to check that fluid n solid 1 Although ρ fluid and ρ solid may be constant (undeformable water and soil particles), the porosity n is not constant (soil is deformable) It is easier to write the conservation equation separately for each of the phases n

11 Continuity + Darcy Water: Solid: Where: fluid t n fluid fluid solid solid n vsolid t div nv 0 1n div 1 0 v fluid = v solid + v rel v rel is the relative velocity of the fluid with respect to the solid particles q

12 Continuity + Darcy Elimination of ρ fluid and ρ solid (because they are constant in time and space) leads to: n n divnv fluid 0 div n solid rel 0 t t v v 1n div 1n solid 0 t v Adding the two equations results in: div nv div v 0 rel solid

13 Continuity + Darcy On the other hand: v solid : absolute velocity of solid (spatial derivative) u solid : displacement vector of solid particles Therefore: v solid u solid solid vol div vsolid div divusolid t u t t t changing the order of the differential operators definition of volumetric deformation and Soil Mechanics sign criterion

14 Continuity + Darcy Resulting in: div vol nvrel 0 t Darcy s law provides the relationship between nv rel and the piezometric head: nv K grad h h z unit flow vector q defined in chapter, but now we consider also the particle s motion p rel ; w piezometric head (potential) w

15 Continuity + Darcy Finally: that is a general form of the continuity equation that includes the motion of the solid skeleton, which is more strict If K = ct. div K grad vol K h t and if there is no deformation, we recover the flow equation h = 0 (chapter ) h t vol (e) + (f)

16 Summary of equations (1) (a) Equilibrium, written in terms of effective stresses because now we do want to know the porewater pressure: x ij j b (b) Compatibility: in general not imposed a priori, but need to check. Often automatically fulfilled. (c) Constitutive law, written in vector form for simplicity: i p x dσddε w i

17 Summary of equations () (d) Boundary conditions: include also initial conditions because the variable time also appears in the equations. (e) Continuity (f) Darcy s law p w t w vol div K grad z We know that (a), (b) and (c) represent 6 PDE with 6 unknown functions σ x, σ z, τ xz, ε x, ε z, ε xz Now we have one more equation, and one more unknown: the porewater pressure

18 Coupling These equations represent a coupling between a mechanical problem (stress strain) and a hydraulic problem (flow): Mechanical problem (a) + (c): b p ij w i x j x i with Hydraulic problem (e) + (f): w vol div z dσ D dε ( 1) p K grad () w t

19 Coupling Equations (1) and () are coupled: in general they cannot be solved separately Only if ε vol does not change with time () is the classical (chapter ) flow equation and in this case it can be solved independently from (1) This happens, e.g., for long term analysis when the problem is time-independent ( ε vol / t = 0)

20 Coupling Therefore, at long term (what we have called drained conditions), equations (1) and () are uncoupled: x ij j b i p x w i p w (1) div K grad z 0 () w Equation () can be solved directly to obtain p w, and then substitute it in (1) working with effective stresses This is how we have worked so far...

21 Coupling But in any other circumstance (1) and () must be solved simultaneously. If the constitutive model (c) is complicated (plasticity, Cam-clay,...) it is not possible in general to obtain an analytical solution and we must resort to numerical methods (i.e. finite elements) In some very simple cases there are analytical solutions

22 One-dimensional consolidation equation (Terzaghi & Fröhlich, 1936)

23 1-D consolidation theory Assume that the problem is one-dimensional and that the soil is an elastic material there is only vertical deformation lateral deformation is zero This is the case in the oedometer: z uniformly distributed load Δσ WT x z H z

24 1-D consolidation theory The relationship between and z z is the one seen for the oedometric test Because we assume the soil to be elastic, there is no need to write this relationship in incremental form: z E z m 1 1 E1 z

25 1-D consolidation theory In this case, also: ε vol = ε z If K is constant, equation () becomes: K h z t E z p w where: p h z p u w ; pw w,hydrostatic w the hydrostatic pressure is constant with time linear with depth, z excess porewater pressure (over hydrostatic pressure)

26 1-D consolidation theory After derivation, the hydrostatic component of the pressure vanishes from the equations: p p w,hydrostatic w,hydrostatic 0 z And so equation () becomes: K w u 1 t z z Em t u

27 1-D consolidation theory If the external load is constant with time, σ z = ct.: z 0 t KE w m K u 1 u w z Em t u u z t This is the one-dimensional consolidation equation usually attributed to Terzaghi. The general theory is from Biot (1941)

28 1-D consolidation theory Definition: KE coefficient of consolidation: cv w can be obtained from tests with the oedometer combines: permeability (flow): K m deformability: E m c v for clays: between 10 - cm /s and 10-4 cm /s

29 Original references Terzaghi, K. v. (193) Die Berechnung der Durchlässigkeitsziffer des Tones aus dem Verlauf der hydrodynamischen Spannungserscheinungen. Akad. Wiss. Wien. Math-naturw. Klasse 13, No. 3/4, Terzaghi, K. v. (194) Die Theorie der hydrodynamischen Spannungserscheinungen und ihr erdbautechnisches Anwendungsgebiet. Proceedings of the International Congress for Applied Mechanics, Delft. Terzaghi, K. v. (195) Erdbaumechanik auf bodenphysikalischer Grundlage. Leipzig and Vienna: Deuticke. (Principles of soil mechanics) Terzaghi, K. v. & Fröhlich, O.K. (1936) Theorie der Setzung von Tonschichten: eine Einführung in die Analytische Tonmechanik, Leipzig: Deuticke. (Theory of the settlement of clay layers: an introduction to the analytical mechanics of clay) Biot, M.A. (1941) General Theory of Three-Dimensional Consolidation. J. Appl. Physics, 1,

30 1-D consolidation theory The equation c v u z u t is a parabolic PDE with analytical solution After a change of variables, we obtain a dimensionless form of the equation: u z t W Z T H Δσ = external load; H = thickness of layer; τ = time (to be defined)

31 1-D consolidation theory Substituting these equations into the original differential equation: Therefore: u W ; u W z H Z t T W cv H Z T W

32 1-D consolidation theory Defining now results finally in H c W W Z T v where T c v t H T = dimensionless problem time t = real (physical) time

33 Boundary conditions z Δσ z = 0; Z = 0 z = H; Z = 1 z= H Þ q = 0 Þ K j z = 0 j = H - z+ p w = H - z+ p hydrostatic w g w g w j z = é z H - z+ p hydrostatic ù w ê ú ëê ûú + 1 =0 g w g w + u g w u g w z = = é z H - z+ g z ù w ê ú+ 1 u ë û g w z = 1 u g w z j z = 0 Þ u z = 0 WT H z 0 u 0 Z 0W 0 u z H 0 z Initial condition : W Z 1 0 Z t 0 u T 0 W 1 Therefore we must solve : W W Z T with conditions : W Z 0 W 0; Z 1 0 Z T 0W 1

34 Solution Using separation of variables, the following solution is obtained: é 4 W(Z,T) = å exp ê - p n+1 n=0 (n+1)p 4 ëê ( ) f (T ) ù é T ú ( sin n+1 ù ê )p Zú ûú ë ê û ú f (Z)

35 Isochrones Z n1 n1 4 W ( Z, T) exp T sin Z n0 (n 1) 4 Z = 0 porewater pressure over the hydrostatic T = W = 0 T 4 T 3 T T 1 Δσ T = 0 + W = 1 isochrones Z = 1 W = 0 W = 1 Porewater pressure dissipation process in time Remember that u = Δσ W is the excess porewater pressure over the hydrostatic

36 Isochrones 1 Z T W

37 Isochrones in terms of the physical variables (z, u, t) z z = 0 hydrostatic pressure t 4 t 3 t t 1 Δσ t = 0 + u = Δσ isochrones z = H H w

38 Terzaghi s hydraulic analogy C 0 P P ' t u t C t t = 0 P P S u u 0 ' C t P t ut S u u 0 t t t = u t P S 0 t u t t

39 Double-sided drainage Δσ z = 0; Z = 0 WT isochrones z z = H; Z = H Double-sided drainage is equivalent to single-side drainage of a layer of half the thickness: the middle plane of the layer behaves as an impermeable boundary. The same set of formulas are used, but the layer thickness is designated as H

40 Degree of consolidation

41 Degree of consolidation at a point The degree of consolidation at a point is defined as: where: U P ( z, t) v 0 zt, z, e a, v av z 1 e 1 e 1 e at the end a z, t z, t 1 e for t a v = coefficient of compressibility (chapter 3)

42 Degree of consolidation at a point z u Δσ Δσ' u U P z, t z, t u z, t u z, t 1 W( Z, T ) U Z, T 1 W Z, T P

43 Degree of consolidation of a layer The degree of consolidation of a layer is defined as: where: s = st Ut () H s surface settlement at time t final surface settlement ò e ( z, )dz = a v Ds H 0 1+ e 0 s t = H ò e ( z,t )dz = a v ò D s ( z,t )dz = e 0 H = a v ò HéDs - u z,t 0 1+ e ë 0 ( ) ù û dz

44 Degree of consolidation of a layer W area U(T) : degree of consolidation of the layer at time T 1 W 1-W U T U ( t) = s t =1- s ò 0 H u( z,t ) Ds ( ) =1- ò W Z,T 1 0 dz ( )dz Z If the layer drains on both sides and its thickness is H, then the degree of consolidation is U ( T) =1-1 ò 0 W( Z,T)dZ

45 Degree of consolidation of a layer U (T) =1- ò 1 W(Z,T)dZ 0 T = 0 U = 0 T = U =1 (100% of consolidation) Solving the integral results in: U (T) =1-8 é 1 å exp ê - p n+1 p ( n+1) n=0 4 ëê ( ) T ù ú ûú

46 Degree of consolidation of a layer Settlement at time t: c st U( T) s T t H v U(T) is tabulated: T = U = (35.6%) T = U = (76.4%) T = 1.19 U = (95%) consolidation is practically finished

47 Degree of consolidation of a layer It can be demonstrated that U(T) can be approximated as: 8 T 0. : U ( T ) 1 exp T 4 T 0. : U ( T) U 100% T parabola first term of the series 0% T

48 Example 10 m clay WT c v = 10-3 cm /s Clay layer, 10 m thick How long does it take to reach 95% consolidation? H 1000 cm 36 years 9 U 0.95 T 1.19 t T s 3 cv 10 cm s If the bottom boundary is permeable and the layer drains on both sides, H = 10 m H 500 cm 9 years t T 1.19 c 10 3 v cm s It takes a long time: consolidation settlements may last many years

49 Consolidation caused by changes in the hydraulic conditions

50 Examples The differential equation is the same, only boundary conditions change. Two examples: Example 1: decreasing piezometric level at the layer s bottom boundary Example : raising the water table Generalization to changes of water pressure at the layer s top and bottom boundaries

51 Example 1 WT sand clay sand well t initial p w (γ w z) isochrones At the sand layer the changes of water pressure are fast; at the clay layer they are slow At the clay layer settlements occur as pressure dissipates: final p w decrease of p w Δp w < 0 Δσ' = Δσ Δp w > 0 Δe < 0 settlement e Δσ'

52 Example sand clay sand increase of p w final p w t initial p w (γ w z) assume that at this point the initial water head is maintained WT fin WT ini isochrones At the sand layer the changes of water pressure are fast; at the clay layer they are slow At the clay layer swelling occurs as pressure increases: Δp w > 0 Δσ' = Δσ Δp w < 0 Δe > 0 swelling e Δσ'

53 General case linear increment p w1 H steady state conditions z p' w1 p' w p w1 initial conditions p w Differential equation: c v p z w p t Boundary conditions: w z 0 p p w z H p p w w w1 p w Initial conditions: t 0 p p H w w1 pw pw 1 H z Steady state conditions: t pw pw 1 pw pw 1 H z H

54 General case linear increment Change of variable u = p w p w,steady state ; then: z 0 u 0; z H u 0 pw pw 1 t 0 ui pw, ini pw, sst pw 1 H z H Changing to a dimensionless form with c v u u z t z c u p p Z ; T t ; W ; u H H u v w1 w 0 0

55 General case linear increment We obtain: W Z W t T And the solution is: Z 0 W 0; Z W 0 Z( 1) p 0 W ; 1 p w1 w n nz n T W Z, T sin exp n1 1 n 4

56 General case linear increment Calculating the degree of consolidation with this function W(Z,T), we obtain: U (t) = s t s = ò ò H 0 H 0 D dz s t D dz s ( ) D s t = s t - s ini = s total - p w - ( s - p ) total w,ini s t s ini = p w,ini - p w = ( p w,ini - p ) w,fin - u u+ p w,fin D s = p w,ini - p w,fin

57 General case linear increment Therefore: U (t) =1+ ò 0 H ò H udz 0 =1- ( p w,ini - p w,fin )dz U (T) =1-1 ò 0 W(Z,T) dz ò 0 H udz H u 0 p w1 H Area : p p H u w1 w H 0 p w

58 General case linear increment Introducing the values of the obtained W(Z,T) results in a degree of consolidation U(T) equal to the one obtained for the case of a uniform load Δσ Therefore: any problem with a linear increment of porewater pressure results in the same degree of consolidation U(T) There are also tables for non-linear increments of porewater pressure

59 U-T Relations H h u 0 u 0 u 0 u 0 u 0 Case I

60 U-T Relations H h h u 0 u H u 0 u 3 u u 0 sin h u u 4H 0 3 sin h H u u 0 0 h uh 4 H u4( H h) H u 0 u 4 de h 0 a h H de h H a h H Case II Case III Case IV

61 Table of T for several values of U U Case I T Case II T Case III T Case IV T Infinite in all cases

62 Table of U for several values of T T Case I U Case II U Case III U Case IV U T Case I U Case II U Case III U Case IV U

63 6. Consolidation with radial flow

64 Introduction Clay layers are slow to consolidate The preload technique allows improving the ground Problem: all loading/unloading processes are slow Why? The main variable controlling the problem is the maximum distance the water must travel to exit the layer, so that pressure may dissipate:

65 Introduction The time depends essentially on the distance: U 95% 0% 1.19 T c T t t T H v w H K Em t t H To decrease the distance we may introduce vertical drains: Vertical and radial flow (small vertical flow) Vertical deformation vertical settlement

66 Radial consolidation Consolidation equation in cylindrical coordinates: u r + 1 r c vr é u ê r + 1 ë r u r + 1 u r q + u z»0,symmetry radial part u r = u t ù ú = u û t ; c = K E r m vr g w é1 c vr ê ër r æ ç è r u r öù ú = u øû t K r : radial permeability

67 Radial consolidation It can be proved that With radial + vertical flows only vertical deformation U 1 1U 1U rz r z The vertical deformation (settlement) is: st Urz s The vertical part is the classical solution already seen: c K E U z U z Tz ; Tz t ; c vz H vz z m w

68 Radial consolidation r e r w r s r e Drain pattern r w : drain radius r s : radius of remoulded zone, with permeability K s r e : radius of the drain influence zone, with permeability K r

69 Radial consolidation Radial degree of consolidation U r : m U r 1 exp T T r n 3n 1 K s1 n 1 n n K n r ln n 1 4 s n r r e w c t r vr e ; s r r r s w m

70 6.3 Time-dependent load

71 Time-dependent load 0 T 0 T 0 T u c 0 t v 0 integration variable H has dimensions (kpa) 4 1 n Z T n T 0 d u Z, T sin exp n T 4 exp dt n1,3,5,... n 4 dt0