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Compressibility & Settlement

Faculty of Engineering – Cairo University Third Year Civil Spring 2015

Soil Settlement

 Settlements are vertical deformations of a soil mass under the effect of an applied stress causing vertical movement of the supported structure  Settlements are one of the safety and performance criteria in geotechnical assessment of structures  Excessive settlements can result in structural damage to a building frame, or loss of functionality.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

1 Soil Settlement

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Soil Settlement

 Uniform settlements: equal settlements across the area of the structure  Differential settlements: unequally settlements in different areas of the structure.  Differential settlement can result in severe structural damage while uniform settlements are of less consequences

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

2 Soil Settlement

 Differential settlements

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Soil Settlement

 Differential settlements

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

3 Define: Compressibility

 Compressibility: is the property through which of soil are brought closer to each other, due to escapage of air and/or water from voids under the effect of an applied pressure.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Settlement of Cohesive

Coefficient of compressibility (av): is the rate of change of (e) with respect to the applied effective pressure (p) during compression. e a  v p e e-p curve

eo = initial void ratio eo po = initial e e 1 p e1 = final void ratio

p1 = final effective stress

e = eo –e1 p = p –p p p 1 o o 1 p

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

4 Settlement of Cohesive Soils

Coefficient of volume compressibility (mv): is the volume decrease of a unit volume of soil per unit increase of effective pressure during compression.

ΔV ΔeVs

Vo (1 eo )Vs 1 Δe a v mv     Δp Δp (1 eo ) Δp (1 eo )

eoVs

Initial Condition Final Condition

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Settlement of Cohesive Soils

For a thin layer, p at mid depth is considered as an average stress within the layer. q ΔH V Δe   H V (1 eo ) Δe Δe Δp ΔH  S  H  H (1 e ) ΔP (1 e ) p o o H a S  v ΔpH (1 eo )

S  mvΔpH e H where: eo S = settlement 1+e mv = coefficient of volume compressibility o H p = stresses at mid depth of layer due to 1 added loads H = total thickness of layer

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

5 Settlement of Cohesive Soils

For a thick layer, the layer may be divided to number of

sub-layers, each of thickness Hi. The stress at mid-depth of each sub-layer is pi. q

S   mviΔpiHi

p1 H1 Clay p2 H2

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Settlement of Cohesionless Soils

 Sand may be considered as an elastic material with Young’s modulus (E)  Young’s Modulus (E): is the slope for the linear portion of the stress-strain curve. q strain 1  stress E H 1 H  p E p H S p  Sand H E 1 S  pH E E 100 (v. loose) -800 (v. dense) kg/cm2

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

6 Example

Compute the settlement due to compressibility of the sand and clay layers at points (a) and (b) of the building shown in the figure. The building has a basement and is founded on raft foundation. The stress from the building at the foundation level is as shown in the figure. Sand:  = 1.70 t/m3 E = 500 kg/cm2

2 Clay: mv = 0.03 cm /kg

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Example

2 qnet =q–h = 20 – 1.70x3.0 = 14.9 t/m For sand layer: z = 5.0 m = 1” = 2.5 cm L = 30 m  30x2.5/5.0 = 15.0 cm, B = 20 m  20x2.5/5.0 = 10.0 cm,

2 na = 191 a,sand =na/200 (qnet) = 191/200 x 14.9 = 14.2 t/m 2 nb =50 b,sand =nb/200 (qnet) = 50/200 x 14.9 = 3.7 t/m

a b

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

7 Example

2 qnet =q–h = 20 – 1.70x3.0 = 14.9 t/m For clay layer: z = 11.0 m = 1” = 2.5 cm L = 30 m  30x2.5/11 = 6.8 cm, B = 20 m  20x2.5/11 = 4.5 cm

2 na = 144 a,clay = n/200 (qnet) = 144/200 x 14.9 = 10.7 t/m 2 nb = 47.3 b,clay = n/200 (qnet) = 47.3/200 x 14.9 = 3.5 t/m

a b

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Example

Sa =Ssand +Sclay

Ssand =1/Ea,sandH 2 2 H = 10.0 m E = 500 kg/cm a,sand = 14.2 t/m

Ssand = 1/500 x 14.2/10 x 1000 = 2.84 cm

Sclay =mv a,clayH 2 2 H = 2.0 m mv =0.03cm/kg a,clay = 10.7 t/m

Sclay = 0.03 x 10.7/10 x 200 = 6.42 cm

Sa =Ssand +Sclay = 2.84 + 6.42 = 9.26 cm

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

8 Example

Sb =Ssand +Sclay

Ssand =1/Eb,sandH 2 2 H = 10.0 m E = 500 kg/cm b,sand =3.7t/m

Ssand = 1/500 x 3.7/10 x 1000 = 0.74 cm

Sclay =mv b,clayH 2 2 H = 2.0 m mv =0.03cm/kg b,clay =3.5t/m

Sclay = 0.03 x 3.5/10 x 200 = 2.1 cm

Sb =Ssand +Sclay = 0.74 + 2.1 = 2.84 cm

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Time of settlement compared to construction time P  Loading Diagram

Construction period Time Time  Settlement of Sand 1 S  pH E

S Time

 Settlement of Clay S  mvΔpH

S

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

9 Causes of Settlement

 Static Loads

 Dynamic Loads

 Groundwater table lowering

 Loads from adjacent buildings

 Capillary forces

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Theory of Consolidation

 Consolidation: is the process of squeezing water out of saturated soil under the effect of loads.

 In sandy soils, high permeability, drainage of water out of the soil under the effect of loading happens immediately.

 In clay soils, low permeability, drainage of water out of the soil under the effect of loading is time dependent.

 Therefore, parameters governing consolidation process include: soil properties, stress (p), time (t), drainage conditions.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

10 Mechanism of Consolidation Process

Analogy:

Confined saturated soil Piston for Valve ~ soil ( and voids filled loading permeability with water)

Versus

Container filled with water that has a spring and a valve Water ~ pore-water

Remember: water is Spring ~ Container Solids incompressible

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Mechanism of Consolidation Process

For Sands: high permeability ~ valve is open, water gets out immediately (t = 0), excess u = 0

Stress = p

At t = 0 (immediately), u = 0 Load carried by spring

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

11 Mechanism of Consolidation Process For Clays: low permeability ~ valve is partially open, water gets out slowly (time-dependent), u decreases with t Stress = p Stress = p

At t = 0 (immediately), u = p At t = t1, 0 u < p Spring carry no load Load carried by water and spring

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Mechanism of Consolidation Process

For Clays:

Stress = p Stress = p

At t = t1, 0 u < p At t = long time, u = 0 Load carried by water and spring Load carried by spring

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

12 Mechanism of Consolidation Process

 If we apply a certain load, and the valve is closed, then water (incompressible material) carries the entire added load.  If the valve is opened, water flows out, and thus its volume is reduced.  As volume is reduced, the spring starts to be compressed and carries part of the load. Meanwhile, excess pore water pressure decreases.  The process continues until sufficient amount of water escapes from the valve, resulting in compression of the spring to carry the entire load.  At this point, the excess pore water pressure decreases to zero, the spring carries the entire load, the system is in equilibrium.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Application on Soil

 Stress distribution before application of load

GWT GS

Sand

Clay

Gravel

σt u σ'

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

13 Application on Soil

 Short time (immediately) after application of q (t = 0)

q GWT GS

Sand q

Clay q

Gravel q q

σt u σ'

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Application on Soil

 Long time after application of q (t = h)

q GWT GS

Sand q

Clay

Gravel q q

σt u σ'

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

14 Application on Soil

 Total stress: σt  Σγh  q  Pore water pressure:

. t=0  u  γwhw  q (clay)  u  γwhw (sand)

. t=h  u  γwhw (clay)

 u  γwhw (sand) At time t ???  Effective stress: . t=0  σ' Σγ'h (clay)  σ' Σγ'h  q (sand)

. t=h  σ' Σγ'h  q (clay)  σ' Σγ'h  q (sand)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Rate of Consolidation

q GWT GS

Sand t=0 q

Clay q

Gravel q q

σt u σ'

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

15 Rate of Consolidation

u

t = 0

t = t1

Clay t = t2

t = t3

t = ∞

q

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Assumptions of Theory of Consolidation

1. Clay is homogeneous, isotropic, and saturated

2. Water and clay particles are incompressible

3. Darcy’s law is valid

4. The clay layer is laterally confined

5. One-dimensional compression, and one-dimensional flow

6. Consolidation parameters from test applies to clay layer from which sample for test was taken

7. Soil properties are constant with time

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

16 Consolidation Test

 Objectives: . Volume change-effective pressure relationship . Stress history of soil . Volume change – time – pore water dissipation relationship

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Consolidation Test

Dial gauge Compression Load cell

Loading frame

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

17 Consolidation Test

Dial gauge

Compression cell Lever Arm

Load

Consolidation Test set-up

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Consolidation Test

Dial gauge

Compression cell Lever Arm

Load =

Po

Consolidation Test set-up

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

18 Consolidation Test

Loading Head Dial gauge Saturated Metal ring sample

Porous stone

Filter paper

Compression Cell/Oedometer Porous stone

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Consolidation Test

Loading Head Dial gauge Load (P ) Saturated 1 Metal ring sample

Porous stone

Filter paper

Compression Cell/Oedometer Porous stone

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

19 Consolidation Test

 Equipment:

1. Compression cell / Oedometer

2. Loading frame to apply weights (Po)

3. Dial gage to measure soil compression

Note:

Load acting on the cell (P1) is magnified by the lever arm

ratio of the loading frame (LAR), where: P1 = LAR x Po

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Consolidation Test

 Procedure:

1. Trim the sample into the metal ring and place filter paper on both sides of sample.

2. Measure initial conditions of sample: ho, eo, wo, Gs

3. Place the ring into the compression cell between the two porous stones.

4. A metal cap (loading head) is placed over the top porous

stone, on which the load (P1) is applied.

5. Set-up the dial gage to zero reading. Fill the compression cell with water until the top porous stone is covered with water.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

20 Consolidation Test

 Procedure:

6. Place the hanging load (Po) such that the pressure on the clay sample is equal to the first loading step.

7. Record readings of dial gage (compression of sample) with time until compression stops (usually within 24 hours).

8. Repeat steps 6 and 7 for subsequent loading and unloading increments.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Consolidation Test

 Loading scheme, for example:

 0.25, 0.5, 1.0, 2.0, 4.0, 8.0, 16.0 (kg/cm2) Loading

 16.0, 4.0, 1.0, 0.25 (kg/cm2) Unloading

 Results: for each loading/unloading increment, we record: dial gage (h) that measures sample compression versus elapsed time (t).

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

21 Consolidation Test Data Reduction

 For each loading increment, plot h-log t:

Time (min) 0.1 1 10 100 1000 0

10

20

30

40 mm) -2 50

h (x 10 60 

70

80

90

100

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Consolidation Test Data Reduction

 At the end of each loading increment:

1. Calculate effective pressure (p) = Load x LAR/sample area

2. Measure h = sample compression during this loading increment e h 3. Calculate void ratio (e):

eo e h  1+eo h 1 e h o o o 1

e = eo – e

 Plot e-p curve

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

22 Consolidation Test Data Reduction

2 Example: LAR = 3, sample area = 41.85 cm , ho = 2.54 cm, eo = 0.636

Load (kg) 0 7 14.5 29 58 116 232

Dial Reading x 10-2 (mm) = h 0 89 135 223 359 550 705

Stress (p) = Load x lever arm 7 x 3/41.85 0 1.04 2.08 4.16 8.32 16.63 ratio/area (kg/cm2) = 0.50

0.89/25.4 (1+0.636) e = h/h (1+e ) 0 0.087 0.1436 0.2312 0.3543 0.4541 o o = 0.0573

0.636-0.0573 e = e - e 0.636 0.5490 0.4924 0.4048 0.2817 0.1819 o = 0.5787

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Consolidation Test Data Reduction Plot e-p curve:

0.7

0.6

0.5

0.4 e

0.3

0.2

0.1

0 0 5 10 15 20 p (kg/cm2)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

23 Consolidation Test Data Reduction

 From e-p curve, for a certain stress range, calculate av

and mv, where:

e e a  v p e o e e1 a v p mv  (1 eo )

p p o 1 p

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Consolidation Test Data Reduction Plot e-log(p) curve:

0.7

0.6

0.5

eo 0.4 e 0.3 e

e1 0.2  log(p)

0.1

0 p p 0.1 1o 101 100 p (kg/cm2)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

24 Consolidation Test Data Reduction

 From e-log(p) curve, the slope of the linear portion is the

Compression Index (Cc):

e C  c  log( p)

 Determine preconsolidation pressure (pc): is the largest effective pressure that has been applied on the soil in its geological history.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Consolidation Test Data Reduction Determination of preconsolidation pressure:

0.7 5 0.6 a b 3 1 0.5 4

0.4 2 e 0.3

0.2

0.1

0 0.1 1pc 10 100 p (kg/cm2)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

25 Determination of Preconsolidation Pressure

1. Determine point of maximum curvature (a).

2. At (a), draw tangent to e-log(p) curve.

3. At (a), draw horizontal line.

4. Draw a bisector to the angle enclosed between the tangent and horizontal lines.

5. Extend the linear portion of the curve until it intersects the bisector at point (b).

6. The preconsolidation pressure (pc) is the x-coordinate of the point (b).

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Types of clays w.r.t. preconsolidation

1. Normally consolidated clay: Clay that has never been loaded in the past by more than the

existing effective (po) pc ~po

2. Overconsolidated (preconsolidated) clay: Clay that has been loaded in the past by more than the existing

effective overburden pressure (po) pc > po Causes: pre-existing structures, of overburden

3. Underconsolidated clay: Clay that has been loaded partially by the existing effective

overburden pressure (po) pc < po

Define: Overconsolidation Ratio (OCR): p OCR  c po

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

26 Determination of Compression Index

 For normally consolidated clays, Cc can be calculated as follows:

 From e-log(p) curve:

e C  c  log( p)

 From liquid limit (empirical):

Cc  0.009(wL 10)

where wL is in (%)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Definitions

0.7 Recompression curve 0.6

0.5 Virgin compression 0.4 curve e 0.3

0.2

Unloading/reloading curve 0.1

0 0.1 1pc 10 100 p (kg/cm2)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

27 Definitions

 From e-log(p) curve, the recompression index (Cr) is the slope of the equivalent linear portion of the unloading/reloading curve.

0.7 e C  0.6 r  log( p) Cr 0.5

Empirical relationship for C : 0.4 r e 0.3 Cc Cr 0.2 Cr  (1/10 1/ 5)Cc 0.1 Cr  (1/ 8)Cc 0 0.1 1 10 100 p (kg/cm2)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Settlement

 Calculate settlement (S) of normally consolidated clay layer

of thickness (H) due to added stress (p)given Cc of clay: 0.7 e p 0.6 Cc  eo  log( p) 0.5

e0  e1 0.4 e Cc  e Cc log( p )  log( p ) 0.3 1 0 e1 0.2 e Cc  0.1 p1 0 log( ) p ~p p p 0.1 1c o 101 100 0 p (kg/cmp(kg/cm2)2) p1 e  Cc log( )...... 1 p0 e h S   ...... 2 1 eo ho H

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

28 Settlement

 From 1 and 2:

p1 S Cc log( )  (1 eo ) po H C p S  c H log( 1 ) 1 eo po

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Settlement

 For overconsolidated clay, (po and p1) < pc:

0.7 e C 0.6 o r e1 0.5

0.4 e e 0.3 p

0.2

0.1

0 p p p 0.1o 11 c 10 100 2 p(kg/cmp (kg/cm2) ) C p S  r H log( 1 ) 1 eo po

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

29 Settlement

 For overconsolidated clay, po < pc and and p1 > pc

0.7 e o C 0.6 r

0.5

0.4 e e e1 Cc 0.3

0.2 p

0.1

0 p p p 0.1o 1c 101 100 p (kg/cmp(kg/cm2)2) C p C p S  r H log( c )  c H log( 1 ) 1 eo po 1 eo pc

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Rate of Consolidation

Short time after application of load “q”

q GWT GS

Sand p

Clay p

Gravel p p

σt u σ'

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

30 Rate of Consolidation Sand u u Sand t = 0 z t = 0 u ' 2H t = t1

Clay Clay t = t2 t = t t = t3 Isochrone t = ∞ t = ∞ Sand Hydrostatic pore p p Sand water pressure Excess pore (GWT) water pressure

At time “t” and depth “z”: p = u + ’ = increase in total stress due to added load u = excess pore water pressure due to added load ’ = increase in effective stress due to added load

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Rate of Consolidation u Sand z t = 0 u ' 2H Clay t = t t = ∞

p Sand

Degree of Consolidation (Utz) at time “t” and depth “z”:  ' p  u u U   1 tz p p p

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

31 Rate of Consolidation

Average degree of consolidation (Ut) for the clay layer at time “t”: u Sand H U  2 U dz z t = 0 t  tz u ' 0

2H umax where: Clay  ' p  u U   t = t tz p p t = ∞ p Sand Ut = (area of ’ diagram)/(area of p diagram)

Ut = (area of p diagram – area of u diagram)/(area of p diagram) where: area of Δp diagram  Δp 2H 2 area of u diagram  u 2H 3 max Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Rate of Consolidation

 Settlement of clay layer of thickness 2H at time t=∞:

Sultimate = mvp(2H)

 Settlement of clay layer of thickness 2H at time t=t:

St = mv’(2H)

St = mvp Ut)(2H) = Ut Sultimate

 Average degree of consolidation (Ut) at time “t”:

Ut = St/Sultimate The degree of consolidation at any time (t) is the percentage of settlement that took place at this time w.r.t. to ultimate consolidation settlement.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

32 Differential equation of consolidation

 Equation governing rate of consolidation

 Consider a differential element of soil

Qin Area = dA Volume = dV Surcharge load dy x ground surface y dx z dz

Qout

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Differential equation of consolidation

(dV ) Q  Q  Q  ( )dz compression and flow 1-D t out in z

Q  kidA Darcy’s law is valid

h  k dA z  u  k ( )dA u = excess p.w.p Qin z  w Area = dA Volume = dV k u dy  dA  z w dx x dz y (dV )  k u  ( dA)dz z t z  w z Qout (dV ) k  2u  2 dV Equation (1) soil is homogeneous t  w z

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

33 Differential equation of consolidation

dV e e v  dV  dV dV 1 e v 1 e dVv e dV (dV ) (dV )  v dV 1 1 t t s  dV  dV dV 1 e s 1 e dVs 1

e ( dV ) (dV ) (dV ) (edV ) e e dV  v  1 e  s  dV  t t t t t s t 1 e (dV ) e dV  Equation (2) t t 1 e

From equations 1 and 2,

k 2u e dV 2 dV   w z t 1 e

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Differential equation of consolidation

k(1 e)  2u e 2   w z t e e  '  (chain rule) t  ' t e  ' e  a : a  t v t v  '

e  u u  (load doesn’t vary with time)  av (  )  av :  0 t t t t t u Sand k(1 e) 2u u t = 0 z u 2  av '  w z t 2H u u k(1 e) 2u max  Clay t a  z 2 v w t = t u  2u k(1 e) k t = ∞  c : cv   v 2 a  m  t z v w v w p Sand

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

34 Differential equation of consolidation

 Coefficient of Consolidation: cv k 2 cv  cm /sec mv w

where: k = soil permeability

mv = coefficient of volume compressibility

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Solution of differential equation of consolidation

 Boundary conditions: Example: a clay layer, of thickness 2H, with top and bottom boundaries freely draining Sand

t = t1  u(z=0,t>0) = 0 as time increases, u decreases 2H  u(z=2H,t>0) = 0 t = t Clay z 2

u Sand  Initial condition: For wide fill, applied instantly to saturated soil, the initial excess pore-water pressure u(z,t=0) is independent of depth and equals the applied stress (p)

 u(z,t=0) = ui = p

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

35 Solution of differential equation of consolidation

 Solution:

 2ui 2 u(z,t)   (sin MZ)exp(M Tv ) ……………… Equation 3 m0 M where:  M  (2m 1) 2

cvt Tv  H 2 Tv = Time factor

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Solution of differential equation of consolidation

Equation (3) was solved chart: u/p = f(Tv, z)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

36 Solution of differential equation of consolidation

 2ui 2 u(z,t)   (sin MZ)exp(M Tv ) m0 M

 Solution in terms of degree of consolidation (U):

u  u u ui U (z,t)  i 1 u ui ui z u ui-u  2 2 U (z,t) 1  (sin MZ)exp(M Tv ) 2H m0 M t = t  Average degree of consolidation for entire layer: U = area / area z By integration: 2H

udz   2 2 0 Ut 1 exp( M Tv ) Ut 1  2 m0 M ui

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Solution of differential equation of consolidation

Equation (4) was solved Table and chart: Tv = f(U)

U (%) 0 10 20 30 40 50 60 70 80 90 95 99 100

0.0 0.008 0.031 0.071 0.126 0.197 0.287 0.403 0.567 0.848 1.129 1.781 ∞ Tv

0

10

20

30

40

50 (%)

U 60

70

80

90

100 0 0.2 0.4 0.6 0.8 1 1.2

Tv

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

37 Solution of differential equation of consolidation

 Time Factor: Tv

cvt Tv  2 hd

where:

Tv = time factor corresponding to degree of consolidation occurring at time t cv = coefficient of consolidation (determined from consolidation test) hd = longest drainage path within clay layer

The above equation holds for consolidation of sample in the lab, and for clay layer consolidating in the

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Solution of differential equation of consolidation

 Drainage Path: hd

Sand Sand

2H Clay 2H Clay

Sand Impervious hd = H hd = 2H

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

38 Determination of cv - logarithm-of-time method

 cv is determined from the consolidation test data

 There is a cv value that corresponds to each loading

increment, i.e. cv is stress dependent.

 Two methods to calculate cv:  Logarithm-of-time method

 Square–root-of-time method

 For the following test data corresponding to a certain

loading increment, find cv.

Time (min) 0 0.25 0.5 1 2.25 4 6.25 10 15 30 60 120 240 885

Dial Reading x 10-2 135 153 155 161 173 184 193 202 209 215 219 220 221 223 (mm)

135-135 153-135 = h x 10-2 (mm) 20 26 38 49 58 67 74 80 84 85 86 88 = 0 =18

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Determination of cv - logarithm-of-time method

Time (min) 0.1 1 10 100 1000 t = t t = 4t t50 = 3.6 0 1 1 min

10 U = 0%

20

30

40

mm) U = 50% -2 50

h (x 10 60 

70

80 U = 100%

90

100

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

39 Determination of cv - logarithm-of-time method

1. Plot the (h) or (e) on the vertical axis (arithmetic scale), and the time (t) on horizontal axis (logarithmic scale) S-shape

2. Determine (h) or (e) corresponding to 100% consolidation:

 Extrapolate the linear portion at the middle of the consolidation curve

 Extrapolate the linear portion at the end of the consolidation curve

 The two lines intersect at a point corresponding to (h) or (e) at 100% consolidation

3. Determine (h) or (e) corresponding to 0% consolidation:

 Consider (h1) corresponding to a small time (t1)

 Consider (h2) corresponding to (4t1)

 Determine the difference between (h1) and (h2) y

 Determine (h) corresponding to 0% consolidation at a distance equivalent

to y above (h1)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Determination of cv - logarithm-of-time method

4. Determine (h) or (e) corresponding to 50% consolidation at mid distance between 0% consolidation and 100% consolidation.

5. Determine t50

6. Calculate cv:

cvt Tv  2 hd

where: T50 = 0.197, hd = ½ sample height (sample drains from both

sides), and t = t50 from consolidation curve

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

40 Determination of cv – square root-of-time method

Square root time (min)0.5 00.5 5 10 15 20 25 30 35 (t90) 0 a 10 U = 0%

20

30

40 mm) -2 50

h (x 10 h (x 60  70 c U = 90%

80

90 b x 0.15x 100

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Determination of cv – square root-of-time method

1. Plot the (h) or (e) on the vertical axis (arithmetic scale), and the square root of time on horizontal axis (arithmetic scale)

2. Determine (h) or (e) corresponding to 90% consolidation:

 Draw a straight line through the initial linear portion of the curve. This line will intersect the vertical axis at a point (a) that corresponds to U = 0%

 Measure distance (x) between the vertical axis and the drawn line

 Measure additional distance (0.15x) point (b)

 Draw a line connecting points (a) and (b) and extend it until it intersects the original curve at point (c) corresponds to U = 90%

0.5 3. Determine (t90)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

41 Determination of cv – square root-of-time method

4. Calculate cv:

cvt90 T90  2 hd

where: T90 = 0.848, hd = ½ sample height (sample drains from both

sides), and t = t90 from consolidation curve

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Types of Compression

Time (min) 0.1 1 10 100h = 0 1000 0 Instantaneous/Immediate Consolidation 10 U = 0% 20

30

40 Primary Consolidation mm) -2 50

h (x 10 60 

70

80 U = 100%

90 Secondary Consolidation

100

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

42 Types of Compression

1. Instantaneous/Immediate Compression: due to compression of entrapped air and machine parts

2. Primary Compression: due to consolidation by escapage of water (relief of excess pore-water pressure) under the effect of pressure (loading)

3. Secondary compression: compression at constant effective stress (no excess pore-water pressure)

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Secondary Compression Index

Time (min) 0.1 1 10t100 100 1000 0

10

20

30

40 mm) -2 50

h (x 10 60 

70 U = 100% 80 h 90 log(t)

100

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

43 Secondary Consolidation Settlement (Ss)

 Secondary compression index (C): e C    log(t) where: h e  (1 e ) H o

 Secondary settlement (Ss):

C t1 Ss  H log( ) 1 eo to

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Example

 A rectangular building 24m  16m is founded on a raft and has its (a) (c) (b) foundation level at 2.0 m below ground surface. If the building exerts a net stress of 2.0 kg/cm2 on soil at foundation level, determine: i.The ultimate settlement of points a, b Sand and c. ii.The time required for these points to settle5.0cm,andthetimerequiredto undergo 90% consolidation.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

44 Example 2 i. qnet =20t/m z=8.0m H=5.0m Using Influence Chart:

Points a & b c L (m) 24 12 B (m) 8 8 m 3 1.5 n 1 1

Iz 0.193 0.203 (t/m2) 0.193x20x2 = 7.72 0.203x20x4 = 16.24

S = mv  H 0.004x7.72x5 = 0.154 0.004x16.24x5 = 0.325 m m

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Example

(b) ii. (a) (c)

2 cv = 0.006 cm /sec

Points a c b

U = St/Sf =5/15.4 = 0.325 5/32.5 = 0.154 =5/15.4 = 0.325 From Chart 0.083 0.019 0.083

(Tv)

cv t 0.006 t 0.006 t 0.006 t T  0.019  0.083  v 2 0.083  2 2 2 h D (250) (500) (500)

t (days) 10.0 9.2 40.0

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

45 Example

(a) (c) (b)

Points a c b

U = St/Sf 0.90 0.90 0.90 From ChartExample0.848 0.848 0.848 (Tv) ii.

2 c t cv = 0.006 cm0.006/sec t 0.006 t 0.006 t v 0.848  Tv  2 2 0.848  2 0.848  2 h D (250) (500) (500) t (days) 102.2 409.0 409.0

St (cm) 0.90 x 15.4 = 13.9 cm 0.90 x 32.5 = 29.3 cm 0.90 x 15.4 = 13.9 cm

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

Potential Sources of Error

 Sample disturbance: due to improper handling and/or trimming. Disturbance affects e-logp curve, and thus preconsolidation pressure. Larger samples help reduce effect of trimming on sample disturbance.

Kontopoulos, Nikolaos S., 2012

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

46 Potential Sources of Error

 Specimen not completely filling the ring: erroneous volume change due to compression of voids.

 Clogged porous stones: porous stones should be cleaned after every test to remove embedded soil particles.

between specimen and consolidation ring. Part of load applied to specimen can be lost by side friction. This effect can be reduced by lining the consolidation ring with Teflon.

Soil Mechanics –Soil Third Mechanics Year Civil (PBW Eng. N302)

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