1. Consolidation GLE/CEE 330 Lecture Notes Soil Mechanics
William J. Likos, Ph.D. Department of Civil and Environmental Engineering University of Wisconsin-Madison

2. Higher compressibility
Construction ( )
Higher compressibility
CPT Tip Resistance Profiles for North and South Sides of Tower
grout injection-
(361 holes)
Ground water pumping
(Burland et al., 1998)

3. Lead weights on North side
(~1993)
33 tons of soil were excavated from under the north side
Moved further toward vertical by in.
Now exhibits a 5-degree tilt
Rate of subsidence reduced to less than a couple of millimeters per year
Soil Extraction
( )

4. Soil Settlement and Compression
Immediate Settlement
Elastic deformation, undrained compression (sands, gravels)
Primary Consolidation
Time dependent settlement in saturated soil as water is squeezed from voids due to increase in vertical effective stress (clays)
Secondary Consolidation
Particle reorientation, creep, organic decay; does not involve expulsion of water (highly plastic clays, organics)
Distortion Settlement
Lateral movements near edges of loaded area

5. Changes in Vertical Effective Stress
Fill Placement
External Loading
Water Table Changes
Combinations of 1, 2, and 3
gfill
Hfill
sz’ P
sz’
sz’
(e.g., water table lowering)
sz’

6. Subsidence from Water Table Lowering
Example:
Initial GWT
g = 120pcf 5’
Final GWT 5’ A
USGS

7. Primary Consolidation: Piston-Spring Analogy
1. The container is completely filled with water, and the hole is closed. (Fully saturated soil)
2. A load is applied onto the piston, while the hole is still unopened. At this stage, only the water resists the applied load. (Development of excessive pore water pressure)
3. As soon as the hole is opened, water starts to drain out through the hole and the spring shortens. (Drainage of excessive pore water)
4. After some time, the drainage of water no longer occurs. Now, the spring alone resists the applied load. (Full dissipation of excessive pore water pressure. End of consolidation)

8. P
Flow
Spring is analogous to effective stress (stress carried by soil skeleton)
Initially, the pore water takes up the change in total stress so effective stress does not change
As excess pore water pressure drains, the effective stress increases (skeleton takes up load)
Consolidation is complete when excess pressure dissipates and flow stops
So consolidation is TIME DEPENDENT because it is a pressure dissipation (flow) process!
Depends on hydraulic conductivity (k) and length of drainage path (Hdr)

9. Example 11.2 (Coduto, 1999)
Before Fill:
Short Term After Fill:
Long Term After Fill:

10. Consolidation (Oedometer) Testing
1-D consolidation test:
Undisturbed saturated soil (clay, silt) – representative of in-situ stratum
Typical specimen size: h = 1”, diam. = 2”-3”
Specimen confined in rigid ring (no lateral deformation, “plane strain”)
Drainage allowed on top and bottom via porous stones
Apply increment of load and measure 1-D compression with time

11. Trimming Procedures
(Bardet, 1997)

12. Assumptions All compression occurs due to change in void ratio
i.e., the grains do not compress
Thus, we can relate change in void ratio (e) to change in volume
2) All strains are vertical (1-D) DH H0 e0

13. Procedures for Incremental Consol Testing
Trimming
Specimen set up and initialization (seating load, s’v0)
Apply an increment of vertical load (s’v = P/A)
Record DH with time, compute De with time
Monitor until volume change ceases (~24 h)
Repeat 3-5 to generate load-compression curve
aka “e-log p” curve
s’v0
s’v1
s’v2
s’vn
log s’v e0
1st increment e0 e1 e1
~24h
2nd e2 e2
nth en en
We will use this data to predict rate of consolidation
We will use this data to predict magnitude of consol settlement
time

14. The “e-log p” curve Loading Loading Unloading Unoading
San Francisco Bay Mud (Holtz and Kovacs, 1981)

15. Cr Cc Cs s’p OC Compression Divide e-log p curve into linear segments
OC = Overconsolidated (stiff response)
NC = Normally Consolidated (soft response)
OC and NC portions separated by s’p
s’p = “maximum past pressure”
Cr = Recompression Index (slope of OC)
Cc = Compression Index (slope of NC)
Cs = Swell Index (slope of unload response) B Cr A
Virgin (NC)
Compression Cc D
0.01 ~ Cr ~ 0.5 C Cs
0.1 ~ Cc ~ 2.6
s’p

16. (Bardet, 1997)

17. Stress History Cr Cc Cs s’p OC Maximum past pressure (s’p ) quantifies
The “stress history” of the soil – it is the largest magnitude of effective stress the soil has been consolidated to in the past.
Overconsolidation Ratio (OCR) quantifies the magnitude of a soil’s existing state of stress relative to its maximum past stress. B Cr A NC Cc
If OCR = 1, then s’p = s’ and the soil is “normally consolidated” (soft response – virgin compression)
If OCR > 1, then s’p > s’ and the soil is “over consolidated” (stiff response – it has been precompressed)
Sources of Overconsolidation:
Extensive erosion
Past glacial activity
Removed structures
Risen water table
Evaporation D C Cs
s’p

18. Disturbance & Empirical Correlations
Disturbance “erases” stress history
For NC or undisturbed specimens:
For OC or disturbed specimens:
(Disturbance decreases virgin compressibility)
In general:

19. Load-Rebound Behavior
Maximum past stress (aka preconsolidation
Stress) is a plastic yield stress.
If unloading occurs at a stress less than s’p, then the soil rebounds elastically.
If unloading occurs at a stress greater than s’p, then a new s’p results and the soil rebounds plastically. A eA
Similar to elastic-plastic response you learned about in Mechanics of Materials ec B C
yield
stress s
s’p
s’B
s’A = s’C
plastic deformation e

20. Casagrande Construction for s’p
(Coduto, 1999)

21. Calculating Consolidation Settlement
Site characterization to quantify thickness of compressible layer.
1-D laboratory consolidation testing to determine stress history and compression indices (s’p, Cc, Cr, Cs). Will also need cv for rate prediction.
Divide compressible layer into sublayers.
Calculate initial (preconstruction) effective vertical stress at midpoint of each sublayer (s’i).
Calculate final post-construction effective vertical stress at midpoint of each sublayer (s’f).
Calculate consolidation settlement for each sublayer (DH) and sum for total settlement.
gfill
Fill H1 1 H2 2
Soft
Clay H3 3 H4 4
Rock

22. Where did that equation come from?Cc Cs Cr
s’p OC
s’i
s’f e0 ef De
Where did that equation come from?
De comes from compression in the OC range and in the NC range.
For OC portion:
For NC portion:
So,

23. What if s’f < s’p ? e0 ef Cr Cc Cs s’f s’i s’pNC Cc Cs Cr
s’p OC
s’i
s’f e0 ef
What if s’f < s’p ?
In other words, what if we put a load on a highly overconsolidated deposit? (maybe significant past glacial activity).
Consolidation will result solely from recompression.
Settlement will be relatively small because response is stiff.
Equation must be modified:

24. What if s’i = s’p ? e0 Cr Cc ef Cs s’i = s’p s’fNC Cc Cs Cr
s’i = s’p OC
s’f e0 ef
What if s’i = s’p ?
In other words, what if we put a load on a normally consolidated (NC) deposit?
Consolidation will result solely from virgin compression.
Settlement will be relatively large because response is soft.
Equation must be modified:

25. Example – Fill Placement
Prop. A B Cc
0.25
0.20 Cr
0.08
0.06 e0
0.66
0.45
s’p
101 kPa
510 kPa
4.5 m A
See Coduto for solution using 7 layers
Let’s try using two layers (A and B)
9 m B
Compare to max past stress: 60
101
233
Log s’ e
Need both terms

26. Compare to max past stress:
Prop. A B Cc
0.25
0.20 Cr
0.08
0.06 e0
0.66
0.45
s’p
101 kPa
510 kPa
Now for layer B:
Compare to max past stress:
188
510
Log s’
Compare to solution for 7 layers…. DHtot = 0.83 m
361 e
Only need
Cr term

28. Rate of Consolidation Settlement
Recall that consolidation is volume change due to pore water being squeezed out
Dissipation of excess pore pressure
So consolidation takes time!!! – depends on:
Hydraulic conductivity (k)
Drainage boundaries (max length of drainage path, Hdr) z
Static Pressure us + =
Excess Pressure
Transient
Profile
gfill
Fill
Soft
Clay
Rock k
Hdr u u
Immediate
Long
Term z z

29. Consider a case of double-drainage (like an oedometer test)Ds
Top porous disk z H0
Saturated
Clay
Bottom porous disk Ds
So there is a hydraulic gradient from middle of sample to boundaries
The magnitude of the gradient decreases with time
So flow rate decreases with time
So rate of volume change decreases with time
Volume
We want to be able to plan for this!
Time

30. Drainage Path Length, Hdr Single Drainage Double Drainage
Flow rate (rate of consolidation) depends on k and dht/dL
Single Drainage
Double Drainage
Ground surface, sand layer, etc.
Ground surface, sand layer, etc.
Hdr = ½ H0
Hdr = H0
Saturated
Clay
Saturated
Clay
Significantly decreases time for consolidation
Impervious Rock,
Aquitard, etc.
Pervious layer, sand, etc.

31. Sand Drains at Kansai Intnl. Airport

32. 1-D Consolidation Theory
We seek a solution for excess pressure a function of location and time
Dissipation of excess pressure is a diffusion process governed by 1-D PDE
cv = “Coefficient of Consolidation”
(determined from lab or field testing) Ds z H0
Saturated
Clay Ds

33. Units of cv = L2/T (e.g., m2/year)
“Coefficient of compressibility” av e
or in terms of strain.….. s’
“Coefficient of
volume compressibility” mv e

34. Solution of the Consolidation Equation
Assumptions:
Darcy’s Law is valid
Soil solids and fluid are incompressible
Sr = 100%
Linear compressibility (const av or mv)
Boundary Conditions: H0
Saturated
Clay
Initial Conditions: z
Solution is infinite Fourier series:
(“Time Factor”)

35. How Can We Apply This Solution?
Consider Kansai Airport….How much consolidation for given time?
Define “Percent Consolidation” U
St = consol settlement at any time t
S∞ = consol settlement at end of primary
Note that U only depends on time (T) – Percent consolidation is independent of load!!!

37. Example 1: How long for 90% consolidation?
20’
From figure, T = 0.848
Clay
cv = 0.05 ft2/day
Sand

38. If S∞ = 4.0’, how long for 2’ of settlement?
20’
Clay
cv = 0.05 ft2/day
From figure, T = 0.196
Sand

39. If bottom boundary is impervious, how long for 90% consolidation?
20’
Clay
cv = 0.05 ft2/day
From figure, T = 0.848
Impervious Rock

40. Compute total consolidation settlement and time for 95% consolidation
OCR = 1, so clay is NC.
20’
Prop. fill
g= 100pcf
Analyze using Point A at midpoint of clay
25’
50’ A
Clay
k = 10-6 cm/s
e0 = 1.2
Cc = 0.40
G = 100 pcf
OCR = 1
Impervious Rock

41. Not given cv, so need to calculate…
20’
Prop. fill
g= 100pcf
25’
50’ A
Clay
k = 10-6 cm/s
e0 = 1.2
Cc = 0.40
G = 100 pcf
OCR = 1
Impervious Rock
If U = 95%, T = 1.129

42. (Bardet, 1997)

43. Log-Time Method for Determining cv
(Bardet, 1997) e0 e1
plot deformation
vs. log time e2 en
time
Select some point near U = 50% (tb, db) (this is an estimate of d50)
Find ta such that tb = 4ta
Calculate (db – da) and find d0 = da – (db – da)
Find d100 graphically with two tangent lines
Calculate actual d50 as ½(d0 + d100); find corresponding t50
Calculate cv using t50 and time factor T

44. Square-Root-Time Method for Determining cv
(Bardet, 1997)
*Preferred method in practice (don’t need to wait for t100)
Extrapolate linear portion backward to find d0
Measure length of segment AB (linear portion)
Draw AC such that AC = 1.15(AB)
Draw line through d0 and C to find d90 and
Calculate cv using t90 and time factor T

45. Secondary Compression
Ca = secondary compression index
Cae = mod. secondary compression index
ts = time of start of secondary comp.
t = time
Additional time-dependent compression after primary consolidation
Not due to dissipation of excess pore pressure (expulsion of water)
Relatively small amount of volume change
Creep, particle reorientation, organic decomposition

46. Secondary Compression