This document discusses soil mechanics and consolidation. It provides background on soil mechanics, explaining that it involves determining soil parameters and properties based on mechanical laws. It then focuses on consolidation, defining it as the process where saturated soil decreases in volume due to expulsion of pore water under pressure. The document outlines the theory of one-dimensional consolidation proposed by Terzaghi, describing how it can be used to determine rates of volume change, settlement, and pore pressure dissipation over time in saturated soils. It also discusses laboratory testing methods like oedometer tests that are used to characterize consolidation properties.

1. Dr. D.Mythili
Assistant Professor
Department of Civil Engineering
Excel Engineering College
Stress Distribution and
Settlement
1

2. Soil Mechanics
2
Soil Mechanics is a very fundamental subject
consisting of determination of various soil
parameters theoretically and experimentally
based on laws of mechanics. Any Civil
Engineering structure needs strong and stable
foundation which depends on proper
understanding of soil behaviour, determination
and interpretation of soil parameters,
determination of stresses in soil. The course on
Soil Mechanics provides the students basic
knowledge on soil properties, testing procedures,

3. Consolidation
Presentation Outline
 Introduction
 1-D Consolidation
 Spring Analogy
 Oedometer Test
 Consolidation Parameters
 Terzaghi Theory
 Consolidation Settlement
 Case Study
 Problems
3

4. Less than 10 years ago (1929) the Foundation
Committee of a well-known Engineering
Society decided, at one of its meetings, that
the word “Settlement” should be avoided in
public discussions, because it might disturb
the peace of mind of those who are to be
served by the Engineering Profession.
Karl Terzaghi, 1939
(Consolidation – Time dependent Settlement)
4

5. Introduction
 Whenever a soil mass is stressed, it deforms.
Deformation may be either in the form of
distortion or a change in volume of the soil
mass
 Since natural soil deposits are laterally
confined on all sides, deformation under stress
is primarily associated with volume changes,
specifically, volume decrease
 The property of a soil by virtue of which
volume decrease occurs under applied
pressure is termed its ‘Compressibility’
5

6. Introduction
 On loading, a soil compress due to:
- Compression of the solid grains
Under typical engineering loads, the
compression of soil solids and pour water in
Negligible
- Compression of pore water or pore air &
Expulsion of pore water or pore air from the
voids
Through Compaction & Consolidation
6

7. Compaction
 The process in which the soil particles are
packed more closely together by mechanical
means,
i.e., dynamic loading such as rolling, tamping
and vibration etc.
 Achieved through reduction of air void.
 Little or no reduction in the water content.
7

8. Consolidation
 The process of gradual compression due to
the expulsion of pore water under steady
pressure
i.e., static loading. It is achieved mainly by
gradual drainage of water from the soil pores
 Occurs to saturated or nearly saturated clays
or other soils of low permeability
8

9.  A time-related process of increasing the
density of a saturated soil by draining some of
the water out of the voids.
 Generally related to fine-grained soils such as
silts and clays
 Coarse-grained soils (sands and gravels) also
undergo consolidation but at a much faster rate
due to their high permeability
Consolidation
9

10. During Consolidation…
GL
saturated clay
q kPa
A
Due to a surcharge q applied at the GL, the
stresses and pore pressures are increased at A

u
’
..and, they
vary with time.
10

11. During Consolidation…
11

12. Factors
 Consolidation may be due to one or more of the
following factors:
1. External static loads from structures.
2. Self-weight of the soil such as recently
placed fills.
3. Lowering of the ground water table.
4. Desiccation ( Draught).
12

13.  Prediction of both the
magnitude and the rate
of consolidation
settlements to ensure
the serviceability of
structures founded on a
compressible soil layer.
 Differential settlements
lead to structural failures
due to tilting should be
avoided.
Why Consolidation Need?
13

14. 1-D Consolidation
 Since water can flow out of a saturated soil
sample in any direction, the process of
consolidation is essentially three-dimensional.
 However, in most field situations, water will not
be able to flow out of the soil by flowing
horizontally because of the vast expanse of
the soil in horizontal direction.
 As a result, the soil layer undergoes 1-D
consolidation settlement in the vertical
direction.
14

15. The Spring Analogy
15

16.  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)
The Spring Analogy
16

17. Oedometer Test
To predict consolidation settlement in soil, we need
to know the stress-strain properties (i.e., the
relationship between the effective pressure and void
ratio) of the soil.
17

18.  Undisturbed saturated soil (clay, silt) –
representative of in-situ stratum
 Typical specimen size: h = 20mm, diam. = 75mm
 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
Sample & Setup
18

19. Assumptions
1)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)
19

20. 1)Trimming
2)Specimen set up and initialization (seating
load)
3)Apply an increment of vertical load
4)Record ΔH with time, compute Δe with time
5)Monitor until volume change ceases (~24 h)
6)Repeat 3-5 to generate load-compression
curve
17-Jun-23
Procedure

21.  The soil sample is contained in the brass ring between two
porous stones about 1.25 cm thick. by means of the porous
stones water has free access to and from both surfaces of
the specimen.
 The compressive load is applied to the specimen through a
piston, either by means of a hanger and dead weights or by
a system of levers. The compression is measured on a dial
gauge.
 At the bottom of the soil sample the water expelled from the
soil flows through the filter stone into the water container. At
the top, a well-jacket filled with water is placed around the
stone in order to prevent excessive evaporation from the
sample during the test. Water from the sample also flows
21
Procedure

22. Procedure
 Loads are applied in steps in such a way that the successive
load intensity, p, is twice the preceding one. The load
intensities commonly used being 1/4, 1/2,1, 2,4, 8, and 16
tons/ft2 (25, 50,100,200,400, 800 and 1600 kN/m2).
 Each load is allowed to stand until compression has
practically ceased (no longer than 24 hours). The dial
readings are taken at elapsed times of 1/4, 1/2, 1,2,4, 8,15,
30, 60, 120, 240, 480 and 1440 minutes from the time the
new increment of load is put on the sample (or at elapsed
times as per requirements).
 Sandy samples are compressed in a relatively short time as
compared to clay samples and the use of one day duration is
common for the latter.
22

23. Procedure
 After the greatest load required for the test has been
applied to the soil sample, the load is removed in
decrements to provide data for plotting the expansion
curve of the soil in order to learn its elastic properties
and magnitudes of plastic or permanent deformations.
 The following data should also be obtained:
 Moisture content and weight of the soil sample before
the commencement of the test.
 Moisture content and weight of the sample after
completion of the test.
 The specific gravity of the solids.
 The temperature of the room where the test is
conducted
23

24. Load Arrangements
24

25. A
dL
dh
k
Q t

Saturated
Clay
Flow rate (rate of consolidation) depends on k and
dht/dL
Hdr = H0
Impervious Rock, Aquitard,
etc.
Ground surface,
sand layer, etc.
Single
Drainage
Saturated
Clay
Hdr = ½ H0
Pervious layer, sand,
etc.
Ground surface,
sand layer, etc.
Double
Drainage
Significantly
decreases
time for
consolidation
Drainage Path
25

26. Pressure-void Ratio Curves
 The pressure-void ratio curve can be obtained if the void ratio of the
sample at the end of each increment of load is determined. Accurate
determinations of void ratio are essential and may be computed from the
following data:
 The cross-sectional area of the sample A, which is the same as that of
the brass ring.
 The specific gravity, Gs, of the solids.
 The dry weight, Ws, of the soil sample.
 The sample thickness, h, at any stage of the test.
 Let Vs = volume of the solids in the sample where
 We can also write
 If “e” is the void ratio of the sample, then
 hs is a constant and only h is a variable which decreases with increment
load. If the thickness h of the sample is known at any stage of the test, the
void ratio at all the stages of the test may be determined.
26

27. 27
Pressure-void Ratio Curve

28. Compression Index
28

29. Co-efficient of Compressibility
The slope of the void ratio versus effective stress
for a given stress increase in void ratio versus
effective stress
Co-efficient of volume compressibility (mv)
29

30. Over Consolidation Ratio (OCR)
30
σz ’ - Present Effective vertical Stress

31.  A soil that has never experienced a vertical
effective stress that was greater than its
present vertical effective stress is called a
Normally Consolidated (NC) soil
 Most NC soils have fairly low shear strength
 A soil that has experienced a vertical effective
stress that was greater than its present vertical
effective stress is called an Over Consolidated
(OC) soil
 Most OC soils have fairly high shear strength
 The OCR cannot have a value less than 1
NC and OC Soils
31

32.  For the soil loaded along
the recompression curve
AB the effective stress
close to point B will be the
pre consolidation
pressure.
 If the soil is compressed
along BC and unloaded
along CD and then
reloaded along DC the
effective stress close to
point C will be the new pre
consolidation pressure.
Preconsolidation Pressure
32

33.  The preconsolidation pressure (Pc) for an over
consolidated soil should not be exceeded in
construction, if possible.
 Consolidation settlements will small if the
effective vertical stress in the soil layer remains
below its Pc
 If effective vertical stress in the soil layer
exceeds its Pc, the consolidation settlements
will be large due to further yielding of the soil
layer.
 The estimation of Pc is greatly affected by the
Preconsolidation Pressure
33

34. To Find Pc
 Field method
 Graphical procedure based on consolidation test
results
 Field Method:
 Based on geological evidence.
 The geology and physiography of the site may
help to locate the original ground level.
 The overburden pressure in the clay structure
with respect to the original ground level may be
taken as the preconsolidation pressure pc.
34

35. Graphical Procedure
The method involves locating
the point of maximum curvature,
B on the laboratory e-log p curve
of an undisturbed sample as
shown in Fig. From B, a tangent
is drawn to the curve and a
horizontal line is also
constructed. The angle between
these two lines is then bisected.
The abscissa of the point of
intersection of this bisector with
the upward extension of the
inclined straight part
To Find Pc
35

36. The parameter used to describe the rate at
which saturated clay or other soil undergoes
consolidation, or compaction, when subjected
to an increase in pressure
To determine the coefficient of consolidation (Cv)
of a clay layer:
– Log Time Method
– Square Root Time Method
Coefficient of Consolidation
36

37. 1)Select some point near U = 50% (this is an
estimate of d50)
2)Find ta such that tb = 4ta
3)Calculate (db – da) and find d0 = da – (db – da)
4)Find d100 graphically with two tangent lines
5)Calculate actual d50 as ½(d0 + d100); find
corresponding t50
6)Calculate Cv using t50 and time factor T
Log -Time Method
37

38. time
e0
e1
e2
en
plot deformation
vs. log time
  
 
  
50
2
2
50
50
197
.
0
197
.
0
t
H
c
H
t
c
T
dr
v
dr
v



Log -Time Method
0.0050
0.0150
0.0250
0.0350
0.0450
0.10 1.00 10.00 100.00 1000.00
Log of Time (min)
Deformation,
d
(cm)
d0
50
2
50
197
.
0
t
H
c D
v 
d100
t50
d50
38

39. Square-Root-Time Method
1)Extrapolate linear portion backward to find d0
2)Measure length of segment AB (linear
portion)
3)Draw AC such that AC = 1.15(AB)
4)Draw line through d0 and C to find d90 and
5)Calculate cv using t90 and time factor T
*Preferred method in practice (don’t need to wait for t100)
90
t
39

40.   
 
  
90
2
2
90
90
848
.
0
848
.
0
t
H
c
H
t
c
T
dr
v
dr
v



Square-Root-Time Method
40

41. ConsolidationTime
Cv
H
.
T
t
2
v

 
%
U
100
log
.
933
,
0
781
,
1
Tv 


%
100
x
S
S
U
c
i
,
c

Where :
t = consolidation time
Tv = consolidation factor depended on consolidation degree (U)
U = consolidation degree in percent, descript as ratio of
design settlement to total settlement
Cv = coefficient of consolidation, get from consolidation test
2
v
100
%
U
4
T 







U = 0 – 60%
U > 60%
41

42. Hc
Porous Layer
Porous Layer
Hc
Porous Layer
Impermeable
layer
Cv
H
.
T
t
2
v

Where :
H = length of water path
H = Hc H = 0.5Hc
ConsolidationTime
42

43. Terzaghi Theory
The theory of 1-D consolidation considers the
rate at which water is squeezed out of an
element of soil and can be used to
determine the rates of
 volume change of the soil with time
 settlements at the surface of the soil with
time
 pore pressure dissipation with time
43

44. Assumptions
 1 Dimensional
 Saturation is complete
 Compressibility of water is negligible
 Compressibility of soil grains is negligible (but
soil grains rearrange)
 Darcy’s Law is valid
 Soil deformation is small
 Soil permeability is constant
 Soil skeleton of each layer is homogeneous, so
isotropic linier elastic constitutive law is valid
44

45.  Permeability (Kv) and mv are assumed constant,
but as consolidation progresses void spaces
decrease and this results in decrease of Kv, Kv
is not constant
 mv also changes with stress level, mv 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
Limitations
45

46. Settlement
A vertical downward movement due to a volume
decrease of the soil on which it is built
46
Uniform Settlement Differential Settlement
Smax
Smax
Smax
Smin
L
OriginalLevel
FinalLevel
Building

47. 47
Types of Settlement

48.  Maximum Settlement : It is the absolute maximum
downward movement of any part of building element.
Maximum Settlement = Smax
 Differential Settlement : It is the maximum difference
between two points in a building element.
Differential Settlement = Smax – Smin
 Angular Distortion : It is another method of expressing
differential settlement.
Angular Distortion = Differential Settlement/Length of
element
= (Smax - Smin)/L
48
Settlement

49. Causes of Settlement
 Bearing capacity failure or instability, including
landslides.
 Failure or deflection of the foundation structure.
 Elastic or distortion of the soil or rock.
 Consolidation (compression) of the soil or rock.
 Shrinkage due to desiccation.
 Change in density due to shock or vibration.
 Chemical alteration of constituents, including decay.
 Underground erosion.
 Collapse of underground openings such as caves or
mines.
 Structural collapse due to weakening of cementation
49

50. Permissible Values
Sand & Hard Clay Plastic Clay
Max.
Settlement
Diff.
Settlement
Angular
distortion
Max.
Settlement
Diff.
Settlement
Angular
distortion
Isolated
foundation
i)Steel
ii)RCC
50mm
50mm
0.0033L
0.0015L
1/300
1/666
50mm
75mm
0.0033L
0.0015L
1/300
1/666
Raft
foundation
i) Steel
ii) RCC
75mm
75mm
0.0033L
0.002L
1/300
1/500
100mm
100mm
0.0033L
0.002L
1/300
1/500
50

51. Components of Settlement
S - Total Settlement
Se - Elastic or Immediate Settlement
Sc - Primary Consolidation Settlement
Ss - Secondary Settlement
S = Se + Sc + Ss

52.  Immediate settlement - Caused by elastic
deformation of dry and moist soil without
any change in moisture content
 Primary or consolidation settlement -
Volume change caused by expulsion of
water from voids in saturated cohesive
soils
 Secondary consolidation settlement -
Volume change after primary
consolidation as a result of plastic
52
Components of Settlement

53. 53
Components of Settlement

54. Elastic Settlement
54
 Net elastic settlement of flexible footing
µ varies from 0.1 to 0.5
For saturated clay, µ = 0.5
If is a function of the L/B ratio of the foundation,
and the thickness H of the compressible layer.

55.  Elastic Settlement of Rigid Footing
55
Elastic Settlement
Sef = CrdfSe
Sef = Elastic settlement
Cr = Rigidity Factor ( Highly rigid footing, Cr =
0.8)
Df = Depth factor
Se = Settlement for a surface flexible footing

56.  Occurs in cohesive soils
 Because of the soil permeability the rate of
settlement may varied from soil to another
 The variation in the rate of consolidation
settlement depends on the boundary conditions.
SConsolidation = SPrimary + Ssecondary
Consolidation Settlement
56

57. Consolidation Settlement







 











o
o
o
c
c H
e
C
S



10
log
1
Sc = Consolidation Settlement
Cc = Compression Index
eo = Initial Void Ratio
H = Thickness of clay layer
σo = Initial overburden pressure at the middle of clay layer
Δσ = Extra pressure due to the new construction

58. Problems
 Ex. 1:
In a consolidation test the following results
have been obtained. When the load was
changed from 50 kN/m2 to 100 kN/m2, the void
ratio changed from 0.70 to 0.65. Determine the
coefficient of volume decrease, mv and the
compression index, Cc.
58

59. Problems
 Ex. 2:
A sand fill compacted to a bulk density of
18.84 kN/m3 is to be placed on a compressible
saturated marsh deposit 3.5 m thick. The
height of the sand fill is to be 3 m. If the
volume compressibility mv of the deposit is 7 ×
10–4 m2/kN, Estimate the final settlement of the
fill.
59

60. Problems
 Ex. 3:
A layer of soft clay is 6 m thick and lies
under a newly constructed building. The
weight of sand overlying the clayey layer
produces a pressure of 260 kN/m2 and the
new construction increases the pressure by
100 kN/m2. If the compression index is 0.5,
compute the settlement. Water content is 40%
and specific gravity of grains is 2.65.
60

61. Problems
 Ex. 4:
The void ratio of clay A decreased from
0.572 to 0.505 under a change in pressure from
120 to 180 kg/m2. The void ratio of clay B
decreased from 0.612 to 0.597 under the same
increment of pressure. The thickness of sample
A was 1.5 times that of B. Nevertheless the time
required for 50% consolidation was three times
longer for sample B than for sample A. What is
the ratio of the coefficient of permeability of A to
that of B ?
61

62. Problems
 Ex. 5:
A saturated soil has a compression index of
0.25. Its void ratio at a stress of 10 kN/m2 is 2.02
and its permeability is 3.4 × 10–7 mm/s.
Compute:
(i) Change in void ratio if the stress is increased
to 19 kN/m2
(ii) Settlement in (i) if the soil stratum is 5 m
thick
(iii) Time required for 40% consolidation if
drainage is one-way.
62

63. Problems
 Ex. 6:
The soil profile at a building site consists of dense
sand up to 2 m depth, normally loaded soft clay from 2
m to 6 m depth, and stiff impervious rock below 6 m
depth. The ground-water table is at 0.40 m depth below
ground level. The sand has a density of 18.5 kN/m3
above water table and 19 kN/m3 below it. For the clay,
natural water content is 50%, liquid limit is 65% and
grain specific gravity is 2.65.
Calculate the probable ultimate settlement resulting
from a uniformly distributed surface load of 40 kN/m2
applied over an extensive area of the site.
(Cont…)
63

64. Problems
In a laboratory consolidation test with porous discs
on either side of the soil sample, the 25 mm thick
sample took 81 minutes for 90% primary compression.
Calculate the value of coefficient of consolidation for the
sample.
64

65. Problems
 Ex. 7:
The void ratio of a clay is 1.56, and its
compression index is found to be 0.8 at the
pressure 180 kN/m2. What will be the void ratio
if the pressure is increased to 240 kN/m2 ?
65

66. Problems
 Ex. 8:
A bed of sand 12 m thick is underlain by a
compressible stratum of normally loaded clay, 6
m thick. The water table is at a depth of 5 m
below the ground level. The bulk densities of
sand above and below the water table are 17.5
kN/m3 and 20.5 kN/m3 respectively. The clay
has a natural water content of 40% and LL of
45%. G = 2.75. Estimate the probable final
settlement if the average increment in pressure
due to a footing is 100 kN/m2.
66

67. LEANING TOWER OF PISA
CASE STUDY
67

68. grout injection-
(361 holes)
Ground water
pumping
Higher
compressibility
Construction
(1173-1399)
(Burland et al., 1998)
CPT Tip Resistance
Profiles for North and
South Sides of Tower
68

69. Lead weights on North side
(~1993)
Soil Extraction
(1999-2001)
• 33 tons of soil were excavated
from under the north side
• Moved further toward vertical by
17.72 in.
• Now exhibits a 5-degree tilt
• Rate of subsidence reduced to
less than a couple of millimeters
per year
69

70. Some unsung heroes of Civil
Engineering..
foundations soil
exploration
tunneling
… hide right under your feet
70

71. 71
THANK YOU