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SOIL MECHANICS
INTRODUCTION
Geotechnical Engineering is that part of engineering which is concerned with the behaviour of soil and rock. Soil Mechanics is the part concerned solely with soils. From an engineering perspective soils generally refer to sedimentary materials that have not been cemented and have not been subjected to high compressive stresses.
As the name Soil Mechanics implies the subject is concerned with the deformation and strength of bodies of soil. It deals with the mechanical properties of the soil materials and with the application of the knowledge of these properties to engineering problems. In particular it is concerned with the interaction of structures with their foundation material. This includes both conventional structures and also structures such as earth dams, embankments and roads which are they made of soil.
As for other branches of engineering the major issues are stability and serviceability. When a structure is built it will apply a load to the underlying soil; if the load is too great the strength of the soil will be exceeded and failure may ensue. It is important to realise that not only buildings are of concern, the failure of an earth dam can have catastrophic consequences, as can failures of natural and man made slopes and excavations. Buildings or earth structures may be rendered unserviceable by excessive deformation of the ground, although it is usually differential settlement, where one side of a building settles more than the other, that is most damaging. Criteria for allowable settlement vary from case to case; for example the settlement allowed in a factory that contains sensitive equipment is likely to be far more stringent than that for a warehouse. Another important aspect to be considered during design is the effect of any construction on adjacent structures, for example the
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excavation of a basement and then the construction of a large building will cause deformations in the surrounding ground and may have a detrimental effect on adjacent buildings or other structures such as railway tunnels.
Many of the problems arising in Geotechnical Engineering stem from the interaction of soil and water. For example, when a basement is excavated water will tend to flow into the excavation. The question of how much water flows in needs to be answered so that suitable pumps can be obtained to keep the excavation dry. The flow of water can have detrimental effects on the stability of the excavation, and is often the initiator of landslides in natural and man made slopes. Some of the effects associated with the interaction of soil and water are quite subtle, for example if an earthquake occurs, then a loose soil deposit will tend to compress causing the water pressures to rise. If the water pressures should increase so that they become greater than the stress due to the weight of the overlying soil then a quicksand condition will develop and buildings founded on this soil may fail.
Soil mechanics differs from other branches of engineering in that generally there is little control over the material properties, we have to make do with the soil at the site and this is often highly variable. By taking samples at a few scattered locations we have to determine the soil properties and their variability. At this stage in a project knowledge of the site geology and geological processes is essential to successful geotechnical engineering.
Soil mechanics is a relatively new branch of engineering science, the first major conference occurred in 1936 and the mechanical properties of soils are still incompletely understood. The first complete mechanical model for soil was published as recently as 1968. Over the last 40 years there has been rapid development in our understanding of soil behaviour and the application of this knowledge in engineering practice. The
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subject has now reached a phase of development similar to that of structural mechanics a century ago and the words of William Anderson in 1893 about structural engineers are relevant today for geotechnical engineering, "There is a tendency among the young and inexperienced to put blind faith in formulas (computer programs), forgetting that most of them are based upon premises which are not accurately reproduced in practice, and which, in any case, are frequently unable to take into account collateral disturbances which only observation and experience can foresee, and common sense provide against."
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1. SOILS AND THEIR CLASSIFICATION
1.1 Introduction
Soil mechanics is concerned with particulate materials (soils) found in the ground that are not cemented and not greatly compressed. These soils usually have a sedimentary origin, however, they can also occur as the result of rock weathering without any transport of the particles. The soil particles can have varying sizes, shapes and mineralogies, although these properties are usually interrelated. For instance the larger sized particles are generally composed of quartz and feldspars, minerals that have high strengths and the particles are fairly round. The smaller sized particles are generally composed of the clay minerals kaolin, illite and montmorillonite, minerals that have low strengths and form plate like particles. One of the most important aspects of particulate materials is that there are gaps or voids between the particles. The amount of voids is also influenced by the size, shape and mineralogy of the particles.
Because of the wide range of particle sizes, shapes and mineralogies in a typical soil a detailed classification of each soil would be very expensive and inappropriate for most geotechnical engineering purposes. However, some form of simple classification system giving information about the engineering properties is required on all sites. Why is this necessary?
Usually the soil on site has to be used. Soils differ from other engineering materials in that one has very little, if any, control over their properties.
The extent and properties of the soil at the site have to be determined.
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Cheap and simple tests are required to give an indication of the engineering properties such as stiffness and strength for preliminary design.
To achieve this continuous samples are recovered from boreholes, drilled to a depth that will depend on the scale of the project. Observation of the core enables the different soil layers to be determined and then classification tests are performed for these different strata. The extent of the different soil layers can be determined by correlating the results from different boreholes and this information is used to build a picture of the sub-surface profile.
An indication of the engineering properties is determined on the basis of particle size. This crude approach is used because the engineering behaviour of soils with very small particles, usually containing clay minerals, is significantly different from the behaviour of soils with larger particles. Clays can cause problems because they are relatively compressible, drain poorly, have low strengths and can swell in the presence of water.
1.2 Particle Size Definitions
The precise boundaries between different soil types are somewhat arbitrary, but the following scale is now in use worldwide.
Gravel Sand Silt ClayC M F C M F C M F C M F60 20 6 2 0.6 0.2 0.06 0.02 .006 .002 .0006 .0002 where C, M, F stand for coarse, medium and fine respectively, and the particle sizes are in millimetres.
5
Note the logarithmic scale. Most soils contain mixtures of sand,
silt and clay particles, so the range of particle sizes can be very large.
not all particles less than 2 m are comprised of clay minerals, and some clay mineral particles can be greater than 2 m. (A micron, m, is 10-6m).
1.3 Broad Classification
1.3.1 Coarse-grained soils
These include sands, gravels and larger particles. For these soils the grains are well defined and may be seen by the naked eye. The individual particles may vary from perfectly round to highly angular reflecting their geological origins.
1.3.2 Fine-grained soils
These include the silts and clays and have particles smaller than 60 m.
Silts These can be visually differentiated from clays because they exhibit the property of dilatancy. If a moist sample is shaken in the hand water will appear on the surface. If the sample is then squeezed in the fingers the water will disappear. Their gritty feel can also identify silts.
Clays Clays exhibit plasticity, they may be readily remoulded when moist, and if left to dry can attain high strengths
Organic These may be of either clay or silt sized particles. They contain significant amounts of vegetable matter. The soils as a result are usually dark grey or black and have a noticeable odour from decaying
matter. Generally only a surface phenonomen but layers
6
of peat may be found at depth. These are very poor soils for most engineering purposes.
1.4 Procedure for grain size determination
Different procedures are required for fine and coarse-grained material. Detailed procedures are described in the Australian Standard AS 1289.A1, Methods of testing soil for engineering purposes. These will be demonstrated in a laboratory session.
Coarse Sieve analysis is used to determine the distribution of the larger grain sizes. The soil is passed through a series of sieves with the mesh size reducing progressively, and the proportions by weight of the soil retained on each sieve are measured. There are a range of sieve sizes that can be used, and the finest is usually a 75 m sieve. Sieving can be performed either wet or dry. Because of the tendency for fine particles to clump together, wet sieving is often required with fine-grained soils.
Fine To determine the grain size distribution of material passing the 75m sieve the hydrometer method is commonly used. The soil is mixed with water and a dispersing agent, stirred vigorously, and allowed to settle to the bottom of a measuring cylinder. As the soil particles settle out of suspension the specific gravity of the mixture reduces. An hydrometer is used to record the variation of specific gravity with time. By making use of Stoke’s Law, which relates the velocity of a free falling sphere to its diameter, the test data is reduced to provide particle diameters and the % by weight of the sample finer than a particular particle size.
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Figure 1 A schematic view of the hydrometer test
1.5 Grading curves
The results from the particle size determination tests are plotted as grading curves. These show the particle size plotted against the percentage of the sample by weight that is finer than that size. The results are presented on a semi-logarithmic plot as shown in Figure 2 below. The shape and position of the grading curve are used to identify some characteristics of the soil.
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0.0001 0.001 0.01 0.1 1 10 1000
20
40
60
80
100
Particle size (mm)
% F
iner
Figure 2 Typical grading curves
Some typical grading curves are shown on the figure. The following descriptions are applied to these curves
W Well graded materialU Uniform materialP Poorly graded materialC Well graded with some clayF Well graded with an excess of fines
The use of names to describe typical grading curve shapes and positions has developed as the suitability of different gradings for different purposes has become apparent. For example, well graded sands and gravels can be easily compacted to relatively high densities which result in higher strengths and stiffnesses. For this reason soils of this type are preferred for road bases.
9
The suitability of different gradings is discussed in some detail by Terzaghi and Peck (1967).
From the typical grading curves it can be seen that soils are rarely all sand or all clay, and in general will contain particles with a wide range of sizes. Many organisations have produced charts to classify soils giving names for the various combinations of particle sizes. One such example is given in Figure 3 below.
1009080706050403020100
100
90
80
70
60
50
40
30
20
10
0100
90
80
70
60
50
40
30
20
10
0
Silt Sizes (%)
Sand
Size
s (%
) Clay
Sizes (%)
SandSilty Sand Sandy Silt
Clay-Sand Clay-Silt
Sandy Clay Silty Clay
Clay
LOWER MISSISSIPPI VALLEY DIVISION,U. S. ENGINEER DEPT.
Figure 3 Classification Chart
Important observations from figure 3 are that any soil containing more than 50% of clay sized particles would be classified as a clay, whereas sand and silt require 80% of the particles to be in that size range. Also any soil having more than 20% clay would have some clay like properties.
10
The hydrometer test is usually terminated when the percentage of clay sized particles has been determined. However, there are significant differences between the behaviour of the different clay minerals. To provide additional information on the soil behaviour further classification tests are performed. One such set of tests, the Atterberg Limit Tests, involve measuring the moisture contents of the soil at which changes in the soil properties occur.
1.6 Atterberg Limits
These tests are only used for the fine-grained, silt and clay, fraction of a soil (actually the % passing a 425 m sieve). If we take a very soft (high moisture content) clay specimen and allow it to dry we would obtain a relation similar to that shown in Figure 4.
As the soil dries its strength and stiffness will increase. Three limits are indicated, the definitions of which are given below. The liquid and plastic limits appear to be fairly arbitrary, but recent research has suggested they are related to the strength of the soil.
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Moisture Content (%)LLSL PL
Volume
Figure 4. Moisture content versus volume relation
(SL) The Shrinkage Limit - This is the moisture content the soil would have had if it were fully saturated at the point at which no further shrinkage occurs on drying.
(1)
In the shrinkage test the soil is left to dry and the soil is therefore not saturated when the shrinkage limit is reached. To estimate SL it is necessary to measure the total volume, V, and the weight of the solids, ws. Then
(2)
where w is the unit weight of water, andGs is the specific gravity
(PL) The Plastic Limit - This is the minimum water content at which the soil will deform plastically
(LL) The Liquid Limit - This is the minimum water content at which the soil will flow under a small disturbing force
(PI or Ip) The Plasticity Index. This is derived simply from the LL and PL
IP = LL - PL(3)
(LI) The Liquidity Index - This is defined as
(4)
The Atterberg Limits and relationships derived from them are simple measures of the water absorbing ability of soils
12
containing clay minerals. For example, if a clay has a very high LI and LL it is capable of absorbing large amounts of water, and for instance would be unsuitable for the base of a pavement. The LL and PL are also related to the soil strength.
Remember that only the fraction finer than 425 m is tested in the Atterberg Tests. If this fraction is only small (that is, the soil contains significant amounts of sand or gravel) it might be expected that the soil would have better properties. While this is true to some extent it is important to realise that the soil behaviour is controlled by the finest 10 - 25 % of the particles
1.7 Classification Systems for Soils
Several systems are used for classifying soil. This is because these systems have two main purposes
1. To determine the suitability of different soils for various purposes (see p8 Data Sheets)
2. To develop correlations with useful soil properties, for example, compressibility and strength
The reason for the large number of such systems is the use of particular systems for certain types of construction, and the development of localised systems.
1.7.1 PRA (AASHO) system
An example is the PRA system of AASHO (American Association of State Highway Officials), which ranks soils from 1 to 8 to indicate their suitability as a subgrade for pavements. The detailed classification is given in the Data Sheets p9.
1. Well graded gravel or sand; may include fines2. Sands and Gravels with excess fines
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3. Fine sands4. Low compressibility silts5. High compressibility silts6. Low to medium compressibility clays7. High compressibility clays8. Peat, organic soils
1.7.2 Unified Soil Classification
The standard system used worldwide for most major construction projects is known as the Unified Soil Classification System (USCS). This is based on an original system devised by Cassagrande. Soils are identified by symbols determined from sieve analysis and Atterberg Limit tests.
Coarse Grained Materials
If more than half of the material is coarser than the 75 m sieve, the soil is classified as coarse. The following steps are then followed to determine the appropriate 2 letter symbol
1. Determine the prefix (1st letter of the symbol)
If more than half of the coarse fraction is sand then use prefix S
If more than half of the coarse fraction is gravel then use prefix G
2. Determine the suffix (2nd letter of symbol)
This depends on the uniformity coefficient Cu and the coefficient of curvature Cc obtained from the grading curve, on the percentage of fines, and the type of fines.
First determine the percentage of fines, that is the % of material passing the 75 m sieve.
14
Then if % fines is < 5% use W or P as suffix> 12% use M or C as suffixbetween 5% and 12% use dual symbols. Use
the prefix from above with first one of W or P and then with one of M or C.
If W or P are required for the suffix then Cu and Cc must be evaluated
If prefix is G then suffix is W if Cu > 4 and Cc is between 1 and 3
otherwise use P
If prefix is S then suffix is W if Cu > 6 and Cc is between 1 and 3
otherwise use P
If M or C are required they have to be determined from the procedure used for fine grained materials discussed below. Note that M stands for Silt and C for Clay. This is determined from whether the soil lies above or below the A-line in the plasticity chart shown in Figure 5.
For a coarse grained soil which is predominantly sand the following symbols are possible
SW, SP, SM, SCSW-SM, SW-SC, SP-SM, SP-SC Fine grained materials
These are classified solely according to the results from the Atterberg Limit Tests. Values of the Plasticity Index and
15
Liquid Limit are used to determine a point in the plasticity chart shown in Figure 5. The classification symbol is determined from the region of the chart in which the point lies.
Examples CH High plasticity clayCL Low plasticity clayMH High plasticity siltML Low plasticity siltOH High plasticity organic soil (Rare)Pt Peat
0 10 20 30 40 50 60 70 80 90 100Liquid limit
0
10
20
30
40
50
60
Plas
tici
tyin
dex
CH
OH
or
MH
CLOL
MLor
CL
ML
Comparing soils at equal liquid limit
Toughness and dry strength increase
with increasing plasticity index
Plasticity chartfor laboratory classification of fine grained soils
Figure 5 Plasticity chart for laboratory classification of fine grained soils
The final stage of the classification is to give a description of the soil to go with the 2-symbol class. For a coarse grained soil this should include:
the percentages of sand and gravel maximum particle size angularity surface condition hardness of the coarse grains local or geological name
16
any other relevant information
If the soil is undisturbed mention is also required of
stratification degree of compactness cementation moisture conditions drainage characteristics
The information required, along with all the details of the Unified Classification Procedure is given in Figure 6. Note that slightly different information is required for fine-grained soils.
Give typical names: indicate ap-proximate percentages of sandand gravel: maximum size:angularity, surface condition,and hardness of the coarsegrains: local or geological nameand other pertinent descriptiveinformation and symbol inparentheses.
For undisturbed soils add infor-mation on stratification, degreeof compactness, cementation,moisture conditions and drain-age characteristics.
Example:
Well graded gravels, gravel-sand mixtures, little or nofines
Poorly graded gravels, gravel-sand mixtures, little or nofines
Silty gravels, poorlygraded gravel-sand-silt mixtures
Clayey gravels, poorly gradedgravel-sand-clay mixtures
Well graded sands, gravellysands, little or no fines
Poorly graded sands, gravellysands, little or no fines
Silty sands, poorly gradedsand-silt mixtures
Clayey sands, poorly gradedsand-clay mixtures
GW
GP
GM
GC
SW
SP
SM
SC
Wide range of grain size and substantialamounts of all intermediate particlesizesPredominantly one size or a range ofsizes with some intermediate sizesmissing
Non-plastic fines (for identificationprocedures see ML below)
Plastic fines (for identification pro-cedures see CL below)
Wide range in grain sizes and sub-stantial amounts of all intermediateparticle sizes
Predominantely one size or a range ofsizes with some intermediate sizes missing
Non-plastic fines (for identification pro-cedures, see ML below)
Plastic fines (for identification pro-cedures, see CL below)
ML
CL,CI
OL
MH
CH
OH
Pt
Dry strengthcrushing
character-istics
None toslight
Medium tohigh
Slight tomedium
Slight tomedium
High to veryhigh
Medium tohigh
Readily identified by colour, odourspongy feel and frequently by fibroustexture
Dilatency(reaction
to shaking)
Quick toslow
None to veryslow
Slow
Slow tonone
None
None to veryhigh
Toughness(consistencynear plastic
limit)
None
Medium
Slight
Slight tomedium
High
Slight tomedium
Inorganic silts and very fine sands,rock flour, silty or clayeyfine sands with slight plasticityInorganic clays of low to mediumplasticity, gravelly clays, sandyclays, silty clays, lean clays
Organic silts and organic silt-clays of low plasticity
inorganic silts, micaceous ordictomaceous fine sandy orsilty soils, elastic silts
Inorganic clays of highplasticity, fat clays
Organic clays of medium tohigh plasticity
Peat and other highly organic soils
Give typical name; indicate degreeand character of plasticity,amount and maximum size ofcoarse grains: colour in wet con-dition, odour if any, local orgeological name, and other pert-inent descriptive information, andsymbol in parentheses
For undisturbed soils add infor-mation on structure, stratif-ication, consistency and undis-turbed and remoulded states,moisture and drainage conditions
ExampleClayey silt, brown: slightly plastic:small percentage of fine sand:numerous vertical root holes: firmand dry in places; loess; (ML)
Field identification procedures(Excluding particles larger than 75mm and basing fractions on
estimated weights)
Groupsymbols
1Typical names Information required for
describing soilsLaboratory classification
criteria
C = Greater than 4DD----60
10U
C = Between 1 and 3(D )
D x D----------------------30
10c
2
60
Not meeting all gradation requirements for GW
Atterberg limits below"A" line or PI less than 4
Atterberg limits above "A"line with PI greater than 7
Above "A" line withPI between 4 and 7are borderline casesrequiring use of dualsymbols
Not meeting all gradation requirements for SW
C = Greater than 6DD----60
10U
C = Between 1 and 3(D )
D x D----------------------30
10c
2
60
Atterberg limits below"A" line or PI less than 4
Atterberg limits above "A"line with PI greater than 7
Above "A" line withPI between 4 and 7are borderline casesrequiring use of dualsymbolsD
eter
min
epe
rcen
tage
sof
grav
elan
dsa
ndfr
omgr
ain
size
curv
e
Use
grai
nsi
zecu
rve
inid
enti
fyin
gth
efr
actio
nsas
give
nun
der
fiel
did
entif
icat
ion
Dep
endi
ngon
perc
enta
ges
offi
nes
(fra
ctio
nsm
alle
rth
an.0
75m
msi
eve
size
)co
arse
grai
ned
soils
are
clas
sifi
edas
foll
ows
Les
sth
an5%
Mor
eth
an12
%5%
to12
%
GW
,GP,
SW
,SP
GM
,GC
,SM
,SC
Bor
deli
neca
sere
quir
ing
use
ofdu
alsy
mbo
ls
The
.075
mm
siev
esi
zeis
abou
tthe
smal
lest
part
icle
visi
ble
toth
ena
ked
eye
Fin
egr
aine
dso
ils
Mor
eth
anha
lfof
mat
eria
lis
smal
ler
than
.075
mm
siev
esi
ze
Coa
rse
grai
ned
soils
Mor
eth
anha
lfof
mat
eria
lis
larg
erth
an.0
75m
msi
eve
size
Silt
san
dcl
ays
liqui
dlim
itgr
eate
rth
an50
Silt
san
dcl
ays
liqui
dlim
itle
ssth
an50
Sand
sM
ore
than
half
ofco
arse
frac
tion
issm
alle
rth
an2.
36m
m
Gra
vels
Mor
eth
anha
lfof
coar
sefr
actio
nis
larg
erth
an2.
36m
m
Sand
sw
ithfin
es(a
ppre
ciab
leam
ount
offin
es)
Cle
ansa
nds
(lit
tleor
nofin
es)
Gra
vels
wit
hfin
es(a
prec
iabl
eam
ount
offin
es)
Cle
angr
avel
s(l
ittle
orno
fines
)
Identification procedure on fraction smaller than .425mmsieve size
Highly organic soils
Unified soil classification (including identification and description)
Silty sand, gravelly; about 20%hard angular gravel particles12.5mm maximum size; roundedand subangular sand grainscoarse to fine, about 15% non-plastic lines with low drystrength; well compacted andmoist in places; alluvial sand;(SM)
0 10 20 30 40 50 60 70 80 90 100Liquid limit
0
10
20
30
40
50
60
Pla
stic
ity
inde
x
CH
OH
or
MHOL
MLor
CL
"A" lin
e
Comparing soils at equal liquid limit
Toughness and dry strength increase
with increasing plasticity index
Plasticity chartfor laboratory classification of fine grained soils
CI
CL-MLCL-ML
17
Figure 6 Unified Soil Classification Chart
18
Example - Classification using USCS
0.0001 0.001 0.01 0.1 1 10 1000
20
40
60
80
100
Particle size (mm)
% F
iner
Classification tests have been performed on a soil sample and the following grading curve and Atterberg limits obtained. Determine the USCS classification.
Atterberg limits:Liquid limit LL = 32, Plastic Limit, PL =26
Step 1: Determine the % fines from the grading curve
%fines (% finer than 75 m) = 11% - Coarse grained, Dual symbols required
Step 2: Determine % of different particle size fractions (to determine G or S), and D10, D30, D60 from grading curve (to determine W or P)
D10 = 0.06 mm, D30 = 0.25 mm, D60 = 0.75 mm
Cu = 12.5, Cc = 1.38, and hence Suffix1 = W
Particle size fractions: Gravel 17%
19
Sand 73% Silt and Clay 10%
Of the coarse fraction about 80% is sand, hence Prefix is S
Step 3: From the Atterberg Test results determine its Plasticity chart location
LL = 32, PL = 26. Hence Plasticity Index Ip = 32 - 26 = 6
From Plasticity Chart point lies below A-line, and hence Suffix2 = M
Step 4: Dual Symbols are SW-SM
Step 5: Complete classification by including a description of the soil
2. BASIC DEFINITIONS AND TERMINOLOGY
Soil is a three phase material which consists of solid particles which make up the soil skeleton and voids which may be full of water if the soil is saturated, may be full of air if the soil is dry, or may be partially saturated as shown in Figure 1.
Solid
Water
Air
20
Figure 1: Air, Water and Solid phases in a typical soil
It is useful to consider each phase individually as shown in Table 1.
Phase Volume Mass WeightAir VA 0 0Water VW MW WW
Solid VS MS WS
Table 1 Distribution by Volume, Mass, and Weight
2.1 Units
For most engineering applications the following units are used:
Length metresMass tonnes (1 tonne = 103
kg)Density (mass/unit volume) t/m3
Weight kilonewtons (kN)Stress kilopascals (kPa) 1 kPa = 1 kN/m2
Unit Weight kN/m3
To sufficient accuracy the density of water w is given by
w = 1 tonne/m3
= 1 g/cm3
In most applications it is not the mass that is important, but the force due to the mass, and the weight, W, is related to the mass, M, by the relation
21
W = M g
where g is the acceleration due to gravity. If M is measured in tonnes and W in kN, g = 9.8 m/s2
Because the force is usually required it is often convenient in calculations to use the unit weight, (weight per unit volume).
= g
Hence the unit weight of water, w = 9.8 kN/m3
2.2 Specific Gravity
Another frequently used quantity is the Specific Gravity, G, which is defined by
It is often found that the specific gravity of the materials making up the soil particles are close to the value for quartz, that is
Gs 2.65
22
For all the common soil forming minerals 2.5 < Gs < 2.8
We can use Gs to calculate the density or unit weight of the solid particles
s = Gs w
s = Gs w
and hence the volume of the solid particles if the mass or weight is known
2.3 Voids Ratio and Porosity
Using volumes is not very convenient in most calculations. An alternative measure that is used is the voids ratio, e. This is defined as the ratio of the volume of voids, Vv to the volume of solids, Vs, that is
where Vv = Vw +Va
V = Va + Vw + Vs
A related quantity is the porosity, n, which is defined as ratio of the volume of voids to the total volume.
23
The relation between e and n can be determined by noting that
Vs = V - Vv = (1 - n) V
Now
and hence
2.4 Degree of Saturation
The degree of saturation, S, has an important influence on the soil behaviour. It is defined as the ratio of the volume of water to the volume of voids
The distribution of the volume phases may be expressed in terms of e and S, and by knowing the unit weight of water and the specific gravity of the particles the distributions by weight may also be determined as indicated in Table 3.
Vw = e S Vs
Va = Vv - Vw = e Vs (1 - S)
24
Phase Volume Mass WeightAir e (1 - S) 0 0Water e S e S w e S w
Solid 1 Gs w Gs w
Table 2 Distribution by Volume, Mass and Weight in Soil
Note that Table 2 assumes a solid volume Vs = 1 m3, All terms in the table should be multiplied by Vs if this is not the case.
2.5 Unit Weights
Several unit weights are used in Soil Mechanics. These are the bulk, saturated, dry, and submerged unit weights.
The bulk unit weight is simply defined as the weight per unit volume
When all the voids are filled with water the bulk unit weight is identical to the saturated unit weight, sat, and when all the voids are filled with air the bulk unit weight is identical with the dry unit weight, dry. From Table 2 it follows that
S = 1
25
S = 0
Note that in discussing soils that are saturated it is common to discuss their dry unit weight. This is done because the dry unit weight is simply related to the voids ratio, it is a way of describing the amount of voids.
The submerged unit weight, ´, is sometimes useful when the soil is saturated, and is given by
´ = sat - w
2.6 Moisture content
The moisture content, m, is a very useful quantity because it is simple to measure. It is defined as the ratio of the weight of water to the weight of solid material
If we express the weights in terms of e, S, Gs and w as before we obtain
Ww = w Vw = w e S Vs
Ws = s Vs = w Gs Vs
and hence
26
Note that if the soil is saturated (S=1) the voids ratio can be simply determined from the moisture content.
Example – Mass and Volume fractions
A sample of soil is taken using a thin walled sampling tube into a soil deposit. After the soil is extruded from the sampling tube a sample of diameter 50 mm and length 80 mm is cut and is found to have a mass of 290 g. Soil trimmings created during the cutting process are weighed and found to have a mass of 55 g. These trimmings are then oven dried and found to have a mass of 45 g. Determine the phase distributions, void ratio, degree of saturation and relevant unit weights.
1. Distribution by mass and weight
Phase Trimmings Mass(g)
Sample Mass, M(g)
Sample Weight, Mg(kN)
Total 55 290 2845 10-6
Solid 45 237.3 2327.9 10-6
Water 10 52.7 517 10-6
2. Distribution by Volume
Sample Volume, V = (0.025)2 (0.08) = 157.1 10-6 m3
27
Air Volume, Va = V - Vs - Vw = 14.9 10-6 m3
3. Moisture content
4. Voids ratio
5. Degree of Saturation
6. Unit weights
If the sample were saturated there would need to be an additional 14.9 10-6 m3 of water. This would weigh 146.2 10-6 kN and thus the saturated unit weight of the soil would be
Example – Calculation of Unit Weights
A soil has a voids ratio of 0.7. Calculate the dry and saturated unit weight of the material. Assume that the solid material occupies 1 m3, then
28
assuming Gs = 2.65 the distribution by volume and weight is as follows.
Phase Volume(m3)
Dry Weight(kN)
Saturated Weight(kN)
Voids 0.7 0 0.7 9.81 = 6.87
Solids 1.0 2.65 9.81 = 26.0
26.0
Dry unit weight
Saturated unit weight
If the soil were fully saturated the moisture content would be
Moisture content
Alternatively the unit weights may be calculated from the expressions given earlier which are on p. 5 of the Data Sheets
29
3. COMPACTION
Compaction is the application of mechanical energy to a soil to rearrange the particles and reduce the void ratio.
3.1 Purpose of Compaction
The principal reason for compacting soil is to reduce subsequent settlement under working loads.
30
Compaction increases the shear strength of the soil.
Compaction reduces the voids ratio making it more difficult for water to flow through soil. This is important if the soil is being used to retain water such as would be required for an earth dam.
Compaction can prevent the build up of large water pressures that cause soil to liquefy during earthquakes.
3.2 Factors affecting Compaction
Water content of the soil
The type of soil being compacted
The amount of compactive energy used
3.3 Laboratory Compaction tests
There are several types of test which can be used to study the compactive properties of soils. Because of the importance of compaction in most earth works standard procedures have been developed. These generally involve compacting soil into a mould at various moisture contents.
Standard Compaction Test AS 1289-E1.1
Soil is compacted into a mould in 3-5 equal layers, each layer receiving 25 blows of a hammer of standard weight. The apparatus is shown in Figure 1 below. The energy (compactive effort) supplied
31
in this test is 595 kJ/m3. The important dimensions are
Volume of mould
Hammer mass Drop of hammer
1000 cm3 2.5 kg 300 mm
Because of the benefits from compaction, contractors have built larger and heavier machines to increase the amount of compaction of the soil. It was found that the Standard Compaction test could not reproduce the densities measured in the field and this led to the development of the Modified Compaction test.
Modified Compaction Test AS 1289-E2.1
The procedure and equipment is essentially the same as that used for the Standard test except that 5 layers of soil must be used. To provide the increased compactive effort (energy supplied = 2072 kJ/m3) a heavier hammer and a greater drop height for the hammer are used. The key dimensions for the Modified test are
Volume of mould
Hammer mass Drop of hammer
1000 cm3 4.9 kg 450 mm
32
collar (mould extension)
Metal guide to control drop of hammer
Handle
Figure 1 Apparatus for laboratory compaction tests
3.4 Presentation of Results
To assess the degree of compaction it is important to use the dry unit weight, dry, because we are interested in the weight of solid soil particles in a given volume, not the amount of solid, air and water in a given volume (which is the bulk unit weight). From the relationships derived previously we have
which can be rearranged to give
Because Gs and w are constants it can be seen that increasing dry density means decreasing voids ratio and a more compact soil.
In the test the dry density cannot be measured directly, what are measured are the bulk density
33
Cylindrical soil mould
Hammer for compacting soil
Base plate
and the moisture content. From the definitions we have
= (1 + m) dry
This allows us to plot the variation of dry unit weight with moisture content, giving the typical reponse shown in Figure 2 below. From this graph we can determine the optimum moisture content, mopt, for the maximum dry unit weight, (dry)max.
Moisture content
Dry
uni
t w
eigh
t
mopt
( )max
dry
Figure 2 A typical compaction test result
34
If the soil were to contain a constant percentage, A, of voids containing air where
writing Va as V - Vw - Vs we obtain
then a theoretical relationship between dry and m for a given value of A can be derived as follows
Now
Hence
If the percentage of air voids is zero, that is, the soil is totally saturated, then this equation becomes
From this equation we see that there is a limiting dry unit weight for any moisture content and this occurs when the voids are full of water. Increasing the water content for a saturated soil results in a reduction in dry unit weight. The relation between the moisture content and dry unit weight for
35
saturated soil is shown on the graph in Figure 3. This line is known as the zero air voids line.
Moisture content
Dry
unit w
eig
ht
Figure 3 Typical compaction curve showing no-air-voids line
3.5 Effects of water content during compaction
As water is added to a soil ( at low moisture content) it becomes easier for the particles to move past one another during the application of the compacting forces. As the soil compacts the voids are reduced and this causes the dry unit weight ( or dry density) to increase. Initially then, as the moisture content increases so does the dry unit weight. However, the increase cannot occur indefinitely because the soil state approaches the zero air voids line which gives the maximum dry unit weight for a given moisture content. Thus as the state approaches the no air voidsline further moisture content increases must result in a reduction in dry unit weight. As the state approaches the no air voids line a maximum dry unit weight is reached and the moisture content at
36
this maximum is called the optimum moisture content.
3.6 Effects of increasing compactive effort
Increased compactive effort enables greater dry unit weights to be achieved which because of the shape of the no air voids line must occur at lower optimum moisture contents. The effect of increasing compactive energy can be seen in Figure 4. It should be noted that for moisture contents greater than the optimum the use of heavier compaction machinery will have only a small effect on increasing dry unit weights. For this reason it is important to have good control over moisture content during compaction of soil layers in the field.
Moisture content
Dry
unit w
eig
ht increasing compactive
energy
Figure 4 Effects of compactive effort on compaction curves
It can be seen from this figure that the compaction curve is not a unique soil characteristic. It depends on the compaction energy. For this reason it is
37
important when giving values of (dry)max and mopt
to also specify the compaction procedure (for example, standard or modified).
3.7 Effects of soil type
The table below contains typical values for the different soil types obtained from the Standard Compaction Test.
Typical Valuesdry )max
(kN/m3)mopt (%)
Well graded sand SW
22 7
Sandy clay SC
19 12
Poorly graded sand SP
18 15
Low plasticity clay CL
18 15
Non plastic silt ML
17 17
High plasticity clay CH
15 25
Note that these are typical values. Because of the variability of soils it is not appropriate to use typical values in design, tests are always required.
3.8 Field specifications
To control the soil properties of earth constructions (e.g. dams, roads) it is usual to specify that the soil
38
must be compacted to some pre-determined dry unit weight. This specification is usually that a certain percentage of the maximum dry density, as found from a laboratory test (Standard or Modified) must be achieved.
For example we could specify that field densities must be greater than 98% of the maximum dry unit weight as determined from the Standard Compaction Test. It is then up to the Contractor to select machinery, the thickness of each lift (layer of soil added) and to control moisture contents in order to achieve the specified amount of compaction.
Moisture content
Dry
uni
t w
eigh
t
39
AcceptReject
Moisture content
Dry
uni
t w
eigh
t
(a)(b)
Figure 5 Possible field specifications for compaction
There is a wide range of compaction equipment. For pavements some kind of wheeled roller or vibrating plate is usually used. These only affect a small depth of soil, and to achieve larger depths vibrating piles and drop weights can be used. The applicability of the equipment depends on the soil type as indicated in the table below
Equipment Most suitable soils
Typical application
Least suitable soils
Smooth wheeled rollers, static or vibrating
Well graded sand-gravel, crushed rock, asphalt
Running surface, base courses, subgrades
Uniform sands
Rubber tired rollers
Coarse grained
Pavement subgrade
Coarse uniform
40
Accept
Reject
soils with some fines
soils and rocks
Grid rollers Weathered rock, well graded coarse soils
Subgrade, subbase
Clays, silty clays, uniform materials
Sheepsfoot rollers, static
Fine grained soils with > 20% fines
Dams, embankments, subgrades
Coarse soils, soils with cobbles, stones
Sheepsfoot rollers, vibratory
as above, but also sand-gravel mixes
subgrade layers
Vibrating plates
Coarse soils, 4 to 8% fines
Small patches
clays and silts
Tampers, rammers
All types Difficult access areas
Impact rollers
Most saturated and moist soils
Dry, sands and gravels
3.9 Sands and gravels
For soils without any fines (sometimes referred to as cohesionless) the standard compaction test is difficult to perform. For these soil types it is normal to specify a relative density, Id, that must be achieved. The relative density is defined by
41
where e is the current voids ratio,emax, emin are the maximum and minimum
voids ratios measured in the laboratory from Standard Tests (AS 1289-5.1)
Note that if e = emin, Id = 1 and the soil is in its densest state
e = emax, Id = 0 and the soil is in its loosest state
The expression for relative density can also be written in terms of the dry unit weights associated with the various voids ratios. From the definitions we have
and hence
The description of the soil will include a description of the relative density. Generally the terms loose, medium and dense are used where
Loose 0 < Id < 1/3Medium 1/3 < Id < 2/3Dense 2/3 < Id < 1
Note that you cannot determine the unit weight from knowing Id. This is because the values of the
42
maximum and minimum dry unit weights (void ratios) can vary significantly. They depend on soil type (mineralogy), the particle grading, and the angularity.
4. EFFECTIVE STRESS
43
4.1 Saturated Soil
A saturated soil is a two phase material consisting of a soil skeleton and voids which are saturated with water. It is reasonable to expect that the behaviour of an element of such a material will be influenced not only by the forces applied to its surface but also by the water pressure of the fluid in the pores.
Suppose that a soil sample having a uniform cross sectional area A is subjected to an applied load W, as shown in Fig la, then it is found that the soil will deform. If however, the sample is loaded by increasing the height of water in the containing vessel, as shown in Fig lb, then no deformation occurs.
WW
Fig(1a) Soil loaded by an applied weight W Fig(1b) Soil loaded by water weighing W
Soil Soil
In examining the reasons for this observed behaviour it is helpful to use the following quantities:
(1)
and to define an additional quantity the vertical effective stress, by the relation
44
(2)
Let us examine the changes the vertical stress, pore water pressure and vertical effective stress for the two load cases considered above.
v uw v´Case (a) 0
Case (b) 0
These experiments indicate that if there is no change in effective stress there is no change in deformation, or alternatively that deformation only occurs when there is a change in effective stress.
Another situation in which effective stresses are important is the case of two rough blocks sliding over one another, with water pressure in between them as shown in Fig 2.
The effective normal thrust transmitted through the points of contact will be
(3a)
45
where U is the force provided by the water pressure
The frictional force will then be given by where is the coefficient of friction. For soils and rocks the actual contact area is very small compared to the cross-sectional area so that U/A is approximately equal to uw the pore water pressure. Hence dividing through by the cross sectional area A this becomes:
(3b)
where is the average shear stress and v is the vertical effective stress.
Of course it is not possible to draw a general conclusion from a few simple experiments, but there is now a large body of experimental evidence to suggest that both deformation and strength of soils depend upon the effective stress. This was originally suggested by Terzaghi in the 1920’s, and equation 2 and similar relations are referred to as the Principle of Effective Stress.
4.2 Calculation of Effective Stress
It is clear from the definition of effective stress that in order to calculate its value it is necessary to know both the total stress and the pore water pressure. The values of these quantities are not always easy to calculate but there are certain simple situations in which the calculation is quite straightforward. The most important is when the ground surface is flat as is often the case with sedimentary (soil) deposits.
46
4.2.1 Calculation of Vertical (Total) Stress
Consider the horizontally "layered" soil deposit shown schematically in Fig.3,
Layer 1
Layer 2
Layer 3
d1
d2
d3
z
Fig 3 Soil Profile
Surcharge q
v
bulk 3
bulk 2
bulk 1
If we consider the equilibrium of a column of soil of cross sectional area A it is found that
(4)
Calculation of Pore Water Pressure
47
Fig 4 Soil with a static water table
Water table
H
P
Suppose the soil deposit shown in Fig. 4 has a static water table as indicated. The water table is the water level in a borehole, and at the water table uw = 0. The water pressure at a point P is given by
(5)
Example
A uniform layer of sand 10 m deep overlays bedrock. The water table is located 2 m below the surface of the sand which is found to have a voids ratio e = 0.7. Assuming that the soil particles have a specific gravity Gs = 2.7 calculate the effective stress at a depth 5 m below the surface.
Step one: Draw ground profile showing soil stratigraphy and water table
48
Layer 1
Layer 2
2 m
3m
Fig 5 Soil Stratigraphy
bulk 1
bulk 2
Step two: Calculation of Dry and Saturated Unit Weights
Distribution by Volume
Solid
Voids Vv=e Vs = 0.7m3
Distribution by Weightfor the dry soil
Solid
Voids
Fig 6 Calculation of dry and saturated unit weight
Distribution by weightfor the saturated soil
Solid
Voids
Vs= 1m3
Ww=0W V kN
kN
kN
w v w
*
. * .
.
0 7 9 8
686
W V G
kN
kN
s s s w
* *
* . * .
.
1 2 7 9 8
26 46
W V G
kN
kN
s s s w
* *
* . * .
.
1 2 7 9 8
26 46
(6)
Step three: Calculation of (Total) Vertical Stress
(7)
49
Water Table
Step four: Calculation of Pore Water Pressure
(8)
Step five: Calculation of Effective Vertical Stress
(9)
Note that in practice if the void ratio is known the unit weights are not normally calculated from first principles considering the volume fractions of the different phases. This is often the case for saturated soils because the void ratio can be simply determined from
The unit weights are calculated directly from the formulae given in the data sheets, that is
50
Effective Stress under general conditions
In general the state of stress in a soil cannot be described by a single quantity, the vertical stress. To fully describe the state of stress the nine stress components (6 of which are independent), as illustrated in Fig. 7 need to be determined. Note that in soil mechanics a compression positive sign convention is used.
yy
yx
zx
xy
zy xz
yz
zz
Fig 7 Definition of Stress Components
x
y
z
xx
The effective stress state is then defined by the relations
(10)
51
Figure 7 Definition of Stress Components
Example – Effects of groundwater level changes
Initially a 50 m thick deposit of a clayey soil has a groundwater level 1 m below the surface. Due to groundwater extraction from an underlying aquifer the regional groundwater level is lowered by 2 m. By considering the changes in effective stress at a depth, z, in the clay investigate what will happen to the ground surface.
Due to decreasing demands for water the groundwater rises (possible reasons include de-industrialisation and greenhouse effects) back to the initial level. What problems may arise?
Assume
bulk is constant with depth bulk is the same above and below the water
table (clays may remain saturated for many metres above the groundwater table due to capillary suctions)
The vertical total and effective stresses at depth z are given in the Table below.
Initial GWL Lowered GWLv
uv´
At all depths the effective stress increases and as a result the soil compresses. The cumulative effect throughout the clay layer can produce a significant settlement of the soil surface.
52
When the groundwater rises the effective stress will return to its initial value, and the soil will swell and the ground surface heave (up). However, due to the inelastic nature of soil, the ground surface will not in general return to its initial position. This may result in:
surface flooding
flooding of basements built when GWL was lowered
uplift of buildings
failure of retaining structures
failure due to reductions in bearing capacity
5. STEADY STATE FLOW
5.1 Introduction
The flow of water in soils can be very significant, for example:
1. It is important to know the amount of water that will enter a pit during construction, or theamount of stored water that may be lost by percolation through or beneath a dam.
2. The behaviour of soil is governed by the effective stress, which is the difference between total stress and pore water pressure. When
53
water flows the pore water pressures in the ground change. A knowledge of how the pore water pressure changes can be important in considering the stability of earth dams, retaining walls, etc.
5.2 Darcy’s law
Because the pores in soils are so small the flow through most soils is laminar. This laminar flow is governed by Darcy's Law which will be discussed below.
5.2.1 Definition of Head
P
z(P)
Datum
Fig 1 Definition of Head at a Point
Referring to Fig. (1) the head h at a point P is defined by the equation
h Pu P
z Pw
w
( )( )
( )
(1)
54
IMPORTANT
z is measured vertically UP from the DATUM
In this equation w (9.8 kN/m3) is the unit weight of
water, and uw(P) is the pore water pressure .
55
Note
1. The quantity u(P)/w is usually called the pressure head.
2. The quantity z(P) is called the elevation head (its value depends upon the choice of a datum).
3. The velocity head (not shown in Equation 1) is generally neglected. The only circumstances where it may be significant is in flow through rock-fill, but in this circumstance, the flow will generally be turbulent and so Darcy's law is not valid.
Example - Calculation of Head
2 m
5 m
X
P
Static water table
Impermeable stratum
Fig 2 Calculation of head using different datum
1 m
1 m
1. Calculation of Head at P
Datum at the top of the impermeable layer
56
Fig 2 Calculation of head using different datums
z (P) = 1 m
2. Calculation of Head at X
Datum at the top of the impermeable layer
z (X) = 4 m
It appears that when there is a static water table the head is constant throughout the saturated zone.3. Calculation of Head at P (Datum at the water table)
z (P) = - 4 m
4. Calculation of Head at X (Datum at the water table)
z (X) = - 1 m
When there is a static water table the head is constant throughout the saturated zone, but its numerical value depends on the choice of datum.
It is very important to carefully define the datum. The use of imaginary standpipes can be helpful in visualising head. The head is then given by the
57
height of the water in the standpipe above the datum
Note also that it is differences in head (not pressure) that cause flow
5.2.2 Darcy’s Experiment
Soil Sample
h
Fig 3 Darcy’s Experiment
L
During his fundamental studies of the flow of water in soil Darcy found that the flow Q was:
1. Proportional to the head difference h
2. Proportional to the cross sectional are A
3. Inversely proportional to the length L of the soil sample.
Thus Darcy concluded that:
58
(2a)where k is the coefficient of permeability or hydraulic conductivity.
Equation (2a) may be rewritten:
(2b)
wherei = h/L is the hydraulic gradientv = Q/A is the Darcy or superficial velocity.
Note that the actual average velocity of the water
in the pores (the groundwater velocity) is where
n is the porosity. The groundwater velocity is always greater than the Darcy velocity.
5.3 Measurement of Permeability
5.3.1 Constant Head Permeameter
59
Manometers
L
inlet
outlet
H
constant headdevice
device for flow measurement
load
porous disk
Fig. 4 Constant Head Permeameter
sample
This is similar to Darcy's experiment. The sample of soil is placed in a graduated cylinder of cross sectional area A and water is allowed to flow through. The discharge X during a suitable time interval T is collected. The difference in head H over a length L is measured by means of manometers.From Darcy’s law we obtain
(3)
The piston is used to compact the soil because the permeability depends upon the void ratio
60
Sample
5.3.2 Falling Head Permeameter
H2
H1H
L
Fig. 5 Falling Head Permeameter
Standpipe ofcross-sectionalarea a
Sample of area A
porous disk
During a time interval t
The flow in the standpipe =
The flow in the sample =
and thus
(4a)
Equation (4a) has the solution:
61
(4b)
Now initially at time t = t1 the height of water in the permeameter is H = H1 while at the end of the test, t = t2 and H = H2 and thus:
(4c)
5.4 Typical permeability ranges
Soils exhibit a very wide range of permeabilities and while particle size may vary by about 3-4 orders of magnitude, permeability may vary by about 10 orders of magnitude.
10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12
Grovels Sands Silts Homogeneous Clays
Fissured & Weathered Clays
Fig 6 Typical Permeability Ranges
Permeability is often estimated from correlations with particle size. For example
This expression was first proposed by Hazen in 1893. It is satisfactory for sandy soils but is less
62
Gravels
reliable for well graded soils and soils with a large fines fraction.
63
5.5 Mathematical form of Darcy's law
Because of their geological history soils tend to be deposited in layers and hence have different flow properties along the layering and transverse to the layering.
z
xx
z
A
B C
O
Fig 7 Definition of Hydraulic Gradients
Suppose that the permeability in the horizontal plane is kH, then the velocity vx in the x direction is approximately given by:
(5a)
The negative signs in these equations have been introduced because flow occurs in the direction of decreasing head.
Similarly if the permeability in the vertical direction is kv then the velocity vz is given by
64
(Horizontal)
(Vertical)
(5b)
Should there also be flow in the y direction this is similarly governed by
65
5.5.1 Plane Flow
In many situations, as in the dam shown below, there will be no flow in one direction (usually taken as the y direction).
Soil
Dam
Flow
Impermeable bedrock
Fig. 8 Plane Flow under a Dam
Cross section of a long dam
5.5.2 Continuity Equation
In order to be able to analyse the complex flows that occur in practice it is necessary to examine the water entering and leaving an element of soil. Consider plane flow into the small rectangular box of soil shown below:
66
A
B
C
D
x
z
vz
vxSoilElement
Fig. 9 Flow into a soil element
Net flow into element =(6a)
For steady state seepage the flow into the box will just equal the flow out so the net flow in will be zero, thus dividing by xyz and taking the limit for an infinitesimal element, it is found:
(6b)
When equation (6) is combined with Darcy’s law it is found that:
(7a)
For a homogeneous material in which the permeability does not vary with position this becomes:
67
Fig 9 Flow into a soil element
(7b)
and for an isotropic material in which the permeability is the same in all directions (kH = kv):
(7c)
For the more general situation in which there is flow in three dimensions the continuity equation becomes:
(8)
The equation governing seepage then becomes:
(9a)
For a homogeneous material in which the permeability does not vary with position (x, y, z) this becomes:
(9b)
and for an isotropic material in which the permeability is the same in all directions:
(9c)
68
69
6. FLOW NETS
6.1 Introduction
Let us consider a state of plane seepage as for example in the dam shown in Figure 1.
Drainageblanket
Phreatic line
UnsaturatedSoil
Flow of waterz
x
Fig. 1 Flow through a dam
For an isotropic material the head satisfies Laplace's equations, thus analysis involves the solution of:
subject to certain boundary conditions.
6.2 Representation of Solution
At every, point (x,z) where there is flow there will be a value of head h(x,z). In order to represent these values we draw contours of equal head as shown on Figure 2.
70
Flow line (FL)
Equipotential (EP)
Fig.2 Flow lines and equipotentials
These lines are called equipotentials. On an equipotential (EP). by definition:
(1a)
it thus follows
(1b)
and hence the slope of an equipotential is given by
(1c)
It is also useful in visualising the flow in a soil to plot the flow lines (FL), these are lines that are tangential to the flow at a given point and are illustrated in Figure 2.
It can be seen from Fig. (2) that the flow lines and equipotentials are orthogonal. To show this notice
71
that on a flow line the tangent at any point is parallel to the flow at that point so that:
(2a)
it follows immediately that:
(2b)
and so
(3)
and thus the flow lines and equipotentials are orthogonal in an isotropic material.
72
6.3 Some Geometric Properties of Flow Nets
Consider a pair of flow lines, clearly the flow through this flow tube must be constant and so as the tube narrows the velocity must increase. Suppose now we have a pair of flow lines as shown in Figure 3.
Q
X
y
z
t
T
Y
Z
X
FL
FL
Q
hh+h
h+2h
EP
Fig. 3 Equipotentials intersecting a pair of Flow Lines
Suppose that the flow per unit width (in the y direction) is, Q, then the velocity v in the tube is given by
(4a)Also let us assume that the potential drop between any adjacent pair of equipotentials is h then it follows from Darcy’s law:
73
x
(4b)
It thus follows that:
(4c)using an identical argument to that used in developing equation(4c) it can be shown that:
(4d)
and hence that:
(5)Thus each of the elemental rectangles bounded by the given pair of flow lines and a pair of equipotentials (having an equal head drop) have the same length to breadth ratio.
Next consider a pair of equipotentials cut by flow tubes each carrying the same flow Q, as shown in Fig. (4)
74
Q
a
b
c
d
D
B
C
A
FL
Q
EP( h )
EP ( h + h )
Fig. 4 Flow lines intersecting a pair of Equipotentials
Then we see that if it is assumed that each of the tubes is of unit width (in the y direction) then the velocity in the tube is:
(6a)
and using Darcy's law:
(6b)It thus follows that:
(6c)It can be similarly shown that:
(6d)Hence again a pair of flow tubes carrying equal flows will intersect a given pair of equipotentials in
75
elemental rectangles which have the same length to breadth ratio.In drawing flow nets by hand it is usual to draw them so that each flow tube carries the same flow and so that the head drop between adjacent equipotentials is equal. In such cases all elemental rectangles will be similar. It is usually most convenient to draw the net so that these rectangles are 'square' (it is possible to draw an inscribed circle). This is illustrated in Fig.(5).
Fig. 5 Inscribing Circles in a Flow Net
To calculate quantities of interest, that is the flow and pore pressures, a flow net must be drawn. The flow net must consist of two families of orthogonal lines that ideally define a square mesh, and that also satisfy the boundary conditions. The three most common boundary conditions are discussed below.
6.4 Common boundary conditions
6.4.1 Submerged soil boundary - Equipotential
Consider the submerged soil boundary shown in Figure 6
76
Fig. 5 Inscribing circles in a Flow Net
Water
Datum
H-z
z
H
Fig. 6 Equipotential boundary
The head at the indicated position is calculated as follows:
(7)That is, the head is constant for any value of z, which is by definition an equipotential. Alternatively, this could have been determined by considering imaginary standpipes placed at the soil boundary, as for every point the water level in the standpipe would be the same as the water level. The upstream face of the dam shown in Figures 1 and 2 is an example of this situation.
6.4.2 Flow Line
At a boundary between permeable and impermeable material the velocity normal to the boundary must be zero since otherwise there
77
would be water flowing into or out of the impermeable material, this is illustrated in Figure 7.
Permeable Soil
Flow Linevn=0
vt
Impermeable Material
Fig. 7 Flow line boundary
The phreatic surface shown in Figures 2 and 8 is also a flow line marking the boundary of the flow net. A phreatic surface is also a line of constant (zero) pore pressure as discussed below.
6.4.3 Line of Constant Pore Pressure
Sometimes a portion of saturated soil is in contact with air and so the pore pressure of the water just beneath that surface is atmospheric. The phreatic surface shown in Figure 8 below is an example of such a condition. We can show from the expression for head in terms of pore pressure that equipotentials intersecting a line of constant pore pressure do so at equal vertical intervals as follows:
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(8)
Fig. 8 Constant pore pressure boundary
6.5 Procedure for Drawing Flow Nets
1. Mark all boundary conditions
2. Draw a coarse net which is consistent with the boundary conditions
and which has orthogonal equipotential and flow lines. ( It is usually
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easier to visualise the pattern of flow so start by drawing the flow lines).
3. Modify the mesh so that it meets the conditions outlined above and so
that the rectangles between adjacent flow lines and equipotentials are
square.
4. Refine the flow net by repeating step 3.
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6.6 Calculation of Quantities of Interest from Flow Nets
6.6.1 Calculation of Increment of Head
In most problems we know the head difference (H) between inlet and outlet and thus:
(9)
15 m
h = 15m
h = 12m h = 9m h = 6mh = 3m
h = 0
P5m
Fig. 9 Value of Head on Equipotentials
For example let us assume that the depth of water retained by the dam is 15 m, and that downstream of the dam the water table is level with the ground surface. For this case it can be seen that the total head drop is 15 m. Inspection of Fig. 2 or Fig. 9 shows that the are 5 potential drops and hence the head drop between each pair of potentials is h = 15/5 = 3 m.
6.6.2 Calculation of flow
The flow net has been drawn so that the elemental rectangles are approximately square thus referring
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to Fig (3) and equation(4) it can be seen that between each pair of flow tubes the flow is:
(10a)
It should be noted in the development of this formula it was assumed that each flow tube was of unit width and so equation (10a) gives the flow per unit width (into the page).
Suppose that the permeability of the underlying soil is k=10-5 m/sec (typical of a fine sand or silt) then the flow between each pair of flow tubes is:
(10b)
there are 5 flow tubes and so the total flow per unit width of dam is:
(10c)
and if the dam is 25m wide the total flow under the dam:
(10d)
The flow per unit width can alternatively be calculated from the formula
(10e)
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This equation (10e) is given in the Data Sheets to calculate the flow per unit width. In this equation Nf
is the number of flowtubes (The number of flowlines – 1), and Nh is the number of equipotential drops (The number of equipotential lines – 1).
Note that there are occasions where this formula (10e) cannot simply be applied, but equation (10a) will always be applicable for individual flow tubes. It is often necessary to determine h from consideration of a single flow tube. If a square flow net has been constructed that value of h will apply to all flow tubes.
6.6.3 Calculation of Pore Pressure
The pore pressure at any point can be found using the expression
(11a)
Now referring to Fig. 9 suppose that we wish to calculate the pore pressure at the point P. Taking the datum to be at the base of the dam it can be seen that z = - 5m and so:
(11b)
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Example – Calculating pore pressures
The figure below shows a long vessel, 20 metres wide, stranded on a sand bank. It is proposed to pump water into a well point, 10 metres down, under the centre of the vessel to assist in towing the vessel off. The water depth is 1 metre.
The sand has a permeability of 3 10-4 m/sec. Assuming that a head of 50 m can be applied at the well point calculate:
1. The pore pressure distribution across the base of the vessel
2. The total upthrust due to this increase in pore pressure
3. The rate at which water must be pumped into the well point.
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Step 1: Choose a convenient datum. In this example the sea floor has been chosen
Then relative to this datum the head at the well point, H1 = 40 mAnd the head at the sea floor, H2 = 1 m.
The increment of head, h = 39/9 = 4.333 m
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Stranded Vessel
Water Supply
Soft Sea Bottom
Well Point
Reaction Pile
Figure 10 Schematic diagram of vessel on sandbank
Figure 11 Flow net for situation in Figure10
Figure 12 Enlarged view of flownet in the vicinity of the base of the vessel
Step 2: Calculate the head at points along the base of the vessel. For convenience these are chosen to be where the EPs meet the vessel (B to E) and at the vessel centerline (A). Hence calculate the pore water pressures.
A B C D EHead (1)
H1 – 4.5 h
H1 –5 h
H1 – 6 h
H1 – 7 h
H1 – 8 h
Head (2)
H2 + 4.5 h
H2 + 4 h
H2 + 3 h
H2 + 2 h
H2 + h
Head (1 and 2) (m)
20.5 18.33 14.0 9.67 5.3
Pressure (kPa)w(h – z)
201.1 179.8 137.3 94.9 52.3
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Stranded Vessel
Water Supply
Soft Sea Bottom
Well Point
Reaction Pile
A B C D E
5 m
2.5 m
1.
Step 3: Measure the lengths off the flow net (Note that diagram must be drawn to scale) and hence calculate force from pressure distribution. For simplicity assume linear variation in pressure between points. Then the TOTAL UPTHRUST (per unit length of the vessel) is
= 3218 kN/m
Without pumping Upthrust = 20 1 9.81 = 196 kN/m
Upthrust due to Pumping = 3218 – 196 = 3022 kN/m
Flow required, =
m3/m/sec
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7. FLOW NETS FOR ANISOTROPIC MATERIALS
7.1 Introduction
Many soils are formed in horizontal layers as a result of sedimentation through water. Because of seasonal variations such deposits tend to be horizontally layered and this results in different
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permeabilities in the horizontal and vertical directions.
7.2 Permeability of Layered Deposits
Consider the horizontally layered deposit, shown in Figure 1, which consists of pairs of layers the first of which has a permeability of k1 and a thickness of d1 overlaying a second which has permeability k2
and thickness d2.
d1
d2
k=k1
k=k2
Fig. 1 Layered soil deposit
First consider horizontal flow in the system and suppose that a head difference of h exists between the left and right hand sides as indicated in Fig. 2. It then follows from Darcy’s law that:
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Fig. 1 Layered Soil
(1a)
(1b)
d1
d2
v=v1
v=v2
Fig. 2 Horizontal flow in a layered soil deposit
h h 0 h h h 0
L
It therefore follows:
Next consider vertical flow through the system, shown in Fig.3. Suppose that the superficial velocity in each of the layers is v and that the head loss in layer 1 is h1, while the head loss in layer 2 is h2
d1
d2
v
v
Fig. 3Vertical flow in a layered soil deposit
h h 0
L
h h h h 0 1 2
h h h 0 1
In layer 1:
so
(3a)
90
Fig. 3 Vertical flow through layered soil
(2a)
Fig. 2 Horizontal flow through layered soil
(2b)
Similarly in layer 2 and
(3b)
The total head loss across the system will be h=h1+h2 and the hydraulic gradient will be given by:
(3c)
For vertical flow Darcy’s Law gives
(3d)and hence
(3e)
Example
Suppose that that the layers are of equal thickness d1 = d2 = d0 and that k1 = 10-8 m/sec and that k2 = 10-10 m/sec, then:
and
Showing that, as is generally the case, the vertical permeability is much less than the horizontal.
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7.3 Flow nets for soil with anisotropic permeability
Plane flow in an anisotropic material having a horizontal permeability kH and a vertical permeability kv is governed by the equation:
(4)
The solution of this equation can be reduced to that of flow in an isotropic material by the following simple device. Introduce new variables defined as follows:
(5a)
the seepage equation then becomes
(5b)
Thus by choosing:
(5c)
It is found that the equation governing flow in an anisotropic soil reduces to that for an isotropic soil, viz.:
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(5d)
and so the flow in anisotropic soil can be analysed using the same methods (including sketching flow nets) that are used for analysing isotropic soils.
Example - Seepage in an anisotropic soil
Suppose we wish to calculate the flow under the dam shown in Figure 4;
x
z
Impermeable bedrock
L
H1H2
Impermeabledam
Soil layerZ
Fig. 4 Dam on a permeable soil layer over impermeable rock (natural scale)
For the soil shown in Fig. (4) it is found that and therefore
(6)
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In terms of transformed co-ordinates this becomes as shown in Figure 5
z
Impermeable bedrock
L/2
H1H2
x Soil layerZ
Fig. 5 Dam on a permeable layer over impermeable rock (transformed scale)
The flow net can now be drawn in the transformed co-ordinates and this is shown in Fig.6
Fig. 6 Flow-net transformed coordinates
5m
Impermeable bedrock
Fig. 6 Flow net for the transformed geometry
94
It is possible to use the flow net in the transformed space to calculate the flow underneath the dam by introducing an equivalent permeability
(7)
A rigorous proof of this result will not be given here, but it can be demonstrated to work for purely horizontal flow as follows:
xx
Natural scale transformedscale
Qt
h h - h h h - h
Fig. 7 Horizontal flow through anisotropic soil
For the natural scale
(7a)
For the transformed scale
(7b)
From Equations 7a and b it can be seen that
Example
Suppose that in Figure 6 H1 = 13m and H2 =
2.5m, and that kv = 10-6 m/sec and kH =4 10-6
m/sec The equivalent permeability is:
95
(8a)
The total head drop is 10.5 m and there are 14 head drops and thus:
(8b)
The flow through each flow tube, Q = keq h = (210-6 )(0.75) = 1.5 10-6 m3/s/m
There are 6 flow tubes and so the total flow , Q= 6 1.5 10-6
= 9.010-6
m3/sec/(m width of dam)
For a dam with a width of 50 m Q = 450 10-6 m3/sec = 41.47 m3/day
7.4 Piping
Many dams on soil foundations have failed because of the sudden formation of a piped shaped discharge channel. As the store water rushes out the channel widens and catastrophic failure results. This results from erosion of fine particles due to water flow. Another situation where flow can cause failure is in producing ‘quicksand’ conditions. This is also often referred to as piping failure.
In order to analyse this situation consider water flowing upwards through the element shown in Figure 8.
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Area =A
(z=z2 , h=h2 , u=u2,)
(z=z1 , h=h1 ,u=u1,)
Elevation
Fig. 7 Analysis of Piping
Plan
u2
u1
(9a)The pore pressure can be calculated from the head and so:
(9b)
For piping to occur the Uplift must be greater than the self-weight of the soil
or alternatively
97
(9d)
Fig. 8 Analysis of Piping
(9c)
Example
Suppose the dam shown in Figure 6 is 39 metres wide (this may be determined from the scale drawing), the water levels are the same as in the previous example (H1 = 13 m, H2 = 2.5 m), and the saturated unit weight of the soil is 18 kN/m3. Piping is most likely to occur at the toe of the dam, the hydraulic gradient there can be obtained from the flow net:
h1 - h2 = h = 0.75 m (calculated from Fig. 6)
z2 - z1 = 1.125 m (scaled from Fig. 6)
thus
Now
(10)
The safety factor against piping failure is thus icrit/i = 0.83/0.67 = 1.25 which is probably not adequate given the potentially disastrous consequences of a piping failure.
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8. ONE DIMENSIONAL SETTLEMENT BEHAVIOUR
8.1 One Dimensional Loading Conditions
Soils are often subjected to uniform loading over large areas, such as shown in Figure 1, from an embankment. Under such conditions soil which is remote from the edges of the loaded area undergoes vertical strain, but no horizontal strain. That is strains, and hence surface settlement, only occur in one-dimension.
x
z
Rock
Soil layer 1
Embankment
Soil layer 2
Figure 1 Embankment loading on a layered soil
The accuracy of this assumption depends on the relative dimensions of the loaded area and thickness of the soil layer. If the area is relatively large and the thickness of the soil layer relatively small then the assumption of 1-D conditions will be reasonable.
It is possible to make approximate estimates of surface settlement using the 1-D approach even when the loaded area is not relatively large. The
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procedures for doing this are discussed in section 9 on the calculation of settlement.
8.2 The Oedometer
The behaviour of soil during one-dimensional loading can be tested using a device called an oedometer, which is shown schematically in Fig. 2. The one-dimensional condition in which the vertical strain, zz 0, and the lateral strains, xx = yy = 0 is also referred to as confined compression.
Cell
Loading cap
Load Displacementmeasuring device
Soil samplewater
Porous disks
Figure 2 Schematic diagram of an oedometerThe following points may be noted:
1. The soil is loaded under conditions of no lateral strain (expansion), as the soil fits tightly into a relatively rigid ring.
2. Uncontrolled drainage is provided at the top and bottom of the specimen by porous discs (two way drainage). In more sophisticated oedometer apparatus control of drainage is possible.
3. A vertical load is applied to the specimen and a record of the settlement versus time is made.
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The load is left on until all settlement ceases (usually 24 hours although this depends on the soil type, impermeable clays may take longer).
4. The load is then increased (usually by a factor of 2, so the vertical stresses might be e.g. 20, 40, 80, 160 kPa). When the maximum load is reached, the soil is unloaded in several increments. If desired reloading can be carried out. At each step time-settlement records are made.
5. The relationships between voids ratio and effective stress, and settlement and time are found from the test. The methods by which these are obtained will be explained in the laboratory classes.
It is conventional to plot the void ratio versus the logarithm of the effective stress in examining the behaviour of soil, rather than plotting the relationship between effective stress and strain as is often done in materials testing. The reason for this is that the relationship between effective stress and voids ratio is fundamental to an understanding of soil behaviour. The relationship obtained is similar to that between effective stress and strain because changes in voids ratio and strain are simply related as shown later.
8.3 Relation of volume strain and vertical strain
The volume strain v of an element of material is defined to be the change in volume V divided by initial volume V0
101
Note: Compressive strains
are positive (1)
The volume strain is related to the vertical (axial) strain. To show this consider Figure 3.
x
z
x xx( )1
z zz( )1
(a) Before Deformation (b) After Deformation
Fig.3 Deformation of a soil element
(2a)
(2b)
(2c)
thus neglecting second order and higher terms
(2d)
For confined compression xx = 0, yy = 0. and thus:
(for confined compression) (2e)
102
Figure 3 Deformation of a soil element
8.4 Relation between volume strain and voids ratio
For most soils the skeletal material is far stiffer than the soil composite and thus referring to Figure 4 it can be seen that the relationship between volume strain and voids ratio is:
(3a)
and thus for confined compression:
(3b)
Skeletal material
Voids
Vs
Vse0
Vs
V e es( )0
V V e
V V e e
s
s
0 0
0
1
1
( )
( )
(a) Before Deformation (b) After Deformation
Fig. 4 Soil undergoing deformation
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Figure 4 Deformation of soil element
8.5 Behaviour of soil under one dimensional loading
The behaviour of an initially unloaded soil under one-dimensional conditions is illustrated in Fig. 5.
A
B
C
D
Voids ratio e
1
2
Log 10 (effective stress)
Fig. 5 Typical Effective stress voids ratio relationship
1. AB corresponds to initial loading of the soil.
2. BC corresponds to an unloading of the soil.
3. CD corresponds to a reloading of the soil.
4. Upon reloading the soil beyond B the soil continues along the path that it would have followed if loaded from A to D.
8.5.1 Preconsolidation Stress (pressure)
The preconsolidation stress, pc, is defined to be the maximum effective stress experienced by the
104
Figure 5 Typical effective stress, voids ratio relationship
soil. For soil at state C this would correspond to the effective stress at point B in Fig. 5.
8.5.2 Normally consolidated soils
If the current effective stress, ', is equal (note that it cannot be greater than) to the preconsolidation stress, pc, then the deposit is said to be normally consolidated (NC)
(normally consolidated)(4a)
During deposition of a soil (which usually takes place through sedimentation), the weight of the soil (which increases with depth below the surface) causes a decrease in void ratio. Suppose that at a particular depth below the surface the soil is represented by point P in Figure 6. If the soil is now subjected to an effective stress increase under 1-D conditions the path that will be followed in the e-log10 plot will be along the extension of the deposition line as shown in Fig. 6. A soil which lies at any point on this line is called normally consolidated, and the line is called the normal consolidation line.
Normally consolidated soils are usually found as recent alluvial deposits, and are mainly composed of silt and clay sized particles. It is extremely rare to find normally consolidated soils inland, away from the rivers or lakes in which they were deposited.
105
e
log10 (’)
Impossible states
Over-consolidated
states
NormalConsolidationLine
Figure 6 The normal consolidation line
8.5.3 Overconsolidated soils
If the current effective stress ' is less than the preconsolidation stress, pc, then the soil is said to be over-consolidated (OC).
(over-consolidated)(4b)
Note(not possible)
(4c)
If a soil after deposition, is normally consolidated to point P and then unloaded (perhaps because of erosion of the surface layers of soil) it may exist in the state indicated by point Q in Figure 7. The path QFR will be followed upon reloading of the soil.
It may be seen that for the same increase in effective stress, the change in void ratio will be much less for an overconsolidated soil (from e0 to ef ) than it would have been for a normally consolidated soil. Hence settlements will generally
106
P
be much smaller for structures built on overconsolidated soils.
Most soils are overconsolidated to some degree; this can be due to the effects of shrinking and swelling of the soil on drying and rewetting, changes in ground water levels, and unloading due to erosion of overlying strata.
Q
P
R
e = e0
e = ef
O
0 f pc logarithmic scale
F
Figure 7 Typical effective stress, voids ratio response
The distance from the normal consolidation line has an important influence on the soil behaviour. This is described numerically by the overconsolidation ratio (OCR). The OCR is defined as the ratio of the preconsolidation stress to the current effective stress
(5)
107
Note that when the soil is normally consolidated OCR = 1.
8.5.4 Estimation of the preconsolidation stress
A distinct change of slope is not generally observed at the preconsolidation pressure, making it difficult to accurately determine its value. Empirical procedures are used to estimate the preconsolidation stress, the most widely used being Casagrande's construction which is illustrated in Figure 8.
A C
B
D
F
pc
e
log (’)
E
Figure 8 Casagrande’s construction for estimating preconsolidation pressure
Steps in the construction are given below:
1. Determine the point of maximum curvature A. (It’s important to draw the graph to a sensible scale)
2. Draw a tangent to the curve at A, i.e. line AB.
3. Draw a horizontal line at A, i.e. line AC.
108
4. Draw the extension of the straight line (normally consolidated) portion of the curve DE.
5. Where the line DE cuts the bisector (AF) of angle CAB, is the preconsolidation stress.
For a normally consolidated soil the preconsolidation stress will be the same as the vertical overburden stress (due to weight of overlying soil) existing at the depth from which the sample was taken. Some unloading of the sample will take place during sampling so that a preconsolidation stress may be detected upon reloading in the oedometer at the point where the soil is loaded back to the stress state existing in the ground.
An overconsolidated soil will exhibit a preconsolidation stress which is much larger than the overburden stress at the level from which it was sampled.
8.5 Idealised soil behaviour
The behaviour shown in Figures 5 to 7 may be idealised by simple linear relationships in a void ratio, e, logarithm of effective stress, ´, plot as shown in Figure 9. This idealisation is based on observations that:
1. the behaviour of most normally consolidated soils can be approximated by straight lines for the range of stresses that are of interest.
109
2. the response of most over-consolidated soils can be approximated by straight lines, and further:
the behaviour is assumed to be reversible, unloading and reloading follow the same path
the slope of the unload-reload response is constant
e
log (’)
Figure 9 Idealised void ratio, effective stress relationship
8.6 Compression and Recompression Indexes
Figure 10 shows a portion of the e - log plot for a normally consolidated soil.
110
I
F
eI
eF
log ( )10 I log ( )10 F
Fig. 9 Idealised behaviour of a normally consolidated soil
Suppose that a soil is in an initial state I and after loading moves to the final state F, as shown in Figure 10.
(6a)
Because the relationship between effective stress and voids ratio can be closely approximated by a straight line, the slope is a constant. The slope constant, Cc is called the compression index.
(6b)
The above equation can be used to calculate the final voids ratio from the known final effective stress and initial conditions as follows:
(6c)
111
Figure 10 Idealised response for NC soil
A similar approach is possible if the soil is over-consolidated and the final stress is less than the preconsolidation stress, this is shown in Fig. 11.
Again suppose that a soil is at an initial state I and after loading moves to a final state F, as shown in Figure 11. As before we have:
(7a)
I
F
eI
eF
log ( )10 I log ( )10 F
Fig. 10 Idealised behaviour of a over consolidated soil
As the relationship between effective stress and voids ratio is approximately linear, thus:
(7b)
The constant Cr is called the recompression or swelling index. Again this equation can be used to determine the final voids ratio provided the final
112
Figure 11 Idealised response of OC soil
effective stress and initial conditions are known, as follows:
(7c)
Sometimes a soil may move from an overconsolidated state to a normally consolidated state. Suppose the initial state of the soil is given by point 1 in Figure 12, the point at which it reaches the preconsolidation stress is denoted by 2 and the final state is denoted by 3. The resulting change in voids ratio as the soil moves from the initial state 1 to the final state 3 can be considered to occur in two distinct stages. Stage 1 in which the soil is oveconsolidated and stage 2 in which the soil is normally consolidated.
(1)
(2)
(3)
e1
e2
e3
log ( )10 1 log ( )10 2 log ( )10 3
Figure 12 Response of soil moving from OC to NC Stage 1 (12)
During stage 1 the soil is over-consolidated and so:
113
(8a)
where 2= the initial value of the preconsolidation stress pc
Stage 2 (23)
During stage 2 the soil is normally consolidated and so:
(8b)
Since the soil is normally consolidated the current state of effective stress will be the preconsolidation stress and thus the final value of the preconsolidation stress (pc) will be 3
If the soil at 3, where it is normally consolidated, is unloaded so that the effective stress drops, the change in void ratio should be determined from equation 7c for over-consolidated soil.
9. CALCULATION OF SETTLEMENT
9.1 Settlement of a Single Layer
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The settlement s of a single relatively thin layer, shown in Fig. 1, can be calculated once the change in void ratio is known.
H
Fig. 1 Settlement of a single layer
S
z
x
For confined compression the horizontal strains are negligible i.e. xx = 0, yy = 0 and thus:
(1)
the settlement of a thicker layer can be calculated by dividing the layer into a number of sub layers as shown in Fig. 2. This is necessary because both the initial and final effective stress vary with depth as do the voids ratio and the OCR.
sub-layer 1
sub-layer 2
sub-layer n
Notation
e voids ratio at the centre of layer i
e increase in voids ratio at the centre of layer i
H thickness of layer i
i
i
i
Fig. 2 Soil profile divided into a number of sub-layers
115
Fig. 1 Settlement of a soil layer
Fig. 2 Division of soil layers into sub-layers
The settlement of the soil layer is calculated by calculating the settlement of the individual sub-layers and adding them, in doing this it is assumed that the voids ratio and the effective stress are constant throughout the sub-layer and equal to their values at the centre of the sub-layer.
(2)
Example - Settlement Calculation
A soil deposit, shown in Fig. 3 consists of 5 m of gravel overlaying 8 m of clay. Initially the water table is 2 m below the surface of the gravel. Calculate the settlement if the water table rises to the surface of the gravel slowly over a period of time and surface loading induces an increase of total stress of 100 kPa at the point A and 60 kPa at the point B. The preconsolidation pressure at A is 120 kPa, and the deposit is normally consolidated at B. The gravel has a saturated bulk unit weight of 22 kN/m3 and a dry unit weight of 18 kN/m3 and is relatively incompressible when compared to the clay. The void ratio of the clay is 0.8 and the skeletal particles have a specific gravity of 2.7. The compression index of the clay is 0.2 and the recompression index is 0.05.
In solving this problem it will be assumed that the gravel is far less compressible than the clay and thus that the settlement of the gravel can be neglected. The settlement of the clay layer will be calculated by dividing it into two sub-layers
116
2m5m
4m
4m
Gravel
Clay
A
B
Fig. 3 Layered soil deposit
dry kN m18 3/
sat kN m22 3/
sat ?
In order to commence the calculations it is first necessary to calculate the unit weight of the clay, this is shown schematically in Fig. 4.
Fig. 4 Determination of Saturated Unit Weight
or
Initial State at A
117
Fig. 3 Layered soil deposit
Total stress zz = 2 18 + 3 22 + 2 19.06 = 140.12 kPaPore water pressure uw = 5 9.8 kPa = 49 kPa
(3a)Effective stress zz = zz - uw = 140.12 - 49 = 91.12 kPa
Notice the initial effective stress is less than pc =120 kPa thus the clay is initially over-consolidated.
Final State at A
Total stress zz = 100 + 2 22 + 3 22 + 2 19.06 = 248.12 kPaPore water pressure uw = 7 9.8 kPa = 68.6 kPa
(3b)Effective stress zz = zz - uw = 248.12 - 68.6 = 179.52 kPa
Notice that the final effective stress exceeds the initial preconsolidation stress and thus the clay moves from being initially over-consolidated to finally normally consolidated.
Settlement of the first sub-layer
The soil in the first sub layer moves from being over-consolidated to normally consolidated and so the calculation of the change in voids ratio must be made in two stages.
Stage 1 Soil over-consolidated ( < pc (initial))
e1 = - Cr log10(pc (initial)/I)
Stage 2 Soil normally consolidated ( = pc)
(3c)e2 = - Cc log10(F/pc (initial))
now
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(3d) Initial State at B
Total stress zz = 218 + 3 22 + 6 19.06 = 216.36 kPaPore water pressure uw = 9 9.8 kPa = 88.20 kPa
(4a)Effective stress zz = zz - uw = 216.36 - 88.20 = 128.16 kPa
Final State at B
Total stress zz = 60 + 2 22 + 3 22 + 6 19.06 = 284.36 kPaPore water pressure uw = 11 9.8 kPa = 107.80 kPa
(4b)Effective stress zz = zz - uw = 284.36 - 107.80 = 176.56 kPa
Settlement of the second sub-layer
The soil in the second is normally consolidated and thus:
e2 = - Cc log10(F/I)(4c)
now
(4d)
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Total Settlement
Total settlement = 0.0911 + 0.0620 m
(5)= 0.1531m
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9.2 Calculation of Stress Changes
The calculation of settlement depends upon knowledge of the initial and final effective stress within each sub layer of the deposit. The initial effective stress state can be determined, from knowledge of the bulk unit and the position of the water table. The increase in total stress can be estimated using the theory of elasticity. (Note the soil is in general not really elastic however in the working stress range this assumption provides reasonably accurate estimates of the stress increases due to the applied loads)
A fundamental solution of the equations of elasticity is Boussinesq's solution. This relates to a point load applied to the surface of a half-space (very deep layer) and is shown schematically in Fig. 5.
Point load of magnitude P
H z
x
Fig 5 Point load on an elastic half-space
Boussinesq found that :
121
Fig. 5 Point load acting on a half space
(6)uz = vertical displacement due to load
The symbol is used to indicate that each of the quantities in equation (6) represents the increase in the particular quantity, due to the applied load.
The solution for a point load is important because it can be used to develop solutions for distributed loads by integration. Some of these solutions are presented in the Soil Mechanics Data Sheets.
9.3Calculation of Stress Changes
9.3.1 Stresses due to Circular foundation loads applied at the ground surface
A circular foundation of diameter 5 m, subjected to an average applied stress of 100 kPa is shown in Fig. 6.
122
5m
p=100 kPa
z
r
A
2m
B5m
Fig 6 Circular loaded area on a deep elastic layer
(a) Calculate the increase in vertical stress at point A
There is a simple analytic expression (given in the Data Sheets) for points on the centre line under a circular load:
(7a)
where
p = the surface stress = 100 kPaa = the radius of the loaded area = 2.5mz = the depth of interest = 2m
(7b)
(b) calculate the increase in vertical stress at point B
123
Fig. 6 Circular loaded area on a deep elastic layer
In this case there is no simple analytic expression and the solution must be found by using the influence charts given in the data sheets, reproduced in part in Figure 7. Note that this chart can also be used for points on the centre line for which r = 0.Now z/a = 2/2.5 = 0.8
r/a = 5/2.5 = 2(8)
using the data sheets zz/p = 0.03 and so zz = 3.0 kPa
10-3 10-2 10-1 10
2
4
6
8
10
10
9
8
7
6
5
4
3
2.52.0
1.51.25
1.00
0.0
z/a
Values on curvesare values of r/a
Fig.7 Influence Factors for a Uniformly Loaded Circular Area of radius a
Ipzz
9.3.2 Stresses due to Rectangular foundation loads applied at the ground surface
124
Fig. 7 Influence factors for a uniformly loaded circular area of radius a
Plan
Elevation
L
B
z
Point immediatelybeneath one of therectangle’s corners
Uniformly distributedsurface stress p
Fig. 8 Rectangular surface loading on a deep elastic layer
Many loads which occur in practice are applied to foundations that may be considered to consist of a number of rectangular regions. It is thus of interest to be able to calculate the vertical stress increases due to a uniformly distributed load acting on a rectangular loaded area. This is shown schematically in Fig. 8.The vertical stress change at a distance z below one of the corners of the rectangular load may be determined from a chart which is given in the data sheets and is reproduced in Fig. 9
125
Fig. 8 Rectangular uniform loading on a deep elastic layer
m=B/z=0.0
0.2
0.4
0.6
0.8
1.0
2.03.0
8
1010.10.01(n=L/z)
0.00
0.05
0.10
0.15
0.20
0.25
Iqzz
Note m & n are interchangeable
Fig. 9 Influence factors for a uniformly loaded rectangular area
This chart can be used to determine the value of stress increase at any point in an elastic layer, the method for doing this is illustrated below.
9.3.2.1 Calculation of Stress below an interior point of the loaded area
This situation is shown schematically in Fig.10. The stress change is required at a depth z below point O.
The first step in using the influence charts is to break the rectangular loading up into a number of components each
126
Fig. 9 Influence factors for uniformly loaded rectangular areas
having a corner at O, this is relatively simple as can be seen in Fig.(10)
It thus follows that at the point of interest, the stress increase zz(ABCD) is given by:
(9)
O Point of interest
O
A B
D C
X
Y
Z
T
Fig. 10 stress increase at a point below a loaded rectangular region
z
Example
Suppose we wish to evaluate the increase in stress at a depth of 2m below the point O due to the rectangular loading shown in shown in Fig. 11, when the applied stress over ABCD is 100 kPa.
127
Fig. 11 Dimensions of rectangular loaded area
Plan
Elevation
For rectangular loading OZCT
m = L/z =1n = B/z =1
thus I = 0.175
and sozz = p I = 100 0.175 = 17.5 kPa
(9a)
For rectangular loading OTDX
m = L/z = 1.5n = B/z = 1
thus I = 0.194
and sozz = p I = 100 0.195 = 19.4 kPa
(9b)
For rectangular loading OXAY
m = L/z = 1.5n = B/z = 1.5
thus I = 0.216
and sozz = p I = 100 0.216 = 21.6 kPa
(9c)
For rectangular loading OYBZ
m = L/z = 1.5n = B/z = 1
thus I = 0.194
128
and sozz = p I = 100 0.194 = 19.4 kPa
(9d)
Thus the increase in stress zz = 17.5 + 19.4 + 21.6 + 19.4 = 78.9 kPa (9e)
This must of course be added to the existing stress state prior to loading to obtain the actual stress zz. 9.3.2.2 Calculation of stress below a point outside the loaded
area
The stress increase at a point vertically below a point O which is outside the loaded are can also be found using the influence charts shown in Fig. 9.
A B
D C Fig. 12 Rectangular loaded area
X
Y
Z
T
O
This is achieved by considering the stress q acting on ABCD to consist of the following:
1. A stress +q acting over OXAY
2. A stress +q acting over OZCT
3. A stress -q acting over OZBY
4. A stress -q acting over OXDTThis is illustrated in Fig. 13.
129
Fig. 12 Rectangular loaded area ABCD and point of interest O
It thus follows that at the point O, the stress increase zz(ABCD) is given by:
and thus(10)
130
Fig. 13 Decomposition of loading over a rectangular area (for stress at external point)
Example
A B
D C
Fig. 14 Dimensions of rectangular loaded area
X
Y
Z
T
O
2m
10m
1m
1m
Suppose the rectangular area ABCD, shown in Fig. 14 is subjected to a surface stress of 100 kPa AND it is required to calculate the vertical stress increase at a point 1.5m below the point O.
For rectangular loading OZCT
m = L/z = 0.67n = B/z = 0.67
thus I = 0.121
and sozz = p I = + 100 0.121 = +12.1 kPa
(11a)
For rectangular loading OTDX
m = L/z = 7.67n = B/z = 0.67
thus I = 0.167
and so
131
Fig. 14 Dimensions of rectangular loaded area
zz = p I = -100 0.167 = -16.7 kPa(11b)
For rectangular loading OXAY
m = L/z = 7.67n = B/z = 2.00
thus I = 0.240
and sozz = p I = + 100 0.240 = + 24.0kPa
(11c)
For rectangular loading OYBZ
m = L/z = 2n = B/z = 0.67
thus I = 0.164
and sozz = p I = -100 0.164 = -16.4 kPa
(11d)
Thus the increase in stress zz = 12.1 - 16.7 + 24.0 + -16.4=3.0 kPa (11e)
9.3.3 Stresses due to foundation loads of arbitrary shape applied at the ground surface
Newmark’s chart provides a graphical method for calculating the stress increase due to a uniformly loaded region, of arbitrary shape resting on a deep homogeneous isotropic elastic region.
132
Newmark’s chart is given in the data sheets and is reproduced in part in Fig 15. The procedure for its use is outlined below
1. The scale for this procedure is determined by the depth z at which the stress is to be evaluated, thus z is equal to the distance OQ shown on the chart.
2. Draw the loaded area to scale so that the point of interest (more correctly its vertical projection on the surface) is at the origin of the chart, the orientation of the drawing does not matter
3. Count the number of squares (N) within the loaded area, if more than half the square is in count the square otherwise neglect it.
4. The vertical stress increase zz = N [scale factor(0.001)] [surface stress (p)]
The procedure is most easily illustrated by an example.
Example
Suppose a uniformly loaded circle of radius 2 m carries a uniform stress of 100 kPa. It is required to calculate the vertical stress at a depth of 4 m below the edge of the circle.
The loaded area is drawn on Newmark’s chart to the appropriate scale (i.e. the length OQ is set to represent 4 m) as shown in Fig. 15. It is found that the number of squares, N = 194 and so the stress increase is found to be
zz = 194 0.001 100 = 19.4 kPa(12)
133
This result can also be checked using the influence charts for circular loading and it is then found that:
z/a = 2, r/a = 1. zz /p = 0.2 and so zz = 20 kPa(13)
O Q4m
LoadedArea
Fig 15 Newmark’s Chart
10. ANALYSIS OF CONSOLIDATION
10.1 Introduction: the consolidation process
134
From the response of soils under one-dimensional conditions it is apparent that when the effective stress increases there will be a tendency for the soil to compress. However, when a load is applied to a saturated soil specimen this compression does not occur immediately. This behaviour is a consequence of the soil constituents, the skeletal material and pore water, being almost incompressible compared to the soil element; deformation can only take place by water being squeezed out of the voids. This can only occur at a finite rate and so initially when the soil is loaded it undergoes no volume change.
Under one dimensional conditions this implies that there can be no vertical strain and thus no change in vertical effective stress. For 1-D conditions we have
(1)
Hence if v = 0 then e = 0 and ´F = ´I.
When the load is first applied the total stress increases, but as shown above for 1-D conditions there can be no instantaneous change in vertical effective stress, this implies that the pore pressure must increase by exactly the same amount as the total stress as:
´ = - u(2)
Subsequently there will be flow from regions of higher excess pore pressure to regions of lower
135
excess pore pressure, the excess pore pressures will dissipate, the effective stress will change and the soil will deform (consolidate) with time. This is shown schematically in Fig. 1.
Excess PorePressure
Fig. 1 Variation of total stress and pore pressure with time
TotalStress
EffectiveStress
Settlement
Time Time
Time Time
10.2 Derivation of the equation of consolidation for 1-D conditions
If we assume that the pore fluid and soil skeleton are incompressible, then:
Volume decrease of the soil = Volume of pore fluid which flows out
and thus
Rate of volume decrease of soil = Rate at which pore fluid flows out
136
Fig. 1 Variation of stress, pore pressure and settlement with time
In deriving the equations governing consolidation we will consider only one-dimensional conditions with purely vertical soil movements and water flows. The solutions obtained will only be strictly relevant to the vertical consolidation of relatively thin soil layers occurring as a result of extensive uniform loading. (This is a common situation). A similar approach can be followed for more general loading but the resulting equations can only be solved numerically.
v(z, t)
v(z z, t) vv
zz
z
Fig. 2 Flow of pore fluid into an element of soil
Soil element
Referring to Fig 2 it can be seen that:
The rate at which water enters the element
The rate of volume decrease of the element
and thus
137
Fig. 2 Flow of pore fluid into an element of soil
(3)where
v = the pore fluid velocity,v = the element volume strain,A = the cross sectional area of the
element.It will also be assumed that Darcy’s law holds and thus that
(4)
In applying Darcy’s law it is only the velocity due to the consolidation process that is of interest, and consequently the head in (4) is the excess head due to the consolidation process (not the total head). The excess head is related to the excess pore water pressure by
(5)
Note that the elevation is not involved in (5) because it only relates the excess heads and water pressures. From (3), (4) and (5) it follows that
(6)
If it is also assumed that the soil element responds elastically to a change in effective stress then:
138
(7)where
= the change in effective stress from the original value
= (8)
with = the increase in total stress over the original value
u = the increase in pore water pressure over the original value (excess pore
water pressure)and
mv = the coefficient of volume decrease,
The value of mv must be determined over the appropriate effective stress range because it depends on the mean effective stress. This can be seen by considering the relation between voids ratio and effective stress:
and hence
now
(9)
Thus mv depends on both voids ratio e, and effective stress, ´.
Combination of equations (6), (7) and (8) leads to the equation of consolidation:
139
(10)The equation of consolidation must be solved subject to certain boundary conditions and initial conditions
10.3 Boundary Conditions
At a boundary where the soil is free to drain the pore water pressure will be constant and will not change during consolidation. For such a boundary the excess pore water pressure will be zero.
u = 0 at a permeable boundary(11a)
At an impermeable boundary the pore water velocity perpendicular to the boundary will be zero and thus from Darcy’s law
at an impermeable boundary
(11b)
10.4 Initial Conditions
At the instant of loading there is no volume strain and thus no change in vertical effective stress. At this instant the excess pore water pressure will be equal to the initial increase in total stress.
at the instant of loading.(12)
10.5 The Equation of Consolidation for a Homogeneous Soil
140
If the soil layer being considered is homogeneous then equation (10) becomes:
(13)
where
is called the coefficient of
consolidation.
The coefficient of consolidation (cv) can be estimated using the oedometer apparatus as can the coefficient of volume decrease (mv). The procedure to do this will be discussed in the laboratory classes. It is difficult (time consuming) to measure the permeability of clays (kv) and so the value of permeability is usually inferred from the values of cv and mv .
10.6 Analytic Solutions to the equations of consolidation
10.6.1 Two-way drainage
Fig. 3 represents a layer of clay of thickness 2H subjected to a uniform surface stress q applied at time t = 0 and held constant thereafter. The clay layer is free to drain at both its top and bottom boundaries. This is called two-way drainage.
141
Uniformly distributed surcharge q
2HZ
Fig 3 Homogeneous Saturated Clay Layerfree to drain at Upper and Lower Boundaries
The increase in stress through out the layer and does not vary with time and so
Equation (10) therefore becomes:
(14a)
The clay layer is free to drain at its upper and lower boundary and so
u = 0 when z = 0 for t > 0(14b)
u = 0 when z = 2H for t > 0(14c)
Initially the excess pore pressure will just match the increase in total stress so that there will be no instantaneous volume change and thus:
142
Fig. 3 Homogeneous clay layer free to drain from both upper and lower boundaries
u = q when t = 0 for 0 < z < 2H.(14d)
It can be shown that the solution of equations (14 a,b,c,d) is:
(15)
where n = (n + ½) Z = , a dimensionless distance
Tv = , a dimensionless time
Notice that H which occurs in both dimensionless quantities is the maximum drainage path length.
The settlement of the soil layer can be determined by summing the vertical (= volume) strains, giving:
(16a)
and the variation of settlement with time can be obtained by substituting in equation (15) which gives the variation of u with time and depth.
giving
143
(16b)
Noting that the final settlement of the layer, S = mv q 2H the settlement may be written:
(16c)
where U is known as the degree of settlement
The variation of excess pore pressure within the layer is shown in Figure 4 (also in data sheets).
T=0.8 0.5 0.3 0.2 0.1
0
1
20.0 0.5 1.0
Z=z/H
u/qFig. 4 Variation of excess pore pressure with depth
The lines on Figure 4 represent the variation of pore pressure with depth at different non-dimensionalised times (T). These lines are known as isochrones. It can be seen that initially the excess pore pressure is constant (u/q = 1) throughout the layer. With time the pore water
144
flows from the interior of the layer to the drainage boundaries, and the excess pore pressures dissipate until after a very long time there are no excess pore pressures.
The variation of settlement with time is most conveniently plotted in the form of the degree of settlement (U) versus dimensionless time Tv, and this is illustrated in Fig. 5 (also in data sheets)
10-3 10-2 10-1 1 10
Dimensionless Time Tv
0.00
0.25
0.50
0.75
1.00
U
Relation of degree ofsettlement and time
Fig. 5 Degree of settlement versus dimensionless time
There are several useful approximations for the degree of settlement, viz:
(17)
145
alternatively Fig. 5 may be used. It is worth remembering that U = 0.5 when Tv = 0.197.
10.6.2 One-way drainage
Fig. 6 represents a layer of clay of thickness H subjected to a uniform surface stress q applied at time t = 0 and held constant thereafter. The clay layer is free to drain at its top boundary but is unable to drain at its base. This is called one way drainage.
Uniformly distributed surcharge q
HZ
Fig 6 Homogeneous Saturated ClayLayer resting on an impermeable base
Impermeable base
The increase in stress through out the layer and does not vary with time and so
Equation (6) therefore becomes:
(18a)
The clay layer is free to drain at its upper boundary and as before
146
Fig. 6 Homogeneous saturated clay layer on an impermeable base
when z = 0 for t > 0(18b)
at the lower boundary
when z = H for t > 0
(18c)
Initially the excess pore pressure will just match the increase in total stress so that there will be no instantaneous volume change and thus:
u = q when t = 0 for 0 < z < H.(18d)
Reference to figure 4 reveals that solution (15) also satisfies the condition
when z = H for t > 0
and is thus also the solution for one way drainage ( the two way drainage problem can be viewed as two one-way drainage problems ‘back to back’). Further examination reveals that although the expression for final settlement differs for the two cases the expression for degree of settlement is precisely the same.
147
Example - Calculation of settlement at a given time
Figure 7 shows a soil profile, it can be assumed that the sand and gravel are far more permeable than the clay and so consolidation in them will have occurred instantaneously.
Gravel
4mClay
Sand
5m
Impermeable
Clay
Fig 7 layered Soil Deposit
Final settlement=100mm cv=0.4m2/year
Final settlement=40mm cv=0.5m2/year
It is assumed that the final settlement has for each of the clay layers has been determined by the methods described in the previous sections and that their values are as indicated on figure 7. It is required to find the settlement after 1 year
(a) Settlement of the upper Layer
In layer 1 there is two way drainage and so the drainage path H = 2m.
148
Fig. 7 Layered soil deposit
Using Figure 5 it can be seen that U = 0.36
and thus the settlement of layer 1 = 100 0.36 = 36mm
(b) Settlement of the lower Layer
In layer 2 there is one way drainage and so the drainage path H = 5m.
Using Figure 5 it can be seen that U = 0.16
and thus the settlement of layer 2 = 40 0.36 = 6.4mm
The total settlement after 1 year is thus = 36 + 6.4 = 42.4mm
Example - Use of scaling
An oedometer specimen reaches 50% settlement after 2 minutes. If the specimen is 10 mm thick calculate the time for 50% settlement of a 10 m thick layer under conditions of one-way drainage.
In order that the test may be carried out as quickly as possible oedometer tests are normally conducted with two way drainage and thus the drainage path in the oedometer = 5mm = 0.005m.
For the oedometer test
149
For the clay layer the drainage path is 10m
Since the degree of settlement for the two case is the same the two values of the dimensionless time, Tv are equal and so:
Example - Calculation of the coefficient of consolidation
The data in the previous example can be used to calculate cv. The dimensionless time for 50% consolidation is Tv = 0.197 (from Figure 5 Tv 0.2) thus:
11. NUMERICAL SOLUTION OF THE 1-D CONSOLIDATION EQUATION
150
The 1-D equation of consolidation cannot be solved analytically except for some very simple situations. For more difficult cases it is necessary to use approximate numerical techniques. One numerical technique that can be used for consolidation problems is the finite difference approach. In this method the solution is evaluated at a number of points at different times as indicated on the figure below.
z
tt
z
t=0 t=t1 t=t2
Fig. 1 Finite difference grid
11.1 Finite Difference Formulae
The 1-D consolidation equation and the boundary conditions are approximated by finite difference formulae. These can be derived by referring to the figure below and taking local axes at B:
151
Fig. 1 Grid showing points at which solution calculated
1
2
3
4
z = - z
z = 0
z = + z
A
B
C
u = uA
u = uB
u = uC
Fig. 2 Excess pore water pressure variation at time t
z
u
Suppose that the excess pore pressure at any time t can be approximated by a parabola
(1a)
The constants in this equation can be related to the values of the excess pore pressures at points A, B, C. Taking B as the origin for z gives:
(1b)so that
(1c)
thus evaluating the slope and curvature of u at the point B (z = 0) it is found:
(1d)
11.2 Finite Difference Approximation of Consolidation Equation
The equation of consolidation is:
152
(2a)
where q is the change in total stress, due to applied loads, from the initial equilibrium situation when the excess pore pressures were zero.
When this equation is evaluated at any point in the soil it is equivalent to evaluating the equation at point B, and hence the finite difference formulae developed above can be introduced so the equation becomes:
(2b)
if the above equation is now integrated from times t to t+t it is found that:
(2c)
where
153
F t dt F t tt
t t( ) ( )
t
F(t)
Fig 3 Approximation of integral
Error in approximation
t t t
If the integral appearing in equation (2c) is now approximated as indicated in Fig. 3, it is found that:
(2d)
where
Or(2e)
Suppose the solution for u has been found up to time t. The applied load will be known at time t + t and so the quantity q is known. This means all the quantities on the right hand side of equation (2e) are known and thus that u at time t + t can be calculated. Thus a knowledge of the distribution of u at time t means that the distribution of u at time t + t can be inferred. Now the initial distribution of u can always be
154
Fig. 3 Approximate integral evaluation
determined and thus the solution can be found by ‘marching’ forward in time.
11.3 Stability
There is an important restriction on the use of equation (2e) to obtain a numerical solution of the equation of consolidation, this is
If this condition is violated the calculation becomes unstable and is invalid.
155
11.4 Boundary Conditions
The solution of the equation of consolidation depends on the boundary conditions.
11.4.1 Fully Permeable Boundary
At a free draining boundary there is no impediment to flow and so the pore pressure remains constant and thus the excess pore water pressure is zero, this is illustrated in figure 4a.
B
Saturated soil
Drainage Boundary u=0
Fig. 4a Finite difference approximation of a drainage boundary
11.4.2 Impermeable Boundary
A
B
C
.
.
z
Saturated soil
Impermeable barrier
Fig. 4b Finite difference approximation of an impermeable boundary
.z
156
Fig. 4b Finite difference approximation of an impermeable boundary
Fig. 4a Finite difference approximation of a drainage boundary
At an impermeable boundary, such as that illustrated in figure 4b there can be no flow in a direction perpendicular to the boundary. As outlined earlier this implies:
(3a)the finite difference analogue of this equation is
(3b)and hence
(3c)An impermeable boundary is modelled by equating the excess pore pressure at C to that at A. To do this a dummy node has to be introduced at C into the finite difference grid. This dummy node has no affect other than to give the correct excess pore pressure at the impermeable boundary.Example - Numerical Solution when =1/2
Suppose that a 4m layer of clay, shown in figure 5, which is free to drain at its upper boundary and rests on an impermeable base, is subjected to a surface loading of 64 kPa.
157
q = 64 kPa
Fig . 5 Clay layer subjected to a surcharge loading
Impermeable bedrock
4m 4 sub-layerscv = 2 m2/year
If = 0.5 the finite difference equation takes a particularly simple form:
(4)
In the case under consideration the surcharge is applied at t = 0 and remains constant thereafter so that q = 0
The solution then proceeds as follow:
Step 1: Divide the deposit into layers - this fixes the value of z.
In this case the deposit is divided into 4 sub-layers (all with the same thickness) and thus z = 1m
Step 2: select = 0.5 this fixes the value of t
For the case under consideration:
158
Fig. 5 Clay layer subjected to a surcharge loading
mv = 0.0003 m2/kN
159
Step 3: Calculate the initial pore pressure
Because there cannot be an instantaneous volume change it follows that
Step 4: Introduce the dummy node to simulate the impermeable boundary
Step 5: March the solution forward using the finite difference equation and introducing the boundary conditions
The solution is shown in the table below:
t(years) 0 0.25
0.5 0.75
1 1.25
1.5
q(kPa) 64 64 64 64 64 64 64z=0 64 0 0 0 0 0 0z=1m 64 64 32 32 24 24 20z=2m 64 64 64 48 48 40 40z=3m 64 64 64 64 56 56 48z=4m 64 64 64 64 64 56 56dummy 64 64 64 64 56 56 48
Step 6: Calculate settlement
The settlement is calculated as follows
(5)
160
In the above equation the integral of the excess pore pressure cannot be evaluated exactly because the excess pore pressures are only calculated at the grid points. However, the integral can be evaluated approximately using numerical techniques. The simplest approach, and that used here, is to use the trapezoidal method:
Thus
(6)
Thus after 1.5 years
The settlement at other times can be similarly calculated from the excess pore pressures hence the values can be determined as shown in the table below.
t(years) 0 0.25
0.5 0.75
1 1.25
1.5
q(kPa) 64 64 64 64 64 64 64z=0 64 0 0 0 0 0 0z=1m 64 64 32 32 24 24 20z=2m 64 64 64 48 48 40 40
161
z=3m 64 64 64 64 56 56 48z=4m 64 64 64 64 64 56 56dummy 64 64 64 64 56 56 48Settlement (mm)
0 9.6 19.2
24 28.8
32.4
36
Example - Numerical Solution when 1/2
Suppose now the previous example is solved using a step size of 2 months but keeping the number of layers the same. If this is the case =1/3. The numerical solution proceeds as above but now using the more complex form of the finite difference equation, viz. equation (2e), the solution is shown in the table below:
t(mth’s) 0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
settlement(mm)
0.00
9.60
16.00
20.27
23.82
26.90
29.67
32.20
34.54
36.73
q(kPa) 64.00
64.00
64.00
64.00
64.00
64.00
64.00
64.00
64.00
64.00
z=0 64.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
z=1m 64.00
64.00
42.67
35.56
30.81
27.65
25.28
23.44
21.92
20.61
z=2m 64.00
64.00
64.00
56.89
52.15
48.20
45.04
42.32
39.92
37.74
z=3m 64.00
64.00
64.00
64.00
61.63
59.26
56.63
53.99
51.39
48.86
z=4m 64.00
64.00
64.00
64.00
64.00
62.42
60.31
57.85
55.28
52.68
162
dummy 64.00
64.00
64.00
64.00
61.63
59.26
56.63
53.99
51.39
48.86
For z = 3 m at 12 months the calculations are
0 + 59.26 + 0.3333 [48.20 + 62.42 - 2 59.26] = 56.63
The results for the two analyses are quite close. After 18 months the settlement predicted in example 1 is 36 mm, which compares well with the settlement calculated in example 2, viz. 36.7mm.
Example - Variable loading
Suppose fill having unit weight 20 kN/m3 is placed at a rate of 0.5 m/month for 12 months after which no more load is applied, the analysis only differs from that in the previous examples in that the value of q needs to be included in the finite difference equation. Choosing = 0.5 with 4 layers gives a time step of 0.25 years as before. The results are shown in the table below.
t(years) 0 0.25
0.5 0.75
1 1.25
1.5
settlement(mm)
0 4.5 13.5
24.75
38.25
48.938
56.813
q(kPa) 0 30 60 90 120 120 120z=0 0 0 0 0 0 0 0z=1m 0 30 45 60 71.
2552.5
46.875
z=2m 0 30 60 82.5
105 93.75
82.5
z=3m 0 30 60 90 116 112 105
163
.25 .5z=4m 0 30 60 90 120 116
.25112.5
dummy 0 30 60 90 116.25
112.5
105
Note that when the load is applied gradually the excess pore pressure at the permeable upper boundary remains at zero. This is because there is no instantaneous change in load.
If the calculation is repeated for the case in which there are 5 sub-layers, and a time step of 0.1 years is adopted, this gives = 0.3125 and the results are shown below:
t (years) 0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
settlement(mm)
0.00
1.44
3.78
6.74
10.21
14.13
18.45
23.13
28.16
33.50
39.15
q(kPa) 0.00
12.00
24.00
36.00
48.00
60.00
72.00
84.00
96.00
108.00
120.00
z=0 0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
z=0.8m 0.00
12.00
20.25
27.09
33.04
38.38
43.27
47.80
52.04
56.05
59.85
z=1.6m 0.00
12.00
24.00
34.83
44.78
54.00
62.64
70.78
78.50
85.86
92.90
z=2.4m 0.00
12.00
24.00
36.00
47.63
58.86
69.66
80.07
90.11
99.81
109.18
z=3.2m 0.00
12.00
24.00
36.00
48.00
59.89
71.60
83.10
94.35
105.33
116.04
z=4.0m 0.00
12.00
24.00
36.00
48.00
60.00
71.93
83.72
95.33
106.72
117.85
dummy 0.00
12.00
24.00
36.00
48.00
59.89
71.60
83.10
94.35
105.33
116.04
164
t (years) 1 1.1 1.2 1.3 1.4 1.5settlement(mm)
39.145
43.6
47.495
50.997
54.236
57.269
q(kPa) 120 120
120 120 120 120
z=0 0 0 0 0 0 0z=0.8m 59.
85151.5
46.697
43.186
40.441
38.191
z=1.6m 92.904
87.7
82.158
77.59
73.681
70.29
z=2.4m 109.18
106
103 99.486
96.068
92.804
z=3.2m 116.04
114
112.6
110.44
108.01
105.41
z=4.0m 117.85
117
115.31
113.61
111.63
109.37
dummy 116.04
114
112.6
110.44
108.01
105.41
Again the settlements at 1.5 years are quite similar. Thus although greater refinement of the grid leads to more accurate excess pore pressures and settlements there is, in practice, little advantage of using 0.5.
165
Example - Abrupt change of load
Suppose that in the case detailed in example 1 a further surcharge of 32 kPa is added after 12 months. The solution in this case is best handled in two stages.
Stage 1 follows exactly the path outlined in example 1 and is detailed in the table below. Just after 1 year the load is abruptly increased, and since there can be no instantaneous volume strain there can be no increase in effective stress and no change in settlement. This means that the increase or decrease in applied stress must be matched by a corresponding increase or decrease in pore water pressure. This enables the excess pore water pressure to be calculated.
t 0 0.25
0.50
0.75
1.00
settlement(mm)
0 9.6 19.2
24 28.8
q(kPa) 64
64.00
64.00
64.00
64.00
z=0 64
0.00
0.00
0.00
0.00
z=1m 64
64.00
32.00
32.00
24.00
z=2m 64
64.00
64.00
48.00
48.00
z=3m 64
64.00
64.00
64.00
56.00
z=4m 64
64.00
64.00
64.00
64.00
dummy 64
64.00
64.00
64.00
56.00
166
Stage 2 of the calculation then proceeds in the same way as in stage 1 or in example 1. This is shown in the table given below:
t 1 1.25
1.5 1.75
2 2.25
2.5
settlement(mm)
28.8
37.2
45.6
51 56.4
60.75
65.1
q(kPa) 96 96 96 96 96 96 96z=0 32 0 0 0 0 0 0z=1m 56 56 36 36 29 29 24.
5z=2m 80 72 72 58 58 49 49z=3m 88 88 80 80 69 69 59z=4m 96 88 88 80 80 69 69dummy 88 88 80 80 69 69 59
12. SETTLEMENTS OF STRUCTURES
12.1 The settlement process
167
An important task in the design of foundations is to determine the settlement, this is shown schematically in Figure 1.
MaximumSettlement
Soil Layer
Fig. 1 Settlement of a loaded footing
As discussed earlier the skeletal soil material and the pore water are relatively incompressible and any change in volume can only occur due to change in the volume of the voids. For the volume of the voids to change, pore water must flow into or out of a soil element. Because this cannot happen instantaneously when a load is first applied to a soil there cannot be any immediate change in its volume. For one-dimensional conditions with no lateral strain this implies that there is no immediate vertical strain and hence that the excess pore pressure is equal to the change in vertical stress. However, under more general conditions both lateral (or horizontal) and vertical strains can occur. Immediately after load is applied there will be no change in volume, but the soil deformations will result in an initial settlement. This is said to occur under undrained conditions because no pore water has been able to drain from the soil. With time the excess pore pressures generated during the undrained loading will
168
Fig. 1 Settlement of a loaded footing
dissipate and further lateral and vertical strains will occur. Ultimately the settlement will reach its long term or drained value.
When the load is first applied to the soil there will be a tendency for the more highly stressed parts of the soil to compress and thus for there to be a reduction in the volume of the voids. The pore water will respond to this tendency towards a decrease in volume by undergoing an increase in pore water pressure and so initial excess pore water pressures will develop. Subsequently there will be a flow of water from regions of high excess pore water pressure to regions of low excess pore water pressure, and the load induced excess pore water pressures will dissipate. This is the process of consolidation, and during this process the soil will undergo a settlement which varies with time. Ultimately after a long period of time all the excess pore water pressures will have dissipated and the settlement of the soil will cease and it will reach its long term or drained settlement (the term drained is used because all excess pore water pressures have dissipated and there will be no further drainage of water from the voids although the voids will still remain saturated). The process of consolidation is shown schematically in Figure 2. It should be stated that the process described above represents a simplification because some soils tend to creep. For such soils there will be additional creep settlements even though the effective stress does not change.
169
TotalStress
Time
Time
ExcessPorePressure
EffectiveStress
Time
Fig. 2a Variation of stress and pore pressure at a typical point under a footing
Settlement
Time
Consolidationsettlement
Initialsettlement
Finalsettlement
Fig. 2b Variation of settlement with time
12.2 Analysis of Settlement under three dimensional conditions
Previously the settlement under foundations has been estimated assuming purely one-dimensional conditions. However, it is clear from consideration
170
of the stress changes (predicted by the theory of elasticity) under the centre and edges of various loaded areas that in general the stress changes may differ significantly from those deduced using the purely one-dimensional assumption.If it is hypothesised that the soil can be treated as a linear isotropic elastic material then solutions for the settlement can be obtained using the theory of elasticity. This assumption involves a considerable level of approximation which is necessary because:
real soil behaviour is highly non-linear the geometry of the foundation is often complex simple models enable calculations to be easily
performed
Linear isotropic elasticity is used because:
closed form solutions which are easily evaluated can be obtained
complicated loadings can be synthesised from simple components using superposition
only 2 material constants are required from (E, , G, K)
the solutions obtained agree with intuition and experience
12.3 Theory of Elasticity for Saturated Soils
In an isotropic elastic solid it is found that Hooke’s law relates the changes in stress to the changes in strain as described in equation (1):
171
(1a)
where xx , yy , zz denote the strains which arise from the changes in stress xx , yy , zz and where E is Young’s modulus and is Poisson’s ratio.
Hooke’s law in this form does not apply to soil except for undrained conditions which will be discussed later. For soil the correct relationship is one between effective stress and strain as shown below:
(1b)
where E´ is called the effective stress, or drained, Young’s modulus and ´ is called the effective stress, or drained, Poisson’s ratio, and where the increments of effective stress are related to the increments of total stress and the increment of pore water pressure by:
(1c)
172
The relationship between effective stress and strain can always be used to calculate the deformation of soils. However, to do so it is necessary to know both the change in total stress and the change in pore water pressure. The change in total stress can usually be estimated using elastic solutions, but the change in pore pressure is, in general, very difficult to determine.
One important case where the effective stresses are known is in the long term. In this situation all excess pore water pressures have dissipated and thus the change in effective stress is equal to the change in total stress. The settlement can then be calculated using the effective stress, strain relations.
Equations (1b) can be modified as follows:
(2)
this alternative form of Hooke’s law is useful as will be seen below.
12.4 Behaviour of an elastic soil under undrained conditions
It was shown above that the long term behaviour of soil can be analysed using Hooke’s law since all
173
excess pore pressures have dissipated and so the effective stress equals the total stress. Another important case which can be analysed using Hooke’s law is immediately after loading when no water has drained out of the soil pores and no excess pore pressures have dissipated, i.e. undrained behaviour. To establish this note that under such conditions there can be no volume change and thus:
(3a)
The volume strain can be calculated using equations (2) and (3a) giving:
(3b)
If the volume strain is zero the change in mean effective stress is zero and thus:
(3c)
This enables the increment in excess pore water pressure to be expressed in terms of the total stress. Using this relation and substitution into equation (2) leads to the following relation between total stress and strain:
174
(4)
This and similar expressions for yy and zz are equivalent to Hooke’s law for undrained loading, which may be written as:
(5)
The quantities Eu , and u are called the undrained Young’s modulus and Poisson’s ratio respectively. By comparing equations (4) and (5) it can be seen that these quantities are related to the drained or effective stress relations as follows:
(6)
It is interesting to note that so far there has been no mention of shear behaviour, for shear stresses and strains Hooke’s law may be written as:
(7)
175
where G´ is a material property called the shear modulus which is related to the effective stress parameters as follows.
(8)
It is interesting to observe that:
(9)
Showing that the shear modulus (and shear strain) is unaffected by the state of drainage in the soil.
It is important to emphasise that the relation between effective stress parameters and undrained parameters is based on many approximations (soil assumed elastic) and should not be expected to be exact. Thus, although the undrained value of Poisson’s ratio will be precisely 1/2 for a saturated soil because of incompressibility, the undrained Young’s modulus should be measured directly rather than determined from the effective E´ value.
12.5 Values of the Elastic Parameters for soils
The selection of parameters to use in elastic analyses of settlement prediction presents considerable difficulties in geotechnical engineering. Soil is not a linear elastic material. In
176
selecting values for the "elastic" parameters consideration must be given to:
The initial effective stresses in the ground.
The values of E´,´ are both dependent on the
mean effective stress, , with the moduli
increasing with stress level.
The soil stress history
OCR for clays Relative density (Id) for sands For a given stress level, the moduli will increase
with increasing OCR or Id
The strain level
It is advisable to use an appropriate secant modulus for the expected strain level under the footing.
12.5.1 Values of E'
Typical values may be selected from the following values given in the data sheets (p. 65)
Soft normally-consolidated clays ( 1400 - 4200 kPa)
Medium clays ( 4200 - 8400 kPa)
Stiff clays ( 8400 - 20000 kPa)
177
Loose normally-consolidated sands ( 7000 - 20000 kPa)
Medium normally-consolidated sands(20000 - 40000 kPa)
Dense normally-consolidated sands(40000 - 84000 kPa)
For over-consolidated sands, double the above values.
12.5.2 Values of v'
Soft clay 0.35 - 0.45
Medium clay 0.30 - 0.35
Stiff Clay 0.2 - 0.3
Medium sand 0.3 - 0.35
These typical values should be used with caution. Soils are extremely variable materials and considerable expertise is needed to determine accurate parameters.
Example - Strains during undrained loading
A cuboidal soil specimen is in equilibrium with a uniform stress acting on all faces of 100 kPa, and no pore pressure, that is u = 0. The vertical stress is then increased by 90 kPa with the stresses on the other faces remaining constant and with the
178
sample prevented from draining. Calculate the vertical and lateral strains if E´ = 10 MPa and ´ = ¼.
Initially: 1 = 2 = 3 = 100 kPa; u = 0
Analysis of undrained loading can be performed in terms of undrained parameters (Total Stress Analysis) or drained parameters (Effective Stress Analysis).
1. Total Stress Analysis
Calculate undrained parameters u = 0.5,
MPa
Now the total stress changes are xx = 0 kPa, yy = 0 kPa, zz = 90 kPa
Use Hooke’s Law in terms of Total Stress
Hence
zz = 90/12000 = 0.0075
xx = yy = - 0.5 0.0075 = - 0.00375
2. Effective stress analysis
179
Changes in effective stress are needed to evaluate the effective Hooke’s Law relations.
Calculate u = m for undrained loading (see above)
=
= 90/3 = 30 kPa
Hence ´xx = - 30 kPa, ´yy = - 30 kPa, ´zz
= 60 kPa
Now using Hooke’s Law
giving the same result as before.
Example – Strains during drained loading
If the same sample from example 1 is now allowed to drain and consolidate, without any change to the applied stresses, what strains will develop.
Only an effective stress analysis is relevant. Total stress analysis cannot be used because the total stress parameters (Eu, u) are only relevant to undrained loading, that is when deformation occurs at constant volume.
180
In this example during consolidation the total stresses remain constant. The effective stress changes are thus ´xx = + 30 kPa, ´yy = + 30 kPa, ´zz = + 30 kPa, they are all equal to the reduction in pore water pressure. Then from Hooke’s Law
xx = yy = zz = 0.0015
Note that the total strains due to the undrained loading followed by consolidation are
xx = yy = - 0.00375 + 0.0015 = -0.00225
zz = 0.0075 + 0.0015 = 0.009
The same total strains are obtained if the load is applied slowly so that no pore pressures are obtained. In this case the pore pressure change is zero and hence the change in total stress is the same as the change in effective stress (´xx = 0 kPa, ´yy = 0 kPa, ´zz = 90 kPa). The strains are then given by
Note that the strains are identical to those determined as a result of undrained loading followed by consolidation. This result is not surprising when it is remembered that this is an elastic analysis.
181
182
13. SETTLEMENT OF STRUCTURES
13.1 Solutions based on the theory of elasticity
Figure 1 represents a surface footing resting on a soil layer of depth H.
Soil Layer
Rigid bedrock
H
P
Fig. 1 Foundation resting on a soil layer
The settlement, s, of any point can be determined from
183
Fig. 1 Foundation resting on a soil layer
(1a)
where for an elastic soil
(1b)
and under undrained conditions:
(1c)
As discussed earlier, to determine the settlement immediately after the application of the load equation (1c) is used, and to determine the long term or drained settlement equation (1b) is used. In the latter case the changes in pore water pressure u are usually zero and so the increment in effective stress is equal to the increment in total stress. Thus, in both cases the settlement can be calculated if both the change in total vertical stress zz and the change in the mean total stress (xx+ yy+ zz ) are known.
It has been shown previously how the Boussinesq solution for the stresses in an elastic half space due to a point load acting on the surface can be used to determine the stress distributions under a variety of shapes of loaded areas (circles, rectangles, arbitrary shapes). The same solution can be used to determine the surface settlements, sr as a function of the distance, r, from a point load Q, as
184
(2)
This is illustrated in Figure 2.
Qr
sr
Fig. 2 Surface deflection of a deep elastic layer
sQ
Err ( )1 2
Because the soil is assumed to be linear elastic it is possible to use superposition to determine the surface settlements for distributed loads using the point load solution. For example, the settlement at the centre of a circular loaded area, radius, a, with uniform stress, q, (flexible foundation), can be determined by considering the effect of the stress, q, acting over an area r d dr (shown in Figure 3) on the settlement at the centre. The settlement is then given by:
185
Fig. 2 Surface deflection due to a point load on a deep elastic layer
H
dr
r
dd
Fig. 3 Stress q acting over a circular area of radius a
(3)
For other positions under the circular load and for other shapes the integration is not so straightforward, and in many cases analytical solutions will not be possible.
Also a limitation of this (Boussinesq) solution is that it assumes the soil layer is infinitely deep. This rarely occurs in practice as more generally a relatively shallow soil layer usually overlies rock.
The procedure adopted in practice is to make use of charted solutions that are available for a number of commonly encountered situations. Some of these are given in the data sheets, and are discussed below. For other solutions the book
186
a
"Elastic solutions for Soil and Rock Mechanics" by Poulos and Davis should be referred to.
13.2 Settlement under a rigid circular load
Soil Layer
Rigid bedrock
h
P a pav 2
rigid
2a
Fig. 4a Rigid circular footing on an elastic layer on a rigid base
The configuration being considered is shown in Figure 4a and the solution is presented in terms of a settlement factor, I. The settlement, s, is given by the expression:
(4)
where
Pav is the average stress on the footing = Load/Area = P/(a2)
a is the radius of the loaded area
187
E is the soil modulus
I is a settlement factor read from Figure 4b (Data Sheets page 45). Note that I depends on the value of Poisson’s ratio .
188
0.0 0.2 0.4 0.6 0.8 1.0 0.8 0.6 0.4 0.2 0.0h/a a/h
1.6
1.2
0.8
0.4
0.0
Fig. 3b Settlement Factor for rigid circular footing on a layer
2a
P a pav 2
h
sp a
EIav
0.00.2
0.40.5
I
Example
Determine the final settlement under a footing 3 m in diameter which is subjected to a load of 500 kN if it rests on a soil layer 9 m thick with properties E' = 5 MPa, v' = 0.3.
13.3 Settlement of square footings
The settlement under a square footing can be estimated to sufficient accuracy by considering the load to act over an equivalent circular area. So if the square footing has sides of length b the
189
Fig. 4b Settlement factors for a rigid circular footing on a soil layer
following equivalent pressure and radius can be used in equation 4:
13.4 Settlement of a Circular Foundation on a non-homogeneous soil
Soils often have a modulus that increases with depth. The soil does not necessarily change its nature with depth, the reason for the increase in modulus is that the mean effective stress increases with depth and, because the modulus increases with the mean effective stress so the modulus varies with depth. Often the variation with depth is approximately linear and so can be approximated by the relation:
(5)
The modulus increases linearly from E0 at the surface as shown schematically in Figure 5..
E mz0
2a
P a pav 2
Fig. 4 Circular footing on a non-homogeneous soil
190
Fig. 5 Circular footing on non-homogeneous soil
p
A charted solution is available for this modulus variation for the case of a flexible circular footing (p constant) resting on an infinitely deep soil layer. The settlement may be expressed in the form:
(6)where I is the influence factor given in Figure 6 (Data Sheets p 47) and
p is the stress on the footinga is the radius of the loaded areaE0 is the Young's modulus at the surface
E mz0
2a
21
10--1
10--2
10--3
10--4
10---4 10---2 1 102
I
0
1/3
1/2
E
ma0
p
Example
An oil tank applies a uniform stress of 75 kPa over a circular area with diameter 20 m. Calculate the immediate settlement if the undrained modulus increases linearly from 2 MPa at the surface, to 5 MPa at 10 m.
191
Fig. 6 Influence chart for flexible circular load on non-homogeneous soil
13.5 Settlement under the edge of a flexible strip load on a finite soil layer
The configuration is shown in Figure 7a. The settlement at the edge takes the form:
(7)
where I is the influence factor given in Figure 7b (Data sheets p 46) and
p is the stress on the strip footing
h is the depth of the soil layer
E is the Young's modulus of the soil.
The value of the settlement at other locations can be found by superposition, as demonstrated below.
192
For a rigid strip footing the settlement can be estimated by averaging the centre and edge settlements of an equivalent flexible footing.
Soil Layer
Rigid bedrock
h
B
p
Fig. 7a Flexible strip on an elastic layer on a rigid base
2.0
1.6
1.2
0.8
0.4
0.00.0 0.25 0.5 h/B
2.0 1.0 0.0B/h
I
0.0
0.2
0.4
0.5
Fig. 7b Settlement factor for edge of flexible strip on a soil layer
193
Example
Determine the final settlement at a point 10 m from the centre of a 16 m wide embankment, assuming that the embankment can be considered as a flexible strip load which applies a surface stress of 50 kPa. The embankment is constructed on a soil layer 15m deep with the properties E´= 9 MPa, ´= 0.3.
Because of the assumption of elasticity superposition can be used. Thus the embankment loading can be simulated as shown in Figure 8.
8m 10m
Embankment
18m
(+)
2m
(-)
15m
2m
Fig. 8 Decomposition of embankment loading to give settlement not under edge
The embankment loading consists of a strip loading of intensity +50 kPa and width of 18 m for which:
194
and a strip loading of intensity -50 kPa and width of 2 m for which:
Thus the settlement is:
195
13.6 The influence of embedment on settlement
If a footing is embedded the settlement will be reduced. Two cases are shown in Figure 9a for which some solutions are available, both for a very deep elastic layer. The settlement reduction factors are given in Figure 9b (Data sheets p 48). To use these solutions the settlement must be found using the previously derived solutions for the load resting on the surface.
D
Z
(a) Uniform circular load at the base of an unlined shaft
D
Z
(b) Uniform circular load within a deep elastic layer
Fig. 9a Loads applied below the surface in a deep elastic layer
196
Set
tlem
ent
of a
dee
p lo
adS
ettl
emen
t of
an
iden
tica
l su
rfac
e lo
ad
1.0
0.9
0.8
0.7
0.6
0.5
0 5 10 15 20Z/D
0.50 (b)
0.25 (a)
0.49 (a)
0.00 (a)
Set
tlem
ent
of a
dee
p lo
adS
ettl
emen
t of
an
iden
tica
l su
rfac
e lo
ad
1.0
0.9
0.8
0.7
0.6
0.5 0.50 (b)
0.25 (a)
0.49 (a)
0.0 (a)
Fig. 9b Depth reduction factors for embedded circular footings
197
13.7 Selection of Elastic parameters
The settlement of any foundation can be split into 3 components
13.7.1 Immediate or undrained settlement
This component is due to deformations in the soil immediately after loading. As has been discussed previously, immediately after load is applied water has no time to drain out of the voids and so there is no volume change. Hence any deformation must occur at constant volume.
In practice deformation at constant volume only occurs for relatively impermeable clayey soils that remain undrained in the short term.
To estimate the initial settlement, si, due to the constant volume deformation the undrained (total stress) parameters Eu, u = 1/2 are used in the analyses described above.
As observed earlier when the load is applied over a very large area the situation approaches one-dimensional conditions, for which the initial undrained settlement is zero.
In principle effective stress parameters could be used to determine the settlement, but because the excess pore pressures generated by the load vary throughout the soil the analysis is not straightforward, and the simple elastic formulae cannot be used.
13.7.2 Consolidation Settlement
198
This is due to deformations arising from volume changes which occur as a consequence of the excess pore water pressures, which have been generated immediately after loading, dissipating allowing the effective stresses to come into equilibrium with the applied loads. Finally all excess pore water pressures will have dissipated and the final settlement, stf, can be determined by using E', v' in the settlement formulae developed previously.
The settlement due to consolidation, sc, can be determined indirectly from the final settlement stf, and the immediate settlement, si, by:
(8)
13.7.3 Creep deformations at constant load.
Settlements due to creep cannot be predicted using the simple elastic formulae, and are usually only significant for soft soil sites.
13.8 Calculation of the settlement at any time
For relatively impermeable clayey soils, in the short term undrained deformations occur. It is normally assumed that construction occurs sufficiently quickly so that no drainage occurs, and the settlement at the end of construction is then the immediate settlement si. For sandy soils, the total final settlement is reached in the short term and there is no time dependent response, thus it is assumed that consolidation is instantaneous. Note
199
that there will be soils that have intermediate properties, and the initial settlement will be partly drained. The extent of the drainage (consolidation) will depend on the boundary conditions and the coefficient of consolidation.
For clayey soils the time settlement behaviour can be visualised as shown in Figure 10
Load
Time
Constructiontime
Settlement
Time
Consolidationsettlement sc
Initialsettlement si
Total finalsettlement sTf
Const.time
Fig. 10 Components of settlement
The settlement at any time t can then be calculated from the three components described above and it is found that:
(9a)
where U is called the degree of consolidation
200
Solutions for U versus T for a variety of boundary conditions are given in the Data Sheets, pages 50 - 58. In general these charts use the non-dimensionalised time factor T given by cv t / h2,
where h is the thickness of the soil layer irrespective of the boundary conditions (Note that this is different from the definition used for 1-D consolidation). Solutions are given for the following boundary conditions:
PTPB Permeable base, permeable top boundary and permeable footing.
PTIB Impermeable base, permeable top boundary and permeable footing. IFIB Impermeable base, permeable top boundary and impermeable footing.
IFPB Permeable base, permeable top boundary and impermeable footing.
Example
Determine the immediate settlement, the final settlement, and the settlement 1 year after the end of construction of a rigid circular footing 5 m in diameter which supports a load of 1.5 MN, and is founded on a 5 m thick clay layer overlying gravel. The clay layer has the following uniform
201
(9b)
properties: E' = 5 MPa, v' = 0.2, cv = 0.5 m2/yr and Eu = 6.25 MPa.
Step 1 Calculation of the Initial Settlement
Using Figure 4 and
The immediate settlement can now be calculated using:
Step 2 Calculation of the Final Settlement
Using Figure 4 and
It thus follows that
202
Step 3 Calculation of Settlement after 1 year
(a) For the case of an Impermeable footing (IFPB)
The consolidation settlement sc = (36.29-19.25) mm= 17 mm
The degree of consolidation can be determined from Figure 11, thus for:
10-5 10-4 10-3 10-2 10-1 1
0.0
0.2
0.4
0.6
0.8
1.0
U
h/a=50 20 10 5 2 1
0
0.5
Tc t
hv2
203
Fig. 11 Consolidation response for circular footing - case IFPB
It is found that U=0.35
This leads to a settlement after 1 year of:
204
(b) For the case of a Permeable Footing (PTPB)
10-4 10-3 10-2 10-1 1
Tc t
hv2
0.0
0.2
0.4
0.6
0.8
1.0
U
h/a=5020
105
21
0.5
0
Fig. 12 Consolidation response for circular footing - case PTPB
The degree of consolidation can be determined from Figure 12, and it is found that U=0.5 and so the settlement after 1 year is:
205
14. SOIL STRENGTH
Soils are essentially frictional materials. They are comprised of individual particles that can slide and roll relative to one another. In the discipline of soil mechanics it is generally assumed that the particles are not cemented.
One consequence of the frictional nature is that the strength depends on the effective stresses in the soil. As the effective stresses increase with depth, so in general will the strength.
The strength will also depend on whether the soil deformation occurs under fully drained conditions, constant volume (undrained) conditions, or with some intermediate state of drainage. In each case different excess pore pressures will occur resulting in different effective stresses, and hence different strengths. In assessing the stability of soil constructions analyses are usually performed to check the short term (undrained) and long term (fully drained) conditions.
14.1 Mohr-Coulomb failure criterion
The limiting shear stress that may be applied to any plane in the soil mass is found to be given by an equation of the form
= c + n tan
206
where c = cohesion (apparent) = friction angle
This is known as the Mohr-Coulomb failure criterion
The parameters c and are not generally soil constants. The Mohr-Coulomb criterion is an empirical criterion, and the failure locus is only locally linear. Extrapolation outside the range of normal stresses for which it has been determined is likely to be unreliable. The parameters depend on:
the initial state of the soilOverconsolidation ratio (OCR) for claysRelative density (Id) for sands
the type of test Drained - slow fully drained, no excess pore water
pressuresUndrained - no drainage, excess pore water pressures
develop
the use of total or effective stresses
In terms of effective stress the failure criterion is written
= c + n tan
c and are referred to as the effective (drained) strength parameters.
Soil behaviour is controlled by effective stresses, and the effective strength parameters are the fundamental strength parameters. But they are not necessarily soil constants. They are fundamental in the sense that if soil is at failure the state will always be described by an effective stress failure criterion. The parameters can be determined from any test provided that the pore pressures are known.
207
In terms of total stress the failure criterion is written
= cu + n tan u = su
cu, u are referred to as the undrained (total) strength parameters. These parameters can only be determined from undrained tests.
The undrained strength parameters are not soil constants, they depend strongly on the moisture content of the soil. The total stress criterion has limited applicability as it is only valid if soil deformation occurs without drainage.
The undrained strengths measured in the laboratory are only relevant in practice to clayey (low permeability) soils that initially deform without drainage, and that have the same moisture content in-situ.
14.2 Strength Tests
The engineering strength of soil materials is often determined from tests in either the shear box apparatus or the triaxial apparatus.
14.2.1 The Shear Box Test
The soil is sheared along a predetermined plane by placing it in a box and then moving the top half of the box relative to the bottom half. The box may be square or circular in plan and of any size, however, the most common shear boxes are square,
208
60 mm x 60 mm, and test
Motor drive
Load cell to measure Shear Force
Normal load
Rollers
Soil
Porous plates
Top platen
209
A load normal to the plane of shearing may be applied to a soil specimen through the lid of the box. Provision is made for porous plates to be placed above and below the soil specimen. These enable drainage to occur which is necessary if a specimen is to be consolidated under a normal load, and if a specimen is to be tested in a fully drained state. The soil specimen may be submerged, by filling the containing vessel with water, to prevent the specimens from drying out. Undrained tests may be carried out, but in this case solid spacer blocks rather than the porous disks must be used.
Notation
N = Normal ForceF = Tangential (Shear) Force
n = N/A = Normal Stress = F/A = Shear Stress
A = Cross-sectional area of shear planedx = Horizontal displacementdy = Vertical displacement
Usually only relatively slow drained tests are performed in shear box apparatus. For clays the rate of shearing must be chosen to prevent excess pore pressures building up. For freely draining sands and gravels tests can be performed quickly. Tests on sands and gravels are usually performed dry as it is found that water does not significantly affect the (drained) strength.
Provided there are no excess pore pressures the pore pressure in the soil will be approximately zero and the total and effective stresses will be identical. That is, n = ´n
210
The failure stresses thus define an effective stress failure envelope from which the effective (drained) strength parameters c´, ´ can be determined.
Typical test results
At this stage we are primarily interested in the stresses at failure. It is observed that for a set of initially similar soil
211
Shear load(F)
Horizontal displacement (dx)
n = ´n
samples there is a linear failure criterion that may be expressed as
= c + n tan
From this the effective (drained) strength parameters c and can be determined.
A peak and an ultimate failure locus can be obtained from the results each with different c´ and ´ values. All soils are essentially frictional materials and continued shearing results in them approaching a purely frictional state where c 0. Normally consolidated clays (OCR =1) and loose sands do not usually show peak strengths and have c = 0, whereas, overconsolidated clays and dense sands have c > 0. Note that dense sands (OC clays) do not possess any true cohesion (bonds), and the apparent cohesion results from the tendency of soil to expand when sheared.
As a soil test the shear box is far from ideal. Disadvantages of the test include:
Non-uniform deformations and stresses. The stresses determined may not be those acting on the shear plane, and no stress-strain curve can be obtained. There are no facilities for measuring pore pressures in the shear box and so it is not possible to determine effective stresses from undrained tests. The shear box apparatus cannot give reliable undrained strengths because it is impossible to prevent localised drainage away from the shear plane.
However, it has many apparent advantages: It is easy to test sands and gravels Large deformations can be achieved by reversing the shear box. This involves pushing half of the box
212
backwards and forwards several times, and is useful in finding the residual strength of a soil. Large samples may be tested in large shear boxes. Small samples may give misleading results due to imperfections (fractures and fissures) or the lack of them. Samples may be sheared along predetermined planes. This is useful when the shear strengths along fissures or other selected planes are required.
In practice the shear box is used to get quick and crude estimates of the failure parameters. It is sometimes used to obtain undrained strengths but this use should be discouraged.14.2.2 The Triaxial Test
The triaxial test is carried out in a cell and is so named because three principal stresses are applied to the soil sample. Two of the principal stresses are applied to the sample by a water pressure inside the confining cell and are equal. The third principal stress is applied by a loading ram through the top of the cell and therefore may be different to the other two principal stresses. A diagram of a typical triaxial cell is shown below.
213
A cylindrical soil specimen as shown is placed inside a latex rubber sheath which is sealed to a top cap and bottom pedestal by rubber O-rings. For drained tests, or undrained tests with pore pressure measurement, porous disks are placed at the bottom, and sometimes at the top of the specimen. For tests where consolidation of the specimen is to be carried out, filter paper drains may be provided around the outside of the specimen in order to speed up the consolidation process.
Pore pressure generated inside the specimen during testing may be measured by means of pressure transducers. These transducers must operate with a very small volume change, since fluid flowing out of the specimen would cause the pore water pressure that was being measured to drop.
214
Porous disc
Porous disc
Rubber membrane
Water supply to cell
Water supply to soil sample
14.2.2.1 Stresses
r r = Radial stress (cellpressure)
a = Axial stress
F = Deviator loadr
From vertical equilibrium we have a rF
A
The term F/A is known as the deviator stress, and is usually given the symbol q.
Hence we can write q = a - r = 1 - 3 (The axial and radial stresses are principal stresses)
If q = 0 increasing cell pressure will result in:
volumetric compression if the soil is free to drain. The effective stresses will increase and so will the strength
increasing pore water pressure if soil volume is constant (that is, undrained). As the effective stresses cannot change it follows that u = r
Increasing q is required to cause failure
14.2.2.2 Strains
From the measurements of change in height, dh, and change in volume dV we can determine
215
u
Axial strain a = -dh/h0
Volume strain v = -dV/V0
where h0 is the initial height, and V0 the initial volume. The conventional small strain assumption is generally used.
It is assumed that the sample deforms as a right circular cylinder. The cross-sectional area, A, can then be determined from
1It is important to make allowance for the changing area when calculating the deviator stress,
q = 1 - 3 = F/A
14.2.2.3 Test procedure
There are many test variations. Those used most in practice are
UU (unconsolidated undrained) test.Cell pressure applied without allowing drainage. Then
keeping cell pressure constant increase deviator load to failure without drainage.
CIU (isotropically consolidated undrained) test.Drainage allowed during cell pressure application. Then
without allowing further drainage increase q keeping r constant as for UU test.
CID (isotropically consolidated drained) test
A = A
1 + dV
V
1 + dh
h
= A 1 -
1 - o o
v
a
0
0
216
Similar to CIU except that as deviator stress is increased drainage is permitted. The rate of loading must be slow enough to ensure no excess pore pressures develop.
As a test for investigating the behaviour of soils the triaxial test has many advantages over the shear box test:
Specimens are subjected to uniform stresses and strains
The complete stress-strain behaviour can be investigated
Drained and undrained tests can be performed
Pore water pressures can be measured in undrained tests
Different combinations of confining and axial stress can be applied
217
Typical results from a series of drained tests consolidated to different cell pressures would be as follows.q
a
The triaxial test gives the strength in terms of the principal stresses, whereas the shear box gives the stresses on the failure plane directly. To relate the strengths from the two tests we need to use some results from the Mohr circle transformation of stress.
13
14.3 Mohr Circles
The Mohr circle construction enables the stresses acting in different directions at a point on a plane to be determined, provided that the stress acting normal to the plane is a principal stress. The Mohr circle construction is very useful in Soil
218
Increasing cell pressure
Mechanics as many practical situations can be approximated as plane strain problems.
The sign convention is different to that used in Structural Analysis because for Soils it is conventional to take the compressive stresses as positive.
Sign convention: Compressive normal stresses are positiveAnti-clockwise shear stresses are positive (from
inside soil element)Angles measured clockwise positive
Let us consider the stresses acting on different planes for an element of soil
(a) (b)
(a) shows the stresses on a plane at angle to the minor principal stress, and (b) shows the relevant lengths.
Now resolving forces gives
l l l 1 3sin sin cos cos
1 3
21 2
21 2( cos ) ( cos )
1 3 1 3
2 22cos
and similarly
1 3
22sin
219
3
1
l
l sin
l cos
which define the Mohr circle relation
13
2
(
From the Mohr Circle we have
= p - R cos 2 = R sin 2
where
2
3
and failure occurs on a plane at an angle from the plane on which 3 acts, and
4 2
14.4 Mohr-Coulomb Failure Criterion (Principal stresses)
Failure will occur when we can find any direction such that
p = ( + )
2 =
( + )
21 3 xx zz
R = ( - )
2 =
1
2 ( - ) + 4 1 3
xx zz2
zx2
220
R
p
c + tan
13
c
c cot p
R
At failure from the geometry of the Mohr Circle
R = sin (p + c cot ) = p sin + c cos
4
5
14.4.1 Mohr-Coulomb Failure Criterion for Saturated Soil
As mentioned above it is the effective strength parameters c, that are the fundamental soil strength parameters. To use these parameters the Mohr-Coulomb criterion must be expressed in terms of effective stresses, that is
= c + n tan
1 = N 3 + 2 c N
1
3
2 + c
+ c =
1 +
1 - =
4 +
2 = N
cot
cot
sin
sintan
1 3 = N + 2 c N
221
with N
1
1
sin
sin
and the effective stresses are given byn = n - u - u 3 - u
Note that the difference between the total and effective stresses is simply the pore pressure u. Thus the total and effective stress Mohr circles have the same diameter and are displaced along the axis by the value of the pore pressure.
14.5 Interpretation of Laboratory Data
It is helpful to distinguish between drained and undrained loading.
14.5.1 Drained loading
In drained laboratory tests the loading rate is sufficiently slow so that all excess pore water pressures will have dissipated. From the known pore water pressures the effective stresses can be determined.
The behaviour of drained tests must be interpreted in terms of the effective strength parameters c, , using the effective stresses. It is possible to construct a series of total stress Mohr Circles but the inferred total strength parameters have no relevance to the soil behaviour.
The effective strength parameters are generally used to check the long term (that is when all the excess pore pressures have dissipated) stability of soil constructions. However, for sands and gravels pore pressures dissipate rapidly and for these permeable soils the effective strength parameters can also be used for assessing the short term stability. In principle the effective strength parameters can be used to check the stability at any time
222
for any soil type, but to do this the pore pressures in the ground must be known and in general they are not.
223
14.5.2 Undrained loading
In undrained laboratory tests it is necessary to ensure no drainage from the sample, or moisture redistribution within the sample occurs. In shear box tests this requires fast rates, but because of the more uniform conditions in the triaxial test undrained tests can be performed more slowly simply making sure that no water can drain from the sample.
The behaviour of undrained tests may be interpreted in terms of the effective strength parameters c, , using the effective stresses. In a triaxial test with pore pressure measurement this is possible. The behaviour may also be interpreted in terms of the total strength parameters cu, u. However, if the total stress parameters are being used they must be determined from Unconsolidated Undrained tests if they are to be relevant to the soil in the ground.
Let us consider the behaviour of three identical saturated soil samples in undrained triaxial tests. No water is allowed to drain and three different confining pressures are applied (Samples are Unconsolidated). The Mohr circles at failure will be as follows
131 3
From the total stress Mohr circles we find that u = 0.
Because all samples are at failure the effective stress failure condition must also be satisfied, and because all the circles have
224
the same radius there must be a single effective stress Mohr circle. The different total stress Mohr circles indicate that the samples must have different pore water pressures.
The explanation for the independence of the undrained strength on the confining stress is that increasing the cell pressure without allowing drainage has the effect of increasing the pore pressure by the same amount (u = r). There is therefore no change in effective stress. As it is the effective stresses that control the soil behaviour the subsequent strength is unaffected. The change in pore pressure during shearing is a function of the initial effective stress and the moisture content. As these are identical for the three samples an identical strength is obtained. As will be shown later the fact that the moisture content remains constant is the most important factor in having a constant strength.
In some series of unconsolidated undrained tests it is found that for different soil samples from a particular site u is not zero, or cu is not constant. If this occurs then either
the samples are not saturated, or
the samples have different moisture contents
The undrained strength cu is not a fundamental soil parameter. The total stress strength parameters cu, u are often used to assess the short term (undrained) stability of soil constructions. It is important that no drainage should occur otherwise this approach is not valid. Therefore, for sands and gravels which drain rapidly a total stress analysis would not be appropriate.
For soils that do not drain freely this approach is the only simple way of assessing the short term stability, because in general the pore water pressures are unknown.
225
Note however, that it is possible to measure an undrained strength for any type of soil in the triaxial apparatus.
Example
In an unconsolidated undrained triaxial test the undrained strength is measured as 17.5 kPa. Determine the cell pressure used in the test if the effective strength parameters are c´ = 0, ´
= 26o and the pore pressure at failure is 43 kPa.
Analytical solution
Undrained strength = 17.5 = 1 3 1 3
2 2
Effective stress failure criterion 1 = N 3 + 2 c N
c´ = 0, N
1
12 561
sin
sin.
Hence ’ = 57.4 kPa, ’ = 22.4 kPa
and cell pressure (total stress) = ’ + u = 65.4 kPa
226
Graphical solution
131 3
26
17.5
227
o
15. STRESS-STRAIN BEHAVIOUR OF SOILS
15.1 The behaviour of sands
In practice sands are usually sheared under drained conditions because their relatively high permeability ensures that excess pore pressures are not generated. This behaviour can be investigated in a variety of laboratory apparatus. We will consider the behaviour in simple shear tests. The simple shear test is similar to the shear box test but it has the advantage that the strain and stress states are more uniform enabling us to investigate the stress-strain behaviour. The name simple shear refers to the plane strain mode of deformation shown below:
dx
H
dz
xz
xz = dx/H z = - dz/H = v
For this deformation there are only two non-zero strain components, these are the shear strain, xz =
228
dx/H, and the normal strain z = dz/H. The volume strain, v = z.
For sands the two most important parameters governing their behaviour are the Relative Density, Id, and the effective stress level, . The Relative density is defined by
6where emax and emin are the maximum and minimum void ratios that can be measured in standard tests in the laboratory, and e is the current void ratio. This expression can be re-written in terms of dry density as
7and hence
8Sand is generally referred to as dense if Id > 0.6 and loose if Id < 0.3.
15.1.1 Influence of Relative Density
The influence of relative density on the behaviour can be seen in the plots below for tests all performed at the same normal stress.
dI = e - e
e - emax
max min
ds w = G
1 + e
ddmin d
dmin dmax
I =
1 -
1
1 -
1
229
= tanult
CSL
v
Dense (D)
Medium (M)
Loose (L)
D
M
L
e
D
M
L
The following observations can be made:
• All samples approach the same ultimate conditions of shear stress and void ratio, irrespective of the initial density
• Initially dense samples attain higher peak angles of friction ( = tan-1 (/) )
230
• Initially dense soils expand (dilate) when sheared, and initially loose soils compress
231
15.1.2 Influence of Effective Stress Level
The influence of stress level can be seen in the plots below where the two dense samples have the same initial void ratio, e1 and similarly the loose samples both have the same initial void ratio e2.
= tanult
CSL
v
D2
L2
D2
L2
e
D1
L1
D1
L1
12
= tanult
CSL
D2
L2
D1
L1
’
The following observations can be made:
• The ultimate values of shear stress and void ratio, depend on the stress level, but the ultimate angle of friction (ult = tan-1 (/) ult) is independent of both density and stress level
232
• Initially dense samples attain higher peak angles of friction ( = tan-1 (/)), but the peak friction angle reduces as the stress level increases.
• Initially dense soils expand (dilate) when sheared, and initially loose soils compress. Increasing stress level causes less dilation (greater compression).
233
15.1.3 Ultimate or Critical States
All soil when sheared will eventually attain a unique stress ratio given by / = tan ult, and reach a critical void ratio which is uniquely related to the normal stress. This ultimate state is referred to as a Critical State, defined by
9The locus of these critical states defines a line known as the Critical State Line (CSL). This may be represented by
= tanult
CSL
d
d =
d
d =
d
d = 0v
ee = e - ln0
ln
Straight line
234
e
CSL
At critical states soil behaves as a purely frictional material
= ult = cs = constant = F (mineralogy, grading, angularity)
e
235
15.1.4 Stress-Dilatancy Relation
During a simple shear test on dense sand the top platen is forced up against the applied normal stress. Work must be done against this external force in addition to the work done in overcoming friction between the particles. Thus the frictional resistance of the soil may appear to be greater than ult. Another way to demonstrate this is to consider a "saw-tooth" analogy.
P
Q
FN
Q F N cos sin
Q
P
F N
F N
cos sin
sin cos
P F N sin cos
Q
P
FN
FN
tan
tan
1
236
NowQ
Pand
F
Nult tan tan
tantan tan
tan tan
ult
ult1
ult
tan
dy
dx
d
dv
237
15.1.5 Peak Conditions
The failure conditions are normally expressed by a Mohr-Coulomb criterion using parameters c´, ´. This is the approach that we will be following in estimating the stability of soil constructions.
’
c’
’
Dense sand - Peakstrengths c’, ’
Ultimate strengthc’ = 0, ’ = ’ult
However, this approach obscures the fact that c´ is only an apparent cohesion. An alternative method of presenting the results is to determine the maximum friction angle ´pk which in shear box type tests is simply given by tan-1(´). The relation between ´pk
and effective stress is then as shown below.
238
’pk Id = 1
Id = 0.5
Id = 0
’ult
The position of the lines in this plot is a function of the mineralogy and angularity of the soil.
Note that even loose sand can have ´pk > ´ult if the stress is low enough. This means that loose sands may expand when sheared.
239
15.1.6 Implications for stability analysis
If you choose to use pk (or c, with c 0) in stability calculations then you are saying that everywhere on the critical failure surface the soil will be dilating at failure. In most practical cases this is unlikely to be realistic. For instance consider the case of a retaining wall.
A
A
A
B
B
B
C
C
CFailureSurface
Wall
It is conservative to use c´ = 0 and ´ = ´ult for stability analyses.
240
15.2 Behaviour of clays
The behaviour of clays is essentially identical to that of sands. The data however is usually presented in terms of the soils stress history (OCR) rather than relative density.
To predict the behaviour of soil we need to combine the CSL with our previous knowledge concerning the consolidation behaviour. Experience has shown that the CSL is parallel to the normal consolidation line and lies below it in a void ratio, effective stress plot.
e
NCL - normalconsolidation line
log ’
CSL
swellingline
We find that normally consolidated clays behave similarly to loose sands and heavily over-consolidated clays behave similarly to dense sands. As the OCR increases there is a gradual trend between these extremes. The response in drained simple shear tests with ´ constant is as follows
241
= tanult
CSL
e
CSL
NCL
= tanult
CSL
OCR = 1
OCR = 8
v OCR = 8
OCR = 1
242
15.2.1 Undrained response
In an undrained test volume change is prevented and therefore the void ratio must remain constant. Because the soil always heads towards a critical state when sheared it is possible to show the path that will be followed in an e, plot. This is shown below for normally consolidated (OCR=1) and heavily over-consolidated (OCR>8) samples having the same initial void ratio. Once the final states in this plot are known, so too are the final states in the , plot. Also if the final total stresses are known then the excess pore pressures can be determined.
= tanult
CSL
e
CSL
NCL
= tanult
CSL
OCR = 1
u
OCR = 8
OCR = 1+ve
-ve
OCR = 8
Knowledge of the Critical State Line enables an explanation for the existence of apparent
243
cohesion (undrained strength) in frictional materials
It is also clear that if the moisture content changes then so will the undrained strength,
because failure will occur at a different point on the CSL
244
15.3 Differences between sand and clay
When considering the behaviour of sands and clays we generally use different parameters. For sands stress level and relative density are considered to be the important parameters, whereas for clays the parameters are stress level and stress history (OCR).
However, the broad patterns of behaviour observed for sands and clays are very similar. To understand why different "engineering" parameters are used it is useful to consider the positions of the consolidation and CSL lines in the void ratio, effective stress plot.
0.1 1 10 100 log ’ (MPa)
e
NCL NCL
Loose
Dense
Clay
Sand
245
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