Based on part of the GeotechniCAL reference package
by Prof. John Atkinson, City University, London
http://environment.uwe.ac.uk/geocal/quiz_frame.htm
Soil description and classification
Soils consist of grains (mineral grains, rock fragments, etc.) with water and air in the voids between grains.
The water and air contents are readily changed by changes in conditions and location: soils can be perfectly
dry (have no water content) or be fully saturated (have no air content) or be partly saturated (with both air
and water present). Although the size and shape of the solid (granular) content rarely changes at a given
point, they can vary considerably from point to point.
First of all, consider soil as a engineering material - it is not a coherent solid material like steel and
concrete, but is a particulate material. It is important to understand the significance of particle size, shape
and composition, and of a soil's internal structure or fabric.
Soil as an engineering material
The term "soil" means different things to different people: To a geologist
it represents the products of past surface processes. To a pedologist it
represents currently occurring physical and chemical processes. To an
engineer it is a material that can be:
built on: foundations to buildings, bridges.
built in: tunnels, culverts, basements.
built with: roads, runways, embankments, dams.
supported: retaining walls, quays.
Soils may be described in different ways by different people for their
different purposes. Engineers' descriptions give engineering terms that will
convey some sense of a soil's current state and probable susceptibility to
future changes (e.g. in loading, drainage, structure, surface level).
Engineers are primarily interested in a soil's mechanical properties:
strength, stiffness, permeability. These depend primarily on the nature of
the soil grains, the current stress, the water content and unit weight.
Size range of grains
The range of particle sizes encountered in soil is very large: from boulders with a controlling dimension of
over 200mm down to clay particles less th
in size which behave as colloids, i.e. do not settle in water due solely to gravity.
In theBritish Soil Classification System, soils are classified into named Basic Soil Type groups according to
size, and the groups further divided into coarse, medium and fine sub-groups:
Very coarse BOULDERS
soils
COBBLES
Coarse
soils
60 - 200 mm
coarse 20 - 60 mm
G
medium 6 - 20 mm
GRAVEL
fine
2 - 6 mm
coarse
S
SAND
Fine
soils
> 200 mm
M
SILT
0.6 - 2.0 mm
medium 0.2 - 0.6 mm
fine
0.06 - 0.2 mm
coarse
0.02 - 0.06 mm
medium 0.006 - 0.02 mm
fine
C CLAY
0.002 - 0.006 mm
< 0.002 mm
Aids to size identification
Soils possess a number of physical characteristics which can be used as aids to size identification in the
field. A handful of soil rubbed through the fingers can yield the following:
SAND (and coarser) particles are visible to the naked eye.
SILT particles become dusty when dry and are easily brushed off hands and boots.
CLAY particles are greasy and sticky when wet and hard when dry, and have to be scraped or washed off
hands and boots.
Shape of grains
The majority of soils may be regarded as either SANDS or CLAYS:
SANDS include gravelly sands and gravel-sands. Sand grains are generally broken rock particles that have
been formed by physical weathering, or they are the resistant components of rocks broken down by
chemical weathering. Sand grains generally have a rotund shape.
CLAYS include silty clays and clay-silts; there are few pure silts (e.g. areas formed by windblown Löess).
Clay grains are usually the product of chemical weathering or rocks and soils. Clay particles have a flaky
shape.
There are major differences in engineering behaviour between SANDS and CLAYS (e.g. in permeability,
compressibility, shrinking/swelling potential). The shape and size of the soil grains has an important
bearing on these differences.
Shape characteristics of SAND grains
SAND and larger-sized grains are rotund. Coarse soil grains (silt-sized, sand-sized and larger) have
different shape characteristics and surface roughness depending on the amount of wear during
transportation (by water, wind or ice), or after crushing in manufactured aggregates. They have a
relatively low specific surface (surface area).
Click on a link below to see the shape
Rounded: Water- or air-worn; transported sediments
Irregular: Irregular shape with round edges; glacial sediments (sometimes sub-divided into 'sub-rounded'
and 'sub-angular')
Angular: Flat faces and sharp edges; residual soils, grits
Flaky: Thickness small compared to length/breadth; clays
Elongated: Length larger than breadth/thickness; scree, broken flagstone
Flaky & Elongated: Length>Breadth>Thickness; broken schists and slates
Shape characteristics of CLAY grains
CLAY particles are flaky. Their thickness is very small relative to their length & breadth, in some cases as
thin as 1/100th of the length. They therefore have high to very high specific surface values. These surfaces
carry a small negative electrical charge, that will attract the positive end of water molecules. This charge
depends on the soil mineral and may be affected by an electrolite in the pore water. This causes some
additional forces between the soil grains which are proportional to the specific surface. Thus a lot of
water may be held asadsorbed water within a clay mass.
Specific surface
Specific surface is the ratio of surface area per unit weight.
Surface forces are proportional to surface area (i.e. to d²).
Self-weight forces are proportional to volume (i.e. to d³).
Surface force
1
Therefore
 
self weight forces
d
area
1
Also, specific surface =

d
 * volume
Hence, specific surface is a measure of the relative contributions of surface forces and self-weight forces.
The specific surface of a 1mm cube of quartz ( = 2.65gm/cm³) is 0.00023 m²/N
SAND grains (size 2.0 - 0.06mm) are close to cubes or spheres in shape, and have specific surfaces near the
minimum value.
CLAY particles are flaky and have much greater specific surface values.
Examples of specific surface
The more elongated or flaky a particle is the greater will be its specific surface.
Click on the following examples:
cubes, rods, sheets
Examples of mineral grain specific surfaces:
Mineral/Soil
Thickness
Grain width
Specific
Surface
m²/N
Quartz grain
100
d
0.0023
Quartz sand
2.0 - 0.06
d
0.0001 - 0.004
Kaolinite
2.0 - 0.3
0.2d
2
Illite
2.0 - 0.2
0.1d
8
Montmorillonite
1.0 - 0.01
0.01d
80
Structure or fabric
Natural soils are rarely the same from one point in the ground to another. The content and nature of grains
varies, but more importantly, so does the arrangement of these.
The arrangement and organisation of particles and other features within a soil mass is termed its structure
or fabric. This includes bedding orientation, stratification, layer thickness, the occurrence of joints and
fissures, the occurrence of voids, artefacts, tree roots and nodules, the presence of cementing or bonding
agents between grains.
Structural features can have a major influence on in situ properties.




Vertical and horizontal permeabilities will be different in alternating layers of fine and coarse soils.
The presence of fissures affects some aspects of strength.
The presence of layers or lenses of different stiffness can affect stability.
The presence of cementing or bonding influences strength and stiffness.
Origins, formation and mineralogy
Soils are the results of geological events (except for the very small amount produced by man). The nature
and structure of a given soil depends on the geological processes that formed it:
breakdown of parent rock: weathering, decomposition, erosion.
transportation to site of final deposition: gravity, flowing water, ice, wind.
environment of final deposition: flood plain, river terrace, glacial moraine, lacustrine or marine.
subsequent conditions of loading and drainage - little or no surcharge, heavy surcharge due to ice or
overlying deposits, change from saline to freshwater, leaching, contamination.
Origins of soils from rocks
All soils originate, directly or indirectly, from solid rocks in the Earth's crust:
igneous rocks
crystalline bodies of cooled magma, e.g. granite, basalt, dolerite, gabbro, syenite, porphyry
sedimentary rocks
layers of consolidated and cemented sediments, mostly formed in bodies of water (seas, lakes, etc.)
e.g. limestone, sandstones, mudstone, shale, conglomerate
metamorphic rocks
formed by the alteration of existing rocks due to heat from igneous intrusions (e.g. marble, quartzite,
hornfels) or pressure due to crustal movement (e.g. slate, schist, gneiss).
Weathering of rocks
Physical weathering
Physical or mechanical processes taking place on the Earth's surface, including the actions of water, frost,
temperature changes, wind and ice; cause disintegration and wearing. The products are mainly coarse soils
(silts, sands and gravels). Physical weathering produces Very Coarse soils and Gravels consisting of broken
rock particles, but Sands and Silts will be mainly consists of mineral grains.
Chemical weathering
Chemical weathering occurs in wet and warm conditions and consists of degradation by decomposition
and/or alteration. The results of chemical weathering are generally fine soils with separate mineral grains,
such as Clays and Clay-Silts. The type of clay mineral depends on the parent rock and on local drainage.
Some minerals, such as quartz, are resistant to the chemical weathering and remain unchanged.
quartz
A resistant and enduring mineral found in many rocks (e.g. granite, sandstone). It is the principal
constituent of sands and silts, and the most abundant soil mineral. It occurs as equidimensional hard grains.
haematite
A red iron (ferric) oxide: resistant to change, results from extreme weathering. It is responsible for the
widespread red or pink colouration in rocks and soils. It can form a cement in rocks, or a duricrust in soils
in arid climates.
micas
Flaky minerals present in many igneous rocks. Some are resistant, e.g. muscovite; some are broken down,
e.g. biotite.
clay minerals
These result mainly from the breakdown of feldspar minerals. They are very flaky and therefore have very
large surface areas. They are major constituents of clay soils, although clay soil also contains silt sized
particles.
Clay minerals
Clay minerals are produced mainly from the chemical
weathering and decomposition of feldspars, such as
orthoclase and plagioclase, and some micas. They are
small in size and very flaky in shape.
The key to some of the properties of clay soils, e.g.
plasticity, compressibility, swelling/shrinkage potential,
lies in the structure of clay minerals.
There are three main groups of clay minerals:
kaolinites
(include kaolinite, dickite and nacrite) formed by the
decomposition of orthoclase feldspar (e.g. in granite);
kaolin is the principal constituent in china clay and ball
clay.
illites
(include illite and glauconite) are the commonest
clay minerals; formed by the decomposition of some
micas and feldspars; predominant in marine clays and
shales (e.g. London clay, Oxford clay).
montmorillonites
(also called smectites or fullers' earth minerals) (include calcium and sodium momtmorillonites, bentonite
and vermiculite) formed by the alteration of basic igneous rocks containing silicates rich in Ca and Mg;
weak linkage by cations (e.g. Na+, Ca++) results in high swelling/shrinking potential
Transportation and deposition
The effects of weathering and transportation largely determine the basic nature of the soil (i.e. the size,
shape, composition and distribution of the grains). The environment into which deposition takes place, and
subsequent geological events that take place there, largely determine the state of the soil, (i.e. density,
moisture content) and the structure or fabric of the soil (i.e. bedding, stratification, occurrence of joints or
fissures, tree roots, voids, etc.)
Transportation
Due to combinations of gravity, flowing water or air, and moving ice. In water or air: grains become subrounded or rounded, grain sizes are sorted, producing poorly-graded deposits. In moving ice: grinding and
crushing occur, size distribution becomes wider, deposits are well-graded, ranging from rock flour to
boulders.
Deposition
In flowing water, larger particles are deposited as velocity drops, e.g. gravels in river terraces, sands in
floodplains and estuaries, silts and clays in lakes and seas. In still water: horizontal layers of successive
sediments are formed, which may change with time, even seasonally or daily.



Deltaic & shelf deposits: often vary both horizontally and vertically.
From glaciers, deposition varies from well-graded basal tills and boulder clays to poorly-graded
deposits in moraines and outwash fans.
In arid conditions: scree material is usually poorly-graded and lies on slopes.
Wind-blown Löess is generally uniformly-graded and false-bedded.
Loading and drainage history
The current state (i.e. density and consistency) of a soil will have been profoundly influenced by the history
of loading and unloading since it was deposited. Changes in drainage conditions may also have occurred
which may have brought about changes in water content.
Loading /unloading history
Initial loading
During deposition the load applied to a layer of soil increases as more layers are deposited over it; thus, it is
compressed and water is squeezed out; as deposition continues, the soil becomes stiffer and stronger.
Unloading
The principal natural mechanism of unloading is erosion of overlying layers. Unloading can also occur as
overlying ice-sheets and glaciers retreat, or due to large excavations made by man. Soil expands when it is
unloaded, but not as much as it was initially compressed; thus it stays compressed - and is said to be
overconsolidated. The degree of overconsolidation depends on the history of loading and unloading.
Drainage history
Chemical changes
Some soils initially deposited loosely in saline water and then inundated with fresh water develop
weak collapsing structure. In arid climates with intermittent rainy periods, cycles of wetting and
drying can bring minerals to the surface to form a cemented soil.
Climate changes
Some clays (e.g. montmorillonite clays) are prone to large volume changes due to wetting and drying; thus,
seasonal changes in surface level occur, often causing foundation damage, especially after exceptionally dry
summers. Trees extract water from soil in the process of evapotranspiration; The soil near to trees can
therefore either shrink as trees grow larger, or expand following the removal of large trees.
Grading and composition
The recommended standard for soil classification is the British Soil Classification System, and
this is detailed in BS 5930 Site Investigation.
Coarse soils
Coarse soils are classified principally on the basis of particle
size and grading.
> 200 mm
Very coarse BOULDERS
soils
COBBLES
60 - 200 mm
Coarse
soils
coarse 20 - 60 mm
G
medium 6 - 20 mm
GRAVEL
fine
2 - 6 mm
S
coarse
0.6 - 2.0 mm
medium 0.2 - 0.6 mm
SAND
fine
0.06 - 0.2 mm
Particle size tests
The aim is to measure the distribution of particle sizes in the sample. When a wide range of sizes is present,
the sample will be sub-divided, and separate tests carried out on each sub-sample. Full details of tests are
given in BS 1377: "Methods of test for soil for civil engineering purposes".
Particle-size tests
Wet sieving to separate fine grains from coarse grains is carried out by washing the soil specimen on a
Dry sieving
and shaken through a nest of sieves of descending size.
Sedimentation is used only for fine soils. Soil particles are allowed to settle from a suspension. The
decreasing density of the suspension is measured at time intervals. Sizes are determined from the settling
velocity and times recorded. Percentages between sizes are determined from density differences.
Particle-size analysis
The cumulative percentage quantities finer than certain sizes (e.g. passing a given size sieve mesh)
are determined by weighing. Points are then plotted of % finer (passing) against log size. A
smooth S-shaped curve drawn through these points is called a grading curve. The position and
shape of the grading curve determines the soil class. Geometrical grading characteristics can be
determined also from the grading curve.
Typical grading curves
Both the position and the shape of the grading curve for a soil can aid its identity and description.
Some typical grading curves are shown in the figure:
A - a poorly-graded medium SAND (probably estuarine or flood-plain alluvium)
B - a well-graded GRAVEL-SAND (i.e. equal amounts of gravel and sand)
C - a gap-graded COBBLES-SAND
D - a sandy SILT (perhaps a deltaic or estuarine silt)
E - a typical silty CLAY (e.g. London clay, Oxford clay)
Grading characteristics
A grading curve is a useful aid to soil description. Grading curves are often included in ground
investigation reports. Results of grading tests can be tabulated using geometric properties of the grading
curve. These properties are called grading characteristics
First of all, three points are located on the grading curve:
d10 = the maximum size of the smallest 10% of the sample
d30 = the maximum size of the smallest 30% of the sample
d60 = the maximum size of the smallest 60% of the sample
From these the grading characteristics are calculated:
Effective size
d10
Uniformity coefficient
Cu = d60 / d10
Coefficient of gradation
Ck = d30² / d60 d10
Both Cu and Ck will be 1 for a single-sized soil
Cu > 5 indicates a well-graded soil
Cu < 3 indicates a uniform soil
Ck between 0.5 and 2.0 indicates a well-graded soil
Ck < 0.1 indicates a possible gap-graded soil
Sieve analysis example
The results of a dry-sieving test are given below, together with the grading analysis and grading curve. Note
carefully how the tabulated results are set out and calculated. The grading curve has been plotted on special
semi-logarithmic paper; you can also do this analysis using a spreadsheet.
Sieve mesh
size (mm)
Mass
Percentage
retained (g) retained
Percentage
finer
(passing)
14.0
0
0 100.0
10.0
3.5
1.2 98.8
6.3
7.6
2.6 86.2
5.0
7.0
2.4 93.8
3.35
14.3
4.9 88.9
2.0
21.1
7.2 81.7
1.18
56.7
19.4 62.3
0.600
73.4
25.1 37.2
0.425
22.2
7.6 29.6
0.300
26.9
9.2 20.4
0.212
18.4
6.3 14.1
0.150
15.2
5.2
8.9
0.063
17.5
6.0
2.9
Pan
8.5
2.9
TOTAL
292.3
100.0
The soil comprises: 18% gravel, 45% coarse sand, 24% medium sand, 10% fine sand, 3% silt, and is
classified therefore as: a well-graded gravelly SAND
Fine soils
In the case of fine soils (e.g. CLAYS and SILTS), it is the shape of the particles rather than their size that
has the greater influence on engineering properties. Clay soils have flaky particles to which water adheres,
thus imparting the property of plasticity.
Consistency limits and plasticity
Consistency varies with the water content of the soil. The consistency of a soil can range from (dry) solid to
semi-solid to plastic to liquid (wet). The water contents at which the consistency changes from one state to
the next are called consistency limits (or Atterberg limits).
Two of these are utilised in the classification of fine soils:
Liquid limit (wL) - change of consistency from plastic to liquid
Plastic limit (wP) - change of consistency from brittle/crumbly to plastic
Measures of liquid and plastic limit values can be obtained from laboratory tests.
Plasticity index
The consistency of most soils in the ground will be plastic or semi-solid. Soil strength and stiffness
behaviour are related to the range of plastic consistency. The range of water content over which a soil has a
plastic consistency is termed the Plasticity Index (IP or PI).
IP = liquid limit - plastic limit
= wL - wP
The plasticity chart and classification
In the BSCS fine soils are divided into ten classes based on their measured plasticity index and liquid limit
values: CLAYS are distinguished from SILTS, and five divisions of plasticity are defined:
Low plasticity
wL = < 35%
Intermediate plasticity
wL = 35 - 50%
High plasticity
wL = 50 - 70%
Very high plasticity
wL = 70 - 90%
Extremely high plasticity wL = > 90%
Activity
So-called 'clay' soils are not 100% clay. The proportion of clay mineral flakes (< 2
affects its current state, particularly its tendency to swell and shrink with changes in water content. The
degree of plasticity related to the clay content is called the activity of the soil.
Activity
P / (% clay particles)
Some typical values are:
Mineral
Activity Soil
Activity
Muscovite
0.25
Kaolin clay
0.4-0.5
Kaolinite
0.40
Glacial clay and loess 0.5-0.75
Illite
0.90
Most British clays
Montmorillonite > 1.25
0.75-1.25
Organic estuarine clay > 1.25
Specific gravity
Specific gravity (Gs) is a property of the mineral or rock material forming soil grains.
It is defined as
Method of measurement
For fine soils a 50 ml density bottle may be used; for coarse soils a 500 ml or 1000 ml jar. The jar is
weighed empty (M1). A quantity of dry soil is placed in the jar and the jar weighed (M2). The jar is filled
with water, air removed by stirring, and weighed again (M3). The jar is emptied, cleaned and refilled with
water - and weighed again (M4).
[The range of Gs for common soils is 2.64 to 2.72]
Volume-weight properties
The volume-weight properties of a soil define its state. Measures of the amount of void space, amount of
water and the weight of a unit volume of soil are required in engineering analysis and design.
Soil comprises three constituent phases:
Solid: rock fragments, mineral grains or flakes, organic matter.
Liquid: water, with some dissolved compounds (e.g. salts).
Gas: air or water vapour.
In natural soils the three phases are intermixed. To aid analysis it is convenient to consider a soil model in
which the three phases are seen as separate, but still in their correct proportions.
Volumes of solid, water and air: the soil model
The soil model is given dimensional values for the solid, water and air components: Total volume,
V = Vs + Vw + Va
Since the amounts of both water and air are variable, the volume of solids present is taken as the reference
quantity. Thus, the following relational volumetric quantities may be defined:
Note also that:
n = e / (1 + e)
e = n / (1 - n)
v = 1 / (1 - n)
Typical void ratios might be 0.3 (e.g. for a dense, well graded granular soil) or 1.5 (e.g. for a soft clay).
Degree of saturation
The volume of water in a soil can only vary between zero (i.e. a dry soil) and the volume of voids; this can
be expressed as a ratio:
For a perfectly dry soil:
Sr = 0
For a saturated soil:
Sr = 1
Note: In clay soils as the amount water increases the volume and therefore the volume of voids will also
increase, and so the degree of saturation may remain at Sr = 1 while the actual volume of water is
increasing.
Air-voids content
The air-voids volume, Va , is that part of the void space not occupied by water.
Va = Vv - Vw
= e - e.Sr
= e.(1 - Sr)
Air-voids content, Av
Av = (air-voids volume) / (total volume)
= Va / V
= e.(1 - Sr) / (1+e)
= n.(1 - Sr)
For a perfectly dry soil:
Av = n
For a saturated soil: Av = 0
Masses of solid and water: water content
The mass of air may be ignored. The mass of solid particles is usually expressed in terms of their particle
density or grain specific gravity.
Grain specific gravity
Hence the mass of solid particles in a soil
Ms = Vs .Gs .w
(w = density of water = 1.00Mg/m³)
[Range of Gs for common soils: 2.64-2.72]
Particle density
s = mass per unit volume of particles
= Gs .w
The ratio of the mass of water present to the mass of solid particles is called the water content, or
sometimes the moisture content.
From the soil model it can be seen that
w = (Sr .e .w) / (Gs .w)
Giving the useful relationship:
w .Gs = Sr .e
Densities and unit weights
Density is a measure of the quantity of mass in a unit volume of material.
Unit weight is a measure of the weight of a unit volume of material.
There are two basic measures of density or unit weight applied to soils: Dry density is a measure of the
amount of solid particles per unit volume. Bulk density is a measure of the amount of solid + water per unit
volume.
The preferred units of density are:
Mg/m³, kg/m³ or g/ml.
The corresponding unit weights are:
Also, it can be shown that
 = d(1 + w) and
 = gd(1 + w)
Laboratory measurements
It is important to quantify the state of a soil immediately it is received in the testing laboratory and just prior
to commencing other tests (e.g. shear tests, compression tests, etc.).
The water content and unit weight are particularly important, since these could change during transportation
and storage.
Some physical state properties are calculated following the practical measurement of others; e.g. void ratio
from porosity, dry unit weight from unit weight & water content.
Water content
The most usual method of determining the water content of soil is to weigh a small representative
specimen, drying it to constant weight and then weighing it again. Drying can be carried out using an
electric oven set at 104-105° Celsius or using a microwave oven.
Example: A sample of soil was placed in a tin container and weighed, after which it was dried in an oven
and then weighed again. Calculate the water content of the soil.
Weight of tin empty
= 16.16 g
Weight of tin + moist soil = 37.82 g
Weight of tin + dry soil
= 34.68 g
Water content, w
= (mass of water) / (mass of dry soil)
= (37.82 - 34.68) / (34.68 - 16.16)
= 0.169
Percentage water content = 16.9 %
Unit weight
Clay soils: Specimens are usually prepared in the form of regular geometric shapes, (e.g. prisms, cylinders)
of which the volume is easily computed.
Sands and gravels: Specimens have to be placed in a container to determine volume (e.g. a cylindrical
can).
Example
A soil specimen had a volume of 89.13 ml, a mass before drying of 174.45 g and after drying of 158.73 g;
the water content was 9.9 %. Determine the bulk and dry densities and unit weights.
Bulk density
 = (mass of specimen) / (volume of specimen)
= 174.45 / 89.13 g/ml
= 1.957 Mg/m³
[1 g/ml = 1 Mg/m³]
Unit weight
 = 9.81m/s² x  Mg/m³
= 19.20 kN/m³
Dry density
d = (mass after drying) / (volume)
= 158.73 / 89.13
= 1.781 Mg/m³
d =  / (1 + w)
= 1.957 / (1+0.099)
= 1.781 Mg/m³
Dry unit weight
d =  / (1 + w)
= 19.20 / (1+0.099)
= 17.47 kN/m³
Field measurements
Measurements taken in the field are mostly to determine density/unit weight. The most common application
is the determination of the density of rolled and compacted fill, e.g. in road bases, embankments, etc.
Note: These methods are covered in detail by BS1377. You should understand the general principle that
density is calculated from the mass and volume of a sample. How a sample of known volume is obtained
depends on the nature of the soil. You are not expected to remember the details of each method.
The core cutter method
This method is suitable for soft fine grained soils.
A steel cylinder is driven into the ground, dug out and the soil shaved off level.
The mass of soil is found by weighing and deducting the mass of the cylinder.
Small samples are taken from both ends and the water content determined.
The sand-pouring cylinder method
This method is suitable for stony soils
Using a special tray with a hole in the centre, a hole is formed in the soil and the
mass of soil removed is weighed.
The volume of the hole is calculated from the mass of clean dry running sand
required to fill the hole.
The sand-pouring cylinder is used to fill the hole in a controlled manner. The mass
of sand required to fill the hole is equal to the difference in the weight of the
cylinder before and after filling the hole, less an allowance for the sand left in the
cone above the hole.
Bulk density
 = (mass of soil) / (volume of core cutter or hole)
Current state of soil
The state of soil is essentially the closeness of packing of the grains in the range:
Closely packed
 Loosely packed
Dense
 Loose
Low water content  High water content
Strong and stiff
 Weak and soft
The important indicators of the current state of a soil are:
current stresses: vertical and horizontal effective stresses
current water content: effecting strength and stiffness in fine soils
liquidity index: indicates state in fine soils
density index: indicates state of compaction in coarse soils
history of loading and unloading: degree of overconsolidation. Eng. operations (e.g. excavation, loading,
unloading, compaction, etc.) on soil bring about changes in its state. Its initial state is the result of processes
of erosion and deposition. It is possible for the engineer to predict changes that could result from a
proposed eng .operation: changes from the soil's current state to a new future state.
Soil history: deposition and erosion
Original deposition
Most soils are formed in layers or lenses by deposition from moving water, ice or wind.
One-dimensional compression occurs as overlying layers are added. Vertical and horizontal stresses
increase with deposition.
Erosion
Erosion causes unloading; stresses decrease; some vertical expansion occurs.
Plastic strain has occurred; the soil remains compressed, i.e. overconsolidated.
Subsequent changes
Subsequent changes may occur in the depositional environment: further loading/unloading due to
glaciation, land movement, engineering; and ageing processes.
Soil history: ageing
The term ageing includes processes that occur with time, except loading and unloading. Ageing processes
are independent of changes in loading.
Vibration and compaction
Coarse soils can be made more dense by vibration or compaction at essentially constant effective stress
Creep
Fine soils creep and continue to compress and distort at constant effective stress after primary consolidation
is complete.
Cementing and bonding
Intergranular cementing and bonding occurs due to deposition of minerals from groundwater, e.g. calcium
carbonate; disturbance due to excavation fractures the bonding and reduces strength.
Weathering
Physical and chemical changes take place in soils near the ground surface due to the influence of changes in
rainfall and temperature.
Changes in salinity
Changes in the salinity of groundwater are due to changes in relative sea and land levels, thus soil originally
deposited in sea water may later have fresh water in its pores, such soils may be prone to sudden collapse.
Density index (relative density)
The void ratio of coarse soils (sands and gravels) varies with the state of packing between the loosest
practical state in which it can exist and the densest. Some engineering properties are affected by this,
e.g.shear strength, compressibility, permeability.
It is therefore useful to measure the in situ state and this can be done by comparing the in situ void ratio (e)
with the minimum and maximum practical values (emin and emax) to give a density index D
emin is determined with soil compacted densely in a metal mould
emax is determined with soil poured loosely into a metal mould
Density index is also known as relative density
Relative states of compaction are defined:
Density index
State of compaction
0-15%
Very loose
15-35
Loose
35-65
Medium
65-85
Dense
85-100%
Very dense
Liquidity index
In fine soils, especially clays, the current state is dependent on the water content with respect to the
consistency limits (or Atterberg limits). The
L or LI) provides a quantitative measure of
the current state:
where
wP = plastic limit and
wL = liquid limit
Significant values of IL indicating the consistency of the soil are:
IL
-plastic solid or solid
0 < IL < 1
1 < IL
Predicting stiffness and strength from index properties
Preliminary estimates of strength and stiffness can provide a useful basis for early design and feasibility
studies, and also the planning of more detailed testing programmes. The following suggestions have been
made; they are simple, but not necessarily reliable, and should be not be used in final design calculations.
Undrained shear strength
su = 170 exp(-4.6 L) kN/m²
[Schofield and Wroth (1968)]
su = (0.11 + 0.37 P) 'vo kN/m²
where 'vo = vertical effective stress in situ
[Skempton and Bjerrum (1957)]
Stiffness
The slope of the critical state line may be estimated from:
 = P .Gs / 461
[After Skempton and Northey (1953)]
The compressibility index may be estimated from:
Cc =  ln10 = P Gs / 200 (where P is in percentage units)
BS system for description and classification
BS 5930 Site Investigation recommends the terminology and a system for describing and classifying soils
for engineering purposes. Without the use of a satisfactory system of description and classification, the
description of materials found on a site would be meaningless or even misleading, and it would be difficult
to apply experience to future projects.
BS description system
A recommended protocol for describing a soil deposit uses ninecharacteristics; these should be written in
the following order:
compactness
e.g. loose, dense, slightly cemented
bedding structure
e.g. homogeneous or stratified; dip, orientation
discontinuities
spacing of beds, joints, fissures
weathered state
degree of weathering
colour
main body colour, mottling
grading or consistency
e.g. well-graded, poorly-graded; soft, firm, hard
SOIL NAME
e.g. GRAVEL, SAND, SILT, CLAY; (upper case letters) plus silty-, gravelly-, with-fines, etc. as
appropriate
soil class
(BSCS) designation (for roads & airfields) e.g. SW = well-graded sand
geological stratigraphic name
(when known) e.g. London clay
Not all characteristics are necessarily applicable in every case.
Example:
(i) Loose homogeneous reddish-yellow poorly-graded medium SAND (SP), Flood plain alluvium
(ii) Dense fissured unweathered greyish-blue firm CLAY. Oxford clay.
Definitions of terms used in description
A table is given in BS 5930 Site Investigation setting out a recommended field indentification and
description system. The following are some of the terms listed for use in soil descriptions:
Particle shape
angular, sub-angular, sub-rounded, rounded, flat, elongate
Compactness
loose, medium dense, dense (use a pick or driven peg, or density index )
Bedding structure
homogeneous, stratified, inter-stratified
Bedding spacing
massive(>2m), thickly bedded (2000-600 mm), medium bedded (600-200 mm), thinly bedded (200-60
mm), very thinly bedded (60-20 mm), laminated (20-6 mm), thinly laminated (<6 mm).
Discontinuities
i.e. spacing of joints and fissure: very widely spaced(>2m), widely spaced (2000-600 mm), medium spaced
(600-200 mm), closely spaced (200-60 mm), very closely spaced (60-20 mm), extremely closely spaced
(<20 mm).
Colours
red, pink, yellow, brown, olive, green, blue, white, grey, black
Consistency
very soft (exudes between fingers), soft (easily mouldable), firm (strong finger pressure required), stiff (can
be indented with fingers, but not moulded) very stiff (indented by sharp object), hard (difficult to indent).
Grading
well graded (wide size range), uniform (very narrow size range), poorly graded (narrow or uneven size
range).
Composite soils
In SANDS and GRAVELS: slightly clayey or silty (<5%), clayey or silty (5-15%), very clayey or
silty(>15%)
In CLAYS and SILTS: sandy or gravelly (35-65%)
British Soil Classification System
The recommended standard for soil classification is the British Soil Classification System, and this is
detailed in BS 5930 Site Investigation. Its essential structure is as follows:
Soil group
Symbol
Coarse soils
Fines %
G
GRAVEL
Recommended name
GW
0-5
Well-graded GRAVEL
GPu/GPg
0-5
Uniform/poorly-graded GRAVEL
G-F GWM/GWC 5 - 15
GPM/GPC
5 - 15
GF GML, GMI... 15 - 35
S
SAND
SILT
M
Very silty GRAVEL [plasticity sub-group...]
15 - 35
Very clayey GRAVEL [..symbols as below]
SW
0-5
Well-graded SAND
SPu/SPg
0-5
Uniform/poorly-graded SAND
5 - 15
Well-graded silty/clayey SAND
5 - 15
Poorly graded silty/clayey SAND
GPM/GPC
Fine soils
Poorly graded silty/clayey GRAVEL
GCL, GCI...
S-F SWM/SWC
SF
Well-graded silty/clayey GRAVEL
SML, SMI... 15 - 35
Very silty SAND [plasticity sub-group...]
SCL, SCI...
15 - 35
Very clayey SAND [..symbols as below]
>35% fines
Liquid limit%
MG
Gravelly SILT
MS
Sandy SILT
CLAY
C
ML, MI...
[Plasticity subdivisions as for CLAY]
CG
Gravelly CLAY
CS
Sandy CLAY
CL
<35
CLAY of low plasticity
CI
35 - 50
CLAY of intermediate plasticity
CH
50 - 70
CLAY of high plasticity
CV
70 - 90
CLAY of very high plasticity
CE
>90
CLAY of extremely high plasticity
Organic soils O
Peat
[Add letter 'O' to group symbol]
Pt
[Soil predominantly fibrous and organic]
Basic mechanics of soils
Loads from foundations and walls apply stresses in the ground. Settlements are caused by strains in the
ground. To analyse the conditions within a material under loading, we must consider the stress-strain
behaviour. The relationship between a strain and stress is termed stiffness. The maximum value of stress
that may be sustained is termed strength.
Analysis of stress and strain
Stresses and strains occur in all directions and to do settlement and stability analyses it is often necessary to
relate the stresses in a particular direction to those in other directions.
normal stress
n/ A
normal strain
shear stress
s/ A
shear strain
o
o
Note that compressive stresses and strains are positive, counter-clockwise shear stress and strain are
positive, and that these are total stresses (see effective stress).
Special stress and strain states
Analysis of stress and strain
In general, the stresses and strains in the three
dimensions will all be different.
There are three special cases which are important in
ground engineering:
General case
princpal stresses
Axially symmetric or triaxial states
Stresses and strains in two dorections are equal.
x
y
x
y
Relevant to conditions near relatively small foundations,
piles, anchors and other concentrated loads.
Plane strain:
Strain in one direction = 0
y=0
Relevant to conditions near long foundations,
embankments, retaining walls and other long structures.
One-dimensional compression:
Strain in two directions = 0
x
y=0
Relevant to conditions below wide foundations or
relatively thin compressible soil layers.
Uniaxial compression
x
y=0
This is an artifical case which is only possible for soil is
there are negative pore water pressures.
Mohr circle construction
Values of normal stress and shear stress must relate to a
particular plane within an element of soil. In general, the
stresses on another plane will be different.
To visualise the stresses on all the possible planes, a graph
called the Mohr circle is drawn by plotting a (normal stress,
shear stress) point for a plane at every possible angle.
There are special planes on which the shear stress is zero
(i.e. the circle crosses the normal stress axis), and the
state of stress (i.e. the circle) can be described by the
normal stresses acting on these planes; these are called the
principal stresses 1
3.
Parameters for stress and strain
In common soil tests, cylindrical samples are used in which the axial and radial stresses and strains are
principal stresses and strains. For analysis of test data, and to develop soil mechanics theories, it is usual to
combine these into mean (or normal) components which influence volume changes, and deviator (or
shearing) components which influence shape changes.
stress
mean
deviator
p' =
s' =
strain
a
a
t' =
r) / 3 ev
)
r / 2
n
a
a
-
r)
r)
es =
/2
a
r)
r)
a
a
a
-
r)
/3
r)
In the Mohr circle construction t' is the radius of the circle and s' defines its centre.
Note: Total and effective stresses are related to pore pressure u:
p' = p - u
s' = s - u
q' = q
t' = t
Strength
The shear strength of a material is most simply described as the maximum shear stress it can sustain:
l be a limiting condition at
f is then the shear
strength of the material. The simple type of failure shown here is associated with ductile or plastic
materials. If the material is brittle (like a piece of chalk), the failure may be sudden and catastrophic with
loss of strength after failure.
Types of failure
Materials can ‘fail’ under different loading conditions. In each case, however, failure is associated with the
limiting radius of the Mohr circle, i.e. the maximum shear stress. The following common examples are
shown in terms of total stresses:
Shearing
Shea
f
nf = normal stress at failure
Uniaxial extension
tf
f
Uniaxial compression
cf
f
Note:
f = 0.
Hence vertical and horizontal stresses are equal and the Mohr circle becomes a point.
Strength criteria
A strength criterion is a formula which relates the strength of a material to some other parameters: these are
material parameters and may include other stresses.
For soils there are three important strength criteria: the correct criterion depends on the nature of the soil
and on whether the loading is drained or undrained.
In General, course grained
soils will "drain" very quickly
(in engineering terms)
following loading. Thefore
development of excess pore pressure will not occur; volume change associated
with increments of effective stress will control the behaviour and the MohrCoulomb criteria will be valid.
Fine grained saturated soils will respond to loading initially by generating excess pore water pressures and
remaining at constant volume. At this stage the Tresca criteria, which uses total stress to represent
undrained behaviour, should be used. This is the short term or immediate loading response. Once the pore
pressure has dissapated, after a certain time, the effective stresses have incresed and the Mohr-Coulomb
criterion will describe the strength mobilised. This is the long term loading response.

Tresca criterion


Mohr-Coulomb (c’=0) criterion
Mohr-Coulomb (c’>0) criterion
Tresca criterion
The strength is independent of the normal stress since the response to loading simple increases the pore
water pressure and not the effective stress.
The shear strength f is a material parameter which is known as the undrained shear strength su.
f
ar) = constant
Mohr-Coulomb (c'=0) criterion
The strength increases linearly with increasing normal stress and is zero when the normal stress is zero.
f
n
In the Mohrthis criterion are known as frictional. In soils, the Mohr-Coulomb criterion applies when the normal stress is
an effective normal stress.
>Mohr-Coulomb (c'>0) criterion
The strength increases linearly with increasing normal stress and is positive when the normal stress is zero.
f
n
c' is the 'cohesion' intercept
In soils, the Mohr-Coulomb criterion applies when the normal stress is an effective normal stress. In soils,
the cohesion in the effective stress Mohr-Coulomb criterion is not the same as the cohesion (or undrained
strength su) in the Tresca criterion.
Typical values of shear strength
Undrained shear strength su (kPa)
Hard soil
su > 150 kPa
Stiff soil
su = 75 ~ 150 kPa
Firm soil
su = 40 ~ 75 kPa
Soft soil
su = 20 ~ 40kPa
Very soft soil
su < 20 kPa
Drained shear strength
c´ (kPa)
Compact sands
0
35° - 45°
Loose sands
0
30° - 35°
(deg)
Unweathered overconsolidated clay
critical state
0
18° ~ 25°
peak state
10 ~ 25 kPa 20° ~ 28°
residual
0 ~ 5 kPa 8° ~ 15°
Often the value of c' deduced from laboratory test results (in the shear testing apperatus) may appear to
Often this is due to fitting a
due to suction or dilatancy.
line to the experimental data and an 'apparent' cohesion may be deduced
Stress in the ground
When a load is applied to soil, it is carried by the water in the pores as well as the solid
grains. The increase in pressure within the porewater causes drainage (flow out of the
soil), and the load is transferred to the solid grains. The rate of drainage depends on the
permeabilityof the soil. The strength and compressibility of the soil depend on the
stresses within the solid granular fabric. These are called effective stresses.
Total stress
The total vertical stress acting at a point below the ground surface is due to the weight of
everythinglying above: soil, water, and surface loading. Total stresses are calculated from the unit
weight of the soil.
Unit weight ranges are:
dry soil
d 14 - 20 kN/m³ (average 17kN/m³)
saturated soil
g 18 - 23 kN/m³ (average 20kN/m³)
water
w 9.81 kN/m³
v) may also result in a change in the horizontal total stress
h) at the same point. The relationships between vertical and horizontal stress are complex.
total stress
Total stress in homogeneous soil
Total stress increases with depth and with unit weight: Vertical total stress at depth z,
v
Simple total stress calculator
z
v
20
3
60
z,
i.e. related to depth z.
The unit weight, , will vary with the water content of the soil.
d
g
total stress
Total stress below a river or lake
The total stress is the sum of the weight of the soil up to the surface and the weight of water above
this: Vertical total stress at depth z,
v
w .zw
where
weight of the saturated soil,
i.e. the
total weight of soil grains and water
weight of water
w = unit
The vertical total stress will change with
changes in water level and with excavation.
Note that free water (i.e. water outside the
soil) applies a total stress to a soil surface.
Simple total stress calculator
z
zw
v
20
3
1
69.81
total stress
Total stress in multi-layered soil
The total stress at depth z is the sum of the weights of soil in each layer thickness above.
Vertical total stress at depth z,
1d1
v
2d2
3(z
- d1 - d2)
where
1
2
3,
etc. = unit weights of soil layers 1, 2 , 3, etc. respectively
If a new layer is placed on the surface the total stresses at all points below will increase.
Layer
1
2
3
Thickness
1.5
2
5
Unit weight
16
19
20
0
stress
0
@
m=
kPa
Enter a value in any box (except the last) then click outside the box to see the effect
Total stress in unsaturated soil
total stress
Just above the water table the soil will remain saturated due to capillarity, but at some distance
above the water table the soil will become unsaturated, with a consequent reduction in unit weight
u)
v
w
. zw
g(z
- zw)
The height
above the
water table
up to
which the soil will remain saturated depends
on the grain size.
See Negative pore pressure (suction).
Total stress with a surface surcharge load
The addition of a surface surcharge load will increase the total stresses below it. If the surcharge
loading is extensively wide, the increase in vertical total stress below it may be considered
constant with depth and equal to the magnitude of the surcharge.
Vertical total stress at depth z,
v
For narrow surcharges, e.g. under strip and pad foundations, the induced vertical total stresses will
decrease both with depth and horizontal distance from the load. In such cases, it is necessary to use
a suitable stress distribution theory - an example is Boussinesq's theory.
Pore pressure
The water in the pores of a soil is called porewater. The pressure within this porewater is called pore
pressure (u). The magnitude of pore pressure depends on:


the depth below the water table
the conditions of seepage flow
Groundwater and hydrostatic pressure
Pore pressure
Under hydrostatic conditions (no water flow) the pore pressure at a given point is given by the
hydrostatic pressure:
w .hw
where
hw = depth below water table or overlying water surface
It is convenient to think of pore pressure represented by the column of water in an imaginary standpipe; the
pressure just outside being equal to that inside.
Water table, phreatic surface
Pore pressure
The natural static level of water in the ground is called the water table or the phreatic surface (or
sometimes the groundwater level). Under conditions of no seepage flow, the water table will be
horizontal, as in the surface of a lake. The magnitude of the pore pressure at the water table is zero.
Below the water table, pore pressures are positive.
w .hw
In conditions of steady-state or variable seepage flow, the calculation of pore pressures becomes
more complex.
See Groundwater
Negative pore pressure (suction)
Below the water table, pore pressures are positive. In dry soil, the pore pressure is
zero. Above the water table, when the soil is saturated, pore pressure will be negative.
u = - w .hw
The height above the water table to which the soil is saturated is called the capillary rise, and this
depends on the grain size and type (and thus the size of pores):
· in coarse soils capillary rise is very small
· in silts it may be up to 2m
· in clays it can be over 20m
Pore water and pore air pressure
Between the ground surface and the top of the saturated zone, the soil will often be partially
saturated, i.e. the pores contain a mixture of water and air. The pore pressure in a partially
saturated soil consists of two components:
· porewater pressure = uw
· pore-air pressure = ua
Note that water is incompressible, but air is compressible. The combined effect is a complex
relationship involving partial pressures and the degree of saturation of the soil. In Europe and other
temperate climate countries most design states are associated with saturated conditions, and the
study of partially saturated soils is considered to be a specialist subject.
Pore pressure in steady state seepage conditions
In conditions of seepage in the ground there is a change in pore pressure. Consider seepage
occurring between two points P and Q.
The hydralic gradient, i, between two points is the head drop per unit length between these points.
It can be thougth of as the "potential" driving the water flow.
Hydralic gradient P-Q, i = -
=
. 1w
Thus
w
But in steady-state seepage, i = constant
Therefore the change in pore pressure due to seepage alone,
For seepage flow vertically downward, i is negative
For seepage flow vertically upward, i is positive.
s
=
w
.s
Effective stress
Ground movements and instabilities can be caused by changes in total stress (such as loading due
to foundations or unloading due to excavations), but they can also be caused by changes in pore
pressures (slopes can fail after rainfall increases the pore pressures).
In fact, it is the combined effect of total stress and pore pressure that controls soil behaviour such
as shear strength, compression and distortion. The difference between the total stress and the pore
pressure is called the effective stress:
effective stress = total stress - pore pressure
-u
Note that the prime (dash mark ´ ) indicates effective stress.
Terzaghi's principle and equation
Karl Terzaghi was born in Vienna and subsequently became a professor of soil mechanics in the
USA. He was the first person to propose the relationship for effective stress (in 1936):
All measurable effects of a change of stress, such as compression, distortion and a change of
related to total stress and pore pressure by 
The adjective 'effective' is particularly apt, because it is effective stress that is effective in causing
important changes: changes in strength, changes in volume, changes in shape. It does not represent
the exact contact stress between particles but the distribution of load carried by the soil over the
area considered.
Mohr circles for total and effective stress
Mohr circles can be drawn for both total and effective stress. The points E and T represent the total
and effective stresses on the same plane. The two circles are displaced along the normal stress axis
by the amount of pore pressure n
n' + u), and their diameters are the same. The total and
effective shear stresses are equal
.
The importance of effective stress
The principle of effective stress is fundamentally important in soil mechanics. It must be treated as
the basic axiom, since soil behaviour is governed by it. Total and effective stresses must be
distinguishable in all calculations: algebraically the prime
Changes in water level below ground (water table changes) result in changes in effective stresses
below the water table. Changes in water level above ground (e.g. in lakes, rivers, etc.) do not
cause changes in effective stresses in the ground below.
Changes in effective stress


In some analyses it is better to work in changes of quantity, rather than in
absolute quantities; the effective stress expression then becomes:
If both total stress and pore pressure change by the same amount, the
effective stress remains constant. A change in effective stress will cause: a change in strength and
a change in volume.
Changes in strength
The critical shear strength of soil is proportional to the effective normal stress; thus, a change in
effective stress brings about a change in strength.
Therefore, if the pore pressure in a soil slope increases, effective stresses will be reduced by
- sometimes leading to failure.
A seaside sandcastle will remain intact while damp, because the pore pressure is negative; as it
dries, this pore pressure suction is lost and it collapses. Note: Sometimes a sandcastle will remain
intact even when nearly dry because salt deposited as seawater evaporates slightly and cements the
grains together.
Changes in volume
Immediately after the construction of a foundation on a fine soil, the pore pressure increases, but
immediately begins to drop as drainage occurs.
The rate of change of effective stress under a loaded foundation, once it is constructed, will be the
same as the rate of change of pore pressure, and this is controlled by the permeability of the soil.
Settlement occurs as the volume (and therefore thickness) of the soil layers change. Thus,
settlement occurs rapidly in coarse soils with high permeabilities and slowly in fine soils with low
permeabilities.
Calculating vertical stress in the ground
The worked examples here are designed to illustrate the principles and methods dealt with in Pore
pressure, effective stress and stresses in the ground. The examples chosen are typical and simple.
Simple total and effective stresses
The figure shows soil layers on a site.
Unit weights are:
d = 16 kN/m³
g = 20 kN/m³
(a) At the top of saturated sand (z = 2.0 m)
Vertical total stress
v = 16.0 x 2.0
Pore pressure
u=0
Vertical effective stress
v
v-u
(b) At the top of the clay (z = 5.0 m)
Vertical total stress
v = 32.0 + 20.0 x 3.0
= 32.0 kPa
= 32.0 kPa
= 92.0 kPa
Pore pressure
Vertical effective stress
u = 9.81 x 3.0
v
v - u = 92.0 29.4
= 29.4 kPa
= 62.6 kPa
Effect of changing water table
The figure shows soil layers on a site. The unit weight of the silty sand is 19.0 kN/m³ both above
and below the water table. The water level is presently at the surface of the silty sand, it may drop
or it may rise. The following calculations show the effects of this:
Water table
Stresses under foundations
From an initial state, the stresses under a foundation are first changed by excavation, i.e. vertical
stresses are reduced. After construction the foundation loading increases stresses. Other changes
could result if the water table level changed.
The figure shows the elevation of a foundation to be constructed in a homogeneous soil. The
change in thickness of the clay layer is to be calculated and so the initial and final effective
stresses are required at the mid-depth of the clay.
Unit weights: sand above WT = 16 kN/m³, sand below WT = 20 kN/m³, clay = 18 kN/m³.
Calculations for
Short-term and long-term stresses
The figure shows how an extensive layer of fill will be placed on a certain site.
The unit weights are:
clay and sand = 20kN/m³ ,
rolled fill 18kN/m³ ,
assume water = 10 kN/m³.
Calculations are made for the total and effective stress at the mid-depth of the sand and the middepth of the clay for the following conditions: initially, before construction; immediately after
construction; many years after construction.
Initially, before construction
Initial stresses at mid-depth of clay (z = 2.0m)
Vertical total stress
v = 20.0 x 2.0 = 40.0kPa
Pore pressure
u = 10 x 2.0 = 20.0kPa
Vertical effective stress
v
v - u = 20.0kPa
Immediately after construction
The construction of the embankment applies a surface surcharge:
q = 18 x 4 = 72.0 kPa.
The sand is drained (either horizontally or into the rock below) and so there is no increase in pore
pressure. The clay is undrained and the pore pressure increases by 72.0 kPa.
Initial stresses at mid-depth of clay (z = 2.0m)
Vertical total stress
v = 20.0 x 2.0 + 72.0 = 112.0kPa
Pore pressure
u = 10 x 2.0 + 72.0 = 92.0 kPa
Vertical effective stress
- u = 20.0kPa
(i.e. no change immediately)
Initial stresses at mid-depth of sand (z = 5.0m)
Vertical total stress
v = 20.0 x 5.0 + 72.0 = 172.0kPa
Pore pressure
u = 10 x 5.0 = 50.0 kPa
Vertical effective stress
v
v - u = 122.0kPa
(i.e. an immediate increase)
v
v
Many years after construction
After many years, the excess pore pressures in the clay will have dissipated. The pore pressures
will now be the same as they were initially.
Initial stresses at mid-depth of clay (z = 2.0 m)
Vertical total stress
v = 20.0 x 2.0 + 72.0 = 112.0 kPa
Pore pressure
u = 10 x 2.0 = 20.0 kPa
Vertical effective stress
v
v - u = 92.0 kPa
(i.e. a long-term increase)
Initial stresses at mid-depth of sand (z = 5.0 m)
Vertical total stress
v = 20.0 x 5.0 + 72.0 = 172.0 kPa
Pore pressure
u = 10 x 5.0 = 50.0 kPa
Vertical effective stress
v
v - u = 122.0 kPa
(i.e. no further change)
Steady-state seepage conditions
The figure shows seepage occurring around embedded sheet piling.
In steady state, the hydraulic gradient,
Then the effective stresses are:
A = 20 x 3 - 2 x 10 + 0.4 x 10 = 44 kPa
B = 20 x 3 - 2 x 10 - 0.4 x 10 = 36 kPa
Drainage and volume change
Solid soil grains are very stiff: their volume change under load can be ignored. Water or air can be
squeezed out of soil under load. The loss of water from the soil is called drainage. The grains are
rearranged and the volume of voids reduced.
Volume compressibility under load
Consider a volume (V) of soil in equilibrium under a
constant total stress:
o and the pore pressure is uo:
o
o
- uo,
Click the hypertext links to change the
diagram
Immediately after loading the total stress is increased by
Howeve
-
Some time after loading drainage will have occurred:
w)
d the volume of the soil has
Finally then:
Drainage under load
If drainage cannot take place when a soil is loaded the
volume cannot change.
In the oedometer test porous stones are placed above and
below the sample, so that drainage is two-way: upward and
downward.
Under a concrete foundation drainage may only take place
downward.
In an embankment layers of sand can be placed to speed up
drainage and thus changes in volume.
The installation of vertical sand drains, called sandwicks, can
further speed up volume change in embankments by
allowing horizontal radial drainage.
Click the hypertext links to change the
diagram
Permeability and time
The rate of drainage of water from soil depends on the permeability. Volume change under load
takes place quickly in sands and gravels, and very slowly in clays. Seepage is driven by the excess
pore pressure and as this is dissipated the rate of seepage slows down. Thus, the rate of volume
change is fast to begin with, but slows down with time.
The volume-change/time curve is exponential. A simple approximate rule is: "half the total volume
change occurs in one-tenth of the total time".
Volume change under constant effective stress
In saturated soil volume changes can only occur as drainage occurs and as effective stresses
change. In unsaturated soils volume change is due to changes in water and air
volume; both of which can change without change in effective stress.
Compaction, in which air is expelled, can occur due to vibration (e.g. from
traffic, machinery, piling, etc..); also, loosely-placed fill can compact under its
own weight.
Shrinking and swelling can occur in some clays near the surface due to climatic
changes (shrinking in summer, swelling in winter).
Creep occurs in some clay soils due to gradual changes in fabric.
Drained and undrained loading
The relative rates of the increase of total stress and drainage are of critical importance in
determining soil behaviour and predicting future conditions and changes.
If the rate of drainage is quicker than the rate of loading, effective stress and volume changes
occur quickly - these are called drained loading conditions. In drained loading the pore pressures
are always in equilibrium - if construction stops the pore pressures will remain constant and there
are no more volume changes.
If the rate of drainage is slower than the rate of loading, the pore pressure increases and the
effective stress and volume remain unchanged - these are undrained loading conditions. In
undrained loading of saturated soil there is no volume change - if construction stops the excess
pore pressures dissipate, consolidation occurs then the volume changes.
Drained loading conditions
Under fully drained conditions the pore pressure does not change,
Thus, volume decrease will follow loading increase, i.e. increase in total stress.
Drained loading conditions may be assumed to occur when either the soil has a high permeability
(e.g. in sands and gravels), or the loading rate is slow (e.g. natural erosion).
drained or undrained conditions.
Undrained loading conditions
Under undrained conditions there can be no volume change, since water cannot
escape.
From a practical point of view undrained loading occurs: in a laboratory test
(e.g. triaxial) when drainage is prevented; and in field situations where loading
changes occur quickly on soils of low permeability.
In a saturated soil, the increase in total stress produces an equal increase in pore
pressure:
u = uo
As drainage occurs, u decreases and so does the volume.
At an elapsed time t:
ut = uo
t,
- Dut and
t
Vt = Vo t
Eventually, when all of the excess pore pressure has dissipated, equilibrium is regained and
steady-state pore pressure conditions prevail:
u = uo and
o
Consolidation
The dissipation of excess pore pressure, accompanied by volume change is called consolidation.
Usually (but not always) the total stress remains constant (e.g. under a foundation) and the pore
pressure and volume slowly change. The rate of consolidation (volume change with seepage) is
dependent on the permeability of the soil and the size of the consolidating layer.
Transient undrained conditions prevail during consolidation, but eventually, when all of the excess
pore pressure has been dissipated, conditions are the same as those for drained loading.
Swelling will occur during unloading as water is sucked back into the soil.
Rates of loading and seepage
In any geotechnical calculation (analysis or design) it is important to distinguish between drained
and undrained loading - soils behave quite differently in the two sets of conditions. In making this
distinction it is the relative rates of loading and seepage that must be considered.
Seepage rates depend on the coefficient of permeability which is related mainly to grain size: For
design purposes it is common to assume quick seepage in coarse soils and slow seepage in fine
soils.
soil type coeff. of permeability (k) seepage rate
gravel
> 10-2
sand
-2
10 ~ 10
silt
10-5 ~ 10-8
-8
very quick
-5
quick
slow
clay
< 10
very slow
Different rates of loading arise from different natural events or construction operations. Very
rapid loading rates may occur in earthquakes, due to piling and due to wave action.
Event/Operation
Duration
Shock wave - piling
<1 s
Shock wave - earthquake
1-2 s
Wave breaking against wall
5 - 10 s
Trench excavation
1 - 3 hours
Small building foundation
5 - 20 days
Large excavation or building
1 - 6 months
Construction of dam or embankment 1 - 3 years
Filling of reservoir
2 - 5 years
Natural erosion
> 50 years
Volume change
Saturated soil contains only mineral grains and water. Both are relatively incompressible
so the volume can only change if water can drain out. In unsaturated soil volume changes
can occur as air compresses or bleeds out. In both cases loading will bring the grains closer
together and the specific volume will reduce.
1.
2.
3.
4.
Reduction in volume leads to
increase in strength
increase in stiffness
settlement of foundations
If soil is unloaded it will swell as the grains move apart. Swelling leads to reduction in strength
and stiffness and heave of excavations.
Compression and swelling
The relationship between volume change and effective stress is called compression and swelling.
(Consolidation and compaction are different.) The volume of soil grains remains constant, so
change in volume is due to change in volume of water.
Compression and swelling results from drained loading and the pore pressure remains constant. If
saturated soil is loaded undrained there will be no volume change.
Mechanisms of compression
Compression of soil is due to a number of mechanisms:
rearrangement of grains
fracture and rearrangement of grains
distortion or bending of grains
On unloading, grains will not unfracture or un-rearrange, so volume change on unloading and
reloading (swelling and recompression) will be much less than volume change on first loading
(compression).
In compression, soil behaviour is:

non-linear

mostly
irrecoverable
Common cases of
compression and swelling
In practice, the state of stress in the ground will be complex. These are simple theories for two
special cases.
Isotropic:
Equal stress in all directions. Applicable to triaxial test before shearing.
a
r) / 3
= mean stress
v
o
= volumetric strain
One-dimensional:
Horizontal strains are zero. Applicable to oedometer test and in the ground below wide
foundations, embankments and excavations.
z = vertical stress
/ Vo
v
o
o)
= volumetric strain
Isotropic compression and swelling


Equations
Overconsolidation

State
Isotropic compression and swelling is applied at the start of a triaxial test.
a
r) / 3
= mean stress
V = Vo w
= volume
v
o
o
= volumetric strain
v = V / Vs
= specific volume
As the mean stress p' is raised and lowered there are volumetric strains and the specific volume
changes.
p'o = initial mean stress
vo = initial specific volume
Note the paths of compression, swelling and re-loading.
Equations
For isotropic compression and swelling there are simple relationships
between specific volume v and (the natural logarithm of) the mean stress p'.
First loading
normal compression line
OAD on the graph
v=NUnloading and reloading
swelling line
BC on the graph
v = vk vk and p'y locate the particular swelling line.
p'y is referred to as the yield stress.
If the current stress and the history of loading/unloading are
known, the current specific volume can be calculated.
Bulk modulus
Isotropic compression can be represented by a bulk modulus K' or by the slope of the normal
ted.
v
v=Ndv / v = Bulk modulus K' depends on v and p'. Both of these will change during compression or swelling
and so K' is not a soil constant.
Typical values for isotropic compression parameters
the nature of the soil.
Typical values
very high plasticity clay
high plasticity clay
intermediate plasticity clay
low plasticity clay
quartz sand
carbonate sand
For clays
p / 170.
wL
80
60
42
30
Ip
50
34
23
12
l
0.29
0.20
0.14
0.07
0.15
0.34
- 0.35) because clay particles can bend and distort.
For sands
pressure).
compression.
Overconsolidation
If the current state of soil is on the normal compression line it is
said to be normally consolidated. If the soil is unloaded it
becomes overconsolidated.
(Soil cannot usually be at a state outside the normal compression
line unless it is bonded or structured).
At a state A the overconsolidation ratio is
Rp = p'y / p'a
(on NCL Rp = 1.0 and soil is normally consolidated).
Note: p'y is the point of intersection of the swelling line through A and the NCL. This is usually
close to the maximum past stress.
State
The current state of a soil is described by the stress p', the specific volume v and the
overconsolidation ratio Rp (for a complete description the shear stress q' is required).
The state at A is given by any two of
va , p'a , Rp = p'y / p'a
All states with the same Rp fall on the lines parallel with the NCL.
ln Rp = ln ( p'y / p'a )
= ln p'y - ln p'a
Many features of soil behaviour, especially shear modulus and
peak strength, increase with increasing overconsolidation.
Back to Isotropic compression: state
Change of state
Loading and unloading
(relevant to all soils)
Change of state A to B can only be achieved by normal compression along CD followed by
swelling along DB. Note that the yield stress corresponding to B is larger than the yield stress
corresponding to A.
Vibration or compaction
(relevant to sands)
or creep
(relevant to clays)
Change of state can occur directly from A to B. Note that the yield stress corresponding to B is
larger than the yield stress corresponding to A.
Critical state
There is a critical overconsolidation ratio which separates states in which the soil will
either compress or dilate during shear. This corresponds to the critical state line CSL.
Look at the possible specific volumes (v) that can occur at a mean effective stress p'.
wet side of critical
(W on the graph)
vw > vc at stress p'
water content ww is larger than critical wc
· loose
· normally consolidated
or lightly overconsolidated
· compress during drained shear
dry side of critical
(D on the graph)
vd < vc at stress p'
water content wd is smaller than critical wc
· dense
· heavily overconsolidated
· dilate during drained shear
Back to Isotropic compression: state
Normalising parameters
Normalising parameters change the current state to a normalised state so that all states with the
same overconsolidation ratio have the same value.
Equivalent specific volume
vl = va + ln p'a
Equivalent pressure
ln p'e = ( N - va ) /
Critical pressure
ln p'c = ( - va ) /
If A is on the wet side of critical
ve
p'a / p'c > 1
If A is on the dry side of critical
ve
p'a / p'c < 1
One-dimensional compression and swelling
One-dimensional loading is applied in an
oedometer and occurs in the ground beneath
wide foundations, embankments or
excavations.
z = vertical effective stress
H = height or thickness
vertical strain = volumetric strain
v
o
o)
where Ho, eo
o are initial values.
z is raised and lowered the top of the sample settles or heaves, or the layer
contracts or expands.
Note that the compression-swelling-recompression curve is similar to that for isotropic
z, e) rather than (p', v).
Equations
For one-dimensional compression and swelling there are simple relationships between the void
z.
First loading:
normal compression line (NCL)
OAD on the graph
e = eN - Cc
z
Unloading and reloading:
swelling-recompression line (SRL)
BC on the graph
e = ek - Cs
z
· eN, Cc and Cs are soil parameters
· ek
y locate a particular swelling line
o and the history of loading and unloading are known, the current void ratio
can be calculated. e.g.
eo = eN - Cc
'o )
y + Cs
y-
One-dimensional modulus and compressibility
The one-dimensional stiffness modulus is the slope of the stress/strain curve:
z
v or
E'o
z
z
h = 0)
The reciprocal of stiffness is compressibility. The one-dimensional coefficient of
compressibility is the slope of the strain/stress curve:
mv
z (1+e))
= 1 / E'o
E'o and mv apply for the normal compression line and for swelling and recompression lines, and
depend on the current state, on the history and on the increment of loading, so they are not soil
constants.
Since mv
z
z = 100kPa.
Overconsolidation
If the current state of soil is on the normal
compression line it is said to be normally
consolidated. If the soil is unloaded it
becomes overconsolidated.
Soil cannot usually be at a state outside the
normal compression line unless it is bonded
or structured.
At a state A the overconsolidation ratio is
Ro
y
a
soil is normally consolidated).
intersection of the swelling line through A
but not always, close to the maximum past
(on NCL Ro = 1.0 and
y is the point of
and the NCL. This is usually,
stress (see change of state).
Horizontal stress in one-dimensional loading
During one-dim
h
h
will change since
= 0) is imposed.
The ratio Ko
h
z is known as the coefficient of earth pressure at rest.
Ko depends on
· the type of soil
· the overconsolidation ratio (Ro)
· the loading or unloading cycle
Approximations
normally consolidated soils:
Konc » 1 c
overconsolidated soils:
Ko » Konc ÖRo
State
ratio Ro
The state at A is given by any two of
ea
a , Ro
y
a
All states with the same Ro fall on the lines parallel with the NCL.
log Ro
y
a)
ya
Many features of soil behaviour, especially shear modulus and peak strength, increase with
increasing overconsolidation.
Change of
state
(relevant to all
Loading and unloading
soils)
Change of state A to B can only be achieved by normal compression along CD followed by
swelling along DB. Note that the yield stress corresponding to B is larger than the yield stress
corresponding to A.
Vibration or compaction
(relevant to sands)
or creep:
(relevant to clays)
Change of state can occur directly from A to B. Note that the yield stress corresponding to B is
larger than the yield stress corresponding to A.
Critical state
There is a critical overconsolidation ratio which separates states in which the soil will either
compress or dilate during shear. This corresponds to the critical state line CSL. Look at the
possible voids ratios (e) that can o
a.
wet side of critical
(W on the graph)
ew > ec
water content ww is larger than critical wc
· loose
· normally consolidated
or lightly overconsolidated
· compress during drained shear
dry side of critical
(D on the graph)
ed < ec
water content wd is smaller than critical wc
· dense
· heavily overconsolidated
· dilate during drained shear
Normalising parameters
Normalising parameters change the current state to a normalised state so that all states with the
same overconsolidation ratio have the same value.
Equivalent void ratio
el = ea + Cc
a
Equivalent stress
e = ( eN - ea ) / Cc
Critical stress
c = ( eG - ea ) / Cc
If A is on the wet side of critical
el > eG
>1
If A is on the dry side of critical
el < eG
a
c<1
a
c
Wet and dry states

State parameters
Soils whose states lie on the normal compression line
(NCL) are normally consolidated. There is a critical
overconsolidation ratio that corresponds with the
critical state line (CSL).
A lightly overconsolidated soil has a state which lies above the CSL.
A heavily overconsolidated soil has a state which lies below the CSL.
States lying above the CSL are said to be on the wet side of critical.
States lying below the CSL are said to be on the dry side of critical.
In the diagrams: va > vb, and yet since the stress at B is greater, state B is on the wet
side of critical, while state A is on the dry side of critical.
Wet and dry states

State parameters
Soils whose states lie on the normal compression line (NCL) are normally consolidated. There is
a critical overconsolidation ratio that corresponds with the critical state line (CSL).
A lightly overconsolidated soil has a state which lies above the CSL.
A heavily overconsolidated soil has a state which lies below the CSL.
States lying above the CSL are said to be on the wet side of critical.
States lying below the CSL are said to be on the dry side of critical.
In the diagrams: va > vb, and yet since the stress at B is greater, state B is on the wet side of
critical, while state A is on the dry side of critical.
State parameters
A measure of the initial state of a soil are the distances it lies at from the CSL, in terms of either
volume or stress. These distances are expressed as state parameters:
Stress state parameter
Ss = pa' / pc'
ln Ss = ln pa' - ln pc'
Volume state parameter
Sv = va - vc
The state parameters are related:
Sv = ln Ss
Normally consolidated state:
Sv
s=0
States on the wet side of critical:
Sv and ln Ss are positive
States on the dry side of critical:
Sv and ln Ss are negative
Consolidation
When soil is loaded undrained, the pore pressures increase. Then, under site conditions, the excess
pore pressures dissipate and water leaves the soil, resulting in consolidation settlement. This
process takes time, and the rate of settlement decreases over time.
The amount of settlement which occurs in a given time depends on the
1. permeability of the soil
2. length of the drainage path
3. compressibility of the soil
If soil is unloaded (e.g. by excavation) the excess pore pressures may be negative.
The process of consolidation and settlement
In coarse soils (sands and gravels) any volume change resulting from a change in loading occurs
immediately; increases in pore pressures are dissipated rapidly due to high permeability. This is
called drained loading.
In fine soils (silts and clays) - with low permeabilities - the soil is undrained as the load is
applied. Slow seepage occurs and the excess pore pressures dissipate slowly, consolidation
settlement occurs.
The rate of volume change diminishes with time; about one-half of the total consolidation
settlement occurs in one-tenth of the total time.
The basic consolidation process and terminology
Consider a site on clay soil with initial steady-state
groundwater conditions. An embankment is built, the
loading is undrained: the pore pressure in the soil increases,
seepage flow and therefore volume changes commences. As
consolidation takes place, settlement occurs, and continues
at a decreasing rate until steady-state conditions are
regained.
Click on the buttons to see the sequence of loading and pore
pressure changes.
Terms and symbols
Seepage refers to the flow of groundwater in a saturated soil.
q = rate of seepage flow
Excess pore pressure ( )
is the difference between the current pore pressure (u) and the steady
state pore pressure (uo).
= u - uo
Hydraulic gradient (i)
is the difference in total head between two points in the soil.
Permeability or the coefficient of permeability (k)
relates to flow in a given direction, i.e. along a given drainage path.
One-dimensional consolidation
A general theory for consolidation, incorporating three-dimensional flow vectors is complicated
and only applicable to a very limited range of problems in geotechnical engineering. For the vast
majority of practical settlement problems, it is sufficient to consider that both seepage and strains
take place in one direction only; this usually being vertical.
One-dimensional consolidation specifically occurs when there is no lateral strain, e.g. in
theoedometer test
One-dimensional consolidation can be assumed to be occurring under wide foundations.
One-dimensional consolidation theory
A simple one-dimensional consolidation model consists of rectilinear element of soil subject to
vertical changes in loading and through which vertical (only) seepage flow is taking place.
There are three variables:
1. the excess pore pressure ( )
2. the depth of the element in the layer (z)
3. the time elapsed since application of the loading (t)



The total stress on the element is assumed to remain constant.
The coefficient of volume compressibility (mv) is assumed to be constant.
The coefficient of permeability (k) for vertical flow is assumed to be constant.
Mathematical model and equation
Consider the element of consolidating soil. In time :
· the seepage flow is 
(q = A k i = A k 
· the change in excess pressure is
· the thickness changes by
 = -mv ´
It can be shown that the basic equation for one-dimensional consolidation is:
By defining the coefficient of consolidation as
this can be written:
Isochrones
Solutions to the one-dimensional consolidation equation can be obtained by
plotting the variation of with the depth in the layer at given elapsed
times. The resulting curves are called isochrones. (Gk. iso = equal;
kronos = time)
The figure shows a set of supposed standpipes inserted into a
consolidating layer.
Before loading, the pore pressure in the drain is zero. At the base of
each standpipe there is some initial pore pressure u= uo, the excess
pore pressure = 0.
Immediately after the loading is applied the standpipes will each
show an initial excess pore pressure of i, thereafter the excess pore
pressure will dissipate.
Click on the following time intervals to observe the changes in across the thickness of the layer
with time.
1. Before loading = 0
2. Initial (after loading) when time = 0
3. 0 < time < tc
4. time = tc (still no change at the bottom)
5. tc < time < t¥
6. Finally at time = ¥
Adjacent to the drain (at the top) the excess pore pressure drops to zero almost immediately
At the bottom of the layer the dissipation is quite slow.
Some properties of isochrones
The gradient of an isochrone is related to the hydraulic gradient (i):
At the drainage surface, isochrones are steepest and = 0.
At the impermeable (k = 0) base the seepage velocity is zero since V = ki; the isochrones will
therefore be at 90° to the impermeable boundary.
Between two isochrones the change in thickness in time
-mv
,
2 - t1
is the shaded area.
Thus, the settlement at the surface of the layer is given by:
v
area OAB
Terzaghi's solution
The basic equation is
(z,t) is excess pore pressure at depth z after time t.
The solution depends on the boundary conditions:
The general solution is obtained for an overall (average) degree of consolidation using nondimensional factors.
General solution
The following non-dimensional factors are used in order to obtain a solution:
· Degree of consolidation at depth z
· Time factor
· Drainage path ratio
The differential equation can now be written as:
If the excess pore pressure is uniform with depth, the solution is:
Putting Ut
t
Drainage path length
During consolidation water escapes from the soil to the surface or to a permeable sub-surface layer
above or below (where = 0). The rate of consolidation depends on the longest path taken by a
drop of water. The length of this longest path is the drainage path length, d. Typical cases are:
An open layer, a permeable layer both above and below (d = H/2)
A half-closed layer, a permeable layer either above or below (d = H)
Vertical sand drains, horizontal drainage (d = L/2)
One-dimensional consolidation theory
Solution using
parabolic
isochrones
Isochrones can be approximated to parabolas, affording reasonably accurate solutions to the
differential equation for one-dimensional consolidation. Solutions must be obtained for two
separate, but adjoining, cases:
· When the elapsed time (t) is less than the critical time (tc)
· When the elapsed time (t) is greater than the critical time (tc)
The critical time is the time that must elapse before the excess pore pressures at the impermeable
base first begin to dissipate.
Solution for t < tc case
Putting time factor
and average degree of consolidation,
the general solution is
This is valid for 0 < t < tc
At t = tc, n = H =
Giving
and
Ut = 0.3333
Solution for t > tc case
Putting time factor
and average degree of consolidation,
the general solution is
This is valid for tc < t < t¥
At t = tc, n = H =
Giving
and
Ut = 0.3333
The oedometer test

Apparatus and procedure
The one-dimensional compression and swelling characteristics of a soil may be measured in the
laboratory using the oedometer test (from the Greek: oidema = a swelling).
A cylindrical specimen of soil enclosed in a metal ring is subjected to a series of increasing static
loads, while changes in thickness are recorded against time.
From the changes in thickness at the end of each load stage the compressibility of the soil may be
observed, and parameters measured such as Compression Index (Cc) and Coefficient of Volume
Compressibility (mv).
From the changes in thickness recorded against time during a load stage the rate of consolidation
may be observed and the coefficient of consolidation (cv) measured.
The test is fully detailed in BS 1377.
Apparatus and procedure
The saturated specimen is usually 75 mm diameter and 15-20 mm thick, enclosed in a circular
metal ring and sandwiched between porous stones.
Vertical static load increments are applied at regular time intervals (e.g. 12, 24, 48 hr.). The load is
doubled with each increment up to the required maximum (e.g. 25, 50, 100, 200, 400, 800 kPa).
During each load stage thickness changes are recorded against time.
After full consolidation is reached under the final load, the loads are removed (in one or several
stages - to a low nominal value close to zero) and the specimen allowed to swell, after which the
specimen is removed and its thickness and water content determined. With a porous stone both
above and below the soil specimen the drainage will be two-way (i.e. an open layer in which the
drainage path length, d = H/2)
Determination of cv from test results
The recorded thickness changes during one of the load stages in an oedometer test are used to
evaluate the coefficient of consolidation (cv).
The procedure
involves plotting
thickness changes
(i.e. settlement)
against a suitable
function of time
[either Ötime or log(time)] and then fitting to this the theoretical Tv:Ut curve.
In this way known intercepts of Tv:Ut are located from which cv may be calculated.
The Root-Time method
The first portion of the curve of settlement against Ötime is approximately a straight line. The U0
(Ut = 0) point is located at the intercept with the Ut axis. A second point is required: suppose this is
U90/Öt90 (point C). The location of this point depends on the equation for the curved portion [See
curve fitting methods: Terzaghi or parabolic isochrones]. Once U90 has been located other values
follow since the Ut axis scale is linear. The coefficient of consolidation is therefore:
where d = drainage path length
[d = H for one-way drainage, d = H/2 for two-way drainage]
Other appropriate time-interval values could be used:
e.g. U50, ÖT50, Öt50 , etc.
Curve fitting based on Terzaghi's equation
From Terzaghi’s analysis, the straight-line portion is:
For 0 < Ut < 0.6,
On the straight line:
ÖT90
On the curved portion:
ÖT90 = AC = Ö0.848 = 0.9209
Thus, a line drawn through points O and C has abscissae 1.15 times greater than those of the
straight line (OB). [0.9209/0.7976 = 1.15]
After the laboratory results curve has been plotted, line OB is drawn, followed by line OC: this
crosses the laboratory curve at point (ÖT90,U90) and locates Öt90
The coefficient of consolidation is therefore:
Curve fitting based on parabolic isochrones
From the parabola equation the straight-line portion is:
For 0 < Ut < 0.333,
On the straight line:
ÖT90 = AB = 0.9 x Ö(3/4) = 0.7794
On the curved portion:
ÖT90 = AC = Ö0.716 = 0.8462
Thus, a line drawn through points O and C has abscissae 1.086 times greater than those of the
straight line (OB). [0.8462/0.7794 = 1.086]
After the laboratory results curve has been plotted, line OB is drawn, followed by line OC: this
crosses the laboratory curve at point (ÖT90,U90) and locates Öt90
The coefficient of consolidation is therefore:
The Log-Time method
An alternative to the Root-Time method, that is particularly useful when there is significant
secondary compression (creep). The Uo point is located by selected two points on the curve for
which the times (t) are in the ratio 1:4, e.g. 1 min and 4 min; or 2 min and 8 min.; the vertical
intervals AP and PQ will be equal.
The U100 point can be located in the final part of the curve flattens sufficiently (i.e. no secondary
compression). When there is significant secondary compression, U100 may be located at the
intercept of straight line drawn through the middle and final portions of the curve.
Now U50 and log t50 can be located.
The coefficient of consolidation is therefore:
Calculation of settlement times
After the coefficient of consolidation (cv) has been determined from laboratory data calculations
are possible for site settlements. It is important to note that cv is not a constant, but varies with
both the level of stress and degree of consolidation. For practical site settlement calculations,
however, it is sufficiently accurate to measure cv relative to the loading range applicable on site
and then assume this value to be approximately constant for all degrees of consolidation (except
for very low values).
The basic equation used is:
where d = drainage path length
[d = H for one-way drainage, d = H/2 for two-way drainage]
Tv and t are coupled to a given degree of consolidation
Prediction of time for given settlement
Example
The final consolidation settlement of a layer of clay 5.0 m thick is calculated to be 280mm. The
coefficient of consolidation for the loading range is 0.955 mm²/min. There is two-way drainage,
upward and downward. Calculate the time required for (a) 90% consolidation settlement, (b) a
settlement of 100 mm.
(a) Drainage path length, d = 5.0/2 = 2.50 m = 2500 mm
For U90, T90 = 0.848. Then
(b) For 100 mm settlement, Ut = 100/280 = 0.357
and since Ut < 0.6, Tv = 0.357² x
Then time for 100mm settlement
Prediction of settlement amount at given time
Example
A layer of clay has a thickness of 4.0 m and drains both upward and downward. A laboratory test
has yielded a coefficient of consolidation for the appropriate loading range of 0.675 mm²/min. The
final consolidation settlement has been calculated to be 120mm. Provide estimates of the
consolidation settlement that may be expected 1yr, 2yr, 5yr and 10yr after construction.
Drainage path length, d = 2000 mm
When Ut < 0.6, use Ut = Ö(4Tv/
When Ut > 0.6,
cv = 0.645 mm²/min = 928.8 mm²/day
time t time t
Tv
Ut
Ut
rc (mm)
(years) (days) = cvt/d² (<0.6) (>0.6) at time t
1
365
0.0848 0.328
39
2
730
0.1965 0.465
56
5
1825
0.4238 0.735 0.715
86
10
3650
0.8475
0.900
108
23.6
8613
2.0
0.994
119
Reliability for design purposes
Laboratory measurements of stress-strain parameters (Cc, Cs, mv) are generally acceptable,
provided sampling quality is good, e.g. minimal disturbance, valid representation of strata,
maintenance of structure and water content, careful preparation, etc.
Measurements of strain/time relationships (cv) and permeability (k) are not so reliable.
Observed rates of settlement are generally greater than values based on oedometer test results.
Reliability is compromised by factors such anisotropy (e.g. silt/sand
layers, varves, fissures, etc), presence of roots, organic matter and
voids, and also the effects of secondary compression.
Loads are not often applied instantaneously, and so due allowance
should be for the gradual application of loading.
Secondary compression or creep
In some soils (especially recent organic soils) one-dimensional compression continues under
constant loading after all of the excess pore pressure has dissipated, i.e. after primary consolidation
has ceased - this is called secondary compression or creep.
It is generally thought that creep is due to changes in soil structure, although no reliable theory has
been proposed as yet.
It is likely that some creep is occurring due primary consolidation, affecting the linearity of the
r/Ötime curve and thus making the accurate prediction of settlement difficult and possibly
unreliable.
For practical purposes, the Log-Time plot (described elsewhere) can be used to estimate a
.
Coefficient of secondary compression
The amount of secondary compression is the settlement occurring after t100, i.e. after full
dissipation of excess pressures
a (or sa).
100 can be approximated to a straight line, the slope of which gives the
coefficient of secondary compression (Ca).
The slope of the laboratory curve is measured over one log-time cycle, e.g.1000 to 10000 mins.
Overconsolidation due to creep
Creep (secondary compression) is basically similar to compaction, except it takes place slowly.
The result of creep is a change in volume (also water content and void ratio).
The soil is in effect further consolidated, and therefore if unloaded is left overconsolidated.
The phenomenon of overconsolidation due to creep is noticeable in soft clays.
Compaction
Compaction is a process that brings about an increase in soil
density or unit weight, accompanied by a decrease in air
volume. There is usually no change in water content. The
degree of compaction is measured by dry unit weight and
depends on the water content and compactive effort (weight
of hammer, number of impacts, weight of roller, number of
passes). For a given compactive effort, the maximum dry unit
weight occurs at an optimum water content.
Compaction purposes and processes
Compaction is a process of increasing soil density and removing air, usually by mechanical means.
The size of the individual soil particles does not change, neither is water removed.
Purposeful compaction is intended to improve the strength and stiffness of soil. Consequential (or
accidental) compaction, and thus settlement, can occur due to vibration (piling, traffic, etc.) or selfweight of loose fill.
Compaction as a construction process
Compaction is employed in the construction of road bases, runways, earth dams, embankments
and reinforced earth walls. In some cases, compaction may be used to prepare a level surface for
building construction.
Soil is placed in layers, typically 75 mm to 450 mm thick. Each layer is compacted to a specified
standard using rollers, vibrators or rammers.
Refer also to Types of compaction plant and Specification and quality control
Objectives of compaction
Compaction can be applied to improve the properties of an existing soil or in the process of
placing fill. The main objectives are to:



increase shear strength and therefore bearing capacity
increase stiffness and therefore reduce future settlement
decrease voids ratio and so permeability, thus reducing potential frost heave
Factors affecting compaction
A number of factors will affect the degree of compaction that can be achieved:




Nature and type of soil, i.e. sand or clay, grading, plasticity
Water content at the time of compaction
Site conditions, e.g. weather, type of site, layer thickness
Compactive effort: type of plant (weight, vibration, number of passes)
Types of compaction plant
Construction traffic, especially caterpillar-tracked vehicles, is also used.
In the UK. further information can be obtained from the Department of Transport and handbooks
on civil engineering construction methods.
Smooth-wheeled roller



Self-propelled or towed steel rollers ranging from 2
- 20 tonnes
Suitable for: well-graded sands and gravels
silts and clays of low plasticity
Unsuitable for: uniform sands; silty sands; soft clays
Grid roller




Towed units with rolls of 30-50 mm bars, with spaces between of 90-100 mm
Masses range from 5-12 tonnes
Suitable for: well-graded sands; soft rocks; stony soils with fine fractions
Unsuitable for: uniform sands; silty sands; very soft clays
Sheepsfoot roller





Also known as a 'tamping roller'
Self propelled or towed units, with hollow drum fitted with projecting
club-shaped 'feet'
Mass range from 5-8 tonnes
Suitable for: fine grained soils; sands and gravels, with >20% fines
Unsuitable for: very coarse soils; uniform gravels
Pneumatic-tyred roller





Usually a container on two axles, with rubber-tyred wheels.
Wheels aligned to give a full-width rolled track.
Dead loads are added to give masses of 12-40 tonnes.
Suitable for: most coarse and fine soils.
Unsuitable for: very soft clay; highly variable soils.
Vibrating plate



Range from hand-guided machines to larger roller combinations
Suitable for: most soils with low to moderate fines content
Unsuitable for: large volume work; wet clayey soils
Power rammer


Also called a 'trench tamper'
Hand-guided pneumatic tamper


Suitable for:
work in confined
Unsuitable for:
Laboratory
trench back-fill;
areas
large volume work
compaction tests
The variation in
compaction with water
content and compactive
effort is first established in
the laboratory. Target values are then specified for the dry density and/or air-voids content to
be achieved on site.
Dry-density/water-content relationship
The aim of the test is to establish the maximum dry density that may be attained for a given
soil with a standard amount of compactive effort. When a series of samples of a soil are
compacted at different water content the plot usually shows a distinct peak.



The maximum dry density occurs at an optimum water content
The curve is drawn with axes of dry density and water content and the controlling values
are values read off:
d(max) = maximum dry density
wopt = optimum water content
Different curves are obtained for different compactive efforts
Explanation of the shape of the curve
For clays
Recently excavated and generally saturated lumps of clayey soil have a relatively high undrained
shear strength at low water contents and are difficult to compact. As water content increases, the
lumps weaken and soften and maybe compacted more easily.
For coarse soils
The material is unsaturated and derives strength from suction in pore water which collects at grain
contacts. As the water content increases, suctions, and hence effective stresses decrease. The soil
weaken, and is therefore more easily compacted.
For both
At relatively high water contents, the compacted soil is nearly saturated (nearly all of the air has
been removed) and so the compactive effort is in effect applying undrained loading and so the void
volume does not decrease; as the water content increases the compacted density achieved will
decrease, with the air content remaining almost constant.
Expressions for calculating density
A compacted sample is weighed to determine its mass: M (grams)
The volume of the mould is: V (ml)
Sub-samples are taken to determine the water content: w
The calculations are:
Worked example
A compacted soil sample has been weighed with the following results:
Mass = 1821 g Volume = 950 ml Water content = 9.2%
Determine the bulk and dry densities.
Bulk density
d = 1.917 / (1+0.092) = 1.754 Mg/m³
Dry density and air-voids content
A fully saturated soil has zero air content. In practice, even quite wet soil will have a small air
content
The maximum dry density is controlled by both the water content and the air-voids content.
Curves for different aird / w plot using this expression:
The air-voids content corresponding to the maximum dry density and optimum water content can
d/w plot or calculated from the expression (see the worked example).
Worked example
Determine the dry densities of a compacted soil sample at a water content of 12%, with air-voids
contents of zero, 5% and 10%. (Gs = 2.68).
Effect of increased
compactive effort
The compactive effort will
be greater when using a
heavier rammer in the
compactive effort:
heavier roller on site or a
laboratory. With greater



maximum dry
optimum water content decreases
air-voids content remains almost the same.
density increases
Effect of soil type



Well-graded granular soils can be compacted to higher densities than uniform or silty soils.
Clays of high plasticity may have water contents over 30% and achieve similar densities
(and therefore strengths) to those of lower plasticity with water contents below 20%.
As the % of fines and the plasticity of a soil increses, the compaction curve becomes flatter
and therefore less sensitive to moisture content. Equally, the maximum dry density will be
relatively low.
Interpretation of laboratory data
During the test, data is collected:
1.
2.
3.
4.
5.
Volume of mould (V)
Mass of mould (Mo)
Specific gravity of the soil grain (Gs)
Mass of mould + compacted soil - for each sample (M)
Water content of each sample (w)
d
/ w curve is plotted together with the air-voids curves.
The maximum dry density and optimum water content are read off the plot.
The air content at the optimum water content is either read off or calculated.
Example data collected during test
In a typical compaction test the following data might have been collected:
Mass of mould, Mo = 1082 g
d
Volume of mould, V = 950 ml
Specific gravity of soil grains, Gs = 2.70
Mass of mould + soil (g) 2833 2979
Water content (%)
3080
3092
3064
3027
8.41 10.62 12.88 14.41 16.59 18.62
Calculated densities and density curve
The expressions used are:
1.84
Water content, w
d
(Mg/m³)
2.00
2.10
2.12
2.09
2.05
0.084 0.106 0.129 0.144 0.166 0.186
1.70
1.81
1.86
Air-voids curves
The expression used is:
Water content (%) 10 12 14 16 18 20
d when Av = 0% 2.13 2.04 1.96 1.89 1.82 1.75
d when Av = 5% 2.02 1.94 1.86 1.79 1.73 1.67
d when Av = 10% 1.91 1.84 1.76 1.70 1.64 1.58
1.851
1.79
1.73
The optimum air-voids content is the value corresponding to the maximum dry density (1.86
Mg/m³) and optimum water content (12.9%).
Specification and quality control
The degree of compaction achievable on site depends mainly on:
 Compactive effort: type of plant + No of passes
-versa
-graded soils; fine soils have higher water contents
End-result specifications require predictable conditions
Method specifications are preferred in UK.
End-result specifications
Target parameters are specified based on laboratory test results:
Optimum water content working range, i.e. ± 2%
Optimum air-voids content tolerance, i.e. ± 1.5%
For soils wetter than wopt, the target Av can be used, e.g.
10% for bulk earthworks
5% for important work
The end-result method is unsuitable for very wet or variable conditions.
Method specifications
A site procedure is specified giving:


type of plant and its mass
maximum layer thickness and number of passes.
This type of specification is more suitable for soils wetter than wopt or where site conditions
are variable - this is often the case in the UK. The Department of Transport publishes a
widely used method specification for use in the UK.
Moisture condition value
This is a procedure developed by the Road Research Laboratory using only one sample, thus
making laboratory compaction testing quicker and simpler. The minimum compactive effort to
produce near-full compaction is determined. Soil placed in a mould is compacted by blows from a
rammer dropping 250 mm; the penetration after each blow is measured.
Apparatus and sizes
Cylindrical mould, with permeable base plate:
internal diameter = 100 mm, internal height at least 200 mm
Rammer, with a flat face:
face diam = 97 mm, mass = 7.5 kg, free-fall height = 250 mm
Soil:
1.5 kg passing a 20 mm mesh sieve
Test procedure and plot







Firstly, the rammer is lowered on to the soil surface and allowed to penetrate under its own
weight
The rammer is then set to a height of 250 mm and dropped on to the soil
The penetration is measured to 0.1 mm
The rammer height is reset to 250 mm and the drop repeated until no further penetration
occurs, or until 256 drops have occurred
The change in penetration ) is recorded between that for a given number of blows (n)
and that for 4n blows
A graph is plotted of  / n and a line drawn through the steepest part
The moisture condition value (MCV) is give by the intercept of this line and a special
scale
Example plot and determination of MCV
After plotting  against the number of blows n, a line is drawn through the steepest part.
The intercept of this line and the 5 mm penetration line give the MCV
The defining equation is:MCV = 10 log B
(where B = number of blows corresponding to 5 mm penetration)
On the example plot here an MCV of 13 is indicated.
Significance of MCV in earthworks
The MCV test is rapid and gives reproducible results which correlate well with engineering
properties. The relationship between MCV and water content for a soil is near to a straight line,
except for heavily overconsolidated clays.A desired value of undrained strength or compressibility
can be related to limiting water content, and so the MCV can be used as a control value after
calibrating MCV vs w for the soil. An approximate correlation between MCV and undrained shear
strength has been suggested by Parsons (1981).
Log su = 0.75 + 0.11(MCV)
Shear strength
Near any geotechnical construction (e.g. slopes, excavations, tunnels and foundations) there will
be both mean and normal stresses and shear stresses. The mean or normal stresses cause volume
change due to compression or consolidation.
The shear stresses prevent collapse and help to support the geotechnical structure. Shear stress may
cause volume change.
Failure will occur when the shear stress exceeds the limiting shear stress (strength).
Common cases of shearing
Back to Shear strength
In practice, the state of stress in the ground will be complex.
There are simple theories for two special cases.
Triaxial (axial symmetry)
Parameters used for analysis:
· deviator stress
· shear strain
· normal stress
· volumetric strain
· specific volume
Direct or simple shear
Parameters used for analysis:
· shear stress
· shear strain
· normal stress
· volumetric (normal) strain
· void ratio
It is not possible to draw a Mohr circle for a shear test unless stresses on vertical planes are
measured.
Strength
Back to Shear strength
In very simple terms, the strength of soil is the maximum shear stress ( f) it can sustain, or the
shear stress acting on a shear slip surface along which it is failing. There are three distinct
strengths: peak, critical (or ultimate) and residual. Shearing may be simple or direct.
Drained direct (ring) shear
Drained simple shear
We explore the relationship between the maximum shear stress and the effective normal stress (
f
Some aspects of the behaviour show up more clearly if we normalise
f
Back to Strength
Peak strength
The peak strength is the maximum value of the shear stress or the maximum value of the ratio of
shear stress to effective mean or normal stress. For drained tests these will occur simultaneously,
for undrained tests they may occur at different points and the definition used here is the maximum
stress ratio.





Peak strengths can only occur at shear stresses above the critical state line and at water
contents below the CSL.
Peak states can occur anywhere in the regions above and below the CSL.
Peak states at the same water content fall on unique smoth envelopes.
The peak states can be represented on a graph in 3 dimensions.
All peak states fall on a surface in this graph.
Peak strength in shear tests
The circle represent the results of a set of shear tests on samples
at the same moisture content but different normal stresses. The
squares represent the results of a second set of tests at a
different moisture content.
We normalise the data by plotti
against /
f/
The basic peak states, before normalisation, fall on different
curves each for a particular water content or void ratio.
After normalisation all the peak states fall on a single unique
envelope.
Equations
At a given water content or void ratio, all the peak states fall on a single
smooth envelope.
This may be represented in one of two ways:
As a power law
p
As a linear (Mohr-Coulomb) envelope
if the curvature is relatively small over a given range.
p = c'p
p
The parameters a, B and c'p p depend on the water content or void
ratio.
Even at a given water content or voids ratio, the parameters c'p
depend on the range of stress for linear approximation.
p
Peak strength in triaxial tests
The basic peak states, before normalisation, fall on different curves each for a particular
water content or specific volume.
After normalisation all the peak states fall on a single unique envelope.
equations
At a given water content or specific volume all the peak states fall on a single smooth
envelope. This may be represented in one of two ways:
As a power law
As a linear envelope
if the curvature is relatively small over a given range.
q'p = Gp + Hp p'
p, Hp depend on the water content or voids ratio.
Even at a given water content or voids ratio, the parameters Gp and Hp depend on the range of
stress for linear approximation.
Peak strength and dilatancy

Peak state and initial state
Stresses and displacements in a
shear sample are analagous to the
forces and movements of a
friction block on an inclined
plane.
At critical state
c
The additional stress ratio (above the critical state) is due to the rate
of dilation
Peak state and initial state
The peak stress ratio depends on the initial state given by the initial
overconsolidation ratio. The maximum rate of dilation increases
with overconsolidation ratio. For the same initial overconsolidation ratio (i.e. A and A') the peak
stress ratio is the same.
Critical state strength
At its critical state soil continues to distort at constant effective stress and at constant volume.
This applies for turbulent flow of the particles: if the flow becomes laminar, as in clays at large
strain, the strength falls to the residual.
When soil is at its critical state there is a unique relationship between shear stress, effective normal
stress and water content (or specific volume or void ratio). Critical states are unique and do not
depend on initial state or stress path.
Critical states correspond to shear strains typically 10% to 40%.
Critical shear stress (critical state strength) increases with increasing effective normal stress and
with decreasing water content.
The critical state line can be represented as a graph in 3 dimensions. For isotropic compression,
shear stresses are zero and the isotropic normal compression line can also be represented.
Critical state strength in shear tests
The graphs show the critical state line. If you know either
c cc and eG are soil parameters.
The one-dimensional normal compression line (NCL) for zero shear stress is also shown.
c or the equivalent
void ratio el. The critical state line and the isotropic normal compression line both reduce to single
points.
Critical state strength in triaxial tests
The graphs show the critical state line. If you know either p' or v at the critical state you can
calculate the critical state deviation stress q'.
M
We should really use subscripts c and e for compression and extension as the values are slightly
different.
The isotropic normal compression line corresponds to zero deviator stress
c
or the equivalent
specific
volume vl. The critical state line and the isotropic normal compression line both reduce to single
points.
Typical values of critical state strength parameters
The critical state parameters are basic soil parameters and they depend principally on the nature of
the soil grains. For fine grained soils the CS parameters are related to the Atterberg limits; for
coarse-grained soils they are related to the mineralogy and shape of the grains.
Typical values
G
M
high plasticity clay
0.16
2.45
0.89
23º
low plasticity clay
0.10
1.80
1.18
29º
quartz sand
0.16
3.00
1.28
32º
carbonate sand
0.34
4.35
1.65
40º
For fineby
Cc = (Ip x Gs) / 200
p x Gs) / 460
c
of the critical state line is related to the Atterberg limits
For many soils the critical state lines all pass
v(W) = 1.25 p'(W) = 10MPa
e(W
W) = 15MPa.
Undrained strength
The critical state
strength is uniquely related to the water content.
If the soil is sheared without change of water content
(i.e. undrained) its strength remains the same. This is
called the undrained strength su. But if the soil is not
undrained and the water content changes the strength
will also change.
The undrained strength is directly related to the
liquidity index IL. Some authors give slightly different values for su but
su at PL (i.e. IL=0) is always 100 times
su at LL (i.e. IL=1)
Residual strength
This is the very lowest strength which occurs after very large displacements. For sands the residual
strength is the same as the critical state strength. For clays the residual is about ½ the critical state
strength. For clays the flat clay particles become aligned parallel to the direction of shear.
The residual strength occurs after very large (>1m) movements and is not usually relevant for
geotechnical engineering where generally ground movements must be small. However, on old
landslides there may have already been very large movements and in such cases the strength may
already be at the residual before construction starts.
Residual strength: equations
Residual strength applies to clays after very large shear displacements when clay particles have
become aligned in well-defined shear zones or slip planes.
Drained case
r
r
= residual friction angle
r
c.
For London Clay,
c
r»10º.
r depends on the quantity of clay present.
Undrained case
ur
sur = undrained residual strength
(depends on water content)
Groundwater
Soils consist of mineral particles in contact surrounded by voids or pores. The voids contain fluid which
may be liquid, gas or a mixture of the two.
The relative volumes can be described by the void ratio e, specific volume v, or porosity n.
In dry soils (Sr = 0) the single pore fluid is air.
In
saturated soils (Sr = 1) the single pore fluid is water.
Discussion of groundwater is usually concerned with saturated soils.








Pore water pressure
Permeability
Analytical solutions
Flow nets
Quick condition and piping
Measurement
Groundwater control
Some case histories
Pore water pressure




Water table
Elevation, pressure and total head
Hydraulic gradient
Effective stress
In general, the water in the voids of an element of
saturated soil will be under pressure,
either due to the physical location of the soil or as a
result of external forces. This pressure
is the pore water pressure or pore pressure u. It is measured relative to atmospheric pressure.
When there is no flow, the pore pressure at depth d below the water surface is: u = w d
Water table



Fine-grained soils
Coarse-grained soils
Perched water table
The level in the ground at which the pore pressure is zero (equal to atmospheric) is defined as the water
table or phreatic surface.
When there is no flow, the water surface will be at exactly the same level in any stand pipe placed in the
ground below the water table. This is called a hydrostatic pressure condition.
The pore pressure at depth d below the water table is: : u = w d
Fine-grained soils
In fine grained soils, surface tension effects can cause capillary water to rise above the water table. It is
reasonable to assume that the pore pressure varies linearly with depth, so the pore pressure above the water
table will be negative.
If the water table is at depth dw then the pore pressure at the ground surface is uo = and the pore pressure at depth z is
w (z - dw)
w.dw
Where the water table is deeper, or where evaporation is taking place from the surface, saturation with
capillary water may not occur. The height to which the soil remains saturated with negative pore pressures
above the water table is called the capillary rise.
Coarse-grained soils
Below the water table the soil can be considered to be saturated. In coarse-grained soils, water will drain
from the pores and air will therefore be present in the soil between the ground surface and the water table.
Consequently, pore pressures above the water table can usually be ignored. Below the water table,
hydrostatic water pressure increases linearly with depth.
With the water table at depth dw
u = 0 for z < dw
u = w(z - dw) for z > dw
Perched water table
Where the ground contains layers of permeable soil (e.g. sands) interspersed with layers of much lower
permeability (e.g. clays) one or more perched water tables may develop and the overall distribution of
pore pressure with depth may not be exclusivelyly linear.
Detection of perched water tables during site investigation is important, otherwise erroneous estimates of
in-situ pore pressure distributions can arise.
Pore pressure conditions below perched water tables may be affected by local infiltration of rainwater or
localised seepage and therefore may not be in hydrostatic equilibrium.
Elevation, pressure and total head
Pore pressure at a given point (e.g. point A in the diagram) can be measured by the height of
water in a standpipe located at that point.
Pore pressures are often indicated in this way on diagrams.
The height of the water column is the pressure head (hw)
hw =
u
w
To identify significant differences in pore pressure at different points, we need to eliminate the
effect of the points' position. A height datum is required from which locations are measured. The
elevation head (hz) of a point is its height above the datum line. The height above the datum of
the water level in the standpipe is the total head (h).
h = hz + hw
Hydraulic gradient
Flow of pore water in soils is driven from positions of higher
total head towards positions of lower total head. The level of the
datum is arbitrary. It is differences in total head that are
important. The hydraulic gradient is the rate of change of total
head along the direction of flow.
hz1 and hz2 above datum.
In the first diagram, the total heads are equal. The difference in pore pressure is
entirely due to the difference in altitude of the two points and the pore water has no
tendency to flow.
In the second diagram, the total heads are different. The hydraulic gradient is
i = (h2 - h1
and the pore water tends to flow.
Effective stress
All strength and stress:strain characteristics of soils can be linked to changes in effective stress
Effective stress (') = total stress () - pore water pressure (u)
u
Groundwater
Permeability




Void ratio
Stratified soil
Seepage velocity
Temperature
The rate of flow of water
Darcy's law
q (volume/time) through cross-sectional area A is found to be proportional to hydraulic gradient i according
to Darcy's law:
v = q = k.i
i=
A
where v is flow velocity and k is coefficient of permeability with dimensions of velocity (length/time).
The coefficient of permeability of a soil is a measure of the conductance (i.e. the reciprocal of the
resistance) that it provides to the flow of water through its pores.
The value of the coefficient of permeability k depends on the average size of the pores and is related to the
distribution of particle sizes, particle shape and soil structure. The ratio of permeabilities of typical
sands/gravels to those of typical clays is of the order of 106. A small proportion of fine material in a coarsegrained soil can lead to a significant reduction in permeability.
Void ratio and permeability
Permeability of all soils is strongly influenced by the density of packing of the soil particles which can be
simply desrcibed through void ratio e or porosity n.
Sands
10)²
m/s where d10 is the effective particle size in mm. This
relationship was proposed by Hazen.
The Kozeny-Carman equation suggests that, for laminar flow in saturated soils:
where ko and kT are factors depending on the shape and tortuosity of the pores respectively, Ss is the surface
area of the solid particles per unit volume of solid material, and w
the pore water. The equation can be written simply as
Clays
The Kozeny-Carman equation does not work well for silts and clays. For clays it is typically found that
where Ck is the permeability change index and ek is a reference void ratio.
For many natural clays Ck is approximately equal to half the natural void ratio.
Stratified soil and permeability
Consider a stratified soil having horizontal layers of thickness t1, t2, t3, etc. with coefficients of permeability
k1, k2 k3, etc.
For vertical flow, the flow rate q through area A of each layer is the same. Hence the head drop across a
series of layers is
The average coefficient of permeability is
For horizontal flow, the
So i1 = i2 = i3 etc. The flow rate through a layered block of soil of breadth B is therefore
The average coefficient of permeability is
Seepage velocity
Darcy's Law relates flow velocity (v) to hydraulic gradient (i). The volume flow rate q is calculated as the
product of flow velocity v and total cross sectional area:
q = v.A
At the particulate level the water follows a tortuous path through the pores. The average velocity at which
the water flows through the pores is the ratio of volume flow rate to the average area of voids Av on a cross
section normal to the macroscopic direction of flow. This is the seepage velocity vs
Porosity of soil is related to the volume fraction of voids
Seepage velocity can be measured in laboratory models by injecting dye into the seeping pore water and
timing its progress through the soil.
Temperature and permeability
The flow of water through confined spaces is controlled by its viscosity
temperature.
An alternative permeability K (dimensions: length² ) is sometimes used as a more absolute coefficient
depending only on the characteristics of the soil skeleton.
The values of k at 0°C and 10°C are 56% and 77% respectively of the value measured at 20°C.
Analytical solutions
Steady one-dimensional flow
Darcy's Law indicates the link between flow rate and hydraulic gradient. For one-dimensional flow,
constant flow rate implies constant hydraulic gradient.
Steady downward flow occurs when water is pumped from an underground
aquifer. Pore pressures are then lower than hydrostatic pressures.
Steady upward flow occurs as a result of artesian pressure when a less
permeable layer is underlain by a permeable layer which is connected through
the ground to a water source providing pressures higher than local hydrostatic
pressures.
The fountains of London were originally driven by artesian pressure in the
aquifers trapped beneath the London clay. Pumping from aquifers over the
centuries has lowered the water pressures below artesian levels.
Analytical solutions
Quasi-one-dimensional and radial flow



Cylindrical flow: confined aquifer
Cylindrical flow: groundwater lowering
Spherical flow
Where flow occurs in a confined aquifer whose thickness varies gently with
position the flow can be treated as being essentially one-dimensional.
The horizontal flow rate q is constant. For an aquifer of width B and varying thickness t, the discharge
velocity
and Darcy's Law indicates that
Hydraulic gradient varies inversely with aquifer thickness.
Quasi-one-dimensional and radial flow
Cylindrical flow: confined aquifer
Steady-state pumping to a well which extends the full
thickness of a confined aquifer is a one-dimensional
problem which can be analysed in cylindrical
coordinates: pore pressure or head varies only with
radius r.
Darcy's Law still applies, with hydraulic gradient dh/dr and area
A varying with radius:
where ro is the radius of the borehole and h0 the
constant head in the borehole.
Cylindrical flow: groundwater lowering
Pumping from a borehole can be used for deliberate groundwater lowering in order to facilitate excavation.
This is an example of quasi-one-dimensional radial flow with flow thickness t=h. Then
Spherical flow
Variation of pore pressure around a point source or side (for example, a piezometer being used for in-situ
determination of permeability) is a one-dimensional problem which can be analysed in spherical
coordinates: pore pressure or head varies only with radius r.
Darcy's Law still applies, with hydraulic gradient dh/dr and area A varying with radius: A=4
where r0 is the radius of the piezometer and h0 the constant head in the piezometer.
Two-dimensional flow, Laplace

Anisotropic soil
Two-dimensional steady flow of the incompressible pore fluid is governed by Laplace's equation which
indicates simply that any imbalance in flows into and out of an element in the x direction must be
compensated by a corresponding opposite imbalance in the y direction. Laplace's equation can be solved
graphically, analytically, numerically, or analogically.
For a rectangular element with dimensions
into the element is
x
y
and unit thickness, in the x direction the velocity of flow
the negative sign being required because flow occurs down the hydraulic gradient. The velocity of flow out
of the element is
Similar expressions can be written for the y direction. Balance of flow requires that
and this is Laplace's equation. In three dimensions, Laplace's equation becomes
Anisotropic soil
For a soil with permeability kx and ky in the x and y directions respectively, Laplace's equation for twodimensional seepage becomes
This can be solved by applying a scale factor to the x dimensions so that transformed coordinates xt are
used
In the transformed coordinates the equation regains its simple form
and flownet generation can proceed as usual. Calculations of flow are made using an equivalent
permeability
It may be preferable in some cases to transform the y coordinates using:
The equivalent permeability remains unchanged.
For many natural sedimentary soils seasonal variations in the depositional regime have resulted in
horizontal macroscopic permeabilities significantly greater than vertical permeabilities. Transformation of
to analysis of seepage in such situations.
coordinates lends itself
Transient flow, consolidation
Since water may be regarded as being essentially incompressible, unsteady flow may arise when water is
drawn into or expelled from pores as a result of changes in the size of pores. This can only occur as a result
of changes volume associated with changes in effective stress.
The time-dependent transient change in pore pressure that occurs as a result of some perturbation, and
associated change in effective stress is called consolidation.
One-dimensional compression tests in an oedometer define the relationship between vertical effective stress
v and specific volume v or void ratio e from which a one-dimensional compliance mv can be defined
Then, under conditions of constant total stress, consolidation is governed by a diffusion equation:
where cv is the coefficient of consolidation having dimensions (length²/time).
Solutions of the consolidation equation are typically presented as isochrones, i.e. variations of pore
pressure with position at successive times, but can also be converted to curves linking settlement with time.
Flow nets





Calculation of flow
Calculation of total flow
Boundary between layers
Boundary conditions
Flow through embankments
Solutions to Laplace's equation for two-dimensional seepage can be presented as flow nets. Two orthogonal
sets of curves form a flow net:
equipotentials connecting points of equal total head h
flow lines indicating the direction of seepage down a hydraulic gradient
If standpipe piezometers were inserted into the ground with their tips on a single equipotential then the
water would rise to the same level in each standpipe. (The pore pressures would be different because of
their different elevations.)
There can be no flow along an equipotential, because there is no hydraulic gradient, so there can be no
component of flow across a flow line. The flow lines define channels along which the volume flow rate is
constant.
Calculation of flow
Consider an element from a flow channel of length L between equipotentials which indicate a fall in total
head
gradient is
and for unit width of flow net the volume
flow rate is
There is an advantage in displaying or sketching
flownets in the form of curvilinear 'squares' so that
a circle can be insrcibed within each four-sided
figure bounded by two equipotentials and two flow
lines. Then b = L and q = k
equipotentials.
Calculation of total flow
For a complete problem, the flownet has been drawn with the overall head drop h divided into Nd equal
intervals:
d
with Nf flow channels.
Then the total flow rate per unit width is
It is usually convenient in sketching flownets to
make Nd an integer. The number of flow channels
Nf will then generally not be an integer. In the
example shown, of flow under a sheet pile wall
Nd ;= 10, Nf = 3.5 and q = 0.35kh per unit width.
Boundary between layers
Flow across a boundary between two layers of soil of different permeability produces a refraction effect.
Consideration of continuity of flow and of continuity of velocity normal to the interface shows that
It is not possible to construct a flow net with curvilinear squares on both sides of the interface unless the
head drop between equipotentials is changed in inverse proportion to the permeability ratio.
If the ratio of permeabilities is greater than about 10, e.g. at the boundary of a drainage layer then
construction of the part of the flow net in the more permeable soil is unlikely to be necessary.
Flow nets
Boundary conditions
A surface on which the total head is fixed (for example, from the level of a river, pool, reservoir) is an
equipotential. A surface across which there is no flow (for example, an impermeable soil layer or an
impermeable wall) is a flow line
For the situation shown, with flow occurring under a sheet pile wall, the axis of symmetry must also be an
equipotential.
Flow through embankments
Seepage through an embankment dam is an example of unconfined
flow bounded at the upper surface by a phreatic surface which
represents the top flow line and on which the pore pressure is
everywhere zero (atmospheric).
Total head changes and elevation changes thus match and for equal
head intervals
intervals between the points of intersection of equipotentials with the
phreatic surface.
Quick condition and piping
If the flow is upward then the water pressure tends to lift the soil element. If the upward water pressure is
high enough the effective stresses in the soil disappear, no frictional strength can be mobilised and the soil
behaves as a fluid. This is the quick or quicksand condition and is associated with piping instabilities
around excavations and with liquefaction events in or following earthquakes.
Seepage force
The viscous drag of water flowing through a soil imposes a seepage force on the soil in
the direction of flow.
Consider the actual distribution of pore water pressure around an element length L and
thickness b taken from a flownet, bounded by two equipotentials with fall in
wbL of water in the element and partly
providing the seepage force. It is found that the seepage force is
wbL
equivalent to a seepage pressure (force per unit volume) in the direction of flow
Critical hydraulic gradient
The quick condition occurs at a critical upward hydraulic gradient ic, when the seepage force just balances
the buoyant weight of an element of soil. (Shear stresses on the sides of the element are neglected.)
The critical hydraulic gradient is typically around
1.0 for many soils. Fluidised beds in chemical engineering
systems rely on deliberate generation of quick conditions
to ensure that the chemical process can occur most efficiently.
Measurement
Laboratory measurement of pore pressure
Laboratory measurements of pore pressure are required in undrained testing where soil properties are to be
measured in terms of effective stresses and in model tests which involve the loading or unloading of beds of
clay.
Traditionally, pore pressures are measured in the drainage lines just outside a triaxial cell. It is essential that
the pore pressure measurement system should be completely free of air and as stiff as possible so that
minimal amounts of pore water flow are required in order to register the changes in pore pressure.
In research testing, miniature pore-pressure transducers may be mounted directly on a soil sample in order
to speed up the expected response time.
Field measurement of pore pressure
Pore pressures can be measured in the ground with different types of piezometer. The simplest piezometer
consists of a open tube or standpipe with a porous tip. The change in measured head of water requires a
large flow of water into the tube, so response is slow.
A Casagrande piezometer comprises a porous tip in a filter zone at the base of a borehole, connected to a
narrow tube. The cross sectional area of the tube is small by comparison with the surface area of the filter,
so the flow required to register a change in pressure is smaller than in a standpipe, and the response is
quicker.
Closed circuit piezometer systems are read remotely by mechanical or electrical means and provide
possibilities for de-airing the pore water circuit. The response time is dependent on the length of connecting
tubing.
Electrical transducers (using strain gauge or vibrating wire techniques) can be placed in the ground. These
are stiff devices which respond rapidly, but can be difficult to keep de-aired particularly if there is a
possibility of the surrounding soil becoming unsaturated.
Laboratory measurement of permeability
Laboratory measurements of the permeability of soils can be made using a permeameter. For fine-grained
soils (clays), the coefficient of permeability can be estimated directly or indirectly during one-dimensional
compression tests in an oedometer.
Permeameter
Constant head test
Recommended for coarse-grained soils.
steady total head drop
L, as water flows through a sample of cross-section area A.
Falling head test
Recommended for fine-grained soils.
Total head h in standpipe of area a is allowed to
fall; heads h1 and h2 are measured at times t1
and t2.
Oedometer
Indirect measurement
Transient consolidation phenomena are controlled by the coefficient of consolidation. With knowledge of
one-dimensional compliance mv, coefficient of permeability k can be estimated from
Direct measurement
Direct measurement of permeability in oedometers is preferable. Flow pumps can be used to maintain a
constant flow rate (q) across the sample and to measure the resultant constant head (h). The coefficient of
permeability is then given by k = q.L / A.h
Field measurement of permeability
Field or in-situ measurement of permeability avoids the difficulties involved in obtaining and setting up
undisturbed samples in a permeameter or oedometer and also provides information about bulk permeability,
rather than merely the permeability of a
small and possibly
unrepresentative sample.
Pumping test
In a well-pumping test, the steady-state
heads h1 and h2 in observation
boreholes at radii r1 and r2 are monitored at flow rate q. If the pumping causes a drawdown in an unconfined
(i.e. open surface) soil stratum then
If the soil stratum is confined and of thickness t and remains saturated then
Constant head and falling head tests with in-situ piezometers can also be used.
Constant head and falling head tests
Field tests equivalent to the laboratory constant head and falling head tests can be performed in which
controlled heads or flows are applied to piezometer tips. In general, conditions around such piezometers are
not ideally cylindrically symmetric or spherically symmetric and an intake factor F (with dimensions of
length) is required for each particular geometry. Values of the intake factor may be deduced from analytical
or numerical studies.
For a borehole open to its base, of diameter D, and lined to the full depth F=2.75D.
If the cased hole is through impermeable soil and the base of the casing is at the interface with a permeable
stratum F=2D.
For an intake formed by a cylindrical filter zone of diameter D and length L in an infinite isotropic stratum
for L/D > 4
Then for a steady state, constant head test in which a flow q is required to maintain a head h:
For a falling head test in which heads h1 and h2 are measured at times t1 and t2 in a borehole of area A:
Groundwater control
Failure to control groundwater adjacent to a construction project may result in



flooding
instability
ground movement

loss of bearing capacity
The available techniques fall into 4 broad categories:




control of surface water
removal from within the works
extraction from the surrounding ground
exclusion
Selection of the most appropriate method depends on cost, which depends on




the nature of the ground
the size of the works
the duration of the works
the level of acceptable risk
Groundwater lowering
Removal of water from the ground will cause the water level to fall. How quickly and by how much
depends on the permeability of the soil and the distance between the adjacent wells.
Drainage of clays is impractical. Silt particles may be removed along with the water causing the formation
of voids in the ground and damage to pumps. The high yield in gravels may make the method impractical.
Lowering the ground water will reduce the pore water pressure and hence increase stability. Removing
water will tend to cause settlement although the effect is likely to be small for sandy soils in which the
technique works well.
Groundwater exclusion
Exclusion methods involve the installation of an impermeable barrier. This may be structural (steel sheet
piles or concrete diaphragm wall) and may form a part of the permanent work.
Other methods include





slurry trench with bentonite or native clay
thin grouted membrane
other forms of grouting
ground freezing
compressed air (for tunnels and shafts)
Excluding water may cause a build up in pore water pressure. Heave is a particular problem where a thin
layer of impermeable soil at the base of an excavation results in a high pore water pressure close to the
base.
SOIL MECHANICS
LECTURE NOTES
LECTURE # 1
SOIL AND SOIL ENGINEERING
* The term Soil has various meanings, depending upon the general field in which it is being
considered.
*To a Pedologist ... Soil is the substance existing on the earth's surface, which grows and
develops plant life.
*To a Geologist ..... Soil is the material in the relative thin surface zone within which roots
occur, and all the rest of the crust is grouped under the term ROCK irrespective of its
hardness.
*To an Engineer .... Soil is the un-aggregated or un-cemented deposits of mineral and/or
organic particles or fragments covering large portion of the earth's crust.
** Soil Mechanics is one of the youngest disciplines of Civil Engineering involving the study
of soil, its behavior and application as an engineering material.
*According to Terzaghi (1948): "Soil Mechanics is the application of laws of mechanics and
hydraulics to engineering problems dealing with sediments and other unconsolidated
accumulations of solid particles produced by the mechanical and chemical disintegration of
rocks regardless of whether or not they contain an admixture of organic constituent."
* Geotechnical Engineering ..... Is a broader term for Soil Mechanics.
* Geotechnical Engineering contains:
- Soil Mechanics (Soil Properties and Behavior)
- Soil Dynamics (Dynamic Properties of Soils, Earthquake Engineering, Machine
Foundation)
- Foundation Engineering (Deep & Shallow Foundation)
- Pavement Engineering (Flexible & Rigid Pavement)
- Rock Mechanics (Rock Stability and Tunneling)
- Geosynthetics (Soil Improvement)
Soil Formation
* Soil material is the product of rock
* The geological process that produce soil is
WEATHERING (Chemical and Physical).
* Variation in Particle size and shape depends on:
- Weathering Process
- Transportation Process
* Variation in Soil Structure Depends on:
- Soil Minerals
- Deposition Process
* Transportation and Deposition
Four forces are usually cause the transportation and deposition of soils
1- Water ----- Alluvial Soil 1- Fluvial
2- Estuarine
3- Lacustrine
4- Coastal
5- Marine
2- Ice ---------- Glacial Soils 1- Hard Pan
2- Terminal Moraine
3- Esker
4- Kettles
3- Wind -------- Aeolin Soils 1- Sand Dunes
2- Loess
4- Gravity ----- Colluvial Soil 1- Talus
What type of soils are usually produced by the different weathering & transportation
process?????????????????????????????????????????????????????????????????????????
???
- Boulders
- Gravel Cohesionless
- Sand (Physical)
- Silt Cohesive
- Clay (Chemical)
* These soils can be
- Dry
- Saturated - Fully
- Partially
* Also they have different shapes and textures
LECTURE # 2
SOIL PROPERTIES
PHYSICAL AND INDEX PROPERTIES
1- Soil Composition
- Solids
- Water
- Air
2- Soil Phases
- Dry
- Saturated * Fully Saturated
* Partially Saturated
- Submerged
3- Analytical Representation of Soil:
For the purpose of defining the physical and index properties of soil it is more convenient to
represent the soil skeleton by a block diagram or phase diagram.
4- Weight - Volume Relationships:
Weight
Wt = Ww + Ws
Volume
Vt = Vv + Vs = Va + Vw +
Vs
1Unit
Wei
ght Density
* Also known as
- Bulk Density
- Soil Density
- Unit Weight
- Wet Density
Relationships Between Basic Properties:
Examples:
_____________________________________________________________________________
Index Properties
Refers to those properties of a soil that indicate the type and conditions of the soil, and
provide a relationship to structural properties such as strength, compressibility, per
meability, swelling potential, etc.
______________________________________________________________________________
1- PARTICLE SIZE DISTRIBUTION
* It is a screening process in which coarse fractions of soil are separated by means of series of
sieves.
* Particle sizes larger than 0.074 mm (U.S. No. 200 sieve) are usually analyzed by means of
sieving. Soil materials finer than 0.074 mm (-200 material) are analyzed by means of
sedimentation of soil particles by gravity (hydrometer analysis).
1-1 MECHANICAL METHOD
U.S. Standard Sieve:
Sieve No. 4 10 20 40 60 100 140 200 -200
Opening in mm 4.76 2.00 0.84 0.42 0.25 0.149 0.105 0.074 Cumulative Curve:
* A linear scale is not convenient to use to size all the soil particles (opening from 200 mm to
0.002 mm).
* Logarithmic Scale is usually used to draw the relationship between the % Passing and the
Particle size.
Example:
Parameters Obtained From Grain Size Distribution Curve:
1- Uniformity Coefficient Cu (measure of the particle size range)
Cu is also called Hazen Coefficient
Cu = D60/D10
Cu < 5 ----- Very Uniform
Cu = 5 ----- Medium Uniform
Cu > 5 ----- Nonuniform
2- Coefficient of Gradation or Coefficient of Curvature Cg
(measure of the shape of the particle size curve)
Cg = (D30)2/ D60 x D10
Cg from 1 to 3 ------- well graded
3- Coefficient of Permeability
k = Ck (D10)2 m/sec
Consistency Limits or Atterberg Limits:
- State of Consistency of cohesive soil
1- Determination of Liquid Limit:
2- Determination of Plastic
Limit:
3- Determination of Plasticity Index
P.I. = L.L. - P.L.
4- Determination of Shrinkage Limit
5- Liquidity Index:
6- Activity:
SOIL CLASSIFICATION SYSTEMS
* Why do we need to classify soils ???????????
To describe various soil types encountered in the nature in a systematic way and gathering soils
that have distinct physical properties in groups and units.
* General Requirements of a soil Classification System:
1- Based on a scientific method
2- Simple
3- Permit classification by visual and manual tests.
4- Describe certain engineering properties
5- Should be accepted to all engineers
* Various Soil Classification Systems:
1- Geologic Soil Classification System
2- Agronomic Soil Classification System
3- Textural Soil Classification System (USDA)
4-American Association of State Highway Transportation Officials System (AASHTO)
5- Unified Soil Classification System (USCS)
6- American Society for Testing and Materials System (ASTM)
7- Federal Aviation Agency System (FAA)
8- Others
1- Unified Soil Classification (USC) System:
The main Groups:
G = Gravel
S = Sand
.........................
M = Silt
C = Clay
........................
O = Organic
........................
* For Cohesionless Soil (Gravel and Sand), the soil can be Poorly Graded or Well Graded
Poorly Graded = P
Well Graded = W
* For Cohesive Soil (Silt & Clay), the soil can be Low Plastic or High Plastic
Low Plastic = L
High Plastic = H
Therefore, we can have several combinations of soils such as:
GW = Well Graded Gravel
GP = Poorly Graded Gravel
GM = Silty Gravel
GC = Clayey Gravel
Passing Sieve # 4
SW = Well Graded Sand
SP = Poorly Graded Sand
SM = Silty Sand
SC = Clayey Sand
Passing Sieve # 200
ML = Low Plastic Silt
CL = Low Plastic Clay
MH = High Plastic Silt
CH = High Plastic Clay
To conclud if the soil is low plastic or high plastic use Gassagrande's Chart
_____________________________________________________________________________
2- American Association of State Highway Transportation Officials System
(AASHTO):
- Soils are classified into 7 major groups A-1 to A-7
Granular A-1 {A-1-a - A-1-b}
(Gravel & Sand) A-2 {A-2-4 - A-2-5 - A-2-6 - A-2-6}
A-3
More than 35% pass # 200
A-4
Fine A-5
(Silt & Clay) A-6
A-7
Group Index:
_____________________________________________________________________________
3- Textural Soil Classification System (USDA)
* USDA considers only:
Sand
Silt
Clay
No. Gravel in the System
* If you encounter gravel in the soil ------- Subtract the % of gravel from the 100%.
* 12 Subgroups in the system
Example: ********
MOISTURE DENSITY RELATIONSHIPS
(SOIL COMPACTION)
INTRODUCTION:
* In the construction of highway embankments, earth dams, and many other engineering
projects, loose soils must be compacted to increase their unit weight.
* Compaction improves characteristics of soils:
1- Increases Strength
2- Decreases permeability
3- Reduces settlement of foundation
4- Increases slope stability of embankments
* Soil Compaction can be achieved either by static or dynamic loading:
1- Smooth-wheel rollers
2- Sheepfoot rollers
3- Rubber-tired rollers
4- Vibratory Rollers
5- Vibroflotation
__________________________________________
_______________________________
____________________
General Principles:
* The degree of compaction of soil is measured by its unit weight,
content, wc.
, and optimum moisture
* The process of soil compaction is simply expelling the air
from the voids.
or reducing air voids
* Reducing the water from the voids means consolidation.
Mechanism of Soil Compaction:
* By reducing the air voids, more soil can be added to
the block. When moisture is added to the block (water
content, wc, is increasing) the soil particles will slip
more on each other causing more reduction in the total
volume, which will result in adding more soil and,
hence, the dry density
will increase, accordingly.
* Increasing Wc will increase
Up to a certain limit (Optimum moister Content, OMC)After this limit
Increasing Wc will decrease
Density-Moisture Relationship
Knowing the wet unit
weight and the
moisture content, the dry unit weight can be determined from:
The theoretical maximum dry unit weight assuming zero air voids is:
I- Laboratory Compaction:
* Two Tests are usually performed in the laboratory to determine the maximum dry unit weight
and the OMC.
1- Standard Proctor Test
2- Modified Proctor Test
In both tests the compaction energy is:
1Stan
dard Proctor Test
Factors Affecting Compaction:
1- Effect of Soil Type
2- Effect of Energy on Compaction
3- Effect of Compaction on Soil Structure
4- Effect of Compaction on Cohesive Soil Properties
II- Field Compaction
Flow of Water in Soils
Permeability and Seepage
* Soil is a three phase medium -------- solids, water, and air
* Water in soils occur in various conditions
* Water can flow through the voids in a soil from a point of high energy to a point of low
energy.
* Why studying flow of water in porous media ???????
1- To estimate the quantity of underground seepage
2- To determine the quantity of water that can be discharged form a soil
3- To determine the pore water pressure/effective geostatic stresses, and to analyze earth
structures subjected to water flow.
4- To determine the volume change in soil layers (soil consolidation) and settlement of
foundation.
* Flow of Water in Soils depends on:
1- Porosity of the soil
2- Type of the soil - particle size
- particle shape
- degree of packing
3- Viscosity of the fluid - Temperature
- Chemical Components
4- Total head (difference in energy) - Pressure head
- Velocity head
- Elevation head
The degree of
soil is expressed by the
compressibility of a
coefficient of
permeability of the
soil "k."
k cm/sec, ft/sec,
m/sec, ........
Hydraulic Gradient
Bernouli's Equation:
For soils
Flow of Water in Soils
1- Hydraulic Head in Soil
Total Head = Pressure head + Elevation Head
ht = hp + he
- Elevation head at a point = Extent of that point from the datum
- Pressure head at a point = Height of which the water rises in the piezometer above the point.
- Pore Water pressure at a point = P.W.P. = gwater . hp
*How to measure the Pressure Head or the Piezometric Head???????
Tips
1- Assume that you do not have seepage in the system (Before Seepage)
2- Assume that you have piezometer at the point under consideration
3- Get the measurement of the piezometric head (Water column in the Piezometer before
seepage) = hp(Before Seepage)
4- Now consider the problem during seepage
5- Measure the amount of
the head loss in the
piezometer (Dh) or the drop in the piezometric head.
6- The piezometric head during seepage = hp(during seepage) = hp(Before Seepage) - Dh
GEOSTATIC STRESSES
&
STRESS DISTRIBUTION
Stresses at a point in a soil mass are divided into two main types:
I- Geostatic Stresses ------ Due to the self weight of the soil mass.
II- Excess Stresses ------ From structures
I. Geostatic stresses
I.A. Vertical Stress
Vertical geostatic stresses increase with depth, There are three 3 types of geostatic stresses
1-a Total Stress, stotal
1-b. Effective Stress, seff, or s'
1-c Pore Water Pressure, u
Total Stress = Effective stress + Pore Water Pressure
stotal = seff + u
Geos
tatic
Stres
s
with
Seep
age
When the Seepage Force = H gsub -- Effective Stress seff = 0 This case is referred as
Boiling or Quick Conditon
I.B. Horizontal Stress or Lateral Stress
sh = ko s'v
ko = Lateral Earth Pressure Coefficient
sh is always associated with the vertical effective stress, s'v.
never use total vertical stress to determine sh.
II. Stress Distribution in Soil Mass:
When applying a load on a half space medium the excess stresses in the soil will decrease with
depth.
Like in the geostatic stresses, there are vertical and lateral excess stresses.
1. For Point Load
The excess vertical stress is according to Boussinesq (1883):
- Ip = Influence factor for
- Knowing r/z ----- I1 can be
the point load
obtained from tables
According to Westergaard (1938)
where h = s (1-2m / 2-2m) m = Poisson's Ratio
2. For Line Load
Using q/unit length on the surface of a semi infinite soil mass, the vertical stress is:
3. For a Strip Load (Finite Width and Infinite Length):
The excess vertical stress due to load/unit area, q, is:
Where Il = Influence factor for a line load
3. For a Circular Loaded Area:
The excess vertical stress due to q is:
The Field Vane Shear Web Page
Dr. R. S. Olsen (www.liquefaction.com)
Information on techniques for CPT estimation of field vane and triaxial strengths
are at the bottom of this web.
Web sites on Vane Shear Test (VST) are located at:




http://www.pagani-geotechnical.com/
http://www.geonor.com/Soiltst.html
http://www.envi.se/products.htm
http://www.geotech.se/Vanes/evt-2000.html
The above sites were provided by Prof


Paul Mayne
http://www.igeotest.com/tema5/vane.htm - in Spanish but good
see below (Olsen, 1994)
This above figure shows how the vane shear tries to cut a larger diameter cylinder
because the blades are cutting the soil at a right angle. This effective cutting diameter is
more pronounced for granular clays compared to pure clays. However, even clays exhibit
this effect. This effect produces larger effective diameter with increased silt or sand
content. (Olsen, 1994, 1995). Because the effective cutting diameter is larger than the
vane shear cylinder diameter, the consequence is a calculated vane shear strength too
large because the wrong diameter is used. This is why we must use the vane shear
correction factor.
{Rick Olsen, March-1999}
If the rod is in contact with the soil then the "rod friction" must be accounted for (see the
above figure) .
The Olsen field vane shear - for verifying CPT estimated strengths. This system used a
Torque wrench to measure an estimate of the peak field vane shear strength. A electro
magnetic couple transmits the torque to a computer data acquisition system for a more
accurate record. {Rick Olsen jan-1999}
CPT based estimation of the normalized vane shear strength and normalized triaxial
undrained strength (Olsen, 1994, 1995). The ratio of these two estimated strengths is
equal to the classical vane shear reduction factor
CPT estimation of field normalized vane shear strength
(Olsen, 1994).
Use this correlation to confirm that the CPT soil
characterization technique can estimate strength. You must
perform field vane shear measurement tests next to CPT
soundings.
CPT estimation of laboratory triaxial undrained strength
(Olsen, 1994).
After the CPT soil characterization techniqiue is confirmed
(using the above correlation for field shear vane
measurements) use this correlation to estimate triaxial strength
for design. Remember, these estimated strengths are "best
estimates" using high quality CPT data. Poor CPT without
good controls can generate estimated strengths much higher
than the true in situ strengths.
Ratio of triaxial strength divided by the field vane shear
measurement (Olsen, 1994)
This chart is not for design purposes. It is only to show that
the two CPT estimated strength trends above match historical
trends
See text for important discussion of this ratio (vane shear reduction factor) This ratio matches the trends and magnitude of the Bjerrum and NGI
correction factor. The original correlation is shown below:
Bjerrum correction factor for the Field vane shear test
(Bjerrum, 1972) - NGI has improved this correlation
NOTE: The information at this web site represents results from
research and experience by Dr. Olsen. These correlations should only
be used by experts in the field of in situ site characterization that
understand the underlying data and know how to acquire quality CPT
data.
Dr. Rick Olsen
This page was last updated on