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Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
SOIL MECHANICS is a branch of engineering mechanics that describes the
behavior of soils. It differs from fluid mechanics and solid mechanics in the sense
that soils consist of a heterogeneous mixture of fluids (usually air and water) and
particles (usually clay, silt, sand, and gravel)
THE NATURE OF SOILS
To the civil engineer, soil is any uncemented or weakly cemented accumulation of
mineral particles formed by the weathering of rocks.
If the products of weathering remain at their original location they constitute a
residual soil. If the products are transported and deposited in a different location
they constitute a transported soil.
The process in the formation of soil may be either physical or chemical.
The physical process may be erosion by the action of wind, water or glaciers, or
disintegration caused by alternate freezing and thawing in cracks in the rock.
The chemical process results in changes in the mineral form of the parent rock due
to the action of water (especially if it contains traces of acid or alkali), oxygen and
carbon dioxide. Chemical weathering results in the formation of groups of
crystalline particles of colloidal size (<0:002 mm) known as clay minerals.
The basic structural units of most clay minerals are a silicon–oxygen tetrahedron
and an aluminium–hydroxyl octahedron, as illustrated in Figure ().
The tetrahedral units combine by the sharing of oxygen ions to form a silica sheet.
The octahedral units combine through shared hydroxyl ions to form a gibbsite
sheet.
Kaolinite consists of a structure based on a single sheet of silica combined with a
single sheet of gibbsite. A kaolinite particle may consist of over 100 stacks.
Illite has a basic structure consisting of a sheet of gibbsite between and combined
with two sheets of silica. In the silica sheet there is partial substitution of silicon by
aluminium.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Montmorillonite has the same basic structure as illite. In the gibbsite sheet there is
partial substitution of aluminium by magnesium and iron, and in the silica sheet
there is again partial substitution of silicon by aluminium.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
PARTICLE SIZE ANALYSIS
Particle sizes in soils can vary from over 100 mm to less than 0.001
mm. In British Standards the size ranges detailed in Figure 1 are
specified. The terms ‘clay’, ‘silt’, etc. use to describe the sizes of
particles between specified limits and also use to describe particular
types of soil. Most soils consist of a graded mixture of particles from
two or more size ranges.
For example, clay is a type of soil possessing cohesion and plasticity
which normally consists of particles in both the clay size and silt size
ranges.
Fine Soils; Soils which its properties are influenced mainly by clay
and silt size particles.
Coarse soils; Soils which its properties are influenced mainly by
sand and gravel size particles.
The particle size analysis of a soil sample involves determining the
percentage by mass of particles within the different size ranges.
For a coarse soil, It can be determined by the method of sieving. The
soil sample is passed through a series of standard test sieves having
successively smaller mesh sizes
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
For a fine soil; It can be determined by;
1- The method of sedimentation which is based on Stokes’ law
which governs the velocity at which spherical particles settle in a
suspension: the larger the particles the greater is the settling velocity
and vice versa. The size of a particle is given as the diameter of a
sphere which would settle at the same velocity as the particle.
2- The measurement of the specific gravity of the suspension by
means of a special hydrometer, the specific gravity depending on the
mass of soil particles in the suspension at the time of measurement.
The particle size distribution of a soil is presented as a curve on a
semilogarithmic plot, A coarse soil is described as well graded if there
is no excess of particles in any size range and if no intermediate sizes
are lacking. In general, a well-graded soil is represented by a smooth,
concave distribution curve.
Examples of particle size distribution curves appear in Figure 2.The
particle size corresponding to any specified value on the ‘percentage
smaller’ scale can be read from the particle size distribution curve. The
size such that 10% of the particles are smaller than that size is denoted
by D10. Other sizes such as D30and D60 can be defined in a similar way.
The size D10is defined as the effective size. The general slope and
shape of the distribution curve can be described by means of the
coefficient of uniformity (CU) and the coefficient of curvature
(CZ), defined as follows:
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The higher the value of the coefficient of uniformity the larger the
range of particle sizes in the soil. A well-graded soil has a coefficient
of curvature between 1 and 3.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
PLASTICITY OF FINE SOILS
The term plasticity describing the ability of a soil to undergo
unrecoverable deformation without cracking or crumbling. In general,
depending on its water content (defined as the ratio of the mass of
water in the soil to the mass of solid particles), a soil may exist in one
state of:
- liquid,
-plastic,
-semi-solid and
-solid .
Water content
If the water content of a soil initially in the liquid state is gradually
reduced, the state will change from liquid through plastic and semi-
solid, accompanied by gradually reducing volume, until the solid state
is reached.
The upper and lower limits of the range of water content over which
the soil exhibits plastic behavior are defined as:
- the liquid limit (wL) and
- the plastic limit (wP), respectively.
The water content range itself is defined as the plasticity index (IP):
IP= wL - wP
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The natural water content (w) of a soil (adjusted to an equivalent water
content of the fraction passing the 425-Mm sieve) relative to the liquid
and plastic limits can be represented by means of the liquidity index
(IL):
IL=(w - wP)/ IP
Activity
The degree of plasticity of the clay-size fraction of a soil is expressed
by the ratio of the plasticity index to the percentage of clay-size
particles in the soil: this ratio is called the activity.
Activity = Plasticity index / Percent clay fraction
Soils have an activity between 0.75 and 1.25. Activity below 0.75 is
considered inactive, while soils with activity above 1.25 are considerd
active.
Soils of high activity have a greater change in volume when the water
content is changed (greater swelling, when wetted and greater
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
shrinkage when drying. Soils of high activity can be particularly
damaged to geotechnical works.
shrinkage limit
The transition between the semi-solid and solid states occurs at the
shrinkage limit, defined as the water content at which the volume of
the soil reaches its lowest value as it dries out.
Note:
1- Liquid index is the indication to nearest the natural water content to liquid
limit.
2- Plasticity is an important characteristic in the case of fine soils,
3- Most fine soils exist in the plastic state.
4- Plasticity is due to the presence of a significant content of clay mineral
particles (or organic material) in the soil. The void space between such
particles is generally very small in size with the result that water is held at
negative pressure by capillary tension. This produces a degree of cohesion
between the particles, allowing the soil to be deformed or moulded.
Adsorption of water due to the surface forces on clay mineral particles may
contribute to plastic behaviour. Any decrease in water content results in a
decrease in cation layer thickness and an increase in the net attractive
forces between particles.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Liquid and Plastic limits determination tests
Only material passing a 425-mm BS sieve is used in the tests.
Liquid limit
1- Penetrometer apparatus. The liquid limit is defined as the
percentage water content (to the nearest integer) corresponding
to a cone penetration of 20 mm.
2- Casagrande apparatus
The liquid limit is defined as the water content at which 25 blows are
required to close the bottom of the groove over a distance of 13 mm.
Plastic limit
The soil sample (approximately 2.5 g) is formed into a thread,
approximately 6 mm in diameter. The thread is then placed on a
glassplate and rolled with the tips of the fingers of one hand until its
diameter is reduced to approximately 3 mm.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Description Details
The basic soil types are boulders, cobbles, gravel, sand, silt and clay,
defined in terms of the particle size ranges; added to these are organic
clay, silt or sand, and peat. They have written in capital letters in a soil
description.
Mixtures of the basic soil types are referred to as composite types.
Soil Classification
For coarse soil; classification depend on size distribution and grading
For fine soil; classification depend on liquid limit and plastic limit
(Atterberg limits)
There are many classification methods such;
1- The unified classification system.
2- The triangular method
3- AASHTO method
1- Unified classification system
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The terms, primary and secondary letters are detailed in Table below.
- Primary letters refer to Soil type
- Secondary letters refer to grading or plasticity
Fine Soils
The liquid and plastic limits are used to classify fine soils, employing
the plasticity chart shown below.
The axes of the plasticity chart are plasticity index and liquid limit.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The plasticity characteristics of a particular soil can be represented by
a point on the chart.
The chart is divided into five ranges of liquid limit.
The diagonal line on the chart, known as the A-line.
The point above A-line, C (clay)
The point below A-line, M Silt)
Coarse Soils
1- Determine the percentage of gravel and sand to assign the soil
type G or S
If the percentage of sand more than gravel, then the soil assign as S
If the percentage of gravel more than sand, then the soil assign as G
2- Determine the percentage of fines
a- Less than 5%
b- More than 12%
c- Between (5-12)%
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The soil in case (a)
Find Cu and Cc in order to check soil graded ( Well or Poor)
For Gravel G
If Cu > 4 and 1< Cc < 3 then well graded (W) and the soil is
GW
Otherwise the graded is poor graded (P) and the soil is GP
For Sand S
If Cu > 6 and 1< Cc < 3 then well graded (W) and the soil is
SW
Otherwise the graded is poor graded (P) and the soil is SP
The soil in case (b)
Use plasticity chart to check the fines within coarse soil,
Clay (C) or Silt (M)
For Gravel G
If the fines clay, then the soil is GC
If the fines silt, then the soil is GM
For Sand S
If the fines clay, then the soil is SC
If the fines silt, then the soil is SM
The soil in case (c)
1- For base soil type (gravel or sand), use procedure used for soil
in case (a)
2- For fines within coarse soil, use procedure used for soil in case
(b)
3- The soil class will be assign by using both as following
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
For gravel For Sand
GW-GC SW-SC
GW-GM SW-SM
GP-GC SP-SC
GP-GM SP-SM
Example: Classify the soil by using unified classification system and
give description for the soil.
Sieve No. Sieve opening (mm) P. passing
1 25.4 100
1/2 12.7 98
No.4 4.74 49
10 2 41
40 0.42 25
No.200 0.074 3
Cu=9, Cc=1.5
Note:
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example: Classify the soil by using unified classification system and
give description for the soil.
Sieve No. Sieve opening (mm) P. passing
1 25.4 100
1/2 12.7 100
No.4 4.74 100
10 2 90
40 0.42 81
No.200 0.074 65
wL=45% , wp=20
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example: Classify the soil by using unified classification system and
give description for the soil.
Sieve No. Sieve opening (mm) P. passing
1 25.4 100
1/2 12.7 81
No.4 4.74 68
10 2 25
40 0.42 21
No.200 0.074 10
wL= 51 %, wp=24%, Cc=3, Cu=2
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
PHASE RELATIONSHIPS
Soils can be of either two-phase or three-phase composition.
The water content (w), or moisture content (m), is the ratio of the mass
of water to the mass of solids in the soil;
------------------1
The degree of saturation (Sr) is the ratio of the volume of water to the
total volume of void space;
---------------2
Sr= 0 for completely dry soil
Sr= 1 for a fully saturation soil.
The void ratio (e) is the ratio of the volume of voids to the volume of
solids;
-----------3
The porosity (n) is the ratio of the volume of voids to the total volume
of the soil;
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
-------------4
The void ratio and the porosity are inter-related as follows:
--------5
-----------6
The specific volume (ν) is the total volume of soil which contains unit
volume of solids;
---------------7
The air content or air voids (A) is the ratio of the volume of air to the
total volume of the soil;
--------8
The bulk density (ρ) of a soil is the ratio of the total mass to the total
volume;
------------------9
The specific gravity of the soil particles (Gs) is given by
-----------10
From the definition of void ratio, if the volume of solids is 1 unit then
the volume of voids is e units. The mass of solids is then Gs ρw and,
from the definition of water content, the mass of water is w Gs ρw. The
volume of water is thus w Gs.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
These volumes and masses are represented in Figure shown. The
following relationships can now be obtained.
The degree of saturation can be expressed as
or from eq 6 and 11
or from Eq. 11
Equations similar to 15–18 apply in the case of unit weights, for example
------------------11
----------------12
-----------------13
----------------14
---------------15
-------------------16
----17
-----------18
----15 a
---15 b
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Where γw is the unit weight of water; (9.8 kN/m3)
When a soil in situ is fully saturated the solid soil particles (volume: 1
unit, weight: Gs γw) are subjected to upthrust (γw). Hence, the buoyant
unit weight (γ') is given by
In the case of sands and gravels the density index (ID) is used to
express the relationship between void ratio (e), and the limiting values
emax and emin.
ID= 1 for a soil in its densest possible state (e=emin)
ID= 0 for a soil in its loosest possible state (e=emax)
H.W
- Express bulk unit weight in term of dry unit weight.
- Express degree of saturation in term of e, w, Gs
- Express bulk unit weight in term of Gs, w, e and γw
- Express bulk unit weight in term of Gs, S, e and γw
- Express Air content in term of e, w, Gs
- Express Shrinkage limit in term of vd,ws,γw
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example:
SOIL COMPACTION
In construction of many engineering structures, loose soils must be
compacted to improve its properties.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Compaction is the densification of a soil by packing the particles
closer together with a reduction in the volume of air
The advantage of compactions
1- Increase shear strength and bearing capacity
2- Decrease settlement
3- Decrease permeability
The factors which compaction depends on
1- Soil type and gradation
2- Compaction energy
3- Water content
The degree of compaction of a soil is measured in terms of dry
density.
Relative compaction, R.C= field dry density / max dry density from
lab test) x 100%
Laboratory compaction test
1- Standard proctor test
- Diameter of mold = 4 in
- Wt of hammer= 2.5 kg
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
- Height of full= 305 mm
- No. of blows= 25/layer
2- Modified proctor test
- Diameter of mold = 4 in
- Wt of hammer= 4.5 kg
- Height of full= 457 mm
- No. of blows= 25/layer
Compaction curve and Optimum water content (wopt)
For a given soil the process of test is repeated at least five times,
the water content of the sample being increased each time. Dry density
is plotted against water content and a curve of the form shown is
obtained.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
This curve shows that for a particular method of compaction there is a
particular value of water content, known as; the optimum water
content (wopt)
Optimum water content (wopt), at which a maximum value of dry
density is obtained.
At low values of water content most soils tend to be stiff and are
difficult to compact.
At high water contents, the dry density decreases with increasing
water content, an increasing proportion of the soil volume being
occupied by water.
To determine dry density, use the equation;
For air content
NOTE; The bulk density and water content of the soil are determined
and the dry density calculated.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The maximum possible value of dry density is referred to as
the ‘zero air voids’ dry density or the saturation dry density
(unattainable in practice ) and can be calculated from the
expression:
In general, the dry density after compaction at water content
w to an air content A can be calculated from the following
expression:
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
In site tests;
1- Nuclear method
nuclear densometer
2- Rubber ballon method
3- Sand replacement method
4- Core cutter method
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Field soil compaction
The following types of compaction equipment are used in the field.
Smooth-wheeled rollers
Suitable for most types of soil except uniform sands and silty Sands.
Pneumatic-tyred rollers
Suitable for a wide range of coarse and fine soils but not for uniformly
graded material.
Sheepsfoot rollers
Most suitable for fine soils.
Suitable for coarse soils with more than 20% of fines.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Grid rollers
Suitable for most coarse soils.
Vibratory rollers
Particularly effective for coarse soils with little or no fines.
Vibrating plates
Suitable for most soil types.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Power rammers
Small areas
Backfill intrenches.
They do not operate effectively on uniformly graded soils.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
SOIL WATER
All soils are permeable materials, water being free to flow through
the interconnected pores between the solid particles.
The level at which the pressure is atmospheric (i.e. zero) is defined as
the water table (WT) or the phreatic surface.
Water table level changes according to:
- climatic conditions.
- constructional operations.
A perched water table can occur locally, contained by soil of low
permeability, above the normal water table level.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Artesian conditions can exist if an inclined soil layer of high
permeability is confined locally by an overlying layer of low
permeability.
Below the water table the pore water may be static, the hydrostatic
Above the water table, water can be held at negative pressure by
capillary tension.
Permeability
In one dimension, water flows through a fully saturated soil in
accordance with Darcy’s empirical law:
V α i or v = k i and v = q/A then v = q/A = k i
So q = k i A
Where q is the volume of water flowing per unit time,
A the cross-sectional area of soil
k the coefficient of permeability (m/s)
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
i the hydraulic gradient ∆h/l
and v the discharge velocity (m/s)
The coefficient of permeability depends primarily on the average size
of the pores, (which is function of 1-distribution of particle sizes, 2-
particle shape and 3- soil structure.)
The coefficient of permeability also varies with temperature, upon
which the viscosity of the water depends.
The coefficient of permeability can be represented by the equation:
(units m2) an absolute coefficient depending only on the characteristics
of the soil skeleton.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The values of k for different types of soil shown below;
For sands, the approximate value of k is given by;
The average velocity at which the water flows through the soil pores
is obtained by dividing the volume of water flowing per unit time by
the average area of voids (Av) on a cross-section normal to the
macroscopic direction of flow: this velocity is called the seepage
velocity (vv),Thus;
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Determination of coefficient of permeability
Laboratory methods
The constant-head test (Figure a); is used for coarse soils.
A steady vertical flow of water, under a constant total head, is
maintained through the soil and the volume of water flowing per unit
time (q) is measured. hydraulic gradient (h/l ) be measured. Then from
Darcy’s law:
k =ql/Ah
The falling-head test (Figure b) is used for fine soils.
The water drains into a reservoir of constant level. The standpipe is
filled with water to fall from h0 to h1. At any intermediate time t the
water level in the standpipe is given by h and its rate of change by
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
_dh/dt. At time t the difference in total head between the top and
bottom of the specimen is h. Then,applying Darcy’s law:
The length of the undisturbed specimen is l and the cross-sectional
area A.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
In-situ methods (More reliable results than laboratory tests)
Well pumping test (suitable for homogeneous coarse soil)
The procedure involves continuous pumping at a constant rate. Steady
seepage is established, radially towards the well, resulting in the water
table being drawn down to form a ‘cone of depression’. Water levels
are observed in a number of boreholes.
unconfined and confined stratums are shown in Figures (a),(b),
respectively.
Analysis is based on Dupuit assumption (hydraulic gradient at any
distance r from the centre of the well is constant with depth and is
equal to the slope of the water table, i.e. ir=dhdr)
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
For unconfined stratum;
For confined stratum;
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Borehole tests
The procedures being referred to as inflow and outflow tests,
respectively. A hydraulic gradient is thus established.
In a constant-head test the water level is maintained throughout at a
given level (Figure (a)).
In a variable-head test the water level is allowed to fall or rise and the
time taken (Figure (b)).
To alleviate the problem the clogging of the soil face at the bottom
borehole may be extended below the bottom of the casing, as shown in
Figure (c) for horizontal permeability and Figure (d) for vertical
permeability.
General formulae can be written, with the above details being
represented by an ‘intake factor’ (F ). For a constant-head test:
where k is the coefficient of permeability,
q the rate of flow,
hc the constant head,
h1 the variable head at time t1.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
h2 the variable head at time t2
A the cross-sectional area of casing or standpipe.
The coefficient of permeability for a coarse soil can also be obtained
by seepage velocity (Figure (e)).
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Determination of k for stratification soil
1- Horizontal flow kx
In this case I will be constant and q=q1+q2+q3+….+qn
q=k i a
For one unit width of soil q=k 1 H*1 =k i H
q= k x1 i H1+ k x2 i H2+ k x2 i H2+…..+ k xn i Hn
q=kxe i H total
where kxe is equivelant k in horizontal direction
q1+q2+q3+….+qn = kxe i H total
k x1 i H1+ k x2 i H2+ k x2 i H2+…..+ k xn i Hn= kxe i H total
kxe= (k x1 i H1+ k x2 i H2+ k x2 i H2+…..+ k xn i Hn) / H total
2- Vertical flow ky
In this case q is constant;
q1 = q2 = q3 =…….= qn
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
q= k i A
q1 = ky1 h1/H1 A, q2 = ky2 h1/ H1 A, …., qn = kyn hn/ Hn A
consider unit area
q= kye htotal/Htotal
htotal= h1+h2+h3+…+hn ------1
from q=kye htotal/Htotal
so htotal= q Htotal/ kye sub. In eq. 1
q. Htotal/ kye = q1. H1/ ky1+ q2. H2/ ky2+……..+ qn. Hn/ kyn
kye = H total / (H1/ ky1+ H2/ ky2+……..+ Hn/ kyn)
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Seepage through soil
It is the steady flow of water is setup under structure owing to the
difference in head, this may lead to an undesirable leakage and with
upward flow of water to the downstream side, damage may occur with
possible failure of structure, the seepage can be studied by use of flow
net.
Flow net
A flow net is pictorial representation of the paths taken by water in
passing through soil and use to determine the quantities of seepage
through soil, it is made up of :
Flow lines: These represent the path of flow through soil, there are
infinite number of flow lines the path which never cross and each line
approximately parallel to the last
Equipontial lines: The lines which jointed the points where the
pressure head is equal.
When drawing flow net it is advisable to choose flow lines and
equipontial lines to give approximately square fields and cross together
at right angles.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Nf= No. of flow channel
Nd= No. of equipontial lines
Let the loss of head from AD to BC = ∆H
∆H= H/ND
The quantity of seepage for one flow channel ∆q.
∆q=k A i
∆q= K a*1 *∆H/b (a=b)
∆q=k∆H
∆q=k H/ND
The total quantity of seepage q
q = ∆q * Nf
q = kH Nf/ND quantity of seepage
Nf/ ND shape factor
Nd=0
Nd=1
Nd=2
Nd=3
Nd=12
Nd=13
a
a
b
a
A
A
D
A C
A
B
A
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example a line of sheet piling driven 6.00 m into a stratum of soil
8.60 m thick, underlain by an impermeable stratum. On one side of the
piling the depth of water is 4.50 m; on the other side the depth of water
(reduced by pumping) is 0.50 m.
Sol.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
In the flow net,
the number of flow channels is 4.3
the number ofequipotential drops is 12;
the ratio Nf/Nd is 0.36.
The equipotentials are numbered from zero at the downstream
boundary.
The loss in total head between any two adjacent equipotentials is
The total head at every point on an equipotential numbered nd is
nd ∆h.
The total volume of water flowing under the piling per unit time per
unit length of piling is given by
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
A piezometer tube is shown at a point P on the equipotential denoted
by nd = 10. The total head at P is
i.e. the water level in the tube is 3.33 m above the datum.
The point P is at a distance zp below the datum, i.e. the elevation head
is -zp.
The pore water pressure at P can then be calculated from Bernoulli’s
theorem:
The hydraulic gradient across any square in the flow net involves
measuring the average dimension of the square. The highest hydraulic
gradient (and hence the highest seepage velocity) occurs across the
smallest square and vice versa.
Example
A river bed consists of a layer of sand 8.25 m thick overlying
impermeable rock; the depth of water is 2.50 m. A long cofferdam
5.50 m wide is formed by driving two lines of sheet piling to a depth
of 6.00 m below the level of the river bed and excavation to a depth of
Soil Mechanics CE 302
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2.00 m below bed level is carried out within the cofferdam. The water
level within the cofferdam is kept at excavation level by pumping. If
the flow of water into the cofferdam is 0.25 m 3 /h per unit length,
what is the coefficient of permeability of the sand?
What is the hydraulic gradient immediately below the excavated
surface?
Sol.
In the flow net there are 6.0 flow channels and 10 equipotential drops.
The total head loss is 4.50 m. The coefficient of permeability is given
by
Soil Mechanics CE 302
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Lecture # Date; / / 2016
The distance (∆s) between the last two equipotentials is measured as
0.9 m. The required hydraulic gradient is given by
Example
The section through a dam is shown. Determine the quantity of
seepage under the dam and plot the distribution of uplift pressure on
the base of the dam. The coefficient of permeability of the foundation
soil is 2.5 * 10 -5
m/s.
The pore water pressure is calculated at the points of intersection of
the equipotentials with the base of the dam. The total head at each
point is obtained from the flow net and the elevation head from the
section. The calculations are shown in Table below and the pressure
diagram is plotted as shown.
Soil Mechanics CE 302
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Example
The section through a sheet pile wall along a tidal estuary is shown. At
low tide the depth of water in front of the wall is 4.00 m; the water
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
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table behind the wall lags 2.50 m behind tidal level. Plot the net
distribution of water pressure on the piling.
Sol.
The flow net is shown in the figure.
The water level in front of the piling is selected as datum.
The total head at water table level (the upstream equipotential) is 2.50
m (pressure head zero; elevation head 2.50 m).
The total head on the soil surface in front of the piling (the
downstream equipotential) is zero (pressure head 4.00 m; elevation
head -4:00 m).
There are 12 equipotential drops in the flow net.
The water pressures are calculated on both sides of the piling at
selected levels numbered 1–7. For example, at level 4 the total head on
the back of the piling is
Soil Mechanics CE 302
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Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Seepage in anisotropic Soil
Most loosely tipped deposits are probably is isotropic which the value
of (k) in the horizontal direction is the same as in the vertical direction.
The deposit characteristics may be changed with the time by
compacting loading to greater permeability in the horizontal than in
the vertical direction (anisotropic conditions). In this case the
principle of flow net can easily be adopted to deal with it by drawing
flow net to distorted scale, in most practical seepage problems the
direction of max.
Permeability is horizontal, so the scale in the horizontal direction is
shortened while the vertical scale depth the same as before.
Horizontal scale xh= x (kz /kx)0.5
Using these scales the flow net is draw in the way previously
described, and the calculation of seepage quantities is exactly as before
except that the effective permeability (k') is used instead of (k).
q = k' h Nf/ND
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Seepage through Earth Dam
This problem is an example of unconfined seepage. In section the
phreatic surface constitutes the top flow line and its position must be
estimated before the flow net can be drawn.
To draw the flow net for seepage through earth dam, the steps below
should be followed:
1-The top flow line must be started at right angle to the upstream face
of the dam.
2- the equipotentioal lines cut the upper line by equal drops in
elevation h
3- The downstream end of the flow line either exists at tangent to the
downstream face of the dam, or if filter of coarse materials is inserted,
take up vertical direction in to the filter.
The flow lines through the dam can be drawn by the parapolic
solution.
The basic parabola can be drawn using the folloing Equation, provided
the coordinates of one point on the parabola are known initially.
Soil Mechanics CE 302
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The initial point of the basic parabola should be taken at G (shown
Figure) where GC = 0.3 HC.
The coordinates of point G, substituted in previous Equation, enable
The value of x to be determined;
The basic parabola can then be plotted.
The top flow line must intersect the upstream slope at right angles; a
correction CJ must therefore be made (using personal judgement) to
the basic parabola.
Example
A homogeneous anisotropic embankment dam section is detailed in
Figure 2.21(a), the coefficients of permeability in the x and z
directions being 4.5 x 10-8
and 1.6 x 10-8
m/s, respectively. Construct
the flow net and determine the quantity of seepage through the dam.
What is the pore water pressure at point P?
Soil Mechanics CE 302
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Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The basic parabola is plotted.
The upstream correction is made.
The flow net completed, ensuring that there are equal vertical intervals
between thepoints of intersection of successive equipotentials with the
top flow line.
In the flow net.
3.8 flow channels.
18 equipotential drops.
Hence, the quantity of seepage(per unit length) is
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Level AD is selected as datum. An equipotential RS is drawn through
point P (transformed position).
By inspection the total head at P is 15.60 m. At P the elevation head is
5.50 m, so the pressure head is 10.10 m and the pore water pressure is;
Example
Draw the flow net for the non-homogeneous embankment dam section
detailed in Figure below and determine the quantity of seepage
through the dam. Zones 1 and 2are isotropic, having coefficients of
permeability 1.0 x10-7and 4.0 x 10-7m/s, respectively.
Sol.
Soil Mechanics CE 302
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Lecture # Date; / / 2016
Seepage Force (Ps)
Whenever water flow through soil, a seepage force is exerted. The
seepage pressure I equal to h γ where:
h, is head at specific point.
h = nd/ ND total head loss
the seepage force Ps = h γw A
for one unit length of the structure
A= a*1=a
The volume of soil under the structure
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
V=A L = b* a*1
For a = b v= a2
The seepage force per unit volume of soil is
Ps = h γw a / volume
Ps= h γw a / a2 = hγw/a = h/b γw
Ps = i γw
Uplift pressure = γw(h-z)
Critical Hydraulic gradient
At critical state the seepage force to equal to submerged weight.
Seepage force = submerged weight
H γw A= γw (Gs-1)/(1+e) A L
When ic =1 ,the sand boiling or quicksand take place (effective
pressure=0) or ic = γ'/γw =( γsat – γw)/ γw
Soil Mechanics CE 302
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Lecture # Date; / / 2016
The safety factor against boiling: S.F= ic / i
i= ∆h/l
( ∆h , and l can be determine from flow net.
Filters
To prevent quick conditions occurring a load should be place on the
surface of the soil thus increasing the effective pressure, this load
should consists of coarser materials ( Filters) than the soil. The Filters
must be subject to the following limiting condition:
1- Fine enough to prevent soil particles being washed through it.
2- Course enough to allow the seepage of water.
For filters Terzaghi developed the following requirments :-
D15 filter ˃ (4 – 5) D15 of base soil
D15 filter ˂ (4 – 5) D85 of base soil
NON-HOMOGENEOUS SOIL CONDITIONS
Two isotropic soil layers of thicknesses H 1 and H 2 are shown, the
respective coefficients of permeability being k1and k2 ; the boundary
between the layers is horizontal. (If the layers are anisotropic, k1 and
k2 represent the equivalent isotropic coefficients for the layers.)
The two layers can be considered as a single homogeneous anisotropic
layer of thickness (H 1 + H2) in which the coefficients in the directions
parallel and normal to that of stratification are kx and kz, respectively.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example
Determine the quantity of seepage under the dam shown in section in
shown Figure. Both layers of soil are isotropic, the coefficients of
permeability of the upper and lower layers being 2.0 x 10-6
and 1.6 x 10-5m/s, respectively.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The two isotropic soil layers, each 5 m thick, can be considered as a
single homogeneous anisotropic layer of thickness 10 m in which the
coefficients of permeability in the horizontal and vertical directions,
respectively, are :
In the transformed section the dimension 10.00 m becomes 6.30 m;
vertical dimensions are unchanged.
The transformed section is shown and the flow net is drawn as for a
single isotropic layer.
From the flow net, N f=5.6 and N=11.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The overall loss in total head is 3.50 m. The equivalent isotropic
permeability is
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Stresses in Soil
The Principle of Effective Stress
The importance of the forces transmitted through the soil skeleton
from particle to particle was recognized in 1923 when Terzaghi
presented the principle of effective stress.
The principle applies only to fully saturated soils and relates the
following three stresses:
1-The total normal stress (σ) on a plane within the soil mass, being
the force per unit area transmitted in a normal direction across the
plane, imagining the soil to be a solid (single-phase) material;
2- The pore water pressure (u), being the pressure of the water filling
the void space between the solid particles;
3-The effective normal stress (σ') on the plane, representing the stress
transmitted through the soil skeleton only.
The relation is:
σ = u + σ'
Consider a ‘plane’ XX in a fully saturated soil, passing through points
of interparticle contact only, as shown below.
A normal force P applied over an area A may be resisted partly by
interparticle forces and partly by the pressure in the pore water.
The normal and tangential components of interparticle forces are N
and T, respectively.
Then, the effective normal stress is interpreted as the sum of all the
components N within the area A, divided by the area A, i.e.
Soil Mechanics CE 302
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Lecture # Date; / / 2016
The total normal stress is given by
For equilibrium in the direction normal to XX
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Note:
- Normal stress resists by:
1- the soil skeleton, and
2- the water filling the voids when the soil is fully saturated
- Shear stress resists only by the skeleton of solid particles,
Effective vertical stress due to self-weight of soil
The total vertical stress at depth z is equal to the weight of all
material (solids+ water) per unit area above that depth:
The pore water pressure at any depth will be hydrostatic at depth z
The effective vertical stress is the difference between the total vertical
stress and the pore water pressure at the same depth.:
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
So alternatively, effective vertical stress may be calculated directly
using the buoyant unit.
Example
A layer of saturated clay 4 m thick is overlain by sand 5 m deep, the
water table being 3 m below the surface. The saturated unit weights of
the clay and sand are 19 and 20 kN/m3 , respectively; above the water
table the unit weight of the sand is 17 kN/m3. Plot the values of total
vertical stress and effective vertical stress against depth.
Sol.
The stresses need to be calculated only at depths where there is a
change in unit weight (Shown Table)..
Soil Mechanics CE 302
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Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Influence of Seepage on Effective Stress :-
When water is seeping through the pores of a soil, total head is
dissipated as viscous friction producing a frictional drag, acting in the
direction of flow, on the solid particles. A transfer of energy thus takes
place from the water to the solid particles and the force corresponding
to this energy transfer is called seepage force.
Seepage force acts on the particles of a soil in addition to gravitational
force and the combination of the forces on a soil mass due to gravity
and seeping water is called the resultant body force. It is the
resultant body force that governs the effective normal stress on a plane
within a soil mass through which seepage is taking place.
Consider a point in a soil mass where the direction of seepage is at
angle θ below the horizontal. A square element ABCD of dimension b
Soil Mechanics CE 302
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Lecture # Date; / / 2016
Therefore, the force on BC due to pore water pressure acting on the
boundaries of the element, called the boundary water force, is given by
and the boundary water force on CD by γw b cosθ
If there were no seepage, i.e. if the pore water were static,
the value of ∆h =0
The forces on BC is γw b2 sinθ
The force on CD is γw b2 cosθ
Their resultant is γw b2, acting in the vertical direction.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The force ∆h γw b represents the only difference between the static
and seepage cases and is therefore called the seepage force (J ), acting
in the direction of flow (in this case normal to BC).
the average hydraulic gradient across the element is given by
The seepage pressure ( j) is defined as the seepage force per unit
volume, i.e.
j = i γw
It should be noted that j (and hence J ) depends only on the value of
hydraulic gradient.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Conditions adjacent to sheet piling
Terzaghi has shown that failure occur within amass of soil adjacent
sheet pile of approximately (d x d/2) and show heave at the surface.
The safety factor against heave =ic/im
im : the average hydraulic gradient = hm/d
In the case of sands, a factor of safety can also be obtained with
respect to ‘boiling’ at the surface. The exit hydraulic gradient (ie ) can
be determined by measuring the dimension ∆s of the flow net field
AEFG adjacent to the piling:
Soil Mechanics CE 302
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Lecture # Date; / / 2016
the factor of safety is
The sheet pile wall problem shown can also be used to illustrate the
two methods of combining gravitational and water forces.
Soil Mechanics CE 302
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Lecture # Date; / / 2016
If the factor of safety against heaving is considered inadequate,
- the embedded length d may be increased or
- a surcharge load in the form of a filter may be placed on the
surface AB.
If the effective weight of the filter per unit area is w' then the factor
of safety becomes:
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Response of Effective Stress with an increment in total stress
In fully saturated soil under undrained condition when the total stress
increase by (∆σ) the (P.W.P) increase by (∆u= ∆σ ) and ( ∆u) , and is
called excess P.W.P.
In drained condition when the excess P.W.P dissipated the effective
stress increase accompanied by corresponding reduction in volume.
Finally when the dissipation of excess P.W.P. is complete ( ∆u=0),
(∆σ) will carried entirely by soil grains (∆σ' = ∆σ).and σ0 '= σ1'+ ∆σ
Example
A 5 m depth of sand overlies a 6 m layer of clay, the water table being
at the surface; the permeability of the clay is very low. The saturated
unit weight of the sand is 19 kN/m3and that of the clay is 20 kN/m
3. A
4 m depth of fill material of unit weight 20 kN/m3is placed on the
surface over an extensive area.
Determine the effective vertical stress at the center of the clay layer
(a) immediately after the fill has been placed, assuming this to take
place rapidly and;
(b) many years after the fill has been placed.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Sol.
At depth 4 m (sand soil)
a. immediately
σ =4x19+20x4=156 kN/m2
u =4x10=40 kN/m2
σ' =156-40=116 kN/m2
b. after many years
same as in a
σ' =116 kN/m2
At depth 8 m (clay soil)
a. immediately
σ =5x19+20x3+20x4=235 kN/m2
u =8x10+20x4=160 kN/m2
σ' =235-160=75 kN/m2
b. after many years
σ =5x19+20x3+20x4=235 kN/m2
u =8x10=80 kN/m2
σ'= 235-80=155 kN/m2
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Capillarity in Soil
Is the phenomenon of water rising in the soil above the water table and
in this manner the water held in a state of suction or negative
pressure by capillary (hc) depends on a nature of the contact surface
between the water and soil particles and on the size of pores and size
of grains, there is an approximate relationship between (hc) and grain
size:
hc = c / (e D10)
c= constant depending on the grains shape
Example
A bore hole on building site has the soil profile as shown in the fig.
below.
a. Find the effective stress at the base of clay soil, the clay damp with
5% moisture content above the water table.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
b. If the W.T.is raised 1m, determine the effective stress at the base of
clay soil.
Sol.
a.
For clay soil
γb = [(1+m) / (1+e) ] Gs γs
= [(1+0.05) / (1+.6)] 2.7 x 10
=17.72
γsat = 10 (2.7+0.6) / (1+0.6) = 20.62 kN/m3
σ = 2x 16 +1x 17.72+4x20.62= 132.2 kN/m2
u = 4x10=40 kN/m2
σ'=132.2-40=92.2 kN/m2
b.
σ = 2x 16 +5x20.62= 135.1 kN/m2
u = 5x10=50 kN/m2
σ'=135.1-50=85.1 kN/m2
Sand, γ = 16 kN/m3
Clay , m 5 % , e = 0.6
Gs = 2.7
W.T
Soil Mechanics CE 302
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Lecture # Date; / / 2016
Shear strength of soil
The shear strength can be defined as the max resistance of the soil to
shearing stress under any given conditions (drained or undrained).
The shearing strength basically consists of:
1. Cohesion (C) between the surfaces of the soil particles.
2. Friction (ϕ) resistance between individual particles.
The shear strength of soils is an important aspect in many
foundation engineering problems such as:
- The bearing capacity of shallow foundations and piles.
- The stability of the slopes of dams, embankments and
excavation.
- Lateral earth pressure on retaining walls.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Mohr–Coulomb failure criteria
Mohr presented a theory for rupture in materials. According to this
theory, failure along a plane in a material occurs by a critical
combination of normal and shear stresses, and not by normal or
shear stress alone.
s = f(σ) …………….1
Where s is the shear stress at failure and σ is the normal stress on
the failure plane. The failure envelope defined by previous Eq. is a
curved line, as shown in Figure low;
Coulomb defined the function s = f(σ) as
s = c +σ tan ϕ ……….…2
Soil Mechanics CE 302
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Lecture # Date; / / 2016
where c is cohesion and ϕ is the angle of friction of the soil.
Equation (2) is generally referred to as the Mohr–Coulomb failure
criteria. The significance of the failure envelope can be explained
using Figure (1). If the normal and shear stresses on a plane in a soil
mass are such that they plot as point A, shear failure will not occur
along that plane.
Shear failure along a plane will occur if the stresses plot as point B,
which falls on the failure envelope. A state of stress plotting as point C
cannot exist, since this falls above the failure envelope; shear failure
would have occurred before this condition was reached.
Mohr–Coulomb failure criteria
To determine the shear strength of soil, Mohr-Coulomb Yield
Criterion is used;
cntan
Where
shear strength
n normal stress
c cohesion
internal friction
Soil Mechanics CE 302
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Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Mohr Circles and Failure Envelope
Soil Mechanics CE 302
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Notes
Soils generally fail in shear,
C and ϕ are measures of shear strength.
Higher the values of C and ϕ , higher the shear strength.
The soil grains slide over each other along the each other along the failure surface.
No crushing of individual grains.
SHEAR STRENGTH TESTS
A- Laboratory Shear tests
1- Direct Shear Test
2- Unconfined Compression Test
3- Triaxial Compression Test
B- In situ Shear tests
1-Vane shear test
2-Penetration test
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Direct Shear Test
The test equipment consists of a metal shear box (square or circular in
plan) into which the soil specimen is placed.
The box is split horizontally into two halves.
Normal force is applied from the top of the shear box by dead
weights.
Shear force is applied to the side of the top half of the box to cause
failure in the soil specimen.
Shear displacement of the top half of the box and the change in
specimen thickness are recorded by the use of horizontal and vertical
dial gauges.
A graph of shear stresses against normal stresses plotted and c and ϕ
can be determined.
Disadvantages:
1- The shear area is changing during the test causing unequal
distribution of shear stress
2- The direction and location of failure plane is at the box split and
parallel to horizontal force practically this condition may not
obtain.
3- The soil is confined.
4- Drainage conditions cannot be controlled and p.w.p cannot be
measured.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Ex.
Sol.
Unconfined Compression Test (qu)
For a rapid check of shear strength of a cohesive soil, the unconfined
test may be used, in this test a cylindrical cohesive soil sample
subjected to an axial load without any lateral pressure (σ3 =0), the
highest compressive strength called the unconfined compressive
strength (qu).
The shear strength in this case is taken to be ( qu / 2).
In this test the sample is sheared rapidly and no drainage takes place.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Cu is unconfined shear
strength
qu= p/A
Cu= qu/2
A=Ao/(1 - ∆L/Lo)
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The Triaxial Test
This is the most widely used shear strength test and is suitable for all
types of soil.
The test has the advantages that
- Drainage conditions can be controlled,
- Enabling saturated soils of low permeability to be consolidated,
if required, as part of the test procedure, and
- Pore water pressure measurements can be made.
A cylindrical specimen, generally having a length/diameter ratio of 2.
The specimen is stressed in two stage to failure under conditions of
axial symmetry in the manner shown in Figure below.
+
∆σ
+
σ3
2
+
f
+
= σ1
At Failure Stage two; loading Stage one; Applying
Confining Stresses
σ1 = Major Principle
σ3 = Minor Prencipal Stress
Confining stress =
Failur
Surface
Deviator stress =
Axial
stress
σ3 σ3
σ3
σ3
σ3 σ3
σ3
∆σ
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Types of test
Shear strength parameters determined by triaxial test procedures are
relevant only in situations where the field drainage conditions
correspond to the test conditions.
There are various types of test; the main types are listed below and
distinguished in the shown Figure.
1- Unconsolidated–Undrained UU: The specimen is subjected to
a specified all-round pressure and then the principal stress
difference is applied immediately, with no drainage being
permitted at any stage of the test.
- Pore pressure develops during shear (Not measured σ' unknown)
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
- Very quick test
- analysis in terms of σ, gives cu and ϕu
- Use cu and ϕu for analyzing undrained situations (e.g., short
term stability, quick loading)
2- Consolidated–Undrained CU: Drainage of the specimen is
permitted under a specified all-round pressure until
consolidation is complete; the principal stress difference is then
applied with no drainage being permitted. Pore water pressure
measurements may be made during the undrained part of the
test.
- Pore pressure develops during shear (Measured, σ' known)
- gives c' and ϕ'
- Faster than CD (preferred way to find c' and ϕ')
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
3- Consolidated – Drained CD: Drainage of the specimen is
permitted under a specified all round pressure until consolidation is
complete; with drainage still being permitted, the principal stress
difference is then applied at a rate slow enough to ensure that the
excess pore water pressure is maintained at zero.
- No excess pore pressure throughout the test
- Very slow shearing to avoid build-up of pore (Can be days! So
not desirable)
- Gives c' and ϕ'
- Use c' and ϕ' for analyzing fully drained situations (e.g., long
term stability, very slow loading)
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The shear strength of a soil under undrained conditions is different
from that under drained conditions. The undrained strength can be
expressed in terms of total stress in the case of fully saturated soils of
low permeability, the shear strength parameters being denoted by Cu
and ϕu. The drained strength is expressed in terms of the effective
stress parameters c' and ϕ'.
Ex.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Sol.
Ex.
Sol.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
A linear
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The vane shear test
This test is used for the in-situ determination of the undrained
strength of intact, fully saturated clays; the test is not suitable for
other types of soil. In particular, this test is very suitable for soft clays,
Generally, this test is only used in clays having undrained strengths
less than 100 kN/m 2. This test may not give reliable results if the clay
contains sand or silt laminations.
The equipment is shown in figure below. The vane and rod are pushed
into the clay below the bottom of a borehole to a depth of at least three
times the borehole diameter; if care is taken this can be done without
appreciable disturbance of the clay. Steady bearings are used to keep
the rod and sleeve central in the borehole casing. The test can also be
carried out in soft clays, without a borehole, by direct penetration of
the vane from ground level; in this case a shoe is required to protect
the vane during penetration.
Torque is applied gradually to the upper end of the rod by means of
suitable equipment until the clay fails in shear due to rotation of the
vane. Shear failure takes place over the surface and ends of a cylinder
having a diameter equal to the overall width of the vane. The rate of
rotation of the vane should be within the range of 6–12 per minute.
The shear strength is calculated from the expression
T=c ᴫ d h x d/2 + 2 x (c ᴫ/4 d2 x 2/3 x d/2)
Where T is the torque at failure, d the overall vane width and h the
vane length.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Cu= undrained shear strength
Undrained shear strength of design = k x undrained shear strength of
vane.
K is vane constant can be determined from curve shown in the fig.
However, the shear strength over the cylindrical vertical surface may
be different from that over the two horizontal end surfaces, as a result
of anisotropy. The shear strength is normally determined at intervals
over the depth of interest. If, after the initial test, the vane is rotated
rapidly through several revolutions the clay will become remoulded
and the shear strength in this condition could then be determined if
required.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Standard Penetration Test (S.P.T)
This test used to estimate the in site shear strength of coarse grains
soil.
A 50 mm diameter tube is driven into the ground at the bottom of bore
hole by 63.5 kg hammer and the number of blows (N) for 300 mm of
penetration given a guide to the resistance of the structure.
There is a fair correlation between (N) and both relative density and
friction angle
N corr= CN x N field
NCorr; No. of blows corrected with respected to overburden pressure.
CN ; Correction factor can be determine from table or curve
Shear Strength of Sand Soil
The sand has the same characteristics for both dry and saturated state
because there is no excess P.W.P in the case of saturated sand.
τf = σn tanυ
=
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Mohr circle of stresses
Mohr circle represented by normal principle stresses along x axis and
shear stress along y axis
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
; Angle of failure plane
then
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Orientation of Failure Plane
From the Figure:
υ + γ + 90 =180
υ + γ = 90……………1
and;
γ = 180 - 2θ………….2
then; θ = υ /2 + 45
Relation Between σ1, σ2, C, υ
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Modified Method (p-q diagram) (stress point)
To determine c and υ, it can be draw q=1/2 (σ1-σ3) against p=1/2 (σ1+σ3) instead
of drawing mohr circle.
From stress point of failure at angle 45o to the horizontal, the intersection with
horizontal axis define σ1, σ3
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example; The following data were obtained from drained triaxial test, determine;
1- Shear strength parameters.
2- The shear strength and normal stresses on the failure plane with respect to
major principle stress.
All round pressure (kN/m2) 200 400 600
Principle stress differences (kN/m2) 410 769 1126
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example; Triaxial test carried out on specimen of sand soil under all round
pressure 100 kN/m2 with principle stress difference 150 kN/m
2. Calculate the shear
strength and normal stress on shear failure plane.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example; An embankment is being constructed of soil whose properties are c'=70
kN/m2, υ'= 20
o, γ=18 kN/m
3. Find the shear strength at the base of the
embankment just after the height of the fill has been raised from 5 m to 8 m.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example; A triaxial test is performed on cohesionless soil sample under cell
pressure 1 kg/cm2, if υ= 37
o. Determine the major principle stress and deviator
stress.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example; The following data were obtained from shear box test on cohesive soil:
Normal stress kN/m2 210 315 420
Shear strength kN/m2 115 142 171
Determine;
1- The unconfined shear strength if the unconfined test was carried out on a
sample of the same soil.
2- State wether τf =160kN/m2
, σn = 25 kN/m2 at point within a mass of that
soil produce failure.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example; Given the shear strength parameters of a soil, c=c'=0, υ'=37, find the
pore water pressure when τ=20kN/m2 and τ=40 kN/m2.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example; A cylinder of soil fails under an axial stress of 8 ton/m2 in an
unconfined compression test. The failure plane makes an angle of 48o with the
horizontal. Calculate the values of the shear strength parameters.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example; A saturated sample of N.C.C was consolidated under a cell pressure of
200 kN/m2. The drainage valve was then closed and the deviator stress was
increased gradually up to failure. Calculate: 1-σ1f 2- uf 3- Af if ccu=c'=0 and
υcu=14o.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Example; A sample of cohesionless soil in a direct shear test fails under a shear
stress (1.6) kg/cm2 when the normal stress is 2.4 kg/cm
2. Find a- υ, b- σ1,σ2 at
failure.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Consolidation settlement in soil
Consolidation vs. Compaction
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Consolidation
The reduction of bulk soil volume under loading due to flow of pore water. For
saturated soils, any increment of loading (∆σ, called surcharge) will be initially
taken up by the pore pressure and result in consolidation until a new equilibrium is
reached where the soil solids (or skeleton) takes up the added load.
Surcharge:
∆σ=∆u+∆σ'
For cohesive soils;
For non-cohesive soils: water drains faster and the load is transferred immediately
"consolidation" does not occur in non-cohesive soils; in non-cohesive soils this
process is called "compression"
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
At time t = 0
At time between t = 0 & ∞
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
At time t = ∞
The Spring Analogy
(a) Initial Loading
Water takes load
Soil (i.e. spring) has no load
(b) Dissipation of Excess Water Pressure
Water dissipating ongoing
Soil starts to be loaded
(c) Final Loading
Water has been dissipated
Soil has load
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Preconsolidation Condition
1. Normally consolidated Soils - present effective overburden pressure =
maximum pressure the soil has been subjected to in the past (pc)
2. Overconsolidated Soils - present effective overburden pressure < maximum
pressure the soil has been subjected to in the past (pc)… i.e. a load has been
removed due to;
- Erosion of materials
- Excavations
- Removal of structures3. Removal of structures
- Groundwater lowering
3. Under Consolidated Soils
A soil deposit that has not consolidated under the present overburden pressure
(effective stress) is called Under Consolidated Soil. These soils are susceptible
to larger deformation and cause distress in buildings built on these deposits.
When a soil is loaded, it consolidates over the virgin consolidation curve (left hand
plot). If the load is removed (or partially removed) it will rebound non-linearly
over a less steep curve (right hand plot).
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Adding a new load (i.e. due to construction) will cause the soil to consolidate over
the less steep curve until it reaches the maximum pressure load from the past (pc
or σc'). Then it will follow the steep curve again. It the new load is less than pc
settlement will be small (following the shallow curve).
6.5 Under Consolidated Soils
A soil deposit that has not consolidated under the present overburden pressure
(effective stress) is called Under Consolidated Soil. These soils are susceptible
to larger deformation and cause distress in buildings built on these deposits.
Preconsolidation pressure determination
(Casagrande, 1936)
1. e-log p is established by lab testing
2. Determine point a at which e-log p has minimum radius of curvature
3. Draw horizontal line from a (line ab)
4. Draw tangent to curve at a (line ac)
5. Draw line ad to bisect angle bac
6. Project the straight-line portion of gh back to intersect ad at f
7. Abscissa of point f is the preconsolidation pressure, pc
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Compression/Consolidation of soil layers due to stress increase by construction of
foundations or other loads. Compression is caused by:
1. Deformation of soil particles
2. Relocation of soil particles
3. Expulsion of water or air from void spaces
Settlement (ρ)
ρ =ρi + ρc +ρs
1. Immediate settlement (ρi) - elastic deformation of dry soil and moist and
saturated soils without change to moisture content.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
a. due to high permeability, pore pressure in clays support the entire added load
and no immediate settlement occurs.
b. generally, due to the construction process, immediate settlement is not
important.
2. Primary consolidation settlement (ρc) - volume change in saturated cohesive
soils because of the expulsion of water from void spaces.
a. high permeability of sandy, cohesionless soils result in near immediate drainage
due to the increase in pore water pressure and no primary (or secondary)
consolidation settlement occurs, only immediate settlement
3. Secondary consolidation settlement (creep) (ρs) - plastic adjustment of soil
fabric in cohesive soils. Volume change is due to the rearrangement of the soil
particles (No pore water pressure change, Δu = 0, occurs after the primary
consolidation)
S Consolidation = S primary + S secondary
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Over Consolidation Ratio (OCR)
It is the defined as the ratio of preconsoliadtion pressure to the present vertical
effective stress
This is indicative of the position of soil away from the normal consolidated line
OCR =1 normally consolidated Soils
For a normally consolidated clay the present effective stress is also the previous
maximum so the OCR=1.
For a heavily overconsolidated clay the OCR may be 4 or more therefore this type
of soil has been subjected to a much greater stress in the past compared to its
present condition.
Note:
-- Soils having higher OCR are less compressible
-- They show elastic behavior to certain extent
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The significance of Pc'for an overconsolidated clay is that if stresses are kept
below this value then settlements can be expected to be small but if the applied
stresses due to loading exceed this value then large settlements will occur as
consolidation will take place along the virgin compression line.
One-Dimensional Laboratory Consolidation Test
The one-dimensional consolidation testing procedure was first suggested by
Terzaghi. This test is performed in a consolidometer (sometimes referred to as an
oedometer) as shown in Figure below. The soil specimen is placed inside a metal
ring with two porous stones, one at the top of the specimen and another at the
bottom. The specimens are usually 64 mm (≈ 2.5 in.) in diameter and 25 mm. (≈ 1
in.) thick.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The load on the specimen is applied through a lever arm, and compression is
measured by a micrometer dial gauge. The specimen is kept under water during
the test. Each load usually is kept for 24 hours. After that, the load usually is
doubled, which doubles the pressure on the specimen, and the compression
measurement is continued. At the end of the test, the dry weight of the test
specimen is determined.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
The general shape of the plot of deformation of the specimen against time for
a given load increment is shown below.
Time–deformation plot during consolidation for a given load increment
From the plot, we can observe three distinct stages, which may be described as
follows:
Stage I: Initial compression, which is caused mostly by preloading.
Stage II: Primary consolidation, during which excess pore water pressure
gradually is transferred into effective stress because of the expulsion
of pore water.
Stage III: Secondary consolidation, which occurs after complete dissipation
of the excess pore water pressure, when some deformation of the
specimen takes place because of the plastic readjustment of soil
fabric.
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
(b) (c)
(a) Schematic diagram of a consolidometer; (b) photograph of a consolidometer;
(c) a consolidation test in progress (right-hand side)
Void Ratio–Pressure Plots
After the time–deformation plots for various loadings are obtained in the
laboratory, it is necessary to study the change in the void ratio of the specimen
with pressure. Following is a step-by-step procedure for doing so:
Step 1: Calculate the height of solids, Hs, in the soil specimen (Figure 1)
using the equation
(1)
where Ws = dry weight of the specimen
Ms = dry mass of the specimen
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
A = area of the specimen
Gs = specific gravity of soil solids
γw = unit weight of water
ρw= density of water
Step 2: Calculate the initial height of voids as
(2)
Where H = initial height of the specimen
Figure 1 Change of height of specimen in one-dimensional consolidation test
Step 3: Calculate the initial void ratio, eo, of the specimen, using the equation
(3)
Step 4 : For the first incremental loading, σ1 (total load/unit area of specimen),
which causes a deformation ΔH1, calculate the change in the void ratio
as
(4)
(ΔH1 is obtained from the initial and the final dial readings for the loading).
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
It is important to note that, at the end of consolidation, total stress σ1 is
equal to effective stress .
Step 5 : Calculate the new void ratio after consolidation caused by the pressure
increment as
(5)
For the next loading, σ2 (note: σ2 equals the cumulative load per unit area of
specimen), which causes additional deformation ΔH2, the void ratio at the end of
consolidation can be calculated as
(6)
At this time, σ2 = effective stress, . Proceeding in a similar manner, one can
obtain the void ratios at the end of the consolidation for all load increments.
The effective stress and the corresponding void ratios (e) at the end of
consolidation are plotted on semi logarithmic graph paper. The typical shape of
such a plot is shown in Figure (2).
Soil Mechanics CE 302
Ass. Prof. Dr. Sa'ad F. Resan
Lecture # Date; / / 2016
Figure (2) Typical plot of e against log
Recommended