gravel) · 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

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.




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.




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




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:




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.




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

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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




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.




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 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




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.




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)%




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




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:







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







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













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;




-------------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.




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




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




Example:

SOIL COMPACTION

In construction of many engineering structures, loose soils must be

compacted to improve its properties.




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




- 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.




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.




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:




In site tests;

1- Nuclear method

nuclear densometer

2- Rubber ballon method

3- Sand replacement method

4- Core cutter method




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.




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.




Power rammers

Small areas

Backfill intrenches.

They do not operate effectively on uniformly graded soils.




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.




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)




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.




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;




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




_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.




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)




For unconfined stratum;

For confined stratum;




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.




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)).




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




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)










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.




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













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.




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




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




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




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.




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




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







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




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.




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?







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




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.




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




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




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.




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.




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.




The overall loss in total head is 3.50 m. The equivalent isotropic

permeability is




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.




The total normal stress is given by

For equilibrium in the direction normal to XX




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.:




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)..







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




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.




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.




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:




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.




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:




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.




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




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.




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




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.




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




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







Mohr Circles and Failure Envelope




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




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.










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.




Cu is unconfined shear

strength

qu= p/A

Cu= qu/2

A=Ao/(1 - ∆L/Lo)







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







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)




- 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')




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)




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.




Sol.

Ex.

Sol.




A linear




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.




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.




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

=




Mohr circle of stresses

Mohr circle represented by normal principle stresses along x axis and

shear stress along y axis




; Angle of failure plane

then




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




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




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




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.




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.




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.




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.




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.




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.




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.




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.







Consolidation settlement in soil

Consolidation vs. Compaction




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"




At time t = 0

At time between t = 0 & ∞




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




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).




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




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.




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




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




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.




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.




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.




(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




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).




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).




Figure (2) Typical plot of e against log

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