1906 - 2006

Republic of the PhilippinesCAVITE STATE UNIVERSITY

(CvSU) DON SEVERINO DE LAS ALAS CAMPUS

Indang, Cavite (046) 415-0973 / (046) 415-0012

E-mail: cvsu@asia.com

Lecture in

SOIL MECHANICS

Prepared by:

BSCE 4-1

Submitted to:

Engr. Rico Asuncion (Instructor)

Soil: It is defines as

“The material naturally obtained from weathering or decomposition or disintegration of rocks”

OR “Soil is an unaggregated or uncemented deposit of minerals and organic particles covering large position of earth crust.”

Soil Mechanics: it is defined as

“It is the discipline in which the soil is studied thereotically and practically: OR

“It is the branch of civil engineering which deals with the properties, behavior and performance of soil as structural material.”

History

The knowledge of use of soil extends into prehistoric times. However, in 1930 “Soil Mechanics was established as a branch of civil engineering.”

Dr. K. Terzaghi’s contributions in the field of soil mechanics have been immerse and he is fittingly called the “Father of Soil Mechanics”. The term soil mechanics was coined by him. In 1942, he is published his book “Introduction to Soil Mechanics” and later on published another book “Soil Mechanics by Terzaghi and Peck.”

Purpose: The purpose of this branch is

To replace old method of construction. To learn and evaluate the properties of soil.

Soil Constituents: A soil mass is commonly considered as to consist of solid particles, enclosed voids or interspaces. Thus there are two constituents of soil.

1. Soil or Solid Particles2. Voids

If we take a sample of soil and compress it we will get voids which may be filled with air, with water, or partly with air and with water and solid particles.

On the basis of constituents the soil types are:

1. Dry Soil: If only air is present in the voids of the compressed soil then it is called dry

soil.

2. Saturated Soil:If only water is present in the voids then it is called saturated soil.

3. Partially Saturated Soil:If water along with air is present in the voids then it is called partially

saturated or 3 phase soil system.

Formation of Soil or Weathering of Rocks“The conversion, decomposition or disintegration of intact mass of rock

mechanically or chemically to small pieces is called weathering.”There are two types of weathering:1. Mechanical weathering

In this type the rock is disintegrated mechanically. i.e. no change in chemical composites. The three actions that happen in mechanical weathering are:

a) Grindingb) Abrasionc) Shattering

The different physical agencies due to which weathering is caused are:

Periodical temperature changes Impact and splitting action of flowing water Wind action Plants and animals Earthquake, etc.

2. Chemical WeatheringChemical weathering may caused due to:

a) Hydrationb) Carbonationc) Nitrationd) Oxidation ande) Leaching by organic acid

Clay minerals are obtained due to chemical weathering.

Types of Soil

According to Origin: The types of soil are

1. Residual soilIt is a soil in which remain in place directly over the parent rock and

relatively shallow in depth.2. Transportation soil

It is a soil which is taken away to other place after decomposition and disintegration.

The deposits of transported soil may be considerable in depth and their homogeneity or heterogeneity depends upon manner of their transportation and deposition. The various agencies of transporting and redepositing of soils are

water, ice, wind, and gravity. Water formed, transported soil is termed as alluvial marine or lacoustine. Material picked up, mixed up, disintegrated, transported and redeposited by glaciers ice on water issued from it is called glacial drift.

Loess is the windblown silt or sand. Soil transported by gravitational forces is termed as colluvial.

ACCORDING TO GRAIN SIZE:

According to grain size these are three main groups:

1.) Coarse grain soil2.) Fine grain soil3.) Organic soil

Coarse grain soil : or cohesion less soil

On coarse grain soil the particles have large size and do not have any intermolecular attractive force.

These particles can be seen with naked eye without using any instrument. Cohesionless soil is formed due to physical disintegration of rocks.

It is subdivided into:

1. Gravel particles – the component of coarse grain soil in which particle size is larger than 2mm

2. San particles – the component of coarse grain soil in which particle size is greater than 0.075mm and smaller than 2mm

Fine grain soil : or cohesive soil

In fine grain soil the particles are small in size and have inter molecular forces of attraction.

These particles cannot be observed with naked eye. It is further subdivided into:

1. Silt – have particle size smaller than 0.075mm and larger than .002mm2. Clay – the particle size is smaller than 0.002 mm

Organic soil : the soil formed by the decomposition of organic materials

- The organic matter may include vegetables and animals- This is a very poor soil having low bearing capacity- It is not good for construction purposes but ideal for agriculture- The example is peat.

IDENTIFICATION OF SOIL

For the identification of soil we may perform the following tests:

1. Visual examination : in visual examination we observe with naked eye the following properties

i. Colorsii. Gradationiii. Angularity

2. Feel test : in feel test we take small amount of soil add a few drops of cc of water, rub it between our fingers and observe that:

i. If large particle then it is sandii. If small particle then it is siltiii. If small and soapy then clay

3. Rolling test : in rolling test we take the sample of soil add few drops of water and make a ball out of it by keeping it on the palm and rolling it. Now convert it into a thread.

i. If it immediately crumbles without assuming thread shape then it sandy soil

ii. If it is converted to thread of 3mm then it is clayey soiliii. If it is converted to a ball and cracks appears then it is silty soil

4. Strength test: take soil and add few cc or drops of water make a cube out of it place it in sun or in oven at 100°C. when all the water is removed then try to break it by hand.

i. In case of sandy soil it will break easilyii. In case if silty soil some pressure is requirediii. In case of clay it can’t break easily

5. Dispersion test: take some dry soil and put it in a glass of water i.e. transparent glass.

i. In case of sandy soil the particles will settle down in a very short time i.e. few seconds

ii. In case of silty soil, it will take some time to settle downiii. In case of clayey soil it will take a lot of time to settle down

All of the above tests are simple and performed to get general idea about soil type.

SIMPLE SOIL PROPERTIES

Unit Weight: is defined as “The weight of substance per unit of volume” it is also called specific weight and represented by “ ”

W = Wa + Ww + Ws

V = Va + Vw + Vs

a. Bulk unit weight : is total weight divided by total volume (include the wt and volume of water and air )

b. Unit wt. of Soil solids: s

c. Dry Unit wt. of soil : d

Va Wa

Www

WATERV

VwW

Soil solidsVs Ws

If dry soil contains only air and soil solid total wt. of air is negligible i.e. total wt. = wt. of air + wt. of soil solids

Ws = Wd

d. Saturated Unit Wt. : sat

sat =

The first one and last are some only difference in that the first one contains air and whereas in last saturated soil i.e. only water

Units: the units of all of the above are

Or

In soil mechanics Mass and Weight are same thus use unit kg instead of newton

Specific Gravity:

It is the ratio of wt. of given volume of substance to ratio of wt. of equal volume of distilled water.

Water Ww

V Soil Solids

W

Ws

- Bulk or Apparent or Mass specific gravity :

- Specific Gravity of soil solids or Absolute Specific Gravity:

NOTE:

is equal to thus here wt. of given volume i.e.

VOLUME RATIOS:

Void Ratio

Void Ratio=e= Volumeof voidsVolumeof soil solids

e=VvVs

Porosity

Porosity=n= Volumeof voidsTotal volume of soilmass

n=VvV

x 100 %

(It is generally expressed in percentage.)

Degree of Saturation – The ratio of volume of water to the volume of void in a given soil sample.

Degree of Saturation=s=Volume of waterVolumeof voids

s=VwVv

x 100 %

WEIGHT RATIOS:

Water Content

Water Content=ω= Weig ht of waterWeigh t of soil solids

ω=WwWs

x 100 %

or

Moisture Content

MoistureContent= Weigh t of moistureWeig ht of soil solids

MoistureContent=WwWs

x100 %

Water Content and Moisture Content are similar and

WwWs

x 100 %=WwWd

x 100 %

(Wd as weight of dry soil)

W d=Wa+Ws;(w here Wa=0)

Wd=Ws

Air Content – is defined as the ratio of volume of air voids to the total volume of voids.

Air Content= Volumeof airVolume of Voids

Ac=VaVv

(as Va=Vv−Vw)

Ac=1−VwVs

Ac=1−s

RELATIONSHIPS:

1. Relation between Void Ratio and Porosity

Void Ratio=e=VvVs

as we know, V=Vv+VsVs=V−Vv

replace Vs, we get

e= VvV−Vv

e=

VvV

V −VvV

=

VvV

VV

−VvV

as we known that porosity=n=VvV

e= n1−n

2. Relation between Void Ratio and Porosity

porosity=n=VvV

= VvVv+Vs

as V=Vv+Vs

n=

VvVs

(Vv+Vs)Vs

=

VvVs

VvVs

+VsVs

as e=VvVs

n= e1+e

3. Relationship between Water Content (ω), Void Ratio (e), Degree of Saturation (s) and Specific Gravity of soil solids (GS).

Water Content=ω=WwWs

x100 -------------- (a)

as, γw=WwVw

Ww=γw ∙Vw --------------- (i) (ɣ = specific weight)

and, γs=WsVs

Ws=γs∙ Vs ---------------- (ii)

putting in equation (a) the (i) and (ii), we get

ω= γw ∙ Vwγs∙ Vs

----------------- (b)

as we know Gs= γsγw

therefore γs=γw Gs ------------- (iii)

as we know s=VwVw

therefore Vw=sVv ------------ (iv)

putting (iii) and (iv) in equation (b)

ω= γw ∙ V wγs ∙Vs

= γw ∙ s ∙Vvγw ∙Gs ∙Vs

as we know that e=VvVs

, thus

ω= sGs

∙VvVs

= e ∙ sGs

ω= e ∙ sGs

------------ (3)

4. Relationship between Bulk Unit Weight (ɣB), Unit Weight of water (ɣw), Specific Gravity of soil solids (Gs), Water Content (ω), and Void Ratio (e) of soil.

Bulk Unit Weigh t=γB¿WV ------------- (a)

TotalWeig h t=W=Wa+Ww+Ws

W =Ww+Ws --------------- (i)

TotalVolume=V =Vv+Vs ---------------- (ii)

putting (i) and (ii) in equation (a), we get

γB=WV

=(Ww+Ws)

(Vv+Vs ) ¿

WsWsVsVs

γB=Ws∙(Ww

Ws+ Ws

Ws )Vs ∙(Vv

Vv+ Vs

Vs )as we know,

ω=WwWs

;e=VvVs

; γs=WsVs

γB=γs(ω+1 )(e+1 )

−−−−−−−(b)

Now as we know that,

SpecificGravity=Gs= γsγw

γs=γw ∙ Gs

putting in (b)

γB=γw ∙ Gs∙(ω+1 )(e+1 )

=γw ∙Gs ∙ (1+ω)

(1+e )

5. Relationship between Bulk Unit Weight (Bال), Unit Weight of water (الw), Specific gravity of Soil Solids (Gs), Voids ratio (e) and degree of saturation of soil (s)

First derive the equation till here i.e. eq (4)

=Bال ال wGs (1+w)(1+e)

------------- (a)

Now according to eq 3

w= eSGs

(1)

putting (1) in eq (a), we get

=BالwGsال [1+ eS

Gs ](1+e)

=Bالwال Gs[ Gs+eS

Gs ](1+e)

=Bالwال [ Gs+eS ]

(1+e)¿(5)

6. Dry Unit Weight (dال), Unit weight of water (الw), Specific gravity (Gs) of soil Solids, Void ratio (e)

First obtain equation ----------- (4)

=Bال ال wGs (1+w)(1+e)

For dry soil w=0 i.e. no water content and الB = dال

d =ال ال wGs (1+0)(1+e)

d =ال wGsال1+e¿

7. Dry Unit Weight (dال), Bulk Unit Weight (Bال) and water content, content of soil (w)

water content=W =WwWs

1=w=WwWs

+1

1+w=Ww+WsWs

As we know that W = Wa+Ww+Ws

W= Ww+Ws

Thus

1+w= WWs

Ws= W1+w

WsV

=( W1+w

V )d =ال

WV

1+w

d =ال B ال1+w¿

Problems:-

Problem No. 1

A soil sample weighs 950 gm and its volume is 510 cc. If dry weight of sample is 890 gm and specific gravity of soil solid is 2.65 determine

a. Water content b. void ratio

c. Porosity d. Degree of Saturation

Data:-

Total wt of soil sample = 950 gm

Total volume of soil sample = 510 cc

Dry weight of soil sample = 890 gm

Specific gravity = 2.65

Required:-

a.) water content b.) Void ratio

c.) porosity d.) Degree of saturation

Solution:-

a.) water content (w) :-

w=WwWd

x100 %=W−WdWd

x100

¿( 950−890890 )x 100

¿6.74 %

b.) Void ratio (e) :-

d =ال wGsال1+e

890510

=1∗2.651+e

e=[ 2.65890510

−1]e = 0.5185

and

d=WdالV

w = 1 gm per ccال

Or

2.4 lb

c. Porosity (n) :-

n= e1+e

¿ 0.5185(1+0.5185 )

∗100

¿34.145 %

d. Degree of Saturation (s) :-

w= e∗SGs

S= wGse

¿ 6.74∗2.650.5185

¿34.45 %

Problem No.2:-

The wet unit weight is 120 lb/ft3 if its specific gravity is 2.67 and specific gravity of soil solid and water content of soil mass is 12.7, find dry unit weight, void ratio, porosity and degree of saturation.

Given Data:-

Wet unit weight of soil = 120 lb/ft3 = الB

Specific gravity of soil solid = 2.67

Water content = 12%

Required:-

a.) dry unit weight b.) Void ratio

c.) Porosity d.) Degree of Saturation

Solution:-

a.) Dry Unit Weight:-

=dال ال B1+w

= 120(1+0.12 )

¿107.14 lb / ft 3

:- wet unit weight is equal to bulk unit weight (الB)

b.) Void ratio:-

=dال wGsال1+e

107.14=62.6∗2.671+e

e=0.555

w = 1 in SI = 62.4 FPSال - :

c.) Porosity:-

n= e1+e

¿ 0.5551+0.555

∗100

¿35.69 %

d.) Degree of Saturation:-

w= eSGs

S= wGse

¿ 12∗2.60.555

¿57.33 %

PROBLEM # 03

A saturated soil has as unit wt. of 116 lb/ft3 and water content of 36% calculate its void ration and specific gravity of soil solids.

Data: Unit wt. of saturated soil = 116 lb/ft3

Water content = 36%

Required:Specific Gravity of soil solids = ?Void ratio = ?

Solution:

a.) Specific Gravity:First using the equation

For saturated soil S = 1 as 100%

Therefore or

Now we use

Gs = 2.6

b.) Void Ratio

e = 0.36 Gs

Semi solid state Solid

state

AB

C

e = 0.36*2.6e = 0.936

ATTERBERG’S LIMITS

These limits are named after Swedish agriculturist Atterberg. Fine grained soil may be mixed with water to form a plastic paste which can be molded into any form by pressure. The addition of water reduces the cohesion until material no longer retains its shape under its own weight, flows as liquid. Enough water may be added until the oil grains are dispersed in a suspension.

In 1911 Attenberg divided the entire range from liquid to solid into four stages.

i. Liquid stateii. Plastic stateiii. Semisolid stateiv. Solid state

He set arbitrary limits in terms of water content. Thus, the consistency limits are the water contents at which soil masses passes from one state to the next. The Attenberg’s limits which are most useful for engineerin purposes are

1 – liquid limit2 – plastic limit3 – shrinkage limit

VL

Vp

Vd

Liquid state Plastcstat

e

Liquid Limit :

It was defined by Casa Granda. The minimum water content at which 25 blows will close a groove of standard dimention for a a distance of ½” when groove is made in the soil paste placed in the liquid limit device.

Plastic Limit :

Of soil is the minimum water content at which soil mass starts crumbling when rolled into a thread of 1/8” diameter

Shrinkage Limit:

The water content at which further loss of water will not caouse any reduction of soil mass.

1/8”

Water content (%)

Flowline

# of Blows (logscale)

Liquid limit (x)

N2

1.)Plasticity Index:

The difference between liquid limit and plastic limit is called P.I.

P.I. = L.L. – P. L.

2.)Liquidity index: It is the difference between water content and plastic limit divided by plasticity index.

3.)Flow index: The slope of the flow curve is called flow index

Log scale

W1

W2

y

N1

x

4.)Toughness Index: the ration of plastic index and Flow index .

VOLUMETRIC SHRINKAGE: (V.S) Is defined as “Ratio of decrease in volume of soil mass to the dry volume of soil when the soil is reduced to shrinkage limit it is usually expressed in percentage.”

OR

The decrease in volume of soil expressed as percentage of its dry volume.

SHRINKAGE RATIO:

Is defined as “ the ratio of the volume decreased expressed as percentage to dry volume to the decrease in water content when the water content is reduced to shrinkage limit.

OR Volumetric Shrinkage = S.R. * (W – Ws)

Shrinkage Limit:

The equation (1) can also be written as:

(V – Vd) Vw

Wd

As we know that

Replace W – Ws by equation (2)

S.R =

And specific gravity of soil in dry state γd = Wd/Vd

Thus we can say that shrinkage ratio is equal to specific gravity at dry state

=

x 100 = W – Ws (2)

(V – Vd)

Vd

(W - Ws)

x 100

(V – Vd)

Vd

(V - Vd) Vw

Wd

1/Vd

Vw/Wd

Wd

(Vd) x (Vw)

x 100

x 100

S. R =

S. R =

S. R = γ d

γ w

= Gd

(V – Vd)

Vd

(W - Ws)

S. R =

x 100 S. R = γ d

γ w

DETERMINATION OF LIQUID AND PLASTIC LIMITS

The liquid limit is determined in laboratory with the help of the standard liquid limit

apparatus designed by casagrande. The apparatus consists of a hard rubber cues

which a brass cup drops.

Through a desired height the brass cup can be raised and lowered to fall on the

rubber base with the help of a cam operated by a handle. The height of fall of cup can

be adjusted with the help of adjusting screws.

Before starting the test, the height of fall of the cup is adjusted to 1cm – two

types of grooving tools are used

Casagranda tool

ASTM tool

The casagranda tool cuts a groove of size 2mm wide at the bottom, 11mm wide

at top and 8mm height.

About 120 gm of the specimen passing through 425micron sieve mixed

thoroughly with distilled water in a dish to form a uniform paste. A portion of paste is

placed in the cup, squeezed down, spread into position and the groove is cut into the

soil pat. The handle is rotated at the rate of two revolutions per second and number of

blows is counted until two paste of soil sample come in contact at the bottom of groove

along a distance of 10mm.

Since it is difficult to adjust the water content precisely equal to liquid limit when

groove should close in 25 blows, the liquid limit is determined by plotting graph between

number of blows at abscissa on a logarithmic scale and corresponding water content as

ordinate.

Such a graph, known as Flow Curve, is a straight line having following equation:

W 1 – W 2 = I f log10n2n1

W 1 = water content corresponding blows n1

W 2 = water content corresponding blows n2

I f = slope of curve, known as flow index

Graph:

35

30

25

20

15

10

5

0 5 10 15 20 25 30 35 40

Water content

(%)

Number of blows (log scale)Number of blows (log scale)Number of blows (log scale)Number of blows (log scale)

Liquid Limit Apparatus

Grooving Tool

Handle

Brass cup

Screws

125 mm

10 mm

80 mm 20 mm

10 mm

SIEVE ANALYSIS

The percentage of various sizes of particles in a given dry soil sample is found by

a particle size analysis or mechanical analysis. The mechanical analysis is performed in

two stages:

Sieve analysis

Sedimentation analysis as wet mechanical analysis

The first stage is meant for coarse – grained soil only, which second stage for fine

– grained soil. The sieve analysis is, however, the true representative of grain size

distribution, since the test is not affected by the temperature.

In BS and ASTM standards, the sieve size is given in terms of the number of

opening per inch. The number of opening per square inch is equal to the square of the

number of the sieve.

Sieving is performed by arranging the various sieves one over the other in the

order of their mesh openings. The largest aperture sieve being kept at the top and the

smallest aperture sieve at the bottom. A receiver is kept on the bottom and a cover is kept

at the top of whole assembly. The soil sample is put on the top sieve and whole assembly

is then shaken. The portion of soil retained on sieve is weighted. The percentage of soil

retained on each sieve is calculated on the basis of total weight of soil sample taken and

from these result, percent of passing through each sieve is calculated.

On the basis of data obtained a graph is made having sieve along x-axis and

percent passing along y-axis.

The graph curve shows what type of soil sample has been taken.

Graph:

The curve is called grain size distribution curve or simply gradient curve. On the

basis of the curve the different types of soil are:

Well graded soil – is good for construction and have

great strength

Poorly graded soil – some of the particles are

missing, also called Gap

Grain size distribution cure or gradient cure

100

90

80

70

40

50

20

10

0

60

30

200 100 30 20 4

Sieve number

percent

passing

D60

D30

D10

3. Uniformly Graded Soil

– consist of particles of same size.

Objects of Sieve Analysis

The main objects of sieve analysis are:

1. To determine which type of soil it is.2. From the curve obtain two coefficient

a. Uniformly coefficientb. Coefficient of curvaturec.

Uniformly Coefficient

Cu=D60

D10

If its value is less than 5, Uniformly graded between 5 and 15, more than 15.

Coefficient of Curvature

Cu=(D 30)

2

¿¿

If its value lies between 1 and 3, then it is a good soil.

60

30

10

Mechanical Analysis

The percentage of various sizes of particles in a given dry soil sample is found by a particle size analysis by mechanics analysis is meant the separation of a soil into its different size fractions. The mechanical is performed in two stages; 1.) Sieve Analysis and 2.) Sedimentation Analysis as wet mechanical analysis.

The sieve analysis is for coarse grain soil where as the sedimentation analysis if for fine grain oil. In general a sample may contain both coarse grained particles as well as fine grained particles. If a portion passed 75 micron size, sedimentation analysis is done for further subdivision particles size distribution.

Sedimentation Analysis

In the sedimentation analysis soil fraction fines than 75 micron size is kept in suspension in liquid ( usually water ) medium.

There are two methods of sedimentation analysis:

a. Hydrometer Analysis/Methodb. Pipette Analysis

The Sedimentation analysis is based on Stoke’s Law

According to Stoke’s Law, if a sphere is allowed to fall in a liquid which is of indefinite extent, its velocity will increase rapidly at first as in start under the acceleration of gravity (g) however a constant velocity (terminal velocity) is reached within few seconds which is maintained indefinitely as long as the condition are not changed- this velocity is given mathematically.

V=(vs−vw)

18 μD2

Where:

V = velocity of settlement or fall of sphere.

vs = unit wt. of sphere

vw= unit wt. of water

D = diameter of sphere

Limitations or Assumptions

Application of Stoke’s Law, to the observation, direct or indirect, of the scale of fall of a large numbers of soil particles suspended in water in a container is inaccurate because;

1. The particles are sieves truly spherical; in fact the shape may bear little resemblance to sphere.

2. The body of water is not indefinite in extent.3. Since many particles are present, the fall of any particles is influenced by the

presence of other particles.4. Similarly, particles near the side walls of the container are affected by the

presence of other particles.5. An average value for specific gravity of grain is used; the value of same particles

may differ appreciably from the average value.

How to make Stoke’s Law accurately

For the first item, the concept of an equivalent diameter is introductively any particle which has same velocity of fall as of sphere of same unit weight and of diameter D will be said to have and equivalent diameters D.

For the second item, if a fairly large container and a relatively small amount of soil are used.

For concentrations of less than 50 gram per litre, it has been shown the influence of particle on each other is not appropriate.

The small quantity of soil in very large container will help reduce the fraction along the wall surface.

As for as last item in concerned the difference in specific gravity is not of major

Validity

The law is applicable as valid for spheres between about 0.2 mm and 0.0002 mm in diameter falling through water.

Hydrometer Method

In this method, a suitable amount of soil is taken and is passed through the specified sieves (different sieve have different requirements). Then it is placed in the oven for drying at about 100-105 °C. The samples to be used are accurately weighed and mixed qith the distilled water in a dish beaker and shake well.

To have proper dispersion of soil, a dispersing agent called deflocculating agent is added to the soil. Same common dispersing agents are sodium sulphate, sodium silicate and sodium polysulphide like sodium hexametaphosphate.

At the commencement sedimentation test, soil particles are assumed to be uniformly distributed.

A hydrometer is then introduced at different time interval in the mixture to determine the density.

The hydrometer consists of a bulb and a calibrated tube called stem. At one end of the bulb (the other end of bulb is closed). The readings on the hydrometer stem give the density of the soil suspension situated at the center of the bulb at any time.

At the beginning, the density is very high because lots of particles will be in suspended form. Slowly the particles start settling down and with it the value of density also decreases. When we will immerse the hydrometer the second time, it will give us a reading less than the first one.

The readings are then in table having the following requirements:

Time elapsed Temperature Hydrometer reading He D Percent finer √

Time elapsed and hydrometric readings are recorded during the experiment.

Temperature may also be obtained by using different instruments at the spat.

He is the effective depth

√ = ¿¿

D = √ 1.8 μ√v s−¿vw

¿ – ω

As we know that

Gs=¿

V s

V w

¿

V s = Gs V w

In G.G.S system V w = 1 g/ccV s=Gs

Degrading other quantities in the equation μ is constant only √ is variable.

√ = Distance

Time

√ = Het

He

Percentage fineness means the percent still remaining in the suspension and is given by:

P = R−a

W x 100%

Where:

R = hydrometric reading (variable)

W = weight of soil solid

a = constant G

G−1 ⟦2.05−1

2.05 ⟧

CORRECTIONS

The corrections are then applied to make the reading more accurate. Some of the corrections are:

Meniscus correction Temperature correction Disposing agent correction

GRAPHA graph is then plotted with percent fineness in the y-axis and particle size on the

x-axis is last two columns of the table.

Particle Size

%finer

Poorly goaded well goaded uniformly goaded

OBJECT: the object of the test is

1) determine which type of soil2) find coefficients

a) uniformity coefficient:

Cu = D60

D10

b) coefficient of curvature

Cc = D30

( D10|D60 )

If Cc = 1 – 3 then, good soil

Pipette Method

The pipette method is a standard sedimentation method used in laboratory. The equipment consists of pipette, a jar and a number of sampling bottles.

Method of Soil Suspension

In this method, a suitable amount of soil is taken…

Sample as in hydrometer method

…soil particles are assumed to be uniformly distributed.

TAKING SAMPLE

In this method a pipette is used to get a sample and then that sample is dried to get the amount of soil suspended in water. The pipette consists of: i) a 125 mL bulb with stopcock for keeping the distilled water. ii) A three way stopcock. iii) Suction and waste water outlets. iv) Sampling pipette of 10 mL capacity. The method consists in drawing off samples of soil suspension 10 mL in volume, by means of the pipette from a depth of 10 cm (He ¿ at various time intervals after the commencement of sedimentation.

The pipette should be inserted in the boiling tube about 10 seconds before the selected time interval and the time taken for sucking the sample should not be more than 10 to 20 seconds. Each sample, so taken, is transferred into suitable sampling bottles and dried in oven. The weight into of solids per ml of suspension is thus found by taking the dry weight and dividing it by 10.Table: The table for pipette method has the following columnsTime Elapsed

wt of soil solids in 10 cc suspension

wt of soil per cc

wt of soil solid per cc of suspension

D T-fines (P)

- time elapsed is noted down with the help of a stop watch or any other time measuring instrument.

- wt of soil solids in 10 cc suspension by drying the sample in oven and noting down the value.

- dividing wt of soil solids in 10cc by 10 will give wt of soil per cc. This will be

constant and W s

10 where Ws is the soil mass obtained for every sample which is

variable so the quantity will be variable.

eq. 1

- wt of solids per cc for suspension is obtained by dividing the whole solid. Solid mass taken in beginning by 500 is total amount of water taken.W d

500 as held constant so it will remain constant.

- is distance covered by time

¿ Het

- D can be calculated by formula

¿(V s−V w

1.8 )D2

D=√ 1.8V s−V w

As we know that

Gs=V s

V w

V s=GsV w

In cgs system Vw = 1 g/ccVs = Gs

So in the experiment Vs is founded from Gs which is founded by experiment, it is held constant quantity. Vw is 1g/cc, is distance divided by time

¿ Het

putting the values in eq. 1 will give us value of D.- percentage fines is obtained by formula

D= wt of soil solid per cc at 10 cm dept hwt of dry soil perr cc∈soil−water suspension

¿

W s

10W d

500

PROBLEM:

After performing laboratory test on two soil samples. A and B the following data was obtained

Given Data:Soil A Soil B

Liquid Limit (%) 36 74Plastic Limit (%) 20 29

Water Content (%) 28 64

Calculate the following for each of the sample

Required:Plasticity Index = ?Liquidity Index = ?

SOLUTION:Soil A

Plasticity Index=PI=36−20PI=16 %

Liquidity Index=ω−PLPI

¿ 28−2016

=0.5

Soil B

Plasticity Index=PI=74−29PI=45 %

Liquidity Index=ω−PLPI

¿ 66−2945

=0.77

PROBLEM:

A saturated sample of clay has weight or mass of 38.76 gm and volume of 19.5 cc. After drying, it’s mass and volume reduces to 30.6 gm and 15.95 cc respectively. Find shrinkage limit, volumetric shrinkage, shrinkage ratio and specific gravity of soil sample.

Given data:

Initial mass / wt of soil sample (W) = 38.76 gmInitial Volume of Soil Sample (V) = 19.5 ccMass/ wt of dry soil (Wd) = 30.6 gVolume of dry soil (Vd) = 15.95 cc

Required:a. Shrinkage Limit =?b. Volumetric shrinkage =?c. Shrinkage ratio =?d. Specific gravity of soil solids =?

SOLUTION:

a. S h rinkage Limit=W s=(ω−(V −V d )

W d

V W )× 100

W s=([ 38.76−30.630.6 ]− (19.5−15.95 )

30.(1 ))× 100

¿15 %

b. Volumetric S hrinkage=VS=(V−VdVd )×100

¿( 19.5−15.9515.95 )×100

¿22.26 %

c. S h rinkage Ratio=SR=( VdVw )= Wd

Vd∗Vw

¿ 30.6015.95∗1

=1.92

d.) Specific gravity:

(Using 2nd eq. of shrinkage limit)

Ws = ( Ɣd / Ɣw ) – ( 1 / Gs ) x 100 = 21 / (30.6/15.95)

0.15 = (1/ (30.6/15.95)) – (1/Gs)

Gs = 2.695

Classification of Soil

The different types of classification of soil are:

a. Grain Size Classification or Particle Size Classification

1. U.S. Bureau Classification:

Clay Silt Sand Gravel

Very fine

fine medium coarse Fine coarse

0.002mm 0.05mm 0.1mm 0.25mm 0.5mm 1.0mm 2.0mm

2. ASTM Classification:

Collidal Clay Clay Silt Sand Gravel

Fine sand Coarse sand

0.001mm 0.005mm 0.074mm 0.25mm 2.0mm

3. MIT Classification: (V. Important)

(Massachusetts Institute of Technology)

Clay Silt Sand Gravel

Fine Medium Coarse Fine

Medium Coarse Fine

Medium Coarse

0.0002mm 0.0006mm 0.002mm 0.006mm 0.02mm 0.06mm 0.2mm 0.6mm 2.0mm

4. Soil Classification of Triangular Chart:

It consists of a triangle divided on each side into 10 equal parts having reading from 0 to 100. The whole triangle into 11 different areas and of soil liemg in the range of any area is given the name of that area.

Example:

Sand = 70%

Silt = 20%

Clay = 10%

From 70 on sand draw line parallel to clay

From 20 on silt draw line parallel to sand

From 10 on clay draw line parallel to silt

These lines will intersect of a point and the point lie in any area will given the name of area.

S . R= ϒ dϒ w

S . R= ϒ dϒ w

ϒ dϒ w

ϒ dϒ w

ϒ dϒ w

S . L={Ws−W (V−VdWd

ϒ w)}x100

S . R= ϒ dϒ w

Ws=[ϒ wϒ d

− 1Gs ]

Ws=[ϒ wϒ d

− 1Gs ]

Permeability and Seepage

Permeability is defined as the property of material to permit the passage as seepage of water (or other fluids) through its pores or voids.

A material having continuous voids is called permeable. Gravels are highly permeable while stiff clay is least permeable and hence such clay may be termed as impermeable.

Impervious is the materials which do not allow the fluid to pass through.

HEAD

The total head on any point is equal to the sum of i.) velocity head ii.) datum head iii.) pressure heat

Total head = datum head + pressure head + velocity head

Types of head

i.) Datum head: - height from the datum line.ii.) Pressure head :- rise of water in tube

iii.) Velocity head :- V 2

2 g

Where: V is velocityG is acceleration due to gravity

pressure A

head DATUM HEAD

Heads always drops as move away from capillary.

Loss of head = h1 – h2

H1= head at pt. A h2 = head at pt. 2

Group index:-

G.I = .2a + .005 ac + .01bd

Where:

a= that portion of percentage of soil passing sieve # 200 which is greater than 35 and not exceeding 75 which is expressed as a whole number from 0 to 40.

b= that portion of percent of soil passing sieve # 200 is greater than 15 and not exceeding 55 which is expressed as whole number from 0 to 40.

c= that portion of liquid limit of soil greater than 40 and not exceeding 60 expressed as whole number from 0 to 20.

d= that portion of plastic index of soil greater than 10 and nit exceeding 30 expressed as whole number from 0 to 20.

Example:

Soil passing sieve no. 200 = 55%

Liquid limit = 60%

A B C

B

Plastic limit = 40

P.I = 60 – 40 = 20 G.I = .2 * 20 + .005 * 20 *20 + .01 * 40 *10

a= 55 – 35 = 20 = 4 + 2 + 6 = 10

b= 55 – 15 = 40 If a exceeds from 75 as equal to 75 – 35 = 40 and 20 -35 = 40

c= 60 – 40 = 20 and also 85 – 35 = 40: same for other

d= 20 – 10 = 10 value.

= P.I = L.L – P.L

d= P.I – 10 = (60 - 40) – 10 = 10

HYDRAULIC GRADIENT

Is the loss of head divided by the horizontal distance as length in which thin loss has taken place.

A A B

L

DANRCY’S LAW

The law of flow of water through soil was first studied by Darcy (1856). This law states that “ the velocity of water is always proportional to the hydraulic gradient”.

Mathematically:

Velocity α hydraulic gradient

Ʋ α i

Ʋ=k i

Multiplying A on both sides

A Ʋ=k i A (1)

Where k is constant of proportionality (1) or q stands for discharge and n equal to

Q=Vt= volume

time

Q= A∗Lt

velocity→

Q=A∗Ʋ (2 )→

Putting (2) in equation (1) we get

Q=k i A

Falling or Variable Head Test:

As constant head test is used for coarse grain soil only where a reasonable

discharge can be collected in a given time falling head test was introduced. The falling

or variable head test is used for relatively less permeable soil where the discharge is

small.

A

L

N1

Filtes

Standpipevalve

Soil

First find the area of soil sample; Area of standpipe then open

valve and let the water come down

When water starts coming down close the valve and nole the water level in stand pipe

h1 is the initial seading i.e before allowing the water to flow and h2 is the seading when

water level has fallen down due to the flow of water, also note the time foe which the

valve was open. In porous soil, the water level will fall vey quickly

or

Where:

a = area of cross section of stand pipe to h2

L = length/height of soil sample

A = area of soil

T = time of fall of water level from h1 to h2

Let water level in standpipe drop by dh in time at velocity of fall = e when water level increase it is taken +e and when fall is taken -e.

Velocity of fall = v = dh dt

K=9.3 x a x L x log

n1n2

A (t 2−t 1 )K=

a x L x logn1n2

A (t 2−t 1 )

when water level increase it is taken +ve and when fall it is taken –ve

Now rate of flow q = Q = av = a * -dh dt

and from Dasey’s law:

Q=q = ki A

q = k h A L

Comparing and we get

k h A = -a dh L dt k A dt = -a dh

L h

Integrating we get

K A t2∫dt = -a h2∫dh L t1 h1 h k A (t2-t1) = -a log h1

L h2 k = ( a-L log h1 ) A (t2-t1) h2

k = 2.303 ( a L log h1 ) h2

At

A

1

1 A

Determination of Co-efficient of Permeability of Soil in Field

The figure given below shows three different wells excavated in the ground to fine the coefficient of permeability. This is a very laborious job.

δ 1

δ 2

Find δ 1 , δ 2 , h1, h2 then fine the rate of pumping i.e. discharge by noting down water collected in containers and the time taken by it.

Q = v/t = volume collected in containers Time taken

Let ;

δ 1 = horizontal distance of observation well no. 1 from main wellδ 2 = horizontal distance of observation well no. 2 from the main well h1 = height of observation well no. 1 h2 = height of observation well no. 2

Discharge = Q =v/t

Then

K1 = 2.303 Q log δ 1 /δ 2

λ (h22 – h1

2)

Test wellObservation well no. 1

Observation well no. 2

h1h2

Cone of depression

Pump

FACTORS AFFECTING PERMEABILITY OF SOILS:

According to Poiseuilles equation

Q = D2 s . δ w . e3 . c . i . A (A)

ŋ 1+e

According to Darcy’s law

Q = k I A (B) Comparing (A) and (B) we get

k. I . A = D2 s . δ w . e3 . c . i . A

ŋ 1+e

k = D2 s . δ w . e3 . c

ŋ 1+e

Effect of Grain Size:

According to Hozen

K α Di o2

means k is dependent on diametes of soil imeasly dependent on Diametes of

soil

EFFECT OF PROPERTIES OF PURE WATER:

K α δ w

ŋ

now if we have two samples of soil then

K1 α δ w1 = K α δ w2

ŋ1 ŋ2

δ w1

k1 = ŋ1 = ŋ1

k2 δ w1 ŋ2

ŋ2

when other factors remain constant the effect of property of water on the values

of permeability are 3. Effect of Void Ratio

K ∝ e3

1+ec

K1∝e1

3

1+e1

c1 K2∝e2

3

1+e2

c2

K1

K2

=

e13

1+e1

e23

1+e2

For courses as sandy soil e1=e2

K1

K2

=

e13

1+e1

e23

1+e2

K1 and e have a linear relationship of K in cease e increase.

Thus, the graph will be a straight line.

K

e ( fog scale)

4. Structural Arrangement of Particles and Stratification

The structural arrangement of the particles may vary at the same voids ratio, depending upon the method of decomposition or compacting the soil mass.

The effect of structural disturbance is much pronounced in fine grained soils. Stratified soil masses have masked variation ρ in their permeability in direction parallel and perpendicular to stratification. The permeability parallel to the stratified being always greats.

Water flow parallel

5. Effects of Adsorbed Water

It water surrounds each particles it will reduces the pose area / space thus decreasing permeability.

The adsorbed water means water taken from the atmosphere.

6. Effects of Gases / Air Etc.

When no air, the permeability will increase and vice versa.

PERMEABILITY OF STRATIFIED SOIL

In mature soil mass may consist of several layers deposited one above the other. The average permeability of the whole deposit will depend upon the direction of bedding. Thus we consider both the case of flow.

1. Parallel to bedding plane.2. Perpendicular to bedding plane.

1.) Average Permeability Parallel to Layers Or Strata

K H q1 , A1 , K1

q1 , A1 , K1 L

q1 , A1 , K1

a. horizontal flowLet q be the total discharge, then,

q=q1+q2+q2+…+qn equation 1

Where:q1 = discharge

A = AreaH = HeightK = permeability

q=K∗c∗A so,

q1=K1 L1 A1 q2=K2 L2 A2 q3=K3 L3 A3

Since i n same too all

q1=K1i A1 q1=K1i A1 q1=K1i A1 equation 2

For a unit length

A=L∗H ¿1∗HA=H

h

L L=1Thus,

A1=H 1 , A2=H 2, A3=H 3 equation 3

Putting equation 2 and 3 in equation 1, we will get,

q 1=KH iH=K1i H 1+K 2i H 2+K3 iH 3

K H H=¿ K1 H 1+K 2 H 2+K3 H 3

K H H =K1 H 1+ K2 H 2+K3 H 3

H

2.) Average Permeability Perpendicular or Vertical to Strata or Layer ( K v )

K v , q

q1 , A1 , K1 , h1 ,i1

hq2 , A2 , K2 , h2 ,i2 H

q3 , A3 , K 3 , h3 , i3

In this case, velocity of flow, and hence unit discharge will be same through each layer – however hydraulic gradient and hence the head loss through each layer will be different. Denote the head loss through layers by h1 , h2 , …hn. The total loss is

h=h1+h2+h3+…+hn

Now as we know,

i= h ead lossdistance (h ere is∈heig h t)

i= hH

And h=iH

So,h1=i1 H 1 h2=i2 H 2 h3=i3 H 3

Putting values of 2and 3 in 1, we get

iH=i1 H 1+i2 H2+i3 H 3

Now as we know that according to

v = ki

i= vk

v is same for all layers

i1=vk1

i2=vk2

i3=vk3

Putting in equation c the values of i, we get:

vkv

H= vk1

H 1+vk2

H 2+vk3

H 3

k v=H

H 1

k1

+H 2

k2

+H 3

k 3

SEEPAGE PRESSURE

The seepage pressure is the pressure exerted by the water on soil through which

it percolates.

To derive an equation for seepage pressure, we will use the figure given below:

Volume of soil

As we know that

Seepage presure=vw h

Multiplying and dividing by z

¿ γ ωhzZ

From equation 1 hz=i, thus

Seepage=vwiz

For seepege force ( P=F∗A)

Seepage force=γ ω∗i∗z∗A and

Az=V

¿ γ ω∗i∗V

Seepage force perunit volume=˙γ ω∗i∗V

V

QUICK SAND

When seepage pressure due to upward flow of water in sand or in sandy soil

balances the downward force of gravity, a condition of instability arises in the sand.

Then sand or sandy soil in this condition is called quick sand. Therefore quick sand is

not a type of material or sand but a condition of instability which coarse grain or granular

soil may attain under certain circumstances. In this condition, the hydraulic gradient is

called critical gradient.

Upward Force

Upward force=(h+L )∗A∗γ ω

Downward Force

Will be due to the weight of the sand.

As we know

γ B=WV

→ 1

And

γ B=γ ω [Gs+e∗s ]

1+e→ 2

Comparing 1 and 2

WV

=γ ω [Gs+e∗s ]

1+eV =A∗L

W =γ ω [ Gs+e∗s ]

1+e( A∗L)

For saturated soil s=1, so

W =γ ω [ Gs+e ]

1+e( A∗L)

At balance

(h+L )∗A∗γ ω=γ ω [Gs+e ]

1+e(A∗L)

Divide both sides by L

hL+1=Gs+e

1+e

hL=Gs+e

1+e−1

hL=Gs+e−1−e

1+e

critical h ydraulic gradient ic=Gs−11+e