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A PREPARATORY COURSE FOR RECRUITMENT EXAMS (CIVIL ENGINEERING) SOIL MECHANICS Web: www.amiestudycircle.com Email: [email protected] Ph: +91 9412903929 1/24 AMIE(I) STUDY CIRCLE(REGD.) A FOCUSSED APPROACH Soil Mechanics Fundamental Concepts PHASE RELATIONSHIP An element of soil is multiphase. A typical element of soil contains three distinct phases, solid, gas(air), and liquid(usually water). Given figure represents the three phases as they would typically exist in an element of natural soil. The phases are dimensioned with volumes on the left and weights on the right side of the sketch. There are three important relationships of volume: porosity, void ratio, and degree of saturation. Porosity is the ratio of void volume to total volume and void ratio is the ratio of void volume to solid volume. Porosity is usually multiplied by 100% and thus the values are given in %. Void ratio is expressed in a decimal value , such as a void ratio of 0.55, and can run to values greater than unity. Both porosity and void ratio indicate the relative portion of void volume in a soil sample. This void volume is filled with fluid, either gas(air) or liquid, usually water. So, we have void ratio(e) = V V s v porosity(n) = V V v x 100 Two relationships between porosity(n) and void ratio(e) are n = e + 1 e Degree of saturation(S r ) = V V v w The degree of saturation indicates the percentage of the void volume which is filled with water. Thus, a value of S = 0 indicates a dry soil, S = 100% indicates a saturate soil, and a value between 0 and 100% indicates a partially saturated soil. The most useful relationship between phase weights is water content, which is the weight of water divided by the weight of solid in a soil element.

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Soil Mechanics Fundamental Concepts

PHASE RELATIONSHIP An element of soil is multiphase. A typical element of soil contains three distinct phases, solid, gas(air), and liquid(usually water). Given figure represents the three phases as they would typically exist in an element of natural soil.

The phases are dimensioned with volumes on the left and weights on the right side of the sketch. There are three important relationships of volume: porosity, void ratio, and degree of saturation. Porosity is the ratio of void volume to total volume and void ratio is the ratio of void volume to solid volume. Porosity is usually multiplied by 100% and thus the values are given in %. Void ratio is expressed in a decimal value , such as a void ratio of 0.55, and can run to values greater than unity. Both porosity and void ratio indicate the relative portion of void volume in a soil sample. This void volume is filled with fluid, either gas(air) or liquid, usually water.

So, we have

void ratio(e) = V

V

s

v

porosity(n) = VV v x 100

Two relationships between porosity(n) and void ratio(e) are

n = e+1

e

Degree of saturation(Sr) = V

V

v

w

The degree of saturation indicates the percentage of the void volume which is filled with water. Thus, a value of S = 0 indicates a dry soil, S = 100% indicates a saturate soil, and a value between 0 and 100% indicates a partially saturated soil.

The most useful relationship between phase weights is water content, which is the weight of water divided by the weight of solid in a soil element.

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Water content w = W

W

s

w

Ws is weight of solid in soil content.

The bulk unit weight(γ) is the weight of entire soil element divided by the volume of entire element.

Bulk density(or bulk unit weight)

= V

W =

e+1

w)+(1G w

Where G is specific gravity(unit weight per unit weight of water), and γw is unit weight of water at 40C. For all practical purposes γw is taken as 1 g/cc or 9.8 kN/m3.

The dry unit weight, often called dry density, is the weight of solid matter divided by the volume of the entire element.

Dry density(dry unit weight)

γd = e+1

G w as w = 0

Void ratio e = S

wG

r

Saturated unit weight

γsat = e+1

e)+(Gw (as e = wG because Sr = 1)

Submerged(buoyant) unit weight

γb = γ - γw = e+1

S)-e(1-1-Gγw

Submerged (saturated soil) unit weight

γb = γ - γw = e+1

1-G γw

You should understand the meanings of these relationships, and add these terms to your active vocabulary. These relationships are basic to most computations in soil mechanics.

RELATIVE DENSITY A usual way to characterize the density of a natural granular soil is with relative density Dr , defined as

Dr = e - e

e - e

minmax

max x 100 % = d,max

d

x d d,min

d,max d,min

-

-

x 100 %

where emin = void ratio of soil in densest condition; emax = void ratio of soil in loosest condition; e = in place void ratio; γd max= dry unit weight of soil in densest condition; γd min= dry unit weight of soil in loosest condition; γd = in place dry unit weight

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SHRINKAGE LIMIT & SHRINKAGE RATIO Shrinkage limit is water content at boundary between semi solid state and solid state. There will be no further reduction of volume by reducing water content. The space left behind due to reduction of water content will be filled by air.

ws = W

)V-V( - )W-W(

d

d1wd1 x 100

where ws = shrinkage limit; W1 = weight of ordinary sample; Wd = weight of dried sample; V1 = Original Volume; V2 = Vd = Volume of sample at shrinkage limit.

Also, ws =

d

w - G

1 where γd = dry unit weight

Shrinkage ratio is defined as ratio of given change in volume expressed as percentage of dry volume, to the corresponding change in water content.( In the region of water content above the shrinkage limit)

SR = w - w

100 x V

V-V

21

d

d1

=

w

d x 100

ATTERBERG LIMITS The limits are based on the concept that a fine grained soil can exist in any four states depending on its water content. Thus a soil is solid when dry, and upon the addition of water proceeds through the semi solid, plastic, and finally liquid states. The water content at the boundaries between adjacent states are termed shrinkage limit, plastic limit, and liquid limit.

In liquid state, shear strength of soil is zero. If water is reduced to liquid limit, then soil starts showing plasticity and some shear strength. In this state(plastic state) soil can be moulded to any desired shape. As the water is further reduced to plastic limit, soil starts crumbling. In this state( semi solid state) soil looses its plasticity. Also, in this semi solid state, when water content is reduced, water spaces diminish and soil grains approach each other. At shrinkage limit they come closer to each other with an extent which is physically possible. When water content is reduced further, voids are filled up with air without any reduction in volume.

Plasticity Index(P.I.) = wl - wp

When a graph is plotted on semi log paper between number of blows(N) on log scale on x axis and water content on y axis on ordinary scale. A straight line is obtained which is called flow curve. Now, slope of this curve is called “flow index(If)”.

If = N - N

w - w

210110

12

loglog

Toughness Index = )Iindex( Flow

)index( Plasticity

f

Liquidity Index(LI) = IL = I

w - w

p

p

Consistency is relative ease with which a soil can be deformed.

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Consistency Index(Ic) = I

w- w

p

l where w is natural water content.

On flow curve, liquid limit is water content corresponding to 25 blows.

UNIFIED SOIL CLASSIFICATION This Classification is based on grain size and plasticity. This system is also adopted by ISI with minor modifications. In this system the group of symbols consist of primary and secondary descriptive letter.

Primary letter Secondary letter

G: gravel W: well graded

S: sand P: poorly graded

M: silt M: non plastic fines

C: clay C: plastic fines

O: organic L: low plasticity

Pt: peat H: high plasticity

Hence GP will be Poorly graded gravel

SM will be Silky sand

ML will be Low plasticity silt

OL will be Low plasticity organic silt

CL will be Inorganic clay; Pt : Peat and other inorganic solids

Example

Prove the relationship e = wG/Sr.

Solution

We know e = V

V

s

v = V

V

s

v .V

V

w

w = V

V

s

w .S

1

r

= V

W

sw

w

.

S

1

r

=

s

sw

w

W.

W .S

1

r

Example

For a saturated soil mass with a void ratio e, and specific gravity of solids, Gs, obtain expressions for water content and saturated unit weight.

Solution

Since only the two ratios i.e. e and Gs are known, solution is obtained by assuming the weight of solids, Ws.

Volume of solids

Vs = ws

s

G

W

Volume of voids

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Vv = e.Vs = e.

ws

s

G

W

Weight of water

Ww = Vwxγw = e.G

W

s

s

Hence

Water content

w = W

W

s

w = W

)G/We.(

s

ss = G

e

s

Saturated unit weight = V+V

W+W

sv

sw = e)+)(1G/W(

W + )G/We(

wss

sss

=

e)+(1

)G+(e s .γw

Example

The undisturbed soil at a borrow pit has a water content of 15 %, void ratio of 0.60 and specific gravity of soil 2.70. The soil from the borrow pit is to be used for construction of an embankment with a finished volume of 40,000 cu m. The specifications for the embankment require a water content of 18% and dry unit weight of 1.76 g/cc. Calculate the quantity of soil required to be excavated for the embankment.

Solution

It is best to solve such problems by determining the volume of solids/weights of solids that will be present in the finished embankment at the specified void ratio. This volume of solids will have to come out of the soil mass in the borrow area. Naturally, the volume of soil needed will be a function of void ratio of the soil in the borrow pit.

Dry unit weight of the embankment soil = VW s = 1.76 g/cc (Or 1.76 t/m3)

Or if V = 1 cu m Ws = 1.76 t

Hence for V = 40,000 cu m, Ws = 40,000 x 1.76 = 70400 t

Dry unit weight of the soil in the borrow pit,

γd = e + 1

Gs .γw = 0.6 + 1

2.7.1.0 = 1.6875 t/m3

γd = Ws/V = 1.6875 t/m3

for Ws = 70,400 t, the volume of soil that has to be taken out

V = Ws/γd = 70400/1.6875 = 41718 cu m

Example (SSC, Junior Engineer, 2017)

The soil from a borrow area having an average in-situ unit weight of 15.5 kN/m3 and water content of 10%, was used for the construction of an embankment (total finished volume 6000 m3). In half of the embankment due to improper rolling, the dry unit weight achieved was

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slightly lower. If the dry unit weights in the two parts are 16.5 kN/m3 and 16.0 kN/m3, find the volume of borrow area soil used in each part and the amount of soil used.

Solution

Borrow area Fill 1 Fill 2

b = 15.5 kN/m3

w = 10%

d = 14.09 kN/m3

e = 0.88

Vv = 0.88Vs

V1 = 3000 m3

d1 = 16.5 kNm3

e1 = 0.604

Vv1 = 0.604

V1 = Vv1 + Vs1 = 1.604Vs1 = 3000 m3

Vs1 = 1870 m3

V2 = 3000 m3

d2 = 16 kN/m3

e2 = 0.654

Vv2 = 0.654 Vs2

V2 = Vv2 + Vs2 = 1.654Vs2 = 3000 m3

Vs2 = 1814 m3

G = 2.7; w = 9.8 kN/m3

d = Gw/(1 + e)

Vs = Vs1 + Vs2 = 3684 m3

Vv = 0.88Vs = 3242 m3

Vborrow = Vs + Vv = 6926 m3

Amount of soil in fill 1

= d1V1 = 16.5 x 3000 = 49500 kN

Amount of soil in fill 2

= d2V2 = 16.5 x 3000 = 48000 kN

Problem (SSC, Junior Engineer, 2010)

An earthen embankment is compacted to a dry density of 1.82 gm/cc at a moisture content of 12%. The hulk density and moisture content are 1.72 gm/cc and 6% at the site from where the soil is borrowed and transported at the site of construction. How much excavation should be carried out in the pit of borrowed area for each cu.m, of the embankment.

Answer: 1.123 cu m.

Example

A soil mass in its natural state is partially saturated having a water content of 17.5 percent and a void ratio of 0.87. Determine the degree of saturation ,total unit weight, and dry unit weight. What is the weight of water required to make a mass of 10 m3 volume to get saturated. assume G = 2.69.

Solution

Degree of saturation

S = e

wG =

0.87

100 x 2.69 x 0.175

Total unit weight

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γt = e+1

w)+(1G w = 0.87+1

10.175)x9.8+2.69(1 = 16.58 kN/m3

Dry unit weight

γd = w+ 1t =

0.175 + 1

16.58 = 14.11 kN/m3

Extra weight of water required for full saturation,

γt = e+1

e)+(Gw = 1.87

0.87)9.81+(2.69 = 18.68 kN/m3

For saturation of 10 m3 of mass, extra weight of water required = (18.68 - 16.58) x 10 = 21 kN

Problem (SSC, Junior Engineer, 2015)

A soil sample in its natural state has, when fully saturated, a water content of 32.5%. Determine the void ratio, dry and total unit weights. Calculate the total weight of water required to saturate a soil mass of volume 10 m3. Assume Gs = 2.69.

Answer: e = 0.874; t = 18.7 kN/m3; d = 14.08 kN/m3; W = 45.8 kN

Problem (SSC, Junior Engineer, 2009)

An embankment was compacted at a moisture content of 15%. Its density was determined with the help of a core cutter and the following data was collected:

Empty weight of the cutter = 1200 gm Weight of cutter when it is full of soil = 3200 gm Volume of the cutter = 1000 cc

Calculate bulk density and saturation percentage of the embankment. If the embankment becomes fully saturated due to rains, then determine its moisture content and saturated density. Take G = 2.70

Answer: Sr = 73.64%; sat = 2.1 g/cc

Problem (SSC, Junior Engineer, 2009)

A sample of soil has a porosity of 35 percent and specific gravity of solids is 2.67. Calculate void ratio, dry density and unit weight if, (i) soil is 50% saturated. (ii) soil is 100% saturated.

Answer: e = 0.54, d = 1.73 g/cc, at 50% saturation = 18.72 kN/m3; at 100 saturation (i.e. S = 1) = 20.45 kN/m3

Example

Tests on a clay sample indicated the following properties of the soil:

(i) natural water content = 45.6 %

(ii) liquid limit = 49.1 %

(iii) plastic limit = 26.5 %

(iv) dia of 60 % size = 0.0060 mm and dia of 10 % size = 0.0005 mm

Calculate (i) liquidity index (ii) uniformity coefficient and (iii) relative consistency

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Solution

Liquidity index = I

w - w

p

p = w - w

w - w

pL

p = 26.5 - 49.1

26.5 - 45.6= 0.845

Uniformity coefficient = D60/D10 = 0.0060/0.0005 = 12

Relative consistency or consistency index = I

w- wL = 0.1545

Example

The natural moisture content of an excavated soil is 32 %. Its liquid limit is 60 % and plastic limit is 27 %. Determine the plasticity index of the soil and comment about the nature of the soil.

Solution

Plasticity index = Ip = wl-wp = 60 - 27 = 33 %

The nature of the soil can be judged by determining its liquidity index, Il

Il = I

w-w

p

pn = 33

27-32 = 0.15

Since the value of Il is very close to 0, the nature of the soil according to following table is very stiff.

Consistency Il

Semisolid or solid state Negative

Very stiff state (wn = wp) 0

Very soft state (wn = wl) 1

Liquid state (when disturbed) > 1

Example (SSC, Junior Engineer, 2010)

The following properties of the soil were determined by performing tests on clay sample:

Natural moisture content = 25%

Liquid limit = 32%

Plastic limit = 24%

Diameter of 60% size = 0.006 mm Diameter of 10% size = 0.006 mm

Calculate the liquidity coefficient, uniformity coefficient and relative consistency.

Solution

Liquidity coefficient

25 24

0.12532 24

N PL

L P

W WI

W W

Uniformity coeff

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Cu = D60/D10 = 0.006/0.006 = 1

Relative consistency or consistency index

32 25

0.87532 24

L NC

L P

W WI

W W

Example

A sample of clay soil has a water content of 40 % at full saturation. Its shrinkage limit is 15 %.Assuming G = 2.70, determine the degree of shrinkage Sr. Comment on the nature of the soil.

Solution

Shrinkage limit ws = M

M

s

w

Ms = GVsρw = 2.7 x 1 x (1) = 2.7 g

Mw = wsMs = 0.15 x 0.40 = 0.40 g

and Vw = (1)

0.40 = 0.40 cm3

Therefore Vd = Vs + Vw = 1.40 cm3

Mw = wnMs = 0.4 x 2.7 = 1.08 g

Vw = (1)

1.08 = 1.08 cm3

V0 = Vs + Vw = 1 + 1.08 = 2.08 cm3

Degree of shrinkage Sr = V

V -V

0

do = 2.08

1.40-2.08 x 100 = 32 %

From following table, the soil is of very poor quality.

Sr % Quality of soil

< 5 Good

5 - 10 Medium good

10-15 Poor

> 15 Very poor

Example

What do you understand by the following terms:

GP, SM, ML, OL, CI, Pt

Solution

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GP : Poorly graded gravel

SM : Silky sand

ML : Low plasticity silt

OL :Low plasticity organic silt

CL : Inorganic clay; Pt : Peat and other inorganic solids

Permeability of Soil Since soils consists of discrete particles, the void spaces between the particles are interconnected and may be viewed as highly complex and intricate network of irregular tubes. In a two phase solid liquid system, these voids are completely filled by the liquid, which is water in most of the cases of soil mechanics. Water in these tubes is free to flow when a potential difference is created in a soil mass. Water flows from zones of higher potential to lower potential.

Gravels are more pervious than sands, sands are more pervious than silts and silts are more pervious than clays. A loose sand is much more pervious then when it is dense.

MEASUREMENT OF PERMEABILITY

Constant Head Test Constant head permeaters are specially suited to the testing of pervious, coarse grained soil. The soil sample is contained in a Perspex cylinder. Water is allowed to flow through the sample from a reservoir designed to keep the water level constant by over flow. The quantity of water flowing out of the soil or discharge Q during a given time t is collected in a vessel and weighed.

Several such tests at varying rates of flow can be performed and the average value of k is determined.

Now, from Darcy's law

q = kiA

Hence k = q/iA = Q.L/A.h.t cm/s

where Q = discharge(cm3) collected in time t(s)

A = cross sectional area of sample(cm2)

h = difference in manometer levels(cm)

L = difference between manometer tapping points(cm)

i = hydraulic gradient = h/L

Falling Head Test This method is used to determine the permeability of fine grained soils such as fine sands, silts, and clays. In such soils, the permeability is too small to enable accurate measurement of discharge using constant head permeaters.

A cylinder containing the soil sample is placed on a base(perforated disc) fitted with a fine gauge. The cylinder is fitted with a rubber stopper on top. A graduated standpipe of known diameter is inserted into the rubber stopper.

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The test is conducted by fitting the standpipe with desired water and allowing flow to take place through the sample. During the test, the water level will continuously drop and the height of water in the standpipe is recorded at several time intervals during the test. Any one pair of measurement, namely the time taken for the head to fall h1 to h2, will yield one value of k. The average value of k can be computed from several such readings.

k = )t - tA(

a.L

12

.logeh

h

2

1 = )t - tA(

2.3.a.L

12

.log10h

h

2

1

FACTORS AFFECTING PERMEABILITY Effect of Grain Size

k = CD10

where k has the unit cm/s, D10 is in mm, constant C varies from 0.4 to 1.2.

Effect of Temperature

k

w

or k

k

2

1 =

1

w1 :

2

w2

Since both viscosity and unit weight vary with temperature, k will be affected by changes in temperature. Viscosity effects are more important. Greater the viscosity, lower the permeability.

Effect of Void Ratio

k

k

2

1 = e+ 1

e

1

1 :e + 1

e

2

2

Another relationship is

k1:k2 = e1:e2

For silts and clays, the above relationships are not reliable. For these, following relationship is valid

log10k1:log10k2 = e1:e2

PERMEABILITY OF STRATIFIED SOILS Horizontal Coefficient of Permeability(kH)

Consider the section of a stratified soil mass. Let the thickness of the layers be H1,H2.....Hn, and let k1,k2....kn be their respective coefficient of permeability.

Or kH = H

1.(k1H1 + k2H2 +.................knHn)

Vertical Coefficient of Permeability (kv)

In this case, the flow is taking place in one layer after another and continuity of flow requires the velocity of flow in each of the layers to be the same. The hydraulic gradients, on the other hand, change from layer to layer.

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kv = )k/H....( + )k/H( + )k/H(

H

nn2211

SEEPAGE PRESSURE Let water is flowing through a soil mass with head difference of "h" at entrance and exit. Evidently this head is lost in the soil mass in friction offered by soil. This friction would be equal to hγw and this is counter balanced by flow force which is called seepage pressure(ps).

ps = hγw

This seepage pressure always act in the direction of flow. For vertical flow, vertical stress may be increased or decreased by this seepage pressure.

effective pressure

σn = γsub x L - ps

but h/L = i where h = head loss

also ps = h.γw = L.i.γw

σn = γsub x L - L.i.γw

Note: Here submerged unit weight sub is taken because water is passing through soil mass i.e. soil is submerged.

QUICK SAND CONDITION If seepage pressure(ps) becomes equal to submerged weight of soil(γsub x L) then net effective vertical stress(σn) would be reduced to zero. In such a case soil would become cohesion less and looses all its shear strength. Now, if ps is slightly increased, then soil particles start lifting and move in the direction of flow. This phenomena of lifting of soil particles is called quick condition, boiling condition or quick sand. The hydraulic gradient corresponding to this situation(just starting) is called critical hydraulic gradient(ic).Obviously, in this condition

γsub x L = L.ic.γw

Or ic =

w

sub =

w

wsat -=

e + 1

1 -G

Example (SSC, Junior Engineer, 2010)

A sample of soil 10 cm diameter, 15 cm length was tested in a variable head permeameter. The initial head of water in the burette was found to be 45 cm and it was observed to drop to 30 cm in 195 seconds. The diameter of the burette was 1.9 cm. Calculate the coefficients of permeability in metre/day.

Answer: 0.975 m/day

Hint: Use 110

2

2.303log

haLk

At h

Example

A loose uniform sand with rounded grains has an effective grain size D10 equal to 0.3 mm. Estimate the Coefficient of permeability.

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Solution

k = CD10 where C may be taken as equal to 100 and D10 in centimetre. Substituting,

k = 100 x (0.3)2 = 9 x 10-2 cm/sec.

Example

Due to rise of temperature, the viscosity and unit weight of the percolating fluid are reduced to 75 % and 97 % respectively. Other things being constant, calculate the percentage change in coefficient of permeability.

Solution

Let k1,γw1 and η1 represent the coefficient of permeability, weight and viscosity at the increased temperature. Dropping the suffix 1 to represent these quantities at the standard(or original) temperature, we have

k = A. w and k1 = A.

1

w1 , where A = constant

kk1 =

1

w1 x

w

k1 = k.

w

w1 .

1

Now, γw1 = 0.97.γw and η1 = 0.75 η

k1 = k.0.75

(0.97)1 = 1.295 k

Increase in k = 29.5 %

Seepage Of Soils

TWO DIMENSIONAL FLOW - LAPLACE'S EQUATION The seepage taking place around sheet pile walls, under masonry dams and other water retaining structures and through earth dams, embankments etc are very often two dimensional, which means that the velocity components in the horizontal and vertical directions vary from point to point within the cross section of the soil mass.

x

h2

2

+

z

h2

2

= 0

This equation is a partial differential equation called the Laplace's equation, for two dimensional flow.

Example (SSC, Junior Engineer, 2012)

A sand deposit is 10 m thick and overlies a bed of soft clay. The ground water table is 3 m below the surface. If the sand above the ground water table has a degree of saturation of

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45%. plot the diagram showing the variation of the total stress, pore water pressure and the effective stress. The void ratio of the sand is 0.70. Take G = 2.65.

Solution

Given data:

Thickness of sand deposit =10 m

Water table below the surface = 3 m

Degree of saturation in sand above W.T. (Sr) = 45%

Void ratio of sand (e) = 0.70

Specific gravity of sand solids (G) = 2.65

t = Unit weight of soil above WT

31

2.65 0.7 0.459.81 17.11 /

1 1 0.7w

G eS xx kN m

e

Z1 = 3.0 m [hw = 0, since no water rise in piezometer tube)

2 = Unit weight of fully saturated soil

32

2.65 0.79.81 19.33 /

1 1 0.7w

G ekN m

e

Z2 = 7.0 m

At level AA'

Total Stress . 0(sin 0)Z ce Z

Pore pressure 0(sin 0)w w wu h ce h

Effective stress

0u

At level BB'

Total stress 21 1 17.11 3 51.33 /Z x kN m

Pore pressure 0wwu h

Effective stress

251.33 /u kN m

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At level CC'

Total stress 22 2 1 1( ) ( ) (19.33 7) (17.11 3) 186.64 /Z Z x x kN m

Pore pressure 27 9.81 68.67 /w wu h x kN m

Effective stress

2186.64 68.67 117.97 /u kN m

Compressibility Of Soil And Consolidation

TOTAL SETTLEMENT The total settlement St, of a loaded soil can be recognised as having three components. The immediate settlement Si, the settlement due to primary consolidation Sc and the settlement due to secondary consolidation or creep Ss.

St = Si+Sc+Ss

The immediate settlement or distortion settlement occurs almost immediately after the load is imposed, as a result of distortion of the soil without any volume change.

The squeezing out of pore water from a loaded saturated soil causing a time dependent decrease in volume is known as primary consolidation. With the passing of time, as the pore pressure dissipate, the rate of flow will decrease and eventually the flow ceases altogether, leading to a condition of constant effective stress. This signifies the end of primary consolidation.

Some soils exhibit further time dependent settlement at constant effective stress in the post primary consolidation period. This is known as secondary consolidation or creep settlement. Secondary consolidation becomes important for certain types of soil such as peat and soft organic clays.

FACTORS AFFECTING COMPRESSIBILITY Soil Type : Granular soil(sand) exhibit a compressibility behaviour quite distinct from that of clay. Because of the very high permeability of sand, it does not take much time for pore water to drain out. This is the reason why a structure on a sand soil experiencing very little settlement after it has been constructed.

For saturated fine grained soil(clay), the major factor is the escape of pore water from the soil. In contrast to sand where the expulsion of pore water takes place very quickly, a much longer time is needed in fine grained soils for pore water to escape. Thus there is a considerable time lag in clay soils between the load application and resulting deformation. Indeed, it may take many months or even years for deformation to be completed.

Stress : When a soil is stressed to a level greater than the maximum stress to which it was ever subjected in the past, perhaps some kind of a break down in the soil structure occurs, resulting in a much higher compressibility.

The maximum value of stress, the soil has ever experienced, is called the preconsolidation stress, σc’

A soil is said to be normally consolidated when the existing effective stress σ' is the maximum that it has ever experienced in its stress history, i.e. σ' = σc'.

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A soil is said to be preconsolidated or over consolidated, if the existing effective stress is less than the preconsolidation stress, σ' < σc’.

Overconsolidation ratio(OCR) is the ratio of the preconsolidation stress to the present vertical effective stress,

OCR = c

Type of soil OCR

Normally consolidated soil

1

Preconsolidated soil >1

Underconsolidated soil <1

COMPRESSION INDEX (CC) For a normally consolidated clay, it is therefore possible to express the compressibility property of the soil by noting the slope of the straight line portion of the e vs. log σ' plot. This parameter is known as the compression index(Cc) and is given by the equation

Cc = 110210

21

-e-e

loglog =

1

2

e

log

High compression index indicates high vertical deformation in the clay.

COEFFICIENT OF COMPRESSIBILITY(AV) When a stress is applied to a saturated soil, there is flow of water out of the pores. Under such condition, volume reduction occurs and effective stress is increased.

If a soil is subjected to a certain increment of stress, Δσ', it will undergo a decrease in void ratio equal to Δe. If the soil is again subjected to the same magnitude of stress increment, that is, Δσ', this time around, the reduction in void ratio will not be equal to Δe1. It will be equal to Δe2, smaller than Δe1(See figure).

Thus, it can be said that the compressibility of a soil decreases as the effective stress increases. A parameter called the coefficient of compressibility(av) is used to indicate the slope of the e - σ' relationship by the equation

av = d

de

av will be negative as the void ratio decreases with increase in stress. When the stress increment is small, the curve can be approximated to a straight line and av can be computed using the relation,

av = e

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Coefficient of compressibility decreases with increase in effective stress.

TERZAGHI'S THEORY OF ONE DIMENSIONAL CONSOLIDATION The rate of settlement is directly related to the rate of dissipation of excess pore water pressure. Therefore, in order to predict the time rate of consolidation of a consolidating layer, a theory which can predict the pore pressure at any elapsed time and space, is necessary. Terzaghi's theory is one such theory and is most popular.

The theory relates the following three quantities:

1. The excess pore water pressure

2. The depth(z) below the top of clay layer

3. The time(t) from the instantaneous application of a total stress increment.

or t

u

= Cv.z

u2

2

This is Terzaghi's differential equation of consolidation, in which

Cv = wvm

k =

w0

v

e+1a

k

= wv

0

a

)e+k(1

Cv is called coefficient of consolidation, suitable units being m2/year.

DRAINAGE CONDITIONS Solution of Terzaghi's equation for a given set of boundary conditions, describes the distribution of excess pore water pressure u with respect to time elapsed, t and location z. H is the maximum distance that water has to travel to reach a drainage face; that is, the length of the longest drainage path. If there are two drainage surfaces, one at the top and another at the bottom of the consolidating layer, it is a case of " double drainage" and H will then be equal to half the thickness of the clay layer. If there is only one drainage surface, it is a case of " single drainage" and H will be equal to the thickness of the clay layer.

In the Terzaghi's solution, three non dimensional factors are provided : one is the drainage path ratio which is related to location, the second is time factor, related to the time elapsed and the third, degree of consolidation, related to the excess pore water pressure.

Drainage path ratio

z = z/H

Time factor Tv = H

tC2

v

Degree of consolidation(Or consolidation ratio)

Uz = u

u-u

i

zi = 1 - u

u

i

z

where ui = initial excess pore water pressure

Time factor Tv will be directly proportional to time elapsed (t) for a particular soil and the drainage conditions when Cv and H are constant.

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Uz indicates the degree of consolidation at a particular location. This is not something of practical interest. What is more significant practically is, how much the consolidating layer as a whole has consolidated. In other words, what the average degree of consolidation U is for the entire stream.

Taylor(1948) gave the following approximate relationships between Tv and U:

For U 60 % Tv = (π/4)U2

For U > 60 % Tv = 1.781-0.933(100-U%)

For a given degree of consolidation U, time factor Tv has a certain value, depending on the boundary conditions governing the problem. The time required for a soil to reach a given degree of consolidation is directly proportional to the square of the drainage path and inversely proportional to the coefficient of consolidation, that is,

t H2/Cv

CONSOLIDATION TEST The test is carried out in a special device called an consolidometer, which consists basically of a loading mechanism and a specimen container known as the consolidation cell. An undisturbed soil specimen, representing the in situ soil layer is certainly trimmed and placed in a metallic confining ring which is the main component of the consolidation cell. Porous stone discs are provided at the top and bottom of the sample to allow drainage in the vertical direction, both ways. Two types of consolidation cells, the floating ring cell and the fixed ring cell, are commonly used.

DETERMINATION OF COEFFICIENT OF CONSOLIDATION There are two methods for finding coefficient of consolidation

Casagrande's Logarithm of Time Fitting Method

Taylor's Square Root of Time Fitting Method

COEFFICIENT OF PERMEABILITY The coefficient of permeability of a fine grained soil can be indirectly determined by the consolidation test. The coefficient of permeability k can be calculated using following equation.

Cv = 0

v

k(1+ e )

a . γw = k.mv.γw

Here, volume of solids Vs is assumed as unity and the void volume equal to e0 - the void ratio of the soil before compression.

Example

A certain clay layer has a thickness of 5 m. After 1 year when the clay was 50% consolidated, 8 cm of settlement had occurred. For a similar clay and loading conditions, how much settlement would occur at the end of 1 year and 4 years respectively, if the thickness of this new layer were 25 m ?

Solution

For the layer of 5 m thickness,

U = 50%, t = 1 year and Sc = 8 cm.

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Since Sc = U.Sf,

Sf = 8/0.5 = 16 cm

For H0 = 5 m, Sf = 16 cm

Hence for the 25 m thick layer, Sf = 16 x 25/5 = 80 cm(since SfH0)

From the equation Tv = Cvt/H2, for the same clay layer and the same degree of consolidation,

1 22 2

1 2

t t

H H

For the 5 m thick layer, t50 = 1 year

Hence for the 25 m thick layer, t50 = 1x(25/5)2 = 25 years

For the 25 m thick layer

)T(

)T(

2v

1v = t

t

2

1

Also, since Tv = (π/4) x U2 for U<60 per cent,

2

1

22

U

U =

)T(

)T(

2v

1v = t

t

2

1

U1 = 50 % for t1 = 25 years; for t2 = 1 year

U22 = U1

2

t

t

1

2 = 0.52 x (1/25) = 0.01

or U2 = 0.1

Sc = U.Sf

Sc = 0.1 x 80 = 8 cm

For t2 = 4 years

U22 = U1

2

t

t

1

2 = 0.52 x (4/25) = 0.04

or U2 = 0.2

Sc = 0.2 x 80 = 16 cm

Example

Time taken for construction of a building above ground level was from March 1962 to August 1963. In August 1966 average settlement was found to be 6 cm. Estimate the settlement in December 1967 if it was known that ultimate settlement will be 25 cm.

Solution

Here loading period is from March 1962 to August 1963, i.e. 10 + 8 = 18 months.

For calculating settlement, time t is taken from the middle of the loading period.

Thus 6 cm of settlement occurred in (18/2) + 12 x 3 = 45 month.

And it is required to know settlement which will occur in (18/2) + 12 x 4 + 4 = 16 months.

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Let us assume that at t = 16 months degree of consolidation U will be 0.6.

Under this condition

U = 1.13 Tv

Let S1 = settlement at time t1

and S2 = settlement at time t2

S

S

2

1 = U

U

2

1

and U

U

2

1 = v1

v2

T

T

S

S

2

1 = T

T

v2

v1 = t

t

2

1

S

6

2

= 45

61

S2 = 6 61

45 =

6.7

7.8 x 6 = 6.98 cm

Degree of consolidation

U = 15

6.98 = 0.465

Since U is less than 0.6, the relationship used is valid. If U > 60% this relationship will not be used.

The required settlement is 6.98 cm.

Example (SSC, Junior Engineer, 2013)

Find out the time required for 50% consolidation in a soil having thickness of 800 cm and pervious strata at top and bottom. What will be the value of coefficient of consolidation if coefficient of permeability = 0.0000001 cm/sec? Void ratio = 1.8, mv = 0.0008 cm2/gm Time factor (Tv) = 0.3, w = 1 gm/cc

Solution

Drainage path d = H/2 = 800/2 = 400 cm

4 20.00000013.33 10 /

(0.0003)(1)vv w

kC x cm s

m

2 2

650% 4

0.3(400)144 10 sec 1668.35

3.33 10v

v

T dt x days

C x

Example (SSC, Junior Engineer, 2011)

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In a consolidation test on a soil, the void ratio of the sample decreases from 1.24 to 1.12 when the pressure is increased from 20 to 40 tonnes/sq.m. Calculate the co-efficient of consolidation in m2/year, given that the co-efficient of permeability of the soil during this pressure increment is 8.5 x 10-3 cm/ sec.

Solution

Change in void ratio (e) = 1.12 – 1.24 = -0.12

Change in pressure ( ) = 40 – 20 = 20 tonnes/m2

Coefficient of compressibility (av) = e/ = (-0.12/20) = 6 x 10-3 m2/tonnes

Coefficient of volume change (mv) = av/(1 + e0) where e0 = initial void ratio.

3

3 26 102.678 10 /

1 1.24v

xm x m tonnes

Coeff of consolidation

5

23

8.5 100.0317 / s

2.678 10 1vvw

k xC m

m x x

= 9.99 x 105 m2/year

Shear Strength Of Soil Shear strength of a soil is the capacity of the soil to resist shearing stress. The shear strength is very important engineering property of the soil. All stability analysis which normally use the limiting equilibrium approach, require the determination of the limiting shearing resistance, that is, the shear strength of the soil.

MOHR CIRCLE OF STRESS In a stressed soil mass, shear failure can occur along any plane and hence it is necessary to study the stress conditions at a point in such a soil mass. At any stressed point, there exist three mutually perpendicular plans on which there are no shearing stresses acting. These are known as the principal planes. The normal stresses that act on these planes are called the principal stresses; the largest of these is called the major principal stress σ1, the smallest, the minor principal stress, σ3 and the third one is called intermediate principal stress σ2. The corresponding planes are respectively designated the major, minor and intermediate principal planes.

Fig (a) shows an element within the stressed soil mass in two dimensional stress system. The directions of the major and minor principal planes are shown in the figure. These are the horizontal and vertical directions respectively. To draw the Mohr circle, the normal stress σ is plotted along the x-axis and the shear stress is plotted along the y-axis.

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(a) (b)

For convenience, compressive normal stress are taken as positive, because most of the normal stresses acting on soils are compressive in nature. Shear stresses that produce counter clockwise couples on the element are considered positive.

Given the values of σ1 and σ3, a circle (called Mohr's circle) is constructed with its centre at C((σ1+σ3)/2,0) and radius equal to ((σ1-σ3)/2). The circle cuts the X axis at two points, A(σ3,0) and B(σ1,0). It is necessary to represent σ and τ to the same scale on this diagram.(Fig b).

If a line is drawn through the point (σ3,0) and parallel to the plane AB of fig (a), the line intersects the Mohr circle at point D whose coordinates represent the normal stress and shear stress on the plane AB.

This can be shown as below:

BCD = 2θ

σ = OE = OC + CE

= 1 3

2

+

2

- 31 .cos2θ

and τ = DE = 1 3-

2

.sin2θ

MEASUREMENT OF SHEAR STRENGTH -TRIAXIAL TEST The triaxial test is carried out on a cylindrical specimen of soil, usually having a length to diameter ratio of 2. the usual sizes are 76mm x 38mm and 100mm x 50mm.

The triaxial test, is carried out in two stages. In the first stage, a cell pressure is applied to the sample, subjecting it only to normal stresses. As soon as the cell pressure is applied, a pore water pressure of equal magnitude builds up in a saturated soil. The drainage line valve can be kept either open or closed. If it is kept open, pore water will drain out of the soil, pore water dissipation shall occur and the soil will eventually be consolidated under an effective stress equal to cell pressure. If the drainage line is closed, the soil remains unconsolidated and no volume change takes place.

During the second stage of the test, when the additional axial stress is applied, shear stresses are induced in the soil. The resulting pore water pressure is not equal to the applied additional axial stress. Again, the test can be conducted under two conditions of drainage. The sample may be tested to failure without allowing pore water pressure to dissipate i.e.

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undrained conditions(valve closed). Alternatively, water may be allowed to drain out of the sample with the dissipation(release) of pore water pressure i.e. drained conditions.

VANE SHEAR TEST This test is used for the in-situ determination of the undrained strength of non fissured, fully saturated clays: the test is not suitable for other types of soil.

The equipment consists of a stainless steel vane of four thin rectangular blades, carried on the end of a high tensile steel rod: the rod is enclosed by a sleeve packed with grease. The length of the vane is equal to twice its overall width, typical dimension being 150 mm by 75 mm and 100 mm by 50 mm. Preferably the diameter of rod should not exceed 12.5 mm.

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.

Example

Two triaxial tests were done on soil samples. In the first test the all round pressure was 2.4 kg/cm2 and failure occurred at an added axial stress of 7.5 kg/cm2. In another test, the all round pressure was 4.0 kg/cm2 and failure occurred at a total axial stress of 16 kg/cm2. Determine the values of cohesive and angle of internal friction at failure.

Solution

For first observation,

σ3 = 2.4 kg/cm2; σ1 = 2.4 + 7.5 = 9.9 kg/cm2

For 2nd observation,

σ3 = 4.0 kg/cm2; σ1 = 16.0 kg/cm2

With the two sets, two Mohr's circles will be drawn and the Mohr's line of rupture(failure) tangent to the two circles will determine the value of c and φ (see figure).

Since the Mohr's line of rupture is passing through origin c=0 and its inclination with x -axis gives value of φ which is equal to 36.50.

c = 0 and φ = 36.50

Example (SSC, Junior Engineer, 2017)

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A sample of normally consolidated clay was subjected to a consolidated undrained triaxial compression test that was carried out until the specimen failed at a deviator stress of 50 kN/m2. The pore water pressure at failure was recorded to be 20 kN/m2 and confining pressure of 50 kN/m2 was used in the test. Determine the consolidated undrained friction angle.

Solution

Deviator stress , (d)f = 50 kN/m2

Pore water pressure at failure, (ud)f = 20 kN/m2

Now 21 3 ( ) 50 50 100 /d f kN m

For normally consolidated clay with C = 0

2 01 3 tan 45

2

Using this eq., we get

= 19.460

Example

A vane, 10 cm long and 8 cm in diameter, was pressed into soft clay at the bottom of a bore hole. Torque was applied and gradually increased to 45 N-m when failure took place. Subsequently, the vane was rotated rapidly so as to completely remould the soil. The remoulded soil was sheared at a torque of 18 N-m. Calculate the cohesion of the clay in the natural and remoulded states and also the value of the sensitivity.

Solution

(a) Normal State:

T = 4500 N-cm; H = 10 cm; d = 8 cm.

T = πd2τf

6

d +

2

H

Putting values

τf = 3.54 N/m2 and c = τf = 35.4 kN/m2 ( as φ = 0)

(b) Remoulded State

T = 1800 N-cm

Putting values

τf = 1.41 kN/m2 and c = τf = 14.1 kN/m2 (kPa)

sensitivity = 35.4/14.1 = 2.5

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