# Lecture – 46 Soil Mechanics

Students we had 3 lectures so far on this

topic of shear strength of soils. Today we will go on to the fourth lecture. Let us take

a look at the first slide which describes what we have seen in the past. We have seen

basically that this topic of shear strength of soil needs to be studied in essentially

3 stages; number one is understanding the behavior under shear depending upon the type

of test or depending upon the type of equipment that is used for testing shear strength. Secondly, the drainage conditions used in

the test depend very much and influence very much the behavior of the soils under shear

and lastly we need to study the shear strength according to the type of soil because behavior

of cohensionless soil is quite different from behavior of cohesive soil under shear. Having

classified this topic of shear strength into these 3 basic categories, we had started looking

at the first category that is the type of equipment or type of test that is used for

determining the shear strength of soil. Under that if we go to the second slide we will

be able to recapitulate that we have already gone into complete detail of the shear strength

known as the direct shear test DST. We had also seen an overview of triaxial shear test

TST of course in one of my early lectures the first one perhaps I had also given you

the general idea about the next 2 test, the unconfined compression test and the vane shear

test. Today what we shall be doing will be to quickly

recapitulate briefly what we saw under direct shear stress test. Then again recapitulate

what we have studied under triaxial shear stress and then go on further to get in to

the nitigrity and details of the triaxial test and also do some examples. If you remember this was the basic schematic

diagram of a typical direct shear apparatus. You can see there that there are 2 parts in

this. The lower part and an upper part separated at the junction here like this and the soil

specimen is also a specimen and this specimen is sheared by applying a tangential load,

a shearing force here. Of course a normal load is applied to keep this specimen confined. Then there are of course other small additions

such as the dial gauge here to measure the volume change, the dial proving ring here

to know the force that is applied, porous stone here to permit drainage and then the

displacement gauge to know the displacement that the sample undergoes. All these together

constitute the entire shear strength apparatus known as the direct shear apparatus. Then we also saw that the results of a direct

shear test can be analyzed and can be interpreted with the help of a Mohr circle. The principle

of the Mohr circle was explained to you in very great detail by me in one of the earlier

lectures. So today we will quickly see a typical Mohr circle particularly the way it looks

for the direct shear stress. You know that a Mohr circle is this, the point A corresponds

to the minor principle test, point B corresponds to the major principle test and the point

OP which is obtained by drawing a line parallel to the plane of sigma1 gives the pole OP because

by definition O is a point on the Mohr circle which has coordinates sigma and tow which

in turn or equal to those on a plane passing through that OP and the point A. It’s so happens that this red line which

passes through the A and OP is nothing but the minor principle plane and therefore point

OP represents the stresses which act on the minor principle plane. And now from this OP

if you draw a line horizontal OPC then since it is horizontal it will be nothing but the

representation of the failure plane. In a direct shear stress as you remember this failure

plane is an imposed plane, a horizontal plane and C will then be having coordinates which

are equal to normal and shear stresses on the failure plane. Now from this, one can

find out the values of cohesion and friction being a direct shear test which is drained

where drainage cannot be prevented, you only get an angle of friction phi, cohesion is

zero. Now let us take a quick look at the triaxial

test, the way we have seen it in the earlier lecture. This is what a typical triaxial apparatus

looks like. There is a loading frame, there is a cell which contain the triaxial specimen

where the so called triaxial stress state is achieved, below that there is a motorized

unit which drives the platen upwards, the base upwards thereby imposing a vertical load

and then of course there is a proving ring to measure the vertical load and dial gauge

and so on. Let us see a little more detailed diagram,

a schematic diagram other than this photograph. This is the schematic diagram of a triaxial

apparatus. You see here starting from bottom, taking the right hand side this is where we

have the pedastal and from the pedastal there is an outlet here rather than an inlet which

is attached to the pedastal you can see here through which we apply this cell pressure.

That cell pressure gets applied on the specimen inside this cell so this perspex cylinder

here thus is so called the cell. On top of this specimen here we have a loading platen

and so also at the bottom. At the bottom the pedastal there is also a

outlet which goes to the pore pressure measuring unit and there is a pipe here which goes into

the soil specimen. So the pore pressure inside the soil specimen gets reflected here in the

pore pressure measuring unit. Then there is rubber membrane which holds the specimen in

place. Then of course we have an air vent to ensure that the water that is stored here

through which the lateral pressure is applied is free of air and then we have the proving

ring to know the vertical load applied. This shows the stresses that act on the typical

triaxial test specimen. The way that the test is conducted as we have seen earlier is first

of all a radial all round pressure is applied that’s known as the cell pressure sigmar

is acting on the circumference of this specimen. And after this the vertical load is increased

until the specimen fails and therefore at the end of test, at the point of failure we

will have an all-round pressure sigmar which incidentally being watered. It also act at

the top and the bottom and then this stress sigmar is increased by an amount known as

the deviatoric amount well, so as to get the total axial stress sigma1 at the top and bottom

at the time of failure. So having seen this quickly let us move on

to today’s lecture. In today’s lecture I plan to go further and talk a little more

about details of the triaxial shear test and see a few problems and then move on to the

unconfined compression test and finally the vane shear test and everywhere we will understand

the principles and the method of utilization of the test through some example. Basically

all these test ultimately are intended to give the shear strength parameters C and phi.

If they are total strength parameter they are C and phi and if they are effective stress

parameter we generally refer to them as C dash and phi dash. So we will see a few examples which involve

determination of the effective strength parameters or the determination of the total stress parameters.

So more about the triaxial tests. Let’s see the components of a triaxial apparatus

by flitting through a few photographs again. What we see here are the major components

of a triaxial apparatus. I have highlighted here the lateral pressure and the vertical

load units, here is we have the lateral pressure unit and here is we have the vertical loading

unit. So using the same terminology which is used in one of the earlier photograph we

have here the cell, we have here the platen, we have here the loading frame and then below

that the motorized vertical load application unit. And on the left you have a compressor which

develops the air pressure that is required to apply the lateral pressure. So this is

the lateral pressure unit and then lastly we have the pore pressure measuring unit,

to know the pore pressure at any stage of loading or any stage in the test thats developed

inside the specimen. Another photograph, a greater detail of the

motorized vertical loading unit. We have here the facility to control the speed of this

motor so that the movement in the platen upwards is controlled in a string controlled manner.

So by adjusting the string in the upward direction we can control the load that is applied in

the vertical direction. Here is the loading frame, the cell and the motorised unit and

the pore pressure unit altogether, one of these views. Here we have slightly closer views of the

compressor and the lateral pressure unit. Compressor develops the air pressure that

is required to drive the water and the pressure into the cell, from here the air pressure

comes through this and the pressure is applied to water which goes out through this into

the triaxial cell. This is the pore pressure measuring unit. Now in a case of a triaxial test we need perfectly

cylindrical specimens whose ends are quiet planar whose sides are perfectly vertical

because it has been found from experience that a specimen which does not enjoy these

benefits or these characteristics sometimes gives errors in the final results. Sample

or specimen preparation is therefore a very important aspect and here is a unit which

is meant for extracting a sample from a core box that’s brought from the field. We extract

a core from the field and then that we take out a sample. From the sample we then cut

out a specimen. Here is where we have this specimen and this specimen is taken out of

the spilt mould that is if you go back to the earlier slide, when you extract with the

help of a plunger the core out of this cylinder then what comes out is collected in a split

mould and then because its a split mould, you can open it out and take the sample out

and once the sample comes out you can cut it to appropriate sizes and together with

this patterns it can be put on the loading machine. Here we have the closer view of this split

mould and the specimen holder in the triaxial apparatus. Now what are the kinds of stresses

and strains that develop in the triaxial test when we apply the so called the radial pressure

and the axial pressures. So first the self-pressure is applied in the form of a radial pressure

and then the pressure is increased in the axial direction until failure take place.

So the radial pressure is say sigmar and the axial stress is say sigmaa. Obviously since no other pressures are acting,

the radial pressure is going to be the minor principle stress and the axial stress that

is applied is going to be the major principle stress respectively acting on a vertical plane

and a horizontal plane. The difference between this two sigmad is

known as the derivator stress and it is therefore equal to the sigma axial minus sigma radial

and since axial stress is nothing but the major principle stress and the radial minor.

We have sigmad taken is sigma1- sigma3. Further the stress sigmar that is applied all around

is nothing but a confining pressure that confines the sample. And the reason why we have applied this is

this helps us to simulate the confining pressure that exists at the depth from which the sample

has been taken out. Since the behavior of a soil under stress is very much controlled

by the stress that it has experienced insitu. We must recreate the insitu stresses in the

specimen that we used in the lab and that’s the reason why we apply a confining cell pressure

sigmar whose magnitude is equal to the confining pressure that is expected to be existing at

that depth from which the specimen has been collected. Now this stress condition inside

the specimen which is basically sigmar and sigmaa depends on the drainage conditions.

If drainage is permitted from this specimen then what happens is as the water expels out

there is a compression taking place. A volumetric compression takes place in the

specimen and as the water goes out therefore effective stresses develop in the specimen.

So as the test progresses if it’s a drained test then the effective stresses in the specimen

go on increasing. So if we are applying stresses sigma1 and sigma3 to start with at the end

of the test we will have the effective stresses sigma1 dash and the sigma3 dash acting on

the specimen. If on the other hand we have the undrained conditions then since drainage

is not permitted, soil volume does not change it remains constant and therefore the pore

pressure get build up in the water inside the specimen and effective stress are not

allowed to develop. So in an undrain test purely undrain test, the initially applied

sigma1 and sigma3 remain the total stresses and the total stresses remain the same at

the end and effective stresses are not mobilized. So this shows the stress conditions through

a Mohr diagram which exists in a triaxial test. In a typical triaxial test as we have

seen we will be having the minor principle stress and the major principle stress. Generally

we conduct at least 3 tests on any particular soil. So here we have the minor principle stress

and the major principle stress and corresponding circle for one pair of these stresses then

we can apply another cell pressure sigma32 get a corresponding failure axial stress sigma12

so also a third one. So the test is conducted on the same soil under 3 different confining

pressures and 3 different Mohr circles are obtained corresponding to the 3 different

failure axial stresses. Now if I take a common tangent to all of these and assume that the

Mohrculum failure criteria is valid for this. Then what we will have will be an intercept

which represents the cohesion in the soil and an angle here, the angle of internal friction

phi. And the stress state in the triaxial test

if that is known then we can plot these graphs and we can calculate the shear strength parameters.

That’s the reason why need to understand correctly the state of stress in the specimen

at any given time. Now what happens to the strain? By measuring the change in height

during the axial loading and change in volume during the lateral loading we can determine

the strains. If we assume that these cylindrical specimen remain like a cylindrical else specimen

even after deformation then we can derive simple expressions for the volumetric and

axial strain. It is very obvious that the volumetric strain will be epsilonv equal to

minus dV by V0, dV is the change in volume. Now the change volume can be easily measured

by attaching the pore pressure unit and the original volume V0 is supposed to be known

for the specimen and therefore epsilonv the volumetric strain can be easily computed. Then on the other hand the axial strain will

be the change in the vertical height of the specimen divided by its original height epsilona

and since on the vertical compression, the area of cross section of the specimen increases

as the specimen gets compressed and its diameter increases. The new area of cross section which

will tell us the new stress state will be given by this expression in terms of the vertical

and volumetric strains and the original area of cross section, we can get the new area

of cross section. So A is the new area of cross section after

deformation, A0 is the initial area of cross section of the specimen and h0 is the initial

height of the specimen. With this we will be able to tell what will be strains at any

given instance in the specimen. So the sum total of all this is during the experiment

there will be some stresses and strain developing. The stress state can be expressed in terms

of the Mohr circle as shown here whereas the strain state can be expressed in terms of

the variation of the deviatoric stress as the function of the axial strain epsilona. This shows that the deviatoric stress goes

on increasing and the axial stress goes on increasing upto a point beyond which the deviatoric

stress comes down because the soil would have already reached its failure strain value and

the stress cannot be build up anymore. Now why do we need to conduct the triaxial test,

why do we need to understand the importance of the distraxial test? It’s because there

are several advantages associated with the distraxial test compared with the direct shear

stress. The advantages are one real situations of confinement in the field can be simulated.

As I mentioned just a few seconds ago a soil is under certain confining stress at a given

depth in the nature, in the field. Now when you extract a sample from a certain

depth and cut a specimen out of it and test it, obviously in order to simulate the conditions

which it experienced in the field, we need to know what was the condition of confinement

and reproduce that in the experiment. And the triaxial test helps us to do precisely

this because we can apply a cell pressure of any known magnitude which will simulate

the confinement that the sample is experiencing in the field at a given depth. Then the manner in which the pressures are

applied gives us the facility to have fairly uniform stresses inside the specimen. The

direct shear if you compare for example as the specimen deforms you find that the area

over which the normal stress is acting goes on changing. Therefore the stress level goes on changing

it doesn’t remain uniform whereas on the other hand in the triaxial test we have the

advantage of imposing fairly uniform stresses on the sample. Then most important, the shear

strain behavior of the soil can be studied in detail. As the test progresses you can

measure the strain level, volumetric strain and the axial strain and compute the shear

strain. Lastly both drained and undrained conditions

can be easily simulated in a triaxial test; in the direct shear stress on the other hand

we found that undrained conditions can hardly be simulated adequately. Then finally pore pressure can be measured

and effective stresses can be determined during the undrained test. This is a tremendous advantage

because the knowledge of the pore pressure exactly tells us what is the kind of effective

stresses that are mobilized during the test and the shear strength is a direct function

of the effective stress and therefore knowledge of an effective stress is very necessary and

triaxial equipment can really help you to get that. Let us quickly take a look at some examples.

To begin with let me take the same examples which we saw in the earlier class we have

called this class example 6. So let us see example 6 which we had seen in the last lecture,

it helps to maintain the continuity. So that example which we saw in the last example was

data about the triaxial test on saturated sample of sand. The cell pressure used was 300 kilo newton

per meter square and in the test it failed at a deviatoric stress of 200 kilo newton

per meter square and pore water pressure was measured and that failure it was found to

be 175 kilo newton per meter square. So with the help of this the shear strength parameters

were determined. To determine the shear strength parameter we can use either the total stresses

or subtract the pore pressure which has been measured and get the effective stresses. And

we can do this by plotting the Mohr circles or by using the formula for the shear strength. Here the example which is given makes use

of the formula but a Mohr diagram can be easily plotted and the student is advised to try

that out himself and compare the results with the formula and find out how right he is.

So this is what the typical Mohr circle looks like in a triaxial test. Here we have the

point corresponding to the major principle stress and the minor principle stress and

this is the failure plane at an angle theta critical. Now in terms of total stresses we

can say that tan square 45 plus phi by 2 is equal to 500 by 300. Why 500 by 300? It is

nothing but you can see here it is cell pressure plus the deviator stress that is sigma1 and

see here 500 so sigma1 upon sigma3 that’s what we are having here. Sigma1 by sigma3 is equal to tan square 45

phi by 2, this is an expression which we had seen in one of the earlier lecture and from

this we can get phi equal to 14.5 degrees. And if we wish to use the effective stressed,

what we need to do is to subtract the pore pressure from the values of sigma1 and sigma3.

And if you see go backwards, the pore pressure is 175 and coming back if you subtract 175

from here you have 325 and 125 which gives you the effective angular friction phi dash

equal to 26.4 degrees. We also saw another example, example number 7 which was very similar. Now in both these example we considered single

test with one sigma3 and sigma1 but as I mentioned a few minutes back in a typical triaxial test

we always make use of several confining pressures and find out the corresponding failure axial

stresses and plot Mohr circles one after the other, for each confining stress and find

out a common tangent which will give us the failure envelop. So let us see the examples

of that, let us go example number 8. In this example we are having data of undrained

triaxial test on clay. So undrained triaxial test in drainage is not permitted, obviously

it means that the stresses which we have are all total stresses. And incidentally in this

example pore pressures were not measured and therefore the stresses which we have are the

total stresses and in order to get the effective stresses, if required we need to have the

pore pressure measurement. Now the values of the sigma1 and sigma3 are given here. The

sigma3 is this, the deviator stress is this. Normally we apply confining pressures of these

ranges, of these magnitudes. So if I take these data and go further I find that since

pore pressures are not measured, we can only determine the total stress parameters. To determine the total stress parameter today

we will see two methods, one is of course the Mohr circle plot about which have talked

earlier, for a long time. Another is so called p q plot and both of them can be used for

determining the total stress parameters. Here we have the typical Mohr’s diagrams that

is we have 3 stress values, minor principle stresses corresponding to 50, 150 and 250

and 3 axial stresses. What we need to do is to plot the 3 different Mohr cycle and then

plot a common tangent. Now that’s not shown here, the Mohr circle plots are not shown

here. However the student may attempt drawing these Mohr circles and draw a common tangent. Alternatively we can use the so called p q

plot. What is this p q plot? We know already this equation, the relationship between sigma1

dash, sigma3 dash, c dash and tan square 45 plus phi by 2 that is n phi or its equivalent

which is this. Now you find this equivalent expression that is the second expression is

a linear expression and if you therefore plot half of sigma1 + sigma3 along the x axis and

half of sigma1- sigma3 along the y axis we will get a linear relationship which is shown

in the next slide. Here we have half of sigma1 + sigma3 and we

have half of sigma1- sigma3 along the y axis. This is known as pu p this is known as q because

it’s a linear relationship the p q plot as it is called will be a straight line. Here

I have an example of a pq plot. What is special about this p q plot is if you go back to the

basic equations. The slope of this will be given by tan phi dash and the intercept will

be given by c double dash. So this intercept c double dash is here and the slope of this

line is given by the angle of inclination which is phi double dash. Now you can compare

this equation with the basic equation which is given here the original one and you can

show that c double dash is related to c dash and phi double dash is related to phi. That relationship is phi double dash is tan

inverse of sing phi and c double is c cos phi. So what this means is ultimately we are

interested in the shear strength parameters c and phi. By taking the straight line plot

we can find out sine phi which is equal to tan phi double dash and then from sin phi

we can find phi. Then we have c double dash which is equal to c cos phi, knowing c double

dash which is an intercept here we can always find out c because phi is already determined

from this. So this is what normally done in order to

determine the shear strength parameters. So now if I take the total stresses that have

been obtained in this experiment, we have sigma3 total stress we have sigma1 the total

stress and corresponding values of p and q will be sigma1 plus sigma3 by 2, sigma1 – sigma3

by 2 and these are the values. These values can be used for drawing the p q plot. Here is the p q plot, so as we saw the data

of the problem is adequate enough to compute p and q values and plot them. So if you plot

p and q this is a kind of a plot you will get for the total stresses and this being

a straight line you can fit a straight line through all the 3 points and you can find

out the intercepts and the angle of inclination from which you can calculate the cohesion

and friction parameters corresponding to the total stress conditions that is the undrained

conditions which are known as usually cu and phiu undrained. So in this particular problem

we get 27.31 kilo Pascal’s and 15.82 degrees. This is the situation for the total stresses

but since in this problem no pore pressure where known, we can only get total stresses

and total stress parameters. Now let’s go further. If you plot the Mohr

circle diagram which is not shown here in the slide you will find that you will get

slightly different values of cu and phiu compared to what we got for the p q plot. In the p

q plot we got 27.31 and 15.82 whereas if you plot the Mohr circle diagram and get the values

of cu and phiu you will find 29.0 and phiu equal to 15 degrees. A slight difference is

always permissible because they are both graphical procedures but by in large the p q plot is

preferred these days because a straight line can be plotted more reliably than a Mohr circle

and a common tangent. And going by that statement, we can say that the values of cu and phiu

which we have observed from the p q plot may be taken as the appropriate value for the

particular soil. One more example here again the triaxial test

and undisturbed clay sample and confining pressure used were same 50, 150 and 250 kilo

newton per meter square as in the previous example but the deviatoric stress are different

41 and 112 and 176 and the pore pressure were measured. The pore pressure at failure are

45, 105 and 160. So this is the situation in which we are in a position to calculate

the effective values of sigma3 and sigma1 and therefore we can, not only get total stress

values p and q but we can also get p dash and q dash and therefore we can get cu phiu

corresponding to the total stress conditions and c dash phi dash corresponding to the effective

stress conditions. Again we can use either the Mohr circle plot

or we can use the p q plot. So let’s proceed. If I consider total stresses first I find

that, I know sigma3 and sigma1- sigma3. It will give raise to three values of sigma1

like this. Since we know the pore pressure we can subtract the pore pressure from the

given values of sigma3 and the computed values of sigma1 and we will get sigma3 dash and

sigma1 dash and also sigma1 dash – sigma3 dash the deviatoric stress. Since the pore

pressure u gets detected from sigma1 as well as sigma3, the difference between sigma1 and

sigma3 and sigma1 dash, sigma3 dash will both be same. So the deviatoric stress either in

the total stress or in effective stress manner they are identical. Once again either we get the shear stress

parameter using the more envelop like this, plot three Mohr circles draw a common tangent

or you can get it from the p q plot. If we use the Mohr circle plots we can draw Mohr

circle plots for the total stresses entirely or for effective stresses entirely. If I were

to plot the 3 Mohr circles using the total stress value and get the intercept and inclination

of the common tangent I will get 5 kilo newton per meter square as the undrained cohesion

and 14 degrees as the angular friction. If on the other hand I use sigma1 dash and

sigma3 dash for each of the 3 tests, plot three Mohr circles draw a common tangent I

will get the corresponding effective stress, shear stress parameters intercepts and angle

of internal friction as c dash is equal to 15 kilo Newton per meters and phi dash is

equal to 24 degrees. You can find that there is a large difference here that’s what happens

when you don’t have drainage and when you have drainage. When you don’t have drainage

and measure the pore pressure this is what you get. Now suppose you want to determine

the same thing using p q plots then the data for plotting the p q straight line in terms

of total stresses it will be this because we already know the values of sigma3 and the

sigma1. P is nothing but sum of the 2 by 2, q is nothing

but the difference between the 2 divided by 2. So if we plot this value of p and this

value of q we get the total stress p q plot. So here is the total stress p q plot for this

test and from this the intercept and the angle of inclination will gives us the undrained

cohesion and undrained angular friction as 3 and 14 approximately. If we were to use the effective stress parameter

then the data that we need for plotting the p q graph is, sigma3 dash is this after subtracting

the pore pressure, sigma1 dash is this after subtracting the pore pressure, p dash will

be the addition of 2 divided by 2, q dash is the difference divided by 2. So we now have p dash which is quiet different

from the value of p we had earlier but q dash however remains the same because it is the

difference between the stresses and as I pointed out a little earlier sigma1- sigma3 is equal

to sigma1 dash minus sigma3 dash. So we have p dash and q dash if we plot them we get the

p q plot for the effective stresses and from here from this intercepts and angle of inclination

we can get the effective cohesion and effective angle of friction as 11 and 26 as shown in

this graph. So to summarize we get undrained cohesion

and undrained friction under total stress conditions from the p q graph like this. The

corresponding Mohr circles gave us 5 and 14 the difference is marginal and this much difference

does occur because of the two ways of plotting. And as I said p q plot always preferred because

plotting a straight line is always easier and more accurate then plotting a circle and

a common tangent. Similarly under effective stress condition c dash and phi dash are like

this for p q plot whereas the corresponding Mohr circles values which we got a little

earlier on 15 and 24 degrees. So this is what a typical triaxial test is all about we conducted

the test, measuring the pore pressure usually so at the end of the test we not only have

the total stresses that we have applied but also the pore pressure and therefore the effective

stresses. So we can calculate both the total strength parameters and the effective strength

parameters. The utility of these parameters we shall be

seeing in the next lecture. When we are going to look at shear strength in a different way

that is from point of view of the effective of drainage undrained and drained conditions

and their effects on shear strength. Let us move further, take another example. Example

10 is a test on rock. The triaxial test can also be conducted on

rocks but what important to notice is the kind of stress that are required to make a

rock sample fail under shear in triaxial conditions are much higher than what we require for soils.

So in a drained triaxial test or volcanic breccia we get values of sigma3 and deviatoric

stresses sigma1 and sigma3 which are much higher than what are usually obtained in the

case of soils. So if sigma3 dash is 5, 10 and 20 mega Pascal’s, deviatoric stresses

are 31.8, 42.2 and 68 mega Pascal’s. Then sigma1 dash will be 36.8, 52.2 and 80 mega

Pascal’s. Now from this data we can calculate p dash and q dash and get the p q plot. Since

this is a rock sample we sell the measure pore pressure in a rock sample because it’s

virtually impossible and we get a p q plot therefore in terms of the effective stresses. The p q plot in terms of the effective stresses

for this rock gives us a c dash of 5 mega Pascal’s and phi dash of 33.34 degrees.

So this is the utility of the p q plot, there is a lot of advantage in drawing a straight

line through the test points than in plotting Mohr circles and drawing a common tangent.

Now let us see one more example of triaxial testing. A clay specimen here is tested under

a condition of full drainage; the cell pressure used is 80. Here these shear stress parameters

are given c dash and phi dash the effective stress parameters. What’s required is the

compressive strength of the specimen that is now from shear strength we want to determine

the compressive strength of the soil. It’s possible from the triaxial shear test data.

From lecture number 3 we know that there is a certain relationship that holds good in

a Mohr circle at failure. Let’s see what is the relationship in the next slide. This is the Mohr circle and from geometry

we can easily derive this relationship here which holds good for a Mohr circle at failure,

c cot phi that’s what you will get here on the x axis when you produce the envelop

and make it cut the x axis here. This is c the angle which will be made will be phi and

therefore the distance here will be c cot phi, c cot phi plus sigma3 is the distance

from the shifted origin up to the point A and c cot phi plus sigma3 plus sigma D the

deviatoric stress or Df at failure is the total length here. So one divided by the other

can be shown to be equal to 1-sine phi by 1 +sine phi. This is simply a modified form

of writing an equation which we saw a little earlier in terms of tan square 45 plus phi

by 2. So from this equation it’s very easy to calculate c and phi and that’s what we

will do. Compressive strength of a soil is nothing

but the deviatoric stress at failure Df. So straight away from this equation the unknown

Df can be calculated because sigma3 is known, phi is known and c is known. So if you substitute

the known values, you will find that c cot phi is nothing but 27.47. This is sigma3,

c cot phi plus sigma3 plus Df. So this is equal to 1- sine 20 by 1+ sine 20 where 20

is the angle of internal friction which is known. This gives us Df equal to 111.86 kilo

newton per meter square that’s the compressive strength of the soil sample. Now there are

two other test, the unconfined compression strength test and the vane shear test which

we will see now. These two tests are primarily meant for a clay soils, they can be done in

the laboratory as well as in the field. So if I take unconfined compressive strength

test this is what the equipment looks like. It’s a simple loading frame with a facility

to keep a soil specimen and load it and test it to failure without applying any confining

pressure. So this is known as the UCS test. An unconfined compression test is a special

case of a triaxial test because the only difference between the two is in unconfined compressive

test sigma3 is zero. Now this is usually employed to test cohesive soils, hence an axial load

is applied rapidly to failure. There is no confining pressure applied at all therefore

undrained condition will prevail obviously because the loading is rapid which means the

angle of internal friction will not get mobilized at all. That means phiu is zero therefore

the corresponding Mohr circle for this will be a single circle which will pass through

the origin. This semi-circle is a Mohr circle, Sigma3

being zero this point will coincide with the origin and sigma1 is the stress axial stress

at failure. If you draw a tangent to this obviously this tangent here will give you

the undrained cohesion and sigma1- sigma3 is nothing but the compressive strength and

therefore what you get here at sigma1 is nothing but the unconfined compressive strength the

specimen that is qu and it is from this that the name UCS test is derived. So once you

have this kind of a Mohr circles you just measure the axial stress sigma1 that’s the

confined compressive strength. Then what happens to the cohesions? The cohesion

is nothing the radius that’s equal to half the diameter that is it is equal to qu by

2. So unconfined compressive strength qu divided by 2 gives you the undrained cohesions. So

this is a very simple and very effective convenient method of determining the shear strength parameter

and the compressive strength of a clay soil can be done both in the lab and in the field.

Let’s take an example. The example would involve a number of parameters

which are described here and their relationships are also given here. A zero is the initial

cross section; there is a vertical axial strain that the sample undergoes. So if you take

that into account and calculate the modified area of cross section of the specimen you

will get Af. So from this you can always find out what will be the final stress that acts

on the specimen which is what required in order to plot the Mohr circles. Now UCS requires simple equipment. It is conducted

in the field and it’s a method for rapid assessment of consistency of soil for its

classification. Clay soils are classified according to their consistency based on the

value of unconfined compressive strength as shown in this slide. Depending upon the value

of the undrained cohesion or the compressive strength qu the consistency of a sample is

defined like this. Now further the unconfined compressive strength

of a soil depends upon the nature of disturbance that the soil is undergone. In the undisturbed

state if it has a compressive strength qu undisturbed and after remoulding if it has

got an undrained compressive strength of qu, the ratio of this is known as sensitivity

that shows how sensitive the soil is to disturbance because there is a reduction in strength due

to disturbance. This value usually lies in between 1 and 8 for clays and for flocculent

marine clays it can even go from 10 to 80. Let us take one quick look at an example.

The clay specimen which was tested in UCS has these dimensions 38 mm and 76 mm. The

vertical deformation was measured to be 10 mm at failure and the load at failure was

0.25 kilo newton. In order to compute cu of this clay all that we need to know is what

is the final area of cross section corresponding to the strain vertical strain and then the

load divided by the final area of cross section gives you the unconfined compressive strength

of 192.31 kilo newton per meter square and the corresponding cohesion of 96.15. Now we

will probably take a look at the vane shear stress in a subsequent lecture. So let us now summarize what we have seen

today. In this lecture today according to the classification of shear strength based

on the type of equipment used we have completed seeing direct shear test, triaxial shear test

and unconfined compression test and a few examples. In the next lecture we will continue

with the type of equipment used and see the vane shear stress and proceed further with

the undrained and drained conditions of testing. Thank you.

thanks sir..

Sir how u cal total stresses

And eff. Stress