# Module 41 – Shear Strength of Soil

welcome back to the course geology and soil
mechanics so in the last lecture we just started uh the discussion on triaxial shear strength
which is another laboratory test by which you can determine the shear strength parameters
for a particular soil right and there we have seen that you have 3 different
kinds of say triaxial test one is the consolidated drained test that is cd test one is consolidated
undrained test that is cu test and another one is the consolidated unconsolidated undrained
test that is uu test okay now first we will discuss what is consolidated
drained test so that is cd test right in cd test the saturated specimen is first subjected
to an all round confining pressure say sigma 3 okay so basically that sigma 3 is i mean
we are putting sigma 3 that is nothing but your minor principle stress and once you once
you apply the all round confining pressure basically that means your sigma i mean 3 is
becoming the minor principle stress because along along the specimen side you do not have
any shear stress applied through the cell pressure right
so cell water will be uh applying the pressure in the radial direction and that is known
as uh your uh that is that will be basically eventually will be becoming one principle
stress now we will see that so will be that will be becoming the minor principle stress
so mean first what we do we take the saturated soil specimen and then that saturated soil
specimen is subjected to an all round confining pressure say sigma 3 right
so at that condition so as confining pressure is applied the pore water pressure of the
specimen increases by u c agreed so as you i mean from your consolidation chapter if
you recall immediately you apply some extra pressure right so i mean soil specimen was
there now you are applying some all round say cell pressure that is sigma 3 once you
apply sigma 3 basically immediately you will be getting some increase in the pore water
pressure right because water is incompressible so in the soil specimen you will be getting
some increase or enhancement in the pore water pressure that is nothing but your excess pore
water pressure now which can be expressed as say b capital b is equal to u c by sigma
3 where b is nothing but the skemptons pore pressure parameter right so i mean we will
we will come to that point so basically for saturated soft soil b is
approximately equal to 1 why b is approximately equal to 1 so because you are not allowing
the drainage right so you are stopping the drainage if you recall that spring uh loading
loading spring analogy so if you apply the load i mean the pressure okay on the soil
specimen immediately water will try to drain out right but the because of the valve is
closed immediately the whatever extra pressure is coming on the soil specimen that will be
taken care of by the pore fluid that is water right
so therefore for saturated soil b is approximately equal to 1 that means u c that is the excess
pore water pressure must be equal to sigma 3 that means there is no enhancement in the
effective stress so whatever sigma 3 you are applying that is completely taken care of
by the uh water right which is which is present inside the soil specimen so therefore b is
approximately equal to 1 however for saturated stiff clay the magnitude of b can be less
than 1 okay now basically this is happening we will we
will come to this figure uh this is happening that you are applying so there are 2 steps
basically happening in this consolidated drained test in the first step you are applying sigma
3 all round okay from top from the radial direction all round sigma 3 that means you
are increasing the cell pressure and that cell pressure will be giving you some sigma
3 all round right and due to that you will be getting enhancement in the u c that is
excess pore water pressure will be building up and at that time you just open the drain
drainage valve right so once you open the drainage valve basically
water whatever water will be there under excess pore water pressure inside the soil specimen
that water will be coming out or will be draining out from the drainage path right or the drainage
pipe now once the excess pore water pressure is getting dissipated basically u c that is
nothing but the excess pore water pressure due to the application of sigma 3 already
we have seen that u c will be becoming zero after complete consolidation under sigma 3
right and then you are applying the deviator stress right uh through the axial loading
ram whatever we have seen in the test setup right
through this you are applying sigma d delta sigma d and this when you are applying delta
sigma d basically at that time you are closing the drainage valve so whatever load you are
applying immediately it will be taken care of by the soil specimen right and then basically
if you open the valve then basically this sigma d whatever excess pore water pressure
will be building up that excess pore water pressure will be draining out
so eventually so your due to this application of sigma delta sigma d you have got the excess
pore water pressure delta u d which will be becoming zero at the final state or the final
when the consolidation will be happening right that soil water will be draining out so that
is why it is known as consolidated drained test because at the first instance when you
are applying sigma 3 all round at that time you consolidate the sample by allowing the
drainage and then during the test basically when you are applying sigma d at that time
also you are allowing the drainage that is why it is known as consolidated drained test
okay now if the connection to drainage is opened
dissipation of the excess pore water pressure happens and thus consolidation will occur
with time so u c will be becoming equal to 0 as i told you right so once you (cons) once
you apply the sigma 3 at that time you are you you have the valve closed drainage valve
is closed and due to that you will be getting the building up of u c and then if you allow
the drainage that means if you open the drainage valve right so at that time your u c will
be dissipating out that means excess pore water pressure will
be dissipating out and ultimately or the eventually u c will be becoming zero next the deviator
stress that is sigma del delta sigma d is increased very slowly very gradually now you
are increasing delta sigma d and you are allowing the drainage now you are not closing the valve
so you are allowing the drainage and therefore the drainage connection is kept
open and the slow rate of deviator stress application allows complete dissipation of
any pore water pressure that developed as a result that is delta u d is equal to 0 now
as i told you that if you keep the valve closed and if you apply delta sigma d what will happen
so some delta u d that is some excess pore water pressure will be building up inside
the soil specimen due to the application of delta sigma d the same amount of i mean whatever
delta sigma d we will be applying the same amount of uh excess pore water pressure will
be building up however you do not do that in the test actually
you keep the valve open drainage valve open so as you apply delta sigma d and this application
is very slow that means slowly you are applying delta sigma d and slowly your water is draining
out so there is no change of building up of excess pore water pressure inside the soil
specimen so therefore the whatever i mean i mean may
have happened actually if you if you keep the valve closed that is delta sigma i mean
building up of delta u d that will be becoming zero because there is no excess pore water
pressure getting built up uh in inside the soil specimen due to the application of delta
sigma d okay so as as we have seen so basically uh u c will be becoming 0 because you are
allowing the consolidation u c will be becoming 0 eventually and then you are applying delta
sigma d gradually and also your delta u d will be becoming 0 so that means there is
no excess pore water pressure uh getting built up in the soil specimen due to the drainage
valve is open now because the pore pressure developed during
the test is completely dissipated right so there is no excess pore water pressure so
the pore pressure developed whatever pore pressure was getting developed inside the
soil specimen due to the application of sigma 3 alone and then sigma 3 plus delta sigma
d along the axial direction so whatever is the situation whatever is the condition you
have allowed the drainage so therefore no pore pressure is getting built up right
so therefore we can say total and effective confining stress that is sigma 3 is nothing
but sigma 3 prime so total must be equal to effective because there is no excess pore
water pressure agreed okay similarly total and effective axial stress at failure it will
be equal to sigma 3 plus because sigma 3 alone is not causing the failure as i told you right
so as you are applying the deviatoric stress delta sigma d and then basically you are initiating
the failure you are initiating the shear failure so as you increase delta sigma d and once
you reach the failure at that time the delta sigma d is expressed is delta sigma d f so
sigma 3 plus delta sigma d will be the axial stress and that is nothing but sigma 1 because
on the top plane you do not have any shear stress getting developed right
so that plane is another principle plane and on that plane whatever stress will be normal
stress will be acting that is nothing but the principle stress and from the from the
uh i mean mechanism you have you have understood now that along the radial direction whatever
stress whatever principle stress was there so that was the minor principle stress and
now you are applying sigma 3 plus delta sigma d f for at the failure and that will be your
sigma 1 that will be your major principle stress and the total major principle stress
must be equal to effective major principle stress because you have allowed the drainage
is that clear okay so now basically we will see that deviatoric
stress versus axial strain this plot for loose sand and normally consolidated soil and dense
sand and over consolidated soil it has been seen from the experience and from the from
the test that loose sand and normally consolidated clay will be behaving in a very similar fashion
whereas dense sand and over consolidated clay will be behaving in a very similar fashion
so what is that what kind of plot you will be getting in case of loose sand and normally
consolidated clay if you perform this cd test basically you will be getting so as you increase
delta sigma d so gradually you are increasing delta sigma d and you will be obtaining or
you will be measuring right axial strength right you can measure that we will come to
that point later on so basically you are applying that load and
because of that you will be getting the shortening or the squeeze or the strain in the soil specimen
right in the in the vertical direction right so slowly you increase delta sigma d your
axial strain is also increasing and then it is becoming almost constant right whereas
in case of dense sand and over consolidated clay you get initial increase and then it
reach it reaches the peak which will be giving you the failure say stress okay the failure
point and after that it will be decreasing okay this is the falling part
now basically this is a this is the typical say mohr coulomb failure envelope obtained
from the consolidated drained testso basically you will be getting uh 2 different regions
one region you will be telling about the over consolidated region another region will be
telling about the normally consolidated region we will come to that point
so over consolidation results when a clay is initially consolidated under an all round
chamber pressure of sigma c and which is eventually equal to sigma c prime because the total stress
and the effective stress both are equal or both are same in case of cd test right because
you are allowing the drainage so there is no excess pore water pressure getting built
up so the pore water pressure is zero so sigma c or sigma must be equal to sigma
prime in every situation okay so over consolidation results when a clay is initially consolidated
under all round uh chamber pressure of sigma c and is allowed to swell by reducing the
chamber pressure sigma 3 which is again equal to sigma 3 prime so what we are doing here
now initially we are consolidating the soil specimen with a with an all round chamber
pressure of sigma c so we are we are increasing the chamber pressure
that is the cell pressure by an amount say sigma c and with that pressure basically we
are trying to consolidate the soil specimen and then we are reducing okay we are reducing
the chamber pressure to from sigma c to sigma 3 okay and in the reduction basically if you
if you recall the consolidation chapter if you increase the pressure and then if you
if you allow the swelling that means if you allow the uh unloading then basically after
reaching the sigma c okay you are allowing the unloading till sigma 3 right and due to
this reduction in the cell pressure or the chamber pressure what will happen it will
swell right so this i mean this swelling is allowed and
i mean therefore you can say the soil is over consolidated under the pressure of sigma c
so that was the soil already experienced sigma c amount of cell pressure all round and now
it is under the pressure of sigma 3 which is lesser than sigma c okay now the failure
envelope obtained from drained test as such of such over consolidated clay specimen show
2 distinct branches ab and bc right so you will be getting 2 distinct branches
ab this is a this is b so ab and bc so you will be getting 2 different branches now what
you what you did basically you consolidated the specimen you have applied all round cell
pressure by an amount sigma c prime right so you you had reached up to this now you
are performing the test by reducing sigma 3 up to this point and then you are you are
actually doing the shear failure i mean triaxial test so basically the soil will fail okay
so this failure envelope this is your failure envelope mohr coulomb failure envelope so
ab will be nothing but your failure envelope that failure envelope will be because of your
over consolidated soil now if you consider the normally consolidated soil if you say
the my soil is normally consolidated soil then the soil should be falling in this region
because this region will be experiencing the pressure which will be higher than sigma c
right so therefore the soil will be behaving as normally consolidated soil so you will
be getting 2 distinct zones one zone will be talking about the over consolidated zone
another zone will be talking about the under consolidated zone okay so ab and bc they will
be basically talking about the failure envelope okay for over consolidated specimen
now the portion ab has a flatter slope as you have seen portion ab has a flatter slope
with a cohesion intercept with a cohesion intercept and the shear strength can be written
as tau f is equal to c prime plus sigma prime tan phi 1 prime okay where phi 1 is this angle
angle of internal friction right now the portion bc of the failure envelope represents a normally
consolidated stage of soil as i told you and shear strength is given by tau f equal to
sigma prime tan phi prime where phi prime is this angle okay and this bc basically
so if you continuously do the test if you if you go on increasing sigma 3 say for example
if you go on increasing sigma 3 so this is your sigma 3 first sigma 3 if you increase
the sigma 3 you will be getting like that if you further increase you will be getting
like that something like that okay so basically you will be once you are within the over consolidated
zone you will be getting the failure envelope which is defined by line ab and you will be
getting the failure envelope which is define by line bc under the normally consolidated
region a consolidated drained triaxial test on a
clayey soil may take several days to complete agreed because as you know as you have seen
from our consolidation chapter that the consolidation itself is a long process it is time dependent
process right so consolidation consolidated drained triaxial test if you want to perform
on the clayey sample it may take several days to complete because first you have to allow
the consolidation okay and then you will be doing the shearing
so there are 2 i mean steps involved in this test first you do the consolidation and then
you shear it and you get the failure right so this consolidation process itself will
take several days to complete so therefore the shearing and all those things and the
total i mean if you want to i mean obtain the shear strength parameters it will take
several days to complete the test for this reason cd test is pretty uncommon
uncommon means people do not i mean we will see that that whatever you are getting from
cd test we can achieve those things by other triaxial test and those tests are comparatively
weaker than cd test because cd test the consolidation is i mean consolidation needs to be uh there
i mean otherwise you will not be getting the proper behaviour so for this reason cd test
is pretty uncommon in the geotechnical community so i will stop here today thank you very much

## 4 thoughts on “Module 41 – Shear Strength of Soil”

1. raviraj singh says:

thankuu so much sir…clear explanation…

2. S Gain says:

Thank you Sir, interesting lecture and wonderfully explained.

3. Jackson Zheng says:

This professor is amazing~

4. Koushiki Pal says:

But sir at first you said that in CD test, during application of deviator stress initially the valve is closed and then opened. Why?