# Module 49 – Problem on shear strength of soil

welcome back so in the last lecture we solved
couple of problems on shear strength of soil today we will be taking another few problems
todays first problem is a sample of dry coarse sand is tested in the lab triaxial apparatus
in the undrained condition under a cell pressure of 2 kg per centimeter square the sample failed
when the deviator stress reached 4.38 kg per centimeter square now determine the shear
parameters of the soil as i told you shear parameters means c and phi at what deviator
stress will the soil fail if the cell pressure be 3 kg per centimeter square okay so under
the cell pressure of 2 kg per centimeter square the soil failed okay when the deviator stress
reached at 4.38 kg per centimeter square and now we are going to solve this problem
the first so problem 22 so here sigma 3 is equal to 2 kg per centimeter square sigma
d that is deviator stress is 4.38 kg per centimeter square therefore sigma 1 that is the major
principle stress will be sigma 3 plus sigma d as we know that gives me 6.38 kg per centimeter
square okay now we are going to draw a mohr circle now a mohr circle is drawn considering
this situation so tau sigma space okay so this is the mohr circle okay now since the
sample is made of coarse sand as it is given in the problem dry coarse sand okay you are
using since the sample is made of dry coarse sand and since it is in the dry state right
so no apparent cohesion will be coming into the picture so you will be getting cohesion
is 0 so therefore the coulomb failure envelope will pass through the origin right if c is
0 so there will be no intercept on the tau axis so your coulomb failure envelope will
pass through the origin right so
now this is the mohr circle this is your sigma 1 this is your sigma 3 now one thing we have
decided because you are using the dry uh coarse sand therefore c is 0 so the mohr i mean mohr
coulomb failure envelope will pass through the origin origin means this point okay and
it will be tangent to this mohr circle right otherwise it will not be i mean the mohr circle
should touch the failure envelope then only it will be the mohr circle corresponding
to the failure of the sample so it will be tangent to the mohr circle at failure so this
is say center of the mohr circle say c This is the point where the mohr circle is touching
the failure envelope is d and this is the origin o okay and so therefore this angle
is phi so you need to find out the magnitude of phi because c is 0 okay
so we can if you look at this figure we can write down oc is equal to sigma 1 plus sigma
3 by 2 right what is oc oc is the distance between the origin and the center of the mohr
circle so that is nothing but sigma 1 plus sigma 3 by 2 which is 6.38 plus 2 by 2 is
equal to 4.19 similarly cd what is cd cd is nothing but your radius of the mohr circle
right cd is the radius of the mohr circle which will be making an angle 90 degree at
the tangent point okay so cd is equal to sigma 1 minus sigma 3 by 2 which is equal to 6.38
minus 2 by 2 is equal to 2.1 okay therefore from the figure i can write sin phi is equal
to cd bar by oc right cd by oc is your sin phi so which is nothing but 2.19 by 4.19
so from this i get phi is equal to 31 degree okay so therefore your c is 0 what are the
shear strength parameters c is 0 and phi is 31 degree okay so now we have we have seen
that how i can find out the shear strength parameters for uh this arrangement like when
your cell pressure is 2 kg per centimeter square and the deviator stress at failure
at is 4.38 kg per centimeter square now we will go to the second part now in the second
part it is asked that at what deviator stress will the soil fail if the cell pressure is
3 kg per centimeter square now the second part we know that sigma 1 is
equal to sigma 3 tan square alpha plus 2 c tan alpha so as your c 0 so this term will
be coming 0 so therefore i can write sigma 1 equal to sigma 3 tan square alpha okay so
which is nothing but sigma 3 tan square 45 degree plus phi by 2 okay now in the second
part it is given sigma 3 is equal to that is the cell pressure is becoming 3 kg per
centimeter square and already we have calculated phi equal to 31 degree because you are dealing
with the same sample just you are varying the cell pressure so phi will not be uh changing
right phi cannot be changed cannot be varied with
the variation of the cell pressure right so phi will be remaining same whatever you have
got with some other cell pressure that is 2 kg per centimeter square earlier okay so
from this i can find out sigma 1 is equal to 3 tan square 45 degree plus
so which gives me 9.37 kg per centimeter square so you have got the major principle stress
corresponding to the cell pressure 3 kg per centimeter square okay at failure
so if this is the major principle stress at failure then you can find out the deviator
stress is nothing but sigma 1 minus sigma 3 so sigma 1 minus 3 which is nothing but
9.37 minus 3 is equal to 6.37 kg per centimeter square so if you go so if you increase the
cell pressure as 3 kg per centimeter square then you have to you have to apply 6.37 as
deviator stress to get the failure okay for the same soil so we have solved this problem
so i hope that you have understood the concept and uh how we are using the parameters and
how we are playing with the parameters i hope that you have understood
now we will take the second problem the problem says the shear strength parameters of a given
soil are c equal to 0.26 kg per centimeter square and phi equal to 21 degree undrained
triaxial tests are to be carried out on specimens of this soil determine first deviator stress
at which failure will occur if the cell pressure be 2.5 kg per centimeter square and the second
part the cell pressure during the test determine the cell pressure during the test if the sample
fails when the deviator stress reaches 1.68 kg per centimeter square okay
so now again this is pretty simple problem already we have solved so we know sigma 1
equal to sigma 3 tan square alpha plus 2 c tan alpha right for the given soil c equal
to 0.26 kg per centimeter square and phi equal to 21 degree therefore tan square alpha will
be 2.117 and tan alpha will be 1.455 okay now hence sigma 1 is equal to 2.117 sigma
3 plus 2 into c c is 0.26 so and tan alpha is 1.455 so that gives me sigma 1 is equal
to 2.117 sigma 3 plus 0.757 so i am calling this as equation 1
now when your sigma 3 is 2.5 kg per centimeter square so that is the first part that determine
the deviator stress at which few failure will occur if the cell pressure be 2.5 kg per centimeter
square so if the cell pressure becomes 2.5 kg per centimeter square then i can find out
sigma 1 from equation 1 as 6.05 right you place sigma 3 equal to 2.5 and you get the
magnitude of sigma 1 from equation 1 so once i know sigma 1 then basically deviator
stress sigma d is nothing but sigma 1 minus sigma 3 which gives me 3.55 kg per centimeter
square okay so hence the required deviator stress is 3.55 kg per centimeter square to
get the failure with cell pressure as 2.5 kg per centimeter square now in the second
part let the required cell pressure be x kg per centimeter square
so therefore sigma 1 is equal to sigma 3 plus sigma d is equal to 1.68 plus x what is 1.68
1.68 is the deviator stress okay so if you if you if you apply if you get the failure
at deviator stress 1.68 then what will be the cell pressure that is the question right
that is the second part of the problem okay so i have got sigma 1 is equal to 1.68 plus
x now i put that thing in equation 1 so i have got sigma 3 which is nothing but x kg
per centimeter square and sigma 1 which is nothing but 1.68 plus x kg per centimeter
square okay so i will put that thing in the equation 1
so 1.68 plus x is equal to 2.117 x plus 0.757 so from this i will get x which is nothing
but your required sigma 3 which is nothing but 0.83 kg per centimeter square okay so
i hope that you have understood the problem very very straightforward problem right so
you have you have got the shear strength parameters and you you are finding out that how much
pressure you need to apply to get the failure now we will go for the next problem
so next problem says the following are the results of a set of drained triaxial tests
performed on 3 identical specimens of 38 millimeter diameter and 76 millimeter height determine
the shear parameters of the soil so you have done some drained test on 3 identical specimens
of this size 38 millimeter diameter and 76 millimeter in height and you need to find
out the shear strength parameters so sample there are 3 samples sample 1 2 and
3 so you have got the cell pressure different cell pressure 50 100 and 150 kg per centimeter
square You have got the deviator load okay so basically you are calculating or you are
you are you are reporting the deviator load at failure is 0.0711 kilo newton and so on
change in volume so when you are applying because you will be you are doing the drained
test so you will be observing some volume change because consolidation is happening
right i mean uh you will be getting the water will
be draining out so you will be getting the volume change during the shear so you have
got uh 3 (diff) uh 2 different say stages right one is the in the first stage you do
the consolidation and second stage you do the shearing okay so you have got the change
in volume and you have got the axial deformation also so that is 5.1 millimeter for sample
1 7 millimeter for sample 2 and 9.1 millimeter for sample 3 now let us let us do this problem
so problem 24 the deviator loads at failure corresponding to each cell pressure
are given okay in order to determine the corresponding deviator stress these loads
are to be divided by the corrected area why corrected area because
when you are considering the shearing that means the volume change you are considering
so the i mean you you may get the uh decrease in the volume or increase in the volume right
whatever may be the situation you will be getting the volume change
so when the volume change is happening so you will be getting the area whether it is
bulging or contracting depending on that you will be getting the different areas at different
deviator stress so you need to correct the i mean you need to find out the actual area
on which the deviatoric load is acting and based on that you can find out the deviator
stress okay so by the corrected area of the sample which can be obtained from this equation
a c is equal to v 1 plus minus delta v divided by l 1 minus delta l okay so where v 1 is
the initial volume of the specimen delta v is the volume change now it could be additive
it could be subtractive depending on whether you are getting the uh increase in the volume
or decrease in the volume so already you have seen that if you if you consider the dense
sand you generally get the increase in the volume right at that time delta v will be
additive so if you get the decrease in the most of
the times you will be getting the decrease in the volume so then your delta v will be
subtractive well l 1 is the initial length of the specimen and delta l is the change
in length or the axial deformation okay so now your v 1 that is the initial volume of
the specimen is pi by 4 3.8 square now 3.8 centimeter is the diameter and 7.6 centimeter
is the length so cc which is coming as 86.19 cc so l 1 is 7.6 centimeter as given in the
problem so for the first sample your delta v is given
as minus 0.9 cc okay and your delta l is given as 5.1 millimeter okay that is given in the
problem you see the table you will see that therefore a c corrected area will be 86.19
minus 0.9 divided by 7.6 minus 0.51 okay which gives me 12.03 centimeter square which is
nothing but 12.03 into 10 to the power minus 4 meter square okay therefore your sigma d
that is deviator stress is now the deviator load by the corrected area so deviator load
is given as 0.0711 kilo newton by 12.03 into 10 to the power minus 4 meter square that
is the corrected area which gives me 59.10 kilo newton per meter square
so therefore sigma 1 is equal to sigma 3 plus sigma d which is 50 plus 50 is the cell pressure
that is nothing but sigma 3 for sample 1 you see that thing from the table plus 59.1 is
equal to 109.10 kilo newton per meter square so now we will complete the the calculations
for other 2 samples so if i make the table something like this where you have sample
then sigma 3 in kilo newton per meter square then deviator load in kilo newton then volume
change in cc then axial deformation in millimeter then corrected area in centimeter square then
sigma d that is the deviator stress in kilo newton per meter square and then sigma 1 kilo
newton per meter square okay so for sample 1 already we have calculated
we are just putting the values so i did not calculate the values for other
2 samples however you can calculate all those things by following the similar procedure
whatever we followed for sample 1 so sample 2 is 100 0.0859 minus 1.3 712.36 69.50 169.50
and for sample 3 cell pressure is 150 deviator load is 0.0956 volume change is minus 1.6
axial deformation 9.1 millimeter corrected area 12.65 centimeter square deviator stress
is 75.61 kilo newton per meter square and sigma 1 is equal to 225.61 okay so this is
the table we have completed so once we get this table now you can calculate the shear
strength parameters by graphically as well as analytically so i will i will i will give
you the hints for the analytical solution however we will see that how you can get this
graphically so tau this is sigma so basically you will
be getting 3 mohr circles so if you draw all the mohr circles okay this
is for sample 1 this is for sample 2 this is for sample 3 this is your sigma 1 for sample
1 this is for sigma 1 for sample 2 this is for sigma 1 for sample 3 and so on right this
is for sigma 3 for sample 1 sigma 3 for sample 2 sigma 3 for sample 3 okay so you will now
this is the these are 3 mohr circles for 3 different samples but they are identical so
they should give you the same shear strength parameters right
so you will be getting a common tangent okay which will be acting as the mohr coulomb failure
envelope so this is your common tangent this is the phi and this is the c value so graphically
if you solve that means to the scale if you plot so you will be getting c equal to 25
kilo newton per meter square and phi equal to 3.80 okay the same problem can be solved
analytically also analytically also means say you know sigma 1 and sigma 3 combination
right for all 3 samples so you will be able to i mean you know sigma
sigma 1 is equal to sigma 3 tan square alpha plus 2 c tan alpha that equation you know
so in that equation if you put okay sigma 1 and sigma 3 for sample 1 for sample 2 and
sample 3 you will be getting 3 simultaneous equations and if you solve these 3 equations
you will be getting c and phi value so already we have solved similar kind of problem earlier
so if you follow that thing without graphical plot you can get the value of c and phi so
i will stop here today thank you very much

## 3 thoughts on “Module 49 – Problem on shear strength of soil”

1. Muskan Sinha says:

very much bhut sara ty

2. Irfan Bhat says:

nice

3. Swarup Das says:

Dear Sir, problem no-24, Shear parameters value are coming different when mathametically solving. And that value is not matching with the graphical solution which you have been mentioned @ 27.47 during your lecture. Please reply