Triaxial Test: Shear Strength Video Lecture | Soil Mechanics - Civil Engineering (CE)

last time we saw the direct shear test
for the determination of shear strength
parameters C and Phi but this test has
many limitations as he saw and some of
those are like we cannot control the
drainage and we cannot find a pore water
pressure and it is only suitable for
sandy soils or coarse grained soils so
because of these limitations we prefer
the triack shelters it overcomes all
these limitations it is the most widely
used test to find out the shear strength
parameters and study the soil in
different conditions so this is suitable
for all kind of soils satis clay soils
or coarse grain soils and drainage
conditions can be controlled it means
that it can be performed for it can be
performed in drained condition as well
as in undrained condition and we cannot
measure the pore water pressure also and
butwe're soils can be tested under
undrained condition and clay soil can be
tested under drained condition so as we
saw if the loading is rapid then
impervious oil also undrained conditions
are possible
so those condition can also be simulated
and similarly the drain condition in
clay soils that can also be simulated
and as I said pore water pressure can be
measured so if we have measured for
water properly so we can find out the
effective stress and volume change can
be measured so if volume change is
measured and using that we can find out
the corrected area that will be the area
at the failure so if we know the area at
failure so we can find out the better
value of all we can say it's in the
correct value of stress so this is one
more benefit and principle planes do not
rotate so in case of direct shear test
when we move the shear box so what
happens the principle planes rotate
during this movement
and so we do not know that principal
place or all the principal stresses
pertain tri-axial test we know the
principal planes also as well as we know
the principal stresses so we can plot
the Mohr circle and we can obtain the C
and Phi parameters for any condition
that is trained or untrained or for any
soil so interaction test what we do we
take a cylindrical expressed specimen
and then we cover it in a rubber
membrane and it is capped on some porous
tour also so that if we want to if you
want the pore water to move out so we
can allow the pore water to move out or
if we want to pour or if you want this
for water to be inside the soil then we
can stop the drainage so we can stop the
pore water to go out so these all these
conditions can be controlled so what
happens in this case and after putting
it in the membrane we put up we put it
in the Tri glial cell and there we apply
the confining pressure losing water so
water will apply equal pressure on all
the sides and this pressure is called as
a confining pressure that is Sigma C or
sigma3 and at this point at this point a
PC so the stress at the top or stress at
this vertical surface it is equal that
is equal to Sigma C and if you want to
draw the Mohr circle so Mars our curve
will just be a point because both the
stresses are that is here also it is
Sigma C or here also it is Sigma C so
both the stresses are equal so
motorcycle will just be a point but
after this stage what we do we apply an
additional stress that is that is called
as the deviator stress from the top it
is denoted with Sigma so because of this
trash difference
so our major principle stress now will
become Sigma a plus Sigma C and minor
principle stress will be equal to sigma3
so because of the stress difference in
all the planes shear stresses will be
induced except these vertical and
horizontal plane so this vertical and
horizontal planes are our principal
planes and this top that is horizontal
plane is our major principle plane and
this for diggle plane is our minor plane
that is minor principle plane so after
conducting this test on many samples we
can plot the Mohr circles on depending
upon various stresses and then we can
plot this tangent that will give give us
the C and Phi values what else is
measured in this test will you measure
the axial strain also axial strain mr
change in length divided by the original
length so as we apply load so this
sample will compress and so we put some
proving ring here so that we can measure
the this axial strain and so this
deviator stress for the deviator stress
it is equal to axial load divided by the
corrected area so this stress this
additional load that we are applying so
because of that the stress which is
coming that we are calling as deviator
stress and this is equal to HL dot
divided by corrected area now how do how
do we find out this corrected area let's
say that corrected area is AC so now
call this for the cylindrical sample we
can write that initially Nia a not
initial height H naught is equal to
initial volume V naught and after
completing the test the final area that
is corrected area time
the change in that is changed lengths
will be equal to H not content eyes and
the change volume will be equal to V
naught plus Delta V and also that
because we apply the load so this length
will be decreasing or the steps will be
decreasing so this area strain is
negative and let's denote it with
epsilon a so here you see we can write
it as V naught the Delta V divided by H
naught plus Delta H and from here
VC is equal to UB not a national if you
take common so we Delta we have all V
naught divided by H naught times 1 plus
1 plus Delta H upon H naught V naught
upon H naught is equal to a naught and I
wrote here V naught upon eyes not will
give me the a naught so AC is equal to a
naught and 1 this Delta V upon V naught
is the volumetric strain and if the
volume is decreasing then this will be
negative so here I can write 1 minus
epsilon V divided by and this is also
negative if the depth is decreasing so
it is me
so with this formula with this formula
we can find out the corrected area and
if the soil sample is undrained so in
undrained condition what happens the
pore water pressure cannot go out and we
know that if pore water pressure cannot
go out so the void ratio cannot decrease
and why they show cannot decrease or
let's say it cannot increase also no
change in void ratio as possible so that
volume change is also not possible
so in undrained condition this
volumetric strain is 0 so this corrected
area is equal to initial area divided by
1
a special strain so this is our traction
test and different kind of trial tests
are there that is consolidated trained
consolidated-undrained or al
consolidated under these type of tests
we will study later in in a good detail
now here like we will see one more test
that they that is uncomfort can confine
compression test it is used for clay
soils that is saturated and in undrained
condition so the load loading and upside
coast lace or drainage is not permitted
and the very specific thing in this test
is this confining pressure is what we
apply using water the Sigma C is equal
to 0 so the Sigma C is also equal to the
minor principle stress so this minor
principle stress is equal to 0 and only
we apply here we apply Sigma 1 let's say
this stress is Sigma 1 so Sigma 1 here
will be equal to applied load divided by
area at failure that is the final area
so as I said in undrained condition we
can write this a CS a naught upon 1
minus 1 minus epsilon a so solve that we
can find out the Sigma and the Sigma is
nothing but the unconfined compressive
strength it is denoted with qu now one
formula that we saw earlier that is
Sigma 1 is equal to sigma3 tan square 45
plus Phi by 2 plus 2c 10 45 plus Phi by
2 now as I said this test we are
conducting for clay soil so for that Phi
is equal to 0 and in undrained condition
we denote it with by you and one more
thing that if Sigma 3 that is minus
principal stress is equal to 0
so seeing these things in this vacation
here we see that Sigma 1 is equal to 2c
because this term will become 1 and this
term will become 0 C is equal to Sigma 1
upon 2 this C the undrained cohesion so
we did notice that c CU here we can
write it as qu upon qu is the unconfined
compressed compressive strength and Cu
is undrained cohesion so this way we can
find out the undrained cohesion of the
soil the same thing we can obtain from
the Mohr circle also so here we are
putting sigma3 is equal to 0 so if Sigma
3 is equal to 0 it means that this minor
principle stress is equal to 0 this here
it will be Sigma 3 that is equal to 0 it
is just Sigma 1 so here we can see that
this envelope for this will be
horizontal and failure plane will be at
45 degree so here we can find out ah
this is the value of Cu and Phi is equal
to Phi U is equal to 0 so from all
circle also same thing can be obtained
like this so if we want to find out the
sensitivity of the clay this test is you
to find out the sensitivity also so that
is the unconfined compressive strength
of undisturbed sample divided by and
confined compressive strength of
remoulded sample so this way we can find
out the sensitivity of the clear