# Lecture – 52 Soil Mechanics

Welcome to lecture three under earth pressure
theories. In the previous lecture we understood active
and passive earth pressures and discussed about Rankine’s earth pressure theory and
then we also discussed about different assumptions which are involved. In this lecture we will try to extend what
we learnt with a special cases for computing earth pressures and with some example problems
and extension of Rankine’s theory for a special cases like wall with a inclined face
and backfill soil which is not horizontal which is having a certain inclination with
a horizontal. So in the previous lecture we saw the development
of these slip lines within the zone of plastic equilibrium through a demonstration experiment
which is carried out at IIT Bombay which is with a retaining wall which is nothing but
a cantilever wall having sand as a backfill. So you can see the sand in dry state having
a relative density about 55% which is constructed by polivating that is air polivation of the
sand, like the way the deposit happens in the field and this is the passive soil because
the wall movements are not adequate so the passive state of plastic equilibrium was not
observed here but you can see here the active state with a movement or rotation about at
the top portion, we can see that different series of ruptures planes which actually shows
the evidence of what actually we are discussing in these lectures. Now we try to derive a universal equation
for determining this earth pressures at different depths and this can be extended to say for
a homogeneous soils having cohesion less soils or cohesive soils or soils with combination
of c and phi in different states are the drained or un drained conditions. So in this slide we are going to discuss about
the relationship between sigma1 and sigma3. Sigma1 is major principle stress; sigma3 is
minor principle stress and where in soil reaches a state of plastic equilibrium can be derived
from the more circle below. So a tow sigma plot is shown here and phi
is the friction angle, sigma3 is the minor principle stress. Sigma1 is major principle stress and if this
intercept is indicated by say cohesion and phi is the friction angle and we can write
from the geometry of the circle, we can write that sin phi is equal to this particular diagram,
we can write sin phi is equal to half sigma1 minus sigma3 divided by half sigma1 plus sigma3
plus 2c cot phi. This deliberation you would have already discussed
by learning about shear strength. So by re arranging this term shear, this we
can write as sigma3 into 1 plus sin phi is equal to sigma1 into 1- sin phi minus 2c cos
phi. So this further simplification in rearranging
the terms we can write sigma3 is equal to sigma1 into 1 minus sin phi by 1 plus sin
phi minus 2c cos phi by 1 plus sin phi. So this now changes to 1 minus sin phi by
1 plus sin phi this term we can recognize now already which is nothing but ka. So sigma3 is equal to sigma1 into 1 minus
sin phi by 1 plus sin phi minus 2c cos phi by 1 plus sin phi, this we can write sigma3
is equal to sigma1 into 1 minus sin phi by 1 plus sin phi into minus 2c into 1 minus
sin phi by 1 plus sin phi. We have written this by using the trigonometry
principle. So now substituting for say in active state
of plastic equilibrium say this equation is to be valid then we can say that sigma3 being
the minor principle stress in active state of plastic equilibrium, the active earth pressure
be minor sigma3 becomes sigmaa, sigma1 becomes sigmav which is actually nothing but a major
in case of a active state of plastic equilibrium. So with these what we get is that sigma a
is equal to gamma z ka minus two c root ka. So this is an equation for active case with
sigma3 is equal to sigmaa and sigma1 is equal to sigmav for an active state of plastic equilibrium
condition. So this equation now helps us say for example
a sandy soil like the physical modeling experiment were we have seen in a dry state when the
cohesion is equal to zero then in that case, the earth pressure equation is nothing but
sigmaa equal is equal to ka gamma z. At z is equal to zero, earth pressure zero
at z is equal to H, the earth pressure is ka gamma H then the total active earth pressure
thrust is half ka gamma H square acting at a distance H by 3 from the base. Now in this case suppose if you are having
a soil then if you substitute at z is equal to soil which is predominantly say cohesive
soil then say at phi is equal to zero, suppose if you are having a soil which is pre dominantly
cohesive soil then you can see that sigmaa is equal to minus 2c that is there is a possibility
that the soil actually is under tension behind a retaining wall. So when the soil attains its ultimate tensoil
stress, it cracks and it actually develops tensile cracks. That is the reason why like some of the retaining
walls having say cohesive backfills, the top portions are subjected to tension cracks. So this particular equation now if you wanted
to generalize for cohesive soils or soils having both cohesion and friction angle. Then we can write sigmaz is equal to gamma
z ka minus 2c root ka and by using the same concepts like, with that when you say for
passive state also when we wanted to derive the relationship between sigmap and sigmav
and sigma cohesion and kp with that we can arrive the condition for equation for passive
case also. In this case the only difference is that the
sigma3 becomes the major principle stress and sigma1 becomes the minor principle stress
which is nothing but sigmav in passive state of plastic equilibrium. So with this now the generalized equations
without any surcharge on the backfill surface if you consider then these are the generalized
equations for computing the earth pressure. Then sigmaa is equal to gamma z ka minus 2c
root ka this is the active state and sigmap is equal to gamma z kp plus 2c root kp that
is for the passive state. So follow the same steps as cohesion less
soils, only difference is that c dash is not equal to zero. So if you look into it sigmaa is equal to
gamma z ka minus 2c root ka and sigmap is equal to gamma z kp plus 2c root kp. So this is for the passive case and this is
for the active case. Now Rankine’s lateral pressure diagram for
a cohesive soil. If you look in to it here, what we saw is
active case with minus 2c root ka is the earth pressure at top because ka gamma z minus 2c
root ka z is equal to zero the pressure is minus 2c root ka here. In case of passive case we have 2c root kp
plus kp sigmav. So this is that equation what we discussed
just now and this total length distance is ka sigmav dash. So if you analyze further, now we have a lateral
pressure diagram for a cohesive soil which we just now have seen. Once again we look into it, if you look into
it here at certain depth z0 the negative pressure changes to positive pressure. So it indicates that for soil there is a certain
depth it actually remains un supported, particularly cohesive soils. So that particular depth is called the critical
height so beyond which the soil cannot remain un supported particularly cohesive soils. That is because there will be a negative pressure
and positive pressure which actually gets cancelled and then the soil will be in equilibrium
and then we can say that it has got safe unsupported height. So this is generally about, for a certain
type of soils having cohesive strength or some apparent cohesion up to 4 to 5 meters. So this is with the passive case, the equation
which is now shown in a diagrammatic way. Now just now we discussed about the cohesive
soil and critical height. So if you look into it, this is the equation
what we derived sigmaa is equal to gamma z ka minus 2c root ka. So z is the depth at z equal to zero here,
top of the wall and where the backfill surface is horizontal, H is the height of the wall
and wall is smooth that is the Rankine’s wall. Now at z is equal to zero, sigmaa is equal
to minus 2c root ka so this being zero it gets cancelled, sigmaa is equal to minus 2c
root ka and z is equal to H, it is ka gamma H minus 2c root ka. So which is nothing but earth pressure at
the base. At z is equal to z0 that is sigmaa is equal
to zero that is this point here then that z0 is nothing but the depth that is 2c gamma
root over ka. So this particular depth is also referred
as depth of the tension crack, so moment the soil cracks then after attaining its ultimate
tensile strength. Even with this expression we can even determine
the possible depth of a tension crack in a particular site. So z0 is equal to 2c by gamma root ka with
a negative pressure which is shown here and positive pressure showing, so 2z0 which is
nothing but the safe that is the critical height of a slope that can actually remain
un supported because of the negative lateral pressure and positive lateral pressure. So with that we can say that z is zc is equal
to 2 z0 is equal to 4c by gamma root ka which is nothing but 4c by gamma tan into 45 plus
phi by 2. Suppose if the soil is say having un drained
cohesion, say friction angle is equal to zero. In such situations the critical height of
a particular vertical cut can be determined as Hc is equal to zc is equal to 4cu by gamma
where cu is equal to un drained cohesion and gamma is the unit weight of the soil. So in case, soil is having both c and phi
then the critical height can be determined by using expression 4c by gamma tan 45 plus
phi by 2. So in this particular slide what we saw, the
possible earth pressure distribution what we get when we have a cohesive soil, purely
cohesive soil with cohesion pre dominantly and then certain friction angle then we can
say that by using this equation we try to derive and we try to derive the depth of the
tension crack and that we said as a 2c by gamma root ka and critical height that is
the height to which actually remains unsupported that is 4c by gamma root ka which is actually
given by expression 4c by gamma into tan 45 plus phi by 2. Now Rankine’s lateral pressure diagram with
uniform surcharge, if you look into it here suppose if you are having a soil say having
unit weight gamma and no water table and then if surcharge is q. So this is possible when you are having a
pavement behind a retaining a wall or any other utility which is behind a retaining
wall then that exerts a unit surcharge on the wall. How this surcharge can be created surface
stress in the…. This surface stress can be transferred as
an earth pressure to the soil that actually we are now trying to discuss in this slide
through a form of a special case where H is the height of the wall which is again a Rankine
wall. At top of the wall the earth pressure is zero,
at bottom it is ka gamma H. So now when we have surcharge at the top then
sigmaa is equal to gamma z ka minus 2c root ka and kaq has to be utilized, so where ka
is the quotient of earth pressure in the active case which is actually has to be multiplied
by q. So with that what we add is that, see if you
can see a parallel line to this surface. So this is nothing but entire q which is actually
transferred with ka. So that is actually given here as kaq. So at z is equal to zero the earth pressure
diagram in case with surcharge then this is with surcharge, it becomes kaq and then at
the top of the wall and at the bottom that is at depth z is equal to H, it is ka gamma
H plus kaq. Say sometimes we also have completely submerged
soil behind the retaining wall or partially submerged soil. So in such situation how the earth pressure
will change? For example if you have got a wall which is
having a sandy soil or a granular soil and say water table is there on both the sides
and the liver is same. Then the lateral pressure due to water will
get cancelled but if there is say due to some tidal fluctuations, if there is water which
actually can cause a differential water pressure which is actually required to be considered
in waterfront structures, basically in the bending structures. So Rankine’s lateral pressure diagram with
partially submerged soil can be discussed. Consider again a cross section of a retaining
wall having height H and in this case what we see is a water table location at a depth
zw from the ground surface. Here the ground surface which is horizontal
and H dash is equal to H minus Hw, H is the total height of the wall. By using the equation that is sigmaa is equal
to, all these examples what we are discussing is for active state the similar concepts can
be used for passive case also. But the Rankine’s earth pressure diagram
for partially submerged soil for active state. So here will be using a quotient of active
earth pressure. So here at z is equal to zero, earth pressure
is zero and at z is equal to Hw at this point then you can see that the earth pressure ordinate
here is ka gamma Hw. So ka gamma Hw which actually gets transferred
here. Now below this the soil is completely saturated
and having say gamma is equal to gamma dash, here gamma which is equal to gamma sat if
you can say then ka gamma dash H is the lateral earth pressure here and gammaw w H dash is
nothing but a water pressure which is actually exerted by a water of height H minus Hw which
is nothing but H dash. So here the earth pressure is now ka gamma
H plus gammaw H dash. Suppose this water table is actually up to
the top surface then what will happen is that then the water pressure which is the earth
pressure diagram which looks like this thing ka gamma dash H which is actually at the till
bottom and then having a water pressure which is gammaw H. So presence of the water behind a retaining
wall actually causes lateral pressure that is the reason why for designing or constructing
retaining walls, we generally use the drainable fills. That is in case if the soil is having inadequate
say drainage facility, the drainage facilities have to be provided so that the building up
of the water pressure behind the wall can be prevented because designing a wall for
the additional water pressure is for a conventional walls is difficult but in such situations
where there is a possibility of tidal fluctuations and then differential water pressure is required
to be considered. This situation can be prevented by providing
adequate drainage provisions so that the water pressure remains zero. Otherwise if it is required, if the water
pressure prevails then this how we have to consider in computing earth pressures. Now let us consider, extend the special case
because we all know that the soil is not an homogeneous but we can extend this Rankine’s
theory for computing the stratified even for earth pressures for the stratified soil deposits. Suppose if you are constructing a retaining
wall and assuming that Rankine’s conditions prevail then in that case we can extend this
theory for determining this earth pressure diagrams. So once we obtain this earth pressure diagram
then that can be extended for designing a retaining wall at a particular site. So here by using the same concept the equation
sigmaa is equal to gamma z ka minus 2c root ka plus kaq where here the surcharge is not
there. So if the surcharge is there for example highway
embankment where one of the sides is being retained by a wall and if it is constructed
with different soils and if it is subjected to a surcharge. Then that surcharge has to be considered but
in this case q is being zero so this particular term vanishes. So at z is equal to zero earth pressure is
zero, so this is the starting point of the earth pressure diagram. In all case actually we are assuming that
wall rotates about a tow and it actually yielded sufficiently to produce active state of plastic
equilibrium condition. So at z equal to H1 just above A-A so let
us assume that in this stratified deposit example, we have got say c is equal to zero
and phi1 dash and gamma1 and phi2 dash gamma2, phi3 dash gamma3. So three soils which are considered having
thickness as H1 and H2 and H3, H is equal to H1 plus H2 plus H3 and A-A is the interface
between these 2 soils that is phi1 and phi2 soils and B is interface between phi2 and
phi3 soils and phi1 is greater than phi2 is greater than phi3 and in this case if z is
equal to H1 above A-A. So we can see above A-A that means this particular
portion then actually it is governed by the soil properties which are in this particular
region that is phi1 dash and gamma1. So here sigmaa is equal to ka1 gamma1 H1 so
observe that coefficient of earth pressure which is actually ka1 which is actually obtain
from the friction angle phi1. So ka1 is equal to 1 minus sin phi1 by 1 plus
sin phi1 so with that this gamma1 is the unit weight of the soil in the H1 region. Now the earth pressure ordinate just below
A-A that means that we just got this particular point say now we wanted to get this point
just below this thing. Now just below A-A the soil is now same height
but the soil is just having properties which is phi2 soil and which yields ka2 is equal
to 1 minus sin phi2 by 1 plus sin phi2. So with that sigmaa is equal to ka2 gamma1
H1, this gamma1 H1 is now for one simple methodology which one should understand here is that for
computing earth pressures just consider the top soil as a surcharge which is lying. Suppose if I assume that my retaining wall
starts from here, now in this case you can observe at depth say at this level A-A you
can see how will you calculate. Suppose gamma1 H1 is the say surcharge which
is acting on this particular level then we calculate that sigmaa is equal to ka2 gamma
H1. So just consider that as a top soil you consider
that as a surcharge and then transfer that as a lateral pressure. So here at z equal to H1 above A-A we said
that ordinate is here and z equal to H1 just below A-A which is ka2 gamma1 H1. At depth z1 H1 plus H2 that is z equal to
H1 plus H2 that is at level B and just above BB. So in this case sigmaa is equal to ka to gamma1
H1 plus gamma2 H2. Now at depth z is equal to H1 plus H2 just
below BB that is this particular portion then now the third soil which is now phi3 soil
which is ka3 which is obtained by 1 minus sin phi3 by 1 plus sin phi3 which is actually
sigmaa is equal to ka3 into gamma1 H1 plus gamma2 H2. Now above cc if z is equal to H1 plus H2 plus
H3 is there then sigmaa is equal to ka3 into gamma1 H1 plus gamma2 H2 plus gamma3 H3. Suppose if the soil is also having cohesion
then we can use the same concept of example where using this equation sigmaa is equal
to ka gamma z minus 2c ka plus kaq which we can determine the earth pressures in a stratified
soil deposit also. So is the identical approach for the c phi
soil also can be adopted for with different stratified, c phi soil with a stratified soil
deposits we can use for computing earth pressures. So here what we need to consider is that the
here at just above AA and just below BB, you try to get this ordinates. So when you try to obtain the resultant pressure
and that is actually gives the resultant earth pressure. that means the resultant of all these, this
particular shape seesaw type this which is obtained saw type diagram and that is nothing
but earth pressure diagram in case of a stratified soils. This ordinate movement this side and that
side depends upon the properties of the soil with the phi1 is less than phi2 is more. So based on that it changes so with that we
can obtain earth pressure diagram and by obtaining the resultant we can actually calculate the
active thrust. In case if you are determining for the passive
case then we can determine the total passive thrust. So the variation of active and passive lateral
earth pressures with depth which are shown here. So in this case we are having a soil, vertical
wall which is actually soil having water up to the top surface and now as we discuss in
this case, this is the earth pressure diagram. The equation of this particular line is ka
gamma dash z, at z is equal to zero the lateral earth pressure is zero, at z equal to H the
lateral earth pressure is say z is equal to H0 in this diagram H is indicated by H0 say
k gamma dash H0. So pa is nothing but half ka gamma dash H0
square that is the lateral earth pressure plus this has got a additional component which
is nothing but if the water is there up to this level then that actually acts as a hydro
static pressure which is half gammaw for this case if the water is up to this level it acts
like a half gammaw H0 square. But if the water table is there on the both
sides of this thing case then what will happen is that say the wall is submerged, say this
part here and the left hand side of the wall there is a water and right hand side of the
wall there is water. Then this case that hydro static pressure
gets cancelled but the effect of water we need to consider by calculating the submerged
weight of the soil and then lateral pressure due to the submerged soil mass. Similarly if you look into here, if that condition
prevails here and the wall move towards the fill then this is the equation kp gamma dash
z and for the earth pressure diagram and kp gamma dash H0 is the ordinate at the base
then pp is equal to half kp gamma dash H0 square is the passive earth pressure thrust. So if you look into it kp is greater than
ka and with that what will happen is that the earth pressure thrust in passive case
is much more than the passive case. Incase if you are having a partially submerged
soil state then you have got this hydrostatic pressure or if the water table is there on
one side only then actually it exerts hydro static pressure that is considered and if
so wall is actually, the surface stress is there, this is how the surface stress is considered
laterally. So if Qs is the surface stress which is actually
acting then ka Qs into H0 which is actually here gets transferred and ka Qs H0 is the
resultant thrust due to surface stress acting at a certain point. If suppose if I have got a different complicated
lateral stresses then it will be considered in the subsequent design. So basically in this slide variation of the
active and passive lateral pressures which are shown and by knowing this, we can actually
obtain this equation of this particular line and then earth pressure diagrams. Once you obtain the earth pressure diagrams
then you can calculate the active thrust and then passive thrust and then by considering
the equilibrium conditions we can actually design the wall for equilibrium. Now let us consider some examples with whatever
we have discussed it just now. In the example one, let us consider for a
smooth and vertical retaining wall of height 8 meters. We need to calculate the total active lateral
force, so active lateral force that means the wall moves away from the soil and its
location from the base in the following cases. The soil is basically fine sand having friction
angle is going to 40 degrees, void ratio 0.85 and in specific gravity of the soil this is
2.65 and no ground water table anywhere and for the same soil with the same backfill but
with a ground water table raising up to the ground surface and for the same soil with
a same void ratio considering the capillarity. So we have the three different conditions
then we will see that how the influence of, suppose if the ground water table raises behind
the retaining wall what will happen to the active earth pressure. So that will actually tell us the influence
of the raising ground water table on the lateral earth pressure magnitude. So in this case the case a if you look into
it 8 meter wall, it’s a smooth wall Rankine’s wall and the earth pressure diagram is now
ka gamma h so using sigmaa is equal to gamma z ka minus 2c root ka plus kaq, kaq is equal
to zero so q that is because q being zero and where c is equal to zero with that ka
is equal to 1 minus sin phi dash by 1 plus sin phi dash with that we will get coefficient
of active earth pressure 0.2174 and for the unit weight of the soil which is obtained
from the given parameters 14.3 kilo newton per mere cube. So vertical stress is about 114.6 kilonewton
per meter square so at a depth of 8 meters the vertical stress is this much, here is
zero. So sigmaa is equal to at depth, z equal to
8 meters soil being uniform which is ka sigmav dash which is equal to 24.91 kilo newton per
meter square. So these retaining walls are generally considered
as per meter walls, the active earth pressure thrust units are kilo newton per meter. Suppose if the wall say of about 5 meters
length then total earth pressure thrust acting on the wall in active case can be calculated
for the given length. So but generally these are considered as a
per meter length walls being a continuous and then being a plain strain structures we
get finally the thrust in per meter length. So here in this case please note pa is equal
to 99.65 kilonewton per meter and its acting at a H by 3 from the base that is 2.67 meters. Similarly case b now the water surface is
at the ground that is at the ground surface here you can see at the top of the wall, so
this is ka gamma dash H now being because of the submerged unit weight which we actually
obtain by using this deliberation. Then we calculate water pressure now we can
see that tremendous amount of water pressure which actually acts on the wall. So as it actually rises from the bottom so
actually induces instability to the wall, if it is not designed if it is not drained
properly. So now the net pressure is now 382.05 that
is about 3 to 4 times there is a increase in the lateral active thrust on the wall with
an raising water table. So the effect of the ground water surface
on the active thrust can be noted here. So a case where suppose you have got a soil
and then which is actually prone for say capillarity saturation. Let us assume that we have got a complete,
the saturation is possible in this particular zone, entire capillarity fringes develop here. So this is 3 meters is the zone and 5 meters
is which is saturated with water, this is the ground water surface. Then in this case here the earth pressure
is, the total stress is here and then pore water pressure for this case is pore water
pressure which is minus 3 gammaw because gammaw is say 10 minus 30 kilo meter per square and
50 kilometer per square. So this is the raising water table above this
is because of the capillarity nature. So the effective stress is sigma dash is equal
to sigma minus u with that we will get this particular distribution for the effective
stress. So this is the stress distribution diagram
for the total stress and pore water and effective stress. So this is that vertical stress, now that
one if you convert into for calculating the lateral thrust that is now ka into 29.43 because
ka which is actually obtained from that for the soil and now this particular portion of
the water induces hydro static pressure on the wall so that is gammawH that H is equal
to 5 meters. So with that what will happen is that the
pa is now 279.81 kilonewton per meter that is this particular situation is that earth
pressure diagram changes and which is actually having ka into 29.43 at the top and ka into
151.36 ordinate at the bottom and the resultant of this trapezium actually is nothing but
the total active thrust which for example which is given here is simplified way, pa
is equal to 279.81 kilonewton per meter acting at a 2.47 meters from the base. So the location of the resultant with respect
to the bottom can be calculated by individual cg. For example the cg of this particular portion
or rectangle located here at H by 2 from the base and this particular portion is located
H by 3 in the bottom and then cg of these particular water pressure diagram, they combine
you will able to get the total active thrust and then acting at a location 2.47 meters
from the base. So in this example what we saw first is that
a case where the dry soil and then raising ground water table and a soil subjected to
capillary saturation. So these conditions actually affect the active
earth pressure and then can induce lateral pressure on the soil. So that is the reason why we generally use
drainable or granular soils as a backfill material. In case the situation is that you are prone
to use low quality fills then adequate measures have to be taken to drain the water behind
the wall. The second example, a retaining wall with
a smooth vertical back is 10 meters retains at 2 layer sand backfill with the following
properties. So assume that you have got zero to 5 meters,
that is c dash is equal to zero, phi dash is equal to 30, gamma is 18 kilo newton meter
per cube, below 5 meters that is a wall of total 10 meters. So below 5 meters the c dash is equal to zero,
phi dash equal to 34, say 2 soils are there now with gamma is equal to 20 kilo newton
meter per cube. So basically we need to show the active earth
pressure distribution assuming that the water table is well below the base of the wall and
here we have 2 types of soils and up to 5 meters and below 5 meters there is another
type of soil. So this particular case is illustrated here
so c dash is equal to zero, phi dash is equal to 30 dash, ka1 is obtained 0.33 from 30 degrees
friction angle 1 minus sin 30 by 1 plus sin 30. So 1 minus sin 34 by 1 plus sin 34 that is
for ka2 it is 0.283. Now by using sigmaa is equal to ka gamma z
minus 2c root ka plus kaq that is the universal equation for determining earth pressure distribution
for the active case. So active pressure distribution for the upper
layer at z is equal to zero that is at this point sigmav is equal to zero so lateral pressure
is equal to zero, at z is equal to 5 meters just above this point then we need to use
the soil properties that is ka1 0.33 into this 90 kilo newton meter per square which
actually is 30 kilonewton meter square which actually yields this ordinate. At z is equal to 5 meters but just below this
line which actually is governed by ka2 that is phi is equal to phi dash is equal to 34
degrees which ka2 is equal to 0.283 which yields now ka2 into sigmav that is same vertical
stress but with a different soil properties. With that what will happen is that this ordinate
that we computed now at z is equal to 5 meters just below this line which is 25.47 kilonewton
meter per square. Now at z is equal to 10 meters we have vertical
stress plus the 90 kilonewton meter per square plus 20 into 5. So this 20 is that unit weight now, please
note that unit weight is also changed here. So with that 190, now entire this is governed
by which is something like surcharge on this level where ka2 which is nothing but 0.283
into 190 which actually yields 53.77 that is ordinate. So the resultant of this actually gives the
total active thrust exerted on this particular type of 10 meter wall. So that is how we use whatever we have learnt
to determine this earth pressure distribution and then active earth pressure at thrust on
the wall can be determined. So third example where retaining wall 6 meter
high with a smooth vertical back say in this case we should be able to push against a soil
mass where this wall is actually move towards the soil and c dash is equal to 40 kilonewton
meter per square and phi dash is equal to 15 degrees and gamma is 19.40 kilonewton per
cube and what is the total Rankine pressure, if the horizontal soil surface carries a uniform
load of 50 kilo newton meter per square. There is a surcharge which is acting on the
top. What is the point of application of the resultant
thrust? This is the problem under cohesion. So what we can do is that surcharge is 50
kilonewton meter per square, wall is height of 6 meters Rankine case but wall moves towards
the backfill. So it indicates the passive case now, sigmap
is equal to kp gamma z plus kp q plus 2c root kp. So with that what will happen is that kp for
a given friction angle, we can obtain at z is equal to zero, sigmap is equal to 1.698
into 50 plus 2 in to 40 into root over 1.698 which actually yields 188.9 kilo newton meter
per square that is at z is equal to zero because this is a passive case wall moves towards
the backfill so 1.698 into 50. That is surcharge at this particular point
plus this particular portion is due to plus 2c root kp that is actually this portion here. At z is equal to 6 meters we get kpq plus
2c root kp plus kp gamma dash that is actually this particular portion when z is equal to
H which actually is a pressure here. So at sigmap is equal to 1.698 into 19 into
6 plus 84.9 plus 104 yields 382.54 kilo newton per square. So total passive earth pressure is this much
kilonewton per meter acting at 2.66 meters from the base. So that is how we actually use for active
case as well as passive case to determine the active thrust or passive thrust. Now this particular case of Rankine theory,
all we discuss can be extended for special cases like, when you have got a backfill but
vertical wall initially let us consider that wall surface is vertical but it is smooth
and in such situations we can actually extend this theory for calculating earth pressures. So in this particular slide what you are seeing
for an extension of the Rankine theory for a sloping soil surface with a smooth vertical
retaining wall. So this particular surface, the wall initially
is at this particular position and when the wall moves away from the backfill, when the
wall movement is this side then is an active case. So here that fill inclination with the horizontal
is beta that is actually shown here and which is having a soil which is shown in this element. So here in this passive case the elements
are now, one thing which is to be noted here is that the surface is not horizontal. In the previous case we have actually assumed
that typical ideal Rankine wall having a smooth interface and as well as backfill surface
is horizontal but in this case, you have got a smooth surface but the backfill surface
is having a certain inclination. So this is a case where passive case which
is shown, the wall portion is initially here and the movement of the wall is towards the
backfill then what will happen is that there is wall movement. This is the wall position which is shown after
pushing adequately inside to create a passive state of plastic equilibrium. Then at point of plastic state of equilibrium
a rhombic element is shown here. So in this case these stresses can no longer
be called as a principle stresses because these surfaces are not horizontal. So they carry some shear stresses so but these
particular stresses what you are showing this is sigmaz at a depth z, this z is measured
here as shown here and sigmaa is the active earth pressure and sigmaz at a depth z and
sigmaz is vertical stress here. So this is a typical rhombic element in case
of active case and this is the case in the passive case where you have got sigmap as
the passive earth pressure and sigmaz is the vertical stress. So as these stresses are not normal to their
respective planes, they are not principle planes, that has to be noted. Now assumptions involved in this case in extending
this theory to this particular case. This existence of the conjugate relationship
between the vertical pressures and lateral pressures on vertical planes within the soil
adjacent to a retaining wall. So existence of conjugate relationship between
the vertical pressures and lateral pressures on the vertical planes within the soil adjacent
to a retaining wall. That means these are the vertical planes which
are there. So what is being shown is that existence of
conjugate relationship between the vertical pressures and lateral pressures on vertical
planes within the soil adjacent to a retaining wall. The effect of wall friction and friction wall,
the soil is neglected. So with this assumption what it means is that
the effect of wall friction or that means or friction between the wall and the soil
is neglected. So with this assumption the Rankine action
theory is extended to calculate the earth pressures for a case. Now what we need to do is that to extend this
theory, sloping the soil surface with smooth vertical retaining wall. So consider this particular horizontal length
is actually having a unit length. Now weight of the column soil above the horizontal
width is gamma z that is at depth z as shown in that rhombic element. Area of the parallelogram is z into 1 that
is z is the depth and one is that unit width. Then volume of the parallelepiped is z is
equal to 1 into 1, the area of the face is 1 by cos beta into 1. So sigma on the face of the element parallel
to the slope surface is gamma z by 1 by cos beta into 1 which actually gets sigmaz that
is gamma z cos beta that is the sigmav on the face of the element parallel to the slope
surface. That is on that inclined which is parallel
to the slope surface means which is having an inclination beta with that sigmaa is equal
to now we can obtain ka gamma z cos beta. Suppose if cos beta is equal to zero then
it actually changes to sigmaa is equal to ka gamma z with some inclination beta what
we get is sigmaa is equal to ka gamma z. In case if it is for passive case then sigmap
is equal to kp gamma z cos beta. Now by further simplifications and deliberations
we can actually obtain quotient of active earth pressures for this case where the fill
surface is not horizontal, is inclined with having a beta with horizontal. For a smooth vertical retaining wall with
vertical face we can actually obtain ka is equal to cos beta minus root over cos square
beta minus cos square phi divided by cos beta plus root over cos square beta minus cos square
phi where if you set beta is equal to zero it actually changes to ka is equal to 1 minus
sin phi by 1 plus sin phi. This is the typical ideal retaining wall case
where smooth interface and backfill surface is horizontal. Similarly for passive quotient of earth pressure
can be obtained kp is equal to cos beta plus root over cos square beta minus cos square
phi by cos beta minus cos square beta minus cos square phi with that what will happen
is that kp quotient of passive earth pressure can be obtained. So when we substitute beta is equal to zero,
kp also changes out to one plus sin phi to one minus sin phi when beta is greater than
phi is not possible because the phi is suppose if it is more than its angle of repose then
there is which is not possible and beta is equal to phi, ka is equal to kp is equal to
one which is incompatible with a real soil behavior. So with the beta is equal to phi, ka is equal
to, it actually these coefficients change to ka is equal to kp is equal to 1 which is
in incompatible with a real soil behavior. Suppose now you have got a case where there
is a vertical wall and there is say vertical wall of height 8 meters and there is a slope
inclination that is of the fill behind the wall is say 20 degrees that is beta is equal
to 20 degrees and soil is having a unit weight gamma then you can determine the active thrust
like this. Suppose if you have got a wall of 8 meters
then if the beta is equal to say 20 degrees then now the condition is that wall is smooth
that is the case. So we can’t use one minus sin phi by one
plus sin phi but we need to use this particular quotient of active earth pressure say ka is
equal to for the sloping soil surface but still Rankine case, with that we can by substituting
for beta is equal to 20 degrees and say phi is equal to some slope inclination say 40
degrees with phi is equal to angle of internal friction say for example for 40 degrees then
we will be able to get ka is equal to cos beta minus root over cos squared beta minus
cos square phi by cos beta plus root over cos square beta minus cos square phi with
that we will be able to get the quotient of active earth pressure for this case. By substituting in sigmaa is equal to ka gamma
z cos beta will be able to get z is equal to H, we will get the earth pressure ordinate
here parallel to the slope surface of the fill then ka gamma H cos beta. The resultant thrust is actually determined
by pa that is which is indicated here. This is how one of the case which we can be
extended like the ideal Rankine wall case two, a case where sloping soil surface and
in this case what you need to do is that use appropriate expressions for quotients of active
earth pressure and passive earth pressure and determine those quotients of active earth
pressure and passive earth pressure correctly and then use the expression for sigma z is
equal to ka gamma z cos beta where beta is the sloping relation of a backfill surface. With that you will be able to get the earth
pressure ordinate or a equation for the earth pressure so from their we will be able to
calculate the active earth pressure thrust due to soil, having a inclined backfill surface. Suppose we have say retaining wall which is
smooth but inclined retaining wall. Suppose initially let us consider, suppose
you have got a wall face, all the time we are actually assuming vertical and smooth
and backfill surface is horizontal. Suppose you have got a wall say typical mesentery
wall say assume that it is smooth, in such situation the earth pressure is the resultant
of the weight of the soil in that wedge portion and the lateral thrust exerted by the soil. So in this case suppose if you are having
an inclined wall surface but a smooth interface but with a backfill surface inclined in this
case. How you need to determine is that, we need
to determine first the weight of this wedge which is w and then the resultant active thrust
along this face, now this face which actually is the frictionless interface and with that
what will happen is that you will be able to get pa. The resultant of w and pa which actually is
somewhere here that is nothing but the resultant active thrust which is needed to be considered
in designing a wall which is having a inclined slope face and having smooth interface behind
the wall. So in this lectures what we try to understand
is that how the Rankine’s theory can be used for computing earth pressures and we
also discussed the problems like a simple homogeneous soil deposit and a cases like
when we have got a partially submerged soil behind the retaining wall and what will be
the influence of raising ground water table on lateral earth pressure, especially for
the active case and then we try to also discuss extension of this problem for a stratified
soil deposits then what we need to do is to follow the equation sigmaa is equal to ka
gamma z minus 2c root ka plus kaq for an active case. In case of passive case sigmap is equal to
kp gamma z plus 2c root kp plus kpq in case of a passive case and follow the case like
when you got a interface say two soils are separated by say interface AA then consider
the soil properties for that particular depth above that interface and the below that interface
and try to get the net pressure diagram for the case. So when we wanted to calculate say for a cantilever
sheet pile wall then you need to calculate the net earth pressures from the active side
and passive side and those things have to be considered in the design as a net earth
pressure diagrams. Further we extended this theory for determining
say a case like having inclined wall face but smooth in nature but having horizontal
surface then in that case what you need to do is that you need to determine the weight
of the soil in that triangular portion, wedge portion and then active earth pressure due
to the horizontal surface, the convention one so that is the resultant. For example if you have got a wall which is
having a vertical face with a inclined surface then we actually discussed about the quotient
of active earth pressure and passive earth pressure with new expressions but when the
beta is equal to zero they again change over to ka is equal 1 minus sin phi by 1 plus sin
phi and kp is equal to 1 plus sin phi by 1 minus sin phi were passive case and we used
these things to calculate the active earth pressure thrust and passive earth pressure
thrust. In the next lecture we will be introducing
the Coulomb’s theory of earth pressure and then graphical methods for determining the
earth pressure and thereafter we will be discussing examples relevant to Coulomb’s and kalmonds
methods.

## 1 thought on “Lecture – 52 Soil Mechanics”

1. Rajwinder singh says:

Thanks Prof.