# 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.

Thanks Prof.