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Mod-01 Lec-06 Shallow Foundation : Bearing Capacity – I

Mod-01 Lec-06 Shallow Foundation : Bearing Capacity – I


Hello, today I will start the second lecture
of this module two that is on shallow foundation. Now, as I have already discussed that the
bearing capacity and the settlement are the two main design criteria of a shallow foundation.
So, that means the the soil the bearing capacity of the soil should be such that that it is
adequate to carry the load that is coming from the super structure as well as the settlement
should be within permissible limit. Now, today in this lecture, I will discuss about the
bearing capacity how to calculate the bearing capacity of a shallow foundation. Now, first before I go to this calculation
or the analysis part, I should discuss some terminology of this bearing capacity that
first one is the gross pressure. Suppose this is the foundation, and this is the ground
level and D f is the depth of foundation and B is the width of the foundation; and then
this gross pressure is the pressure that is coming at the gross pressure that is coming
at this base level of this foundation that means, the pressure that includes the load
that is coming from the super structure, the self-weight of this foundation and the weight
of the soil. Now, next one is the net pressure as the settlement
the soil will settle after because of this net pressure because this soil pressure is
already existing in this soil because then we have to remove this soil, then then we
place this soil. So that means, at the base of this foundation the pressure that is coming
from due to this soil is already there. So, the settlement will start or this after due
to this net pressure, that means, net pressure is the gross pressure minus the pressure that
is coming due to this soil. That means, the if q g is the gross pressure then the net
pressure will be q g minus depth of the foundation into the unit weight of this soil.
Let the ultimate bearing capacity of the soil, the ultimate bearing capacity of the soil
that is the the ultimate load that this soil can carry before it fails. So, that means,
this is the ultimate load carrying capacity of the soil. So, if we notice the q u for
which is the ultimate bearing capacity of the soil. Next, will be the net ultimate bearing
capacity of the soil. So, that means, the same again the net ultimate bearing capacity
would be the ultimate bearing capacity of the soil minus D f into gamma where gamma
is the unit weight of the soil then when we will get the ultimate bearing capacity and
the net ultimate bearing capacity of the soil then we have to go for the net safe bearing
capacity and or gross safe bearing capacity. For this net safe bearing capacity we have
to apply a factor of safety, here F is the factor of safety that F varies either F will
be either two point five or three. So, if we divide this net ultimate bearing
capacity by this factor of safety then we we can calculate the net safe bearing capacity
of the foundation or the soil. Now, the gross safe bearing capacity similar as the if the
net safe bearing capacity is n s then the gross safe bearing capacity will be n s plus
D f into gamma. Now, again this will be n u divided by F plus gamma u. Now, the next
one is the allowable bearing pressure. Now, as I have discussed that from the bearing
capacity consideration basically we will get the two pressure. One is the bearing capacity
consideration that the the net load that maximum load that soil can carry before it fails.
So, that means that term will get the net safe bearing capacity and we have to apply
the factor of safety there. So, we will get the get the net safe bearing capacity and
from the settlement consideration we will get another pressure. That means that pressure
is. So, that is within the permissible settlement. That means, if we apply that that pressure
within the soil then settlement of that foundation will be within that permissible limit.
So, that if there is a permissible limit for the settlement that is a, that is corresponding
that pressure is called the pressure for or the that is the maximum pressure the soil
can carry for that settlement consideration. So, there is basically two criteria, one is
settlement consideration another is bearing capacity consideration. So, that means, bearing
capacity consideration that net safe bearing capacity. That means, it is a net safe bearing
capacity of the soil and then another is the pressure that cause the maximum permissible
limit. So, minimum of these two will give us the allowable bearing pressure that is
q allowable net. So, one is from the net safe bearing capacity wherin bearing capacity consideration
another is the we mean the that settlement consideration that the minimum of these two
bearing and the settlement that will give us the allowable bearing capacity bearing
pressure. So, now. So, these are the different terms
that we will use for this analysis. Now, there are now we have to calculate the bearing capacity
of the foundation. The first expression that I will the analysis I will I will explain
that is given by the Terzaghi. So, that means, this is the Terzaghi’s Bearing Capacity
Theory. Now, Terzaghi proposed this bearing capacity
theory. So, it is, this is the this bearing capacity theory. So, there is the some assumptions
are there. So, before we start this analysis part or the bearing capacity analysis part,
we will go for this assumptions or the that means, the footing is loaded, footing is a
long strip or continuous footing resting on a homogeneous soil having shear parameters
c and phi. So, that means, the soil is homogeneous with shear parameter c and phi and it is a
strip footing or continuous footing. Now, as this is the strip footing or continuous
footing. That means, the analysis is 2D analysis. Now, soil fails in general shear failure.
So, last class I have discussed the different types of failure. One is general shear failure
then the local shear failure and the punching shear failure. Now, this particular analysis
of Terzaghi Bearing Capacity Theory is valid for the general shear failure for this one.
Now, the the load is vertical and concentric, the ground surface is horizontal, the base
of footing is laid at a shallow depth. Now, this is for the shallow foundation. So, that
means, this is the base of the footing laid at the shallow depth. Now, shearing resistance
of the soil between the surface and the depth of footing D f is neglected. So, the footing
is considered as a surface footing with uniform surcharge equal to depth of footing into unit
weight of the soil at a level of the base of footing. So, the the resistance coming
from the the soil that is within between the surface ground surface and the base of the
footing is basically neglected, in place of that the it is considered that one surcharge
that is equal to the pressure that is coming due to that soil part is acting at that base
level. So, now if we go for this Terzaghi Bearing
Capacity part now, if we draw the failure surface for this. Suppose, this is the ground
surface and this is the foundation width say. So, this is ground level now where width is
B, B is the width of the footing and this is the base level of the foundation. Now,
this depth of foundation is D f, D f is the depth of foundation, B is the width of foundation
and say we are talk calculating this pressure that is acting of the base of the foundation
that is this pressure is q u. That means ultimate bearing capacity of this soil; so q u that
is the ultimate load that this footing can carry. So, now as in the first assumption
this is the footing is the strip footing. So, that means, this is the 2D analysis.
Now, soil fail in general shear failure, now load is vertical. So, that load that is applying
this is the vertical and concentric that means it is centrally loaded. Now, suppose this
is the center line. Now, base of the footing laid is the shallow depth. So, D f is the
shallow depth and the shearing resistance. So, as I have mentioned the shearing resistance
of the soil between this zone that means, the ground this is also a ground level ground
level in the base of soil base of the footing is neglected. So, in place of that it is assumed
that this the load, that means, the total pressure that is acting in this point or the
base of the foundation that will be equal to q, that will be gamma D f into gamma where
D is the depth of foundation and gamma is the unit weight of the soil. So, that is the
pressure that is acting at the foundation base level.
If we draw the failure surface that is assumed for this analysis that is general shear failure.
So, this is the triangular portion, then there is a logarithmic spiral and then it is a straight
portion. Similarly, here this triangular portion, then logarithmic spiral portion and the straight
portion. So, this is the failure surface that is considered for this analysis. Now, basically
in this this is the symmetric. So, this part, so we can say it is a three zone. So, the
first zone or we can say this is three zone, this is first zone. Now, this is second zone
this is also second zone and this one is the third zone.
So, three zones are there. So, one is this triangular portion this is the first zone.
So, zone one this is called the triangular zone. Now, similarly zone two this is called
zone of radial shear. Now, the zone three is Rankine Passive Zone. So, there is a three
zone one is, first zone is triangular zone, this this triangular part. The second zone
is zone of radial shear from this zone and this zone and third zone is the Rankine Passive
Zone. So, now the angle which is considered for
this analysis, that means, this angle is 45 degree minus phi by 2 where phi is the friction
angle of the soil. Now, this is also 45 degree minus phi by 2. Similarly, this angle is also
45 degree minus phi by 2 and this angle is also 45 degree minus phi by 2. Now, this angle
is considered phi and this angle is also consider as phi. So, now here this portion this part
is logarithmic spiral part, it is considered. This is logarithmic spiral in the straight
portion. So, triangular part then the logarithmic spiral then the straight portion. So, this
straight portion is passive Rankine Passive Zone, this angle is 45 minus phi by 2 and
this angle is also 45 minus phi by 2. Similarly, these two angles are 45 minus phi by 2. In
the triangular part this angle is phi, this angle is also phi.
Now, if I draw this free body diagram of this triangular portion. So, if I draw this free
body diagram of the triangular portion. So, here we can say this is a, b and d. So, and
this is the center line and this is the width of the footing that is B. So, this triangular
portion we are drawing here and this load that is acting that will be q u ultimate bearing
capacity or load carrying capacity. So, I have this considered this angle is phi and
this angle is also phi. Now, here a d and b d act as a rough back of the rigid wall.
Suppose, it is considered here a d and b d, it acts as a rough back of rigid wall. So,
where the c and the phi this two parameter c and phi are the equivalent of the wall addition
c is equivalent to the this wall addition and phi is equivalent to the angle of wall
friction. So, now for this mean that suppose this is
the free body diagram and this is the perpendicular line, perpendicular to this a d and perpendicular
to b d. So, this dotted line. So, the perpendicular line perpendicular to a d and perpendicular
to b d. Now, here that is the P p, P p is the passive resistance that is coming from
this zone 2 and 3. So, that means, this portion resistance in this triangular portion is coming
from this zone two and these two other zones and so, this is the passive resistance that
is coming from this zone that is a P p, P p is the passive resistance. So, which is
acting at an angle phi with this vertical line?
Similarly, here also P p is acting at a angle phi and this C a. So, because as now this
phi is equivalent to this wall friction angle; so now, P p is acting as a angle of phi with
this perpendicular line of this wall. Similarly, here also P p is acting as a angle phi the
perpendicular line of this wall. Now, this C a is the addition that is acting
for a d and b d line. So, where C a is the addition it is acting this two line. Now,
we can say this C a that is equal to C into a d or C into b d where C is the addition.
So, similarly C will be the cohesion and P p is equal to the total passive resistance.
So, P p is the total passive resistance that is acting here. So, this is the passive resistance,
C is the cohesion. Now, in this figure if we calculate the weight of this triangular
portion, that means, weight of wedge this triangular portion a b d, if you want to calculate
the weight. So, that weight will be this is the half the area this means this is the B
and then this height. Now, this height is basically B by 2 into
tan phi. So, with this base of this triangle is B height is B by 2 into tan phi and then
the unit weight of the soil. So, this will be the weight. So, finally, we can write 1
by 4 gamma B square into tan phi. So, this is the weight of this triangular wedge. The
next we have to calculate the different components of this triangular wedge. So, if we take the
different components then we can write. So, if we take this figure again we can see.
So, there is a in the vertical direction if we take the different components. So, one
is P p that is acting in the upward direction, then q u it is acting in the downward direction,
then addition it has one component in the upward direction and one component in the
horizontal direction. Similarly, here also one component in the upward direction, and
one component in the horizontal direction. So, if we take these vertical components of
the all the forces then we can write let q u into B. So, q u is the total load that is
acting in the downward direction that will be equal to it is the load bearing capacity
of the soil, that means, two P p this passive resistance both are acting in the upward direction
plus C a this addition in the vertical component this C a into sin phi as it is acting in the
both the sides. So, this will be two into C a into sin, sin phi sin phi we multiply
with this a, we will get the vertical components; then this weight of this edge that is acting.
So, this is the passive resistance component, this is the ultimate load carrying capacity
of the bearing capacity of the soil and this is the addition components and this weight
of the wedge that is acting in the downward direction.
So, this this this will be minus 1 by 4 gamma B square into tan phi. So, these are the total
forces here this these two this one and this one was acting in the downward direction and
this P p and this C a components this is acting in the upward directions. So, now we can see
now here as we have mentioned that C a is equal to C into d a or C into b d. Now, from
this figure we can write this d a is equal to b d that will be equal to B by 2 divided
by cos phi. So, here in this figure we can write this d a as this height is d by two
into tan phi similarly this side a d or b d is B by 2 divided by cos phi. So, we can
replace this value here in the C a. So, we will get C a is equal to C into B by 2 divided
by cos phi. Now, if we put this C a value here in this equation. So, this will be q
u into B equal to 2 P p plus 2 into C into B divided by 2 divided by cos phi minus 1
by 4 gamma into B square
into tan phi. So, further simplify we can get this two P
p plus B into C there is another term here that is sin phi because this is two C a is
replaced by C into B by 2 cos phi and sin phi is there. So, this two two cancel. So,
this will be this B C into sin phi by cos phi this will be tan phi, so minus one fourth
gamma B square into tan phi. So, now just consider this P p, this total passive resistance
it has three components basically. So, this P p is the summation of P p gamma plus P p
C plus P p q. So, these resistance are coming from three parts one is for the weight of
the soil this gamma C is due to the cohesion and q is due to the surcharge. So, that surcharge
we have considered here for this part. So, this q is the surcharge that we have considered
here. So, that means, this q that is replaced the surcharge, that means, gamma D f D f into
gamma. So, now here P p will get this P p gamma that
will get. So, first component is P p gamma. So, this is produced by the weight of the
soil in which it is assumed the soil is cohesionless and negligible surcharge. So, this is the
contribution because of the weight of the soil this is produced by the weight of the
soil in shear zone and it is assumed that soil is cohesionless, that means, C is equal
to 0 and negligible surcharge. So, q is also equal to 0. So, that q and C both are 0 in
this condition the resistance that is produced by the weight of the soil is called as P p
gamma. Second one, this second resistance this is P p C which is the produced by the
soil cohesion, this is produced by the soil cohesion assume that this soil is weightless
and negligible surcharge. So, assuming that weightless soil and negligible surcharge.
Similarly, the third component that is P p q it is also produced by the surcharge assuming
that the soil is weightless and also cohesionless. So, here in this third condition soil is weightless
and cohesionless that contribution due to the surcharge is called P p gamma. So, now if we put this value in the final
expression, so our expression was q u B is equal to 2 P p plus B C into tan phi minus
1 by 4 gamma B square into tan phi. So, if P p is the summation of three parts. So, if
we replace this equation by this is two P p gamma plus P p C plus P p q plus B C into
tan phi minus 1 by 4 gamma B square into tan phi
Now, let we consider that two P p gamma minus one by four gamma B square tan phi is equal
to B into half gamma B N gamma. So, we replace this term two P p gamma minus 1 by 4 gamma
B square into tan phi by this expression B half gamma into B into N gamma and two P p
C plus B C into tan phi is replaced by B into C N c where C is the cohesion, arrange two
P p q is replaced by B into q into N q. Now, where this N c N q N gamma these are Terzaghi’s
Bearing Capacity factor. So, this is the Terzaghi’s bearing capacity factors. So, finally, if
we write this, replace this expression by these terms then the final expression will
be q u. So, if we put this, in these expressions. So, this will be q u is equal to C N c plus
q N q plus half gamma B N gamma. So, where C is the cohesion of the soil, N c is the
bearing capacity factor Terzaghi Bearing Capacity, q is we can write that q is equal to gamma
D f if D f into gamma. So, gamma is the unit weight of the soil and
this is half B is the width of the foundation. So, this is the three components. So, three
parts this is the contribution due to the cohesion, this contribution due to the surcharge
and this is the contribution due to the weight of the soil. So, ultimately this is the final
expression where we will get the ultimate load carrying capacity of the foundation.
So, this q u is the ultimate load carrying capacity of the foundation that is q u equal
to C N c plus q N q plus half gamma B N gamma. Now, where we can write this terms, this bearing
capacity terms that we can write that N c Terzaghi has given this bearing capacity factors
value this is cot phi into a square divided by 2 cos square pi by 4 plus phi by 2 minus
1, this is N c. Similarly, N q would be a square divided by 2 cos square pi by 4 plus
phi by 2 and N gamma you can write this is half tan phi K p gamma minus K p gamma divided
by cos square phi is total minus 1. So, where this a is equal to e to the power 3 pi by
4 minus phi by 2 into tan phi and K p gamma is equal to passive earth pressure coefficient.
So, this is passive earth pressure coefficient. Now, if
phi is equal to 0, then N C will be 5.7, N
q will be 1 and N gamma will be 0; so from this expression we can write q u will be N
C 5.7 into C plus N q part is 1. So, into q and the N gamma part is 0. So, this is the
ultimate load carrying capacity of the soil and then q net ultimate that will be q ultimate
minus q. So, this will be 5.7 into C or 5.7 into C u and then correlation of the soil.
So, now by using this expression we can determine what will be the this load varying capacity
of this soil or this foundation. Now, this is the basic expression. This expression now
where C q gamma B these we can this gamma is the unit weight of the soil.
So, B is the width of the if we know this. So, from this expression we can see this bearing
capacity of the soil, this depends on the soil property that is C and phi of the soil,
then this surcharge of the depth of the soil, depth of the foundation and width of the foundation,
that means, the geometry of the foundation then the soil properties also. So, now this
is the basic expression which is derived for the vertical and concentrating loading condition,
it is for homogeneous soil and so, this is simplified expression. So, now, in the next
lecture I will discuss about the different condition the general type. So, for incline
loading or for if if there is a moment is there then how we will determine this bearing
capacity. So, this is the simple expression that we
can use. So, here one thing that this N C N q N gamma expressions are given, because
this are the N C N q N gamma expressions. So, by using this expression we can calculate
the this bearing Terzaghi’s Bearing Capacity factors for different see different phi values.
Now, Terzaghi has also proposed or given this bearing capacity factor in a tabular form
for different phi values. So, that expression values are given here. So, here we can see this is the Terzaghi’s
bearing capacity factors this is the table for different phi’s starting from 0, 0 this
is what as I have mentioned for the 0 this is phi value this N C is 5.7 and Nq is 1 and
N gamma is 0. So, it starts from 0 to 50 degree; so corresponding N C value N q value N gamma
value; so these values are given. So, by using these charts we can determine. So, if we know
the geometry of this foundation and width of the foundation then the depth of the foundation
and we know those if we know the soil properties then by using this table we can determine
this bearing capacity factor then we can use this bearing capacity factor in this our general
expression and then we can calculate the ultimate load bearing capacity of this foundation.
Now, another thing is given that this expression that is derived is valid for general shear
failure. So, that are mentioned. Now, if soil fails in local shear failure then how to incorporate
those effects in these expressions. Now, that is also possible in these expressions. So, suppose this is for the for the previous
one was for the general shear failure and this is for local shear failure. Now, local
shear failure Terzaghi’s as recommended that the shear strength parameters, this parameter
which is C is converted to C m and phi is also converted to phi m. Now, how this? So,
this is this C is for cohesion for the general shear failure and C m is the cohesion for
local shear failure and phi is the cohesion friction angle for general shear failure and
phi m is the friction angle for local shear failure.
Now, how to convert this thing? Now, C m will be two third of C and tan phi m that will
be equal to two third of tan phi. So, phi m value, it will be tan inverse two third
of tan phi. So, now, in place of phi and C you have to use this phi m and C m when we
calculate the load bearing capacity of the soil in case of local shear failure. So, now,
we have to use this expression in case of local shear failure, that means, q u ultimate
this will be two third of C N c dash plus q N q dash plus half gamma B N gamma dash.
So, let this N c dash N q dash N gamma dash are the bearing capacity factors in case of
local shear failure whereas, whereas, N C N q N gamma is where the bearing capacity
factors for general shear failure. So, this N c dash N q dash N gamma dash are determined
using the same expression, but you have to replace phi by phi m and then we can determine
this N c dash N q dash N gamma dash by suing the same expressions.
Here also Terzaghi’s has given this N c dash N gamma dash N q dash values in a tabular
form. So, here, so this is the Terzaghi bearing capacity factors under this local shear failure
where this is the phi value from 0 to 50 degree. So, corresponding phi value what will be the
value of N c dash, N q dash and N gamma dash. So, these values are given here. So, if you
know the phi value and if the soil fails in local shear failure then we need to use N
c dash, N q dash, N gamma dash instead of N C N q and N gamma. So, here we will get
the different value of N c dash N q dash and N gamma dash. Now, generally it is observed
that if phi is greater than equal to 36 degrees then it indicates that this is general shear
failure and if phi less than equal to 29 degrees then
it is local shear failure. So, if soil fails in general shear failure
or if it is a dense soil or strip then we can use this previous expression for the general
shear failure and if soil fails in local shear failure, then this medium dense or moderate
sign then we have to use this expression we have to convert this C to C m and phi to phi
m then we have to use this expression. So, this final expression is for the local shear
failure. Now, the next thing is that these expressions
are derived for strip footing. So, these expressions are derived for the strip footing. Now, if
soil this footing is circular and square footing then also we can use this expression, but
with some modifications. So, Terzaghi has recommended that how we will use, because
these expressions general expression is valid for the strip footing. So, how we will use
this expression for the circular footing or square footing? Now, if the so that the general
expression for any type of footing. So, we can write this is alpha 1 C N c plus q N q
plus alpha 2 gamma B N gamma. Now, as I have already mention that for the strip footing
this alpha 1 is equal to 1 and alpha 2 is equal to 0.5 fine now for the circular footing
this alpha 1 is equal to 1.3 and alpha 2 is equal to 0.3.
So, alpha 1 is 1.3 and alpha 2 is 0.3 now. So, for the square footing alpha 1 is also
1.3 and alpha 2 is equal to 0.4. So, for the square footing the expression will be 1.3
C N c plus q N q plus 0.4 gamma B N gamma. So, N q N c N gamma values that we will get
from this table if we know the phi value. So, here also for different other type of
footing the circular square we can use this expression also. Now, one thing that in these
expressions that we have derived, I have derived so in this expression we have not considered
the effect of water table. So, that effect we have to consider here. Now, here see if we take the effect of water
table suppose this is the footing with width is B and depth is D f. Now, these are the
position of water table suppose this is the. So, water table can be above the base of footing
or below the base of footing. So, there now we have taken the different we will consider
the different cases where we put this water table. So, that can be above the base of footing.
So, this is the position of the water table. Well, that can be below the base of footing.
So, this is the base of the footing that can be below the base of the footing.
So, this is the another position of the ground level. So, here this ground level we can look
we can look at this ground level at distance of D w from the ground level and if it is
below the base of the footing then we can look at this ground level position by D w
dash is measured from the base of the footing. Now, we will consider the different cases.
Now, for the first cases or first case now here we consider that above this soil level. So, suppose if we take this figure. So, this
is the ground level, this is the base of the footing, where B is the width, and D f is
the depth of the foundation. Now, suppose this is the position of the ground level here,
this is D w; and this is the another position of the ground level, ground water level; so
this is D w dash. Now, first case that if case one, the first care if D w dash greater
than equal to B. So, first case is that means, this position of this ground water table below
the base is equal to or greater than width of the foundation.
So, now in this general expression, we can write our expression is q u C N c plus q N
q plus half B gamma N gamma. In this N q term, this is D f is gamma. So, we can see that
here this D f into gamma is there. So, these gamma is basically the unit weight of the
soil above the foundation base; and this gamma, because it is two gammas in two parts. So,
this gamma is the unit weight of the soil below the foundation base. So, this is the
second first term, second term and third term. So, first term is unit weight that means,
gamma less and in the second term, there is one gamma that gamma is basically above the
foundation base and the next one is the below the foundation base, second one. In the case
one if D w dash is greater than equal to B, then gamma will be gamma t or gamma bulk.
So, gamma will be gamma bulk whether this natural unit weight condition. So, both the
case, that means, the water table will not effect these bearing capacity expressions.
So, we will is the natural natural gamma is natural unit weight of the soil is gamma t
or gamma bulk. So, in the both the cases where this expression also we have to use the gamma
bulk and this expression also we have to use the gamma bulk. Now, if case two they will
consider this D w dash is equal to 0. So, if D w dash is equal to 0 this case 2. That
means, the water ground water table level is at the base of the footing then we have
to consider the gamma sub that is gamma sat minus gamma w. So, this is gamma submerge
gamma sat. So, this gamma we have to use for second term for half B gamma N gamma portion
and gamma t or gamma bulk for q N q part. So, here if first case D w dash greater than
equal to B then we will have to use gamma t for both the cases this part and this part.
Now, if D w dash is equal to 0 then we have to consider gamma sub for this this part this
part we have to use the gamma sub and this part we have to use gamma bulk. Now, for the
case three that D w dash is less than equal to B or less than equal to B or you can say
this is greater than B then we have to use gamma is equal to gamma sub plus D w dash
divided by B into gamma t minus gamma sub for half B gamma N gamma and gamma t for q
N q. So, if D w dash is less than equal to B then you have to use this gamma value in
this part and gamma t for this part. Now, case four that if water table is at G L. So,
if water table is at G L then both the parts we have to use the gamma sub, here also we
have to use the gamma sub here also we have to use the gamma sub; so here also for both
the parts. Now, for the case five that D w is greater
than equal to 0 and less than equal to D f then gamma will be gamma sub plus D w by D
f into gamma t minus gamma sub for q N q and for the second part, for this part we have
to use the gamma sub and gamma sub for half B gamma N gamma. So, if the water table is
greater than 0 and less than D f then we have to use this gamma for this this part and gamma
sub for this part. So, these are the five cases, where we can improve the water table
effect putting this water table position at different location and we can incorporate
the water table effect with the different unit weight that is used in the expression.
In the next lecture, I will explain the other effect that is the inclination loading inclination
effect, then the then the shape and factor of shape of the foundation effect, depth of
the foundation effect in this expression through general time. Those things I will explain
in the next lecture, Thank you.

Comments (36)

  1. please be a bit clear about concepts like radial shear zone and how it forms ,…….its NOT WHAT I expect from iits ,…….its like refering punmia text book, nothing was made general ,…………

  2. Terrible teaching

  3. wat a rapeing english….. still u r in IIT WOOOO sir…. i dnt belive … plz dnt make kachara of the english
    ……

  4. rather than that… cncpt was clear

  5. It's pretty good lecture just put on subtitles on and it's a lot easier to understand.

  6. shear jones….brother of indiana jones

  7. This is absolutely pathetic idiotic and dumb teaching I have ever experienced seriously no offence but You should not be in IIT.

  8. its a very good video, my all concept get clear

  9. tommaro is my geotech exam and I feel tired during study… so I came here and feel good

  10. Why is angle of wedge Phi? What is the explanation?

  11. 37:27 for friction angle = 0. Nq = 1 but why isnt Nc = 0 then? you have 5.7 from nowhere ?

  12. Even the terminology at the beginning was explained terribly

  13. What is the reason behind the "let us consider part" at 29:22?

  14. Super sir….. Crystal clear

  15. In the effect of Water table case Dw'<= B and Dw'>=B formula as per you is confusing …
    What if Dw' = B, which formula do we use???
    Please clarify my doubt sir.

  16. this is how the iitians teach….RIP

  17. Check out the online bearing capacity calculator at: http://civilengineeringbible.com/calculators/bearingcapacity.php
    Its the best one you can find. Fast, easy, reliable for both metric and british

  18. Very good lecture. Very well explained. Thanks

  19. simplistic video..thnx sir

  20. wtf hocam what are u talkin aboutt!!?

  21. sir explain how Nc=5.7 at phai=0

  22. I don't want to know how to drive equations they are well explained in textbooks…..please explain how and why these 3 zones are formed in these manners……and if you fails to explain please don't upload these kind of videos which waste our time…..i want concept…..

  23. Thanks team of NPTEL it's very useful for all civilians

  24. Subtitles create disturbance pls remove this

  25. You might be a genius but not a good teacher.

  26. Sir the text above this topic is disturbing please avoid it because topic is hidden by this text.

  27. C*ad kyu hua 🤔🤔

  28. hi ! sir, may god bleess you.

  29. please explain us the considerations given at 29:22

  30. but why u considered to put m instead of n, does n doesnt exist in india

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