In last class, I have discussed about the

settlement calculation for granular soil by using the field test data, then I have discussed

how to calculate the settlement for the by using plate load test data, then for the SPT

test and for the CPT test or static confrontation test.

Now, this class I will explain a few other techniques to determine the settlement for

the granular soil, and then I will solve couple of problems to show how to use those; how

to calculate the settlement by using various field data. Now, first that I have discussed about the

technique for the using the CPT. So, there is another technique by using the SCPT data,

so by using the SCPT data. So, this technique is proposed by Schmertmann and Hartman 1978,

well according to their procedure that your settlement that we will get, the settlement

calculation that is c 1 c 2 this is q bar by q minus q into summation of I z by E into

del z. So, this is the expression, where c 1 is given

by this expression 1 minus 0.5 by q minus q bar minus q, and c 2 is another expression

that is given 1 plus 0.2 log 10 t divided by 0.1.

Now, here this q is the effective overburden pressure, where q is overburden pressure at

the foundation base level or basically we can say that q is basically gamma into D f.

Now, q bar is the pressure or the coming as a load; that means the footing pressure, which

is coming from the super structure. Now, here t is the time in year. So, suppose

if we want to determine the settlement at one year, then we will put t equal to 1 if

we want to put at settlement at 5 years, then t will be 5 now z del z is a thickness of

soil layer. So, where del z is the thickness of the soil

layer, E is the elastic modulus of the soil layer and I z that value will use by the we

will get one value that we will get from the chart. So, this I z we will get from the chart.

So, this chart we have to prepare I will explain how to prepare this chart.

So, that means thus the settlement the total settlement that we will get for the different

layer soil if this layer soil and c 1 into c 2 into q minus q bar minus q into summation

of I z, E and del z, where del z is the thickness of each soil layer, E is the elastic module

of the soil layer, I z that will prepare this chart. And then from there we will get the

I z value and c 1, we will get by using this expression and c 2 by using this expression

where t is the time in year, q is the effective overburden pressure at the foundation base

level and q bar is the footing pressure. Now to determine the I z; so suppose this

I z will get at the base of the footing, suppose this is the footing base; this is the ground

surface, ground level and this is footing base and this is D f is a depth of the footing.

So, at this level and this is the B is the width of the footing, D f is the depth of

the footing, D is the width of the footing and this is at the footing base.

Then here below this footing base, if we draw this line suppose this is for the B; this

is for the 2B; this is for 3B; and this is for 4B. So, this is B; this is 2B line; this

is 3B line and this is 4 B line. So, we will get this chart suppose this is 0.1; this is

0.2; this is 0.3; this is 0.4 and this is 0.5; 0.1, 0.2, 0.3, 0.4, 0.5. So, that means,

here this is 0; this value is in this direction; this is I z that coefficient this is 0.5;

this is 0.4; this one is 0.1 so that means, this is 0.1; this is 0.2; this is 0.3; this

is 0.4 and this is 0.5. Now, this one is the point at the B distance,

now for the axisymmetric condition this graph we have to draw from 0.1 to this 2B and for

plain strain condition this will be the graph that means this is for the plain strain condition,

where L by B is this greater than 10 and this is for the axisymmetric condition, similarly

for the square or circular footing. So, we will get the two graphs; one that is

starting from 0.1, I z and then this is the maximum at B by 2 distance that is for the

axisymmetric graph this is corresponding to B by 2 distance, then up to 0.2B if up to

2B this is minimum and then for the plain strain condition this is graph to start from

0.2, then at B this is maximum, then it will go up to 4B. So, this is plain strain condition

this is axisymmetric condition. So, now from this graph we will get the I

z value suppose the for at any distance because the we will we have to consider the center

point of the each soil layer, then at that point what will be the I z that we have to

determine. So, now if we solve one example and then we will get how to calculate or how

to get the this value suppose this is the one footing that we are getting this value

this is the footing width. So, footing width is we are considering that

this footing width is D. So, suppose this is the footing width we have footing we have

to place it here. So, this is the B which is footing width now here that intensity of

the loading that we are applying here for the footing that intensity is q bar at the

footing base we are talking over this is the footing base this line is basically the footing

base. So, intensity that suppose this is 150 kilonewton

per meter square and this depth of the footing depth of the footing is 2 meter and that due

to this 2 meter soil the intensity of the footing of the load that is coming at this

point that means q is 30 kilonewton per meter square, that means q bar is 150 kilonewton

meter square, that means the footing pressure at the base of the footing that is 150 kilonewton

per meter square. And this q effective overburden pressure at the base of the footing that is

gamma D f is it coming out to be 30 say this is 30 kilonewton per meter square, where D

f is 2 meter and that footing intensity is 150 kilonewton per meter square and depth

we are getting. So, this is the footing width is taken the

given is 2.5 meter. So, the B of the footing width of the footing is 2.5 meter and L of

the footing is 30 meter. So, this is we can say L by B is greater than 10. So, we have

to go for the plain strain condition. So, this is for plain strain condition

strain condition.

Now, first we will consider I will calculate and we have to calculate the settlement

at 5 years. So, now what are the soil that

I will give later on first to prepare that chart to determine the I z. So, suppose this

point if I take from this point. So, this is say 0 0. So, this is B, 2B; this is 0,

B, 2B, 3B. So, this is B where B is 2.5 meter this is 3B, 2B that is 5 meter; this is 7.5

meter

and this one is 10 meter. So, these are the this is the depth in meter. So, this is B,

5B, 7.5B, B, 2B, 3B and 4B. So, we have to go for up to 4B depth. So, this is 4B.

Next suppose this point is 0.1; this is 0.2; this is 0.3; this is 0.4 and this is 0.5.

Now as this is plain strain condition, so that means that means at this condition we have to go

for B depth starting from 0.2, then if I go for up to the B and then from here up to 4B.

So, this is the chart from here this is showing the I z; this value is 0.1, 0.2. So, this

is 0.1; this is 0.2, 0.3, 0.4 and this one is 0.5.

So, once we have prepared this chart, now we will give the. So, this chart we will use

to determine the I z value. So, this is this B, 0 is the base of the footing from here

it will start B, 2B, 3B and 4B, because for this we have to go for the 4B depth. Now, if I give the other properties that now

first we have to calculate the c 1 and c 2 value that c 1 value is 1 minus 0.5 into that

c 1 value that value that will give this is in terms of q by q bar minus q. So, this value

is coming 1 minus 0.4 q is here 30 q bar 150 q is 30 because q is 30 kilonewton per meter

square and q bar is 150 kilonewton per meter square.

So, c 1 value is coming 0.875. Similarly, that c 2 value 1 plus 0.2 log 10 because we

are calculating at the 5 year this is 5 by 0.1; this is coming 1.34; c 1 and c 2. So,

this is after 5 years. Now, we have to prepare one table. So, suppose

this is the table we have to prepare the first column is the layer, this is layer; this is

the del z or the thickness of the each layer in meter; this is the q c value q c is the

static cone resistance value the each layer E s is kilonewton per meter square elastic

modulus, then z value at the of each layer. So, z is the; z at the center of each layer

and I z that we will get at the center of each layer and then we will calculate I z

by E and del z for the each layer. So, this is the total table that we will prepare this

is I z by E into del z. So, this for this for the first one for the layer one the thickness

that is given is thickness of the soil for the layer soil first layer is 1 meter whose

q c is 2500, because these are the measured value 2500 kilonewton per meter square.

And one thing that it is mentioned that how to calculate the E at the base is the aspect

of q c, now if this q c we will E we will get E s it is 3.5 q c for plain strain condition

for plain strain

plain strain condition. Now, E s will give this value so that means, the E s and similarly,

we will get E s equal to 2.5 into q c for square or circular footing. So, q s equal

to 2.5 q c for the square and circular and q c is the 3 point q c for the plain strain.

So, here the recommendation this year; this is for the circular or for the square; this

is 2.5 q c for the square and circular and q c is 3.5 into q c that E s for the plain

strain. Here this is for the plain strain. So, we

will use this 3.5 into q c. Now, here q c value is given 2500. So, if I multiply the

3.5 we will get 8750 is the E value kilonewton per meter square. So, this is kilonewton per

meter square. Similarly, at the center because this is the

first layer whose thickness is 1 meter. So, the center will be 0.5 meter this is the center

of the first layer that is 0.5 meter. Now from the chart that I have prepared, so this

is the chart now here this is 2.5. So, this distance; so that means, this will be 0.15,

0.5. So, 0.5 corresponding I this value is around 0.23. So, this is the 0.5 depth corresponding

to this I z value this is 0.5 corresponding to this graph; this I z is 0.23.

So, in this shall we will get the I z value 0.23 and then we will calculate this term;

this is coming 2.63 into 10 to the power minus 5. Now, we will go for the second layer whose

thickness is given 1.5, q c value the measured value 3500 now if I consider three 0.5 q c.

So, this is coming 12250 is the E s value, then z center this will be. So, this is for

center of this layer is; that means; this z value we have to measure from the top.

So, this is 0.5 meter from the ground surface or from the base of the footing. So, this

will be center will be 1.5 by 2.75 plus 1. So, this will be 1.75 from the base of the

footing. So, corresponding I z we will calculate here,

so 1.75. So, this somewhere here, so one 0.75 corresponding value will be 0.385. So, here

this will be 0.385 corresponding value this is 0.385. So, this value is 4.71 into 10 to

the power minus 5. So, similarly for the third layer this thickness

is 2 meter q c value 6500. So, this is coming 22750 the thickness will be 2.5 plus 1; 3.5

the z value form the center the I z from the graph is 0.45. So, this value is 3.96 into

10 to the power minus 5. Similarly, for the fourth layer this is 0.5

say and for the fifth layer this thickness is 2 meter for the sixth layer the thickness

thickness of the layer is 2 meter, and seventh layer the thickness of the layer is 1 meter.

So, this is the 7 layers are represent 1, 2, 3, 4, 5, 6. So, first layer is 1 meter,

second layer is 1.5 meter thickness, third layer is 2 meter thickness say, fourth layer

is 0.5 meter thickness, fifth layer is 2 meter thickness, sixth layer is 2 meter thickness,

seventh layer is also 1 meter thickness. Now, corresponding q c value here this value

is 2000 say and E we will get into 2000 into 3.5 this is 7000 for this 2 meter layer this

is 10000 q c. So, this E value is with 35000 for the sixth layer this is 4000 corresponding

E will be 14000 kilonewton per meter square for the last layer this is 6000 corresponding

E will be 21000. And the center for this thing this will be

1.5, 2.5, 4.5 plus 0.25. So, 4.75 this will be 4.75. Similarly, this layer this is 1,

2.5, 4.5, 5 plus 1 this is two layer thickness. So, center will be one. So, 5 plus 1 this

is 6 meter similarly, this is also 8 meter and this will be 9.5 meter.

Now, corresponding I z value we will get from the chart is 0.35 this is 0.265; this is 0.13

and this is 0.06. So, this value we will get from the chart this is the center z thickness

from the base of the footing at the center of each layer at the center of each layer

this is the thickness. Corresponding this value is coming 2.5 into

10 to the power minus 5; this is 1.51 into 10 to the power minus 5; this 1.86 into 10

to the power minus 5 and this is 286 into 10 to the power minus 5; 0.286 into 10 to

the power minus 5. Now, if we sum this last column and if I sum this last column value

now the summation of this last column value this will give us the value 17.46 into 10

to the power minus 5, the summation of this last column the summation will give us 17.46

into 10 to the power minus 5. Now, next here now we will know the c 1 value;

we know the c 2 value; we know this summation term. So, this z, so now we will put. Now,

we will calculate the settlement. Now, the settlement that we will get, so this

settlement is c 1 into c 2 into q bar minus q into summation of I z by E into del z. Now,

c 1 value is 0.875 c 2 is 1.34 this is after 5 years q bar is 150 kilonewton per meter

square q is 30 kilonewton per meter square and summation of I z; E into del z that is

equal to 17.46 into 10 to the power minus 5.

Now, if we put this value here S is 0.875 into 1.34 this is 150 minus 30, then into

17.46 into 10 to the power minus 5. So, this value is; that means, 25 millimeter. So, the

settlement of this total soil layer under this loading condition is 25 millimeter after

5 years. Now, by

using this difference, so suppose if I want

to calculate the settlement after one year, then we have to put this we have to modify

the c 1, c 2 coefficient by putting t equal to 1 year, then we can multiply this settlement

by this new c 2 and divided by the old c 2 then we will get the settlement of the soil

and the one year also. So, this is the calculation and then how there

this measure this is the another method by where you are using SCPT value and we are

getting the settlement for the granular soil. Now, the next method that we want to discuss

that is another method which is this is the semi-empirical method that is presented by

Buisman, 1948. Here the settlement is calculated as a summation

of 2.3 into p 0 bar by E into H into log 10 p 0 bar plus del p by p 0 bar or we can say

this is sigma 0 bar or del sigma p both are same, where p 0 bar is effective overburden

pressure and del p is the stress increment due to footing load.

E is the elastic modulus of the soil and H is the thickness of the soil layer. So, by

using this expression also we can determine the settlement of the granular soil by using

this semi-empirical expression. So, these are all for the granular soil. Now, if we summarize the different techniques

to determine the granular soil settlement calculation, then we can say that the settlement

of foundation on sand or granular soil the first one method

is by the elastic theory method. So, by this method we calculate the settlement is by using

q n, B, E, 1 minus mu square into I f. So, this expression I have already given when

we discussed about the immediate settlement. So, this is the immediate settlement or the

by using the elastic theory this q n is the net footing pressure this is the net footing

pressure B is the width of the foundation E is the elastic modulus mu is the Poisson

ratio I f is the influence factor now I have presented one table and here I have explained

how to calculate this I f value. The next method that is the semi-empirical

method that I have just now what I have explained it is a semi-empirical method and this is

given by Buisman, 1948. So, this is the where S i we can calculate summation of 2.3; this

is p 0 by E into H log ten p 0 bar; this is also p 0 bar into del p 0 bar. So, next one

is the plate load test, where this expression we are using S f by S p equal to B f, B p

plus 30 into B p, B f plus 30 to the power whole square this is for the granular soil.

Now, next method that we are using that from the from the SPT chart it is presented in

I s 8009 part one 1976, by using the SPT chart also we can determine the settlement and then

the from SCPT value. So, here the expression S i we can calculate by 2.3 into H divided

by c into log ten; this is p 0 bar plus del p into p 0 bar, where c we will get by using

this expression 1.5; q c by p 0 bar; this is proposed by De-Beers this expression is

proposed by De-Beer and Morten or we can use c equal to 1.9; q c, p 0 bar that is proposed

by Meyerhof. So, now the last method or the next one is

given by; this is also based on SPT value Schmertmann Schmertmann and Hartman; this

is 1978. So, here we will get this value by c 1 into c 2 this is q minus q 0 as the footing

base the summation of z equal to 0 to 2B or 4B. So, this is summation of I z; E into del

z. Where this is 2B or 4B and for if it is axisymmetric

condition that is square of circular, then this z is from the base of the footing from

0 to 2B the summation up to 2B if it is a plain strain condition, then we have to go

for 0 to 4B up to 4B condition, then we will get the if I sum these things it we will get

this value of the settlement. So, these are the methods by which we can

determine the settlement of the granular soil under different loading condition. Now, if

in the next section that now we will solve one problem and then we will compare the we

will determine this value or settlement value by using this different techniques, then we

will compare about the what is the settlement that we are getting by using this different

methods. Now, first if I the settlement calculation

for this different methods and there is a problem that we are taking. So, suppose this

is the foundation; this is ground line; this is the foundation width say 4 meter; this

is square foundation, square footing with dimension 4 meter cross 4 meter.

Now, the this depth of foundation is 1 meter and the position of the water table is as

the base of the footing. Now, this is the layer one and this is layer two; this is for

the medium sand that gamma we are taking 18 kilonewton per meter cube, q c value we are

taking 10000 kilonewton per meter square, E value for the first layer we are taking

this is 25000 kilonewton per meter square. For the second layer also we assume the same

gamma it is 18 kilonewton per meter cube, q c value we are taking 12000 kilonewton per

meter square, and E for the second layer is 30000 kilonewton per meter square. Now, the

total load that is coming, the total load or load Q is 2000 kilonewton.

So, this is the layer. So, this layer here the up to the, this layer thickness first

layer thickness is 4 meter and second layer thickness is 6 meter. So, we are taking the

influence zone if we use the different method and all the method because it is in square

footing even in the last method also the influence zone will be up to 2B

So, here up to 8 meter that means, up to this 2B this will be influence zone. So, we are

taking two points the center of the each layer because this is the influence zone up to 2B

so; that means, this layer is 2 meter from the base of the foundation and this layer

is also 2 meter from the end of the first layer.

Now, we will use those different methods; first we will use the elastic theory. So,

we will use the elastic theory first. So, here the E value elastic theory the S i settlement

that is q n B by E into 1 by mu square into I f influence factor.

Now, q n here we can calculate this is 2000 divided by 4 cross 4 this is; this value is

coming out to be 125 kilonewton per meter square, similarly B is here 4 meter E we are

taking the weighted average value because here 4 meter influenced on 4 meter for the

first layer and 4 meter for the second layer. So, we can take the average one or this E

value that we are taking that is 25000 into 4 plus 3000 into 4 divided by 8 or we can

simply take the average of two layers. So, this is coming out to be 2750 kilonewton per

meter square. Now, here we assume that mu value is 0.3 for

this tan and I f, we can calculate this I f is value is one 0.12 from the table; this

table I have already given. So, from this table this I f value is coming 1.12. So, now this if I put this all the calculation

value this then we can get the S i will be 125 into 4 divided by 27500 1 by 0.3, 1 minus

0.3 square into 1.12. Now, in this calculation we have to use the correction factor because

here this is the isolated footing. So, this is rigid this is not rigid. So, rigidity correction

is not required. So, this is a sandy soil. So, consideration correction due to the consolidation

that is also not required. So, only the depth correction we have to apply here.

Now to calculate the depth correction, so far the depth correction or the fox correction.

So, now the D root L B that value is 0.25 here and L by B equal to 1. So, the correction

factor

is around 0.94 that will give from the chart. So, if L by B equal to 1 and D this value

0.25 we will get 0.9. So, now we have to multiply here 0.94. So,

we will get the corrected value. So, S i immediate corrected that is 17.42 millimeter. So, this

is the settlement by using the elastic theory approach. Now, this settlement is coming 17.42

up to the correction because we have applied the correction factor this is the correction

factor we have applied. So, this will coming 17.42 millimeter.

Now, the next method is by the semi-empirical method if I use the same expression same problem

by using the semi-empirical method. So, here we are taken two points because this is the

two points or the two center of this two layers; this is for the first layer thickness is 4

meter and the center is A point the first second layer thickness up to the influence

zone it is we have taken the 4 meter and this is of the center.

So, A and B two points, so now the we have to calculate the sigma 0 at A that is p 0

or sigma 0 bar is 18 plus 1; 18 into 1 plus 2 into 8 because for the below ground water

table we have the base of the footing. So, this is 8 into 2. So, this is coming 34 kilonewton

per meter square. Now, del p if I consider one is to two distribution

del p will be 125 into 4 into 4 divided by 4 plus 2 into 4 plus 2 because we have considered

one is to two distribution. So, this is coming 55.55 kilonewton per meter square.

Similarly, at point B; p 0 bar is 18 plus 1 18 into 1 plus 4 into 18 plus 2 into 8 4

into 8 plus 2 into 8. So, this is 66 kilonewton per meter square and del p by using same one

is to two distribution this is coming 20 kilonewton per meter square. So, now the final expression for this method

S i we will get this is 2.3 into 34 is the p 0 bar E is 25000 for this layer E and 4

is the thickness of this layer, then log 34 plus 55.55 divided by 34 plus 2.3 into 66

divided by 3000, because this is the elastic modulus of the second layer thickness is 4

meter for this second layer also this is 10; 66 plus 20 divided by 66.

So, here we will get the settlement 5.3 plus 2.33. So, the total settlement is 7.63 millimeter,

and after the correction is been multiplied the depth correction here that is 0.94 the

S i after correction this value is coming 7.2 millimeter.

Now, if I go for the next method that is from this SPT value that is from this is method

from SCPT value. So, here first we consider the Meyerhof expression, where c for the first

layer first layer this is 1.9, then this is this value is q c by p 0 bar. So, 1.9 is 10000

E 0 bar is 34. So, this value is 558.8. Now, see for the second layer is also 1.9

q c is here 12000 and this value is 66. So, this is 345.5 now if I calculate this S i

the expression is 2.3 into 4 that is the h by c by c 558.8 for the first layer into 34

plus 55.55 divided by 34 plus 2.3 into 4 divided by 345.5 into log 10 66 plus 20 divided by

60. So, this value after the calculation we will

get this value is coming 6.92 plus 3.06. So, that is 9.98 millimeter after correction if

I apply the correction if I multiply the depth correction factor that is 0.94. So, this is

9.98 into 0.94 this is around 9.4 millimeter. So, now if I convert this is by this De-Beer’s

method and Morten method, then this settlement is 9.4 into 1.9 divided by 1.5. So, this is

12 millimeter. Similarly, by using the next technique that is the D technique using the

Schmertmann and Hartman technique Hartman technique in 1978, that we have done this

is by because here this is the axisymmetric condition; this is square footing though it

will vary the graph that we have to draw this will start from 0.1, then up to B by 2 this

will give the maximum, then it will go up to twice B.

Now, if the because this method I have already explained how to use this method, then calculate

the this I have explained for the plain strain condition, here it is the axisymmetric condition

and the only difference is that here instead of this graph when we calculate the influence

this factor I z, then this point will start from the 0.1 then it will maximum at points

maximum that value is 0.5 at B by 2, then it will go 0 at 2B. And then by using the

at the center it is two points at the center it is one center is point is at the B, A another

center is at the B and they are, so this we have to calculate the I z value, then c 1

value, c 2 value here. We have calculated the c 1 at the one year.

So, this value at this S i is coming out to be 11.78 this is after one year, and this

coming say 9.7 meter this is after one month. So, we can if I compare the value. So, so

far the elastic method method one this value for the elastic method the value this is coming

out to be this is 17.42 millimeter, for the second method or semi-empirical method this

value that is coming 7.2 millimeter, and for the c for the Meyerhof this is coming 9.4

millimeter, this is one and for De-Beers and Morten it is coming 12 millimeter and d method

is coming 9.7 or you can say this is 11.78 millimeter for the one year

and 9.7 millimeter after one month.

So, from this things we can say that this elastic theory is giving higher settlement;

that means, it is overestimate this the meeting the settlement value; this is the giving the

higher settlement value and the b method by using the semi-empirical method that is giving

the lower settlement value. And other values are more or less same, but only the maximum

one is given by the elastic theory and minimum is given by the semi empirical method, and

whereas the CPT by using SCPT value these values are almost same, and although this

it is expected that it is obvious that the Meyerhof theory is giving lesser settlement

by if I consider the c for the De-Beers or Morten, but the maximum was given by the elastic

settlement elastic theory. So, in the next class I will discuss that

how to calculate the how to calculate the bearing capacity and the settlement of the

foundation and based on this bearing capacity, and settlement council here, until now we

have considered the settlement calculation we have done the settlement or either bearing

or settlement. In the next class, I will consider the how

to using the, this both criteria and then how to design or choose the dimension of a

footing or the depth of a footing by using the considering settlement and bearing criteria

both. And then I will also explain how to apply the depth correction when we will calculate

the settlement based on SPT or based on plate load test data.

Great lecture, thanks for explaining the computation of Iz.

Thank You!

Amazing lecture and explanation of Strain Influence Factor. Thank you very much as this helped me

thank you so much !

Awesome work Doc!!!

wasting of time, I don't understand how did you separate the layers and how did you calculate the (qc).

Great Sir……There is no alternate of you

Correction: De-Beer method Si = 9.4 *1.5/1.9 = 7.42 mm