 In last class, I have discussed about the settlement calculation for granular soil by using the field test data, when 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 combination test. Now, this class I will explain a few other techniques to determine the settlement for the granular soil and then I will solve a 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 techniques for the using the CPT. So, there is another technique by using the CPT data. So, by using the SCPT data. So, this technique is proposed by Scamartman and Hartman in 1978. According to their procedure that your settlement that 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 1 minus 0.5 by Q minus Q bar minus Q and C 1 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 mean the footing pressure which is coming from the super structure. Now, here T is the time in E R. So, suppose if you want to determine the settlement at 1 E R, then you will put T equal to 1. If you want to put at settlement at 5 years, then T will be 5. Now, del z is the thickness of soil layer. So, where del z is the thickness of the soil layer is the elastic modulus of the soil layer and I z that value will use by the will get one value that will get from the chart. So, this I z will get from the chart. So, this chart we have to prepare. We will explain how to prepare this chart. So, that means the settlement the total settlement that we will get for the different layer soil. If this layer soil then C 1 into C 2 to Q minus Q bar minus Q in summation of I z E and del z where del z is the thickness of each soil layer. E is the elastic modulus 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 you will get by using this expression and C 2 by using this expression where T is the time in E R. Q is the effective order pressure at the foundation base level and Q bar is the footing pressure. Now, to determine the I z. So, suppose. So, this I z will get at the base of the footing. Suppose, this is the footing base and this is a ground surface ground level and this is footing base and this is d f is the 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 b 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 and b is this is for the 2 b this is for 3 b and this is for 4 b. So, this is b this is 2 b line this is 3 b 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 5. So, that means here this is 0 this value is in this direction this is I z that coefficient this is 0.5 0.5 this is 0.4 this one is 0.1. So, that means this is 0.1 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 axis symmetric condition this graph you have to draw from 0.1 to this 2 b and for plane strain condition this will be the graph. That means this is for the plane strain condition where l by b is greater than 10 and this is for the axis symmetric condition equally for the square or circular footing. So, we will get the 2 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 axis symmetric graph this is corresponding to b by 2 distance then up to 0.2 b up to 2 b this is minimum. And then for the plane strain condition this graph will start from 0.2 then at b this is maximum then it will go up to 4 b. So, this is plane strain condition this is the axis symmetric condition. So, now from this graph we will get the I z value suppose the for at any distance because the we will 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 b. So, suppose this is the footing width we have footing we have to place 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 about this is the footing base this line is basically the footing base. So, intensity that suppose this is 150 kilo Newton per meter square and this depth of the footing depth of the footing is 2 meter and the due to this 2 meter soil the intensity of the footing of the of the load that is coming at this point that mean q is 30 kilo Newton per meter square. So, that means q bar is 150 kilo Newton meter square that mean the footing pressure at the base of the footing that is 150 kilo Newton per meter square and the q effective over burden pressure at the base of the footing that is gamma d f is it coming out to be 30 say this is 30 kilo Newton per meter square where d f is 2 meter and that footing intensity is 150 kilo Newton 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 plane strain condition. So, this is for plane strain condition strain condition. Now, first we will consider or we will calculate and we have to calculate the settlement at 5 years. So, now, what are the soil data that I will give later on first I have to prepare that chart to determine the I z. So, the post this point if I take from this point. So, this is say 0 0. So, this is b 2 b this is 0 b 2 b 3 b. So, this is b where b is 2.5 meter this is 2.5 meter. So, this is 2.5 meter this is 3 b 2 b that is 5 meter this is 7.5 meter and this one is 10 meter. So, this is the depth in meter. So, this is b 5 b 7.5 b 3 b 2 b 3 b and 4 b. So, you have to go for up to 4 b depth. So, this is 4 b. 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 plane strain condition. So, that means that means at this condition I 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 4 b. 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 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.3 0.4 and this one is 0.5. So, once we have prepared this chart now will give the. So, this chart 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 2 b 3 b and 4 b because for this you have to go for the 4 b depth. Now, if I give the other properties that now first you 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 1 minus 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 30 kilo Newton per meter square and q bar is 150 kilo Newton 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 of the 5 here this is 5 by 0.1. So, 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 you 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 in the each layer E s is kilo Newton 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 and into del z. So, this for the first one for the layer 1 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 measure value 2500 kilo Newton per meter square. And one thing that it is mentioned that how to calculate the E with respect of q c. Now, if this q c will get E will get E s it is 3.5 q c for plane strain condition for plane strain plane 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 3.5 q c for the plane strain. So, here the recommendation this here this is for the circular or for the square this is 2.5 q c for the square and this is square and circular and q c is 3.5 into q c that E s for the plane strain. Here this this is for the plane 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 8 7 5 0 is the E value kilo Newton per meter square. So, this is kilo Newton 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 for this distance. So, that means the distance 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 version 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 major value 3500. Now, if I consider 3.5 q c. So, this is coming 12250 is the E s value. So, this is the second layer. So, then z center this will be this is for center of this layer is that means this z value you have to measure from the top. So, this is 5.5 meter from the ground surface from the base of the footing. So, this will be center will be 1.5 by 2 is 0.75 plus 1. So, this will be 1.75 from the base of the footing. So, corresponding I z value is 0.5. So, this I z we will calculate here. So, 1.75. So, this somewhere here. So, 1.75 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 z value from 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 layer is 1 meter. So, this is the seven layers are present 1 2 3 4 5. 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 2 0 0 0 say and e will get into 2000 into 3.5 this is 7000. For this 2 meter layer this is 10000 q c. So, this e value will be 35000. For the sixth layer this is 4000 corresponding e will be 14000 kilo Newton per meter square. For the last layer this is 6000 corresponding e will be 21000 and 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 2 layer thing. So, from center will be 1. 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 will get from the chart is 0.35 this is 0.265 this is 0.13 and this is 0.06. So, this value 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 is 1.86 into 10 to the power minus 5 and this is 286 into 10 to the power minus 1 0.286 10 to the power minus 1. Now, if you sum this last column 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 the 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 to 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 kilo Newton per meter square q is 30 kilo Newton per meter square and summation of i z e into del z that is e to the power minus equal to 17.46 into 10 to the power minus 5. Now, if you 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 is value is coming is 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 suppose if I want to calculate the settlement after 1 year then we have to put this but to modify the c 1 c 2 coefficient by putting t equal to 1 year. Then we can multiply this settlement by this mu c 2 and divide it by the whole c 2 then we will get the settlement of the soil at the 1 year also. So, this is the calculation and then how this is another method by where you are using s c p t 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 Boiseman 1948. Here the settlement is calculated by the summation of 2.3 into p 0 bar by e into h into loctem p 0 bar plus del p by p 0 bar or you can say this is sigma 0 bar or del sigma p both are same where p 0 bar is effective over burden 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. Now, 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 will discuss about the immediate settlement. So, this is the immediate settlement or 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 is the elastic modulus mu is the position ratio i f is the influence factor. Now, I have presented one table and I have explained how to calculate this i f. The next method that is the semi empirical method that I have just now I have explained it is a semi empirical method and this is given by Pussman in 1948. So, this is the where I will calculate summation of 2.3. This is p 0 by e into h log 10 p 0 bar this is also 3 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 S p t chart that is presented in I s 8009 part 1 1976. By using the S p t chart also we can determine the settlement and then the from S c p t value. So, here the expression S i we can calculate by 2.3 into h divided by c into log 10. 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 D B R this expression is proposed by D B R M Mottin or we can use c equal to 1.9 into h log 10. q c p 0 bar that is proposed by Mirov. So, now the last method or the next one is given by this is also based on S p t value Skemterman, Skemertman, Hartman this is 1978. So, here we will get this value by c 1 into c 2. This is q minus q 0 at the footing base the summation of z equal to 0 to 2 b or 4 b. So, this is the summation of i z e into del z. So, this is the summation of z where this is 2 b or 4 b and for if it is axisymmetric condition that is square a circular then the z is from the base of the footing from 0 to 2 b the summation up to 2 b. If it is a plane strain condition then you have to go for 0 to 4 b up to 4 b condition then you will get the if I sum these things you will get this value of this settlement. So, these are the methods by which we cannot 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 that we will determine this value of this 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 method. Now, first if I the settlement calculation for this different methods and the problem that we are taking 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 lower layer 1 and this is layer 2 this is for the medium sand that gamma we are taking 18 kilo Newton per meter cube Q c value we are taking 1000 10,000 kilo Newton per meter square e value for the first layer we are taking this is 25,000 kilo Newton per meter square. For the second layer also we assume the same gamma that is 18 kilo Newton per meter cube Q c value we are taking 12 kilo Newton per meter square and e for the second layer is 30,000 kilo Newton per meter square. Now, the total load that is coming the total load or load Q is 2000 kilo Newton. 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 you use the different method on all the method because it is in a square footing even in the last method also the influence zone will be up to 2 b. So, here up to 8 meter that means up to this 2 b there will be influence zone. So, we are taking 2 points the center of the each layer because this is the influence zone up to 2 b. 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 the different method 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 of 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. So, this is this value is coming up to be 125 kilo Newton per meter square. Similarly, b is here 4 meter e we are taking the weighted average value because here 4 meter influence zone 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 2500 into 4 plus 30000 into 4 divided by 8 or we can simply take the average of 2 layers. So, this is coming up to be 27500 kilo Newton per meter square. Now, here we assume that mu value is 0.3 for this stand and i f we can calculate this i f is value is 1.12 from the table this table I have already given. So, from this table this i f value is coming 1.12. So, now this value 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 is not rigid. So, rigidity correction is not required. So, this is a sandy soil. So, consolidation 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, for the depth correction of 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 we will get from the chart. So, if l by b equal to 1 and d this value 0.25 we will get 0.94. 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 after 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 have taken two points because this is the two points of the two center of these 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 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 the ground water table is at the base of the footing. So, this is 8 into 2. So, this is coming 34 kilo Newton per meter square. Now, del p if I consider 1 is to 2 distribution del p will be 125 into 4 into 4 divided by 4 plus 2 into 4 plus 2 because we have considered 1 is to 2 distribution. So, this is coming 55.55 kilo Newton per meter square. Similarly, at point b p 0 bar is 18 plus 2 into 4 into 18 plus 1 18 into 1 plus 4 into 18 plus 2 into 8. So, 4 into 8 plus 2 into 8. So, this is 66 kilo Newton per meter square and del p by using same 1 is to 2 distribution this is coming 20 kilo Newton per meter square. So, now the final expression for this method S i will get this is 2.3 into 34 is the p 0 bar e is 250 0 0 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 4 into 66 divided by 3 0 0 0 because this is the elastic modulus of the second layer thickness is 4 meter for the 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, this total settlement is 6.5 plus 2.3 plus 2.3 7.63 millimeter and after the correction if we multiply the depth correction here this 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 next method from SCPT value. So, here first we consider the mayor of 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 q c is 10000 p 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 558.8 for the first layer into 34 plus 55.55 divided by 34 plus 2.3 into 4 divided by 3.5 plus 345.5 into log 10 66 plus 20 divided by 60. So, this value after the calculation will get this value is coming 6.92 plus 3.06 so that is 9.98 millimeter after corrections 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 Beers method on Morton 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 Scamatman and Hotman technique Hotman technique this is 1978 that we have done this is why because here this is the axisymmetric condition versus 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, it is because this method I have already explained how to use this method then the calculate the this I have explained for the plane strain condition here it is the axisymmetric condition and the only difference is that here instead of this graph the when you calculate the influence factor this factor i j then this point will start from the 0.1 then it will maximum at point maximum that value is 0.5 at b by 2 then it will go 0 at 2 b and then by using the at the center here is 2 point the center is 1 center is point is at the b a another center is at the b and there. 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 1 here. So, this value at the this s i is coming out to be 0 at 11.78 this is after 1 year and this coming say 9.7 meter this is after 1 month. So, we can if I compare the value. So, for the elastic method method 1 this value for the elastic method the value this is coming out to be this is a 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 mirror of this is coming 9.4 millimeter this is 1 and for d b s and Motton it is coming 12 millimeter and d method is coming 9.7 or you can say this is 11.78 millimeter for the 1 year and 9.7 millimeter after 1 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 lowest 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 a CPT value these values are almost same and although this it is expected that it is obvious that the mirror of theory is giving lesser settlement by if I consider the c for the d b s or Motton but the maximum one is 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 constant here are until now we have consider the settlement calculation you 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 CPT based on plate load test data.