 Now, last class I have discussed about the footing which is loaded eccentrically, I have discussed about the one way eccentricity and two ways eccentricity. Now, in this class I will discuss the how to calculate the bearing capacity of the soil in layer condition. Now, as we know that soil is not a homogeneous soil because, but, but till now the most of the things that I have discussed is to calculate the bearing capacity of homogeneous soil. Now, soil is a layered type of material. So, now the properties of the soil according to depth is not same, it is changing. Now, if the soil is the condition is the layered condition then how to calculate the bearing capacity of this layered soil that I will discuss in this class. Now, suppose if we go for the bearing capacity of a layered soil system, now I will discuss the bearing capacity of layered soil. Now, the derivation or the expression which is available for this to calculate the bearing capacity of the soil which is basically for the homogeneous soil. Now, if the soil is in this form that this is the ground surface, here we are applying this footing, this is a ground surface and this is width of the footing and this is the depth of the footing d f. This is we are considering the two soil layer, this is one layer which is the strong soil, which is stronger soil or layer one or layer and layer two which is weaker soil. So, this is the stronger soil and this is the weaker soil. Then how to calculate the bearing capacity of this soil condition and suppose the this is the properties gamma 1 c 1 and phi 1 for the layer layer 1 and gamma 2 c 2 and phi 2 for layer 2. Now, if the stronger soil, this is the interface is the stronger soil and one thing that the depth of the stronger soil from this base of the foundation is h, it is the up to this the up to the h depth from the base of the foundation is the stronger soil and the weaker soil. Now, if the depth h is relatively small, then what will happen then then there will be a punching type of shear failure for this stronger soil layer and then general shear failure for the weaker soil. So, suppose this is the punching type of failure surface, punching shear failure type of failure surface for this strong soil and then general failure surface which is expected for the weaker soil. So, this type of failure surface that will develop. So, now if I draw one line vertical line. So, what are the forces which is acting suppose in this because here it is the general shear type of failure. So, that means we can calculate this is the q bottom or the bottom layer and for this punching shear failure what are the forces that is acting. So, first is the adjacent c a, this is which is acting to side of this layer because suppose this is the one block which is falling or that means this is the size. Now, another thing the passive resistance the surrounding soil that will provide for this zone or this zone that is the passive resistance p p which is acting at angle of delta. Similarly, this side also this is the p p which is acting as angle delta. So, these two types of force which is which are acting here both this is the center line. So, one is the addition in the both side of this layer and passive resistance coming from the soil of this side and soil from this side which is acting as angle of delta to the perpendicular of this line this is p p p p is the passive resistance. Another thing that if this h value here if this h which is the distance which is depth of this layer or from the base of the footing is relatively small then you will get this type of failure surface. Now, if the distance is relatively large then we will get the most of the failure surface will be in the stronger layer. So, the failure surface or the effect of the loading intensity of the footing load will not act suppose this is the footing load that is q ultimate. So, that effect will not go to the second layer or the weaker soil layer. So, most of the failure surface or will be located at the stronger layer in that case the general expression for a shear failure expression general shear failure expression for the for homogeneous soil or single layer soil that we can use to determine the loading of bearing capacity of the soil. Now, if it is relatively small then you will go for this approach. Now, one thing is that that this is the value. So, if I go for the different loading condition. So, what will be the different loading condition for this case? So, first we will calculate the q u for the loading condition the vertical load q u which is acting in the downward direction that is equal to the perpendicular load that is acting. So, the resistance that q u is the total load bearing capacity. This is the some portion of the resistance that we will get from the bottom layer and some portion we will get from the top layer. So, this is the summation of the resistance or the load which is carrying by some portion by the strong layer and some portion by the weak layer or the bottom layer and the top layer. Suppose in the bottom layer it is as usual the general shear failure. So, we can write this is the resistance that is q bottom. So, q bottom is the load carrying capacity of this bottom layer or the resistance that is coming from this bottom layer plus the resistance that is given by this top layer that will be 2 into C A that is the addition 2 means 1 from this side another from this side plus because here P P is acting as a delta. So, the sign component that will act in the upward direction. So, that will give the upward resistance. So, that is P P into sin delta. So, that means these are the upward forces divided by B because these things acting as a width of P minus gamma 1 into H because this is the load or the q that is acting here gamma 1 into H that is the surcharge below the base of the foundation up to the second layer or up to the starting of the second layer or up to the end of the first layer. So, that is the total for that mean the contribution of the q is the contribution from the bottom soil plus the contribution from the top layer and minus this value surcharge that we are deducting because this surcharge we will not consider as a contribution because it is it was existing previously. So, now, this are the total load. Now, here P P we can write that is passive force per unit area. So, that is the passive force per unit length. Here we can write that C A is equal to small C A into H because C A is the total load that is total resistance due to addition and small C H is the addition which is per meter. So, that is the total and then as it is acting as a this length H. So, this will be C A into H. So, if I put if we put this value in this final expression then the form of this expression will be because our original expression is q u is equal to q b that is the total load that is the bottom resistance into 2 into C A plus P P into sin delta divided by b minus gamma 1 into H. Now, here C A is basically small C A into H. Now, if we write put this small C A into H. Now, if we write put this small C A in this expression then q u that will be q b plus 2 small C A into H divided by b plus. Now, we can write P P on the other terms in in the final form is gamma 1 H square 1 plus 2 d f b k p H into tan delta divided by b into gamma 1 into H. Now, this expression again we are taking from this book by B M Dars. This is Dars B M 1990. Now, this expression is the final form of this expression that we are using where k p H horizontal component passive earth pressure coefficient. So, this is the final expression that we are getting to calculate the ultimate load carrying capacity of this layer soil k p H is the horizontal component of the passive earth pressure coefficient. Now, let that k p H tan delta that is equal to k s into tan phi 1, where k s is equal to punching shear failure or coefficient. As I have mentioned that for the first layer it will be punching shear type of failure. So, this is the punching shear coefficient. Now, if we put this k p H tan delta is equal to k s tan phi then the q u expression will be q b plus 2 C A H divided by b into tan gamma 1 H square 1 plus 2 d f into H into k s tan phi 1 by b minus gamma 1 into H. Now, in this form we will get this. So, here this expression this is also d f by H. So, this is 2 d f by H and this is also 2 d f by H and then we just replace this k p H tan delta by k s tan phi 1. Now, here now k s this punching shear coefficient is a function of two things. One is q 2 by q 1 value another is phi 1. Now, here it is mentioned that phi 1 is the friction coefficient of friction angle of the first layer and q 2 and the q 1, where q 2 or q 1 is a C 1 N C 1 for the first layer plus half gamma 1 b N gamma 1 and q 2 will be C 2 N C 2 plus half gamma 2 b N gamma 2. So, here we can see that the one second component that is due to the surcharge which is not present here. So, that mean the q 1 and the q 2 are the bearing capacity if the footing is placed at the surface. So, that means this is for the surface loading conditions where q is 0. So, second term that is also 0. So, that is if we for the same material properties, this q 1 is the bearing capacity of the second layer and q 2 is the bearing capacity of the second layer and q 2 is the surface loading conditions where q is 0. So, second term that is also 0. So, that is if we for the same material properties, this q 1 is the bearing capacity for the first layer for the surface loading and this is q 2 is the bearing capacity of the second layer for the again for the surface loading. Here we are not applying any corrections factor. This is the general form. So, let me q 1 and q 2 that we will get by this thing. So, this k s is the function of q 2 by q 1 and phi 1. So, if we know this q 2 by q 1 and phi 1 then we can calculate k s. Now, another thing that this is applicable if h is relatively small. Now, this expression we can use in different form then if h is relatively large as I have mentioned that if h is relatively large then all the failure surface will be on the first layer itself. So, that means our q u will get that is equal to q t where q t is the bearing capacity of the top layer. So, that means we will get c 1 n c 1 plus q n q 1 plus half gamma 1 b n gamma 1. So, by using these expressions we can calculate the bearing capacity ultimate bearing capacity of the top layer because if h is relatively large. So, now we can conclude that that if because the bearing capacity as the first layer or the top layer is stronger layer and the second layer is weaker layer. So, bearing capacity of the combined contribution coming from the combined of first layer and the second layer that cannot be greater than the contribution or the bearing capacity of the top layer itself because the second layer is the weaker layer. So, bearing capacity of this first layer is q t and if all the failure surfaces are located in the top layer then that is the maximum load carrying capacity that is q t because the first layer the top layer is the stronger layer and second layer or bottom layer is the weaker layer. Now, if the h is relatively large then this q t will give you the load carrying capacity of the soil and that is the maximum. Now, if h is relatively small then the contribution of the first layer that will reduce and the contribution of the second layer that will that will introduce in this ultimate load carrying capacity and that part will increase and that the contribution of the second layer that cannot be greater than the contribution of the first layer as the second layer is weaker layer. So, if I get the combined bearing capacity that is contribution from the first layer and the second layer that should not be greater than the contribution or the first layer itself. The contribution I am telling that if h is relatively high that the all the contribution all the bearing capacity is due to the resistance of the top layer. Now, if h is relatively small then in some the bearing capacity ultimate bearing capacity that you will get that is the some part is coming from the resistance of the top layer and some part is resistance from the bottom layer. Now, summation of these two the some part of the top layer and some part of the bottom layer that cannot be greater than the total contribution or the contribution coming from the top layer itself. So, in that conclusion we can write that q u is equal to that q b bottom layer plus that is the contribution from the bottom layer plus 2 c a h b plus gamma 1 h square 1 plus 2 d f by h into k s tan phi 1 divided by b minus gamma 1 h that cannot be greater than q t. So, this is always less than that can be equal. So, this is less than equal to q t. So, in this now further it is given that for the rectangular footing for the rectangular foundation we can write that q u is the contribution from the q b plus 1 plus b by l. It is introduced for the rectangular footing into 2 c a h divided by b plus gamma 1 h square 1 plus b by l plus into 1 plus 2 d f into h into k s tan phi 1 by b minus gamma 1 h that cannot be greater than q t. So, this is for the rectangular footing. Now, again we we can calculate the what will be the value of q b and the q t where q b is the ultimate load carrying capacity of this bottom layer. So, this will be c 2 n c 2 into s c 2 s c 2 is the safe factor plus gamma 1 d f plus h that is the surcharge which is coming from the top layer into n q 2 and s q 2. Then plus half gamma 2 b n gamma 2 into s gamma 2. So, this n c 2 n q 2 into n q 2 into n q 2 and n gamma 2 these are the bearing capacity factor and s c 2 s q 2 and s gamma 2 these are the correction factor for safe or safe factor. Now, similarly q t for the top footing that will be c 1 n c 1 s c 1 plus gamma 1 here d f this is the surcharge. So, this surcharge into n q 1 s q 1 plus half gamma 1 b n gamma 1 into s gamma. So, here s c 1 or 2 s q 1 or 2 s gamma 1 or 2 are safe factor. So, this way we can do this. Determine the q b for the bottom layer loading ultimate load carrying capacity q 2 is the ultimate load carrying capacity of the top layer which is we are including the safe factors also. Here this d f plus h is the surcharge for the bottom layer and d f into gamma 1 is the surcharge for the top layer and d f plus h into gamma 1 is the surcharge for the bottom layer. Now, the next thing we will discuss few special cases that or the different condition then how to get the different conditions. The special cases that is first case this condition is the top layer is strong sand that means this is purely sandy soil whose that means c 1 is equal to 0 and as the top layer is a strong layer. So, this will be strong sand and the bottom layer is saturated soft clay that means phi 2 is 0. So, this is the weaker soil. So, that means for the top layer c 1 is 0 and the bottom layer phi 2 is equal to 0. So, now if I calculate the q b with the safe factors then we can write 1 plus 0.2 b by l that is for the safe factor and as for the bottom layer phi 2 is equal to 0. So, this is the that means c 2 and the n c will be 5.14 and this is c 2. So, this is basically c c 2 n c 2 and s c 2 s c 2 is 1 plus 0.2 b by l c 2 is c 2 is the cohesion of the second layer or the bottom layer and n c will be 5.14 if because we know that n c 2 is equal to 5.14 if phi 2 is equal to 0 that we know and again that n q that part is also 1 and n gamma part will vanish if phi is equal to 0 that we know because this is the common things for the all the cases. So, now we will get only the surcharge at gamma 1 into d f plus h that is the bottom layer and similarly the top layer that will give you that this is the c 1 is 0 top layer. So, first part is c 1 is 0. So, first part c c 1 n c 1 and s c 1 that is 0. So, in second part we can write this is gamma 1 d f into n q 1 and s q 1 plus half into gamma 1 b n gamma 1 into s gamma 1. So, then in this fashion we can determine the q t and the q b. Now, as for the q 2 and the q 1 q 2 is basically here corresponding to the second layer or the bottom layer and it is in the surface. So, if it is in the surface then this surcharge part that is also 0. So, q 2 we can write that q 2 1 is 1 plus 0.2 into b by l that is divided by half into gamma 1 b n gamma 1 into s gamma 1. And we can write that q 2 is 1 plus 1 we can because this part this and again this 5.14 into c 2 because the surface part as it is the surface footing. So, this part will be 0 and this part will be also be 0. So, only this here is the surface part if I if we neglect. So, this one will be 5.14 c 2 divided by half into gamma 1 into b into n gamma 1. So, here we will get the q 2 and q 1. Now, how to use this q 2 and the q 1 that we will explain by using the because charts are available to determine this q 2 with the help of this q 2 and q 1 and I will discuss these things after you complete the different cases. Now, next one that means here for the first case q u will be 1 plus 0.2 b by l 5.14 c 2 plus gamma 1 h square plus gamma 1 h square 1 plus b by l 1 plus 2 d f by h into k s tan phi 1 divided by b plus because here gamma gamma 1 d f plus gamma 1 h is there. But in the original expression also this q u this minus gamma 1 h is there. So, minus gamma 1 h and plus gamma 1 h these are cancelled. So, only gamma 1 and d f that is present. So, only gamma 1 and d f that should be less than equal to the q t where q t value is gamma 1 d f n q 1 s q 1 plus gamma 1 d f half gamma 1 b n gamma 1 s gamma. So, this will give us the ultimate load carrying capacity of the soil for this condition. Now, in the next case or the case 2 that if top layer is stronger sand that means c 1 is equal to 0 and bottom layer is weaker sand. Here c 2 again is 0. In first case the c 1 was 0 and phi 2 was 0. In the second case c 1 is equal to 0 and c 2 is also equal to 0. So, here the q b that will be equal to gamma 1 because c 1 and c 2 both are 0. So, first term will be 0 for the both the cases gamma q b and q t. So, gamma 1 d f plus h into n q 2 is equal to 0. So, s q 2 plus half gamma 2 b n gamma 2 and s gamma 2. Similarly, q t will be gamma 1 into d f into n q 1 s q 1 plus half gamma 1 b n gamma 1 plus half gamma 1 b n gamma 1 plus half gamma 1 s gamma. So, similarly q 2 by q 1 here we can write because which is in the surface footing. So, this part is 0 because surface footing both the cases it is surface footing. So, second part is it is not that because this is here d f is 0, but this is top case is 0, but here also d f and h both are 0. That means for the bottom layer also it is placed at the surface considering that there is no surcharge basically. So, this is 0 this is this part is 0. So, q t by q 2 by q 1 will be half gamma 2 b n gamma 2 s gamma 2 divided by half gamma 1 b n gamma 1 and s gamma 1. So, this b n gamma s gamma the safe factor we are considering this is same. So, ultimately this will be gamma 2 into n gamma 2 and gamma 1 into n gamma 1. So, now how to calculate the ultimate load carrying capacity that is q u that means q b is gamma 1 d f plus h n gamma 2 n q 2 s q 2 plus half gamma 2 b n gamma 2 s gamma 2. Again where s q s gamma these are the safe factor plus because this here the c is 0. So, this c is 0. So, this ultimate the general expression the c term is also 0. So, that part is gone. So, gamma 1 h square into 1 plus b by l 1 plus 2 d f divided by h into k s tan phi 1 divided by b minus gamma 1 h that is less than q t where q t expression is given by this formula. So, that means we will get. So, that would not be get. So, we have to check first we calculate the q u and then we will check by using this q t whether q u is less than equal to q t or not. So, this should be less than equal to q t. If it is not then we have to use this q t as our ultimate bearing capacity of the 40 under this layer soil condition. Now, the second the third case that is the case 3. In the third case the condition is the top layer is stronger saturated that means phi 1 is 0 and the bottom layer weaker saturated clay that means phi 2 is also 0. So, in the case 3 phi 1 is 0 and phi 2 is 0. Then that means and q t. So, you have to calculate q b and q t both. So, q b as this phi 1 phi 2 is 0. So, this part 1 plus 0.2 b by l into 5.14 c 2 plus gamma 1 into 2 d f divided by d f plus h sorry this is h. So, gamma 1 d f plus h because this term this n gamma 2 that will be 1 and as phi 2 is 0 that n q 2 is 1 and n gamma 2 is 0 and here n c 2 is 5.14. Similarly, q top it will be 1 plus 0.2 b by l this is 5.14 similarly this is plus gamma 1 into d f. So, ultimately we can write that q 2 by q 1 that ratio where it is in the surface. So, if it is in the surface then this part is gone. So, this part is 0 again. So, that means this will be 1 plus 0.2 b by l 5.14 c 2 sorry this one will be c 1 c 2 divided by 1 plus 0.2 b by l 5.14 c 1. So, this part is 0 again. So, this part is 0 again so that means the ratio will be c 2 by c 1. So, now the ultimately the q u that will be the q b expression is 1 plus 0.2 b by l to 5.14 c 2 again this gamma 1 d f plus gamma 1 h is present and 1 is minus gamma 1 h. So, both are cancels only gamma 1 into d f will be present. So, now plus 1 plus b by l 2 c a h by b plus gamma 1 d f that is equal to less than q t because as this phi 1 is also 0. So, that general term that tan phi 1 will be 0. So, that part is gone. So, only this part will be present. So, this we can write in this form for the rectangular footing. Now, the question is how we will calculate this suppose for this case 4 we know that q 2 by q 1 is equal to c 2 by c 1. Then how we will use this charts and how we will use this is q 2 or this is q 2 by q 1. So, that is c 2 by c 1. Now, if we go for this chart that this is the chart which is presented by mayor of in 1978 is also the sources this book. Now, here this is as I have mentioned that this punching shear coefficient case is function of q 2 by q 1 and phi 1. So, punching shear coefficient function of q 2 by q 1 and phi 1. So, if we know the phi 1 value this is the top layer frictional coefficient and if we know the q 2 by q 1 that is equal to. So, these are the different value 1, 0.4, 0.2 and 0. So, by if we know this phi and corresponding different q 2 by q 1 we can determine the case. Now, this q 2 by q 1 in the different cases this case 1, case 2, case 3 I have a different explain the how to determine this q 1 and q 2 value this ratio. So, first we have to calculate this q 2 because in the final expression this case term is present. So, that means in the case we have to determine. So, to determine the case value we should know how to calculate this case. So, first step is because we know this gamma 1, c 1, phi 1 this is the properties of the top layer and then the gamma 2, c 2 and phi 2 these are the properties of the second layer. So, these parameters are known to us. So, the help of these two parameters or these things first we will calculate the for the safe factor for different because the dimension of the protein that is also known to us. So, that means with the help with the if it is not known then by trial and error method we have to determine these things that how to use the trial and error method for any design of this foundation that part I will explain when you study the design of this foundation part and then once we get this dimension. So, suppose here we assume this dimensions are known. So, once we get the dimension we can calculate this with the help of phi value we can calculate bearing capacity factors for top layer and the bottom layer then with the help of dimension we can determine the safe factors for the top layer and the bottom layer. So, if these things are known to us then we can determine what is the ratio of q 1 and q 2. So, once we get determine the value of q 1 by q 2 this ratio and then we know the phi 1 value that is the top layer then we can by using this table or this chart q 1 corresponding to phi 1 we can determine. So, suppose this is q 1 by q 2 is 0.4 and theta is 35 degree. So, that means the case value will be 5. So, that means here this case value will be required to calculate the ultimate load carrying capacity of this layer soil. So, that case term is there as in the as you have seen then in this final expression. So, there we will use the case. So, that is for the if this phi 1. So, top layer if it is stronger sat because it is we are getting the case for the phi 1. Now, if the top layer is cohesive soil in the last case because the top layer is stronger saturated clay where the phi 1 is 0 then how to calculate this property because here the phi 1 is 0 then you have to use this chart because in that case we should know the CA value because this CA is CA value but for the if the strong the top layer is a stronger sand then this CA part or the cohesion part that is 0 because we are talking about the purely cohesive soil or cohesion less soil. So, if that is stronger sand then q phi 1 is present but C 1 is 0. Now, that means the cohesion or the adhesion is 0 then we have to use for the stronger sand is K s value. So, by using the first chart if both the cases we have to determine this q 2 by q 1. So, once you determine the q 2 by q 1 if the top layer is stronger sand then we should know the value of K s. So, in the stronger sand we can calculate q 2 by q 1 and we know phi 1 value. So, by using the first chart or first graph we will get the K s value. Now, in the second case if the top layer is stronger saturated clay then K s term is not required but CA term or adhesion term that is required because here we know that C 1 value but q phi 1 is 0. So, once we get the C 1 value and here also we can calculate because in the third case we have seen that q 2 by q 1 is equal to C 2 by C 1. So, now we know the C 1 and C 2 value. So, we can calculate the C 1 and C 2 value ratio. Once we get that the C 1 by C 2 value that ratio that is equal to the ratio of q 2 by q 1. So, that q 2 by q 1 is C 2 by C 1. So, this C 2 by C 1 is known because q 2 by q 1 is C 2 by C 1. So, once we get the C 2 by C 1 we will get the q 2 by q 1. So, and now using this chart for this case also we can determine q 2 by q 1 then by this using this chart then we know the C 1 value. So, only unknown is C A. So, suppose for the q 1 by q 2 is 0.4. So, C A by C 1 is 0.9. So, that means C A will be 0.9 into C 1, but here C 1 is also known. So, in this way we can determine the value of different parameters. So, once we get the C A then in the final expression that C A term is required. So, we will get the put the C A value there and we will calculate the ultimate load carrying capacity of the soil. Because all the other terms the thickness, dimension or the type of properties those are known only main unknowns are case and C A. So, those two things we can determine by using this chart. So, by in different cases we can we have to judge whether where we will use the first chart and where we will use the second chart. If it is stronger sand in the top layer then we will use the first chart because we need to find the case value. If it is stronger clay then we have to need to determine the C A value then we will use the second chart, but both the cases we have to determine Q 2 by Q 1 and I have explained how to determine this Q 2 by Q 1 in different cases. So, in this fashion we can when in this way we can determine the ultimate bearing capacity of the layer soil. In the next class I will discuss how to determine the ultimate load carrying capacity of foundation in slope and then we will go for the analysis or the because you know most of the up to this the most of the bearing capacity calculation we have done for the isolated footing. Or next one we will go for the raft foundation then how to determine the load carrying capacity of the raft foundation. Then we will go for the settlement calculation because we are always talking about the bearing capacity then the next criteria design criteria settlement. So, those things we will discuss in the next classes. Thank you.