 and how to calculate the bearing capacity of the foundation in layered soil and today I will solve one example on that particular type of bearing capacity calculation and how to calculate the bearing capacity of that particular foundation if it is resting or it is on the layered soil. Now next then I will discuss about the bearing capacity calculation on foundation if it is placed on the top of the slope. Now first I will go for the problem for this bearing capacity calculation in the layered soil. Now in this example that is example 10.1 this lecture 1 first example that if a square footing this is ground line or ground surface the depth of foundation df is equal to 1 meter and the it is a foundation it is not square it is a rectangular footing of dimension say width of the footing is 1.5 meter and dimension is 1.5 cross 2 meter. Now this footing is placed on a layer which is stronger clay whose cu value cu1 is equal to 100 kilo Newton meter square and gamma 1 is equal to 18 kilo Newton meter cube. Gamma is the unit weight of the soil cu is the cohesion of the layer 1. So, this is layer 1 which is stronger clay and this is layer 2 which is softer clay whose cu2 is equal to 30 kilo Newton meter square and gamma 2 is 17 kilo Newton meter cube. Now distance from this base of the foundation up to the layer 1 is 1.5 meter. So, this is the depth of the soil from the base of the foundation. So, this is a layer 1 layer 2 layer 1 is stronger clay layer 2 is softer clay the cohesion of the layer 1 is 100 kilo Newton meter square and gamma 1 is 18 kilo Newton meter cube cohesion of layer 2 is 30 kilo Newton meter square and gamma 2 is equal to 17 kilo Newton meter cube and dimension of the footing 1.5 cross 2 meter depth of foundation is 1 meter and the depth of the lower of this of this first layer is 1.5 meter from the base of the footing. Now we have to determine the bearing capacity of this soil of this foundation in the layer soil system. Now as I have already derived the expression for this ultimate bearing capacity calculation for this type of this is a stronger clay and softer clay both the layers are clay then this ultimate bearing capacity with considering the size effect will be 1.2 B L into 5.14 Cu 2 plus 1 plus B by L into 2 C A into H B plus gamma 1 D F and Q T bearing capacity of the top layer is 1.2 B by L 5.14 Cu 1 plus gamma 1 D F and condition is I have already explained this thing that Q U 2 should be less than equal to Q T. So, Q U ultimate load carrying capacity is less than equal to Q T. Now the calculation part here from this it is derived that Q 2 by Q 1 that if the footing is placed in the surface and this is the bearing capacity of that surface footing in the second layer and this is the bearing capacity of the surface footing in the first layer. So, under this both the clay condition if both the layers are clay in Q 2 by Q 1 is equal to C 2 by C 1. Now here C 2 or Cu 2 or we can write Cu 2 by Cu 1. Now here Cu 2 is equal to 30 divided by Cu 1 is 100. So, this is 100. So, this value is equal to 0. Now in the in this previous lecture I have shown this graph where this is the graph that is presented. So, this is Q 2 divided by Q 1. Now from this particular problem that Q 2 divided by Q 1 is equal to 0.3. So, this is approximately 0.3. So, corresponding C A divided by C 1 is equal to 0.85. So, this is the value 0.85. So, corresponding this value we can say from the graph C A by C 1 is equal to 0.85 where C A A is the adhesion of the soil. So, from here we can calculate or we can write that C A is equal to Cu 1 into 0.85. So, Cu 1 is 100 into 0.85. So, this is coming out to be 85 kilo Newton per hour. So, this is equal to 0.5 kilo Newton per meter square. Now, so from this expression we know B L. So, Cu 2 is 30. Gamma 1 is 18 kilo Newton per meter cube. D f is 1 meter and Cu C A we have calculate this is 85 and H value this H value is equal to 1.5. This H value is basically the depth of the first or starting point of the second layer from the base of the foundation that is equal to H. So, from this particular case H is equal to 1.5 meter. So, that the starting point of this second layer is 1.5 meter below the base of the foundation. So, this H value is 1.5 meter. So, now if we calculate. So, there we will get the Cu U value and Cu T is the top layer bearing capacity. This is the total bearing capacity considering both the layers. Now, if we put this value in our calculation that Cu U that is equal to 1 plus 0.2 into 1.5 by 2 here into 5.14 into Cu 2 that is 30 plus 1 plus 1.5 divided by 2 and 2 into C A is 85. H is 1.5 divided by B is 1.5 plus gamma 1 is 1.5 plus gamma 1 is 18 into df is 1. So, now if just we are putting this value B is equal to 1.5, L is equal to 2, Cu 2 is equal to 30, then C A is 85, H is 1.5, B is 1.5 and gamma 1 is 18 kilo Newton per meter square gamma df is 1 meter. So, the total value after the calculation we are getting 492.83 kilo Newton per meter square. Similarly, Cu T we can calculate just putting this value 1 plus 0.2 into B is 1.5 divided by L that is 2 into 5.14 into Cu 1 divided by B that is Cu 1 is 100 plus gamma 1 is 18 into 1, 1 is the df. So, the after the calculation this value Cu T is coming 609.1 kilo Newton per meter square. So, from this calculation we can say that Cu U is less than Cu T. So, it is satisfying this condition. So, we can write that Cu U that load ultimate load carrying capacity of this foundation resting on this layered soil is 492.83 kilo Newton per meter square. So, this is the total value after the calculation we are getting. Now, if we apply the factor of safety that is Cu safe or Cu allowable that 492.83 we consider the factor of safety 3 this 3 is the factor of safety. So, we will get this value is 164.3 kilo Newton per meter square. So, ultimate load carrying capacity or Cu allowable for this total load carrying capacity will be 164.3 into 1.5 into 2. So, this value is in 492.83 kilo Newton per meter square. So, here we can calculate the bearing capacity of the foundation for the layered soil system. So, similarly for the other conditions also if the one layer is clay and one layer is sand if both the layers are sand in those cases also we can calculate the bearing capacity of the foundation in similar process. So, next thing is that if bearing capacity we want to calculate for the here we have calculated the bearing capacity of the foundation for layered soil system. Now, if the foundation is placed on the top of a slope then how to calculate the bearing capacity of this soil. So, now for this we can write that is bearing capacity of foundation for the placed on suppose this is the slope. So, beta is the slope angle and H is the slope height and if one foundation is placed near the slope. So, this is the foundation which is placed near the slope. Now, this foundation width is B and Cu is the ultimate load carrying capacity of this foundation and D f is the base of the footing or depth of the footing on top of this slope. Now, if we can extend this H. So, this distance is equal to B and you can write this or this distance from the H of the slope to the H of this footing. So, this distance say B small b. So, small b is the distance between the H of the slope to the H of the footing and capital B is the distance or the width of the footing. Now, first we have to calculate few things that first we can calculate the NC which is stability number or we can write this is gamma H divided by C where gamma is the unit weight of the soil and C is the cohesion. Now, next condition that we can we know that Cu U is C NC plus half gamma B N plus half gamma B N. For N gamma Cu for the surface footing. Now, if C is equal to 0 then Cu U will be half gamma B N gamma Cu for surface footing. Now, if phi is equal to 0 then Cu will be C NC. So, this is also Cu. So, Cu is the this N gamma Cu NC Cu these are the also bearing capacity factors under this condition. That means, when the foundation is placed on the or near the slope. So, now how to calculate the bearing capacity. So, now for those two conditions if C is equal to 0 that means, if this is a granular soil that means, this case ultimate load carrying capacity is half gamma B N gamma Cu and if phi is equal to 0 then Cu U will be C NC Cu. So, here we know the cohesion value that is the properties of the soil this embankment soil and B is the width of this footing gamma is the unit weight of this soil. So, only unknowns are this N gamma Cu or NC Cu. So, these two unknowns or two bearing capacity factor we have to determine under two different conditions that means, C equal to 0 and phi equal to 0. Now, for this calculation we have to calculate that different parameters that D F divided by B. This is one ratio that we have to calculate and another one is B small B divided by capital B. So, we have to calculate the stability numbers value that we can calculate then D F by B and then B by B and we have we should know the value of phi also. So, if we know this parameter. So, then we can use this chart. So, these are the two different conditions that is one for C equal to 0 and one for N gamma Cu and another one phi equal to 0. So, may have proposed this bearing capacity factor for foundation resting on slope. So, now for if C equal to 0 then you have to calculate this N gamma Cu. Now, here for to calculate the N gamma Cu you should know the value of the ratio between the small B by capital B and these charts are presented for various values of phi. So, this phi and beta beta is the slope angle. So, this varies from 0 degree to 40 degree. So, 0 degree 20 degree and 40 degree. Similarly, phi is for three different phi values phi equal to 30 degree phi equal to 40 degree and these two different phi values these charts are presented and this firm line presented that D F by B is equal to 0. That means, if the depth of foundation is 0 that means, it is placed at the top of the slope and another this dotted line which represents the D F by B is equal to 1. If the depth of foundation is equal to width of foundation. Now, for first one the dotted lines for the if the depth and width of foundations are same and these are the beta value from 0 degree. So, for if phi is 40 degree and beta is 0 this is the line. So, if we know that B small B by capital B value then N corresponding to phi and beta value. So, what are the conditions that D F by B is 0 or 1 say we can calculate what would be the value of N gamma Q or bearing capacity factor which is proposed by the mayors. So, once we get this value then we have to calculate we can easily calculate the ultimate load carrying capacity of the foundation under this condition. Now, these values are for 0 degree this is for 20 degree and this is for 40 degree corresponding to phi equal to 40. So, similarly for this firm lines corresponding to phi equal to 40 this is beta equal to 0 beta equal to 20 and beta equal to 40 degree. Now, if phi equal to 30 degree then this dotted lines this is for beta equal to 0 degree and this line is beta equal to 30 degree. Now, for the phi equal to 30 degree then this line is beta equal to this line is beta equal to 30 degree and this corresponding beta equal to 30 degree and this is beta equal to 0 degree. So, this is 0 degree and this line is for 30 degree. So, similarly by using this chart we can determine this bearing capacity factor of the foundation. Next that if phi is equal to 0, then in this case we have to calculate the stability numbers. Now, b by f. So, here again these are the corresponding to different stability numbers that n s equal to 0, this n s equal to 2 and n s equal to 4 and these are the beta value 0 degree, 15 degree, 30 degree, 45 degree, 60 degree and 90 degree and correspondingly here 30 degree, 60 degree, 90 degrees, here also 30 degree, 60 degree, 90 degree, here also 30 degree, 60 degree, 90 degree. So, different from 0 to 90 degrees, this is also 0 degree. So, these are the curves. So, again the dotted line represents the df by b equal to 1 and firm line represents df by b equal to 0 degree. So, once we know stability value then b by b by capital B value then this stability number or b by h value then we will get calculate the bearing capacity factor n c q which is proposed by Mirov by using this chart. So, once we get these two bearing capacity factors for two different types of soil then we can by using the expression we can easily calculate the ultimate load carrying capacity of the foundation. So, these are the different bearing capacity calculations. So, now I have calculated this bearing capacity, how to calculate the bearing capacity of the soil in layered soil if it is placed on the slope which is proposed by Mirov. So, now if I want to summarize the what are the factor by which the bearing capacity or on which bearing capacity ultimate bearing capacity of the foundation highly dependent. So, first one is the factor influence the bearing capacity we can write. So, here we are dividing these factors in two different types of soils. One is for phi equal to 0 and one is for c equal to 0. So, now if phi equal to first case for c u equal to 0 this is case 1 or. So, here we can write that ultimate load carrying capacity value or q u will be q into n q plus half gamma b n gamma. So, here we know q is equal to gamma. Now, from these expressions we can say that now this n q and n gamma these bearing capacity factors depend on the phi value. So, first influence factors we can say under this condition that for c equal to 0 that is relative density or phi value. Now, if relative density increases then phi value will also increase and if phi value increases then n q n gamma value those values will also increase. So, our ultimate load carrying capacity or q u that will also increase. Now, the next one is we can say this b term is present there with that is b. So, b also influences the bearing capacity calculation that means bearing capacity of the soil. Now, if here b is the factor if b from this expression we can say the b also influences the bearing capacity. Next factor that is our d f or depth of foundation or d f value. So, that means it also influences the bearing capacity. The next one is unit weight of the soil that is gamma this is the gamma values are there that is gamma. So, here we can see this gamma represent the soil or the unit weight of the soil above the foundation base and this gamma representing the unit weight of the soil below the foundation base and then next one is the position of ground water table. So, I have already explained that if the ground water table position changes then this q u value that will also change or the ultimate load carrying capacity or bearing capacity value that will also change. Now, if this position of the ground water table is at the ground surface. So, that will give the minimum bearing capacity of the foundation and now if this position of the ground water table is below the width equal to the below the width of the foundation from the base. So, that means in that case the no influence of the ground water table is observed in the bearing capacity calculation. So, that means the if we position of the ground water table is greater than or equal to the width of the foundation from the base of the footing then the influence of this ground water table on bearing capacity is neglected. So, next case for phi u equal to 0 then you can write that q u is equal to q plus c u n c and this ultimate load carrying capacity. Similarly, we can say that net ultimate will be c u n c. Now, n c value is given that 5.14 to 5.7. So, this is the range of this n c value if phi is equal to 0. Now, this 5.14 that smooth base of the foundation and this is 5.7 is for the rough base of the foundation. So, here we can say from this expression that for first we consider this is the two expressions one is ultimate load carrying capacity of the soil and the next one is equal to ultimate bearing capacity of the soil. So, in these two conditions we will that this is the net ultimate bearing capacity expression and this is the ultimate bearing capacity expression. So, there are two expressions that we are considering. What will be the factor that is affecting that we can calculate considering these two bearing capacity expressions? So, first we consider that relative density or the phi value. So, here from these two expressions one is q u and one is q n q that is q u q ultimate and q net ultimate. Now, if the relative density of the soil that means, the here n c is 5.14 for smooth base and 5.7 for rough base. So, phi value is not influencing this not neither q or nor c u. So, that means it is not even for this other expressions also that means the relative density if it is a cohesive soil the relative density has not any influence on the bearing capacity. Now, second one is the width of the footing from these two expressions we can say that width of the footing is not influencing the bearing capacity of the foundation if it is resting on the clay soil because here this B term is not present. Now, next one is the depth of foundation here we can say that net ultimate capacity is independent of depth of foundation. Now, depth of foundation is not influencing that to calculate the q in u or net ultimate bearing capacity of the clay soil, but if we want to find the ultimate bearing capacity of the soil then q term is there which is where q is equal to gamma into d f and their depth is influencing. So, depth is influencing for q u not q n u. Similarly, unit weight of the soil the same thing it is influencing this bearing capacity calculation for q u not q n u because this q value is gamma into d f. And for net ultimate position of ground motor table. So, when we calculate the q value this position of ground motor table will also influence the this q calculation that means it will also influence the bearing capacity. So, that means the position of ground motor table influences the bearing capacity calculation for clay soil if we want to calculate q u but not on q n u. So, from this observations we can say that q n u only depends on the dependent on c u, c u value because as n c varies from 5.14 to 5.7. So, only parameter that influences the bearing capacity calculation for n u that is c u. So, q u q n u only dependent on c u not any other factor. So, now from this completion we can say that these are the different factors which is influencing the bearing capacity of the sandy soil if the c u is equal to 0 these are the factors and for 5 u equal to 0 for clay soil. So, next we will start new section that is the settlement calculation because this now till now we have discussed that what are the different methods by which we can calculate the bearing capacity of the foundation. Now, next how to calculate the settlement of the foundation because as I have mentioned that settlement and the bearing capacity two are the most important factors or criteria for design of any foundation. So, then how to calculate the settlement of the foundation that we will discuss in the next section. So, the settlement calculation of shallow foundation. So, this is the how to calculate the settlement for the shallow foundation first that part we will discuss. So, when you go for the settlement of the shallow foundation there are different types of settlement that we will get for the shallow foundation that one is first settlement is immediate settlement or elastic settlement that we can write this is yes. So, this is the immediate settlement that when we apply the load on a soil through the foundation the just after the application of load the settlement that immediately will occur that is called the immediate settlement generally it up to or less than 7 days or we can say this is for the 7 days it is that this immediate settlement basically it is completed within the 7 days time. So, this is the change due to the change in shape of the soil without change in the volume. So, this settlement is due to the change in the shape of the soil without change the volume. The next one is that is the primary consolidation this is S c and this is due to the consolidation process. So, so that means once we will get the immediate settlement then we will go for the consolidation settlement. So, then this is the time dependent settlement. So, next the third category settlement is secondary compression settlement or SS. This secondary first is due to the change of shape of soil without change in the volume this is due to the consolidation and this is the volume change occurring due to rearrangement of the soil particles. So, this secondary compression settlement is due to rearrangement of the soil particle and volume change occurring due to this rearrangement of the soil particle. So, the total settlement that we will get is a summation of immediate settlement consolidation settlement and secondary compression settlement. Now, from this different types of soil so that means the total settlement that we will get this is the combination of this three or the summation of this three settlements. Now, from if the contribution of this total settlement the contribution of each parts on this total settlement that depends on the soil type. For example, if it is a granular type of soil then immediate settlement or elastic settlement is much higher compared to this other two types settlement. Similarly, for if it is a clay soil especially for inorganic clay then primary consolidation settlement is the maximum amount of settlement compared to the other two types of settlement. Similarly, if it is a organic clay then secondary compression settlement is more compared to the other two types of settlement. So, in short we can say that for the granular soil immediate settlement contribution is more and for the clay soil that primary consolidation settlement is more. Now, if we draw this one graph that we can say that this is the settlement and this is the time and this is settlement this is settlement this is the time t. So, immediate settlement suppose this is the point of application. So, immediate settlement you get within 7 days. So, you can say this is the time requirement is very less after that. So, we can say this primary consolidation settlement and secondary compression settlement both are time dependent settlement, but immediate settlement that is not time dependent settlement. So, if I draw this graph we will get this type of settlement. So, that means from up to here this is equal to the immediate settlement and suppose this point which is the 100 percent consolidation or t 100. So, from here to here this is the consolidation settlement and this is secondary compression settlement. So, this is the three parts one is immediate settlement then primary consolidation settlement then compression settlement. So, this is that you will give the total amount of settlement or s t. Now, we have to calculate these three settlements or the total settlement or different parts of this immediate settlement then consolidation settlement to find the total settlement. So, next one is that that when we calculate the settlement that means we are applying suppose this is the foundation. So, before we start the settlement calculation then you should know how to calculate the increment of load due to application of increment of stress due to application of this load. Suppose this is the foundation and we if we consider that at a point say A. Now, if in this condition if we can neglect the water temperature table effect then this stress at A point that is gamma into H. Suppose this height of this point from the ground surface is H. So, this is the stress at point A before the application of the load or we can say this is the effective overburden pressure at point A before the application of the load. So, and we are not considering water table effect here. So, now one once we apply this load through footing then definitely at A point an additional load that will act due to this foundation load. So, this is the increment of this increment of load due to apply application of this foundation load at any depth below the base of the foundation. So, from this foundation base suppose this is the df depth of the foundation. So, from this zone this stress increment will occur because of this foundation load application. Now, there is a influence zone that we consider when we calculate the bearing capacity and the settlement. Generally for the bearing capacity calculation we consider the influence zone up to B for the bearing capacity calculation. So, for the bearing capacity calculation we consider generally this influence zone is up to B which is width of the foundation. So, suppose this is width of the foundation. So, this is for the bearing capacity calculation. Now, for the settlement calculation we consider this zone which is twice B from the base of the foundation. So, for the bearing capacity calculation this influence zone is B and for the settlement calculation we consider this influence zone is twice B. So, that means B plus this is for the bearing capacity. So, this is for the 2B for the settlement calculation. And in this lecture I will consider this same thing that for the bearing capacity I will calculate up to zone B and for the settlement calculation I will consider that zone 2B in general case. So, that means now if any point within this we want to calculate the settlement of this total soil that means we have to calculate the stress increment at this zone twice B from the base of the footing. So, that means at any point within this zone if we want to calculate the increment of stress due to application of this footing load then how to calculate. There are few methods by which we can calculate this increment of load or stress due to the increment load then how to calculate this. Now, first concept that will apply that is a stress due to a concentrated. Now, Bruxelles equation we can use 1885 propose this expression that if one load suppose this is the ground surface. And here this is x axis this is y axis and this this is z axis this is on the surface of the ground x axis y axis z axis. And if one load is applied here at this point so what will be the load or stress increment at this point that we have to calculate. So, now this point is at a depth of z. So, that means the coordinate of this point x y z then how to calculate this load. So, that means this is y and this is x and radially this is r. So, and from this point this is the position of the point from the application of the load. So, that means this point is x distance and y distance and z distance from the origin. Origin means where this concentrated load is applied. So, now according to Bruxelles expression that del p the increment of load at this p point say at p point this is the del p we can say that 3 p divided by 2 pi z square 1 plus r by z 3 power square divided to the power 5 by 2. Now, where you can say r will be given by root over x square plus y square. So, now if we know this x and y. So, and similarly we will get the r value root over x square plus y square and we know the z value. So, if we apply this expression put this value then we will get the stress due to the concentrated load at any point below this point of application of the load. So, similarly by using the Bruxelles expression we can use the we can calculate the stress increment for circular loaded foundation. Then you can calculate for the rectangular loaded region then at any point what would be the stress below this loaded region. So, this we will get in the soil mechanics course that because you have already done this thing then how to calculate this stresses for a circular loaded area or a rectangular loaded area for the point load. Then that means for the footing if it is circular footing or steep footing or rectangular footing or square footing then using those technique by Boussineq proposed by this Boussineq we can determine the stress at any point below the loaded region. So, that you will you have done for any soil mechanics course that technique you can use to calculate this stress. The next technique that the approximate method which is very useful for the use to calculate this stress at any point. Suppose this is the foundation and this is ground surface this is q 0 is the load and d f is the depth of the foundation. Here we consider this load is distributed by 2 is to 1. So, at any depth if you want to determine the stress at any depth suppose this as a center of this point. So, at any depth so this will be the stress and the stress amount will reduce as we increase the depth. So, if as depth is increased then the stress which is acting at this foundation base level that will reduce. And here it is assumed that this distribution is 1 is 2 that means 2 vertical and 1 horizontal. Now, if that depth is z from the base of the foundation then the stress increment will be q 0 into B. If this is the width of the foundation is B and dimension of this foundation B cross L is the dimension of the footing. So, d p will be p 0 into L divided by B plus z into L plus so that means in this way we can determine the increment of stress due to this footing. So, either we can use the procedure or by you can use this approximate method this is one approximate method to calculate the increment of stress at any point below the foundation base. So, in the next class I will explain how to calculate the settlement that immediate settlement and consolidation settlement what are the corrections that you have to apply to calculate those settlement for different types of soils. So, that settlement calculation I will explain in next class. Thank you.