 Now, today I will discuss about the design of shallow foundation, because till now I have discussed about the bearing capacity calculation and the settlement calculation, because these two are the main criteria of the design of foundation. Now, today I will explain how to use these two criteria as a same time, that is bearing capacity as well as the settlement to design of a shallow foundation. Basically, this lecture I will try to design isolated footing. Now, here design means that I will not be discussing about the reinforcement detailing and reinforcement part of the design. Here, I will basically discuss about the how to choose a dimension of the footing and the depth of the footing based on the soil parameters considering and satisfying the settlement as well as the bearing capacity calculation. Now, before I start about the design part, now here I will give you an example, then how to apply the calculation or how to use the plate load test data, because last class I have discussed about the plate load test and how to get the settlement and bearing capacity from this plate load test. Here, I will give one example, where we can see that from the plate load test data, this is the plate load test. Using this plate load test data, how to determine the settlement of a real footing. Now, suppose after the plate load test, this is the load versus settlement graph, this is load, this is say ton per meter square or any other unit and this is the settlement, this is in millimeter. Now, suppose we will get this type of graph from the plate load test. Now, as I have already explained for this type of graph, we can determine the ultimate load carrying capacity of this plate by using this double tangent method. So, the extend the state initial state portion and extend the final state portion, the intersection point the corresponding load intensity is the ultimate load carrying capacity of the plate. Now, from this plate load test data, we have to determine the settlement of a real foundation. Suppose, the here the size of the plate or plate size was a 60 centimeter and the plate is placed the depth of 1 meter below g l, the plate is placed at a depth of 1 meter in level. That means, the condition is plate size is this one and depth of the plate is 1 meter, which is plate is under the real footing, the footing size is say 2.5 meter cross 2.5 meter. So, real footing dimension is 2.5 meter cross 2.5 meter and depth of footing that is d f is 2.5 meter, this is also 2.5 meter. So, d f also 2.5 meter, total load on footing is say 150. So, in the real case the total load that is acting on the footing is 150 ton, size of the footing is 2.5 cross 2.5 square footing and depth of the footing is 2.5 meter. The footing depth is 2.5 meter, where the plate is placed at a depth of 1 meter below g l and size of the plate is 60 centimeter, this is also square plate. Now, the load intensity actual load intensity that we can. So, this is on the footing that will be 150 is the total load divided by the dimension is 2.5 and 2.5. So, this is coming 24 ton. So, here these are the points of the plate load test that we are conducting. So, now, from here the 24 ton per meter square the loading intensity. Now, from here, this is the load is 2150 ton now, where we have to determine what would be the settlement of this real footing based on this plate load test, because we will do the plate load test whose dimension is 60 centimeter depth is 1 meter and we will get this type of load versus settlement curve. And in the real footing, this is the real footing data, all is the real footing data, this is size and here the total load is 150 ton. So, that on the footing the loading intensity is 24 ton per meter square. Now, we have to determine what will be the settlement of the real footing under this 150 ton load. Now, this 24 ton per meter square loading intensity. Now, from this graph of this plate load test, we have to determine what is the settlement of this plate corresponding to this 24 ton per meter square loading. Suppose, this is the 24 per ton per meter square value. So, corresponding suppose this is 24 ton per meter square, this point, this is the loading intensity. Now, corresponding settlement that we have to determine from this plate, say suppose this settlement is 12 millimeter. So, in the real footing, this settlement corresponding to 24 ton per meter square is 12 millimeter. So, now we can say the settlement of plate that is S p corresponding to 24 ton per meter square loading intensity is 12 millimeter. Now, we have to determine, so S p, so S p is 12 millimeter. Now, if I use this expression that S f is equal to S p into B f B p plus 30 B p B f plus 30 to the power square. So, this is the settlement of the real footing here. So, here the known value S p is 12 millimeter, B f with of the footing is 250 centimeter and B p is 60 centimeter. Now, if we put this value, we will get this is 12 250 into 60 plus 30 divided by 60 250 plus 30 to the power square. So, this value is going to be 21.5 millimeter. So, corresponding to this 150 total load or 24 ton per meter square, this 1 per meter square loading intensity, the footing settlement is 21.5 millimeter. Now, here we have to apply the depth corrections because depth corrections will be required because there is a difference of the depth of the plate and the depth of the footing, because if the plate was placed at a depth of 3 meter from the G L, then this depth correction is not required. Now, as this depth is the depth of the plate is 1 meter. So, there is a difference of 2 meter from the 2 depth. So, now, we have to apply the depth correction. So, suppose D root over L by B, this value is because depth correction is depth, difference of depth is 2.5 minus 1. So, this is the difference of the depth. So, this is the difference of the depth to root over 2.5 into 2.5 because this depth correction we are doing that basically because of the difference of the depth because here the settlement we are calculating is we are considering that it is at the surface level. So, at that the but the real footing is placed at a depth certain depth D f. So, that D f amount we have to apply some depth correction because of this depth difference from the surface and the real footing depth. But here the plate is placed at a 1 meter depth below G L whereas the real footing depth is 2.5 meter below G L. So, there is a difference of 1.5 meter from the plate and the footing. So, that 1.5 meter due to that 1.5 meter we have to apply this depth correction. So, here D is 1.5 meter root over L. So, this value is 0.6 and L by B here is is 1. So, the correction factor factor is coming 0.83 this is from the chart. So, the chart I have already explained or given. So, now the corrected settlement at would be say 21.5 into 0.83. That is 17.85 millimeter. So, the corrected settlement of the or actual settlement of the real foundation corresponding to the total load 150 ton is 17.85 millimeter is calculated based on the plate load test data. So, in this way we can determine the settlement of the found real foundation based on the plate load test data. The next part that I will discuss about the design of one isolated footing. So, for this design purpose we should know some permissible value. So, that is the first criteria is our bearing capacity criteria and second one is the settlement criteria. So, we have to satisfy both these criteria when we choose the dimension of a footing and depth of the where we place that depth of the foundation. So, that we will discuss. So, now for the first criteria is the bearing capacity criteria. So, there we have to apply one factor of safety ranging from 2.5 to 3. So, factor of safety means the load carrying capacity of this soil under this footing or loading or footing condition and the load intensity that is applied on this footing. So, that means the load carrying capacity of the footing is 2.5 to 3 times that should be 2.5 to 3 times higher than the load that is coming on this soil. So, that means we have to apply one factor of safety is 2.5 to 3. Next is settlement calculation. So, IS code recommends that. So, for the different types of foundation whether it is a isolated footing or it is a WAP footing and the different types of soil whether it is sandy soil or steep clay or it is a soft clay or plastic clay. So, the IS code recommends few permissible settlements. So, that means we have to go for the permissible settlement that means when we do the design have to keep in mind that the settlement of the foundation that should not be beyond this permissible settlement. So, the settlement of this foundation should be within this permissible settlement. So, I will consider in this design the IS code recommendation for this permissible settlement. So, that means that element should not be within should be below this permissible settlement as well as the bearing capacity that here that factor of safety minimum 2.5 have to satisfy. Now, here so there are different types of settlements are we have to mention. So, that one is our total settlement the another one is the differential settlement then the next one is the tilt or the angular distillation. So, then I will first discuss about the different types of settlement and what are the recommendation of IS code on that regarding the permissible value of these different types of settlement then I will start the design. So, first if I go for this different types of settlement. So, now for the settlement part now suppose this is the footing. So, one is first one is total settlement or S, S t means total if it is uniform that means this is the ground level a load is applied. So, there is a uniform settlement of this footing. So, that means total footing will come down here. So, this is the new position of the footing after the load is applied. So, this amount is the total settlement S t if it is uniform. So, that means first one is the total settlement if there is a uniform settlement of the footing that is the total settlement. The next one is the differential settlement. Now, suppose this is the two footings or column this is footing one this is the ground surface and this is another footing. So, load are applied here this two and this is the distance between the column is L. So, this is also G L. So, suppose there is a deformation of the footing suppose this is the deflected shape of deflected position of the footing for the first footing. So, this is the amount of the settlement for the first footing. So, this is S one is the total settlement of the first footing this is and suppose there is a further deformation of this footing. So, this is the amount of deflection. So, this is S two a total settlement of the footing or the second footing. So, there is a difference between the after the deformation there is a deflection difference. So, that amount is this one. So, this is the delta or the deflect differential settlement of this system. So, that means, this is the settlement because as we know that for the different types of footing the loading condition may be different and when you are calculating this footing the soil beneath this footing one and footing one they may not be same. So, if there is a different type of soil and the loading condition if the soil condition is not same between the two footings if the loading condition is not same on this loading that is coming on the footing then there is a chance there is the amount of settlement for these two footings may not same. So, that means, there is a differential settlement between these two footing. So, that is called the differential settlement. So, that we have to also calculate now if you want to make these two settlements same then you have to design our foundation such that under the soil condition beneath the two different types of footing or do different footing because the soil condition may not same. So, that means, we have to design our footing such that the under the two different soil condition and the two different loading condition also the settlement would be same. So, in that case there will be no differential settlement. So, if you want to avoid the differential settlement then you have to consider or design that two system such that there will be no such differential settlement. So, that means, the total settlement of one column and total settlement of the other column that should be same or within the permissible limit. So, that means, here the differential settlement is delta is S 2 minus S 1 and if there is a tilt or the if there is a tilt or the this total system deform uniformly then this one is a settlement and the last one is the another one is the angular distortion that also we can suppose the total system suppose this is the initial position of one system or one building. Now, due to the difference about the here suppose there is a this is the deformation or the settlement of this building or the deformation shape of this building after the deflection. So, this is the deformation shape. So, that means, we can say suppose this is the settlement of the first or the this corner is S 1 and this is the settlement of this corner is S 2. So, again here also there is a differential settlement. Now, the difference of this differential settlement because here the total system is tilted or this is the angular distortion or you can sell this is the tilt. So, because of this tilt this building there is a angle that we will get. So, here also that delta is we will get S 2 minus S 1. So, the angular distortion all the tilt that we will get delta by R if the total length of this two corners are this is L. So, that means, there is a three different types of settlement. One is our total settlement this is uniformly settlement is S t and this is the differential settlement where the two columns there is a two different settlement. The difference of these two column settlements will give you the differential settlement and the angular distortion of the the total system tilts any any direction. So, I am there is because of this settlement difference the two different corners. So, this here the first corner difference is S 1 and the second corner is S 2. So, tilt will be delta by L where delta is S 2 minus S 1 L is the length between two corners. So, that means, when you design this all the principal value for the all the total settlement, differential settlement and angular distortion of the tilt we have to satisfy. Now, on the next thing is I will give the total permissible limit of this different settlement. So, first for the IS code recommendation this is for the IS code 1904 to 1978. So, according to this IS code if I draw this table. So, this is the footing and this is the type of sand we can say this is sand and hard clay and another one we can say this is the plastic clay. So, here also we will give for the maximum settlement or the total settlement then we will go for the differential settlement then we will go for the angular distortion. Here also we will go for the total differential and angular. So, these are the different permissible value that we will give for this is the footing. So, first if is the isolated footing. So, this is isolated footing that if this is resting on if it is resting on sand or hard clay then the permissible total settlement is 50 mm differential settlement is 0.0033 L and angular distortion is 1 by 300 where L is the length where is the length of deflected part of wall or wrapped or center to center distance between column. So, either L is the total length of deflected part of the wall or wrapped or center to say the distance between the columns. So, this is for the isolated footing. Now, if it is resting on the plastic clay this isolated footing then the maximum settlement this permissible value is 50 mm differential settlement again 0.0033 L and angular is 1 by 300. Now, here this is for the isolated it is isolated steel structure. Now, these conditions for the isolated steel structure. But next one is this is isolated this is isolated RCC structure I mean this is steel and this is concrete reinforcement concrete RCC structure. Now, again this permissible value for the maximum settlement or the total settlement for the sand or hard clay is 50 mm and then this is 0.0015 L is the differential settlement and angular distortion is permissible 1 by 666 and for the plastic clay or the clay which is 75 mm and this is 0.0015 L and this is 1 by 666 for the angular distortion. So, this is for the angular distortion. Similarly, the next one the raft foundation or mat foundation in steel for steel structure raft foundation the permissible total settlement on the sand it is 75 mm differential settlement limit is 0.0033 L angular distortion 1 by 300 and for the clay this is 100 mm this is 0.0033 L and this is also 1 by 300. Similarly, for the raft foundation or mat foundation RCC structure this permissible value for the total settlement on sand is 75 mm differential settlement is 0.002 L angular 1 by 500 on the clay plastic clay this is 100 mm this is 0.002 L the differential settlement and angular is 1 by 500. So, these are the table for the permissible settlement or the footing isolated steel on the sand and clay isolated RCC on sand and clay wrapped steel on sand and clay and wrapped RCC on sand and clay. So, total settlement difference for the isolated steel is 50 mm and RCC also 50 mm on sand wrapped for steel and RCC both on sand is 75 mm and isolated steel on plastic is 50 mm RCC is 75 mm and wrapped for both steel and RCC on the plastic clay for maximum settlement limit is 100 mm. So, now, we will design this isolated footing considering this total design consideration. So, suppose we will design this footing and this is the isolated footing we will design a particular footing. So, this is the footing so, here we will design or choose the dimension of the of a single footing or isolated footing and we will consider only the here the maximum settlement consideration because we are not considering the other footing and then if we are considering what 2 or 3 footing or column then you have to go for the differential settlement or that angular distillation. But, here we will consider we will design the only one section one particular column footing. So, that is why we will consider the permissible settlement of maximum settlement only and as well as the bearing capacity calculation. So, this is the footing which is the ground level or plus 0 meter this is ground level. Now, suppose the depth of the footing we assume is 2 meter and the position of the water table is also at 2 meter. Now, this is the first layer whose thickness is from the ground level is 4 meter and this is the second layer whose thickness of this is this would be minus because we are in downward direction. So, thickness is 6 meter and this is the third layer whose thickness is I mean 10 meter I mean 4 meters this thickness is 2 meter and this is the 4 meter. So, total 10 meter soil strata we have this is the layer 1 this is layer 2 and this is layer 3. So, the condition that is given this is for the silty clay. So, here c c by 1 plus e 0 value is given 0.07 for this layer gamma unit weight is given 18 kilo Newton per meter cube and this c u value is given 30 kilo Newton per meter square and e value we have calculated here all the clay soil because this is the all the clay strata we calculated e a considering 600 c u if I consider that thing that mean e value is given 18000 kilo Newton per meter square. Similarly, for the third strata the c c 1 plus e 0 value is given 0.15. So, these are the soil parameter after the testing these values are given for that and we will use these value for the design purpose. So, 0.15 then gamma is 19 kilo Newton per meter cube c u is 20 kilo Newton per meter square and e for the second layer is 12000 kilo Newton meter square c u also in meter square. For the third strata the c c for the third layer e 0 value is 0.12 gamma is also 19 kilo Newton per meter cube c u value is 50 kilo Newton per meter square and e for the third layer is given 30000 kilo Newton per meter square. Now, the pore water pressure coefficient for the all the average pore water pressure parameter a. So, so we can write the average pore water pressure parameter a is given 0.6. Now, we have to choose a proper dimension of the footing and depth of the footing. So, here for the first trial we choose that our dimension is 3 meter by 3 meter a square footing. So, that means the dimension that we will choose. So, that is b cross l is equal to 3 meter cross 3 meter and d f as I have already mentioned the d f value I have taken that at the 2 meter. So, depth of the footing is 2 meter. So, now considering the first trial we will consider this value we assume this is our assume value assume value for the first trial. Now, we will design this thing and we will say whether this assumption or this dimension is satisfying the bearing as well as settlement calculation or not that is our task. So, now for this thing this is the dimension of the footing. So, that means the for the bearing calculation zone will be up to b and for the settlement calculation zone will be up to twice b. So, twice b means up to 6 meters the influence zone for the settlement calculation up to 6 meters. So, this is the influence zone up to 6 meters and for the bearing capacity this is the influence zone up to 3 meter. So, I have mentioned for the bearing calculation we will consider the influence zone up to b. So, this is for up to b this is for b and this is for twice b. This we will use for the settlement and this is for the bearing. Now, bearing up to b means 3 meter and 2 b means 6 meter. So, now for the our as I have mentioned this is the assume dimension. So, now the net load that is acting and another thing the total load that is acting here the total load is 600 kilo Newton. So, total load that is acting on this footing or this column is 600 kilo Newton. So, that means the net footing pressure is 600 and our assume dimension is 3 by 3. So, this will be 66.7 kilo Newton per meter square. This is the assume this is the total loading intensity is coming on the footing base. So, first and we have taken the d f is also 2 meter. So, first we will calculate the bearing capacity. So, here we will use the Skempton bearing capacity expression that is for N net ultimate is C u N c. This is for the bearing net ultimate load using the Skempton expressions where N c is given 5 1 plus 0.2 b by L into 1 plus 0.2 d f by b or as it is a square footing. So, b by L is equal to 1. So, directly we can write this is 6 1 plus 0.2 d f by b. So, here d f we can write. So, this is 6 1 plus 0.2 into d f is 2 meter b is 3 meter. So, this is coming 6.8 which is less than 9 it is ok. Now, for the when you calculate the C value this is for the Skempton expression Skempton expression that we have used. Now, for the C u part we have taken the weighted C u value because from this figure we can see for the bearing capacity calculation this is the b is the influence zone. So, 2 this total 3 meter 2 meter is in the first layer and 1 meter is in the second layer. So, we have taken the weighted C value instead of taking this the first layer value because although the most of the influence zone is within first layer, but we have taken the weighted average value C value C u value that is for the 2 layer for the first layer 2 meter and 1 meter for the second layer. So, first layer C u is 30 and second layer C u is 20. So, weighted C u value will be 30 into 2 plus 20 into 1 divided by 3 because our width is 3 meter. So, influence zone will be up to 3 meter for the bearing. So, this value is coming 26.67 kilo Newton per meter square. So, q net ultimate will be C u value is 26.67 N c is 6.8. So, this value is 26.67 N c is 6.8 so this is 181.33 kilo Newton per meter square. Now, the factor of safety that we will calculate our net ultimate is 181.33 then divided the total load intensity that is coming is 66.7. So, the value is 2.7 is the factor of safety that is coming which is greater than 2.5 so you can say this is safe against bearing. So, the load carrying capacity net ultimate load is 181 or ultimate stress is 181.33 kilo Newton per meter square where the the so that and the load that is coming is 66.7 kilo Newton per meter square. So, factor of safety will be 181 divided by 33.3 181.3.3 divided by 66.7 kilo Newton per meter square. So, 2.7 which is greater than 2.5 so it is safe against bearing. So, next we will go for the settlement calculation or the settlement calculation and we will consider the same dimension of the footing. First we will calculate this is the B is the settlement calculation first we will calculate the immediate settlement. So, now here q N u is 181.33 kilo Newton sorry q net intensity is 66.7 kilo Newton per meter square that means this is the load intensity coming on the footing and B we have assumed is 3 meter mu for the clay soil have taken 5.5 that is we have taken in I f is 1.12 that we have for the I f is 1.12 that we have for the I f at the center that is from the table. So, this is from the table for the influence factor table that I have given. So, from there we will get the I f value is 1.12. So, here also that we have when you calculate the E because the immediate settlement expression is q N into B divided by E 1 by mu square into I f. So, we know q N is 66.7 kilo Newton per meter square B is 3 meter mu is 0.5 I f is 1.12. So, we have to calculate the E value when you calculate the E value here also we have taken the weighted average E value because for the bearing capacity we have taken the influence zone is for the bearing capacity we have taken influence zone is up to B for the settlement calculation we have taken influence zone is up to twice B. So, here first 2 meter for the first layer second 2 meter for the second layer and third 2 meter is for the in the third layer. So, for the calculation of E we have taken the weighted value. So, this is E will be 18000 into 2 plus second layer E value is 12000 that is also to third layer E value is 30000 that is into 2 divided by total thickness 6. So, that E value is coming 20000 kilo Newton per meter square. So, our S I immediate settlement that is 66.7 into 3 meter. So, we have to calculate the E divided by 20000 into 1 by 0.5 1 minus 0.5 square into 1.12. So, this value is coming is 8.4 millimeter. So, now in the immediate settlement we have to apply the corrections. Now, these corrections are the here isolated footing. So, rigidity correction is not required. So, only correction for the immediate settlement is the depth corrections. When you calculate the consolidation part then we have to apply the depth correction as well as the consolidation correction. For the immediate settlement this correction is the depth correction. Now, for the depth corrections. So, our D by root L by L B. So, this D is 2 meter L is 3 B into 3. So, this is coming 0.67 and L by B is 1. So, correction factor that is equal to from the chart because the this correction factor chart that I have already given that from the chart corresponding to D by root L B 0.67 and L by B equal to 1 the correction factor is coming 0.81. So, this immediate settlement corrected that will be 8.4 into 0.81. So, this is coming 6.8 millimeter. So, immediate settlement total immediate settlement is 6.8 millimeter. So, next we will calculate the consolidation settlement or the second part is the consolidation settlement. So, when you calculate this is for the consolidation settlement our expression is C C 1 plus E 0 this is summation H thickness of each layer into log P 0 bar plus del P into P 0 bar. So, for the consolidation corrections we have taken the 3 points 1 point the center of the first layer that is the 2 meter that is A second point the center of the second layer that is B and third point is the center of the in will zone within the third layer. So, distance of this A point from the base of the footing is 1 meter distance of B point from the top of the first layer is 1 meter and distance of C point from the top of the third layer also this is 1 meter this is also 1 meter and this one is also 1 meter because this is the center of this this layer first layer thickness below the base of the foundation is 2 meter. So, this 1 meter is A point is 1 meter below the base of the foundation. Now, this B point is the center of the second point the second layer second layer thickness is 2 meter. So, this is also 1 meter from the top of the second layer. Now, the influence zone within the third layer is 2 meter. So, the point is 1 meter from the this starting of the third layer. So, when you get this 3 point. So, we will calculate at A this p 0 bar we will calculate this 2 is the 18 that because the first 2 2 meter the water table as the 2 meter from the ground level. So, this will be 2 into 18 18 is the density we consider these are the saturated density. We consider this saturated and the bulk also use the this 18. So, 2 into 18 plus last 1 meter below the water level. So, this will be 18 minus 10. So, this value is 44 kilo Newton per meter square. Similarly, del p if you consider 2 is to 1 distribution this will be 67 into 9 into 3 divided by 3 plus 1 and 3 plus 1. So, this is 37.52 kilo Newton per meter square consider 2 is to 1 distribution. Similarly, at B the p 0 bar that will be 2 into 18 plus the first layer 2 into 8 plus the second layer 1 meter 1 into 19 is the density unit weight. So, 19 minus 10 this is 9. So, this is 61 kilo Newton per meter square. Similarly, del p considering this again 2 is to 1 distribution 3 into 3 is the density. So, 3 and this thickness is from the base is 2 meter for the first layer plus 1 meter from the second layer this is 3 meter. So, this is plus also 3 meter square footing. So, this is 16.7 kilo Newton per meter square at C p 0 bar that calculation is 2 into 18 plus 2 into 8 plus 2 into 9 in the second layer plus 1 into 9 in the third layer in the third layer also unit weight is 19. So, this is in the below water level. So, this will be 9. So, 79 kilo Newton per meter square similarly we can calculate del p is 66.7 into 3 into 3 divided by 3 because total distance of this C point from the base of the footing is 2 meter from the first layer 2 meter from the second layer plus 1 meter in the third layer. So, this will be 5. So, this is also 3 plus 5. So, this is 9.4 kilo Newton per meter square. So, once you get this value then this then finally, if I put this value in this consolidation settlement calculation. So, then we will get this is for C c for the first layer is 0.07 thickness is 2 meter this is log 10 44 plus 37.52 divided by 44 then plus C c for the first second layer is C c by 1 plus 3 into plus E 0 is 0.15. Thickness is 2 meter this is log 10 del p 0 is 61 plus 16.7 divided by 61 plus C c divided by 1 plus E 0 for the third layer is 0.12 thickness is 2 meter thickness 2 meter is means only the influence zone thickness in the third layer not the total thickness of the third layer. This is why considering only influence zones of the third layer log 10 79 plus 9.4 divided by 79. So, the total settlement will get 80.75 millimeter. Now, the depth correction factor is here 0.81 as it is for the immediate settlement and the pore water pressure correction pressure correction that we will get from the chart is 0.7 because A is taken 0.6 and A c by B influence zone that is 2. So, the chart I have given for the to calculate the pore water pressure correction factor from that chart corresponding to A and A c by 2 will get 0.7. So, A c corrected is 80.75 into 0.81 into 0.7. So, this is 45.78 millimeter. So, the total settlement will be the immediate settlement corrected plus the consolidation settlement. So, this is 52.58 millimeter which is less than 75 millimeter that is permissible. So, this is safe. So, between the clay and this is RCC structure. So, permissible value is 75. So, that is safe for the RCC structure. So, in this fashion we have designed the isolated footing and we have calculated first we have chosen the dimension and the depth of the footing and based on that we have designed and all the things are within the permissible limit. So, our design is safe. If it is not safe then you have to go for the next trial and then you have to choose the new dimension and the depth and this fashion we have to determine the dimension and the depth of the footing to design this isolated footing. So, next class I will design the raft foundation and other techniques also I will discuss how to design any foundation system.