 class on pile foundation lecture, I have discussed about the bearing capacity calculation of a single pile and then group pile. Now, today I will discuss about the bearing capacity or calculation for the under rim pile and then how to calculate the settlement of the pile. And the first I will go for the load bearing capacity of the under rim pile. Now, this is the load carrying capacity of under rim pile. So, that means the load carrying capacity of the under rim piled in clay. As I have already discussed the under rim piled means, where we will provide a bulb in the pile or in the base to increase the load carrying capacity of the pile. So, this is the bulb on different diameter. Suppose this is the diameter of the shaft or this is the diameter of the shaft and this is the diameter of the. Similarly, this is a single bulb. This can be 2 bulbs also. Suppose this is the again diameter of the shaft or the pile shaft. This is the diameter of the bulb TB and this is again the diameter of the bulb. So, this is the single bulb under rim piled and this is double bulb under rim piled. So, this is the 2 bulbs. Now, first I will discuss about the single bulb under rim piled. That first I will go for the single bulb. Now, for that purpose ultimate load carrying capacity of the pile Q u will be that C u cohesion at the base level at the base into N c into A at the base plus alpha cohesion at the surface into A s. So, that means the C u base is the cohesion at the base level because this is the contribution from the tip and this is the contribution from the friction or surface friction. Now, here this cohesion at the base level N c is the bearing capacity factor and area A B is the area of the base. Now, if this bulb is very close to this tip of the pile, then we can consider this area of the base is basically the area of this enlarged portion. If this is this bulb is very close to the tip of the pile, then we can neglect this portion resistance and we are considering thus this tip resistance is basically in this portion which is the area of this enlarged portion and C u is the cohesion at this base level N c is taken as 9 as I mentioned and now finally, if I write this expression that is C u at the base into N c is 9 into area of the enlarged base plus this alpha is the adhesion at the surface of between the pile and the soil. So, this is the adhesion factor. So, this alpha value is chart I have already given. So, from there we can calculate the alpha, alpha is the adhesion factor and C u this cohesion and that cohesion is at the surface level of the bulb. So, this is the cohesion at the surface into A s is the area of the pile shaft surface area. A s is the surface area of the pile shaft. So, this is the friction portion, this portion is the area of the surface area of that shaft and this is the C u is the base cohesion and A b is the area of the enlarged base. So, this is for single bulb if the bulb is very close to the. So, first one if this bulb is very close to the tip. This is the first case single bulb pile where bulb is very close to the tip. Now, if the single this is the single bulb pile now if the bulb is not very close to the tip then how what will be the bearing capacity for the load carrying capacity of that pile. This is also single bulb if the bulb quite high from the tip. So, in that case suppose this is the condition a bulb is quite high say from the base of the tip from the tip. So, here the friction resistance at this bulb level you are getting and another tip resistance we are getting from the pile tip. So, here we are getting suppose this is the again this is the diameter of the shaft and this is the bulb diameter d b. Now, we are getting this tip resistance at the tip of the pile and at this bulb level also from this enlarged portion. Thus when this bulb is very close to the tip of the pile then we are getting this we are considering we are getting resistance only this enlarged portion of the bulb. Now, for in that case we can write that q u ultimate load carrying capacity of that portion this is the a b area of the base then c u at the base level into n c. This is the tip resistance we are getting from the base then plus we are getting another tip resistance from this enlarged portion. So, that is the a bulb or bulb level this is a base means this is the base level this is the bulb level again c u at the bulb level into n c plus the alpha into c u at the surface into a s. So, in that case we are getting the surface resistance or the friction resistance of the pile as usual then the tip resistance we are getting at the tip level or the base level or at and the bulb level also. So, base level the area will be if it is a circular pile then pi by 4 into d square then into n c is 9 into c u at base level and the area of the bulb level is pi by 4 into d b minus d s or this is also you can say this is d s because this one is also d s d b by d s. So, this d b square d s square d s square into 9 into c u at bulb level plus alpha c u at surface level into s area of the shaft this outside area of the shaft. Now, this because this is the difference because this here we are getting only this area this resistance we are getting from this enlarged portion not the centre one because we are already this pile shaft is existing. So, that is why this is d b square minus d s square into 9 because here where n c taken is equal to 9. So, in alpha is again the addition factor. Now, here that as the pile settle generally when we calculate the a s between the calculation of a s generally it is recommended the length of the shaft equal to 2 d above the bulb it was a neglected. So, neglect the length equal to 2 d s above the bulb. This is this that means when you calculate the a s we neglect the 2 d shaft diameter 2 times the shaft diameter above the length of the bulb that portion is neglected because the reason is that due to the settlement settlement of the or the during the settlement of this pile pile it is sometimes there is a separation possibility of separation because of this bulb during this region. So, during this region above the bulb portion there is a separation between the soil and the pile during the settlement because of this enlarge bulb. So, that is why it is recommended neglect it is better to neglect this portion because here there is no contact between the soil and the bulb and the pile. So, there will not get any resistance from that portion. So, it will be better if we neglect this portion when you calculate the a s during the calculation of friction resistance. So, next one is the double bulb or the two or more bulbs this case this is for the two or more bulb. So, in this case this is the suppose one bulb this is another bulb again this is the diameter of the shaft d s and if we consider same bulb diameter for the two bulbs this will be d b. So, this is d b and this is the bulb shaft. So, this is the two or more bulb we are considering two bulb case. So, bulb diameter d b and shaft diameter d s. So, in such case q u. So, here we are getting the resistance one is at the tip level and it is it is suppose this bulb the first bulb the base bulb is quite high from the tip of the pile. So, that means you will get a resistance from here and another resistance we will get from this enlarge portion. So, here also the lower bulb is quite high from the tip of the pile. So, first we will calculate this resistance. So, this is pi by 4 d s square again n s is 9 into c u at base level. So, this is the c u at base level this tip level plus we will get this resistance from this enlarge bulb portion. So, that is pi by 4 into d b square minus d s square into 9 into d b square into 9 into d u at lower bulb level. So, this is the base level or the tip level this c u is at the lower bulb level then you will get the resistance. So, here we will get the this is the for the tip resistance or the base resistance. Now, we will get the friction resistance. So, friction resistance as usual for the shaft we will get. So, this is for the alpha into c u at the surface into a s as usual. So, this is the addition factor plus here additional one if I take this is the cylinder one cylinder you consider. So, that tip resistance we are getting this tip and another resistance we are getting at the due to this enlarge bulb and the shaft resistance we are getting from the upper shaft and this from the lower portion during the two bulbs. If we consider one cylinder we are getting the resistance from these two cylinders this cylinder for the due to these two bulbs movement. So, that means another resistance we will get that is c u at the surface between the two bulbs. This is the surface c u at the surface between the bulb then we will get the surface a s at bulb between the bulb. So, this is the area of the a this area is basically this one is a s b here this is a s this is the base and this one the enlarge portion. So, again when we calculate the here the c u dash we take the average cohesion during this shaft and average cohesion or a s b during the c u the surface between the two bulbs here this is the average cohesion in this region. So, again when we calculate this a s we neglect the two d s length when we calculate this a s during the a s calculation then again neglect the two d s length above the higher bulb. So, of the higher bulb we will getting this a s basically the resistance we are getting from the shaft above the uppermost bulb or the higher bulb or the uppermost bulb and this a s this resistance a s b we are getting between the two bulbs and then we get the tip resistance and the from this enlarge base resistance. So, this is the total four parts one is from the tip one is from the enlarge base resistance one is for the shaft resistance above the upper most bulb and another is the shaft resistance between the two bulb this. So, four part we are getting when we calculate the ultimate load carrying capacity of the underring pile for the two or more piles for this type of arrangement. So, if there is a three bulb then also we will get the this lower bulb is all the things when same when we calculate the a s b then we will get the this cylinder between the three bulbs. So, in this way we can determine the load carrying capacity of the underring piles. The next section we will go for the settlement calculation of the pile you know we have to calculate how to calculate the settlement of the pile. So, that means the settlement of the pile in group. Now, first one for the first case suppose for the displacement piles or friction piles in homogeneous clay. So, first pile first the pile is suppose this is the homogeneous soil this is the pile cap this is the q in the group and here this is the length of the pile l which is installed at the pile. So, this is say l. So, now this is the length of the pile this arrangement is say this is the 3 by 3 pile group arrangement. So, and this one is b say now here the centre to centre spacing between the pile say s and this is the diameter of the pile this is the diameter of the pile. So, the b value will be. So, b value is basically for if I take the total block. So, b value will be for this particular case is twice s plus. So, this is the half d by 2 plus d by 2. So, this will be twice s plus d this is the diameter of the pile s is the spacing between the 2 piles. So, now if we in such case how to calculate. So, this is the homogeneous clay which is the piles are installed in this homogeneous clay. Now, here it is recommended that you have to take the as if this total b load is transfer at this level. So, distance from this level is from the top is l 2 l 2 by 3 into l and this distance is l by. So, as if all the load all this q load is transfer at this level then it is distributed into the ground. So, this distribution can be 2 is to 1 distribution or this angle can be 30 degree. This is the 1 is to 2 distribution or this angle can be 30 degree. So, this one is also 30 and this one is also 30 degree. Now, up to how much you have to consider this. Now it is the same technique of this settlement calculation for the say suppose 1 wrapped foundation or mat foundation which is placed at the depth of 2 third l from the top of the pile and then the load is acting on the wrapped is q g with of this wrapped is b and which is distributed this load. Now, this extension will be up to twice b for the settlement calculation as it is mentioned during the calculation of settlement for the wrapped foundation. So, this extension of the b will be up to twice extension of this influence up to twice b of the pile. Now, this is the now we will calculate the spacing if there is a another pile is there then this will be 3 s plus d. So, then this total influence zone will be twice b and then for if it is a then every other center point of that influence zone or we can divide that part into number of segment and each segment center we have to calculate the set stresses and then you will calculate the settlement as we have discussed for the same as for the wrapped settlement. As if that wrapped is placed at a depth of twice 2 third l from the top of the pile and then it is this load is distributed at this 2 is to 1 distribution or at angle 30 degree. Now, the calculation the settlement the same as for the immediate settlement s i this value will be again this q n q n into b by e to the power of e 1 by mu square into i f is the influence factor. So, those things are same as the shallow foundation wrapped bearing capacity calculation. Now, here we have to again multiply it by the depth factor say alpha where alpha is the depth factor. So, depth factor you have to multiply it by the depth factor in case of immediate settlement and in case of consolidation settlement you have to multiply it by the depth factor as well as the pore water pressure correction. And for the consolidation settlement as this is the summation c c 1 plus a 0 h log sigma 0 bar or del 0 bar into del p by sigma 0 bar. So, the sigma 0 is the effective over burden pressure at any point and del p is the increment of stress due to the application of this load. So, again here we have to multiply it by alpha and then the mu. So, is there alpha is the alpha is the depth factor correction depth correction factor and mu is the correction factor due to pore water pressure due to pore water pressure. So, these are the due to. So, these two corrections are applied. Now, this how to calculate this depth factor this is mu, how to calculate this depth factor correction factor and mu correction factor due to over burden or pore water due to pore water pressure. These things I have already explained during the calculation settlement calculation of the RAP foundation the same way you have to determine this depth correction factor and correction factor due to over burden pressure. And then the same process have to multiply this correction factor with the correction settlement that you will get for the consolidation settlement. And then for the immediate settlement you have to apply only the depth correction factor because here we will not apply the consolidation correction factor due to the pore water pressure. So, then the total settlement will be the immediate settlement plus consolidation settlement then that should be within the permissible limit. So, in this way we have to now if there is any heart stratum within any layer. So, in that case we have to go up to that heart stratum not up to twice because within that soil will settle this heart stratum will not be affected. So, we have to go up to the heart stratum from this two third L from the top of the pile up to that heart stratum that will be the influence zone same as the RAP foundation settlement calculation. Now, this is where the first case if the friction pile is in the homogeneous clay. Now, the second case will go for the next one the second case now this is the pile pile cap which is installed in the soil. Now, this is layered soil this is different layer this is one layer and this is this soil is the clay layer or soft clay layer and this is the second layer this is strong layer. Now the strong layer or the higher strength layer the soft layer strong layer. So, these are the two different layer layer one and layer two now in the strong layer or the bearing layer suppose the pile is length is L or L 1 now the total length is say L from that L L 1 is within that strong layer or layer two this is the soft layer or this is the high strength layer we can say this is higher strength layer this is the higher strength layer now here the distribution will be at this level that that is this will be the distribution layer. So, this distance is L 1 by 3 and this distance is two third by L 1. So, previous case if the pile is inserted in a homogeneous soil in the clay layer. So, in that case this is the distribution point it will start two third of the total length of the pile from the top. Now, if it is the upper layer is soft soft and lower layer is strong layer or this is the higher strength layer then the distribution level will start the two third of the top of the stronger layer from that point the distribution is same as two is to one distribution or this angle is say 30 degree and then the process is same as I have discussed for the first case. So, only difference here only it is the two third of the L 1 not two third of total length. In third case suppose it is with a board pile or this is this pile is resting on a strong layer or this is the end bearing piles. So, these are the say end bearing pile previous cases was pile in the soft clay. Now, these are the end bearing piles for. So, now this is the firm strata. So, this one is firm strata. So, now that firm strata now this equivalent wrapped because here equivalent wrapped is acting on this level. Now, here equivalent wrapped will act again this one is B. Now, here this equivalent wrapped will act at the base of the pile then the distribution is same two is to one distribution or this angle is 30 degree. So, this will be either two is to one distribution two is to one distribution. So, these are the cases by which we can determine the settlement of the pile group and then you have to check whether that settlement within the permissible limit or not. So, next we will go for some empirical expression by which we cannot determine the settlement of the pile group if we know the settlement of the single pile. Now, first one this is the if the pile group is in sand because in the previous calculation in the previous calculation the pile groups are mainly in the clay soil. Now, here the if the pile groups are in the sand then how to calculate the settlement from that on this proposed expression. Now, the Scampton 1953 suggested that for the driven pile driven pile for same load per pile that mean per pile. All the piles are taking same load in that case S G settlement for the group divided by individual settlement is equal to 4 B with the width of the pile group plus 2.7 into B plus 3.6 whole square. So, where B is the of the pile group in meter. So, here B is the width of the pile group in meter and S G is equal to pile group settlement S i is the individual pile settlement. Now, here the advantage of this expression if I from the single pile load test if I know the settlement of the pile individual pile then by using this expression we can determine what will be the settlement of the pile group. Now, second expression is given by the mayor of 1959 because for the square driven pile in sand and this is again S G divided by S i is S 5 minus S i divided by S i is S by 3 divided by 1 plus 1 by R to the power square. Now, where S is equal to a ratio of pile spacing to pile diameter. So, that means this is spacing versus diameter ratio and R is the number of pile or rows in the pile group. So, R is the number of rows in a pile group and S is the ratio of pile spacing in the pile diameter. So, this is the we can say that if I replace this S by this is the small r or this is the S is the ratio of the pile spacing in the pile diameter. So, these are the different expression by which we can determine the settlement of the pile group in sand. So, next ones now till now I have discussed about the ultimate load carrying capacity of the single pile group pile then for the under rim pile then the settlement calculation. Next this chapter I will explain the negative skin friction of the pile and how to calculate the negative skin friction of the pile. Negative skin friction we are talking about now the load carrying capacity of the pile and then we are talking about the friction resistance that we are getting from the surrounding soil of the pile. And here the resistance that we are getting the if the pile load is acting downwards the resistance will act in the upward. So, that means that will give us the resistance of the load against the load. But the negative skin friction it will act the same direction of the load. So, that means here we will not get any resistance it will act as a load on the pile. So, in suppose in this is the if I go for the negative skin friction suppose you see here this is the pile normally if you apply the load this is the you apply the load q u here. So, here we will get the tip resistance and here the friction resistance that will act in the upward direction because the pile will settle in the downward direction and the resistance that will act in the upward direction. So, that means the this is the friction part q f and this is the tip resistance. So, our q u will be q f plus q p this is the traditional pile load carrying capacity calculation. Now, if one pile is this one and this is the load is q u is applied and suppose up to this level from here to here this is very compressible layer is existing. So, that means this is the very compressible layer. So, what will happen that here the that means the if the pile is moving downward direction so the resistance you are getting in the upward direction, but here due to very compressible nature of this pile suppose this portion is filling soil is there which is very loose. So, during this installation of the pile this there. So, if when we will apply the load now the this soil the settlement of this soil this compressible layer is more compared to the settlement of this pile. So, what will happen the if I consider the relative displacement between the this soil and this pile then compared to this as the settlement of the soil layer is more compared to the settlement of this pile. So, as if the relative settlement the soil will settle more pile will settle less. So, the friction resistance that we are getting it will not act in the upward direction it will act in the downward direction because with respect to soil and this pile as if the pile is moving in upward direction because the soil settlement is more compared to the pile settlement. So, the relative displacement we will consider as if in this region compared to the soil the pile is moving in the upward direction because of this high settlement of this compressible layer. So, here in this layer is as usual the soil layer. So, here we will get the resistance that will act in the upward direction and this is as will the tip resistance. So, here we will get the resistance in the upward direction, but here we will get the resistance in the downward direction as compared to these two soil layer in the pile the pile is moving in the downward direction. So, we will get the resistance in the downward direction. So, now if the resistance of this part is q f for this resistance q p is the tip resistance and this is for the q f 1 is the resistance we are getting from the resistance and this is the fix negative skin friction is q f 2. So, this one this friction now it will not give any resistance it will act as a load in the pile because it is acting in the downward direction. So, we can calculate this q u will be q p plus q f 1 minus q f 2. So, that means we are losing some resistance or load from due to this negative skin friction. So, this q u f 2 is called the negative skin friction of the pile. So, now we will calculate. So, we know how to calculate q p for different soil that we have already explained how to calculate q f 1 that is the friction resistance here q f then how to calculate q f 2 that we will discuss in this section. Suppose, this section this l 1 and from this l 1 how to calculate the negative skin friction in single pile. So, first we will go for the negative skin friction for the single pile this is for the first we will go for the single pile or to calculate this negative skin friction. So, then first case we will go for this for cohesion less soil or cohesive soil. So, this one is the cohesive soil where we are considering that phi is 0. So, the suppose negative skin friction a q f n, n is the negative skin friction. So, this is the negative skin friction. So, this is we can write suppose this is the pile up to this layer up to this length say l 1 there is a negative skin friction zone and this is as usual the pile tip as usual resistance we are getting. So, this is this length is l minus l 1. So, this negative skin friction we will get this is the perimeter and l 1 and c a where p is the perimeter of the pile l i l is the length of the pile in negative skin friction zone alpha is the adhesion factor and c u. So, that means is c a is basically alpha into c u. So, here c a is the adhesion and c u is the undrained cohesion of the compressible layer. So, suppose this one is compressible layer. So, up to this this l 1 is compressible layer. So, that means this is the negative skin friction we are talking about negative skin friction or initially as we have mentioned that is q f 2. So, that is the negative skin friction we are calculating this is the perimeter of the pile within this compressible layer. So, this is the perimeter of the pile. So, this is the perimeter of the pile and then l 1 is the length and this is the adhesion for this layer and alpha is the adhesion factor and c u is the undrained cohesion of this compressible layer. So, that friction resistance we are getting that is the negative skin friction. So, this one we have to add with q u or that total resistance we have to subtract this portion from the other part. So, this is the for the cohesion cohesive soil. Now, for the second one if the cohesion less soil or c u is equal to 0. So, in that case q f n will be half into p is the perimeter l 1 square into gamma into k into tan delta. This expression is same as the calculation of friction resistance for the cohesion less soil. So, here k is same as lateral earth pressure coefficient delta is the angle of friction between pile and that compressible soil. Now, generally delta is taken half phi or two-third of phi where phi is the friction angle between the of the soil of the soil and p is the perimeter as of this pile as I mentioned l 1 is the length of that compressible layer and gamma is the unit weight of the compressible soil the soil. So, this is the expression same as. So, this one is the negative skin friction that will act downward not aqueous. So, this will not give any resistance this will act in the downward direction. So, now the next one we will calculate this is for the single pile. So, we will give o for the group pile or negative skin friction in pile group. So, next one is in the pile group. So, again this negative skin friction f n in the pile group that will be equal to n into friction f n. So, this is we can write group. So, this is the for the group negative skin friction this is for the individual pile pile negative skin friction and n is the number of pile in the group. So, same as this group friction negative skin friction we can calculate this is the c u l 1 this is the p g of the group plus l into l 1 into a g. So, this is c u for the undrained cohesion. So, here p g is the perimeter of the pile group then gamma is the unit weight of the soil within the pile group up to the depth of l 1 into a g and a g is the area of pile group that is within the perimeter p g or within the p g. So, that means, this group friction negative friction we can calculate this is the if I consider the group as a block then this is the perimeter of that pile group then this is the length of this compressible layer into c u and then because here we will not consider any adhesion factor because adhesion factor is basically is equal to 1 because this is the interaction between the soil and soil because this we are considering as a group like the pile foundation in the group and then this is the a g is the area of the pile group within the perimeter p g and l 1 is the length of the compressible layer and the gamma is the unit weight of the soil within the pile group. So, this will give us the negative skin friction or the pile in group. So, then we have to calculate the factor of safety. Now factor of safety is the ultimate load carrying capacity of the pile in single this is the single pile or u g or group of pile divided by the working load plus the negative skin friction that is u f n or q f n g. So, this is for negative skin friction this is for the individual pile and this is for the group. So, if this is the single pile factor of safety we want to calculate then this will be q u divided by working load plus q f n. This is the ultimate load or this is we can say this is the ultimate load and if we are talking about the group pile then this will be q u g group pile ultimate load divided by working load plus q f n g negative skin friction in group. So, in this way we have to calculate the factor of safety now if this then this again if it is a single pile failure or the block failure that you have to consider based on that you have to decide which one you will choose either q u or q u g or q f n or q f n g. So, the same concept you have to use when you the that thing we have discussed for the pile group. So, these are the this way we have to incorporate the negative skin friction effect into our pile load carrying capacity design now when you calculate the load carrying capacity. So, this part up to this we have discussed over the load carrying capacity of the pile under compressive load and then settlement calculation based on that. Then the next section on the next class I will discuss over the load carrying capacity of the tension pile or that means the uplift capacity of the pile. Thank you.