 In the last class, I have discussed the load carrying capacity of a single pile, but generally pile these are used in a group. So, now today I will discuss about the load carrying capacity of the pile or in group or the group action of the pile. Then I will discuss the other method by which we can determine the load carrying capacity of the pile. Now, first I will discuss about the group action of the pile. So, in the group action of the pile, when pile is used in a group then we can calculate the efficiency of that pile group. Now, that efficiency if we write that this is the efficiency of group or the group efficiency that we can write in this form this q u g divided by n into q u. Now, where q u g is load carrying capacity of pile group and q u is the you can say this is the ultimate load carrying capacity of the pile group and q u is the ultimate of single pile and n is the number of piles. So, that means is here n is the number of piles in the group q u is the ultimate load carrying capacity of the single pile and q u g is the ultimate load carrying capacity of the pile group. So, here to get this group efficiency you have to calculate the ultimate load carrying capacity of the pile in group consider this pile as a whole and then this q u is the single pile. Now, this now for this smaller spacing between piles generally efficiency or the group efficiency less than u 1 or less than 100 percent. So, that means so that it indicates that that that means here these in a group the pile that can fail failure can be occurred as a individual pile failure or failure can be occurred as a group or the block type of failure or the considering the total all the piles that as a group and that can fail as a whole or is a block type of failure. So, that is one type of failure another is individual pile failure. Now, if the spacing is very small then this will occur this group type of failure generally will occur and in that case the efficiency is less than 1. Now, for the larger spacing this group efficiency is equal to 1. So, now for the largest type of spacing the individual pile failure will occur. So, we will get the group failure is equal to the individual pile load carrying capacity into the number of piles. So, that means u u g will be equal to is n into u g. So, that means the contribution from each pile will summation of contribution of each pile will give the group load carrying capacity of the pile. In that case this efficiency of group will be 1 and if it is less spacing is very small then we will get the because the overlap due to the overlapping overlapping of the stress zone or the influence zone for the single pile individual piles. Though that means the group failure will occur and that will less than the load carrying capacity of the individual pile. If we sum the all the load carrying capacity of the individual pile that will be generally more than the pile load carrying capacity as a whole. So, in that case if the spacing is very small then the group load carrying capacity is smaller or the lower than the summation of all individual piles load carrying capacity. So, in that case this efficiency will be less than 1. Now another thing that for the driven pile lose to medium sand lose to medium sand this efficiency can be greater than 1. This is because there is when pile is driven into a lose to medium sand then sand become dense. So, in that case because of the installation of the pile the group carrying capacity of the pile surrounding the soil surrounding this pile that get dense. So, that is why the efficiency that will increase. So, these are three possible cases that for the smaller spacing between the piles where the summation of all individual pile load carrying capacity that is more than the group load carrying capacity of the pile. So, in that case the efficiency will be less than 1. Now if the largest spacing in that case basically the individual pile failure will occur and if the spacing is very less then the group type of the block type of failure will occur. So, if the spacing is very large. So, individual pile failure will occur in that case the group load carrying capacity is equal to the summation of the all individual pile load carrying capacity. Now, in third case in the driven pile if it is driven in the lose to medium sand in that case the density of the sand become dense. So, because of this nature the group carrying capacity or the efficiency of the pile that will increase and that efficiency is sometimes even greater than 100 percent or greater than 1. Now in we will discuss about how to calculate the group carrying capacity of the pile. So, in that case we can write that pile group in clay first we will calculate the group carrying capacity of the pile if it is in the clay. So, as I have mentioned there are two types of failure one is that block failure and second one a individual pile failure. So, those two types of failure will occur if the spacing is generally more spacing between the two piles or more then this individual pile failure may occur and if it is less then this is the block failure will occur generally. But in then now for the another thing that is the condition for the clay that generally block failure will occur. Now, if the spacing is less than 2 2 3 d now if the spacing less than 2 2 3 d it is the diameter of the pile and if the larger spacing the individual pile failure will occur. In that case if we calculate u g that is the group load carrying capacity of the pile that is c u u b at the base then n c then a b at the base plus p into l into c u u s. So, in that case here the similar to the individual this expression. So, this is the from the tip resistance and this is the frictional resistance. So, this is the tip resistance and this is frictional resistance. So, now we can write that c u b is the undrained strength of the soil the base of the pile of the clay at the base of the pile. So, this is tip resistance is coming from the base. So, this is at the base of the pile. So, similarly q u s this is the undrained strength of the clay at the base along the surface length of the pile group or block. Now, here also we can write this is pile group. Similarly, n c values we can write is equal to 9 like the single pile then we can write a b is equal to cross section area of the block this is the cross section area of the block l is the embedded length of the pile pile and p b or p that is the perimeter of the block. So, here this u g we are calculating by considering the block failure because here if you consider this is the block failure and then we calculate this is the tip resistance and this is the frictional resistance. So, q u b at the undrained strength at the base of the pile n c is 9 a b is the cross section area of the block p is the perimeter is the embedded length and q u s is the undrained strength along the surface or the pile group or block. So, same as the single pile. Now, if we can similarly we can write we can determine the or we can write the expression for the pile in the sand the same by using the same type of expression for the sand because this is similar to the single pile in but only the thing is the here the cross section area the expression of same in the cross section area in case of single pile we have consider this is the cross section area of the single pile. Here the cross section area we have to consider the block when you calculate the tip resistance. Similarly, in the calculation of the single pile frictional resistance we use the area is the surface area of the single pile here you have to consider the surface area of the block that means the perimeter and the length of the block. So, that is the only difference when you consider the group and the individual failure individual pile load carrying capacity. So, this is the load carrying capacity for the clay. Similarly, by using the same expression like the single pile in the sand we can determine the load carrying capacity of the pile in the sand. Generally and sometimes as I have mentioned that in the driven pile this efficiency is greater than 1. So, but when we design these things we can consider that efficiency is equal to 1 and we can design. So, now that means we are now giving the expression for the clay and similarly the same expression as given for the single pile that we can used by slight modification for the cross section area for the base and the cross section area of the surface. Here we have to consider the cross section area of the block and when you calculate the tip resistance and when the cross section area of the surface area of the block when you calculate the frictional resistance that is the difference between the single and the group. Others the expressions are almost same. So, now we will solve one example by which we can determine that how we can calculate the load carrying capacity of the pile and the single pile and the group piles and then we will determine then the other factors also and the spacing also. Suppose a problem is that that we have one block. So, this is the arrangement of the pile in that thing this is the total 16 piles are there. So, 4 columns and 4 rows. So, total 16 piles now it is in the clay soil. Now diameter of the pile or D of the pile is 300 millimeter. This is the diameter of the pile. Now the thing is that this is the total number of 16 piles with diameter is 300 millimeter. Length of the pile or embedded pile is 10 meter. Now undrained q u undrained shear strength of the soil is 50 kilo Newton. This is unconfined compressive strength of the soil this q u. So, that means c we can calculate c u undrained shear strength of the soil that will be q u divided by 2. So, that is 25 kilo Newton per meter square. So, c u u. So, we are considering the same c u for the surface as well as at the base. So, that means c u b and c u s. Similarly, c u b is equal to c u s equal to c u equal to 25 kilo Newton per meter square. So, q u equal to 50 kilo Newton per meter square c u will be q u by 2 25 kilo Newton per meter square. Now we have to determine the spacing. What will be the spacing such that that group efficiency will be exactly 1. You can take any other value also. So, you can take the 0.9, 0.8. So, here we will design this thing such that the group efficiency will be exactly 1. What will be the spacing? So, as I mentioned that here we have to consider the single and as well as the block failure. So, this is the block basically for the group. So, suppose we consider this is the spacing between the space this is s and similarly this one is also s. This is s, this is s, this is s. Now distance from this because this is the center of the each pile. So, distance from the center to this edge is 0.15 meter. Similarly, this one is also 0.15 meter as similarly this one is also 0.15 meter and this one is also 0.15 meter because the total diameter is 0.3 meter. So, half of this is 0.15 meter. So, as we have calculated the group efficiency q u g is n into q u. Now here our q u g is 1 n is equal to 16. It is the number of pile. So, now if q u g is equal to 1. So, that means q u if efficiency is equal to 1. So, that means q u g will be n into q u or q u g equal to 16 into q u. So, now first we will calculate the q u g and here another thing the condition is that you can neglect the tip resistance of this pile. That means the resistance we are getting. So, we can neglect neglect neglect the bearing at the tip of the pile because these are the in the clay. So, that in the these are the friction pile. So, the resistance coming from the friction will be more as compared to the resistance coming from the tip. So, the one condition that we can we are neglecting that friction components. So, that means q u g will be equal to that I have mentioned that is the 16 q u. So, the single pile capacity is 16 into alpha into c u into a s area at the surface. So, now here this is 16 into alpha into c u is 25 area is pi into 0.3 into 10 is the length. So, area is pi d l. So, pi into 0.3 into c u into a s area at the surface. So, pi into 0.3 into pi into l is the 10 meter. So, I am thinking this alpha we have to calculate. So, in the last class I have given one chart to calculate the how to calculate the table how to calculate this alpha value. Now, here based on the c u value I am giving you another figure by which we can determine the value of alpha. Now, that figure is proposed this is proposed by the Tomlinson 1979 and this is for the driven pile. So, in this figure, so this is so this point we can say this is alpha this is 1.5 this one is 1 this is 0.5 and this is 0 and this side this is 0.5 sorry this is 0.5. So, this is 25, 50, 75, 100, 125 this is the value of c u it is in kilo Newton per meter square. So, the points that we are getting. So, these are the points. So, even if we join these points. So, this is the chart. So, this is from here we can determine the value of alpha corresponding to 25 because our c u value is 25 kilo Newton per meter square. So, now what will be the value of alpha as corresponding to 25 kilo Newton per meter square. So, in this value we can calculate this alpha is coming 0.95. So, we can calculate the value of q u g in terms of 16 into q u and this is 16 into alpha is 0.95 into 25 is c u into pi into 0.3 into 10. So, this is this part is 16 and this basically this part is 0.95 alpha into 25 into pi into 0.3 into 10. This is this total thing is equal to the individual pile load carrying capacity ultimate pile load carrying capacity of the individual pile. So, the total load carrying capacity is 3581.42 kilo Newton that is the load carrying capacity of the pile in group. So, here all the calculations we are doing for the compressive load carrying capacity of the pile. So, next one we will we can calculate this q u g in terms of block failure. This is this is in terms of individual failure. Now, other times in terms of block failure we can write this is P perimeter into L into c u. So, when we are calculating this block failure here we consider that our alpha is basically 1. The reason is that here is the alpha is the addition. So, that means this here the addition here this is the addition factor and this addition we are considering when it is a single pile. That means, here the addition is between the pile surface and the soil. So, that is why you have to consider alpha value because you have two different materials we are considering. This is the pile different materials and the soil is another different materials, but when you are considering the block as a block failure. So, there is blocks this side this is the interaction between the soil and soil. So, both the same material that is why you are considering here when you consider this block failure you consider alpha equal to 1, but when you consider the individual pile failure that is the interaction between the pile surface or pile material and the soil. So, that is two different material that is why you consider different alpha value, but here interaction between the two same soil. Here we will consider the alpha equal to 1 because it is the interaction between the soil and soil when it is a block. So, now here we can write this p is the perimeter is 4 into 3 s because if I look this block total. So, here we are considering the block failure this most of the surface here interaction is soil to soil individual pile is interaction between the soil and pile. So, that is why we are considering alpha in the block, but the side 1 is 3 s plus 0.15 plus 0.15. So, 3 s plus 0.3 is one side another is also 3 s plus 0.3 because here this is the square type of arrangement. So, that means we can write the perimeter is 3 s plus 0.3 into 4 because this is one side another 4 side into l into 25. So, this is 3000 s plus 300. Now, for efficiency is equal to 1 we can write that 3000 s plus 300 that is equal to 3581.42. So, now from here we can calculate this is that s is coming 1094 millimeter or s is equal to 3.65 d, where d is the diameter of the pile this is the diameter of the pile. Now, again another thing that we have to check it that is code recommends that 2 i s 2 9 1 1 this is part 1 1979 that for the minimum spacing or s minimum that is equal to 2.5 d for point bearing pile piles and equal to 3 d for the friction pile. So, here it is minimum spacing here you are designing it is for the friction pile the minimum requirements is 3 d, but here our calculation is coming 3.65 d that means it is ok. So, to get a efficiency equal to exactly 1 we have to provide a spacing 3.65 d. So, that spacing we can provide to get a spacing exactly 1 exactly efficiency exactly 1. Now, if we want to design it for the different efficiency then you have to put that value and then corresponding spacing you have to calculate. So, first you assume the diameter that diameter we choose and based on that we can determine what will be the spacing required to get a particular amount of efficiency group efficiency. So, in the next section I will discuss about these are the. So, in the first class I have mentioned about the there is a 4 different ways by which we can determine the load carrying capacity of the pile. The first one the by the static expressions or the formulae that part I have finished. The next one that I will discuss about the pile load test. So, by pile load test also we can determine the load carrying capacity of the pile. So, next one is the pile load test that we will discuss in that section. So, now in the pile load test. So, this part we will discuss about this load test or the on piles. So, by pile load test also we can determine the load carrying capacity. Now, this is basically this again this pile load test on any field test suitable for cohesion less soil. Now, here we will do the different types of pile load test we can perform one is compression test one is pull out or tension test and one is lateral load test. So, here we will discuss about the compression test how we can do the compression test to determine the load carrying capacity of the pile. So, here basically you are discussing the compression test in addition to this we can do the pull out test or the tension test or lateral load test to know the lateral load carrying capacity of the pile. Now, before we go to the pile load test there are few things that we should know that is one is initial test and next one is the routine test. So, initial test on term that is this is carry out on the test pile is the carry out on the test pile to estimate the allowable load carrying capacity of the pile or to know the settlement of the pile corresponding to the working load that is the test pile. So, this is the test pile that we carry out on the pile to estimate the allowable load and to predict the settlement at the working load. Now, routine test carried this carried out to check the working pile load to and to know the corresponding settlement of the pile corresponding that working load. So, that in the routine test we will perform on the working pile and initial test that we will perform on the test pile. So, now this thing is the more than the condition is that more than 200 piles the minimum 2 test is required. So, more than 200 piles minimum 2 initial test is required and the for the routine test this minimum number of test is generally half percent of the pile used or that can vary up to 2 percent or. So, that means, the initial test that is conducted on the test piles and the routine test that is conducted on the working piles. Now, in test piles means these piles are constructed to for the testing purpose. This is not constructed for the load carrying capacity or load carrying purpose of the super structure. It is not constructed for the working condition it is just constructed for the testing purpose and then once the test is over then these piles are not used. So, these are not used to carry the load which is coming from the super structure and routine piles are routine test are conducted on the working piles. See working piles are the piles where the actual load of the super structure that will come it will work. So, that means, the routine test will conduct for the working piles on the working piles and initial test will conduct on the test pile. Now, more than 200 piles minimum 2 test piles or initial test these are required and for the minimum number of the routine test is generally half percent of the pile used or it up to 2 percent or generally more. Now, when this test pile. So, now that thing is that the test pile and working piles. So, used only to load test does not carry load of the super structure that I have already mentioned these things. Now, the minimum test or the load that these piles are taken the minimum test load on the test pile that is equal to 2 times the safe load. The minimum test load on the minimum test load on the test pile is 2 times the safe load. So, this safe load we can determine by using the static expression that I have already explained. So, now these load test are attained a value of this load test is generally 2 to the safe load or the load at which the total settlement of the pile is 10 percent of pile diameter in case of single pile and for the minimum 40 millimeter in case of group pile. So, that mean this minimum test load on the test pile is 2 times the safe load or the load at which the total settlement of the pile attains the 10 percent of the pile diameter for the single pile or 40 millimeter for the group pile. Similarly, for the working piles that mean the routine test the test load is generally up to 1.5 times the safe load or the load at which total settlement is 12 millimeter again for the single pile 40 millimeter for group pile whichever is lesser here also this is also whichever is lesser. So, that means these are the all the information regarding the initial test routine test then that means the initial test is conducted on the test pile the test piles of the pile which is constructed for the testing purpose not for the it will not take the load which is coming from the superstructure and the routine test is conducted on the working pile. Now, these working piles are the piles which will take the load which is coming from the superstructure. Now, more than 200 piles minimum 2 test piles are required initial test that is required and mean minimum number of test is half percent for the pile used the routine test or 2 percent and more. In the test pile the minimum test load is 2 times the safe load or the load at which the settlement attains a value of 10 percent of the diameter for single pile or and 40 millimeter for the group pile whichever is lesser. And working pile the test load up to which 1.5 times of the safe load or the load at which a total settlement is 12 millimeter for the single pile or 40 millimeter for the group pile whichever is lesser. So, these are the condition for the different test and the different type of piles which are used for the pile load test. Now, next we will discuss about the how to calculate the load carrying capacity of the pile using the pile load test this 2 9 1 1 this is part 4 1979. So, this I S code according to this I S code. Now, load is applied on the RCC cap over the pile and the applied with an increment of 20 percent of the safe load and corresponding settlement of the pile is recorded by using the at least 3 dial gauge attached in the pile cap. So, pile cap is used and where the load is applied with an increment of 20 percent of the safe load and as I mentioned the safe load is calculated based on the static expressions and the settlement corresponding to the each incremental load is recorded which is measured by using at least 3 dial gauges. Now, the allowable load how will get the allowable load of the pile. So, now how we can determine the allowable load on a single pile. So, one first condition is that two third the final load at which total settlement attains a value of 12 millimeter. So, if that means, the two third of the final load at which the total settlement attains a value of 12 millimeter. So, that means, if nothing is specified then we can consider the permissible settlement of the single pile is 12 millimeter and the load at which this 12 millimeter is attained we have to consider two third of that final load. Now, if any permissible settlement is mentioned other than this 12 millimeter then we have to calculate the load two third of that final load at which the total settlement attains that permissible limit which is specified. Now, in or in second case the 50 percent of the final load at which the total settlement 10 percent of the pile diameter. This is for uniform diameter pile and 7.5 percent of bulk diameter is for under rimmed pile. So, that means, this is the conditions is the one is first one and second one is the 50 percent of the final load at which the total settlement is 10 percent of the pile diameter for the uniform diameter pile and 7.5 percent of the pile diameter for under rimmed pile. Now, another condition is that now this we can use the third one or the third condition this is the single pile. Now, for the cyclic test this is the for the static test compressive now if we can do the cyclic test if we want to know the tip resistance as well as the friction resistance of the pile separately. Now, if we go the static test for the single pile that will give us the total resistance of the pile. Now, if we know want to know the tip resistance and the friction resistance separately then we have to go for the cyclic test. Now, for the cyclic test pile load test so that means, the two-third of the load final load at which the total settlement is equal to 6 millimeter. So, here we can replace this one is a it is a similar to the number one condition, but here this is two-third of the final load at which total settlement is equal to 6 mm. So, the minimum of all this condition that we will we will consider as our allowable load on a single pile that is determined from the based on pile load test. Now, for the group pile or the group pile same thing we can use some condition. So, that the first one is similar that the first one is final load at which settlement equal to 25 millimeter. So, now this load test we can perform on the single pile as well as on the group pile. In generally in the group piles the permits is nothing is specified then the permissible limit is settlement is 25 millimeter. Now, the final load at which this settlement is attained that will one condition. So, now if any other permissible settlement is mentioned then you have to calculate that final load corresponding to that specified permissible settlement. If nothing is specified then we will consider for the group permissible settlement of the pile is 25 millimeter. Now, b is that the two-third of final load at which the total settlement attains a value of 40. So, this is the second condition the two-third of the final at which total settlement attains a value of 40 millimeter. So, the minimum of these two will consider the allowable load carrying capacity of that group. So, these are the conditions that to satisfy when you calculate when you determine the load carrying capacity of the single pile as well as the pile group. So, these are the conditions. So, I have discussed about the initial test the routine test then test pile and the working pile and what are the condition by which we can by using those condition we have to determine the load carrying capacity of the single pile as well as the group pile. Now, this up to this we have discussed about the load carrying capacity of the pile. So, this will give us the total load carrying capacity of the pile allowable load carrying capacity of the pile. Now, if we want to determine or want to know the the resistance that we are getting in individually from the tip as well as from the friction then you have to go for the cyclic pile load test. In the next next class I will discuss about the cyclic pile load test and by which we can determine what is the contribution from the friction part and what is the contribution from the tip resistance of the pile individually. Thank you.