 I have discussed about the settlement calculation of shallow foundation and how to calculate the immediate settlement and the consolidation settlement of soil. Now, today I will start the calculation of settlement based on the field test data and it is basically for the granular soil. So, what are the different field test that I have already discussed in the first section. Now, by using those field data, how to calculate the settlement of the soil for the shallow foundation, those things I will discuss in this lecture. Now, this here I will mainly concentrate on the calculation of settlement for granular soil and mostly the test that data that we will use that SPT and the CPT data and that is the standard penetration test or the cone penetration test data. In addition to that, I will explain the another field test that is a plate load test and how to use the plate load test data to calculate the settlement and later on I will explain how to use the settlement or the plate load test data to calculate the bearing capacity of the soil when I will discuss about the allowable bearing capacity calculation. So, first today's class that I will discuss mainly on the settlement calculation by using different field test data. Now, first if I discuss about the settlement of foundation on granular soil. Now, the first thing that for the disadvantage of the for the calculation of settlement based on the on the field test for the clay soil is that for the clay soil these all the settlement or the field test that generally we conduct that is for the for short term test and for as the in the consolidation calculation for the clay soil this is for the long term test long term calculation. So, that means due to the consolidation so that is why for the cohesive soil the using the field data to calculate the settlement is not a good idea. So, it will be better if we use the this field test data to calculate the settlement for the granular soil because this granular soil it is the most of the settlement for the immediate settlement and that is for the short term and the test that we are conducting the field that is also short term. So, as for the cohesive soil as it is the long term settlement due to the consolidation. So, if that if we use the short term test data for the calculation of consolidation settlement or the calculation for the cohesive soil settlement then let be there we will we will not get the appropriate or accurate results. So, that is for this field test data we mainly we will use for the settlement calculation of granular soil. Now, first test that we have already discussed about the SPT and the CPT test or the cone penetration and standard penetration test and today we will discuss the another test that is called the first is called the plate load test plate load test method that we will discuss and here we will discuss about the IS method India standard method 188 to 1982. Now, plate load test is conducted on the soil and it is ideally you should conduct on the as the foundation of the foundation level. That means, the here we will instead of footing we will use a small plate to simulate the footing we we are and then that plate should be placed at the foundation level. That means, you have to construct one dig or the then you have to remove the soil sample up to the foundation depth and then we where there you can place the plate and then we can conduct the test on that plate. So, that means, here directly we can calculate the what the bearing capacity of the soil by using this plate load test data. So, that means, the here for the plate load test suppose this is the depth of foundation here this is the pit or the test pit and this is the depth of foundation and here this is the ground surface is here. So, this is GL or the ground surface and this is the arrangement that here this is the d f or the depth of footing or the foundation and the width of this portion of the pit that should be that means, width is should be 5 times b p or the width of the plate. So, here we will place the plate. So, this is the plate of width b p b p is the width of the plate. So, here first in the test pit this is the depth of the foundation or the real footing at the base of this pit will place the plate here then by hydraulic jack suppose this is the hydraulic jack and this is the proving ring. So, this is hydraulic jack and this is the proving ring dial gauge. So, here we will by this proving ring we will apply the load and here we will place one beam for the section on the ground surface which is attached with the ground by this anchors. So, these are the anchors. So, these are the anchors that we will place in the soil to fix this beam or the loading frame and by the use of this loading frame or the beam we will apply the pressure on this plate by this hydraulic jack and the load that we are applying and corresponding settlement we have to measure by using this dial gauge. So, these are the dial gauges. So, these are the test argument that means this is the pit here at the base of the foundation we will place the footing or the plate here. This is the d f depth of the foundation then by using this hydraulic jack and this we will apply the load and this is the fixed loading frame. By using the hydraulic jack we apply the load and corresponding load that we will apply we can measure this load by this proving ring and corresponding settlement of this plate will get by using this dial gauges. So, this is the total arrangement of the plate load test and the settlement and after the test is done then we will plot the load versus settlement plot. Now, first these are the what are the different sizes of the plate that basically this is the mild rough mild steel plate. This dimension of the size is 30 centimeter, 45 centimeter or 60 centimeter or 75 centimeter. So, these are the different sizes of the plate. Now, this generally the square type of plate is used. Now, before the plate is placed 1 5 millimeter fine sand is placed before placing the. So, before placing the plate 1 5 millimeter fine sand is placed at the base level this level. So, now the condition is the if the soil is very stiff or dense then you will use the smaller plate and if the soil is very loose and soft then you will use the larger plate. Now, for during the testing you have to remove the water from here this test pit that means dewatering have to conduct or you have to done the dewatering to remove the water within this pit a pit and then we will apply the load. Now, the loading application first step that we will place one sitting load the first step of this loading application that we will place first a sitting load load 70 k g per centimeter square is first applied and it is applied for a time. So, some for this for the sometimes it is applied this sitting load and then the next load is applied at a depth of one fifth of the estimated safe load. So, loading applied. So, first we have to estimate the safe load that this foundation or the soil can carry. Now, the first increment of the load will be one fifth of this safe safe load. So, that means, was one fifth one fifth increment we have to apply the load and we have to apply the load until the failure of the soil or at least the 25 millimeter of the settlement. So, at least for the 25 millimeter settlement or up to the failure load which ever is earlier. So, whichever is earlier either 25 millimeter settlement or up to the failure of the soil. So, whichever is earlier up to that we have to conduct the test. Now, next one is that one increment suppose when we apply the increment then you have to record the settlement. Now, settlement you have to record or record the settlement at an interval at time interval of 1 minute 2.25 minute 4 minute 6.25 minute 9 minutes 16 minutes and 25 minutes and then at 1 hour interval. So, that means, the when we apply the first we apply the first increment of load say one fifth of the load then we will record the settlement at 1 minute 1 minute 2.25 minute 4 minute 6.25 minute 9 minute 16 minutes and 25 minutes interval. Then at every 1 hour interval we have to apply the load up to record the settlement. Now, for the for the any other soil other than clay the rate of loading is 0.02 millimeter per minute. So, other than the clay when this rate of settlement that we will get because we will calculate the settlement at different interval. Now, if the rate of settlement is less than 0.02 millimeter per minute then we will apply the second increment or for at least for 60 minutes you have to apply for any load increment. That means, any load increment that we have to apply at least at least for 60 minute that is compulsory and then when we will observe other than the clay soil if this the rate of settlement that we are measuring that 0.02 millimeter per minute that means, this is less than 0.02 millimeter per minute then we will apply the next increment. And for the clay soil if the settlement exceeds 70 to 80 percent of the ultimate settlement at the end of 24 hours. That means, for the clay for the sand of the soil other than clay the condition is that for any increment we have to apply for at least 60 minutes and if or if the settlement increment is 0.02 millimeter per minute then we have to apply the next increment of loading. Now, for the clay soil if the settlement has exceeded 70 to 80 percent of the ultimate settlement or at the end of the 24 hours. But in the clay soil the condition is if the settlement has exceeded 70 to 80 percent of the ultimate settlement or at the end of 24 hours we have to take the reading. Now, these are the different condition of the test that we will conduct. So, now when we will conduct so there are few points. So, what are the few advantages of this plate load test is that that this from this plate load test directly we will get the ultimate settlement load carrying capacity of the soil. Because for the other type of field test over the SPT or CPT then we will get some value and based on the correlation that we are getting the parameters of the soil or ultimate load carrying capacity of the soil. That means, we have to use the some parameters then we will get the ultimate load carrying capacity of the soil. But the advantage of this plate load test that in the field itself we will get the ultimate load carrying capacity of the soil. So, that is one advantage. Now, thing is that there are some limitations also the some comments that we will should note that some comments that for any condition this BP should not be less than 30 centimeter. That means, for any condition this BP should not be less than 30 centimeter. Now, another condition is that that if the soil is not homogeneous it will give incorrect results. Now, one thing that I have already explained that for the clay immediate settlement is not the main settlement but plate load test will give the immediate settlement. So, this plate load test will give settlement. However, for clay is not the main settlement. So, that means, these are the condition that BP should not be less than 30 centimeter. Now, if the soil is not homogeneous then it will give incorrect results. It means that suppose because the plate dimension is 30 centimeter that is the minimum dimension maximum is that we are using that 75 centimeter. So, that means, the influence zone for this plate if we use the 75 millimeter centimeter plate also that means, the influence zone will be up to the 2B because as I have already mentioned that for the bearing capacity calculation we will consider the up to B depth and for the settlement calculation mainly we will consider up to the 2B zone below the depth of the foundation or the base of the foundation. Now, if we are using the plate which is a smaller size plate and that means, the influence zone will be very less. Now, actually in the prototype or the real field that dimension of the footing is much more than the plate that you are using. Suppose, the plate you are within 30 to 75 centimeter or in the field that dimension may be say 300, 3 meter or 4 meter dimension. So, if we take the 4 meter real footing dimension then the influence zone up to say 3 meter into 2 that means, 6 meter. Whereas, if I use the 75 millimeter centimeter plate that means, influence zone will be 150 centimeter. So, you can see that the zone of influence for the plate and the real foundation is very there is a difference. And now if the soil is not homogeneous, if the soil is homogeneous that means, the influence zone for that soil whether it is the 150 centimeter or 600 centimeter that will be same. But if there is a layers if it is the layer soil and if suppose the one layer is thickness is say 2 meter or 200 centimeter. So, then what will happen if we use the 75 millimeter 75 centimeter plate that means, influence zone will be 150 centimeter. And if I use the same plate load test data for the real foundation that with you say 3 meter or that means, influence zone will be say 600 centimeter. So, that means, in actual case the influence zone in the second layer which is present below the 200 centimeter from the base of the footing that will influence actually in the real footing. But whereas, in the plate load that influence that means, the second layer influence will not come into the picture. So, basically you are not actually taking the care of the second layer if it is a layer soil in case of plate load test. Whereas, in the footing that second layer soil is influenced by the footing load. So, in that case we will get some erroneous result. So, if the soil is homogeneous then this plate load will give the correct result or otherwise if the layer soil is there then it will give erroneous or incorrect result as compared to the real footing. Another condition is that the clay this plate load test mainly it will give the immediate settlement, but the for the clay soil the immediate settlement is not the main contribution from the settlement. The main contribution of the settlement is from at the consolidation settlement. So, that is why for the clay soil this plate load test data if we use the settlement calculation that is not will not give the correct result it may give the incorrect or erroneous result. And another condition is the capillary action due to the capillary action of the sand soil. So, that the strength of the sand that may also increase. So, those things we have to consider when we will do the plate load test and then when you use this data for the our real footing calculation then you have to consider all this situation whether soil is homogeneous or not what is the size of the plate that we have used what are the type of the soil that we are testing. So, those things have to consider before we use this data for our real design. Now, the next one that we have done the plate load test then how to use those data. Now, for the settlement calculation this is from the plate load test that plate load test once we get the plate load test first we will get this type of. So, here we may get say this type of graph or you may get this type of graph also. So, there is a different types of graph that we will get. So, suppose this is our load intensity and this is settlement. So, if we get this type of graph some then some from the if we extend this state portion we will get the ultimate load carrying capacity of this plate. If we get this type of graph that by double tangent method. So, this initial state portion and the final state portion. So, this one will give us the q u p ultimate load of the plate. So, if we will get this also be this two method we will get the ultimate load carrying capacity of the plate. So, this is the load settlement graph of the plate and if we will get this type of graph the extension of this state portion will give you give us the ultimate load of the plate and from here for by double tangent method we will get the ultimate load of the plate. And from here then for corresponding to any loading condition we will get the settlement of the plate or vice versa corresponding to any settlement we will get the load from the plate. So, now for the granular soil if I want to calculate the settlement then we can use the expression which is proposed by the Tarzakhi and Peck 1948. So, that is the S F divided by S P that is equal to B F into B P plus 30 to B P by B F plus 30 to the power square. Now, where S F is the settlement of a foundation or footing, foundation of width B F. So, this width is in centimeter. So, here here this width is in centimeter then S P is the settlement of a foundation which is also in centimeter. So, here S F is the settlement of the footing real footing S P is the settlement of the plate now B F is the width of the footing which is in centimeter and B P is the width of the plate which is in centimeter. So, here suppose if we know the settlement of the plate and we know the settlement width of the plate and width of the footing then we can by using this expression we will get the settlement of the real foundation. Now, for the clay soil this is for the granular soil now for clay soil clay soil with the same expression you can use the S F by S P that is equal to B F by 30 to the power B B. And after that we will we have to use the Fox corrections for due to the depth correct due to the depth. That means, if the depth of the plate and depth of the foundation real foundation are same then no correction is required for the depth. But, if there is a difference between the depth of the foundation and the depth of the plate then we have to apply the depth corrections for both the when we calculate the settlement of the real footing by using the Fox method that I have explained in the last class. So, this is the settlement calculation based on the plate load test for granular soil and clay soil. Now, if I want to calculate the bearing capacity of the soil then how will calculate the bearing capacity of the soil by using the plate load test. So, this is the bearing capacity this is also by plate load test. Now, for the granular soil that Q U F ultimate load of the foundation is Q U P into B F divided by B P. Same that Q U P how will get the Q U P that I have explained from this plate load settlement graph for the plate load test you will get the ultimate load carrying capacity of the plate. And if you know the width of the plate and the footing then you will get the ultimate load carrying capacity of the real footing. This is for the granular soil now for the cohesive soil or clay soil that Q U F is equal to Q U P as in case of cohesive soil this is independent of B. The cohesive soil the ultimate load carrying capacity of the plate is equal to the ultimate load carrying capacity of the footing. So, that means by using the plate load test will get the settlement of the foundation as well as the load carrying capacity of the footing. Now, we will get the safe load carrying capacity. Now, here how to calculate the safe bearing capacity of the foundation by using plate load test data. So, now if the permissible settlement suppose the S F is the permissible settlement of the foundation. Now, we have to determine the corresponding S P settlement of the plate of. So, that means from here suppose this is the load versus settlement plot this is the load intensity and this is the settlement and this is the plate load test graph. So, first we will suppose S F is the permissible settlement of the foundation of width B F and the width of the plate that we are using is B P. Now, by the giving expression S F by S P. So, that expression that I have given that by using that this expression. So, this is B F B P plus 30 B P B F plus 30 to the power square. So, here S F is known that is the permissible settlement B P is known width of the plate B F is also known width of the foundation. So, only unknown is S P. So, corresponding to the permissible settlement of the foundation we will get the S P value. Now, once we get the S P value corresponding to the permissible settlement S F then from this chart suppose this is the S P value that we will get this is the S P value. Now, from this chart the intensity that we will get that will give us the Q S F. So, this Q S F will give us the S F bearing capacity of the footing. Similarly, this is the S F bearing capacity that we will get. Now, similarly how to calculate the actual settlement from this plate load test? Suppose the actual settlement that I will give actual settlement that the foundation that is give the settlement computed from plate load test divided by correction factor or C W. This correction factor is basically for the water table position correction factor because in the actual settlement that we are getting from the plate load test where we are not considering the water table effect. So, actual settlement will be the settlement computed from the plate load test by the correction factor. So, this correction factor C W we can calculate by using this expression 0.5 plus 0.5 D W divided by D F plus B. This is given by Peck Hansen and Thorban where D W is the depth of water table below the ground surface and D F D F is the depth of foundation and B is the width of foundation. Now, this calculation we can use or we can use the I S recommendation that is C W is 0.5 plus 0.5 D W dash by B and that should be less than equal to 1. This is given by I S 8009 where D W is the depth of water table from the base of the footing. Now, if you want to incorporate the water table effect in the actual. So, we will get the actual settlement then we have to use the correction factor then the settlement computed from the plate load test by the correction factor. Now, this correction factor either we can use the Peck Hansen and Thorban expression where D W is the depth of water table below the ground surface and suppose this is the footing width and this is the ground surface and this is the base. So, the water table position from this expression this depth is D W is from here that mean from the ground surface and D W dash is basically from the base of the footing. So, D W dash is from is measure from the base of the footing where a D W is measure from the ground surface this is ground surface from the ground surface. These are the two different. So, D W is from the ground surface and D W dash is from the base of the foundation. This is the only difference of this expression. So, this expression either we can use this one or this one to calculate the water table correction. Now, the next step that how to use this data to calculate the plate load test thing. So, that means the next one here we are using a single plate to plate load test data to calculate the settlement and bearing capacity of the foundation. Now, another process is proposed by the Housel in 1929 that here we will use the two plates. Suppose here say load Q 0 is the total load corresponding to the permissible settlement S A or is allowable. So, permissible settlement S E allowable. So, corresponding total load that is Q 0. So, we will conduct the two plate load test. So, one plate dimension is B 1 and B 2. This is the width of two plate one is B 1 and B 2. Now, from the load settlement plot this is the width of the two plates say B 1 and B 2 or the two plates B 1 and B 2 dimension and from the load settlement curve determine the load Q 1 and Q 2 corresponding to the settlement S E. So, that means we will get load settlement curve for two different plates. So, from these two different plates corresponding to the permissible settlement S E allowable will get the total load that this plate can carry corresponding to this permissible settlement. So, that will be say Q 1 and Q 2 are the two load that this two plates is carrying corresponding to the permissible settlement S E allowable. So, suppose these two loads are Q 1 and Q 2, Q 1 and Q 2 corresponding to S E allowable. Now, what we will do? So, suppose this is Q 1 total load is A 1 m plus P 1 n and Q 2 is A 2 m plus P 2 n where A 1 and A 2 are the areas of plates one and two. So, A 1 is the area of first plate and A 2 is the area of the second plate. Similarly, P 1 P 2 of the perimeter plates one and two P 1 is the perimeter of the first plate and M and N are two constant. These are the two constants. So, the value of these two constant we can determine by using this equation number 1 and equation number 2. So, here we will get the Q 1 and Q 2 from the graph A 1, A 2, P 1, P 2 are the known things. So, from this if you solve these two equation we will get the M and N value. Once we get the M and N, then finally this we will get the Q 0 value that is A into M plus P into N. So, where A is equal to the area of the foundation. Now, this is foundation area and P is perimeter of the foundation. So, here if we real foundation, if you know the area of the foundation and the perimeter of the foundation, then this known value is M and P, M and N corresponding. So, we will get the Q 0 is the total loss corresponding to the permissible settlement AC. So, corresponding to the permissible settlement AC what will be the load this foundation can carry is Q 0. So, in this fashion on this way we will determine the load that foundation can carry corresponding to this permissible settlement. This is one process where we are using the two plates. So, next one that we will discuss about the calculation of different other techniques, then how we will calculate the load settlement for the different other techniques that we will discuss here. So, next one is the next field to test that we are using that is our SPT value. This is also for the granular soil. So, first test that I have explained that is for the plate load test the SPT test I have already explained. Now, here we will get one data. So, here from the IS code is recommend that this is for the IS code recommendation this is part 1 1976. So, IS code produced one chart in this fashion. So, this is this side this is width of the foundation E in meter. This is the settlement in meter this is settlement in meter divided by unit pressure. This is 1 kg per centimeter square. So, that means, the ratio that we will get it is the settlement in meter per unit pressure. So, suppose this is the 1 meter width this is second 2 meter width this is 3 meter width this is 4 meter width this is 5 meter width this is 6 meter width. So, we will get this type of chart. So, this is for the 10 to the power minus 3 on the middle 1 is 10 to the power minus 2 or the top 1 say 10 to the power minus 1 and corresponding this value we will get from here. So, this is the this is in the log scale. So, from corresponding to the n value. So, this is the one chart n equal to 5 corresponding this is n equal to 10 this is the log scale this is 15 this is 20. So, this is n equal to 10 n equal to 20 15 this is n equal to 20. So, in this way up to the n equal to say 60 we will get different chart in between these two this value. So, if I know the width of the foundation of footing and then corresponding n value we will get this the settlement in unit pressure. So, that means we have to calculate the pressure or the pressure at that level then we have to multiply that ratio with that pressure here that pressure we have to calculate in terms of kg per centimeter square. So, this is one process by using the IS code or IS chart we will get the settlement for the granular soil. So, next one we will get the third one that we will use the SCPT or static cone penetration test value this is for also granular soil. Here the settlement that we will get the expression is 2.3 h by c log 10 sigma 0 bar divided by sigma 0. So, this is L sigma by sigma 0 bar. Now, here this coefficient c we will get in terms of 1.5 into q c by sigma 0 bar this is proposed by DBA Martin's 1957 where sigma 0 bar is the effective overburden pressure at the depth at which the test is carried out and q c is the static cone resistance q c is the static cone resistance h is the thickness of this layer and del p is the increment of the stress due to the application of the load. Now, further Meherov suggested this c the compressibility coefficient is 1.9 q c by sigma 0 bar this is suggested by Meherov in 1965. So, first expression is suggested by DBA and Martin's 1957 where they proposed this thing where h is the thickness of the soil layer c is the compressibility coefficient which is given by this expression is by DBA and Martin and this expression by Meherov and this that this sigma 0 by effective overburden pressure at the depth at which the test is carried out and del sigma is the increment of stress due to the applied of floating load. So, how to calculate the stress increment that we have already explained. So, again we will solve one example in the next class then we will get how to calculate this sigma del sigma. So, these are the different methods by which we can determine the settlement of the soil for the granular soil that is one first one is by the plate load test then by the SPT value where the charts are available by which we will determine the settlement for the granular soil then by using the static cone penetration test we can also determine the plate load settlement of the granular soil. In the next class we will solve few examples on this how to calculate the settlement by using these methods and we will discuss few more other techniques by which we will determine the settlement of the granular soil. Thank you.