 In the last few classes, I have discussed about the various methodology of soil structure interaction or soil foundation interaction, then I have explained about the modulus of sub grade reaction. So, now the modulus of sub grade reaction, how it is determined by the lab test or it is also explained, then why it is used, how it is used those things also explains. So, today I will just explain that based on this modulus of sub grade reaction application, now how the lateral loaded pile is been analyzed and then how the deflection of the pile, then the slope bending moment, shear force all this thing can be done determined by using this modulus of sub grade reaction techniques. And then I will also explain the tension pile or the uplift capacity of the pile. So, now all the other classes of pile foundation have explained only the compression, piles are subjected by compressive load. Now, today I will explain that the tension pile, the tension pile piles are not only subjected to compressive load, piles are sometimes subjected to tension load also depending upon the situation. For example, if it is a very tall structure or it is a tower, then chimney, then the piles are in that case are subjected to both tension as well as compressive load. So, for example, if we can write or I can explain in this form that the tension pile, so the tension pile or uplift capacity of the piles. So, for example, if it is a very, it is a tower, then the that the foundation say these are the foundation, these are piles. Now, if these are the pile foundation, then if I draw the six plan, then suppose this is one pile cap or this is another pile cap, then these piles when there will be a wind load subjected to a lateral direction, then there is a possibility that this thing will rotate in this form, then these piles will be subjected by compressive load and this pile will be subjected by tension load. So, that means these piles are subjected by tension and this pile is subjected to by compression and if the wind direction changes in this opposite direction, then these things will reverse. Now, this is a possibility it will rotate in opposite direction. So, these piles will be subjected to by compressive load and these piles will be subjected to by tension load. So, that means in that case piles, we have to design the piles for both it can take the compressive load as well as it can take the tension load also. So, now for this how to calculate the tension capacity or uplift capacity of a piles. Now, the first the piles in clay. So, in the piles with the uplift capacity U L that will be friction for the tension F S in tension into A S plus W P. So, now where W P is the weight of the pile. Now, here that if we are considering a single pile or group of pile, suppose if this is one pile which is subjected to tension, then the resistance this pile will get that here there will be no tip resistance because piles are pile is subjected to tension. So, that means the weight of the pile that will act in downward direction and when this pile is going upward direction then this friction will act in the opposite direction. So, this is the friction this is F S T into the area of the surrounding area of the pile. So, this friction. So, that is why this friction that is acting in the opposite direction of the loading. So, loading in the upward direction. So, this will act in the downward direction and the weight of the pile that will also act in the downward direction. So, the ultimate load carrying capacity of the pile in a tensile condition that is the q ultimate in tension. So, that will be F S T friction force in the tension and area of the surrounding area of the pile and the weight of the pile. Now, here we can write if it is in the clay soil that F S T is adhesion of the pile soil and the pile and area plus the pressure of W P where C A is the average adhesion along the pile shaft. So, along the pile shaft as I have explained the adhesion means the this is the interaction between the soil and the pile material. So, that is the average adhesion of the pile and this is F S T is the unit skin friction in tension. A S is the embedded area of the pile shaft. So, now for this particular pile if D is the diameter of the pile. So, A S and L is the length. So, A S will be pi D into L. So, pi is the cross section this periphery of the pile pi D into L is the length. So, this will be the A S value and W P is the weight of the pile. So, F S T is equal to F S T is equal to F S in compression. So, F S T is equal to F S in compression when for the undrained condition of the pile of the soil. So, this F S T will be equal to F S that is a friction that is a skin friction in compression for undrained condition of the pile. Now, sometimes pile has a enlarged base to increase this friction capacity. So, that means if pile has a enlarged base. So, that means if that pile it has this is a shaft of the pile and it has a enlarged base to increase this tension. So, in that case this will be in the tension opposite direction. So, in that case the ultimate load carrying capacity will be C u again A S into K plus weight of the soil plus weight of the pile. So, that means here this is given by the mayor of and Adams that is for undrained condition. So, where again so that means again this will be another expression is available that 2.25 into pi D B square minus D square into C u plus weight of the pile. So, from here the lesser of this two will give the in this case ultimate tension carrying capacity of the pile. So that means we can say that here that A S that is the surface area of the vertical cylinder above the base. Suppose this is the base. So, A S will be from here to here the outside area of this outside cylinder of the pile. So, that means here and then A S will be D B is diameter of the base while shaft diameter. So, that means here this is D B diameter of the base and this is the D pile shaft diameter. And K is one coefficient coefficient that depends on for depends on the soil type then for the soft soil this K is equal to 1 to 1.25 and for the medium clay soil this K is 0.7 and for the stiff soil K is 0.5 and again weight of the soil this W A S is weight of the soil included included the annulus between the pile shaft vertical cylinder. So, that means here we can write the weight of the soil here the vertical cylinder that is considering above the base this is the vertical cylinder. So, this portion this is weight of the pile and this portion the weight of the soil included this annulus between the pile shaft and the vertical cylinder. So, this is the this is the vertical cylinder above the pile and this is pile shaft. So, within this soil this will give the weight of the soil. So, why this expression? So, we have this two expression so lesser of these two will give the uplift capacity of the pile if it has a enlarge base. Now, the next section that if the soil has a C phi soil if it is a C phi soil then what would be the uplift capacity of the pile. So, here we have explained about if the soil is a totally C soil or the cohesive soil. Now, if the soil is a C phi soil then how the uplift capacity of the pile is can be determined. In that case again it is given by the mayor of an Adams 1968 that for a limited, limiting height of failure surface above base suppose for a limiting height surface above base say that is H. So, so here we have considered that if this is the pile and it is subjected by uplift force then if this L is the length of the pile and for a another limited height limiting failure surface that means it is considered that for that means the failure surface of the height of failure surface in the pile is H. So, that means here that is the failure surface and this pile will fail along this surface. So, that means here it is say intact that means this up to this portion there is a separation between the soil and from here it has been failed. So, that means the height of failure surface say H. So, that means that is the limited height of failure surface and in that case if L is less than H that means the shallow depth that means the length of the pile within the height of the failure surface then q ultimate pi C d B into L plus pi by 2 gamma S d B L square K u tan phi into plus W p. So, W p is the weight of the pile now if L is greater than H then q ultimate will be pi C d B into L plus pi C d B into H plus S pi by 2 gamma d B 2 L minus H to H K u tan phi plus W p. So, that means this W p is the weight of the pile. Now where S is equal to safe factor which is 1 plus M L by d B with maximum value of 1 plus W p. So, that means this W p is the weight of the pile now where S is equal to safe factor which is 1 plus M L by d B with maximum value of 1 plus M H by d B. So, that means this cannot be greater than 1 plus M H by d B and M is a coefficient which is function of phi in a friction angle of the soil gamma dash is the effective unit weight of the soil of the soil q is the earth pressure coefficient which varies from 0.92, 0.95 if phi varies from 25 degree to 40 degree. So, we can write in this expression where C is the coefficient d B is the diameter of the base. So, this is the diameter of the base depending upon and S is safe factor and K is the earth pressure coefficient and W p is the weight of the pile. Now that again this M value is given in this form that if we have a phi value say in degree is 20 degree, 25 degree, 30 degree, 35 degree, 40 degree and 45 degree this is the phi value. So, we have this H by d B a ratio is 2.5 say near 3, 4, 5, 7 and 9 then M that value has 0.05, 0.1, 0.15, 0.25, 0.35, 0.50. So, this as the M value that coefficient that we will get. So, that is safe factor similarly we can determine the S max what will be the S max value. This is taken from Ranjan and Rao 2000. So, now similarly that q ultimate max that will be pi by 4 is 0. So, this is the M value that coefficient that we will get. So, that is safe factor similarly we can determine the S max what will be the S max value. This is taken from Ranjan and Rao 2000. So, now similarly that q ultimate max that will be pi by 4 is 0. So, this is the M value that coefficient d B square minus d square into C N C plus sigma V into N q plus A S F S plus W P. So, that means q ultimate max will be this value as N C N q is the bearing capacity factor those are determined by the table which is proposed earlier. So, F S will be the ultimate shaft resistance and sigma V bar will be the effective vertical stress at the base of the pile. So, that means the q ultimate x that the maximum ultimate force that is the W P is the weight of the pile. So, this portion is the weight of the pile. This is the friction resistance and this is the bearing or the resistance or this C N C value. So, that means your q ultimate of the maximum force that is given this friction ultimate shaft resistance and bearing capacity factor C N C and this is given for the total maximum ultimate load of the pile. So, this way we can determine the tension capacity of the pile. So, now this so pile is again can be subjected to lateral load also. So, depending upon if it is subjected to wind load or if it is subjected to seismic load it can be subjected to water pressure or if it is a water fond structure. So, this way and if it is a retaining wall below the retaining wall if pile are placed then the piles can be subjected to lateral load also. So, we have to know how to calculate this lateral load capacity of the pile. So, then that means process here the one method is been explained. So, that is for the lateral loaded pile. So, when piles are subjected to lateral load or moment. So, that means here we can say say suppose this is one pile. So, this is ground surface. So, that can be subjected to a lateral load which is applied q at the ground and that can be subjected a moment at the ground level at the top. So, that means from because of these two moment and this lateral load pile has some deflection at any depth z. So, suppose this is the deflection y of the pile at any depth z. Now, this deflection of the pile is combination of two thing one is due to the moment another is due to the lateral load. So, that means here the process the method this is proposed by Ries and Matlock which is applicable for long pile. So, first condition this theory is applicable for long pile when length of the pile is greater than 5 times of the T where T is the relative stiffness factor can be determined using this expression that E i by small k where E and I are modulus of elasticity and then E i by small k where moment of inertia of pile material. Now, small k is the unit modulus of sub grade reaction. So, this is modulus of sub grade reaction. So, that means here during the analysis this soils are replaced by spring and then based on that got the basic differential equation and then using the finite difference technique these equations are solved and then ultimately the shear of this all the other unknowns that in the deflection the slope or shear force bending moment can be determined. So, that means, but it is applicable only for the long pile because in the if pile is very very length is high when the long pile then it is assumed that a very high depth that means the deflection of the pile is negligible. So, in that is applicable here. So, that is why it is only applicable for the long pile condition where condition that at the higher length or depth the pile deflection of pile may be 0 or negligible. So, now if you have considered that pile is subjected to lateral load as well as a moment at the top and then we will get a deflection of the y at any depth z. So, that means we have a different kind of deflection pattern depending upon the type of load. So, that means this y is a combination of this load due to the q g and due to the m g also. Now, if we can draw the figure which is due to the both deflection. So, that means if I superimpose these things suppose we have a pile which is subjected to lateral load q g and a moment and it will have a deflection like this. So, at any depth z say deformation is or deflection is y. So, that can be equal to this is ground surface. So, you can separately take the two cases one is due to the lateral load due to the lateral load another is due to the moment. So, this is due to the lateral load again we will have some deflection at the same depth z you have a deflection say y 1. So, this is case one. Where deflection is y 1 at the same depth z due to the lateral load only and then plus we have the same pile this is ground surface. So, we will have a moment applied at the top and then again at the deflection that will act here. So, that means we have a same depth we have a deflection y 2. So, that means y 1 is the deflection of the pile at a depth z due to the q g or that horizontal load applied at the top and y 2 is the deflection at the same depth due to the applied moment m g. So, finally we can write that y will be y a plus y y 1 plus y 2. Now again we can say that any deflection if I case the general once any deflection that is function of the load which is applied again which is function of the moment which is applied again the deflection is function of z also because as the depth changes the deflection value will also change it is not uniform to throughout the pile. So, that is deflection is function of z deflection is also function of l depending upon the length of the pile deflection will change deflection will also function of t that the relative stiffness of the pile deflection is also function of E i that means the material properties or the geometry of the pile. So, these 6 things are and again this is also the another one that is a modulus of sub grade reaction that is key. So, that k is the modulus of reaction so that means there is 1 2 3 4 5 6 7 factors. So, deflection depends on this 7 factors. So, say from here these 2 are the loading or the moment and this z is z is that any depth of the pile l is the length t is the relative stiffness. So, this is material properties or the geometry of the pile and k is the soil properties that is the modulus of sub grade reaction of the soil surrounding soil. So, k is also the soil properties or the material properties. Now, if I consider that in the 2 case where case 1 and case 2 then this write this case a for the case a for the 1 that in this form y 1 divided by q g and then that is function of z l t E i and k. So, the first case so here y g is function of again z l t E i and k. Similarly, for the case 2 y 2 by m g that is also function of z l t E i and k. So, we can see that both the cases case 1 and case 2. So, there are in general 6 factors first when you take the case 1. So, and write this y a divided by q g then that is a function of this 5 parameters and similarly, actually this y 1 is function of 5 parameters or including the q g. So, now in this form you have to write and you have some non-dimensional factor that is proposed in for different cases. Now, for the case 1 so it is represented in this form y 1 into E i by q g t cube is 1 factor z by t another factor l by t another factor and k to the t 4 by E i is another factor. So, that means, these 4 thing are expressed in this form then for case 2 y 2 into E i then for case 2 y 2 into E i then for case 2 y 2 by m g into t square. So, this factor. So, again z by t is 1 factor l by t is 1 factor and k t to the power 4 E i is another factor. So, we are represented in this is 4 factor for all the cases. So, here so this form here is z by t is called as a depth factor or depth coefficient then the soil modulus function k t to the power 4 E i is called soil modulus function. Then another one is the deflection coefficient which is for the load it is a y. So, for the y a E i q g into t square t cube and for the deflection this is for the deflection load lateral load and b y that is y b into E i by m g into t square that is for moment. So, these are the factor. So, this is a depth factor or depth coefficient soil modulus function then deflection coefficient all the force are. So, this is represented in this non dimensional form. Now, the finally the deflection. So, once it is so after the so finally this deflection y is given y 1 plus y 2 and that is expressed in this form q g t cube divided by this by E i into a y plus m g into t square divided by E i into b y. So, now the slope s is represented by s a plus s b similarly that is q g t square by E i into a y plus m g into t square divided by E i into a s plus m g t divided by E i into b s. Similarly the moment m is written as m a plus m b. So, a due to the lateral load and b means it is due to the moment. So, again it is written in this form that q g into t to a m plus m g into b m. This is moment. Similarly shear force v is v a plus v b or it is written q g into a v plus m g into a v plus m g into plus m g divided by t into b v. And finally the soil reaction is p is written p a plus p b and that is q g divided by t into a p plus m g divided by t square into b v plus m g b p. So, that means here this is the deflection expression. So, we have deflection, slope, moment shear force and soil reaction. So, here we have all unknowns a y b y a s b s a m b m a v b b a p b p. So, all these unknowns are then also based on this modulus of subcarre reaction techniques this final basic differential equation is solved using the finite difference method. And then these unknowns all these 1 2 3 4 5 6 7 8 9 10 this 10 unknowns are determined and then it is given in a tabular form. So, that means this form this stable we can determine we can pick any value at any depth. So, at any depth based on this any depth in non-dimensional form these values are given. So, that means one a y b y. So, if we know this a y b y at any depth for different condition a s b s. So, we can draw the deflection pattern along the slope of the pile along the depth and the bending moment and shear force and soil reaction of the pile along the depth also. So, now one thing that we have the different pile head condition. So, if we have we can it can be have a free hand pile it can be a fixed end pile also. So, now if it is a free and free head pile or free hand pile then there will be no moment will develop due to the fixity, but if it is a purely fixed end pile then if apply the lateral load then a fixed end or the moment that will develop due to the fixity. So, for the fixed end pile or completely fixed end pile. So, due to the fixity a moment that will develop minus g is point which is equivalent to 0.93 q g into t and for completely free pile this m g or this is that means this is minus we can write here. So, that is 0. So, that means free end pile no moment due to develop due to the fixity. So, here this will develop fixity of the pile. So, that means if it is a free head pile so that means no moment you will develop due to the fixity of the pile, but if it is a fixed end pile then a moment that will develop in the opposite direction of the direction of the loading. So, that is it is given a additional rigidity of the fixity of the pile. So, that due to that is equivalent to 0.93 q g t and if the fixity of the of the pile head is within this value 1 is 0 I mean the free end are they perfectly fixed and if it is within that range then by interpolating from 0 to 0.93 q g t from 0 to this value we can determine what will be the moment due to the fixity if it is fixity is within that range. Now, this is the completely how we can determine this load. So, now far we will calculate solve a example problem that we have a length of the pile L is equal to 20 meter and it is in uniform sand. Now q lateral load that is applied is 25 kilo Newton and the modular unit modulus of sub grade reaction that is the reaction of the soil is equal to is taken a 10 into 10 to the power 6 Newton per kilo Newton millimeter then E I is given 3.7 into 10 to the power 7 Newton meter square. Now, what is the deflection deflection one condition is the free head and another condition completely fixed. So, first we will calculate what is the t value of the pile that is E I by k. So, root over 5 E I is 3.7 into 10 to the power 7 and k is 10 into 10 to the power 6. So, we have a value 1.299 meter. Now here we can say L is greater than 5 times of the t. So, we can apply these things for the long pile. So, we can apply this theory. Now, we can know that y is here only no moment is applied if it is a free head and only the q g is applied. So, the expression for the y value is y is y 1 and y 1 is q g into t cube divided by E I into A y. So, q g is 25 into 3 Newton and t is 1.299 cube divided by 3.7 E I into 10 to the power 7 and it is we are determined the deflection at topmost point of the pile head. So, we are determined the what is the deflection at pile head condition. So, that means at z equal to 0 at z equal to 0 that A y value is given 2.435. So, 2 points that we will get from the table. So, that is 2.435. So, A y value at z or the non dimensional the def factor is equal to 0 at top of the pile. So, ultimately we will get a deformation that is 3.61 millimeter. So, that is the deformation if so that is the deformation if pile head is completely free. So, we will get a 3.61 millimeter deformation at the pile top. But again if we have a fixed head pile, fixed head pile then one additional moment that will generate because of the fixed tip it is a completely fixed at pile 0.93 q g into t. So, this fixed tip will develop in the opposite direction. So, point that value is 93 into 25 into 10 to the power 3 to 1.299. So, ultimately this value is coming minus 0.320 into 10 to the power 3 Newton meter. So, due to this moment which is acting in the opposite direction. So, we have some deflection and basically that will reduce the deflection of the free end condition. So, now here the total deflection will be y 1 plus y 2. So, here y 1 is 3.61 millimeter and plus. So, this will be the minus because it is minus 30. So, minus mg it is due to the moment t square divided by e i into p y. So, this is mg t square there is over the slope expression that is given. So, mg t square e i into p y. So, ultimately 3.61 minus mg is 30.20 into 10 to the power 3 t square is 1.299 square divided by 3.7 into 10 to the power 7 into by 1.023. So, this is the b y value at top of the pile. So, ultimately you will get 3.61 minus 1.41. So, that is so that means this is 2.2 millimeter. So, we have a deflection. So, now we can say for the y for free is 3.61 millimeter y fixed is 2.20. So, this is the b y value at top of the millimeter. So, we can see that because of this fixity we are getting a moment which is basically reducing the deflection of the pile head. So, that means here this is a 3.61 millimeter was the deflection for the free end pile, free end pile and for the fixed pile to reduce another deflection is 2.20 millimeter. So, now this way that if it is a free end pile and if it is a fixed end pile, the fixed end pile due to fixity we can get the less amount of the deflection at the top. So, similarly for this is the deflection at the top. So, from the table if we know the other coefficient values and then we can determine the any moment deformation deflection or deformation or deflection slope bending moment shear force reaction at any depth for the pile for different loading condition or different pile boundary condition also. So, in the next class we will discuss about the pyrofinition and then that will be the last section of this course. Thank you.