 And last class I have discussed about this sheet pile design. So, I have discussed cantilever sheet pile. Now, I was discussing about this anchor sheet pile and the fix support method that I was discussing. Now, in the fix support method or fix earth support method that is discussed here. Actually, the proposed earth pressure distribution that was taken here and then that was simplified and we consider two different beam during the design. That means, this was the simplified form of earth pressure distribution. So, if this is sheet pile, this was the simplified earth pressure distribution. Now, this is the position of the anchor if this is A is the top, D is the position of the anchor, E is the dredge level, I is the point of puncture where moment is 0 and G is the point where this force is this stress is maximum. So, these were the assume earth, this was the simplified earth pressure distribution that was considered. So, these things was discussed in last class. So, now here if I draw the bending moment diagram of this portion, then this is the, this here it is maximum. Then at I point this value is 0, then this will follow this path. So, this is the G again this I point, this is E point and this portion, this is where it is maximum D point, this is A point. So, this is the pressure distribution from this is BMD. Then two different beam was considered one from A to I, this is A to I, this is the same E point and this is some D point where the force F is acting, this F is the anchor force and another one is from I to G. So, this is beam 1 and this is beam 2. Now, what are the forces which is acting here? The reaction force R 1 is acting the same force in opposite direction R 1 is acting in beam 2 and then R G is acting in the G point. So, this R G is the reaction for at G point, R 1 is the reaction force at the G I point for the beam 2, then R 1 is the reaction of the I point in for beam 1 and F is the anchor force. So, these are the forces. Now, we have to start design of this method. Now, one thing that you can notice that here this value where this point is 0, sorry this is the I point where bending moment is 0. So, this will be the I point. So, this is I point not this one. So, this point is where value is A and the I point is the point where bending moment is 0 and then this distance is A, A where the stress is 0 from the gauge level and I is the point where BMD is 0 from the gauge level. So, I is the point where bending moment is 0. So, now, if I consider this two things and then I consider this two things and then we simplify how to calculate these forces, then first if I take that this is our again if I consider the that same distribution that is this is gauge level, then this is the diagram G, this is may be the I point, this is E point, this is the anchored force is acting D point, this is A point. The same force diagram we are considering here, the forces are acting is in this direction, here it is acting in this direction. Now, if this distance or stress is 0, this then this is A and this I point from the gauge level is say I, distance of the I point from the gauge level is I. Now, so if I take two beams, then we have this is our first beam, this is beam 1 and this is from A to I and that beam if I draw the diagram, so this is the gauge level is here say this is say E point and then this is lower part portion. So, this is the pressure distribution diagram for beam 1 from A to I and what are the forces is acting here? So, one is force which is F acting at D point, this is A, this is E, this is I. So, this is F, then the reaction force is acting here R 1, then the P force which is the force due to this pressure is acting here, this is P 1. So, these are the forces, this P 1 is the external force which is acting here, say this P 1 is at a distance of Y bar from D point. Now, this P 1 can be calculated, the P 1 is the total force acting due to this zone from A to I. So, we can divide it one triangle then another triangle here and this one rectangle and then we can calculate what are the forces here. So, this P A E say suppose this portion P A E, so P A E is equal to active pressure at E level or E point. So, that we can determine, so for this one this will be the P A E will be K A into gamma into H, if H is the height of this point I, the point E from the A. So, H is equal to A E distance, then this is I point, this is I point, this is I point and these are the forces. Now, similarly for beam 2 which is I to G, where the force diagram is this I to G, so this small portion and this one. So, this is a small portion and this one. So, here forces is acting in this direction, here forces are acting in this direction. So, now if consider the forces which is acting for this I to G and this is the reaction force R G is acting, this is the same R 1 is acting in opposite direction and then this P 2 that will act for this small portion stress this P 2 and here for this one another force P 3 will act. So, these are the pressure distribution and forces in beam 1 and beam 2. Now, we have to determine the values of this forces. Suppose, this distance is from here to 0 point is A E 2 this point suppose this is O, similarly this is also O. So, E 2 O is that mean E O is equal to A and E I is equal to I then we can write that I O is equal to A minus I. So, this distance is A minus I. Similarly, total distance if I consider that from E to G is D that E to G that is equal to D and this D. So, we can write that this total 1 E 2 D is G and this E 2 O E 2 O is A. So, we can write O to G is equal to D and this E 2 O D minus A. So, this distance also D minus A. Now, what are the forces that will act? So, first if I consider that beam 1 consider this beam 1 and taking the moment at anchor level or odd level that is equal to 0. So, if I consider in that form then we can consider we can write that P 1 into Y bar that is equal to R 1 into H and if this distance from E A to D is E. So, this distance A to D that is equal to E. So, that means this distance total 1 is from up to E point it is H minus E H minus E then E 2 I E 2 I is I. So, plus I. So, we can write that R 1 value is P 1 into Y bar divided by H minus E. So, we can write this is this R 1 is P Y bar H minus E plus I. Now, here is P is known P 1 is known Y bar is known H is known E is known because we know the position of this anchor, but I is unknown. So, how to calculate the I? So, that first we will decide. So, to calculate the I we have to take the help of one graph. So, that graph we can draw here. So, this is 20 degree this is 25 degree this is 30 degree this is 35 degree and this is 40 degree. Similarly, this is 0.1 this is 0.2 this is 0.3. Now, in this axis this is phi is in this direction and I by H is in this direction. Now, this if I draw the graph. So, this will be the graph. So, here if I know the value of phi if I know the value of H then from this graph we can determine the I. So, any problem the phi value is known for a particular site H is also known. So, we can determine this I value. So, so that means the expression that we have derived that R 1 is equal to P 1 into Y bar divided by H minus E plus I. So, in this expression now I is known from this graph then E is known H is known. So, we can determine R 1. So, that is known. So, now from this beam 1. So, that means what are the forces that is. So, this is the beam 1 again if I draw. So, this is the F force acting at the D level this is A and then the P that will act for this external pressure. So, that is Y bar. So, this is P 1 and then will act this R 1. So, here this R 1 is known P is known now this F we have to determine. So, from this expression we will get the R 1 P 1 is known we will get the distribution now. So, this way what will be the pressure at similarly if I draw the pressure distribution diagram of this beam 1. So, at this point this is the diagram. So, this is D may be in this E point is this one and this is I point. So, at E point. So, this is the diagram. So, as I have discussed that at the E point this is P A E. Similarly, this value at the I point we can determine that is P 0. So, now we can determine P 0 linearly interpolating this value that is our P A E divided by I. So, this is this is P A E this is divided by total A that is equal to P 0 divided by A minus I because the total distribution if we see this is up to I is equal to P 0 distance. So, now if I draw the total distribution. So, this is E point. So, this will go up to here. So, at this point this is P 0 at I point. So, this is the point where this stress is 0 that is O point. So, this is E point this is D point and this is A point. So, now if I linearly interpolate this thing this is P A E. So, P A E this total distance is A and this up to I is I. So, P A E divided by A that is equal to P 0 divided by A minus I. So, from this expression we can determine P 0 value is P A E divided by A minus I. So, for this expression P 0 is also known. So, once we get this I value. So, what we have to do that our P A E is known then A is known. Now, we will calculate this A A is simply that expression P A E into K P minus K into gamma. So, that thing is that thing is already we have already discussed that we can determine this point where the distance of the point where the stress value is 0 from the dredge level that is A. So, that A we can determine this P A E the active pressure at dredge level divided by difference up to coefficient of the earth pressure and passive earth pressure minus active earth pressure into gamma. So, that means from this expression the A we can determine and if I put this A and if I determine this value of I from this graph. So, we will determine the I from this graph then we use a determine the A value from this expression. So, if this all the quantities on the right side are known. So, we will get this A value we will get I value from this graph. So, we will put these things in this expression we will get the P 0 value. So, that means this is our expression 2 this is our expression 3. So, we will from this expression 2 we will get the P 0 value. Once we get the P 0 value then we can determine what is the total force P 1 is acting due to this pressure diagram. So, then we can determine the P 1 as well as the y bar. So, once we get the P 1 and y bar then from equation 1 we can determine the r 1. So, now these are the values that we know. So, from this when you go for the second beam that r 1 should be required because r 1 will act this in the second beam also as the reaction force. So, now in the second beam r 1 and P 0 will be required. So, now if I go for the second beam. So, this is the second beam or beam 2. So, this is the expression this is P 0 at I point this is O point and this is G point this is say P 2 or stress. So, now forces that is acting in this beam that is R G reaction force similarly r 1 is coming from the top beam 1 and then the P 3 total force and then the P 2 for this triangle small triangle and this P 3 for this bigger triangle. So, this will act in this direction and this stresses will act in this direction. So, distance is here as I have mentioned that is A minus I and this distance from O to G is D minus A. So, now for the second beam if I write the expression that moment at G level is 0. If I take the moment at the G level that is 0. So, now if I take the moment then the P 3 value is the area of this triangle that means half into P 2 D minus A and that will act the distance of D by D minus A by 3. This is also D minus A divided by 3. So, that is equal to minus r 1 into D minus I because the total one for the beam 2 is D minus I. So, this is D minus I then minus P 2 is the force. So, that is acting the distance of D minus A then plus two third of A minus I. So, that is equal to 0. So, two third of A minus I of this small triangle. Now from this expression your D is unknown which you have to calculate A is known then r 1 is known from the expression 1 then I is known. So, P 2 we can determine that P 2 value the stress P 2 we can determine that if this is P 0 and the stress P 2 from these two similar triangle we can determine that P 2 force I mean this value we can determine that capital P 2 this P 2 capital P 2 is equal to half into P 0 into A minus I. So, half into P 0 into A minus I. Similarly P 3 capital P 3 is equal to half into gamma into small P 2 into small P 2 half into small P 2 into D minus A half into small P 2 is this value. Now small P 2 is gamma in this value is gamma into k p minus k into D minus A. So, that this value have to calculate small P 2 that is also been discussed in the previous classes. So, that is gamma of this level into difference of the two coefficient into that high D minus A. So, if I put this value the capital P 3 value is half into gamma into k p minus k into D A D minus A whole square. So, in the this previous expression number 4 if I put this value in expression 4 then this value is half into gamma into k p minus k into D minus A to the power 3 divided by 3 minus R 1 into D minus I minus P 2 into D minus A plus 2 third of A minus I. So, that is equal to 0. So, this expression will get by putting all this value in expression 4 and final expression that we will get that 1 by 6 gamma k p minus k into D minus I. So, that is equal to 0. So, this expression will get by putting all this value in expression 4 and final expression that we will get that 1 by 6 gamma k p minus k into D minus I into minus A to the power cube minus R 1 D minus I minus P 2 into 1 third into 3 D minus A minus 2 I that is equal to 0. So, then we will get this final fifth expression. So, in this expression or only unknown is D because I is known from the graph A is known from the expression then P 2 is known from this expression capital P 2 where P 0 is known A is known I is known then R 1 is known from the previous beam 1 I is known. So, all this expression in this is all the parameters are known except this D. So, by using expression 5 we determine this D value. So, using expression 5 determine the D value. So, determine D value then increase it by 40 percent to 20 percent. So, we will increase this D value by 20 to 40 percent. So, now this depth required depth we have calculated. Now, we have to calculate the forces which is acting here. So, the F forces F from this beam 1 we can determine if I see the beam 1. So, this is our D position which is F or anchor force. So, this is A position this is the F 1 or R 1 position and this is the force P 1 it act. So, F we can determine is equal to P 1 minus R 1. So, from this P 1 is known as a forces and R 1 we can determine by equation 1. So, once we get this value we can. So, one is R 1 is known P 1 is known then we can determine the F force also and we can determine by using the expression 5 we can determine the depth required for this sheet pipe. So, these are the two different methods one is of free earth support method another is fixed earth support method. So, these two method we can determine the required depth of the sheet pipe. So, in the next section here I will discuss that one another important thing that is our base excavation by using this sheet pile we can construct one very important geotechnical structure that is our base excavation. So, this type of base excavation is required for the excavation purpose or very deep excavation purpose. Where these lateral support are provided by sheet pile and in between the sheet pile we can provide some additional component that is called start to give additional support to the system. Now, what is sheet base excavation? Suppose if we have one excavation level in this form this is say existing round level and here this is the bottom of the excavated area. So, this portion is excavated where soil is taken out. So, now for this lateral support we can provide the sheet pile here for this base excavation. So, we can provide the sheet pile we can provide the sheet pile this side also and this is the existing this is the ground base excavation ground excavated and then in this area we can provide a lateral support or that is this horizontal components. So, depending upon the depth of this base excavation area we can determine how what is the number of this horizontal support will be required. So, this type of this is the sheet pile these components are called compressive member or this is called start. So, these starts are providing additional support to this sheet pile so that they can sustain or they can take the load lateral which is coming from this to side. So, now what are the things that we have to determine during the design of this base excavation? One thing that is very important that first we have to decide the depth of the excavated area. So, this is suppose the depth of the excavation this is the width of the excavation. So, now the very important four parameters that we have to so these things this is the width of the excavation B width D is the depth. So, these things suppose these D and B are fixed before the design. So, that means we have to provide this B and we have to provide this D that is fixed. Now, what are the very important things for design parameters we say. So, design parameter we can say that what will be the thickness of the sheet pile or that is the thickness of the sheet pile. So, that means what is the thickness of the sheet pile or support lateral I should say that what is the thickness of the lateral support I mean based on this thickness what or what is the stiffness of this lateral support or the stiffness. So, this is one design parameter another design parameter the position of this start. So, where we will place this start what is the spacing between two starts. So, this is our another very important design parameter. So, these are the design parameters we can say. So, that means the position of the start position means the spacing then the number of start. So, these are the very important another design parameter the next one is this depth of the lateral support below the base of the excavation. So, that means the small D below the base excavation and another one is the stiffness of the start. So, these are the things we have to decide before the during the design. So, that means what would be the position of the start what will the spacing between two starts what will the number of the start. So, that depends upon that this depth D depending upon this D what will be we can design the number of start we can design the spacing of the between the start. Then the next one in the what would be the thickness or stiffness of the this lateral support what will be the stiffness of the starts and what would be the depth of this lateral support below the base of the excavation. So, these things we have to decide when during the design of this excavation. Now what are the basis of then on what basis we can decide these parameters. So, one thing is that when you are talking about this base excavation suppose if I draw the same figure here. So, if I draw this same figure here this is the start. So, now the things that we have to observe that should be within permissible limit that this support will deflect. So, there is a lateral deformation of the support. So, lateral deformation that should be within permissible limit that means we cannot provide a huge lateral deformation of this system that should be within permissible limit. So, these things we have to check. So, then next one is due to this lateral deformation or this thing this ground surface will deform in this pattern this ground surface will deform. So, vertical deformation of the ground. So, we will get this type of ground surface deformation. So, this ground surface deformation or vertical settlement of the ground vertical deformation of the ground and that is also important because if there is an existing structure is there. So, we have to design this system such that there is no such influence of this excavation on that existing structure then there should not be any deformation. So, for that this lateral support lateral deformation of the support then ground surface deformation these are very important things. So, that we have to taken care during the design. The next one what is the BM bending moment lateral support. So, we will choose this section then what is the maximum bending moment of this lateral support then what is the maximum force or forces in the start. So, start will be subjected when these two wall is deforming in this direction and this is also deforming in this direction. So, this will give a compressive load or force to this starts. So, that means this maximum force in the start. So, these things we have to check during this design. There should not be a huge lateral deformation that lateral deformation of the support should be within permissible limit vertical settlement of the ground within permissible limit. The maximum bending moment what is the maximum bending moment of this lateral support and then what is the maximum struts of this forces acting in the start. So, these things we have to take check during the design and based on that what to design what to decide those design parameters that means the what is to be the stiffness of thickness of lateral support then the position of the starts then depth of the lateral in support below the base of the excavation then stiffness of this starts and then of course the this is the section of any base excavation. So, there is if I draw this another view. So, this side this is another. So, these starts are provided to certain interval. So, that interval one is as a vertical spacing between the two start another is that this is continuous but this starts are not continuous. So, these are placed with certain interval in the longitudinal direction also. So, this is the spacing between in the vertical spacing another is the horizontal spacing between the this starts. So, that that thing we have to also taken care during the design. So, now these are the design things that we have to decide now for this simplified analysis of this type of system. So, now we have to go for the rigorous analysis when we are taking all these parameters during the design and then we have to take we assume that mean first we have to check every time that if I consider this type of configuration then what will be the our design design steps what will be the our settlement in the lateral direction in the vertical direction. So, every time we have to check all these things. So, now that then our aim of the design is that you should design at the system such that that the lateral deformation should be minimum the vertical settlement should be minimum and maximum bending moment is minimized can be minimized and start force is also can be minimized. So, all these things we can minimize. So, that that is our design should be an economical one. So, now for this one first we will discuss a simplified method of during this design suppose if this is our lateral support and this is base excavation then this is our sheet pile and these are the starts. So, we are taking the half portion of this ring. So, then this is the base of the excavated ground and we consider this value is h h or d. Now, for this different type of soil condition we can draw the pressure distribution I mean the this pressure because this this side is soil the soil is giving pressure to this lateral support and then this lateral support is giving pressure to this start. So, that mean there will be a force that will develop in the starts. So, that means we can for the different type of soil we have some different stress diagram. So, that depends upon the speed at which excavation is advanced and then the type of the soil. Suppose if I we have a soil that is sandy soil. So, first soil is sandy sandy soil. So, here we can get this is the distribution pressure 0.65 gamma h k a. So, that means this pressure diagram this pressure this is 0.65 gamma a gamma is the unit weight of the soil gamma h into k a that is the lateral pressure distribution which is acting here suppose here this will be the direction. So, that depends upon the type of soil and the speed at which the excavation is going on. So, here k a we can write 1 minus sin phi divided by 1 plus sin phi. Now, if the soil is firm clay or stiff clay then the distribution will be different. So, then the total distribution pattern will be different. So, this is the this value is 0.22, 0.4 into gamma h. So, this is for your firm clay or stiff clay or stiff clay or stiff clay whose gamma into h then Cu is less than equal to 4. This is another condition. So, that means here this value is h by 4 this is also h by 4. So, that means this one is h by 12. So, this force will also this stress will also act in this direction. So, if the soil is soft clay then the distribution will be different. So, this is the stress that will act. So, that means here 0.2 to 0.4 gamma into h and here the distribution value this is gamma into h minus 4 Cu. So, this is for soft clay medium clay. So, that means gamma h divided by Cu greater than 4. So, this is another distribution this is h by 4 and this value is two-third of h by 4. So, that means these are the three different distribution for the different type of soil. One is if it is purely sandy soil then this is for the firm clay and this is for the soft to medium clay. Now, how to analyze these things to determine the forces in the start and this part and suppose we have a base excavation system like this. This is our ground level this is the base of the excavation. This is another side and then we are taking the number of start is 4 at 4 different level. So, now suppose if it is sandy soil so we will get this type of distribution. This is the sand. So, this is 0.65 into gamma into h into k. Now, we can divide this total diagram or total this sheet perl lateral support in different number of beam or segment where. So, that means here if I consider one beam from up to the second support and then from second to third. So, this is first start, this is the second start, the third start, this is the position of the fourth start. So, we are taking one beam from top to second start and this is the second start. So, this taking one beam from top to second start and then another one second to third another one third to bottom. So, there is three beam where decided one is from top to second start position another is second to third third to the bottom. So, we have the reaction for this second first start that is one and the reaction some reaction of this second start that is R 2 dash and then the second beam this is another reaction for the second beam that is R 2 double dash some portion of the R 3 dash for the second beam and for the third beam that is R 3 double dash and then for this beam R 4. So, these are the forces acting now if I get one by one this beams. So, this is the first beam or beam one. So, where we can draw the distribution or U D L whose density is 0.65 whose value is gamma H into K A that is equal to P A. So, what are the reactions? So, this is the reaction that is R 2 dash and then here this is R 1. Similarly, for the second beam we consider this is P A again the same stress is acting then the reaction is R 2 double dash then R 3 dash this is for beam 2. Now for the third beam the same P is acting then this is R 3 double dash and then the R 4 is acting here. So, these are the this is beam 3. So, this we are taking three beams in this way we can take the more number of beams if the number of starts more and you have to take different diagram for different stress distribution. So, now we have to take these three then you have to solve this determine this R 2 dash R 1 because we know this distance between these two starts and this value that is also known. So, now we have to take determine the R these reactions separately by considering this beam. So, this is one beam whose this spacing this distance between two start is known this U D L is known. So, you can determine these two reactions similarly, for this beam 2 also similarly, for the beam 3. So, once we get all the reactions then the force of the first start will be R 1 force of the second start will be R 2 dash plus R 2 double dash force of the third start will be R 3 dash plus R 3 double dash and force of the fourth start will be R 4. So, then we can determine what is the forces of different starts. This is one method and then we can determine what would be the movement maximum movement at this lateral support also. So, those things. So, we will discuss and for this is what U D L is assigned for the clay will get different type of variation. So, those things I will discuss how to calculate these forces in the next class with some example problem. Thank you.