 So, you must have realized that I introduced the concept of the tie rod with the anchorage. These systems are known as anchored bulkhead alright. Sometimes we also call them as the propped sheet piles alright. What is propping? Propping is nothing but putting a tie rod or a prop material or in simple possible forms these are also known as anchored sheet piles. Now the issue is how I am going to analyze this type of situation. First of all understand why propping has been done or why tie rod has been included. You must have realized that the bending moment is a function of h cube. So the more high you go what is going to happen the bending moments are going to become extremely high. We want to avoid this type of situation in the field clear. So the moment I put a tie rod this point it starts behaving like a pin okay. So that means what I am doing is this becomes a pin joint and by propping it from here I am reducing the bending moments that is the mechanistic model. So the moment you pull it from this side the deflections can be controlled that is what the whole philosophy is all about. Now you must have seen the bow and arrows alright. So a typical bow will look like this and then somewhere here would be the arrow. So what normally we will do we will try to pull it back. What is happening to the bow it deflects correct inside. The same is the logic over here. The moment you apply a tension from the backside you are stopping the deflections. Now there are two ways you must have seen some of the sportsmen in archery what do they do. They will put a leg over here alright if the height or the length of the bow is too much and that is advisable. That means by putting a leg over here what they are doing they are inducing a resistance watch it closely. In the second situation we do not apply any constraint over here there is no restraint for the motion or the movement and we say r equal to 0. So depending upon these two logics two methods have been created for analyzing the propped sheet piles or anchored bulkheads. The second one which I have talked about here is known as a free earth support method and the first one which I talked about is known as fixed earth support method. So these are the two methods which are normally applied to obtain or to analyze the anchored bulkheads. In this case which is free earth support method the value of r is going to be 0. In this case the sheet pile is free to rotate at the bottom point at this point. In the second case r is not equal to 0 it is fixed there is a reaction you are putting a constraint alright. So this is free to deflect and rotate this one is fixed because of the reaction which is imposed on this. So these are the two methods. Now let us see how the analysis will be done. So for analyzing this situation if I consider this is the ground surface this is the dredge 11 somewhere here you have a tension component which is coming over because of the tie rod or the propping. We normally assume this to be a depth a sometimes you have to assume a when you are designing or sometimes a might be provided there is no issues. Now the question is draw the deflected shape of this type of a sheet pile and draw the pressure distribution. So the pressure distribution is very simple yes that is what we did active earth pressure and then how about the passive earth pressure. Pressure diagram is done this is active earth pressure this is passive earth pressure at this point r is 0 why because we are assuming that this is a free earth support method fine. Draw the deflected shape in this case it should start like this because once you are pulling it back the deflection is going to be inside the backfill okay is this okay deflected shape have you understood this. So this is the deflected shape of the sheet pile correct. What we do is as I we were discussing earlier we put a factor of safety of f over here why we are assuming that the full passive earth pressure is not getting mobilized intentionally. So we are designing it for the worst possible situation. Now suppose if I take moment about point T so that means is this okay or not I can put a condition because this happens to be a pin joint is this okay I have shown deflection to be 0 at this point. So the more and more anchors you add the more and more pin joints would get created. What about the second condition the second condition I can get by equilibrating the forces. So that means T plus TP over F equal to PA this is okay what is the principle unknown the value of D and this is the height of the wall. So the problem statements could be like this that I want to design a retaining wall of height H prefixed because H happens to be a finished product alright this is our main objective. So for a given H value what would be the value of D and we are analyzing this situation by assuming 3 years of ordinary for a probed system. The advantage of propping is I can go for a very high higher heights as compared to the initial heights. So suppose if this is H1 and if I keep it as H, H is going to be higher than H1 this is the advantage which I am going to get is this fine. Any questions regarding this can you draw the bending moment diagram try drawing the bending moment diagram for this case where is the prop here how would bending moment diagram will look like good. It gets it reverses this direction alright why this is a there is no constraint on this there is no restriction on this correct. So it is a free or support the pile is free to rotate bending moment is going to be 0 at this point and then you are seeing that there is a point of contraplexure also which is going to come which we can analyze subsequently this is okay. Now if you have followed this the fixed earth support method becomes very easy to follow what we do is in free earth support method we do not impose a factor of safety but yes that D value will be incremented by 20% so in this case this is the propping which has been done draw the deflected shape first there will be a concept of point of contraplexure will come somewhere over here this is normally at a certain depth below the dredge level so we define this as y and y is dependent upon friction angle and I will give you the values so if 5 prime is known you can get the value of y as if it is 25 degree sorry 20 degree is the standard value 20 degree this is 0.25 h if it is 30 degree then it is 0.08 h and if it is 40 degree it is minus 0.007 h what is that you are observing the more the compacted the material would be below the deadline the point of contraplexure as a tendency to pop up in fact for 40 degree what you are observing is the point of contraplexure would be above the deadline fine so what is the significance of this if there is no surcharge over here that is why you must have noticed when we construct the sheet pile walls to take care of the point of contraplexure popping out of the dredge line normally this portion is overloaded with surcharge so the moment surcharge is overloaded over here the y value will still remain as positive this is part okay you have to practice it a bit so if you draw the deflected shape now how it will look like try doing it yourself this portion remains common point of contraplexure what is going to happen the bending moment is going to be 0 at this point somewhere here is the point d which we have taken alright and then this portion is 0.2 times the value of d so if I complete this this is how it will look like this continues but what should have happened at point d where you are imposing r if you are imposing r then at point d the deflection should be 0 correct so this is where your net resultant r is acting you can draw the pressure diagrams this portion up to r is passive earth pressure from here up to r is active earth pressure and this point contraplexure which is at depth of y is somewhere in between now the question is how are you going to solve this of course draw the bending moment diagram also here so what will happen to the bending moment diagram in this figure this will get slightly modified because we will be having a point C which is the point of contraplexure so at point C the bending moment becomes 0 yes fine and where is the point C somewhere here is this correct can make it smooth this corresponds to the r value now there is something known as equivalent beam method you must have studied also so the only difference between fixed earth support method and free earth support method in terms of the mobilization of passive earth pressures is here we have put a penalty on passive earth pressure in the form of macro safety here we do not put okay and then we use the concepts of equilibrium of the forces which are acting on the beam and now I have an advantage I can cut the beam to make an equivalent beam about point C so I can cut the beam about this point so this becomes my section number 1 this is the section number 2 of the beam this is what is fixed earth support method equivalent beam method so what we have done we have taken 2 portions of the beam one is from A this is let us say B C D E and F so from A to A to C we have 1 beam draw the free body diagram of A to C or I can rub this so this is the complete section and this is the point D this is C this is somewhere D this is B this is A this is 1 active earth pressure this is the y value the depth of contra flexure passive earth pressure if I cut the beam over here now what is going to happen the moment you cut there will be 1 reaction R1 moment is 0 so moment about C is also 0 and then what I can do I can go up to point number this point and I can take another section of the pressure which is acting like this and this is our extra 20% so I can get rid of this at the bottom portion you have external imposed R if this is R1 acting over here we will have equal and opposite R1 acting over here this is the PP rest of the portion of this length that is CE and this one is going to be PA of CE you can create different conditions for moments at point B there is a tension occurring I can take moment about point B 0 moment about point C as 0 clear so depending upon how many unknowns you have and then force equilibrium I can have PA1 equal to PP plus R1 and similarly here you have PA CE minus PP CE plus R1 plus R equal to 0 and hence you can solve this equation we will have to practice this to understand how the analysis is done.