 In the last class, I have discussed about the seismic retaining wall. Now, in the last section of this previous class, I have discussed the hydrodynamic force during the seismic condition. So, today I will discuss about the how to determine the location of that hydrodynamic force in different condition. So, now for this section, now this pore water hydrodynamic effect of the pore waters as I have discussed in the last class that if this is the retaining wall and this side is water and this side this is soil. And assume that water surface or ground water surface. So, this is ground line or ground level the ground water table and the height of this water table is h. So, this is retaining wall, this side this is water free water and this side soil is there inside the soil pores water represent and water surface is again here also it is height is h on the base of the retaining wall. So, as I have discussed that if I consider now this is the variation and this variation is p 1. Similarly, we will get another variation here that is p 2 as it is shown that this condition and this condition are different because it is the free water which is moving and here it is the water is within the soil pores. So, here we consider this variation same variation those values are different here we consider the p 1 at any depth from the top of the water table is p 1 stress and here p 2. And last class and it is mentioned that p 1 is 7 by 8 k h gamma w h to the power half y to the power half here if it is small h then this will be small h to the power half if it is capital H then this will be capital H. Now, it is also derived that the total dynamic water force per unit length was given by 7 by 12 k h gamma w h to the power half gamma w h square where h is the height of the water and gamma w is the unit weight of the water k h is the coefficient of horizontal seismic zero static seismic coefficient. So, now this was derived for the last class for water in a free condition. Now similarly now we can we have to determine the location of the resultant water pressure. Now here we know the total force that is p capital 1 per meter that is in 7 by 12 k h gamma w h square now where it will act that we can determine that that location y w that will be equal to 1 by p 1 integration 0 to h p 1 d 1 d y into y. So, as we have consider a small segment of thickness d y with a height y from the water surface. So, that means the total force means y bar that is the location of this water pressure that will be 1 by p 1 p 1 is the total dynamic water force to 0 to h p y small p y is the pressure at any depth from the top into d y y. So, now we will write that p 1 into capital 0 to h small p 1 d y into 7 by 8 k h gamma w h to the power half y to the power half into y into d y. So, finally we can write p 1 into 7 by 8 k h gamma w h to the power half 0 to h into y to the power half into y into d y. So, after the integration we will get 1 by p 1 into 7 by 8 k h gamma w h to the power half into small h to the power 5 by 2 into 2 by 5 and we will get that 1 by p 1 into 7 by 20. So, this will be 7 by 20 k h gamma w h cube. So, finally if I put the value of p y p y that is 7 by 12 k h gamma w h square. So, we will get this value of 12 by 20 h this 7 by 12. So, this will be 12 by 20 into h k h gamma w and 1 h square that will be cancel out from this value. So, now we will get this value is 0.6 of h. So, that means the location of the water force is 0.6 of h from the top of the water level and the total force is 7 by 12 k h gamma w h square. So, now once we get this value next thing is that. So, if I draw the same figure again. So, this is the water free water level and this is small h and this is ground water surface sorry ground level and the same water level here ground water table. This is the base of the retaining wall. So, here we will get a variation that is p 1. So, similarly we will get another very similar variation that is p 2 at any depth. Now, as it is mentioned that is given that p 1 is equal to 7 by 8 k h gamma w h to the power half y to the power half. Similar expression is proposed that p small p 2 is 0.7 times of p 1. So, this is given Matso and Ohara 1960. So, here it is proposed that p 2 is 0.7 times of p 1. So, that means the stress distribution here is what a hydrodynamic pressure of water within the soil is less than the water pressure applied by free water. So, here it is free water. So, here this is p 1 small p 1. Now, here the water is within the void. So, this water pressure is less than the water pressure applied by the free water. So, now here if this expression is applied. So, that means this will be 7 and p 1 is 7 by 8 k h gamma w h to the power half y to the power half. So, this value will be 0.613 approximately k h gamma w h to the power half y to the power half. So, now if we integrate this force also then we will get the total dynamic hydrodynamic force applied by the water within the soils. So, that is p 2 per unit length that will be similar to 0.7 times of 7 by 12 k h gamma w h to the power half. So, that means this is basically p 1 per unit length. So, now we will get this expression that this will be 0.408 k h gamma w h square. So, the total force applied by this free water is 7 by this value is 7 by 12. So, now we know that 7 by 12 k h gamma w h square and then p 2 per unit length that is the 0.7 times of p 1. So, this will be 0.408 k h gamma w h square. Now, during an earthquake the force per unit area on the seaward. So, suppose this is seaward sea side and this is the coast area. So, now we can write the resultant force on a retaining wall during earthquake is p 1 plus p 2. So, the resultant force we are taking the maximum force on a retaining wall that is equal to p 1 plus p 2 as this one side force is applied and another side it is released. Now, here the resultant one will be p 1 plus p 2. So, this value will be if I take p 1 plus p 2 then this will be p 1 plus 0.7 plus 7 times of p 1. So, this is also p w per unit length. So, you can write this is 1.7 times of p 1 and p 1 value is 1.7 times 7 by 12 k h gamma w h square. So, that value will be 0.992 k h gamma w h square. So, that value will be 0.992 k h gamma w into a square. So, this is the force total force applied on a retaining wall due to water pressure. If this side is water then and this side also water both are in the same level. If it is a different level then this h value will be different and you have to adjust this value according to that. So, now, if it is same level and this is 0.992 k h gamma w into h square. So, this is the total resultant force per unit length due to hydrodynamic pressure. So, this is the force that is applied on a retaining wall. So, during the retaining wall you have to consider this force also during the design. Now, these things are done for the hydrodynamic case. Now, this all the analysis that we have done is for the active case condition. So, similarly we can determine these values for the passive case condition also. The expression will be only the coefficient expression will be different otherwise the process is same. So, now for the passive case that p p earthquake or seismic that will be half gamma a square 1 minus k v into k p e where k v and all the other times are remain same as the active case only this k p e expression that will change. So, this k p e expression is cos k square phi plus beta minus theta this is cos theta cos square beta cos delta minus beta plus theta 1 minus sin phi minus delta sin phi plus i minus theta sin phi plus i minus theta cos i minus beta cos delta minus beta plus theta to the power half total square. So, this is the expression of k p e where theta is equal to tan inverse k h 1 minus k v. So, other things are remain same only this k p e expression that will change. Now, these are the design steps for the seismic condition of retaining wall. So, these are our normal traditional retaining wall. So, what are the additional force that you have to consider during the earthquake condition and if water is present or if you not if water is present then you have to consider the hydrodynamic force. So, those things I have discussed. Now, these things are for the our normal retaining wall design. Now, what will happen? What are the changes that we have to measure? We have I have already discussed about the design step of reinforced retaining wall under static condition. Now, what are the additional steps or additional things we have to consider during the earthquake condition that we will discuss in the next section. So, now here so for the reinforced retaining wall in the seismic condition. So, as I have mentioned that in the reinforced retaining wall design we have to consider two condition. One is for the internal stability check another is for the external stability check. So, internal stability basically we have to design the spacing between the reinforcement and the length of the reinforcement how we will provide. So, these things have already been discussed. So, now what are the additional things that we will consider for this part that I will discuss. So, first we will consider the seismic condition, the retaining wall design that for suppose this is the retaining wall and we are assuming we are taking the uniform spacing and the length of the reinforcement. So, this is uniform spacing S B and this is L is the total length of the reinforcement layer including the anchorage length and the L R and L E including the anchorage length also. Suppose this height of the retaining wall is H. So, these are reinforcements. Now, we can consider in the two parts this one one is this is the H and this one is the up to the reinforced zone. This is the reinforced zone we can write this is the reinforced zone. So, in the retaining wall design. So, this is the reinforcement. So, the reinforcement will provide up to this length. So, up to this one this is the reinforced zone and after that this one there also there is soil, but it is not reinforced one. This one is the unreinforced because required length of the reinforcement may be up to this. So, we will call this is reinforced zone and this one is the unreinforced zone. So, what are the forces additional force. So, suppose this is the weight of the reinforced zone that will act here. Now, first step that how will decide this reinforced zone and unreinforced zone that thing first we will design this retaining wall under static condition. So, first we can design this retaining wall under static condition then we can we can determine how much length will provide here and what is the S B value. So, that is the first step for the we should say this is the first iteration that we design this retaining wall under static condition. So, in the how to design the retaining wall how to calculate this length spacing that those things have already been discussed. So, now we condition design this retaining wall under static condition then we will get this reinforced zone say and. So, therefore, this is the reinforced zone this is the length and height and here we will apply at horizontal pressure that is serious static force that is P in the reinforced zone in the seismic condition. So, here P R S is the horizontal force in the reinforced zone for the seismic condition. Now, here the P A that will act for the active pressure and that will act at a height of H by 3. Now, similarly another force del P active in the seismic condition that will act at a height of 0.6 H. So, these are the forces and in additional this is the T is the shear force and this N is the normal force. So, these are the forces that will act in a reinforced retaining wall structure. So, during the design of static condition then we consider this weight and this P A active force under static condition only these two forces we consider during the design of the retaining wall under static condition. For the seismic condition we have to add two additional force this is horizontal force and neglecting the vertical one this is the horizontal force P reinforced zone in seismic condition and this is the additional force due to the earthquake P A E S in the seismic condition. So, as it is known that the seismic condition static force is acting H by 3 from the base of the retaining wall and seismic force which act as a height of 0.6 H from the base of the retaining wall. So, these are the force that will apply. Now, we will design for the external design and the internal design. So, two design step one is external and the internal design. So, and it is assumed it is already been designed for the seismic condition. So, we know that this retaining wall is safe under external and internal design criteria under static condition. Now, let us see what are the additional steps that we have to consider for the seismic condition. For the external design the step one that the determine the peak horizontal ground surface acceleration that is A max. So, during the design you have to determine what the peak horizontal ground surface acceleration for a particular earthquake or that no. So, A max we will determine first. So, that value we have to determine. Now, steps two once you know that so these are the process by which we can determine the A max. So, those things are the beyond the scope of this lecture. So, now here we assume that we know the A max. So, this A max we have to determine for a particular site. And so, now once we know this A max for the seismic condition, then we have to determine peaks acceleration at the centroid of the reinforced zone. So, this A max is the peak ground acceleration. Now, next we will calculate the peak acceleration as the centroid that means this centroid of the reinforced zone by this expression S e that is equal to 1.45 minus A max divided by G into A max divided by G into A max. So, first we will determine the A max the peak ground acceleration for any particular area. So, once we get that A max for the design area, then we will calculate that centroid peak acceleration at the centroid of the reinforced zone by using this expression 1.45 minus A max by G into A max. Now, the step three that once we get the A max then calculate dynamic force this delta P a s. So, now what is the additional force that due to this earth square condition this delta P a s we can calculate by using this expression 0.375 A c gamma b A square minus A max by G into A max. Now, what is gamma b? Now, gamma b is the here if we consider this is reinforced zone and this is the unreinforced zone then we have two different maybe two different densities. So, this densities or unit weight is for the reinforced zone and this is for the backfill zone. So, this gamma b is the unit weight of the backfill zone and here gamma r is the unit weight of the reinforced zone. So, in this expression that A c is the peak acceleration horizontal acceleration at the centroid of the reinforced zone, gamma b is the unit weight of backfill zone, h is the height of the retaining wall, G is the acceleration due to gravity. So, now once we get this value then the step four this will be that internal force that we will calculate. So, here another force the internal force acting on the reinforced zone. So, now once we get this dynamic force P delta P a s active in seismic condition then we will get the inertial force acting on the reinforced zone. Step four to calculate inertial force acting on the reinforced zone and that is P r s actually this inertial force. So, that means what are the things in step one we will calculate the A max then step two calculate the A c in the step three we calculate this delta P a s in step four we will calculate this force actually this is P r s this is the inertial force in the reinforced zone due to seismic condition. So, this is P r s P reinforced zone in the seismic condition this is the inertial force acting on the reinforced zone with the expression of sigma c gamma r h into L divided by G which is very simple that this is this is this gamma r h L by G this is m and this is a c is the acceleration. So, m into acceleration mass into acceleration that is the inertial force. So, here acceleration is in the reinforced zone is a c then here gamma r is a unit weight in the reinforced zone h is the height of the retaining wall L is the length of the retaining wall and G is the acceleration due to gravity. So, now step five. So, now we know that dynamic force you know the inertial force. Now the step five is add P a s that is equal to P a plus delta P a s. So, this is the total force during the seismic condition P a s. So, we have to add P a s and 50 percent of P r s to the static force on the reinforced zone and we have to check the all the possibilities and then check overturning sliding again. So, that means, so as it is it was designed for that purpose that we have already checked all these overturning sliding for the static condition. So, now in this with that the static condition force we have to add this delta P a s to P a s with the static condition force. So, that means, this P a is only for the static condition. So, now we add delta P a s. So, now we can write this is delta P a s and 50 percent of P r s. So, total dynamic force and 50 percent of inertial force to the static force. Now static force is P a. So, we have to add delta P a s. So, with that was and 50 percent of P r s acting on the reinforced zone. And now once these force are added due to this dynamic force dynamic force and this inertial force then we have to check again this overturning sliding similar to the static condition. Because these things when we check this overturning a slide you consider only the static forces that is P a. Now we have to add delta P a s and 50 percent of inertial force that is P r s within the reinforced zone. And then we have to check again all the forces that mean overturning and the sliding again. Now the question is why we have taken 50 percent of P r s. The reason is that that it is a fact that the maximum value of delta P a s and P r s are unlikely to occur at the same time. That means, the same time the full value of delta P a s and this P r s will not exist. So, that is why we will take 50 percent of P r s and then we will check this overturning and sliding. So, once we check this overturning and sliding if it is safe then fine if it is not safe then you have to make you have to change the our properties that means length of the reinforcement other things to make this condition safe for the seismic condition. So, that is the safe this reinforced retaining wall under seismic condition also that means you have to redesign these things you have to change the dimension of the retaining wall and all the other parameters. So, that it can be safe under seismic condition also. So, this is the external checks. Now next part is for the internal checking of the reinforced wall. So, that means the internal stability checking. So, and it is already been discussed that for the design criteria. So, this is the failure pattern for the 45 degree plus 5 by 2 and this is another failure pattern for this two different types of reinforcement. So, this one is H by 2 this one also H by 2 and this total value is H and this value is given 0.3 of H. Now this one is valid for inextensible reinforcement and this is extensible reinforcement. So, now based on that we have designed for the static condition also and we determine what are the spacing and the length. Now let us check for the internal design steps for the check the seismic condition stability. Now furiostatic internal force acting on the unstable internal failure zone. So, we will calculate that P R S. So, that force we have already calculated that is sigma C S e into gamma R L into H divided by G. So, this force you have already calculated. So, this force once we calculate this force this is for the P R S that is our external design criteria. This is for the external stability check. So, when external stability so, there is a difference between two forces. So, P R S we used for the external stability check condition. So, we there we consider that for the forces acting here for this reinforced zone where this is L L. H and this is so, this weight and the force that P R S acting within the reinforced zone. So, that was used for the external checking condition. Now for the internal one we will calculate similar type of internal failure zone that is P R R. So, or we can write that is our pseudo static inertial force acting on the external stability. So, internal failure zone so, that means, in previous case we consider that the total weight of this zone. Now here we will consider only the failure zone. If this is for the failure zone for the inextensible reinforcement and if it is is extensible reinforcement then we will get this type of failure. So, now either we will so, we will consider only the weight of this failure zone. So, now here for the internal checking that is this P R R that will be A C again divided by G into W A, where W A is the internal weight of the failure mass. So, that means, here so, that means, external you consider the total weight of this zone, but in the internal stability check we consider only the weight of this failure mass. So, we consider this weight of this failure mass for two different reinforcement condition. So, this is the only difference. Now once we get this P R R now the step two that will be the distribute this P R R P R R to each reinforcement. So, once we determine this P R R then you distribute this P R R to each reinforcement depending upon its area or you can distribute this reinforcement equal this force this extra additional force due to seismic condition equally to each reinforcement or you can we can distribute is proportionately according to the area this reinforcement is covering to this because it is not always mandatory that reinforcement spacing and length would be same it may be different. So, if it is different then proportionately we can distribute this additional force to the reinforcement and then and then another condition that as I have mentioned that this is the length which is if we this is failure zone we can provide reinforcement up to this. So, that means, there is a anchorage length and that if it is varying then it is not same for all the reinforcement. So, here this anchorage length that we have provided is more compared to the anchorage length that we have provided here. So, depending upon the area of the anchorage length and the spacing we can distribute this force to the each reinforcement layer because then this design will be more economical because here more this force it can resist more if the anchorage length is more. So, we can distribute according to proportionately to the reinforcement. Then the step 3 then add the dynamic force this additional force with the static force. Now, add now here the previously we have considered the static tensile force and based on that we consider the. So, previously based on the tension force developed with the reinforcement we design our spacing and then the length. So, now this static tensile force due to this dynamic force there will be the additional force that will developed within the reinforcement. So, this additional tensile force will develop within the reinforcement. So, we have to add this additional tensile force with the static tensile force in the reinforcement that is develop within the reinforcement. Now, step 4 now that is once we add this dynamic and the static reinforcement. Now, this total tensile force that should be at least 75 percent of tau allowable. Now, that means the total tensile force that will this is due to the dynamic and the static that should be around the 75 percent of the tau allowable because at least 75 percent of the tau allowable. So, the now the thing is that. So, suppose if we design it considering the total tensile force within the reinforcement during the static design if we take the total tensile allowable tensile force that is equal to and this is fully it will be utilized during the static condition then definitely if we design that thing for the seismic condition it would be unsafe because we have already taken the total allowable tensile force of the reinforcement during the design of the static condition. So, the additional force during the dynamic condition if it will act then definitely this will be a unsafe design. So, the option is that initially we do not if we do not consider the total allowable force during a design that is one option and another option we can change the spacing and the length of the reinforcement so that we can account the static and the dynamic condition both. So, that means when we consider the static condition then we redesign our internal. So, initially we consider that basically our static tensile force and now due to this dynamic force there will be a additional force act will act within the reinforcement. So, this additional dynamic force also we have to incorporate within this allowable tensile strength of the reinforcement. So, we have to redesign we have to design the spacing and the length of the reinforcement so that it can account this additional tensile force. Now, step five we have to check about the anchorage length of the reinforcement also. So, once we take check the tensile strength of the reinforcement then there is another possibility you have to check about the anchorage length of the reinforcement. So, that it can account both static and the dynamic condition so that we have to apply proper anchorage length beyond the failure surface so that it can account the both tensile and the static and the tensile condition. So, now here we have to design this if I consider the overall summary of this additional design. So, that means first we have to design it for the static condition. So, then you will get an idea what to be the required length and that thing we can use for the our first trial condition. So, what would be the length of the reinforcement and what is the height and spacing. So, based on that we will design or determine what would be the area of the reinforced zone. So, then there will be two different density one is in the reinforced zone another in the outside the reinforced zone. So, once we get the reinforced zone and based on that we will calculate that inertial force within the reinforced zone and then once then this inertial force and the dynamic force that will act that we have to add with static force to during the external check of the stability. So, once we get add these things then we have to check it again whether this is safe against overturning or sliding not. If it is safe fine otherwise we have to redesign the reinforcement length and other parameter so that it can take the dynamic load also. So, then I mean there is two part one is external another is internal. So, the internal one we have to again calculate the internal inertial force and that is for considering within the failure mass only. So, one and then we have to distribute that force within the reinforcement and so that in the proportionality also and then we have to check whether the tensile strength or tensile force which developed within the reinforcement due to the static condition and dynamic condition that is within the allowable limit or not. So, if it is not within the allowable limit then we have to change the design criteria and then we have to change the spacing and the length of reinforcement and as well as we have to check whether this anchorage length that we have provided during the static condition that is sufficient or not during the dynamic or seismic condition or not. If it is not such a sufficient then we have to provide more anchorage length so that this system can take the seismic load also. So, these are the steps of the design of the retaining or reinforced retaining wall under seismic condition. So, now the next step that another possibility is that so suppose another example how to save the our structure for the reinforcement suppose if there is a slope and this slope is unstable during the suppose this is the slope failure line and this is the unstable it is stable during static condition but it is unstable during seismic condition. So, there also if I provide a reinforcement layer here then we can save the design against seismic condition also. So, that means there is a possibility that these are the techniques that we can make this design save for the seismic condition. One example that I have already explained that how to save this design against retaining wall then we can account this seismic condition also then also we can make the our slope stable if I provide reinforcement also in the seismic condition. So, that is one option of making the structure stable against seismic condition. So, then the next section on the next class I will explain about the how this soil foundation interaction and how this soil and found is the interact each other in during the loading condition and different other condition those things I will explain in the next few classes. Thank you.