 I have discussed about the reinforced earth, and how this behavior of reinforced earth beneath the foundation. So, those things I have discussed in the last class. Now, in the last section of the last class, I have discussed about the reinforced retaining wall. Now, today I will discuss how to design a reinforced retaining wall and what is the behavior of this structure. Now, first if I go for that reinforced retaining wall. So, I have discussed that suppose if this is a existing ground, and above that we will construct this reinforced retaining wall. So, we place to want reinforcement here, then we will place another reinforcement here. So, this is we will fold there, and then within that two layers, this is the compacted sand, then this is the next layer, this is the next layer, this is the next layer, this is sand. So, this is our reinforced retaining wall. Now, if we consider the failure surface, if it is fails like this, then this is your 45, this angle is 45 degree plus phi by 2. If this is the phi is the friction angle of this compacted field with reinforcement, this filling material. Now, here as I have discussed that the main design components of this type of structure is one is SV, is the vertical spacing between the reinforcement. Then next one is the length of the total length of the reinforcement, and then this folding portion, length of this folding portion L G O. So, L is the total length of the reinforcement, then SV is the vertical spacing between the reinforcement, and again the T this tension, mobilized tension that is developed that is the material property. So, T is the mobilized tension in the reinforcement. Now, this T cannot be greater than the axial strength of the reinforcement, because otherwise this total infosystem that will a possibility that this will fail. Now, the when you are talking about the SV and the length, length we can see that this is the failure surface of the reinforced earth, then up to we can divide this L into two parts. One is L R, and another is we can divide into L E. So, L is equal to L R plus L E. Now, the question is what is L R plus L R, and what is L E? L R is the length from the face, suppose this is a face of the reinforcement to the failure surface. So, this much of reinforcement is within this failure zone. So, that is L R and L E is basically the anchorage length, which is from this failure surface to the end of the reinforcement. Now, this L E is very important. As I have mentioned, then every reinforced structure we should have some we have to provide some anchorage length beyond this failure surface. The reason why if I consider or if I consider this reinforced retaining wall, then if this is the failure surface, this inclined line is a failure surface, then what is happening actually this total soil mass along with the reinforcement is trying to slide along this failure surface. It is failure. Now, somebody has to resist this force, that means when it is moving downward direction. So, some resistance we have to offer so that it stands in a stable condition. So, who is giving this resistance? Now, actually when we are providing sufficient anchorage length of this reinforcement, now this reinforcement anchorage because of this anchorage length, it is getting some resistance because of the soil and the reinforcement interaction. So, in the when this total system or the soil mass is trying to move downwards, now there is a interaction between the soil and the reinforcement. Because of that the friction will develop and when this friction and because of this friction, there is a bonding between the reinforcement and the soil and this bonding will allow to hold this total system. As if this total system mass is sliding and because of this bonding between the soil and the reinforcement, this reinforcement is holding this soil mass so that it cannot slide. So, now this bonding depends on the how much length we are providing in the reinforcement. If we do not provide a sufficient anchorage length for this type of structure, then what will happen? This total system will slide along this surface. So, now for this purpose, this providing of the sufficient anchorage length is very important. Now, these things we will discuss the how to calculate this provide the sufficient anchorage length and what are the different expression which is available to get this sufficient anchorage length. So, that means we have to provide this anchorage length such that there is a sufficient bonding between the soil and the geosynthetic so that a tension can develop within the reinforcement and this because of this bonding total system this will it will not slide along this surface, it can hold this surface this total mass. Now, if I so first we have to decide the what is the SV in the vertical spacing, then this anchorage length because L R we can easily calculate. If we consider this is 45 minus 5 by 2, then definitely this will be 45 minus this is 5 by 2 this is 45 degree minus 5 by 2 this angle and if we consider this is the Z is in the downward direction from the top and if this is H is the height of the reinforced retaining wall. So, then at any depth L R we can easily calculate. Now, the question is how we can calculate the L E. Now, when you are talking about this SV as mentioned that when this force is developing here so that means because of this bonding or because of this friction between the soil and the reinforcement as I mentioned in the last class also that the tension will mobilize so tension will develop. Now, and the question is when this lateral earth pressure is acting in this direction because lateral earth pressure is acting this is the sigma H at any depth it is lateral earth pressure is acting from this direction. Now, who is resisting this lateral earth pressure here in case of traditional retaining wall that retaining wall is resisting that lateral earth pressure. Here that lateral earth pressure it counterbalance by this tension forces. So, now because this is acting in this from right to left and tension is acting in the opposite direction. So, these two force will counterbalance each other. So, now if I consider in that way and if we consider a segment suppose if I consider a segment between the two reinforcement wall suppose if I consider this segment. So, between so if I draw this segment so this is the segment where reinforcement is here this is T reinforcement. So, this segment I am drawing here. Now, here at any depth this is at any depth z from the top. So, this depth is z. So, now the amount of lateral pressure at this point will be K into that is sigma H amount of lateral pressure in this point K into gamma into z where K a is earth pressure coefficient. Gamma is the unit weight of reinforced soil gamma is the unit weight of this reinforced soil and K a is the. So, now this sigma H is basically equal to taken by this T tension force. So, if I consider this sigma H and which is acting in this region. So, this is sigma H the unit of T is kilo Newton per meter. Now, this is the force per unit length and now here the unit is kilo Newton per meter. So, we have to multiply it with some length. So, that is equal to the S v because this is S v because here we have taken this segment from the centre of between two reinforcement wall to centre of another two reinforcement wall. So, this is S v plus 2 by 2 this is also S v by 2. So, S v by 2 plus S v by 2 this is S v. So, we can write that sigma H into S v that is equal to T allowable or that is equal to T allowable. Now, why this is T is the mobility tension as I have mentioned that this T mobility cannot be greater than T allowable. So, we provide this is T allowable with some additional factor of safety F s. So, this is F s is the factor of safety. Now, finally, if I write this expression that our T allowable divided by factor of safety that is equal to sigma H into S v. So, S v is equal to T allowable divided by sigma H into factor of safety. So, that is equal to T allowable divided by K A into gamma Z into F s. This F s is generally taken as 1.5. So, now, we can write S v is equal to T allowable divided by 1.5 into K A gamma into Z. So, we can see that if we keep all the other parameter constant then T allowable constant K A constant and gamma constant then this S v is inversely proportional to Z. So, that means at the top required spacing is more compared to the at the bottom because it is inversely proportional. So, that means S v value is more at the top and it is less at the bottom. So, now, if I consider in this concept, so this will give us the, that means if I consider this retaining wall and this is the spacing of different retaining wall. This is the failure surface. So, this will be the L r and this is L e. So, this is S v. So, required spacing here will be very less compared to the top. So, next one to calculate the length, total length. So, total length L is equal to L r plus L e. So, L r we can easily determine. So, this is 45 minus 5 by 2. This is any Z. So, L r we can determine in this expression that J h because this is h minus Z into tan 45 degree minus 5 by 2. So, at any depth this is Z h minus Z is this one. So, into tan 45 minus 5 by 2. The next one you have to calculate the L e. So, L r we can calculate by this expression. Now, when you are talking about this modulus tension is developed. So, the L e is basically the anchorage length. So, that means this resistance that we are getting is because of this interface action of the soil and the geosynthetics. Now, if I draw a particular soil geosynthetics section. So, this is the soil geosynthetics section where T, this is the line, this is L e beyond the failure surface. This is failure surface. This is failure line. So, how this L e T is the mobilized tension which is developed and how this is counter balanced. So, if T is acting this direction. So, there will be a shear stress that will act between the soil and the reinforcement. Now, here the sigma n that is acting here is the normal stress. So, this is acting normal stress and this is tau is the shear stress. So, now we can write that our T allowable this is also we can consider T allowable that is equal to tau that is the shear stress into L e and this is acting both the sides. So, into 2. So, tau is the shear stress which is acting opposite to the direction of the T. So, tau into L e that is tau into L e that is equal to T allowable as this is acting both the sides. So, that is into 2. Now, how we will get the tau? So, that is 2 into tau is we can get the tau is expression of the shear strength that shear strength C plus sigma n into tan delta into L e where delta is the angle interface angle between soil and reinforcement. So, this is the interface angle between the soil and the reinforcement. So, finally, if I go for this expression of this L e. So, this is T allowable is equal to 2 into C sigma n tan delta into L e. So, L e expression is T allowable divided by 2 C plus sigma n into tan delta and sigma n is at any depth the normal stress this is 2 C plus gamma into Z into tan delta. So, here also you can see if Z is equal to 0 or if Z is very small then L e is more. So, that means the required length of the geosynthetics at the top surface is more as compared to the bottom surface bottom of the reinforcement. Now, generally in the field soil the C value if C value is 0 then we have to consider gamma Z into tan delta. Now, here we can say so that means the most surface design that it is observed from these two expressions that is the expression 2 and previously sigma v expression that is the expression 1. So, these are the two main expression 1 is to how to determine the sigma v and this is how to determine the L e anchorage length. See here we can see that at the bottom of this reinforced wall requires spacing is more and here L e at the top of the reinforcement wall the required length of the reinforcement is more. So, that means here if I put the spacing which is required at the bottom of the reinforcement and the length which is required at the top of the reinforced wall. So, bottom of the reinforced wall that is spacing at the top of the reinforced wall the required length then design is safe. But we can see that this spacing requirement is not uniform throughout the depth of the reinforcement that means the spacing at the top region requires spacing is less compared to the bottom region. So, if I provide the uniform spacing that is required at the bottom then that is somehow is the wastage. So, that means we have to design it properly so that we can make it economical this design. Similarly, the length which is required at the top region of the reinforcement if I provide if we provide that is throughout the depth that is also not economical. So, we have to provide the spacing and the length according to that so that we can make the design more economical. So, now here another consideration the L 0 which is the folding length required that is basically T allow well into divided by 4 into c plus gamma z into tan delta. So, these are the two lengths required L e and L 0. So, now we will design a reinforcement wall retaining wall. Now, before we go for the design purpose actually what are the loading condition in the reinforced retaining wall. So, one is our dead load or the load of the reinforced soil itself. So, that is at any z that is gamma into z into k a that is the dead load. Similarly, if there is a surcharge so that is surcharge load that surcharge load is q into k a that is surcharge load then we can go for the live load. So, this is live load then this will give us the total load this is total load. So, these three we can consider once again is the dead load or the load of the reinforced retaining wall itself this is surcharge load this is live load. So, these are the lateral pressure we are talking about this is the lateral pressure due to the dead load this is lateral pressure due to surcharge this is lateral pressure due to the live load and this is the combination of these three. So, this is the total lateral pressure which is active. Now, we will solve one problem and to see how these things are working. So, now this design a reinforced retaining wall with h height of the reinforced wall is 6 meter gamma of the reinforced soil is 19 kilo Newton per meter cube phi is 36 degree and t allowable is 16 kilo Newton per meter. So, we are taking this one here. So, as I have mentioned that this is existing soil and above that the retaining walls are constructed. So, the properties of these two soil can be different. So, here the phi of the this field is 36 degree gamma is 19 kilo Newton per meter cube h is given 6 meter and delta is assume equal to phi, but that can be anything that can be delta can be 70 percent of the phi or 80 percent of the files here it is taken equal to phi and now k a value we can calculate is 0.26 k a value we can use this expression 1 minus sin phi divided by 1 plus sin phi. Similarly, the foundation soil it has the properties gamma is 20 kilo Newton per meter cube phi is 15 degree, c is 0.6 kilo Newton is 50 here for this field soil c is taken as 0. So, c equation is 50 kilo Newton per meter square delta for the foundation soil is taken as 0.95 of phi. So, that is 14.2 degree and c a is taken 80 percent of the c that is 40 kilo Newton per meter square. So, now these are the properties this is the properties of the foundation soil and this is the properties of the reinforced soil. Now, we have to design this reinforced earth or reinforced earth foundation. So, now this retaining wall when you consider the if I put this S v expression that is S v equal to t allow well and then this is the divided by 1.5 into k a into gamma into z. So, this t allow well is 16 k is 0.26 1.5 is the factor of safety gamma is 19 and this is z. So, this is the so now at different level here the height of the retaining wall is taken 6 is 6 meter. So, first as the spacing is not uniform that is because is the function of z spacing is the function of z l e is also function of z. Then what we will do we will take this at different section we will determine the spacing. So, here we are taking at 2 meter at 4 meter and at 6 meter see different position we will determine the what is the required spacing. So, first if I consider at 2 meter depth S v is 16.26 into 1.5 into 19 into 2. So, that is 1.08 into 2. So, this is meter. So, at 2 meter depth S v is 1. So, at 4 meter depth S v is 16 into 0.26 into 15 into 19 into 4. So, that is 0.54 meter. Similarly, at 6 meter depth S v is point 36 meter. So, we can see at the 6 meter depth 0.36 at 4 meter depth 0.54 at 2 meter depth 1.08. So, for this case we are taking that I am taking that use S v is equal to 0.5 meter for z equal to 0 to 4 meter. And S v equal to 0.33 meter for z greater than 4 meter that means 4 to 6 meter. So, we have to choose this spacing such that because here required spacing is 0.36. So, we can provide less we cannot provide more. So, that means 36 we are taking 0.33. Now, it is at 4 meter depth liquor spacing is 0.54 we are taking 0.5. Now, you can another question that at 2 meter we can provide 1. So, that is also possible at 2 meter 1 we can provide up to 2 is 1 meter then 2 to 4 is 0.5 then 2 to 6 is 0.33. So, there is there will be so many variations. So, to just minimize the variation I have taken 2 to only 2 types of spacing. So, now here why I have taken 5 because here we have to decide the spacing such that we can place them properly. Because here from 2 to 6 meter if it is required we cannot place point if I take 0.35 that will be very difficult to place here equal spacing. So, to make it equal spacing and place within this 2 meter we have taken 0.33. So, the number of reinforcement will be so that means here at 0.5 meter we will provide 1. So, point so this is 1. So, at 1 meter we will provide another 2 1.5 we will provide third number of spacing reinforcement. So, 2 meter this is the fourth number of reinforcement then 2.5 fifth number of reinforcement. So, 0.5 1 1.5 2.5 then 3 3.5 2.5 then 3 3.5 4 then 4.33 4.67 5 5.33 5.676 within the top portion z equal to 0 and not providing any spacing because we will fold this one and we will keep this one open. So, that means the number of reinforcement layer will be 1 2 3 4 5 6 7 8 9 10 11 12 13 14. So, number of reinforcement layer is 14 number. So, that is decided because this is the x b equation and we have taken. So, here the spacing is greater than 4 to 6 is 0.33 meter and from 0 to 4 is 0.5 meter. So, that means the spacing part is over. Now, here what we can do now next step is to determine the length. So, first we take that at what depth we have provided the reinforcement. So, the depth that we have provided at 0.5 meter depth we have provided the first reinforcement then at 1 meter depth that at 1.5 meter then at 2 meter then 2.5 meter then 1.5 meter then 3 meter then 3.5 meter then 4 meter. So, up to 4 it is at the rate of. So, up to 4 this is at the rate of 0.5 meter center to center. Now, next one is 4.33 then 4.67 then 5 meter. So, next one is 5.33 then 5.67 then 6 meter. So, total 14 number here it is point at the rate of 0.33 meter center to center. So, these are the total 14 number of reinforcement and their corresponding depth. Now, at every depth we have to calculate the length. So, if I consider the first layer length. So, that is here h minus z into tan 45 degree minus phi by 2 then plus t allowable divided by 2 as c is 0. So, this is gamma into z into tan delta. So, now if I put z equal to 0.5 and every other factors and here another thing that we should mention that the minimum anchorage length is 1 meter. So, for this particular factor that cannot be less than 1 meter. So, L e cannot be less than 1 meter L 0 also cannot be less than 1 meter. So, that is the condition. So, that means if by calculating if it is coming 0.8 then you have to provide 1 by if it is calculation by it is coming 2 meter then you have to provide 2. So, that means this anchorage length cannot be less than 1 meter. So, 1 meter at least we have to provide L e and L 0. So, now here if we put z equal to 0.5 then ultimately required length at 0.5 meter depth is 3.80 meter. So, by putting this value in this expression at 0.5 meter depth required L is 3.8 meter. Similarly at 1 meter required L is 3.55 meter at 1.5 meter required length is 3.3 meter by calculation at 2 meter if I put z equal to 2 then the required length total length we are talking out that is 3 meter. So, at 2.5 that is 2.8 meter. So, at 3 meter that is 2.5 meter at 3.5 meter that is 2.3 meter at 4 meter that is 2.0 meter. So, at 4.33 meter that is 2.5 meter that is. So, these are the calculation that at 0.5 meter, 1.5 meter, 2 meter, 2.5, 3 meter, 3.5, 4 meter. So, these are the requirement and 4.33 it is 1.85 meter and then finally at 6 meter the required length is 1 meter. So, now we have to decide which length because as length is varying from 3.8 meter to 1 meter. So, it is not uniform. So, now as in the during the spacing I have taken the spacing at the rate of 0.5 meter from 0 to 4 meter and 0.33 meter from 4 to 6. Here also at here the maximum spacing requirement up to 4 meter is 3.8. So, in place of 3.8 we have taken that up to 4 meter we will provide total length is 4 meter and here minimum spacing requirement maximum spacing requirement is 1.85 for 4 to 6. So, that we will provide from 4 to 6 is total length is 2 meter. So, this but as we have mentioned that if I take up to 2 meter a different spacing from 2 to 4 or different 4 to 6 then according to that you have to provide the length here up to 2 meter we can provide 4, then from 2 to 4 we can provide 3, then 4 to 6 we can provide 2 meter also. But just to avoid the so many variation we have taken from 0 to so that means, from 0 to 4 meter L is equal to 4 meter and from 4 to 6 meter L is equal to 2 meter. So, this is the variation that we are using. So, now as an again from the calculation part we can calculate that L 0 is equal to basically L e by 2. So, now if L e calculation of L 0 is coming up to be 0.1 meter. So, that is less than 1 meter. So, that at that is the minimum value. So, that is is not satisfying the condition. So, L 0 value here we will provide 1 meter. So, L 0 is 1 meter total length is 4 meter from 0 to 4 and from 4 to 6 total length is 2 meter. So, next one so next category is the. So, in the this during the design of reinforced retaining wall. So, in the other design of reinforced retaining wall when you are talking about the cantilever retaining wall or gravity retaining wall basically we consider the various factor of safety that means, factor of safety for sliding factor of safety for overturning factor of safety for bearing and no tension zone. Here also these things that we have done till now based this is the spacing and the length that is the internal stability of the reinforced retaining wall. Now, you have to check for the external stability also. So, those external stability as again similar to the other type of retaining wall that means, you have to check whether this reinforcement will slide or not because the sliding may take place between the foundation soil and the reinforcement layer. If there is no sufficient bondage between the reinforcement and the foundation soil. So, there is a possibility that total reinforcement reinforced wall may slide along the surface. So, we have to provide sufficient length of the reinforcement. So, that there is a proper bondage or between the soil and the reinforcement. So, that the sliding can be restricted or sliding can be stopped. The next one is the overturning. So, we have to provide such that this way here also weight of the reinforcement is giving the resistance. So, we have to make the reinforcement retaining reinforced retaining wall such that there is no overturning. And another is that the such amount of the soil of the reinforced coming from this reinforced retaining wall that will give a pressure on the existing soil. Now, if existing soil is very poor then there is a possibility that there is a very bearing capacity failure of the soil. So, you have to check this three again for this reinforced retaining wall. Now, how to check this three? So, that means, here if I consider that this is our total length of the is equal to 6 meter reinforced retaining wall. So, from there up to 220 to 400 to 4. So, suppose this is 4 meter, we have provided the reinforcement whose length is 4 meter. And from here to here, I mean 4 meter to the 6 meter, next 2 meter we have provided reinforcement length is 2 meter. So, this is 2 meter. Here the spacing is 0.5 meter center to center, here is 0.3 meter center to center. So, this is the reinforced retaining wall. So, that means, this is one is if I consider this is also. So, here up to 2 meter spacing of 0.3 center to center this is up to 4 meter. Now, here the passive this active earth pressure that is acting here, you can consider this is acting as a P active which is acting at a height of H by 3. So, that is 6 by 3 that is 2 meter from the base. So, now, when you consider the sliding and the overturning and bearing capacity failure. Now, first you consider calculate P a, P a is half gamma a square into k a. So, half gamma here will consider the 19 H is 6 square k a is 0.26. So, this is 88.92 kilo Newton per meter. So, if I consider this form then first we have to consider then weight first to calculate, then when you consider the factor of safety for overturning factor of safety for overturning that means, here this force P a is acting this side. Now, the weight of the reinforced retaining wall is giving the resistance. So, here you will consider the weight W o 1 and W 2 both are the weight which are acting. So, this is the W 1 and the W 2. So, here first you consider the weight of this lower portion. So, that is the weight of this lower portion is 19 into 2 is the weight and 2 is the height. So, that is the weight. So, overturning means that is the summation of resisting moment divided by summation of overturning moment or diverging moment. So, here summation of resisting moment that means, this weight into if I take the moment with respect to this toe. So, that W 2 is acting at a distance of 1 meter from this toe because this is 2 meter. So, W 2 of the center of this 1 meter. So, lever arm is 1. So, weight is 19 into 2 into 2 19 into 2 into the width and 2 is the height into the lever arm is 1. For the next one W 1 the total weight is 19 into this is 4 into 4 and lever arm is half of this 4 is 2 into 2. So, this is the resisting moment and the overturning moment here is acting here P a which is 88.92 into which is acting at the distance of 2 meter. So, we have to calculate these things and it should be greater than 3. So, if it is not greater than 3 then we have to redesign it. Although it is internal stability wise it is fine, but we have to increase the width of the or length of the reinforcement. So, that the weight of the reinforcement can increase. Similarly, the factor of safety for the sliding again this is the summation of resisting force divided summation of the driving force. So, when we are talking about the sliding here in this case within the reinforcement we have taken c equal to 0 because that was the soil properties, but here c is not equal to 0. So, we have to consider the c when we are talking over the sliding. So, sliding mainly will take place between the this layer the last layer and the existing for this is existing foundation soil. So, this is the reinforcement. So, for only this length and the existing soil. So, now the c a addition into 2 2 is the length here is taken plus this is w 1 plus w 2 this is the weight into 10 del. So, this is delta of what here delta of foundation soil you have to consider this is not a delta of the field soil here delta of the foundation soil because this is the this is the sliding between the soil foundation soil and the reinforcement. So, this is delta 2 divided by the p a that should be always greater than 3. If it is not then you have to increase the length again. So, now, here c a is 80 percent of c and w 1 and w 2 you can calculate this is stand delta into p a. Similarly, for the factor of safety. So, this is one factor of safety this is another factor of safety and this is another factor of safety the factor of safety for bearing capacity b c. Now, how to calculate the bearing capacity? That means the q ultimate divided by q acting. So, how to calculate the q ultimate and q acting? So, we can consider the last factor of safety that is bearing capacity and q ultimate by q acting. So, q ultimate will be c n c we can consider into q n q divided by q acting plus 0.5 gamma b n gamma. So, that property you have to consider for the foundation soil based on the foundation soil 5 you have to calculate the c n c n q n gamma and b width of the foundation you will consider the width or length of the reinforcement that is the width of the foundation b and gamma will consider the gamma of the foundation soil. Here another thing is that here it is mentioned that when you talking about this is the this will act as a total load or that reinforcement. So, that means and this is the footing which is resting on the foundation it is not below the foundation not at the surface it is not below the surface. So, that means here q value is 0 basically because when you are talking about this is the foundation if it is resting with some depth if it is ground surface then if it is g f then we can consider q equal to d f into gamma if it is placed at a depth of d f, but if it is placed at the surface then q is 0. So, here also it is placed at the surface. So, it can consider that q value is 0 in this case. So, this is the expression c will provide the c of the foundation n c will calculate based on the foundation soil 5 gamma is the foundation soil unit weight n gamma will calculate based on the foundation 5 and b is the length of the reinforcement. And q acting that is basically the weight of the reinforced soil here we can consider the weight which is acting either we can consider 19 into 6 that is the weight that is acting here. And then we put and we put this value b is the length of the reinforcement and here this is 19 into 6 then it should be also greater than 3. If it is not then we have to redesign this section. So, we can see here based on the external internal stability we have to determine the spacing between the reinforcement and the length of the reinforcement as it is not uniform throughout the depth of the reinforced retaining all. So, you have to make it more economical. So, any different depth different spacing and length we can provide and then we have to check the external stability that is sliding overturning and bearing capacity failure. So, if although the structure is safe in within the internal stability consideration but if it is unsafe based on the external stability then you have to redesign it you have to actually you have to increase the length of the reinforcement you have to design it so that it is satisfy the external stability criteria also. So, this way we can design this reinforced retaining wall and the next class will talk about the other foundation on the on improved ground based on the other improvements. Thank you.