 In last class, I have discussed about the various aspects of design of retaining wall and then different types of retaining wall that is gravity retaining wall, semi-gravity retaining wall, then cantilever retaining wall, then counter for retaining wall. Now, in this class, I will basically solve one example on cantilever retaining wall to show that how this the dimension of this retaining wall is decided and what are the factor safety or the safety that you have to check and during the design of this retaining wall. Now, first if I go for the problem, now suppose if we take one retaining structure which is a cantilever retaining wall in this form, suppose this is the retaining wall and ground surface is this one. Now, this ground surface or the backfill side which is making an angle I equal to 15 degree with horizontal and this is backfill side and this one is the existing ground surface. Now, this first for the first trial based on the guidelines that I have given in the last class, I have chosen this dimension of this retaining wall that for this top portion I have taken this is 0.3 meter, then on this height of this retaining wall from this here to here that height is 4.8 meter and this thickness that I have taken thickness of this base slab is 0.6 meter, 0.6 meter. So, below this ground level the total distance below this ground level is 1 meter basically to the base of the slab. Now, what are the other dimensions that is taken like if we extend this this triangular portion. So, the dimension of this portion as I mentioned that this distance from here to here is taken as 0.6 meter that means from here to here this is the toe then this distance if I extend this triangular part. So, this distance is taken 0.2 meter now as usual as I have mentioned that top portion is 0.3 meter. So, this one is also 0.3 meter and this total slab distance or length is taken as 3 meter. So, the total base width of this retaining wall or base distance is 3 plus 0.6 plus 0.2 plus 0.3 that is 4.1 meter. So, that is basically 0.3 plus 0.6 plus 0.2 plus 0.3 which is equal to 4.1 meter. Now, the other distance or this things we can take we can take the extend this vertical lines which is passing through this edge of this retaining wall then we divide this total retaining wall as we will solve this problem particular this problem by using this Rankine's theory that we will solve this problem by using Rankine's theory. So, as I have mentioned that we will consider the weight of this soil also. So, first we consider that our active earth pressure which will act at an angle of 15 degree with horizontal and the point of application that will act at a distance of this total vertical line to the one third. Now, the distance of this total vertical line before we start we can determine the distance of this portion. So, this portion is this is tan 15 into 3. So, distance from this point to this point is 0.8 meter that is will come from tan 15 into 3. So, the total height of this vertical line from say a to b. So, distance a b that will give us the distance of this height is 0.8 for this one then 4.8 for from here to here then additional 0.6 meter the thickness of glass slab. So, total distance that is a height is 6.2 meter. So, the point to application of this lateral earth pressure is 6.2 divided by 3 which is 2.07 meter from the base of the retaining wall. Now, as from this design we are neglecting neglect the effect of this passive resistance, but for this this can also be included in the calculation, but for the first case this problem will be solved by neglecting this passive pressure. So, now if I want to solve this problem by Rankin's theory. So, as we have to check whether Rankin's theory can be applicable or not. So, if I determine this angle. So, this angle we can determine the expression is 45 degree plus i by 2 minus phi dash by 2 minus sin inverse sin i divided by sin phi dash. So, now the values which is given for this problem that phi dash is 32 degree delta is 23 degree delta is the friction angle between the this concrete retaining wall and the soil. And here for the simplicity that it is assumed that this is the foundation soil, this is backfill and this is a foundation soil. And assume the properties are same for both the soils foundation soil and backfill soil, but if some cases the this property may not be same. So, in that case we have to consider different properties for different conditions. So, for the foundation soil calculation we have to then the delta of the base of the retaining wall and the soil that friction angle will be based on the foundation soil properties. And this friction angle delta which is with backfill and the retaining wall that will also change and that will depend on the backfill properties soil properties. So, then that case the delta of these two may not be same. So, that consideration we have to taken care into our calculation if this soil properties are different for the foundation soil and backfill soil. But for this problem we consider the both the properties are same for this soil total system. So, now the delta for all the system is 23 degree, the unit weight is 19 kilo Newton per meter cube, I which is is 15 degree and the allowable bearing pressure is 400 kilo Newton per meter square. Now, we have to solve this another the unit weight of concrete is taken 24 kilo Newton meter cube and unit weight of soil is taken as 19 kilo Newton per meter cube. So, first we have to determine this angle so that we can check whether this value ranking theory can be applicable or not. This angle we can determine by 45 degree plus 15 degree by 2 minus 32 degree by 2 minus 15 degree sin inverse sin 15 degree divided by sin 32 degree. So, this value is coming out to be 7.26 degree. So, if this value is 7.26 degree so this line will pass through the backfill soil it will not pass through the vertical steam of the retaining wall. So, as this angle is very small so we can consider that the rankings theory can be. So, this is the first check we have done that then we can use the Rankine theory. Now, we will proceed for the other sections. Now, first we will calculate what is the active earth pressure that will act into this soil or into this retaining wall. So, first we will calculate this P a that is equal to half into k a into gamma of soil into h square. The h we are taking this as a mention that h we are now taking the total vertical distance from a to b the total height we are taking and that a b is 6.2 meter. So, if I take this 6.2 meter then k a by using Rankine theory if we can consider this k a is cos i into cos i minus cos i into cos i minus root over cos i square minus cos phi dash square divided by cos i plus root over cos i square minus cos phi dash square. So, if I put this value so then this is cos 15 degree into cos 15 degree minus root over cos 15 degree square minus cos 32 degree square divided by cos 15 root over cos 15 degree square minus cos 32 degree square divided by cos 15 degree plus root over cos 15 degree square minus cos 32 degree square. So, this k a value if I solve this thing this k a value is coming 0.34. So, this is 0.34. So, the p a value is equal to half into 0.34 into gamma is 19 into h is 6.2 square. So, this is 0.34. So, this is this value is coming 124.2 kilo Newton per meter as we are calculating this thing per meter distance. So, this is per meter. Now, as from this figure we can see that p a is acting angle of 15 degree. So, we can take the vertical components and horizontal components of this p a that we can write like this is p v and this is p h. So, this is the vertical component of p a p v and p h is the horizontal component of p a. So, we can calculate that p v is p a into sin 15 degree that is equal to 32.15 kilo Newton per meter. Similarly, p h is p a into cos 15 degree which is equal to 120 kilo Newton per meter. So, the value of this k a value is equal to the vertical force and horizontal force that is coming due to the lateral earth pressure of the backfill that we can have calculated. Now, we have to calculate the other parts of this problem. So, now for the other parts suppose if I take the previous figure our main figure then in this total retaining wall we can take several parts of the retaining wall. So, we can take this is number one. This number one means this rectangular is our first component then this triangle is number two. That means this triangular zone is number two then this total base slab is third zone and this total rectangular soil portion is fourth zone and this triangular soil portion is zone five. Now, we have to determine the individual weight of these five zones they are centroid where these weights are acting actually and then the moment with respect to two. So, all these things we have to calculate because this five zones this load are giving the vertical force and that will also give the resistive moment and this p a due to this p a will get the overturning moment. So, first we have to calculate the weight of different portion of the of this total system. Now, first if we calculate this weight we can take the for the weight calculation we can take this table we can make this table that is our serial number. Then weight calculation that is kilo Newton per meter then the force that we will calculate that is also kilo Newton per meter and this force can be vertical and this can be horizontal. So, this is for the weight calculation table that we will calculate the first serial number one the first segment weight that we will calculate. So, for this first segment this is for the first segment the weight will be this area into the unit weight of the concrete. So, this area will be 4.8 meter into 0.3 into 24. So, if in this way if we calculate the weight of every section then weight one for the first section is 0.3 into 4.8 into 24 that is the unit weight of the concrete. So, and this will act in the vertical direction. So, in vertical direction we can write the total weight w 1 that is equal to 34.56 kilo Newton per meter. Similarly, if we calculate the weight of second portion that is a triangle. So, that we can write half into 0.2 into 4.8 height and then 24. So, we will get this value is 11.52 and this will also act in a vertical direction. So, we will get this is also 11.52. Similarly, the third weight if we calculate the weight of the third segment. So, that is 0.6 into 4.1 into 24 because for the third segment means this total base slab. So, that is 4.1 into 0.6 into 24. 0.6 is this distance and to 4.1 is total base slab with and then for 24 is the unit weight of the concrete. So, if I use this value then it is come 59.04. Then for the fourth section that is for the soil fourth and fifth portion that is the fourth and fifth portion. Fourth this rectangular portion and fifth this triangular this is for the weight of the soil. So, for the fourth portion we can write this is 3 into 4.8 into 19 as the unit weight of this soil is 19 and for the triangular section it will be half into 0.8 into 3 into 19. So, in that way if I write fourth section that is 3 into 4.8 into 19. So, vertical force again this soil pressure will act in vertical direction the 273.6. Now, for the fifth segment for the triangular one for the soil that is equal to half into 0.8 into 3 into 19. So, that value is 22.8 kilo Newton per meter. So, these are the vertical loads vertical force that is acting due to the concrete and the soil. Now, the we can write the next two forces that is the horizontal and vertical forces acting due to the lateral earth pressure. So, sixth segment we can write this is p v that will also act in vertical direction. So, that p v value is 32.15 then the seventh one is pH horizontal force. So, that means pH will act in horizontal direction. So, that is 120 kilo Newton per meter. So, we can write the total vertical force summation of v that is equal to the summation of this vertical column is 433.67. Similarly, summation of this horizontal force total 102 kilo Newton per meter. So, summation of total vertical force is 433.67 kilo Newton per meter and summation of horizontal force is 120 kilo Newton per meter. So, this is force table. Now, we will calculate the moment table. So, once we calculate the moment table. So, this table we will calculate. So, this is the force table the next one is the moment table. So, again this is also serial number. Then we will consider the vertical and horizontal force that is force. So, now we can write the next column is this is serial number. We can write the next column is our liver arm with respect to toe. The next one is the moment. Now, this moment can be resistive moment M R and this can be overturning moment M O. This is kilo Newton per meter per meter. So, first we will go for again this 7 force that we have taken first, second, third, fourth, fifth, sixth, seventh. So, from the third force table we can write the value of these forces. So, for the force table first vertical force for the first segment is 34.56. Then second one is 11.52, then third one is 59.04, then fourth one is 273.6, fifth one is 22.8, then the sixth one is 32.6, then third 15 and seventh one is 120 which is horizontal. Now, for the each segment we have to calculate the liver arm. So, now if I consider the first segment, this is the first segment. So, it is a rectangle. So, this will act the center of this rectangle. So, liver arm from the toe will be 0.6 plus 0.2 plus 0.3 divided by 2. So, that as it will act the center of this rectangle. So, that is 0.6, 0.2, 0.6 plus 0.2 plus 0.3 divided by 2. So, now if I write in this form, so the liver arm will be for this first segment is 0.6 plus 0.3, sorry 0.2 plus 0.3 divided by 2. So, this will give us 0.6 plus 0.3 divided by 2. So, the value 0.95. Similarly, for the second one, second is a triangle. So, second segment is a triangle. So, here liver arm will be 0.6 plus 2.3 of 0.2. So, that will give you the liver arm or the second triangle portion. So, for the second portion the liver arm is 0.6 plus 2.3 of 0.2. So, that will give 0.73. Similarly, for the third section that is the base of this rectangle. So, that will give us the straight forward. This will act as a center of this rectangle. So, 4.1 divided by 2. So, liver arm for this third portion is 4.1 divided by 2 that is equal to 2.05. Now, liver arm similarly for the fourth section that is the rectangle of this backfill soil. So, that is 0.6 plus 0.2 plus 0.3 divided by 2 that is equal to. So, that means here this is 0.6, 0.2 plus 0.2 plus 0.3 plus it will act the center of this 3. So, plus 3 by 2. So, now this liver arm is 2.6. Similarly, for the fifth segment that is for the triangular zone. So, that liver arm is 0.6 plus 0.2 plus 0.3 then it is a triangle. So, triangle means plus 2 third of 3. So, it will give 3.1. So, these are the liver arm value 0.951, 0.73, 2.05, then 2.6, then 3.1 and for the sixth one that is acting vertically the distance of 4.1 meter. So, that is 4.1 meter for the sixth one and as calculated that the horizontal one that is the vertical one will act 4.1 meter from the toe and the horizontal will act 2.07 meter from the toe. So, that distance is 2.07, 2.07. So, these are the all rounded values are the liver arm. Now, this from this this is vertical forces corresponding to that moment will give the resistive moment and the moment corresponding to this horizontal force that will give the overturning moment. Now, if I calculate this moment. So, this is simply 34.56 into 0.95. So, this will give resistive moment is 32.83. So, this is actually 34.56 into 0.3. So, this is actually 34.56 into 0.95. So, that give you 32.83. Similarly, for the second one is 11.52 into 0.73. So, that value is 8.41. Similarly, for the third one is 59.04 into 2.05. So, this will give 121. The fourth one is 273.6 into 2.6 that is given 711.6 into 2.5 into 2.5 into 2.5 into 3.6. The fifth one is 22.8 into 3.1. So, this will give 70.68. Sixth one is 32.15 into 4.1. So, this will give 131.8 and the seventh one is 120 into 2.07. So, this will give 248.4. So, these are the resistive moments and the overturning moments. Now, if I take the summation of all these moments. So, that means, summation of M R that is equal to summation of this column is 1076.1 and summation of overturning moment is 1076.1. So, this is 248.4. So, we can write that M R is 1076.1 kilo Newton per meter kilo Newton meter per meter and M O is 248.4 kilo Newton meter per meter. Now, we have completed the post table, done the moment table. Then, we will calculate the, we will check the factor of safety for different condition. The first factor of safety that we will check that is for the sliding. So, factor of safety for sliding and that is the expression is total vertical force into tan delta divided by all horizontal force. Now, here is total vertical force summation is 433.6867 and tan delta that we have taken is 23 degree. So, into tan 23 degree and summation of all vertical horizontal force is 120. So, this value is equal to 1.53 which is greater than 1.5. So, that means, it is safe. So, now, next check that we will do for the sliding. So, we will do for the sliding overturning moment, factor of safety for the overturning and that is summation of resistive moment divided by summation of overturning moment. So, summation of all resistive moment is 1076.1 divided by 248.4. So, that will give us 4.33 which is greater than 1.5. So, that means, it is safe. The first two checks we have done and it is safe against the sliding and the overturning. Next, we will calculate the E value or the no tension condition. So, first we calculate the X bar that expression I have given that X bar is summation of M R minus summation of M O divided by summation of all vertical force. So, summation of M R is 1076.1, M O is 248.4 divided by vertical force is 433.67. So, it is coming 1.9 meter. So, E we can essentially we can calculate it is the B divided by 2 minus X bar. B is the total base width and here total base is 4.1 divided by 2 minus 1.9. So, it is coming 0.15 meter. So, and definitely which is less than B by 6. So, this no tension condition will occur. So, this is 5. So, next check we will calculate for the bearing capacity calculation. We have done the three checks that is the sliding overturning and the no tension condition. Now, for the bearing for bearing calculate first P max and that expression is all vertical force divided by base width 1 plus 6 E divided by total base width. Now, all vertical force is 433.67 divided by base width is 4.1 into 1 plus 6 into E is 0.15 and divided by 4.1. So, this P max value is 129 kilonewton per meter square. So, now factor safety for the bearing is the allowable force that is 400 kilonewton per meter square divided by this allowable bearing pressure is 400 kilonewton per meter square divided by 129. So, that value is 3.1 which is greater than 3. So, safe. So, in this way we have checked the all the factor safety that is for the sliding overturning then for the no tension condition and a factor safety for the bearing. So, all the checks are satisfied. So, the dimension that we have chosen that is for all the condition here it has satisfy all the factor safety. So, we can use this dimension for our purpose. Now, there are few things that we have to mention that suppose this problem is partially unsafe or slightly unsafe by this sliding. In that case one thing we can do that if this problem is not satisfy suppose this is our retaining wall structure. So, this is the retaining wall structure ground surface this is ground surface. Now, say this retaining wall is unsafe against the sliding then for one thing we can do that we can provide one key here this in this form just to increase this is another portion that we can extend which is called key. The idea of this key this will give the additional resistance against the sliding. So, and another thing is that when you calculate the passive resistance that passive resistance part that depth will also increase because here we will get calculate the passive resistance say P p. So, that passive resistance depth that will also increase, but one thing for the factor safety. So, this will give increase the passive resistance for this design which we have although we have not considered in our previous calculation. But this passive resistance we can that can increase and this will give the additional resistance against the sliding. So, if it is marginally unsafe against sliding we can provide keys. Now, when we can provide key one thing for this situation that the properties for the soil for this portion say phi 2 is the base soil property and C 2 is the foundation soil property and say phi 1 is the backfill soil property C 1 is the backfill soil cohesion this is friction this is the foundation soil. Now, instead of taking the full in the situation when key is used instead of taking full phi 2 and C 2 generally it is reduced by half to two-third of the value of for the extra safety as for the pool passive resistance because in this case as this key we are applying for this portion only. So, it is doubtful that whether the full passive resistance will develop or not. So, that is just to the for additional factor of safety this soil property C and phi is reduced by half to two-third for the additional factor of safety. So, now this is the function of key. Now, another thing is that if we solve this problem that we have solved this problem and where this the soil properties are same for both the cases and another one that when we are talking about this theory that we have used the Rankine's theory. Now, we can solve this problem for Coulomb's theory also. Now, this is the home assignment for you if you want to you can try to solve this problem by Coulomb's theory by applying the Coulomb's theory and you can see the what is the difference we are getting by if you use the two different theory. Another one that we are neglecting this passive resistance. Now, if we we can solve this problem by using this passive resistance that also you can try to solve this one with use of passive resistance. You can see how much extra factor of safety you will get when you consider this passive resistance. These two things by considering passive resistance and using a different theory we can try you can try to solve this problem same problem. Now, when you talking about the retaining wall the next segment that that when talking about the retaining wall it is most of the cases this the backfill material which is used mostly which is used for the granular soil to avoid the water pressure. Because it when you design the retaining wall you always try to avoid the induce of water pressure. So, that this additional water pressure will not come into retaining wall. So, that to avoid this water pressure there are few things we can do that we can provide some holes in the retaining wall. So, that water can pass through through that hole. So, that that water pressure can be neglected. Suppose, if this is one retaining wall structure that suppose this is one retaining wall structure this is base this is the ground surface. So, we can provide a hole or water body can pass through this portion this is called we folds W W P we folds. So, we can provide some filter material we can provide some filter material here in the we folds. So, suppose this is backfill material. So, this is filter material now this filter material this we folds this diameter is around 0.1 meter spacing is around 1.5 meter to 3 meter in horizontal direction. So, now this use of this we folds is to reduce the pore water pressure develop due to the water. Now, this pore water pressure if we can reduce this. So, if all the water can pass through this we folds then we can reduce the pore water pressure. So, that pore water so that the additional lateral earth pressure that we can reduce by this way. Now, another one that that we can consider this different types of we folds that this is suppose the retaining wall. We can provide a perforated pipe here with filter material surrounding this. So, that water can collect here and it can pass through from the retaining structure. So, that is our perforated pipe. And this is filter material now if this is for the granular soil or sandy soil. Now, for the clay soil when fine-gain soil is used because we always try to use this is for the sandy soil or granular soil. Now, case 2 if for the clay soil or fine-gain soil we always use we always try to avoid the use of this fine-gain soil. But sometimes if it is not possible to avoid this fine-gain soil then what we can do suppose this is our retaining structure this is existing ground this is the ground surface. Then we can provide a filter material along the vertical surface or we can provide and then we can provide a path. So, that water can collect here and it can pass through. So, this is we fold. So, this is our filter material or vertical filter. Now, this filter material can be inclined also suppose this is the material retaining wall. So, that material can be inclined and then we can provide the we folds here. So, this vertical material can be this is inclined material.