1 00:00:15,230 --> 00:00:20,199 This is lecture number 31, in this lecture would be talking about reinforced soil retaining 2 00:00:20,199 --> 00:00:25,070 walls. In fact, this is one of the biggest contributions of reinforced soil, because 3 00:00:25,070 --> 00:00:31,429 of which you are able to construct many retaining walls made of this reinforced soil. This is 4 00:00:31,429 --> 00:00:38,429 a significantly contributed to infrastructure construction. 5 00:00:38,920 --> 00:00:45,920 The way that the reinforced soil, we have already seen and its applications to reinforced 6 00:00:47,780 --> 00:00:54,050 soil walls or the retaining walls is something that is very unique. And in fact, this is 7 00:00:54,050 --> 00:00:59,609 the one that we had in a classical retaining walls in which has, say for example, you call 8 00:00:59,609 --> 00:01:06,259 the gravity retaining walls in which the weight of the soil, exerts some sort of a lateral 9 00:01:06,259 --> 00:01:10,840 pressure component and if by the weight of this soil alone the the retaining structure 10 00:01:10,840 --> 00:01:15,970 you try to resist the horizontal force. So, that is a concept of gravity wall in which 11 00:01:15,970 --> 00:01:22,200 the weight of the soil the retaining system, and it is, you try to have some geometrical 12 00:01:22,200 --> 00:01:29,200 propositions; in such a way that the it it it will have, it will withstand all the pressures 13 00:01:30,729 --> 00:01:34,090 both lateral pressures and the vertical components of pressures.. 14 00:01:34,090 --> 00:01:38,789 So, we design this, and in fact, sometimes when this is required we try to provide shear 15 00:01:38,789 --> 00:01:44,179 keys also to increase the shear resistance, because when there is a lateral force acting 16 00:01:44,179 --> 00:01:48,429 the there should be some sort of resistance that needs to be provided, either it can be 17 00:01:48,429 --> 00:01:53,429 a shear resist keys here and it can be even at the bottom sometimes like, so, that is 18 00:01:53,429 --> 00:01:59,250 one concept for gravity retaining walls. Next thing is a reinforced concrete types 19 00:01:59,250 --> 00:02:04,890 in which you have you know when bending and bending develops you know. So, you try to 20 00:02:04,890 --> 00:02:11,849 put some sort of you this is a reinforced concrete wall in which because of the, you 21 00:02:11,849 --> 00:02:16,500 try to understand the bending moments developed and try to provide the steel to withstand 22 00:02:16,500 --> 00:02:21,340 the bending and all that and try to have a suitable design methodology for that which 23 00:02:21,340 --> 00:02:25,430 is well established. So, you have essentially by the you try to 24 00:02:25,430 --> 00:02:32,430 resist the earth pressures by this cantilever action here and we also have what you call 25 00:02:33,599 --> 00:02:38,049 the buttress and counter fort retaining wall in which you try to provide you know when 26 00:02:38,049 --> 00:02:41,019 you want to reduce the thickness you know the thing is that what happens is that the 27 00:02:41,019 --> 00:02:47,189 base width of this wall and there are certain propositions in our retaining wall literature 28 00:02:47,189 --> 00:02:51,659 in which you may you must have sort of dimensions here some dimensions here, some dimensions 29 00:02:51,659 --> 00:02:56,349 here and then when the height is beyond seven meters or eight meters the possibility is 30 00:02:56,349 --> 00:03:00,799 that this widths become little bigger.. So, we try to provide some short of counter 31 00:03:00,799 --> 00:03:07,799 forts here which will see that the concrete areas are reduced the quantities of concrete 32 00:03:08,999 --> 00:03:14,189 are reduced in this system .The other one that we have in the retaining walls is that 33 00:03:14,189 --> 00:03:20,290 you try to compartmentalize the systems like you know in in the form of a bin or a you 34 00:03:20,290 --> 00:03:25,010 try to, this is a system in which you can place all this materials and it is like it 35 00:03:25,010 --> 00:03:31,499 acts as like a confined material. In a confined material that though there is a pressure from 36 00:03:31,499 --> 00:03:36,900 all the sides this confinement that we have it takes care of the pressures developed. 37 00:03:36,900 --> 00:03:42,249 So, that way it is a simple system that we have. In fact, this has been used effectively 38 00:03:42,249 --> 00:03:48,079 in some of the flyovers in Bangalore which is very cost effective. 39 00:03:48,079 --> 00:03:53,609 And the final one was that what we would be talking today is the reinforced that walls 40 00:03:53,609 --> 00:03:58,889 in which you have facing element, you have a geosynthetic reinforcement and you have 41 00:03:58,889 --> 00:04:03,909 a back fill. So, this we call it see in in there are two 42 00:04:03,909 --> 00:04:10,879 types here, One is actually mechanically stabilized earth walls. We call it m s e with metal reinforcement, 43 00:04:10,879 --> 00:04:14,150 one can have metal reinforcement otherwise you can also have a geo-grid reinforcement 44 00:04:14,150 --> 00:04:16,100 or a geosynthetic reinforcement or whatever.. 45 00:04:16,100 --> 00:04:22,680 So, these are the categories and this is the way that it looks like that you have a reinforced 46 00:04:22,680 --> 00:04:25,990 soil and then you have retaining wall and foundation soil. 47 00:04:25,990 --> 00:04:30,139 And you have different elements here you have a segment these are all called actually this 48 00:04:30,139 --> 00:04:35,630 type of wall is called segmental retaining wall, in which you have a you know this called 49 00:04:35,630 --> 00:04:39,670 segmental facing. We will see that what is the difference, you can have a full there 50 00:04:39,670 --> 00:04:45,450 are different types of facing. One one can have as I just mention earlier these retaining 51 00:04:45,450 --> 00:04:51,890 the reinforced elements are designed to take care of the lateral pressures. 52 00:04:51,890 --> 00:04:56,350 Once you know, what should be the length and the spacing of that? We have seen this idea 53 00:04:56,350 --> 00:05:02,070 in the reinforced soil slopes and once you know this design comfortably then, whatever 54 00:05:02,070 --> 00:05:06,840 you get here, the pressures are somewhat going to be minimum minimum and we try to provide 55 00:05:06,840 --> 00:05:11,030 some short of segmental facing types,it can be full facing or it can be even concrete 56 00:05:11,030 --> 00:05:15,750 panel that is what we see in most of the flyover projects in India and we also have; what you 57 00:05:15,750 --> 00:05:21,160 call a shear key or a mechanical connector between the two facing elements. And then 58 00:05:21,160 --> 00:05:27,140 do a geosynthetic reinforcement and geo you have a geo-textile drain here and you have 59 00:05:27,140 --> 00:05:31,540 a granular leveling pad and all that. So, this is a simple way of reinforced reinforced 60 00:05:31,540 --> 00:05:37,880 earth retaining wall system. We will see its design and how it can be constructed in field 61 00:05:37,880 --> 00:05:43,320 and all that in this lecture and the coming lectures also.. 62 00:05:43,320 --> 00:05:48,480 As I just mentioned vidal is one person, who developed this technique and when he developed 63 00:05:48,480 --> 00:05:55,480 this technique he had this precast concrete panels then this is a back fill this is actually 64 00:05:56,460 --> 00:06:02,070 back fill this we call it a wall fill wall fill means this is a reinforced soil wall. 65 00:06:02,070 --> 00:06:06,680 So, whatever you are trying to use it as a wall fill, because it should have good friction 66 00:06:06,680 --> 00:06:13,140 right good friction should be there and that is what we call it the reinforced earth. In 67 00:06:13,140 --> 00:06:17,130 fact, the same company has been continuing and then construct a lot of reinforced earth 68 00:06:17,130 --> 00:06:22,560 walls in abroad and india and it has been quite very useful technique. 69 00:06:22,560 --> 00:06:28,580 And you can have steel strips as a reinforcement then have geo-textile materials you can have 70 00:06:28,580 --> 00:06:34,070 you know the conventional geotextiles can be you know, it can be nonwovens woven’s 71 00:06:34,070 --> 00:06:39,580 knitted stretch bonded many types of materials that we saw earlier can be used as a reinforcement 72 00:06:39,580 --> 00:06:43,930 material. Sometimes ,even geo-textile special geotextiles 73 00:06:43,930 --> 00:06:50,180 like say for example, geo-composites like as I said, when the pore pressures are going 74 00:06:50,180 --> 00:06:56,870 to be higher or if the clay is somewhat poor quality, we can use a geo-composite in which 75 00:06:56,870 --> 00:07:01,080 you have a reinforcement as well as a drainage material attached to it. 76 00:07:01,080 --> 00:07:08,080 So, the reinforcement is good enough to take care of the lateral pressures and the drainage 77 00:07:08,610 --> 00:07:13,039 action of the geo-composite helps in draining the back fill. 78 00:07:13,039 --> 00:07:16,960 . What are the principal requirements of the 79 00:07:16,960 --> 00:07:23,680 reinforcement?We have seen that strength of the reinforcement. In fact, we did calculations 80 00:07:23,680 --> 00:07:29,280 on slope stability and other things strength of the geo-textile or the geo-grid is important 81 00:07:29,280 --> 00:07:34,020 and its stability like it should not creep to much of course,, the polymeric materials 82 00:07:34,020 --> 00:07:41,020 do creep and this strength and stability are important then durability like how long you 83 00:07:41,620 --> 00:07:45,080 know, because you are trying to design for 100 years or 50 years or whatever. 84 00:07:45,080 --> 00:07:48,900 Then ease of handling reinforcement should be able to, you should be able to handle in 85 00:07:48,900 --> 00:07:52,740 a very easy manner like you know, you can bring them to a in a lorry or you know in 86 00:07:52,740 --> 00:07:59,180 some short of containers and put them on the field very importantly high coefficient of 87 00:07:59,180 --> 00:08:03,460 friction or the [adhe/adherence] adherence with the soil like, you know we need good 88 00:08:03,460 --> 00:08:07,570 friction between the soil and reinforcement that is very important, because that is a 89 00:08:07,570 --> 00:08:13,270 whole concept of reinforced soil. Now, it should have low cost also like you 90 00:08:13,270 --> 00:08:16,680 know why are you using this?You need to have you know apart from technical advantages like 91 00:08:16,680 --> 00:08:23,680 we say that, we can increase the width of this reinforced concrete you know the thickness 92 00:08:24,460 --> 00:08:30,780 of the stem and all that in the case of reinforced concrete it involves you can say that. 93 00:08:30,780 --> 00:08:35,959 But we essential want low cost readily available, readily availability. You know ready availability 94 00:08:35,959 --> 00:08:41,050 is something very important why because, it should be done faster; it should be comfortably 95 00:08:41,050 --> 00:08:43,979 in the way that it can influence the whole system. 96 00:08:43,979 --> 00:08:50,709 So, the whole construction should be faster you know the,. So, we we saw that advantages 97 00:08:50,709 --> 00:08:55,649 of the reinforced soil earlier. It should be able to should have this some of this principle 98 00:08:55,649 --> 00:09:02,649 requirements and see that it is also intended function. So, it should have among them the 99 00:09:04,319 --> 00:09:08,620 important properties as I just mentioned tensile strength. Like say for example, there is. 100 00:09:08,620 --> 00:09:13,569 So, much earth pressure that is developed with some spacing and length of the reinforcement 101 00:09:13,569 --> 00:09:17,410 we get this tensile strength like that is one thing we will see that. 102 00:09:17,410 --> 00:09:21,540 Tensile modulus is very important, because if there is a deformation, you know movement 103 00:09:21,540 --> 00:09:28,540 is very important and reinforced soil walls also deform and tensile the modulus of deformation 104 00:09:29,060 --> 00:09:33,819 like the tensile when you get what you get from tensile strength is very important. And 105 00:09:33,819 --> 00:09:37,449 the third one is interface shear strength is also very important these 3 components 106 00:09:37,449 --> 00:09:40,240 are very important in the design. 107 00:09:40,240 --> 00:09:47,240 So, how do you go about design particularly when you are trying to use the the the reinforced 108 00:09:49,680 --> 00:09:56,240 soil technique in this is something that we use a limit equilibrium approach. We need 109 00:09:56,240 --> 00:10:02,139 to look for stability, external stability, internal stability and the properties of the 110 00:10:02,139 --> 00:10:08,160 back fill will be the friction angle at constant volumes phi and c. 111 00:10:08,160 --> 00:10:12,709 You know this shear strength parameters should be obtained corresponding to law sates. This 112 00:10:12,709 --> 00:10:17,540 what we saw in reinforced slopes also that we try to conduct say direct shear test or 113 00:10:17,540 --> 00:10:23,920 triaxil test say for example, better to do triaxial test in which you get c and phi parameters 114 00:10:23,920 --> 00:10:30,740 for effective stress parameters and for say for example, 3 or 4 normal stresses and then 115 00:10:30,740 --> 00:10:37,410 you get CV from this large strain. So, essentially CV means Constant Volume strains and design 116 00:10:37,410 --> 00:10:42,889 strength of the geo grids is what we have seen that the company gives some property 117 00:10:42,889 --> 00:10:48,040 say for example, we are seen one example, where you can say its 50 kilo newton per meter. 118 00:10:48,040 --> 00:10:54,870 And then you have various factors like the partial factors like f m the manufacturing 119 00:10:54,870 --> 00:10:59,709 constant environmental factor, damage factor and all that all this factors are there one 120 00:10:59,709 --> 00:11:04,389 can use and then finally, you should just use this as a design strength ok. 121 00:11:04,389 --> 00:11:09,209 . So, what are the external stability conservations?These 122 00:11:09,209 --> 00:11:16,110 are all you know standard even for regular retaining walls. We know that, we design the 123 00:11:16,110 --> 00:11:23,110 retaining walls for bearing overturning and sliding and deep deep stability like the retaining 124 00:11:25,509 --> 00:11:30,829 wall whether it is a reinforced earth wall or a any other like conventional retaining 125 00:11:30,829 --> 00:11:35,949 wall it should not slide right. That is one thing. So, sliding resistance 126 00:11:35,949 --> 00:11:41,240 is important overturning is also important, then bearing capacity failure should not occur 127 00:11:41,240 --> 00:11:45,569 say for example, it has some it will once you construct a structure structure like this, 128 00:11:45,569 --> 00:11:50,999 since there is nothing here there is some sort of pressures and moments that get developed. 129 00:11:50,999 --> 00:11:57,050 And the pressure bearing pressure should satisfactory here deep seated stability like you may construct 130 00:11:57,050 --> 00:12:00,920 a retaining wall. But it should not it should not be in a slip 131 00:12:00,920 --> 00:12:07,069 circle you know if this is a you try you should construct, you know you should really do a 132 00:12:07,069 --> 00:12:11,819 some sort of analysis in which wherever you are going to place a reinforced earth wall 133 00:12:11,819 --> 00:12:16,449 the area should not lead to some stability like issue like this global global stability 134 00:12:16,449 --> 00:12:22,449 we call it. So, all this things have to be satisfied. 135 00:12:22,449 --> 00:12:28,110 In fact, to to design the reinforced earth walls we have two methods, one is called Tie 136 00:12:28,110 --> 00:12:35,110 Back Wedge method I will just explain what it is and see we what tie is a tension member 137 00:12:35,279 --> 00:12:41,720 and then this is a wedge that is forming right. So, you try to put reinforcement like a tie 138 00:12:41,720 --> 00:12:47,029 member and then keep the wedge back in position. So, that is why we call it Tie Back Wedge 139 00:12:47,029 --> 00:12:54,029 method and the assumptions some of the things are that like, you have a rotation is above 140 00:12:55,319 --> 00:13:00,999 this thing and the earth pressure distributions like this. It is like 44 minus phi by 2 is 141 00:13:00,999 --> 00:13:07,999 a act failure zone like, this is a failure zone at which the failure surface and this 142 00:13:09,149 --> 00:13:14,999 essentially we assume that the wall is cohesion less and it rotates above this and the earth 143 00:13:14,999 --> 00:13:19,670 pressure variation know we assume that, it is same everywhere like we assume active active 144 00:13:19,670 --> 00:13:23,259 earth pressure conditions ok. Activate earth pressure conditions we assume 145 00:13:23,259 --> 00:13:30,259 and, because the there are two conditions state of in the beginning I said the state 146 00:13:31,809 --> 00:13:37,410 of soil is it a K Naught condition or a K A condition like is it the coefficient of 147 00:13:37,410 --> 00:13:40,949 earth pressure thrust is there or is it for the active earth pressure we should design 148 00:13:40,949 --> 00:13:45,470 is something very important. So, this material like you know, you put a 149 00:13:45,470 --> 00:13:48,470 geo-textile say for example, what you are trying to do is that we later will put some 150 00:13:48,470 --> 00:13:54,110 geotextiles here and then there will be some deformation. So, because of which there will 151 00:13:54,110 --> 00:14:00,110 be an active condition. So, we assume that the variation earth pressure is constant and 152 00:14:00,110 --> 00:14:03,420 the earth pressure lateral earth pressure diagram will be like this you know. 153 00:14:03,420 --> 00:14:10,379 Earth pressure diagram in terms of this thing this is just a k a constant this is K A gamma 154 00:14:10,379 --> 00:14:16,019 h right K A gamma h is the pressure and now you have a surcharge also W S so, plus you 155 00:14:16,019 --> 00:14:23,019 add you know. So, this is a actually the earth pressure at any point due to it is a with 156 00:14:23,569 --> 00:14:28,319 a its variation with depth as well as due to the surcharge charge as well. This is called 157 00:14:28,319 --> 00:14:29,740 Tie Back Wedge method. 158 00:14:29,740 --> 00:14:35,679 We have another method which is something which people have you know based on observations 159 00:14:35,679 --> 00:14:42,050 of many of the RE walls you know they have observed that you can in the earth pressure, 160 00:14:42,050 --> 00:14:46,009 if you really ob]erve it will not be you know an active condition. 161 00:14:46,009 --> 00:14:53,009 It will be in a K Naught condition for some about at least 5 to 6 and based on that. In 162 00:14:53,300 --> 00:15:00,050 fact, see they they say that, In fact, K Naught they just put here. You can see K Naught into 163 00:15:00,050 --> 00:15:05,569 that is one thing K Naught will be its about 6 meters up to say for example, if you are 164 00:15:05,569 --> 00:15:10,579 trying to measure small high retaining walls like 3 meters or 4 meters if you measure earth 165 00:15:10,579 --> 00:15:16,029 pressure it will be K Naught like I have measured. In fact, in one of the cases in Delhi, it 166 00:15:16,029 --> 00:15:22,829 is about 7 or 8 meters,, but I was measuring measuring with depth at some points the earth 167 00:15:22,829 --> 00:15:27,579 pressure was K Naught. You know we have we have a measurement of by using some diaphragm 168 00:15:27,579 --> 00:15:32,129 and all that there is a method of measuring earth pressures we I used it and found that 169 00:15:32,129 --> 00:15:35,819 the earth pressure is corresponding to K Naught condition. 170 00:15:35,819 --> 00:15:41,629 Like which means that soil has not moved like there is no lateral strain is 0 that is what 171 00:15:41,629 --> 00:15:48,629 the point is. So, when you have that. So, then. So, K Naught is always 1 minus sine 172 00:15:49,149 --> 00:15:55,709 phi and K A is 1 minus sine phi by 1 plus sine phi. So, as the height is increased it 173 00:15:55,709 --> 00:16:00,670 will become K Naught to K A. It will decrease actually ok. Say for example, 174 00:16:00,670 --> 00:16:06,019 if the wall height is 8 meters. So, it will be this. So, you have to assume a distribution 175 00:16:06,019 --> 00:16:12,569 diagram like this ok. Earth pressure distribution diagram like this and same thing also here. 176 00:16:12,569 --> 00:16:19,569 Like, there the the difference is only that you have the other one superimposed on this. 177 00:16:20,439 --> 00:16:27,439 So, this is a the coherent gravity method is what it means is that, people have observed 178 00:16:27,449 --> 00:16:34,300 that many of the retaining walls of particularly smaller heights or having a state of rest 179 00:16:34,300 --> 00:16:39,769 and then the even the failure diagram. In fact, people have measured. 180 00:16:39,769 --> 00:16:43,699 You know how do you measure the earth pressure diagram in fact, you can monitor the tensile 181 00:16:43,699 --> 00:16:49,059 strengths in the reinforcement instead of the 45 minus phi by 2 that what we have seen 182 00:16:49,059 --> 00:16:53,790 this is the earth pressure the failure zone that they have taken. In fact, even if you 183 00:16:53,790 --> 00:16:58,399 just do a finite element analysis it comes like this. It is this point you know like 184 00:16:58,399 --> 00:17:01,119 in the previous case. 185 00:17:01,119 --> 00:17:06,100 It is like this you know. So, whereas, you are just cutting it here only what it means 186 00:17:06,100 --> 00:17:11,579 is that now the advantage is that, earth pressure that you have to coincident design is much 187 00:17:11,579 --> 00:17:16,500 less in the in the next in the coherent gravity method compared to this case. 188 00:17:16,500 --> 00:17:22,010 Here you are trying to consider all this pressure whereas, in the next this case the earth pressure 189 00:17:22,010 --> 00:17:27,850 is only like this which is which means that this method is somewhat economical is it not, 190 00:17:27,850 --> 00:17:32,830 we are trying to make this if you make based on the observations this is a method that 191 00:17:32,830 --> 00:17:39,350 was used developed by you know some group of people in France and particularly it is 192 00:17:39,350 --> 00:17:42,590 valid for steel reinforcement. You know, because they are initial working 193 00:17:42,590 --> 00:17:48,769 for steel reinforcement and steel reinforcement is stiffer we have seen that. And, whereas, 194 00:17:48,769 --> 00:17:52,919 a geo-textile is a flexible reinforcement at it needs some strain. So, I cannot assume 195 00:17:52,919 --> 00:17:57,399 this sort of a distribution I should only go for the previous distribution. 196 00:17:57,399 --> 00:18:02,500 In this distribution, what people have observed is that this is just a. So, it also rotates 197 00:18:02,500 --> 00:18:08,299 about this point not in other case it was somewhat reverse like you can see in the previous 198 00:18:08,299 --> 00:18:08,700 case. 199 00:18:08,700 --> 00:18:15,700 It is the rotation is about the toe whereas, here the rotation is at the top. So, that 200 00:18:17,049 --> 00:18:20,409 is what the assumption that they have and this is they have measured actually they have 201 00:18:20,409 --> 00:18:27,409 measured the earth pressures and even measured the strains. Like you know you can place you 202 00:18:29,539 --> 00:18:33,470 know you can measure the strain along the reinforcement entire length of the reinforcement. 203 00:18:33,470 --> 00:18:40,470 And in this zone what happens is that the if this is the lateral force acting the tensile 204 00:18:41,470 --> 00:18:43,929 strains direction will be in the opposite direction ok. 205 00:18:43,929 --> 00:18:50,799 The strain measurements they will be like this and beyond a failure point the they try 206 00:18:50,799 --> 00:18:57,799 to hold it ok. They direction of strain should be different on both sides. So, direction 207 00:18:58,240 --> 00:19:02,970 based on that actually shear stress distribution along this reinforcement you know, because 208 00:19:02,970 --> 00:19:08,679 this is a shear stress mobilization right. Shear stress mobilization along the reinforcement 209 00:19:08,679 --> 00:19:15,679 based on that, you can locate this actual failure surface, because at this point you 210 00:19:16,919 --> 00:19:23,919 know the beyond you know this point will constitute one location where the shear strain that distribution 211 00:19:25,140 --> 00:19:28,669 will be different on either side you know. So, that is what you know the thing is that 212 00:19:28,669 --> 00:19:35,669 what it means is that. You can construct this like you know you what people have done is 213 00:19:37,460 --> 00:19:43,710 using finite element analysis also or using experiments also people have measured using 214 00:19:43,710 --> 00:19:46,230 you can put reinforcement and measured its tensile strengths. 215 00:19:46,230 --> 00:19:52,039 So, at this point actually the point at which it is maximum you know we have seen one even 216 00:19:52,039 --> 00:19:58,309 in the Binquit Lee method , also they assume some maximum strain distribution and we we 217 00:19:58,309 --> 00:20:04,049 made some assumptions there similar to that we have the location of strains here. So, 218 00:20:04,049 --> 00:20:07,200 it will be like that right. So, if you construct this particular line 219 00:20:07,200 --> 00:20:13,950 and based on the observations they found they observed that for a h by 2distance it is about 220 00:20:13,950 --> 00:20:20,950 10 inverse 0.6, 0.6 and another h by 2 the height of that it will be straight line and 221 00:20:21,039 --> 00:20:28,039 this particular things is 0.3 h. So, actually this is the line of maximum tension reinforcement. 222 00:20:29,960 --> 00:20:35,450 So, if you observe for all the reinforcement elements where is the maximum tension and 223 00:20:35,450 --> 00:20:40,149 then locate them and join them you know the that actually, you know that there is a you 224 00:20:40,149 --> 00:20:43,889 can find out the distribution of tension on the reinforcement. 225 00:20:43,889 --> 00:20:50,889 So, as friction develops along this there will be a strain mobilization and then all 226 00:20:50,970 --> 00:20:56,880 this points you can join even in flack and flack’s analysis you can get that. So, this 227 00:20:56,880 --> 00:21:00,370 is one important method that one should understand. 228 00:21:00,370 --> 00:21:07,370 So, these two are some design philosophies that people have and another important thing 229 00:21:09,610 --> 00:21:15,269 that is that ok.- You have seen that the there are three different ways of assessing the 230 00:21:15,269 --> 00:21:19,320 external stability. We just saw the pictorially the diagram and 231 00:21:19,320 --> 00:21:23,789 how do you really calculate that.We will see that say for example, this is earth pressure 232 00:21:23,789 --> 00:21:29,929 diagram that we have to consider like half K A b gamma h square half K A b like point 233 00:21:29,929 --> 00:21:34,620 5 gamma k a active, because this is for a geo-textile I am what I am telling is a geo-textile. 234 00:21:34,620 --> 00:21:41,620 Say, half K A gamma h square and then this is h by 3 we assume and then this is the sub 235 00:21:41,820 --> 00:21:48,820 the other one is W S is what other one. So, you have to, like the other important thing 236 00:21:50,889 --> 00:21:54,950 is that. The bearing for the bearing pressure we assume 237 00:21:54,950 --> 00:22:01,950 that, there are two types of earth pressure distribute[on]- the distributions of pressure 238 00:22:02,010 --> 00:22:06,909 contact pressure actually. You know that we have a trapezoidal putting 239 00:22:06,909 --> 00:22:10,970 Trapezoidal distribution you know where you have a maximum tension developed and the minimum 240 00:22:10,970 --> 00:22:17,840 tension developed and we have that b by 6 rule right b by 6 rule we have. So, that b 241 00:22:17,840 --> 00:22:22,529 by 6 rule is very important and here also similar thing we measure and that is called 242 00:22:22,529 --> 00:22:29,029 trapezoidal distribution that is valid for stiff footings I mean rigid footings like 243 00:22:29,029 --> 00:22:34,799 concrete footings and all that and whereas, this is since reinforced soil is a flexible 244 00:22:34,799 --> 00:22:41,260 structure and you will not have the development of maximum pressure and minimum pressure .It 245 00:22:41,260 --> 00:22:45,809 gets adjusted in such a manner that, if l is the length of the foundation like in R 246 00:22:45,809 --> 00:22:52,760 E wall length is about l over a distance l minus 2 it gets adjusted sigma v is what you 247 00:22:52,760 --> 00:22:57,179 get. And. So, instead of a trapezoidal you can 248 00:22:57,179 --> 00:23:02,320 have trapezoidal is one type of distribution we can assume and this is called Meyerhof 249 00:23:02,320 --> 00:23:08,330 distribution ok. In Trapezoidal distribution we assume that, you have a maximum value you 250 00:23:08,330 --> 00:23:14,450 have a minimum value and the distribution varies is in the form of A the the trapezoid. 251 00:23:14,450 --> 00:23:20,000 But then that is valid in the case of rigid foundations,, but reinforced soil being a 252 00:23:20,000 --> 00:23:24,830 flexible structure what happens is that any pressure gets redistributed and you will have 253 00:23:24,830 --> 00:23:29,130 only constant value. So, that is a maximum value sigma v you get 254 00:23:29,130 --> 00:23:34,809 and this is ok. This is a usual thing. So, this one another important thing it is called 255 00:23:34,809 --> 00:23:36,639 Meyerhoff distribution. 256 00:23:36,639 --> 00:23:43,639 So, how do you get this factors of safety like we have seen that. For sliding, there 257 00:23:45,919 --> 00:23:51,029 is a sliding force which is, because of K A b gamma h square by 2 and surcharge and 258 00:23:51,029 --> 00:23:56,230 all that that you just put here resisting force is nothing,, but the weight of the that 259 00:23:56,230 --> 00:24:00,440 2 at bottom comes here. So, 2 into mu is nothing but, the mu is nothing 260 00:24:00,440 --> 00:24:06,950 but, the coefficient of friction on the base soil of the reinforced soil and it is the 261 00:24:06,950 --> 00:24:11,710 total resisting force you know gamma w into height plus W S is a total resisting force 262 00:24:11,710 --> 00:24:16,730 right. So, the fact of, So, you try to calculate this and the fact of safety has to be two 263 00:24:16,730 --> 00:24:18,429 right this is one thing. 264 00:24:18,429 --> 00:24:24,830 Now, even the for the overturning you have to calculate resisting moment and also the 265 00:24:24,830 --> 00:24:27,970 overturning overturning moments. So, this is actually you should know the base 266 00:24:27,970 --> 00:24:32,830 of that like one third where it acts and all that and then you these are all expressions 267 00:24:32,830 --> 00:24:37,029 actually details are you can simply derive this equations which are straight forward 268 00:24:37,029 --> 00:24:41,630 ,considering the equilibrium of forces and the factor of safety should be somewhat say 269 00:24:41,630 --> 00:24:46,850 two or something and most of the cases it is the sliding that governs the design. So, 270 00:24:46,850 --> 00:24:51,620 in this case overturning is not a serious issue if you are able to satisfy the stability 271 00:24:51,620 --> 00:24:57,059 issue, the sliding stability issue automatically this gets shorted out. 272 00:24:57,059 --> 00:24:57,309 .   273 00:24:57,059 --> 00:25:02,799 Then bearing pressure bearing pressure is that you know I just showed you about I told, 274 00:25:02,799 --> 00:25:07,009 you about the two types of distribution. One is called Trapezoidal distribution you can 275 00:25:07,009 --> 00:25:10,000 do that also. And also you can do Meyerhoff distribution 276 00:25:10,000 --> 00:25:15,889 which is nothing but, like there is simple expression that one can get based on the moments 277 00:25:15,889 --> 00:25:21,149 and the forces acting on the at the you know at that below the foundation you will get 278 00:25:21,149 --> 00:25:26,130 a simple expression one can derive this considering considering the weights and all that. And 279 00:25:26,130 --> 00:25:33,130 earth pressure coefficients, the maximum pressure will be this and from meyerhoff distribution 280 00:25:33,820 --> 00:25:39,799 even you can get from trapezoidal distribution also we know how to get maximum pressure and 281 00:25:39,799 --> 00:25:46,799 minimum pressure. So, actually that this pressure whatever pressure should be the less than 282 00:25:49,769 --> 00:25:52,990 the bearing pressure of the soil that is what is very important. 283 00:25:52,990 --> 00:25:57,799 Whatever is the pressure that is created because of the this pressure should be less than the 284 00:25:57,799 --> 00:26:01,259 bearing capacity of the soil. So, usually an allowable bearing pressure of half the 285 00:26:01,259 --> 00:26:06,440 ultimate is taken ultimate pressure is taken. So, we normally normally have a factor of 286 00:26:06,440 --> 00:26:13,440 safety of 2 ,normally you can see that bearing bearing capacity calculations we know and 287 00:26:13,500 --> 00:26:18,210 we calculate say for example, 200 k p a is a bearing pressure of the soil and if this 288 00:26:18,210 --> 00:26:23,019 pressure is say, 150 it is alright like that right. 289 00:26:23,019 --> 00:26:28,070 Then we have to say as I said, we are looking at all this global stability issues we should 290 00:26:28,070 --> 00:26:33,490 look for you may construct a retaining wall,, but is should not be located in a slip surface. 291 00:26:33,490 --> 00:26:37,610 So, all potential slip surfaces should be investigated and the target factor of safety 292 00:26:37,610 --> 00:26:38,730 1.5 is used. 293 00:26:38,730 --> 00:26:43,059 In this case that is what is ours External Stability. 294 00:26:43,059 --> 00:26:49,100 Then, what is Internal Stability, like now, you have introduced a reinforcement. So, it 295 00:26:49,100 --> 00:26:54,039 should not fail by tension or it should not come by pull out. It should not just come 296 00:26:54,039 --> 00:26:58,899 out like a pull out or it should not fail by tension like this is a tension failure 297 00:26:58,899 --> 00:27:03,200 where there is a clear breakage where this pull out you know reinforcement just comes 298 00:27:03,200 --> 00:27:04,919 out this pull out. 299 00:27:04,919 --> 00:27:11,350 For that we have a method, how do you calculate this tension failure and all that you will 300 00:27:11,350 --> 00:27:18,350 see that this is a diagram. You know there is a surcharge acting and you take a a particular 301 00:27:19,960 --> 00:27:26,039 reinforcement that is located at a distance h i and this is what I said, is a actually 302 00:27:26,039 --> 00:27:33,039 the Meyerhoff distribution type you know the pressure actually, the foundation pressure 303 00:27:33,110 --> 00:27:39,049 acts here right. You can assume that the same pressure acts this is for the maximum height. 304 00:27:39,049 --> 00:27:44,789 If the height of the wall is h and this is the maximum pressure that acts here and then 305 00:27:44,789 --> 00:27:49,870 as you go up the pressure comes down. So, the at any level h i you can always calculate 306 00:27:49,870 --> 00:27:56,870 the vertical pressure acting total vertical pressure and for that you you are actually 307 00:28:01,330 --> 00:28:07,669 calculating horizontal force you put k a times vertical pressure you will get the lateral 308 00:28:07,669 --> 00:28:14,669 pressure and and then design for T i. What we say is that at this V i you know V i is 309 00:28:27,370 --> 00:28:34,370 a horizontal. We assume that at this I th level which is at at a distance for the top 310 00:28:35,620 --> 00:28:42,620 H i and V i is a spacing you know the reinforcement is effective, it can take care of the you 311 00:28:42,759 --> 00:28:45,639 know it can provide force within that spacing right. 312 00:28:45,639 --> 00:28:52,639 So, how do you calculate that reinforcement force, it is actually sigma V i you calculate 313 00:28:52,899 --> 00:28:59,899 and sigma then K A times sigma v we will get that that we will see there ok. 314 00:29:00,219 --> 00:29:00,469 . 315 00:29:00,220 --> 00:29:04,990 So, the grids carry tension as a result of the self weight of the fill and the surcharge 316 00:29:04,990 --> 00:29:08,990 acting on the top of the block, that is what why that is what we did like this is what 317 00:29:08,990 --> 00:29:14,370 I said, you know this is actually if we have seen the previous figure, we have seen the 318 00:29:14,370 --> 00:29:19,090 expression for you know this is nothing but, the vertical pressure into k a times ok. 319 00:29:19,090 --> 00:29:26,090 And actually we if the soil has cohesion we introduced cohesion term also otherwise it 320 00:29:26,299 --> 00:29:33,299 is not required and this acts over a distance V i ok. T is nothing but, it is like K A times 321 00:29:36,850 --> 00:29:42,100 into this vertical pressure minus this actually whatever is this thing and then into V i. 322 00:29:42,100 --> 00:29:49,100 You know as I said, it acts over a distance V i, the T i is nothing but, it acts over 323 00:29:49,200 --> 00:29:53,490 this distance you know in this area which is with a spacing V i ok. 324 00:29:53,490 --> 00:29:55,389 This is how we have this equation. 325 00:29:55,389 --> 00:30:02,389 So, what we do is that, we calculate for different reinforcement types you know like I will just 326 00:30:03,950 --> 00:30:08,149 show you that. 327 00:30:08,149 --> 00:30:14,419 This is an expression and we will see the different you will have as a higher as you 328 00:30:14,419 --> 00:30:18,840 you can calculate this as we go down this spacing has to be little higher. And depending 329 00:30:18,840 --> 00:30:24,490 on the type of reinforcement you can develop the short of curves like as I said, see this 330 00:30:24,490 --> 00:30:30,179 is the V i spacing . You can have different V i’s and also their depth actually h i 331 00:30:30,179 --> 00:30:30,830 is here ok. 332 00:30:30,830 --> 00:30:37,830 H i is a everything is known here and once based on this you can calculate the T i required 333 00:30:39,779 --> 00:30:45,110 otherwise otherwise we know what type of reinforcement materials I have. You know I have say for 334 00:30:45,110 --> 00:30:48,210 example, in this case 55 grade and 80 grade. 335 00:30:48,210 --> 00:30:53,210 I can construct back what should be the spacing , if I have this short of reinforcement at 336 00:30:53,210 --> 00:30:56,009 different depths is what is given by this equation. 337 00:30:56,009 --> 00:31:01,309 So, this is that say for example, 40 grade 40 grade I do not want to use,, but I can 338 00:31:01,309 --> 00:31:05,470 say that forty divided by all the partial factors right. So, maybe you will get some 339 00:31:05,470 --> 00:31:11,929 20 number or 15 number here 15 15 is equal to this and then you have the only variable 340 00:31:11,929 --> 00:31:17,389 you assume this as c equal to 0. So, this term vanishes and all the terms are known 341 00:31:17,389 --> 00:31:21,580 and only unknown is. So, you put different values of V i right. It can be from 1 meter 342 00:31:21,580 --> 00:31:24,799 to 0.5 meters to 0.3 meters and all that. 343 00:31:24,799 --> 00:31:29,340 You will get a diagram like this 0.3, 0.4, 0.5 like this and then you can get a diagram 344 00:31:29,340 --> 00:31:36,340 like this. So, it is called spacing versus depth diagram and, this will help us to we 345 00:31:39,610 --> 00:31:43,220 will see, how it helps us in calculating a spacing ok. 346 00:31:43,220 --> 00:31:47,519 That is one thing we know how to calculate now, some you know you know the reinforcement 347 00:31:47,519 --> 00:31:50,809 and particularly particularly in the case of geo grades say for example, some grade 348 00:31:50,809 --> 00:31:57,259 of geo grid is available you know how to work back what should be the spacing, that needs 349 00:31:57,259 --> 00:32:04,259 to be provided so, that that spacing in that spacing ,the lateral pressure is really handled 350 00:32:04,830 --> 00:32:11,830 properly or taken care of. Then once you provide this reinforcement we have some more considerations 351 00:32:15,019 --> 00:32:22,019 like, it should not fail so, we have to calculate tension using the wedge pull out failure also 352 00:32:22,350 --> 00:32:26,629 consider the possibility of failure planes passing through the wall and forming unstable 353 00:32:26,629 --> 00:32:29,779 wedges. We try to have number of wedges like this 354 00:32:29,779 --> 00:32:35,850 and what we do is that say for example, in this case , you have in this wedge of this 355 00:32:35,850 --> 00:32:42,850 particular thing you have 1, 2, 3, 4, 5, 6 layers of reinforcement here like 1 1. So, 356 00:32:43,720 --> 00:32:50,720 you have six layers of reinforcement and that. They have to be in the position that wedge 357 00:32:52,179 --> 00:32:57,440 should be stable which means that the sum of all this tension forces reinforcement should 358 00:32:57,440 --> 00:33:04,440 be such that it is stable like the t, should be equal to the you know it is a component 359 00:33:04,779 --> 00:33:09,789 of this weight ok. Which you can get from. In fact, I will show 360 00:33:09,789 --> 00:33:16,789 you that it can be you know from graphics element. We know the basics of graphics, engineering 361 00:33:17,080 --> 00:33:22,149 graphics, one can get that you know from, what is called polygon force polygon, you 362 00:33:22,149 --> 00:33:24,519 can get that ok. I will see like you know, if you know the 363 00:33:24,519 --> 00:33:29,149 weight acting you know the horizontal they actually this is the horizontal force acting 364 00:33:29,149 --> 00:33:32,679 you know imagine that there is a curvature and there is a vehicle goes like this there 365 00:33:32,679 --> 00:33:36,720 is a component component of f 1 that is acting there is a vertical load acting. 366 00:33:36,720 --> 00:33:42,669 So, it is like a retaining wall and a flyover imagine that ok. So, what we should do we 367 00:33:42,669 --> 00:33:47,730 are trying to take a wedge here at an angle beta and then there is a because of the weights 368 00:33:47,730 --> 00:33:54,159 acting W and all that, there is a resultant is a force resultant, you have a resultant 369 00:33:54,159 --> 00:34:01,159 which has a an angle phi w with this and then you have a theta tension force t which will 370 00:34:03,019 --> 00:34:08,250 be required to keep all these things in equilibrium that is what we try to find out. 371 00:34:08,250 --> 00:34:15,250 What we do is that, we have to we will make some assumptions, we each wedge behaves as 372 00:34:15,450 --> 00:34:20,290 a rigid body friction between the facing and the fills is ignored we investigate a [se/series] 373 00:34:20,290 --> 00:34:24,290 series of wedges as shown below. Like you know ,we take a number of wedge failure conservations 374 00:34:24,290 --> 00:34:30,250 here, like you know see even very important is that see say for example, you have So, 375 00:34:30,250 --> 00:34:36,589 much reinforcement 1, 2, 3, 4, 5, 6, 7, 8 we have all one two this is all. 376 00:34:36,589 --> 00:34:42,919 So, in this wedge, if we are trying to look at this wedge stability I should see that 377 00:34:42,919 --> 00:34:46,419 there should be the tensile force should be. So, much that this wedge should be in under 378 00:34:46,419 --> 00:34:50,829 stable conditions that is what we will investigate in a series of wedges. 379 00:34:50,829 --> 00:34:57,829 This is what we do, I said the force polygon right, you can see here and, these are all 380 00:34:58,119 --> 00:35:05,119 the forces that can act and for a simple case of only surcharge acting and you can simple 381 00:35:07,920 --> 00:35:12,339 expression can be you know just based on force polygon one can calculate this expression 382 00:35:12,339 --> 00:35:17,890 imagine that, you know resolving into vertical forces and horizontal forces in the direction 383 00:35:17,890 --> 00:35:23,780 of you know appropriately you can do that. So, for a simple at any level by changing 384 00:35:23,780 --> 00:35:30,780 beta you know what we do is that ,we investigate this T max for different betas like as I just 385 00:35:30,880 --> 00:35:32,750 showed my in my previous diagram. 386 00:35:32,750 --> 00:35:37,930 Betas can be vary from. So, may betas here say then there could be many betas here. 387 00:35:37,930 --> 00:35:44,930 So, all that betas we investigate is a simple simple excel program one can write, to have 388 00:35:46,020 --> 00:35:53,020 all this information and then plot like this ok. And then once you plot that you will have 389 00:35:53,319 --> 00:35:58,660 the maximum value. So, you take the maximum value as the tensile force required to keep 390 00:35:58,660 --> 00:36:03,839 all the wedges in equilibrium. That is what is required it is a very simple case where 391 00:36:03,839 --> 00:36:07,910 you try to calculate the maximum tensile force required to keep all the wedge the you know 392 00:36:07,910 --> 00:36:12,970 all the. So, that the wedge failure do will not occur. That is a principle here like all 393 00:36:12,970 --> 00:36:19,970 this wedges or all contain you know put in this form right ok. 394 00:36:20,020 --> 00:36:27,020 So. In fact, for a simple case where, there is only a surcharge you know one can it is 395 00:36:27,319 --> 00:36:34,319 a simple equation you will get t t is nothing but, h tan beta into gamma h and if h is the 396 00:36:36,190 --> 00:36:40,930 maximum, you know say for example, height height of the retaining wall is say 7 meters 397 00:36:40,930 --> 00:36:45,619 I put 7 meters here and surcharge I know surcharge is nothing but, say for example, 20 k p a 398 00:36:45,619 --> 00:36:51,280 as per IRC or 22 k p or whatever. So, you can put that number you will get the maximum 399 00:36:51,280 --> 00:36:56,619 tensile force required in the its actually total reinforcement force total reinforcement 400 00:36:56,619 --> 00:37:00,970 force required we will see that. 401 00:37:00,970 --> 00:37:07,970 So, now, the other important thing is that, the wedge pull out failure you know it is 402 00:37:10,270 --> 00:37:17,270 also a function of the overburden, right the thing is that the...This is all that Overburden 403 00:37:18,230 --> 00:37:24,460 pressure and whatever is that you know there should be a sufficient length beyond the failure 404 00:37:24,460 --> 00:37:28,390 actually if this is the failure surface that, we assume, and there should be enough length 405 00:37:28,390 --> 00:37:34,089 to see that the tensile failure you know we have see fact of safety you know the tensile 406 00:37:34,089 --> 00:37:37,970 force we calculate and the pull out resistance, also we calculate there should be adequate 407 00:37:37,970 --> 00:37:42,650 factor of safety there otherwise I mean the pull out resistance should be higher than 408 00:37:42,650 --> 00:37:49,380 the tensile force mobilized maybe two times ok. 409 00:37:49,380 --> 00:37:55,450 This is another one that one can have you know. So, what is the Anchorage force you 410 00:37:55,450 --> 00:38:00,130 can develop you will get is also given by you know actually that length that you have 411 00:38:00,130 --> 00:38:06,559 you know, the length this is called the Anchorage length you know. That length at any point 412 00:38:06,559 --> 00:38:13,520 ,one can calculate and that is given by this expression and the Anchorage force or the 413 00:38:13,520 --> 00:38:18,470 pull out resistance there are two terms, one is one is called a tensile resistance tensile 414 00:38:18,470 --> 00:38:23,369 force mobilize in the reinforcement, now, you have to calculate the Anchorage force, 415 00:38:23,369 --> 00:38:27,300 anchorage force means pull out resistance so, there is nothing the nothing but, it is 416 00:38:27,300 --> 00:38:32,880 a function of the length of the reinforcement below the beyond the failure zone and also 417 00:38:32,880 --> 00:38:37,990 there is a coefficient alpha then tan phi w depends on the surcharge also. 418 00:38:37,990 --> 00:38:43,369 For each layer of the reinforcement cut by the wedge ,the lower of the design values 419 00:38:43,369 --> 00:38:50,369 T design or the Ta is used to determine the contributions from reinforcement. So, what 420 00:38:50,640 --> 00:38:57,640 we do is that we we calculate whichever is lower and then we try to then again compare 421 00:38:58,050 --> 00:39:02,980 the mobilizing force with the resisting force we have to see that we may call this calculations 422 00:39:02,980 --> 00:39:09,980 of say, what is said earlier previous section was that, you calculate the no this is in 423 00:39:11,020 --> 00:39:14,470 terms of the again the wedge wedge failure stability.. 424 00:39:14,470 --> 00:39:21,030 And there are,this is in terms of the anchorage ok; and the previous case was in terms of 425 00:39:21,030 --> 00:39:28,030 the tension So, what we try to see is that all the forces we try to compare and see that 426 00:39:28,859 --> 00:39:35,329 the sum of all the forces should be more than or equal to the total tension mobilized, the 427 00:39:35,329 --> 00:39:41,250 total tension is nothing but, the total tensile force required for the wedge stability right. 428 00:39:41,250 --> 00:39:48,250 So, this is we will see that in an example that you will get much more clarity on that 429 00:39:48,260 --> 00:39:55,260 and this is a typical case of a GRS wall constructed with face you know, like pre precast concrete 430 00:39:56,660 --> 00:40:02,050 facing and you have a wrapped geo-textile also one can have, you know there are many 431 00:40:02,050 --> 00:40:05,040 types of facings are possible. 432 00:40:05,040 --> 00:40:12,040 So, there are number of design methods. In fact, as I said GRS walls have been very useful 433 00:40:13,920 --> 00:40:20,920 and they are they have been very cost effective. And; So, that design approaches that are developed 434 00:40:21,069 --> 00:40:26,160 and essentially the two design methods that we follow, actually Rankine approach is one 435 00:40:26,160 --> 00:40:31,839 thing like 1 minus sine phi that type of approach we use. And we have a Federal Highway approach 436 00:40:31,839 --> 00:40:38,200 ,which is by you know one standard code then we have NCMA approach another one and we also 437 00:40:38,200 --> 00:40:44,740 have BS 8006 UK code these are all from UK, US actually and then you have a US code BS 438 00:40:44,740 --> 00:40:50,599 8006 is another standard code and many of most of the codes in the world, they follow 439 00:40:50,599 --> 00:40:55,319 many of this codes and then sometimes they follow in a in different countries they have 440 00:40:55,319 --> 00:40:56,770 their own codes also. 441 00:40:56,770 --> 00:41:03,770 So, before I go further I want to illustrate a few more points here that as I said facing 442 00:41:06,750 --> 00:41:13,750 is can be anything. So, people have tried different facing blocks, because people these 443 00:41:14,210 --> 00:41:20,010 are all very simple actually, you know the one can carry it with hand like if you are 444 00:41:20,010 --> 00:41:25,220 familiar with some wall structures that are constructed you know in a concrete panel that 445 00:41:25,220 --> 00:41:30,010 we have you know it needs lot of cranes cranes and all that here we do not need cranes you 446 00:41:30,010 --> 00:41:36,140 know, we just have simple equipment and also the a simple company that manufactures this 447 00:41:36,140 --> 00:41:43,140 types of facing blocks ,it is called Segmental Retaining Wall units like they are quite useful 448 00:41:44,750 --> 00:41:47,240 like you know these are all facing blocks you can see that. 449 00:41:47,240 --> 00:41:52,170 You know you can construct like this simple facing like this is a typical example, that 450 00:41:52,170 --> 00:41:56,690 we have seen in many place in India , some place at least this is a facing you know it 451 00:41:56,690 --> 00:42:02,579 is called Segmental Wall design this is another type. So, you the advantage is that, it is 452 00:42:02,579 --> 00:42:09,240 easy to construct the you can have a nice finish and the see it is also cheaper, that 453 00:42:09,240 --> 00:42:13,369 is one thing, because the overheads you know, because of the high heavy heavy equipment 454 00:42:13,369 --> 00:42:16,119 are not there. 455 00:42:16,119 --> 00:42:19,500 This is another type, you can see simply blocks only are kept compared to you have panel of 456 00:42:19,500 --> 00:42:26,500 1.2 or 1.3 meter you know the panel will be 1 meter to one 1.2 meters it depends you know. 457 00:42:26,859 --> 00:42:33,859 This is another one you can just see many types of structures one can have. 458 00:42:38,079 --> 00:42:42,970 So, the way that we construct is that see it is like these are all simple blocks they 459 00:42:42,970 --> 00:42:47,950 can be handle and this weight is you know shear key as I just mentioned ,and this is 460 00:42:47,950 --> 00:42:54,950 another one you know the type of thing we have, So, this is how it is done ok. 461 00:42:55,569 --> 00:43:02,569 So, very importantly actually the you are since you are putting face face shear blocks 462 00:43:04,290 --> 00:43:09,119 one should understand that it can they can fail also. Like it is it is not easy to simply 463 00:43:09,119 --> 00:43:14,170 stack one over the other. So, like a similar to a regular retaining walls, you should do 464 00:43:14,170 --> 00:43:20,200 for base sliding, overturning and bearing capacity calculations, internal stability 465 00:43:20,200 --> 00:43:23,890 calculations. Pull out, tensile resistance, internal sliding 466 00:43:23,890 --> 00:43:28,460 particularly like this I have seen couple of failures like this, because we think that 467 00:43:28,460 --> 00:43:34,490 simple construction is easy,, but its not easy, then facing, there could be a connection 468 00:43:34,490 --> 00:43:41,490 failures, there could be column you know it it fails as a column here, you know it then 469 00:43:42,030 --> 00:43:48,670 there could be some toppling failures many things are possible. So, one though they facing 470 00:43:48,670 --> 00:43:52,859 looks very impressive, one should be very careful, because one should look at all this 471 00:43:52,859 --> 00:43:54,390 possibilities. 472 00:43:54,390 --> 00:43:57,339 So, this is what I just mentioned 473 00:43:57,339 --> 00:44:02,280 Like the it can fail and all that this another type failure. 474 00:44:02,280 --> 00:44:08,420 So, this is a column shear failure toppling we have seen many cases in India, actually 475 00:44:08,420 --> 00:44:14,950 because what they do this is all simple precast block. So, I can get from they try to do well 476 00:44:14,950 --> 00:44:19,410 poor quality then the possibility is that it can lead to failure ok. 477 00:44:19,410 --> 00:44:24,490 So, this is a simple example of Global Stability like you know you have to investigate may 478 00:44:24,490 --> 00:44:31,490 be this is a nice say for example, this is a typical what is called segment Segmental 479 00:44:32,770 --> 00:44:36,839 Wall, what is called Modular Concrete Blocks you can have it. 480 00:44:36,839 --> 00:44:43,839 This is a reinforcement layers ; and you have a drainage here ,drainage pipe is here and 481 00:44:45,470 --> 00:44:51,619 geo grids are placed and this is how it is there right I just want then, we have to investigate 482 00:44:51,619 --> 00:44:58,619 the like you know, because it is forming this part of the slope right so, you have to investigate 483 00:45:00,640 --> 00:45:05,569 all the failure surfaces you can see that 1.33 one point you you have all this numbers 484 00:45:05,569 --> 00:45:11,299 definitely its more than 1.3. So, it is ; like that. So, you have to do some sort of analysis 485 00:45:11,299 --> 00:45:12,520 like this.. 486 00:45:12,520 --> 00:45:19,520 So, the typical factors that we recommended are , you know it could be like this you know 487 00:45:19,599 --> 00:45:24,819 base sliding 1.5, bearing capacity, tensile over stress all that you have to verify you 488 00:45:24,819 --> 00:45:31,109 know toppling and one should calculate and do that global stability 1.3 to 1.5 in the 489 00:45:31,109 --> 00:45:34,619 previous case its more than one point three. So, it is ok. 490 00:45:34,619 --> 00:45:37,339 So, the construction details yes, be 491 00:45:37,339 --> 00:45:43,230 This one way, this is other way like this is a drainage element.. 492 00:45:43,230 --> 00:45:48,900 Actually you know this is another blocking Locking Bar. 493 00:45:48,900 --> 00:45:54,540 This is a General view of the Wall Construction. 494 00:45:54,540 --> 00:46:01,540 Placing blocks see you have a thread here varying wall ties fixing false facing, you 495 00:46:10,140 --> 00:46:14,520 can have you know false facing also.. 496 00:46:14,520 --> 00:46:20,030 This maybe actually required facing this can be a nice facing whichever you want like people 497 00:46:20,030 --> 00:46:27,030 have false roofing, locking geo grid between blocks these are one type.. 498 00:46:28,309 --> 00:46:32,240 So, Sometimes you know Safety Barriers on the Walls at the top of the wall also should 499 00:46:32,240 --> 00:46:36,740 be done, because once you construct the wall people should not just you know at least should 500 00:46:36,740 --> 00:46:41,160 not lead to any problems. So, this is like that. 501 00:46:41,160 --> 00:46:46,119 So, many you know one can have a Nice Fencing and the anything’s are possible like examples 502 00:46:46,119 --> 00:46:47,500 of finished structures. 503 00:46:47,500 --> 00:46:51,200 This is another one. 504 00:46:51,200 --> 00:46:58,200 This is a nice treatment that one can give this is I just want to show, you the influence 505 00:47:01,400 --> 00:47:07,099 of the because of the Earth quake, the possibility is that you know one of the Railway lines 506 00:47:07,099 --> 00:47:13,640 in Japan has yielded and that was reconstructed reconstructed later with a horrible technique 507 00:47:13,640 --> 00:47:15,160 ok.. 508 00:47:15,160 --> 00:47:19,819 And geo grid reinforced soil wall along the JR you know Japanese Railway Kobe Line you 509 00:47:19,819 --> 00:47:25,940 know Kobe Line is an Earth quake that occurred and the thing is that it is before before 510 00:47:25,940 --> 00:47:27,480 the Earth quake occurred. 511 00:47:27,480 --> 00:47:33,329 And you can see the same photo, after the Earth quake you can see that there is a collapse 512 00:47:33,329 --> 00:47:40,329 of all the structures,, but RE wall is still same ok. 513 00:47:41,490 --> 00:47:48,490 So, this is another one damaged [mason/masonry] masonry wall reconstructed to GRS with a Full 514 00:47:49,220 --> 00:47:53,069 Height Facing, you know this is one masonry wall type of construction that again it did 515 00:47:53,069 --> 00:47:58,089 not fail in the Earth quake. But they reconstructed with you know, Full 516 00:47:58,089 --> 00:48:02,280 Height Facing I told you know full height facing; that means, the Facing is fully there. 517 00:48:02,280 --> 00:48:08,950 In fact, we did many one is full height facing is also possible like a Simple Facing, fully 518 00:48:08,950 --> 00:48:15,950 put a one number no do not worry about simple blocks like that, right. 519 00:48:16,539 --> 00:48:21,160 Another example of poor quality in Bangalore. This is one case where you know the it was 520 00:48:21,160 --> 00:48:24,740 not well done you can see that. So, you you this technique is quite simple. 521 00:48:24,740 --> 00:48:29,869 I would like to just now illustrate with a simple example this is where I would like 522 00:48:29,869 --> 00:48:35,180 to end. We have an 8 meter high wall to be built using the sand fill and polymer grid 523 00:48:35,180 --> 00:48:40,410 reinforcement the sand has a friction angle 30 degrees, bulk density is 18 meters and 524 00:48:40,410 --> 00:48:47,410 surcharge load of 15 kPa is there and bearing capacity is about 300 kPa you have two geo 525 00:48:48,170 --> 00:48:52,520 grids available grid A 20 kilo newton per meter and B equal to 40 kilo newton per meter 526 00:48:52,520 --> 00:48:57,200 both have a bond coefficient of 0.9 ok. These are all we known and then the fill has 527 00:48:57,200 --> 00:49:03,510 250 centimeters of compaction which you know why this is required is that, you compact 528 00:49:03,510 --> 00:49:07,950 the sample and put a geo grid. So, next one should not be you know it should be multiples 529 00:49:07,950 --> 00:49:14,780 of 250 only any placement should be in terms of 250 only, because the thickness is important 530 00:49:14,780 --> 00:49:19,200 here. And the location of the geo grids you have 531 00:49:19,200 --> 00:49:23,369 to redesign say for example, you have even your facing block, if you are using a modular 532 00:49:23,369 --> 00:49:30,369 blocks it would be 250 mm then put to all heights this is a very important parameter, 533 00:49:30,390 --> 00:49:37,390 if you decide this then spacing design of your facing blocks all depend on that ok. 534 00:49:40,230 --> 00:49:45,329 As I just mentioned this is all the once you know the quantities and fact of safety I give 535 00:49:45,329 --> 00:49:51,700 an expression for the sliding stability like, if you use previous formula that we discussed 536 00:49:51,700 --> 00:49:57,930 to avoid sliding stability, the factor of safety of minimum is 2. 537 00:49:57,930 --> 00:50:03,170 So, use that number and all the quantities like Kab height and surcharge everything is 538 00:50:03,170 --> 00:50:07,490 known you substitute back in the equation and the minimum length required is about 5.8 539 00:50:07,490 --> 00:50:14,490 meters. So, you can take 6 meters, the height of the wall is 8 meters and 6 by 8 is 75 percent. 540 00:50:15,710 --> 00:50:21,559 So, 7.75 is the length of the 0.75 L is a length of the reinforcement ok.. 541 00:50:21,559 --> 00:50:27,710 Actually, this very important you know this was sliding stability is taken care similarly 542 00:50:27,710 --> 00:50:31,160 you can do the overturning stability like you know you can put all this numbers it is 543 00:50:31,160 --> 00:50:35,630 actually given in my book also, this particular examples and the factor of safety with respect 544 00:50:35,630 --> 00:50:39,630 to overturning is you know if you just put all the numbers its 4.26 little more than 545 00:50:39,630 --> 00:50:40,619 2. 546 00:50:40,619 --> 00:50:46,730 Now, using Trapezoidal distribution you calculate, you know you I said you can do Meyerhoff distribution 547 00:50:46,730 --> 00:50:51,390 or even Trapezoidal distribution and use for example, in this case I would like to show 548 00:50:51,390 --> 00:50:57,450 the use of Trapezoidal distribution one can see that and you get maximum vertical stress 549 00:50:57,450 --> 00:51:03,559 as 271 kPa which is less than the bearing capacity of the soil. And minimum value is 550 00:51:03,559 --> 00:51:08,609 159 ,maximum minimum value is its more than 0 which means that there is no tension developed. 551 00:51:08,609 --> 00:51:15,480 Actually, if you are using why you use this Trapezoidal distribution is that, we would 552 00:51:15,480 --> 00:51:20,819 like to satisfy the no tension criteria also but, remember the reinforced soil is foundation 553 00:51:20,819 --> 00:51:25,950 is a flexible structure. So, we better use a Meyerhoff distribution that leads to actually, 554 00:51:25,950 --> 00:51:29,250 if you use a Meyerhoff distribution then this value match will be much lower. 555 00:51:29,250 --> 00:51:36,049 So, will be instead of telling you telling that its 271 is less than 300 kPa this value 556 00:51:36,049 --> 00:51:42,619 will be much lesser, because there is a readjustment and this sigma v minimum will be zero it will 557 00:51:42,619 --> 00:51:45,380 also come down, because there is a readjustment. 558 00:51:45,380 --> 00:51:52,380 Now, as I just mentioned how to calculate the tensile forces right, this is that horizontal 559 00:51:54,260 --> 00:52:00,630 force into Sv is the spacing right; and this is say for example, we assume a spacing of 560 00:52:00,630 --> 00:52:05,900 Sv and this is a horizontal force, this is a tensile force required to keep that place 561 00:52:05,900 --> 00:52:11,849 in position Ka this is what the this thing and sigma v is nothing but, this formula is 562 00:52:11,849 --> 00:52:18,849 there right. So, you substitute all the quantities and then you know put in terms of the Sv like 563 00:52:19,099 --> 00:52:23,869 you know see the Ti is nothing but, the design strength values like you know I just have 564 00:52:23,869 --> 00:52:29,000 geo grid materials and that design strengths I know like 20 and 40 like and the previous 565 00:52:29,000 --> 00:52:36,000 example I in the statement of the problem I just mentioned. The two grids of different 566 00:52:37,480 --> 00:52:44,480 design strengths are available grid grid at 20 and 20 and 40 ok. This point nine are available 567 00:52:44,900 --> 00:52:47,230 this we should remember ok. 568 00:52:47,230 --> 00:52:53,079 So, you put it here, using the see your putting it here that number and then try to just put 569 00:52:53,079 --> 00:53:00,079 this sort of equation and you will get a grid like this grid A, grid B you just get depending 570 00:53:00,740 --> 00:53:05,710 on the spacing and all that you will get some sort of simple diagram like this and; actually, 571 00:53:05,710 --> 00:53:12,710 we are only going for space close spacing’s only what it means is that, if you are using 572 00:53:15,640 --> 00:53:21,450 a geo grid of A you know which is somewhat weaker compared to this you know 40 kilo newton 573 00:53:21,450 --> 00:53:28,450 is bigger stronger. So, you can have 0.25 spacing right up to then you know up to a 574 00:53:31,339 --> 00:53:38,339 depth of about this this much length. So, here grid a you know 20 kilo newton 20 kilo 575 00:53:40,299 --> 00:53:46,660 newton per meter we can just it starts with you know you you have to go for lesser than 576 00:53:46,660 --> 00:53:53,660 which is not possible here. So, what it means is that Grid A you cannot use bottom at the 577 00:53:54,680 --> 00:54:01,180 bottom, because its less than 0.25 it is going here. So, I can start using 0.25 only from 578 00:54:01,180 --> 00:54:08,180 here at 6.5 between 5 and 6 say 5.5. That is what it means ok. So, once you do 579 00:54:11,829 --> 00:54:18,780 this then this helps you know in the fashion. So, one can construct you know 0.5 you know 580 00:54:18,780 --> 00:54:23,780 just put in a simple excel program and then get this this thing and plot like this, you 581 00:54:23,780 --> 00:54:29,430 will get that information,, but essentially we are going for this case and Wedge stability 582 00:54:29,430 --> 00:54:36,430 check can also be done and we can trial wedges can be done and calculate the total required 583 00:54:37,049 --> 00:54:39,109 force is the one thing I just mention. 584 00:54:39,109 --> 00:54:44,079 And you have to calculate checks with and without surcharge and as I just mentioned 585 00:54:44,079 --> 00:54:48,660 this is an equation for a critical angle that you have and this is an expression, that we 586 00:54:48,660 --> 00:54:53,500 have,and for any reinforcing layer at depth z below the top of the wall the pull out resistance 587 00:54:53,500 --> 00:54:58,339 is given by this number. So, this is you know behind the retaining 588 00:54:58,339 --> 00:55:04,720 wall the behind the failure zone or the you know in the Anchorage length we call it you 589 00:55:04,720 --> 00:55:07,819 [calu/calculate] calculate the passive resistance. 590 00:55:07,819 --> 00:55:14,819 So, you get all this information and what we do is that we try to make all this calculations 591 00:55:16,130 --> 00:55:22,180 and say for example,this is the diagram that I would like to show here wedge depth you 592 00:55:22,180 --> 00:55:28,470 have wedges at different depths 1, 2, 3, 4, 5, 6, 7, 8 meters and force to be resisted, 593 00:55:28,470 --> 00:55:33,940 total force that you have to resist will be, if it is 0 surcharge is 8 and if it surcharge 594 00:55:33,940 --> 00:55:40,079 is 3 so, you calculate all these numbers this is all the force to be resisted by the members. 595 00:55:40,079 --> 00:55:44,690 Then, you know the geo grids you know like you know I just mention you have to just put 596 00:55:44,690 --> 00:55:51,690 some numbers here in the sense that so, if I have 2 2 geo grids, we have A and B and 597 00:55:51,900 --> 00:55:58,900 you have a JAJ grid is a twenty kilo newton’s this I put to 40 and this is one member. 598 00:55:59,510 --> 00:56:04,329 And, pull out resistance is we have an equation that I just mentioned pull out resistance 599 00:56:04,329 --> 00:56:08,900 you are getting, you know with and without surcharge we calculate without surcharge pull 600 00:56:08,900 --> 00:56:15,900 out resistance is 42 with surcharge is this and whatever is minimum say for example. 601 00:56:19,500 --> 00:56:26,500 You have to take here right ,and then see that whatever is available ok. See the that 602 00:56:27,160 --> 00:56:32,369 you have to compare you know both of them you have to compare available force, minimum 603 00:56:32,369 --> 00:56:39,369 of Pd and Pp 40. So, here its 80 ok. So, all this things we maintain and see that. 604 00:56:47,650 --> 00:56:54,650 This is now this is this is a minimum this is what is we have and, the design strength 605 00:56:56,510 --> 00:57:03,510 you compare actually ok. So, they have to be more than that you know 606 00:57:05,339 --> 00:57:12,339 any stage like, pull out resistance have to be higher. This pull out resistance have to 607 00:57:16,260 --> 00:57:23,260 be higher say for example, the at 0 can be w s equal to 0 say, this is 5639 is a pull 608 00:57:24,829 --> 00:57:28,680 out resistance available the design tensile force is 700. 609 00:57:28,680 --> 00:57:35,680 So, that is what we see that at all the time, the design requirement is satisfied here in 610 00:57:35,690 --> 00:57:42,690 terms of the tensile forces and the force to be resisted say for example, here ,192 611 00:57:43,859 --> 00:57:50,859 you can get from this combination only 15 into 20 plus 10 into b b is 40. So, that comes 612 00:57:51,829 --> 00:57:58,829 to totally 700 and that 700 will be higher than this. You know this is this will higher. 613 00:58:01,980 --> 00:58:08,980 So, essentially one one needs to formulate sort of you know evaluate like this and; In 614 00:58:10,089 --> 00:58:12,960 fact, there are many software available on to calculate. 615 00:58:12,960 --> 00:58:17,329 In fact, if you just do this you know this is the way that it should be provided like 616 00:58:17,329 --> 00:58:22,559 you know 1 meter 2 meters and all that and you know spacing can be avoided and [de/design] 617 00:58:22,559 --> 00:58:27,770 design spacing it should be 0.25 it should start here and then with you can have spacing 618 00:58:27,770 --> 00:58:33,150 and length is l is a length of reinforcement. So, one should be able to design a suitable 619 00:58:33,150 --> 00:58:37,210 reinforcement layout and there are many number of software that are available they they are 620 00:58:37,210 --> 00:58:41,849 company specific they are also, one can design systematically using an excel program and 621 00:58:41,849 --> 00:58:45,420 all that. So, essentially, we what we discussed. So, 622 00:58:45,420 --> 00:58:52,420 far is that ,yes some short of design. If reinforced soil walls is possible, using a 623 00:58:53,589 --> 00:58:58,770 given type of reinforcement or you can even recommend a given type of recommend you know 624 00:58:58,770 --> 00:59:04,010 check the stability of this reinforced Earth walls considering Internal Stability as well 625 00:59:04,010 --> 00:59:08,910 as External Stability and all that and with this I will conclude Thank you.