 Today, I will discuss about the different other conditions of the loading and how to calculate the bearing capacity foundation under such loading condition, because in other classes in the last classes I have discussed that about the bearing capacity calculation and other bearing capacity calculation. And then they are the safe factor, depth factor, all those are introduced in the loading condition to get the bearing capacity of the ultimate bearing capacity of the foundation. Now, today's class I will discuss about the, if the loading is not applied at the center of the foundation, if the eccentrically loaded footing is there and then how to calculate the bearing capacity. And then when we calculate the safe factor and the depth factor or basically the safe factor and the ultimate load carrying capacity of the footing. Then how to use those loading condition, that eccentric condition in the foundation, those things I will discuss in this lecture. Now, first I will go for the one way eccentric loading and then I will go for the two way eccentric loading. Now, first case suppose if I go for this, now this is for the eccentrically loaded foundation. In all other cases that suppose this is the footing and this is the center of the footing. Now, suppose if loading is applied at the center or if I take the cross section of this foundation or footing, that this is the cross section and the loading is applied at the center of the milling. In this center of the footing, in this case the previous formula we can apply, but if there is any moment or if the loading is eccentrically. Then how to calculate this bearing capacity of the foundation, first thing we can discuss in this section. Suppose if the loading is, there is a moment which is applied here and this Q is the loading which is applied at the center and M is the additional moment which is applied. And suppose this is the B width of the foundation and dimension of the footing say this is B cross L. So, this is the dimension of the footing or we can say this is B cross L. So, this is this direction is B and this direction the dimension of the footing is L. Now, first under this loading condition if the loading if the loading is applied at the center or there is no moment, then the uniform distribution of loading is observed. Now, if the moment is applied, so the footing is distribution of the load below the footing it will follow this pattern. So, this is the reaction of the soil which is giving to the foundation and so now as this moment is applied in this direction. So, this will give you the maximum reaction that is Q max and this will give you the minimum reaction that is Q min. So, now this is one condition where this is following this pattern. Now, there is another situation where Q max is this value and Q min is the negative value or suppose this is the reaction this is Q max. And now you can see that if we draw this portion that means here the this this side the reaction is this is negative that mean the tension will develop. So, we will not as we know that that soil cannot take the tension cannot able to take the tension. So, that means there is a provision to give the separation between the foundation and the soil if the reaction is negative or if the tension force is developed. So, we will we will neglect the second case suppose this is our case one and this is the case two. So, we will neglect the second case. So, that means there is a case one which is developed if our E value is less than B by 6 and this is developed if E value is greater than B by 6. So, that means our one condition that E should not be greater than B by 6 if it is one way essentially loaded foundation. So, now we can calculate we have to calculate this E max and E mean value. So, one condition. So, no tension will be developed and so under no tension condition this is one condition. So, this is for the no tension condition condition. So, this is our one condition the next thing is we have to calculate. So, that means this E should not be greater than B by 6. So, that means always E will be less than equal to B by 6. So, if E is greater than B by 6 then this tension will develop that is not acceptable for this foundation these are now we will avoid this kind of foundation. So, now we can calculate the E max and E mean. So, suppose E max will get by using this expression that q divided by B L plus 6 M B square L where q is the load which is acting in the foundation and M is the moment which we are applying. And similarly that q mean we will get this value is q B L minus 6 M B square L. So, q is equal to total vertical load and M is moment on foundation. Now, the eccentric value is we are talking about that E is the eccentricity. So, that eccentricity we can calculate E is equal to M divided by q. So, first step we will calculate this if we know the M and q value we calculate the E value that is M by q and then we will check whether this A is within this limit or not. If it is within this limit that means less than equal to B by 6 then we can proceed otherwise we have to redesign our dimension of the foundation we have to change the dimension or we have to redesign this foundation. So, that means that is the first step. Then the next step if we put this E value in our equation 1 and equation 2 then we will get E max is equal to q B L plus 6 into E q divided by B square L or if we take q by B L common then this will be 1 plus 6 E divided by B. Similarly, q mean we can calculate by this equation q divided by B by L 1 minus 6 E divided by. So, that means by putting this E value in equation 1 and 2 we will get the q max and q mean in terms of q B L and E. So, this next first we will check this condition then we will calculate E max and E mean. Another thing is that suppose this the footing is in this fashions that if we put this E value in equation 1 and 2 we will get the q max and q mean in terms of q B L and E. So, this next first we will check this condition then we will calculate E max and E mean. Another thing is that suppose this the footing is in this fashions that this is our footing and loading itself is applied is not applied at the center it is applied at a distance of say E from the center. So, this is the loading q we are applying which is applied at a distance of E from the center and this is also one condition. So, in so we can draw this type of figure. So, this is this type of figure we can draw. So, that means this hatch area because as the loading is applied at a distance of E from the center. So, the effective area if loading is applied at the center then the effective area will be equal to L into B. So, that is the total effective area L into B. Now, because of this eccentricity or if there is moment is applied of both the cases because of this type of loading condition the load is itself is applied at a distance E from the center of it is the moment. So, because of this moment and loading eccentricity the effective area of the footing that mean that now it is considered that this hatched zone is effectively taking the load and this white portion is not taking load. So, when you calculate the ultimate load carrying capacity and use this expression. So, instead of using B we have to use this value B dash and here this value L dash as it is one way eccentricity. So, L dash will be equal to L, but B dash is reduced by some amount. So, here this white portion is written as 2 into E. So, this white portion is written this is 2 into E. So, white portion is 2 into E and this hatched portion is B dash. Now, here suppose this is the point where loading is applied where E is this value. So, now here our effective area is basically reduced because of this loading condition. So, now when we calculate the this effective area. So, instead of B dash we can calculate that will be B minus 2 E because this 2 E is the white portion. So, B dash will be B minus 2 E and L dash is as usual as it is one way eccentric loading. So, L dash will be equal to L. So, when we calculate the effective area instead of using A we will use the A dash that is B dash into L dash because if the effective area without any eccentricity is A into B into L A is equal to B into L then similarly A dash will be B dash into L dash. So, for this first condition is for centrally loaded footing and this second condition is for eccentrically loaded footing. So, in this fashion. So, first step if I say again. So, first step we by use if we know the M moment and the load which is acting. So, we will calculate E value or if I know the directly this eccentricity of this loading then we will calculate this E value. From here we will check whether this E is less than equal to B by 6 or not. If it is less than equal to B by 6 then otherwise the tension will develop and ideally we will not allow the tension to develop. So, we will redesign the foundation then by using these two expressions we will calculate the E max and E min and then next job we will determine the effective width of the foundation that is B dash equal to B minus 2 E 2 E and L dash is equal to L and then effective area that is A dash will be B dash into L dash. Now, when we calculate this bearing capacity of this loading then instead of using this B we will use the B dash and as this is one way essentially loaded foundation then we will use L and L dash these are both are same. And when we calculate the this factors there also we will use B dash not B. Now, the next one is that foundation. So, first case was the foundation with one way eccentricity. Now, the next case foundation with two way eccentricity. So, suppose this is the cross section of the footing this is ground line and this is B and dimension of the footing B cross L this is the dimension. So, this one is B and this is L L is the length of the footing and B is the width of the footing. Now, if q is applied here. So, this is say q ultimate and one moment is applied here like the one way. So, this is one way eccentricity. Now, if the suppose this is x x and this is y y section. Now, in the same footing this is y y and again q ultimate is this one this is the load q ultimate and one moment that is m y and another moment that is m x both are acting in this foundation. Here only one moment is acting that is one way eccentricity. Now, if the m x and m y moment with respect to x x axis and moment with respect to y y axis both are acting in this footing. So, then this is two way eccentricity. Now, under this condition suppose this is the E where this one is equal to E L because this is in terms of length and this is width and this distance is equal to E B. So, this one E L and this one is E B or you can write this is E x E y and this is E x. So, here in this this lecture we will use E L and E B. Now, where this is the length and this is the B is the width. So, that means here again how to calculate this because here also in the previous cases when it was a one way eccentricity then our A dash was B dash and L dash where B dash is equal to B minus 2 E and L dash was equal to L that is for one way. Now, again for the two way eccentricity also effective area L dash will be L B dash into L dash, but here again so this is B dash. So, here B dash will be B minus 2 E and L dash will be A minus 2 E. So, here we will use in case of when you are talking about 2 E. So, we will use this is 2 E B and this is 2 E L. So, this is for 2 A eccentricity again how to calculate this E L and E B. So, E L we can calculate by E max M x divided by U ultimate and E B will be M y divided by U ultimate. So, in two way eccentricity. So, when you calculate E L we use M with respect to x x axis that is M x divided by U ultimate and E B that is M y divided by U ultimate. Now, when you calculate the Q U value. So, then we will use the expression C N C S C D C I C plus Q which is equal to D F into gamma N Q S Q into gamma N Q S Q into gamma N Q D Q I Q plus half into gamma B dash N gamma S gamma D gamma I gamma. So, here C is the cohesion of the soil. So, here we know this is the expression suppose this is the general expression. So, here N C N Q and N gamma these are the these are the bearing capacity factors and S C S Q S gamma are the say factor D C D Q D gamma as a depth factor and I C this is I Q and I gamma are the inclination factor. So, when you calculate this factor instead of using B or L we will use B dash and L dash and here also we are using B dash not B. So, finally when you calculate the suppose we know this Q ultimate Q ultimate load that means Q U into A dash. So, that will give us Q U into B dash into L dash. So, when you calculate this value we will use B dash into L dash. So, this is a two way eccentricity and then one way eccentricity. Now, we will this I will discuss about the various cases suppose the first case that I will discuss that is case one. In this case the condition is that our E L divided by L that is greater than equal to 1 by 6 that is one condition and E B by B that is also greater than equal to 1 by 6. So, suppose if this condition arises that E L by B that is greater than equal to 1 by 6 and E B by B that is greater than equal to 1 by 6. In such case what will happen suppose if in such case if I do not able to avoid these things then if this case is still arising then how we will get the I will design this condition. So, in that case suppose this is the foundation and this is B and this is L and that case we will get this type of effective area. So, this hatched zone will give us the effective area. So, that means here we will get this is B 1, this is B and this is L 1. Now, here this is the eccentric point E this is the point where this one is E L and this is E B. So, now when we calculate the from this area because when we calculate the ultimate load then we will calculate determine this effective area. Now, this effective area A dash will determine by using this expression that half B 1 into L 1 this is half B 1 into L 1 and then I will calculate this this B 1 is given by B into 1.5 minus 3 E B divided by. Now, this thing is been explained is in B M Das book that is Das B M 1999. So, where this value we will get that we can directly use this value and then L 1 we will get similar expression L 1.5 minus 3 E L divided by L. So, by using this expression we will get B 1 and by using this expression we will get L 1. Now, the effective length L dash will be the larger value of these two which one is larger that is the effective area. So, effective length L dash is equal to larger value of B 1 and L 1. So, once we get the L dash then the B dash will be because we will get A dash directly by this expression. So, this mean B dash will be A dash divided by L dash. So, in this the larger value of B 1 and L 1 will give the L dash value and B dash will be A dash divided by L dash. So, we will get the B dash and L dash. So, once we get the B dash and L dash then we can use this B dash and L dash to get the others factors bearing capacity factors or the safe factors and then to calculate the ultimate load carrying capacity of the foundation and then we will get the A dash this is half into B 1 into L 1 then we will get the total ultimate load that the foundation can take. So, this is one case if this condition is this E L divided by L is greater than equal to 1 by 6 or E B divided by B is greater than equal to 1 by 6. Now, the next case or case 2 that I will explain this case 2 thing that this case 2 then the case 2 that E L by L that is less than 0.5 and another condition that E B by B that is greater than 0 and less than 1 by 6. So, E L is less than 0.5 and E B by B greater than 0 1 and less than 1 by 6 in such case the effective area will be as follows. Now, this is the footing now this hatch zone will give us the effective area. Suppose this is the B is the width of the foundation L is the length of the foundation and here this one is the L 1 and this distance is equal to L 2 and this is the A E value say this is E L and this is E B for condition is E B is greater than equal to 0 and less than equal less than 1 by 1 by 6 B and E L is less than 0.5 L. Now, this is the effective area under this second case now here the effective area we can calculate by this half into L 1 by plus L 2 into B. So, this is the effective area now the effective length L dash is larger value of L 1 and L 2. So, about this L 1 and L 2 is equal to the larger value will give us the L dash. So, similarly B dash will get from A dash divided by L dash. So, once we get this L dash which is the larger value of L 1 and L 2 whichever is larger and then we will get the B dash by L dash by A dash by L dash. Now in the case 3 where the condition is that E L L which is less than equal to 1 by 6 and E B by B which is less than 0.5 and greater than 0. So, E L by L less than 1 by 6 and E B by B greater than 0 less than 0.5. So, in such case in the case this is for case 2 if I want to draw the condition for the case 3 suppose this is length this is B. So, in such case effective area will be this one. So, this hatch portion will give us the L effective area. So, this one is equal to now B 1 and this value will give us B 2 this is L and this one B. So, this will be the point. So, this is E B and this one E L. So, this point this is key ultimate. Similarly, this one also this is key ultimate. So, now in this condition. So, area effective area of this hatch zone will get that is half into B 1 plus B 2 into L and L dash that will be equal to L and B dash is equal to A dash divided by L dash. So, here we will get this A dash and L dash is equal to L and B dash is A dash by L dash. So, this is for case. So, next case that is the case 4. So, next case case 4 where condition is that E L by L which is less than 1 by 6 and E B by B which is also less than 1 by 6. So, E L by L less than 1 by 6 and E B by B less than 1 by 6. So, in such case the loading loaded foundation the effective area will be suppose this is L this is B. So, effective area will be like this. So, this one this hatch zone give us the effective area. So, this one is again B this value is equal to L 2 and this value is equal to L and this value is equal to B 2 and this is the E I mean this is E L and this is E B. So, in this case. So, this is B 2 this is B this is L 2 and total is L. So, in this case effective area A dash the effective area A dash will get that A dash is equal to L 2 into B this L 2 if I divide into different portion. So, this one will be L 2 into B this area and the lower part will be plus into B plus B 2 then this additional portion that is L minus L 2. So, this will give us the effective area again the L dash is equal to L and B dash is equal to A dash by L dash. So, these are the four cases. So, by this using this four cases first we will calculate this L dash B dash and then by using this expression we can you calculate this effective area. Now, the question is that here this is in the except the first case case 2 case 3 and case 4 all the cases this L 1 B 1 then B 2 L 2 these terms are involved because here L dash is equal to L because L is known to us. So, we will can determine the L dash, but unless we do not know the A dash it is very difficult to find the B dash. So, once we get the L dash we should know the A dash for that we can determine the B dash. Now, why suppose in this expression we do not know how to what is the value of A dash because here B we know and L we know, but we do not know L 2 and B 2. Now, to use this to determine this L 1 L 2 and B 2 L 2 now the different charts are available. So, now I will show you the charts for this case because in the first case this values L 1 and this L 2 these values are directly we can determine by using the given expression, but in the case 2 because here we should know the value of L 1 and L 2. So, in that case if I use this chart then we can determine this value, this is taken from this Heiter and Anders 1985 they have proposed originally the source it is taken from this again BM Das book this chart. So, we can use this chart that so here we this axis represent E L by L and this axis represent L 1 L or L 2 L. So, from this point towards this direction the value that we will get of this curves that we will give for L 1 L and towards this direction the value that we will get that is for L 2 by L. Now, here this value this is E B dash into B. So, once we get this value so this is for 0.167 0.1. So, we will this side we will get L 1 by L and this side we will get L 2 by L because this E L by L we can calculate suppose E L by L and E B by E B by B that we will also calculate. So, once we get this value we can calculate L 1 by L and L 2 by L and as we know this L so we can calculate L 1 and L 2. So, here we will determine this things I have already explained. Next is the case 3 here also similar charts are available this is for E B by B and this is for E L by L. Similarly, for the first case this is E L by L this is E B by B this is E B by B this is also E B by B. So, here also we will get in this side we will get B 1 and B 2 here from this side onwards this graph this will give us the value of B 1 by B and here this side point point from this point towards this side will give us B 2 by B. So, here we know the E B by B value corresponding E L by L value. So, suppose this is 0.8 and corresponding this value we will get this B 1 by B and B 2 by B. So, as we know B value so you can calculate B 1 and B 2. Similarly, for the case 4 here we should know the B 2 and L 2. So, this is this axis represent E B by B and this axis represents the B 2 B or L 2 L. So, here this is E L by L. So, corresponding so obtaining this so this chart E L by L. So, this chart will use to get this L 2 by L value and this E L by L because here 2 charts here also point this chart represent the point 0 2 that is the value of E L by L and this chart represent again the point 0 2 that is also E L by L. But, so these charts representing the by using this chart we can determine the value L 2 by L and using this chart this one we can use to determine B 2 by B. So, once we get B 2 by B and L 2 by L then we know as we know L and B value. So, we can determine this B 2 and L 2. So, once we get this B 2 and L 2 then we can determine the effective area and L is equal to L dash for the case 4 then once we get the L dash then by using the expression A dash divided by L dash we can determine the B dash. So, in this L dash and B dash we will use when we calculate the ultimate load carrying capacity of the footing instead of using B dash or B or L. Now, we will solve one problem. So, that we help us to understand this things very clear. So, suppose this example problem suppose this is the example 8.1. So, this problem will solve for one way eccentricity similar case by using this chart the presented charts and the expression that mention we can determine the ultimate load carrying capacity of the footing for two way eccentricity also. Now, this condition suppose this is the foundation. So, this is ground line and one we are applying load and one moment. So, now, this dimension of the footing is 2 cross 2 meter and the depth of the footing is 0.5 meter from the ground line. Now, the density that is 19 kilo Newton per meter cube 5 value is 32 degree and C value is 0. So, one way eccentricity this E value directly it is given say 0.18 meter. So, we have to determine what will be the ultimate load cube of this foundation that is the problem. So, this is the footing condition is placed at a depth of 0.5 meter. So, D F is 0.5 meter and this dimension is 2 cross 2 meter density unit weight of the soil 19 kilo Newton per meter cube 5 value is 32 degree C is 0. Now, has this is E value. So, now, in this question the C is 0. So, what Q ultimate that C N C part the first part as C is 0. So, that part is also 0. Now, we can write this Q N Q this is S Q D Q and I Q plus half into gamma here we will use the B dash because I is it is a one way eccentricity and we will not use B we will use the effective with B dash. So, B dash into N gamma into S gamma into D gamma into I gamma. So, E S gamma S Q as the safe factor D Q D gamma is the depth factor I Q and I gamma is the inclination factor. So, first we will calculate the E Q value. So, Q value is D F into gamma here D F is 0.5 meter into gamma is the 19. So, it will coming out to be 0.9.5 kilo Newton meter square. Similarly, corresponding to phi is 32 degree. So, if I use the mayor of table. So, we can again show you the mayor of tables. So, this is the bearing capacity factor tables of which is presented by the mayor of. So, here we will use the mayor of expressions or mayor of bearing capacity factor. So, corresponding to phi is 32 degree our N Q is 23.2 and N gamma is 22. So, that value we will use. So, our N Q is 23.2 and N gamma is 22. Here we calculate again the B dash. B dash is B minus 2 E. So, that is equal to 2 minus 2 into 0.18. So, this is values coming 1.64 meter. So, these are mayor of bearing capacity factor. Now, again by using the mayor of and here this L dash will be L equal to 2 meter. So, again by using the mayor of corrections factor. So, S Q that is equal to S gamma that we can get 1 plus 0.1 into B dash divided by L dash into tan square 45 degree plus phi by 2. So, that value we can we will get the chart from the chart that is presents. So, we can show you the chart that is the mayor of corrections factor. So, different factors is there and this is the expression. So, here for this footing we will use this expression that is for S Q S gamma that is 1 plus 0.1 here B by L, but here we will use the B dash and L dash, but L dash is equal to L and this tan square 45 plus phi by 2 for phi greater than 0 degree. Here because in other factor also use from this table here phi is greater than 0 degree. So, once we put this L dash equal to 2 meter B dash equal to 1.64 meter and phi is 32 degree we will get this bearing capacity factor value 1.26 L. Similarly, D Q is equal to D gamma that is equal to 1 plus 0.1 D F divided by B into tan 45 degree plus phi by 2 into phi by 2. So, here also we will this value from the table also here if we put D F equal to 0.5 B B here in case of B we will use the B dash. So, this B dash equal to 1.64 meter phi is 32 degree. So, this is coming 1.055 as this inclination part inclination is not present. So, we can write that I Q is equal to I gamma that is equal to 0.25. So, once you get this things then by using this Q u expression. So, this expression will just put all the value because in this expression Q is 9.5 kilo Newton per meter square N Q is 23.2 S Q is 1.267 D Q is 1.055 then I Q is 1 here then gamma is 19 B dash 1.255 0.64 N Q N gamma is 22 S gamma is 1.267 D gamma is 1.055 I gamma is 1. So, once we put all the value in this expression. So, we will get this value that is coming 5752.767 kilo Newton meter square because this B dash here we will put 1.64 meter. So, finally, the Q ultimate the load that is coming out to be that is B dash into L dash into Q u. So, this value is coming 1.64 into 2 into 752.767. So, finally, the value that is coming out to be 2469.1 kilo Newton. So, the answer this Q ultimate that is coming out to be 2469.1 kilo Newton. So, the answer this Q ultimate that is coming out to be 2469.1 kilo Newton. So, this is the answer of this question. So, that is we are getting this Q ultimate. So, now, if you want to find the net safe bearing capacity of the footing or ultimate safe bearing capacity of the footing then you have to apply the factor of safety and I have already discussed how to calculate the net bearing capacity. So, here up to the factor of safety here also. So, in this way we can determine the value for the isentically loaded footing. So, in the next class I will discuss that if this is for isentically loading and now if it is a layered soil because now for the up to this we have discussed all the soil condition is homogeneous. Now, if it is a layered soil then how to calculate the bearing capacity if it is in the slope then how to calculate the bearing capacity those things I will discuss in the next class. Thank you.