DRCS Oral Footing Part 5

Added: 2026-05-12

[00:00:03]Good evening.

[00:00:05]Now we are going to see the uh oral questions that is on footing.

[00:00:17]So now what are the different types of footing?

[00:00:21]So we have actually so many types mainly.

[00:00:27]So if we will say we have one type that is uh isolated footing.

[00:00:34]So that is first, isolated footing.

[00:00:41]Isolated footing.

[00:00:43]Second we have that is in isolated actually we have two types.

[00:00:48]One we have sloped footing.

[00:00:52]And another we have that is pad footing. So uh we have seen pad footing in which our depth is same throughout.

[00:01:02]Now we have second footing that is combined footing. Isolated means for single column.

[00:01:09]So this is for single column.

[00:01:13]Then if we will go for combined footing, this is generally for two columns, combined footing.

[00:01:20]In this again we have two types. One is slab type which we have seen. And another we have slab beam type.

[00:01:38]Uh slab beam types. And then we can have third mat footing or mat foundation.

[00:01:50]In which uh dimensions of footing overlap uh each other. And in that case we are going to provide foundation below all the area. So, if we have this will be our plan of the building, then we are going to provide foundation below all the uh dimension of the plan.

[00:02:15]Then, what is the difference between actually loaded column footing and footing for actual load plus uniaxial moment? For actually loaded, we are going to find out for PU only. So, whatever area are required, so that we are going to calculate by using 1.1 PU, where 10% we are going to consider self weight of footing divided by ultimate soil bearing capacity, and then we will get area of footing required. Now, if we will go, this is for actual. Now, if we will go for actual plus uniaxial, so if we will go for actual plus uniaxial, and in this, we are going to find out area actually for this, that is AF equal to same 1.1 PU upon ultimate SBC, but area we are going to increase about 70%.

[00:03:22]70 AF required is increased about 70% than required. So, AF we can write it as 1.7 into AF, which is calculated from this formula. Why we are going to multiply by 0.7 uh 70%? So, reason to accommodate the stresses developed accommodate stresses developed due to moment. Accommodate stresses of moment also. And therefore, we are going to increase area required by 70%. What is criteria for deciding the dimension of footing in actual load and uniaxial bending. So, actually we have uh P max and mean to find these stresses, we have PU upon area of footing provided. So, we can write BF into LF plus minus we have MU upon BF into LF squared.

[00:04:45]LF squared by six. So, how we will get MU by Z. This is MU by Z. How we will get value of Z? So, Z we have B D cubed by 12 divided by Y we have LF by two. And from this we will get value BF into LF squared by six. And from this we will get P max value which is PU upon BF into LF plus MU upon BF LF squared by six should be less than or equal to ultimate soil bearing capacity. And P mean if we will write, so this will be PU upon BF LF minus MU upon BF LF squared by 6. And this value should be greater than or equal to zero. So, critical condition we can say that will be greater than zero. And from which we will get BF upon PU upon BF LF equal to PU into MU divided by BF into LF squared by 6. And this BF get cancelled. One LF get cancelled. PU get cancelled. And this we can write greater than or equal to. And from which this PU cancelled, and we will get E value.

[00:06:29]Uh this value should be PU minus MU. This will be greater than equal to zero. So, we have PU that will be greater than equal to zero. And then we will get value of uh E, and this will go on this side equal to LF by 6.

[00:06:52]So, your value of E that will be greater than So, this is less than, this is greater than. So, we have value of LM should be greater than 6E.

[00:07:06]So, E we can easily calculate by using MU upon PU.

[00:07:13]And then we are going to decide value of LF. And then knowing LF, we can find value of BF. So, likewise we have change in uh deciding the dimensions of footing.

[00:07:30]There we have direct LF BF where we are going to consider CX and CY we are going to find, and then we are going to consider uh uh value that is cxx equal to cyy, and then from that we are going to provide the dimension in case of uh loading, that is actual load. Now, we will go with next question.

[00:08:05]How intensity of upward pressure is calculated in footing for actual load and in the actual moment? So, just I have explained this. So, we have calculated p max, which is pu upon bf into lf plus mu upon bf into lf squared by six.

[00:08:27]And value of p minimum we are going to find pu upon bflf minus mu upon bf into lf squared by six.

[00:08:42]And if we will draw this, then we are going to get this in this fashion.

[00:08:49]This value is greater than zero, and our values will be in this fashion.

[00:08:59]This is our footing.

[00:09:03]So, likewise we have to calculate p max.

[00:09:06]So, this is my value of p max, and this is my value of p min.

[00:09:14]Then, how explain two-way shear check?

[00:09:17]Now, see very carefully. So, critical section we have at d by two from face of the column. So, now you can see we have these values, that is 0.3 and 0.5. These are of the column dimensions. So, now I have taken values.

[00:09:37]So, 0.43 So, we have 430 mm value of dy.

[00:09:44]dy we have 430 mm. So, we have added dy by 2 dy by 2. Means we have 0.5, which is dimension of the column plus 0.43 dy by 2 dy by 2. So, it becomes d plus dy and it becomes this dimension becomes b plus dy. And from this we are going to get this will be critical dimensions of the critical section. So, now we have calculated perimeter of this and then we are going to multiply it by depth. So, keep in mind for resistance area, perimeter is considered as length. So, this will be considered as length.

[00:10:30]And then we have depth that is perpendicular to it. So, perimeter into depth of footing, that will be my resisting area. And we have this will be resisting area, that is perimeter multiplied by dy. So, now we know tau uc ks tau c. So, 0.5 and then we can find this value. Now, resistance we have this resistance stress and resistive area. So, we will multiply resistance by resistive area, we will get resistance is 1598.1991 kN. Now, while finding the shear force, so what we have, we have to subtract this area from area of footing. So, this area we are going to subtract from area of footing and then we are going to multiply it by intensity of pressure.

[00:11:29]So, this is also known as already I told you in class punching shear.

[00:11:38]So, this area goes down and remaining part will remain up and therefore we have shear area at critical section. So, we have area we have multiplied this so this is area now.

[00:11:52]In a resistance area we have perimeter into depth and area at critical section we have these values. That is uh 930 so this is 930 so we can directly go with 0.6789 square and then we have to subtract this area that is from area of putting and then multiply it by intensity of pressure.

[00:12:16]That is the 192 that is intensity of pressure. This is area of putting and we have subtracted area of this critical section and we will get this value which is less than 1.598.

[00:12:29]1.91 kN and therefore safe. So, this criteria is checked. Now, we will go for check for uh one-way shear. Now, we have reinforcement provided.

[00:12:44]So, reinforcement provided we have 31 bars and that is of 10 mm. So, this is my value of AST.

[00:12:55]Now, this is 2600 this is 430. So, we have provided this reinforcement along this direction.

[00:13:10]And then we have this will be 2600 that is width perpendicular to this reinforcement into depth that is depth of dy value. This is dy. Gives us 0.22 value. Then we have calculated tau uc table 19.

[00:13:29]So now try to understand. Now we are going to find out dy. This will be my critical section from face of the column and then we have this portion. So now we have area whatever we have taken 2600 will be width and 430 depth. We have multiplied it by 0.34 and then we got this value as 380 kN which is my shear resistance of concrete I can say. So astx divided by area gives you tau uc and multiply by that area only you will get value of vuc. Now which shear force?

[00:14:11]So shear force in this portion. So this portion which I made it as shaded. Now b we know we have b value 2.6 m.

[00:14:21]Now we are interested with this value.

[00:14:23]How we will calculate now vlc? We have this as 2.8 m.

[00:14:30]Now subtract this 1.36. So 2.18 minus 1.36.

[00:14:35]So we have 4 4 1. Now divide it by two because we have half portion on this side, half portion on this side. 1.36 is this portion. Total 2.8. So 2.8 minus 1.36 divided by two I will get this value which is 0.72. And we are going to find out a shear force of this area. Do not take depth in this case and then somewhere in video I may taken this as but that is wrong one. So make correct this value. So we have shear force 192.31.

[00:15:12]2.6 we have this dimension and this is actually 0.72.

[00:15:19]So you just calculate I will check it. I have taken 0.72 or 0.62.

[00:15:25]So, 192.31 into 2.6 into 0.72. So, yes, I have taken 0.72 only. This will be 360, which is less than 380 kN, therefore safe. So, area you have to take to calculate shear force, that is at this at a distance dy from face of the column. And this is for check for one-way shear. Now, check for bearing pressure. So, we have A1 B F into L N because we have formula, that is 0.45 FCK into root A1 upon A2. And this value should be greater than PU upon area of column.

[00:16:14]This value should be less than 0.45 FCK root A1 A2. A1 we have two values. So, here you can see one vertical with two horizontal. So, if we will go with this, so one vertical with two horizontal, so we have this as depth of footing dy, and one vertical two horizontal, so we have one vertical, we have two horizontal. So, 2 dy.

[00:16:45]Similarly, we have on this side also 2 dy. So, we will get this as 4 dy into depth of 4 d on both sides. That means if I will consider this, so we have on both sides, and we have this value, that is B plus 4 dy. And on this side, that is depth of column plus 4 dy. And this will be my area A1. And minimum we have to consider from one and two, and that will be my A1 value. Then we have root A1 upon A2 should be less than equal to two. If greater than two, then we are going to consider A1 upon A2 equal to two. If this value comes greater than two.

[00:17:35]And then this will be bearing pressure, which should be greater than PU upon area of column. Then we have explained calculations for size of footing. So we have already seen this. So we have This will be my footing. Just I will explain this in 1 minute. So we have These are the two columns.

[00:17:55]So this dimension we can assume as 1 m or something. Then we have This will be value PU1. This will be value PU2.

[00:18:06]The distance between two columns we have let us X.

[00:18:09]Then we are going to take moment about this, and that will use me X bar, that is equal to PU1 into I will mention it as something L1.

[00:18:22]Plus PUB into X plus L1 and divided by PU1 plus PU Sorry, two. So PU2.

[00:18:36]So PU2 we have. And from this I will get X bar, so which will be from this side, that will be X bar, which is equal to L by two. And then we will get L equal to 2X bar. And then we will have X bar from this, and area of footing we have same, that is 1.1 into PU1 plus PU2 divided by ultimate SBC.

[00:19:05]And from this I will get width of footing equal to AF upon LF. And by this we are in position to calculate dimension of the combined footing. Then we have draw loading diagram. So, already you know loading diagram. So, we have this will be my column load. This will be my column load. And we have this will be uplift pressure.

[00:19:29]So, we are in this fashion.

[00:19:35]And then if we will draw shear force diagram, then we have shear force diagram. It will go up.

[00:19:43]Then it will come down. It will go up, come down, and then it will be zero over here. So, while calculating this uplift pressure, so W U, we have to use it as P U 1 + P U 2 upon area provided.

[00:20:07]So, that you have to keep in mind. It is not 1.1. And if we will draw bending moment diagram for this, I will get bending moment diagram in this fashion.

[00:20:17]So, maximum where we have zero shear force, and then we have zero at both ends. And this is my bending moment diagram.

[00:20:26]This is my shear force diagram. This is loading.

[00:20:30]So, likewise we have to go for the questions which are related with your footing. So, now we will stop here only.

[00:20:39]That is beam that is remain, that I will send it tomorrow. Now we will stop here only. Thank you.