 Hello friends, I am Prashant Vishwanath Dinshati, Assistant Professor, Department of Civil Engineering from Walchand Institute of Technology, Solapur. Today I am here to explain you about the influence line diagram for maximum bending moment under the chosen wheel loads. The learning outcome of today's lecture is the students will be able to find out the condition for the maximum bending moment for the chosen wheel load and they will be able to calculate the maximum bending moment for the system of load under the chosen wheel load. Now the maximum bending moment under the chosen wheel load. Let us consider a load system moving here. So it will be moving from A to B. Let P be the chosen load under which we have to find out the maximum bending moment. So for what will be the maximum bending moment under this chosen wheel load? For that first we have to find out the resultant of all wheel loads. So this will be the resultant of all wheel loads that is W and WL will be the resultant of all wheel loads to the left of the chosen wheel load. And D1 is the distance between the resultant of left hand side load and the chosen wheel load and D2 will be the distance between chosen wheel load and the resultant load. And let X be the distance of the resultant load from the middle center of the girder. VA is the left hand reaction and VB is the right hand reaction. Now for this we have to take a bending moment about B to find out the reaction VA. So for finding out the reaction we will just consider this two reaction and the resultant load W. So for that taking moment about B this VA into L. So this total distance is L is equals to this resultant load. So that is resultant of all the loads. So we are not going to take all the loads and distance we are going to take resultant load that is summation of W into this distance. So this distance is L2 L by 2 minus X. So therefore VA is equals to this summation of W divided by L. So L from this it will come into denominator into L by 2 minus X. Now to find the bending moment under chosen wheel load. So for this bending moment I will consider again the left hand section. So here VA into this reaction into this distance now. So this distance so that is L by 2 plus X this up to this it is minus D2 means I will get this distance. So this M is equals to VA into L by 2 plus X minus D2 minus this one is load is there that is the resultant load in the left hand side into the distance it is D1. So now therefore M is equals to now I will put this value VA. So VA we have calculated here so it is summation of W by L into bracket L by 2 minus X into L by 2 plus X minus D2 bracket close minus summation of WL into D1. So this we will treat as equation number 1. Now this is equation number 1. So to find out the maximum bending moment for that so we have to take Dm by dx is equals to 0. So for that again I will multiply this now. So this both the bracket I will multiply summation of W by L into now this L by 2 into L by 2 it is L square by 4 plus L by 2 into X it is L by 2 into X then L by 2 into minus D2. Then again I will multiply with this minus X term minus X term into L by 2 I will get minus L by 2 into X then minus X into X again I will get minus X square then minus X into minus D2 I will get X into D2 and minus WL summation of WL into D1 will remain as it is. So now if I multiply again this with the all this so I will get M is equals to summation of now this L and from L square 1 L will get cancelled. So it is summation of WL by 4 plus summation of W by 2 into X minus summation of W by 2 into D2 minus summation of W by 2 into X minus summation of W by L into X square plus summation of W by L into D2 X minus WL summation of WL into D1. So now if I am taking the derivative of this in terms of x dm by dx is equals to so this is constant term derivative is 0 again this one I will having this minus w l by 2 into 2x. So now before that so this term and this term will get cancelled as because this is having positive sign this is having negative sign. So my dm by dx will be so this constant is equals to 0 then this constant is equals to 0 then minus this summation of w by l into x square derivative is 2x plus summation of w l by 2 into d2. So derivative of x is 1 and this constant is again 0. So now summation of w by l into 2x is equals to summation of w by l into d2. So I will take this negative term on the right hand side therefore x is equals to d2 by 2. So now as x is equals to d2 by 2 now we know the condition so what it should be for maximum bending moment under sojourn will load the load system is to be so placed that on the girder that the chosen will load and the resultant of all will loads are equidistance from the middle of the girder. Now this chosen will load and this resultant should be equidistance from this center of the girder so it is again d2 by 2 d2 by 2. Now we will see a problem the load system moves from left to right on a girder of a span 10 meter and these are having the wheel loads and we have to find out the bending moment under 8 kilo Newton wheel load. Now for this first we have to find out the resultant force or resultant load so for that I will take moment about 12 kilo Newton load so 12 kilo Newton so now resultant into x so we do not know this x so we have to find out x so resultant into x is equals to the moment that is 12 into 0 plus 8 into this distance 1 plus 10 into this distance 1.2 plus 1 that is 2.2 plus again 10 into this distance that is 3 so from this I will get x is equals to 1.5 meter so now this resultant will be 1.5 meter away from this 12 kilo Newton load. Now what will be the condition to have a maximum bending moment under 8 kilo Newton here pause the video and try to write answer on a paper. Now for the condition to be maximum under a chosen wheel load it should be so placed that the resultant load and the chosen that is 8 kilo Newton load should be placed at an equidistance from the center point of the girder so now we will place that position so now the distance between the resultant load and 8 kilo Newton load so now this is 1.5 meter from the 12 kilo Newton 1.5 minus 1 I will get 0.5 so the distance between 8 kilo Newton and resultant load is 0.5 so this I have to place at an equidistance so 0.5 divided by 2 I will get 0.25 so I will place this resultant 0.25 on the left side and this 8 kilo Newton load 0.25 on the right side to have a maximum bending moment under this chosen wheel load. Now again taking moment about A to find the reaction VB so VB into 10 this is the distance total distance is equals to 40 into so this is the resultant into 4.75 distance so this from this I will get VB is equals to 19 kilo Newton therefore VA is equals to this total load minus VB that is 21 kilo Newton so for the maximum bending moment under 8 kilo Newton so it is VB into this distance so this is 3.75 plus 1 that is 4.75 and this is anticlockwise and 12 kilo Newton load will be clockwise it is minus 12 into 1 so this 19 into 4.75 because 19 is your VB reaction then 19 into 4.75 minus 12 into 1 you will get 78.25 kilo Newton meter. Now you will solve the same by influence line diagram method so now if you have seen my previous video to find out the bending moment so the bending moment where you want to find the bending moment so that ordinate will be AB by L means A means this distance that is 5.25 into 4.75 divided by L L is 10 meter so it is 2.494 and depending upon the similarity of triangle you have to project this 10 kilo Newton loads position on this ILD so this also load on ILD so every load I have to take on the projection on the ILD influence line diagram so I have to find out the ordinate or the height of each this with the help of similarity triangle so you know this value this height you know this height and you know this value first from similarity triangle you can find out this value so the load into this ordinate addition of all this you will get the bending moment so 10 into this ordinate then this 10 into this ordinate plus 8 into this ordinate plus 12 into this ordinate so this addition I will get the bending moment that is 78.25 kilo Newton so the maximum bending moment under chosen will load is 78.25 kilo Newton meter so these are my references thank you thank you thank you for watching my video