 Hello friends, I am Prashant Vishwanath Dinshati, assistant professor, department of civil engineering from Walchand Institute of Technology, Solapur. So, today I am here to explain about the influence line diagram for maximum end share of a girder. The learning outcome of today's lecture is students will be able to study the effect of moving load on maximum end share, they will be able to calculate the value of maximum positive and negative end share for a given system of loads. Now, for maximum positive end share force for a moving wheel load, suppose the wheel loads that is W1, W2, W3, W4 and W5 are moving on a bridge or a girder that is from B to A. So, to find out the maximum positive share force, so we have to first draw ILD for reaction A and in the first trial we will see that W is placed at A and the other loads are placed relative to the W1. Now, influence line diagram for maximum end share as per the sign convention of share force. So, for a simply supported girder which end of the girder will have positive share. So, pause the video here and try to write answer on a paper. Now, the greatest left end reaction will be the maximum positive share whereas the greatest right end reaction will be the maximum negative share. Now, as the loads moves from B to A, so the first trial we will take is W1 is placed at A. So, for that we will draw an influence line diagram for VA that is the reaction at A. So, influence line diagram this is the influence line diagram having a unit displacement here and this will be connected to the B. So, you are having this ordinate for every wheel load on the influence line diagram. So, let the wheel load move by an elemental distance dx towards the left. Now, if a wheel load moves to the left by dx, so there will be increase in the ordinate of the influence line diagram by dy. But we know that dy by dx is tan phi, but tan phi we can also take it as 1 divided by length of the girder. Therefore, dy is equals to dx by L. So, when the wheel load W1 is at A, so we can find the reaction VA by using ILD that is load into the ordinate of ILD. So, this W1 into Y1 plus W2 into Y2 plus W3 into Y3 plus W4 into Y4 plus W5 into Y5. So, we will get the reaction VA1. Now, let the load system move again towards the left, so that W2 will come exactly or it is placed on A and the distance between W1 and W2 is small a. Therefore, dx is equals to A. So, if this moves by dx, then dy is equals to A by L. So, now, if this moves by A, then every ordinate will be increased by dy. So, dy is equals to A by L. Therefore, the ordinate of ILD for W2, W3, W4 will be increased by A by L. So, therefore, now this has rolled off W1. So, now the remaining loads we can find out the reaction at A due to the new loading position that is VA2 is equals to W2 into Y2 plus A by L, W3 into Y3 plus A by L, W4 into Y4 plus A by L plus W5 into Y5 plus A by L as each ordinate is increased by this amount that is A by L dy. The reaction at A due to the new load position VA2, so after multiplying this and solving this, what I will get? W2 into Y2, W3 into Y3, W4 into Y4, W5 into Y5 and taking A by L common, so I will get W2 plus W3 plus W4 plus W5. So, this is reaction VA2. So, we have already calculated the reaction VA1 if W1 is placed at A, so that we have already given in equation 1. So, now if you compare these two equations, so A1 will be greater than A2 if W1 Y1 is greater than A by L into bracket W2 plus W3 plus W4 plus W5, because now if you see this both equation, so this W1 Y1 is equals to this, you have to compare this because W2 Y2 is already there, W3 Y3 is there, W4 Y4 is there, W5 Y5 is there. So, for VA1 to be greater than VA2, this W1 Y1 should be greater than A by L into sum of the other loads, but Y1 is 1 as the ordinate is 1. So, therefore VA1 is greater than VA2 if W1 is greater than A by L into the sum of other loads. Then VA1 is greater than VA2 if W1, so I will take this as denominator W1 by A is greater than this sum of remaining loads divided by length or we can say that VA1 is greater than VA2 if the load rolled off divided by succeeding wheel space that is wheel load rolled off means W1 as rolled out divided by succeeding wheel space means this A space is greater than the sum of the remaining load on the span that is W2 plus W3 plus W4 plus W5 divided by span length. In fact with the experience we can easily choose 2 or 3 trial position to calculate the maximum reaction. Now, we will consider an example a wheel load system as shown in figure moves from B to A of a span of 5 meter and the loads of the wheels are given here and the distance between them is given here 0.81 and 0.8 meter. So, as this moves so we have to find out the maximum positive and negative shear force. So, in the first trial what we are doing is the load is exactly placed at A. So, this and we have drawn the ILD for that and we have to calculate the ordinates for that. So, by using similar triangle you can calculate this ordinate using the spacing. So, now here the reaction VA1 is equals to 12 into 1 plus 40 into 0.84 plus 15 into 0.64 plus 25 into 0.48. So, this VA1 is equals to 67.2 kilo Newton. So, now as our condition is that wheel load rolled off divided by the distance that should be greater than sum of all loads remaining load divided by span. So, W1 divided by A. So, W1 is 12 kilo Newton divided by A is 0.8 it is 15. So, the remaining load is W2 plus W3 plus W4 divided by L that is 40 plus 15 plus 25 divided by 5 is 16. Now, since W1 is W1 divided by A is less than W2 plus W3 plus W4 by L. So, we have to go for second trial. So, when this is greater that share force will be maximum. So, for second trial position to find out the maximum positive share. So, first wheel will be rolled off and the second wheel load is placed exactly at A. So, now here again we will find the ordinates for ILD for reaction VA. So, now again to find the reaction VA2. So, that is load into this ordinate 40 into 1 plus 15 into 0.8 plus 25 into 0.64. So, this VA2 is 68 kilo Newton. Now, we will check the condition W2 by B means this wheel load divided by this distance is 40 and the remaining load is 15 and 25 that is 15 plus 25 divided by span. So, it is 8. Now, our condition satisfies that W2 divided by B is greater than W3 plus W4 by L. Therefore, there is no need of third trial. Therefore, the maximum positive share is 68 kilo Newton. Now, we will see the maximum negative share. So, maximum negative share is at VB. So, we will draw the ILD for this VB and now the motion is towards right. So, now the 25 kilo Newton load will be placed exactly on B. So, now again we will find VB1. So, this load into this ordinate 25 into 1 plus 15 into 0.84 plus 40 into 0.64 plus 12 into 0.48. So, VB1 is 68.96 kilo Newton that is negative share. So, we will again check the condition that is W4 divided by C that is 25 divided by 0.8 that is 31.25 and the remaining load divided by span that is 15 plus 40 plus 12 divided by 5 is 13.4. So, now this W4 by C is greater than the remaining wheel load divided by span length. Hence, there is no need for second trial. Therefore, the maximum negative share is 68.96 kilo Newton. So, these are my references which I have referred. Thank you. Thank you very much for watching my video.