 Hello friends, I am Prashant Vishwanath Dinshati, assistant professor, department of civil engineering from Walchand Institute of Technology, Sulapur. So, today I am here to explain you about the influence line diagram for finding out the bending moment at a given section of a girder. The learning outcome of today's lecture is the students will be able to study the effect of moving load on the bending moment of structural member. They will be able to construct influence line for bending moment. They will be able to calculate the value of bending moment for a given system of loads. Influence line diagram for bending moment function. So, it is a curve or a graph that represents the function of a bending moment at a section of a structure for various position of a unit load on the span of the structure. So, it is the influence line diagram for that function. The use of influence line diagram is very important in the design of structure subjected to greater live loads. So, with influence line diagram, we can be able to determine the location of the live load which will cause the greatest influence on the bending moment of that structure. Bending moment at a given section using ILD for moving loads. Now, can you define the bending moment here, pause the video and try to write an answer on a paper. Bending moment, it is the reaction induced in the structural element when an external force or a moment is applied to that element causing the element to bend. So, the bending may be hogging or sagging. So, now the bending moment at a section is considered positive. If bending moment at that section is such that it tends to tend the beam to a curvature having concavity at the top means the top fibers are subjected to compression whereas bottom fibers are subjected to tension at that point we have to consider the sign convention as positive. So, before dealing with the influence line diagram for moving load, we will see the bending moment diagram for static loading condition. Now, static load means a load is applied at this fixed point. So, the load is W. So, we will find the reaction R A and R B. So, it is W B by L and W A by L. So, bending moment at this section is reaction into this distance that is W B by L into A. So, you can take the bending moment by considering the right section also this reaction into this distance. So, it is W A B by L this is the bending moment diagram for static load condition. Now, bending moment diagram for moving unit load condition. So, now let a unit load moves from A to B on a girder. So, let A D be the distance A and D B be the distance B when the unit load is at a distance X from left support that is from A. So, this distance will be L minus X. So, at this time what will be the reaction? So, the reaction V A is L minus X upon L and V B is X by L. So, now when the unit load is between A and D portion when this is between this portion. So, our bending moment. So, bending moment will be reaction R B into this distance. So, wherever it may be here our reaction our bending moment is R B into this distance. So, R B. So, bending moment at D is V B into L minus A. So, L minus A will be this distance. So, X upon L into L minus A. So, we can rearrange this M D is equals to L minus A upon L into X. So, X will be this distance. So, now this relation is true when the loading when the load is between A and D. So, when the unit load is exactly at A that is X is equals to 0 when we put this X is equals to 0. So, the bending moment at D is 0 when the unit load is at exactly at point D then the bending moment is M D is equals to L minus A upon L into A. So, now M D is equals to. So, we will again rearrange this A L minus A upon L where L minus A is B. So, it is A B by L. So, our bending moment varies from 0 to A B by L when the unit load is between A and D portion when the unit load is between D and B portion. So, now here again the value of X will be now from the support the unit load this will be again L minus X. So, now at this time what we will do is we will take the moment. So, now again we have to find out the bending moment at this D. So, we will consider left section that is V A into A. So, now M D is equals to V A into A. So, V A we know that V A is L minus X upon L into A. So, again rearranging this I will get M D is equals to A upon L into L minus X. Now, this relation is true for all the loading position between D and B. So, now we will consider when the unit load is at D means when this unit load is at D exactly. So, M D is equals to A upon L into L minus A. So, it is. So, now I have replaced this X by A. So, it is A B by L when the unit load is at point B that is X is equals to L. So, M D that is moment at D is A upon L into L minus L that is 0. So, now here again the bending moment varies from A B by L at point D to 0 at point B. So, as this unit load moves. So, this bending moment decreases and it will be 0 when the unit load is exactly at point B. Now, we will determine the bending moment at a section D 6 meter from left hand for the two wheel load conditions. So, any wheel load can lead the other. So, this any wheel load can lead the other means sometime this 20 kilo Newton will be on the left side or it will be on the right side of 8 kilo Newton. So, the bending moment maximum bending moment. So, at section D. So, now we will first draw the influence line diagram as explained earlier. So, this the at this section. So, the bending moment at this point will be A B by L. So, we know A value B value and L value. So, from that I will get this ordinate 3.75 and this I L D varies from 0 to this again from it decreases to 0 here. So, now depending upon this diagram. So, we have to study that. So, bending moment will be the load into the ordinate at that position. So, now if the maximum load is placed at the right side or exactly at point D. So, then and the 8 kilo Newton to the right side then I will get the maximum bending moment. See this condition. So, this distance is fixed 2 meter, but if I keep the 8 kilo Newton on the left side I will get the ordinate here lesser value. So, the our condition is that we have to find out the maximum bending moment. So, whenever you place this 8 kilo Newton on the right side of this 20 and the 20 kilo Newton exactly at D then only we will get this maximum bending moment. So, this 20 into this ordinate plus 8 into this ordinate I will get 99 kilo Newton meter as a maximum bending moment. Now, bending moment for uniformly distributed loading condition. So, there will be 2 cases when the UDL is greater than the span when the UDL is shorter than span. So, we will see the first case now when the length of the load is greater than the span. So, again we will draw this ILD for the bending moment for this section D. So, now so seeing this ILD and the maximum that is UDL. So, we can easily come to know that for maximum bending moment at D the whole span must be loaded. So, the bending moment for UDL is intensity into area covered by UDL on the ILD. So, this area will be a triangle that is one half base into height. So, bending moment is W into one half base into height it is WAB by L. So, now maximum bending moment at a section for UDL for the second case when the load is smaller than the span. So, now again studying the ILD. So, we have to see that how the load should be placed. So, the head of the load be at a distance Z from D. So, we will consider this as a Z this is head of the UDL. So, bending moment is again intensity of load into area covered by UDL. So, the condition to find Z is such that average load on this AD portion should be equal to average load on BD portion. So, now again W into L minus Z upon A is equals to WZ by B. So, this WW will get cancelled L1 minus Z upon A is equals to Z by B is equals to. So, I will add this L1 minus Z plus Z upon A plus B. A plus B is L. So, I will equate Z by B is equals to L1 upon L. Therefore, Z is equals to L1 upon L into B. So, the bending moment is again intensity of load into area covered by UDL on ILD. Now, I will find the maximum bending moment at a section D 4 meter from the left hand for UDL of 5 kilo Newton load having 8 meter length. So, 8 meter is L1 and for the span girder of 10 meter. So, L is 10 meter. So, Z is equals to as we have found it. So, it is L1 upon L into B. So, this ordinate Z this Z is equals to 4.8. So, the remaining portion 3.2 will be on the left side of the section. So, bending moment is again intensity into area of ILD covered by this UDL. So, we will find this ordinate here. So, this UDL into area of this portion. So, area of this trapezoidal is one half sum of parallel side into height. So, again multiplying this area into the intensity I will get the bending moment that is 57.6 kilo Newton meter. So, these are my references. Thank you. Thank you very much for watching my video.