 I am Prasanth Vishwanath Dinshati, Assistant Professor, Department of Civil Engineering from Vuljan Institute of Technology, Solapur. So, today I am here to explain you about the influence line diagram for a shear force at a given section of a girder. The learning outcome of today's lecture is students will be able to study the effect of moving load on shear force of structural member, they will be able to construct influence line for shear force and they can calculate the shear force for the system of loads. Influence line diagram for shear force function, so it is a curve or a graph that represents a function like the shear force at a section of a structure for various position of unit load on the span of the structure is called influence line diagram for that given function. Influence lines are very important in the design of structures subjected to a greater live loads. With influence line it is easy to determine the location of the live load which will cause great test influence on the structure for a given function. Moving or live load moving load are the loads applied to a structure with the points of application including their magnitude can vary as a function of position on the structure including their magnitude means the magnitude of the moving loads are also variable. Shear force at a given section using ILD for the moving loads. Can you define a shear force here, pause the video and try to write answer on a paper. Shear force it is the algebraic sum of vertical forces at any section of a beam to the right or left of the section is known as shear force. It is also called as unbalanced vertical force. Shear force diagram is one which shows variation of shear force along the length of the beam. Static shear force diagram so static shear force how we define so here beam is there and a point load is applied here so to draw a static shear force so we have to consider the sign convention though the left of the section upward it will be positive and right of the section downward will be positive. So now here I have taken the reactions here WB by L and WA by L so when I consider a shear force on the left of this section W so I will get this only one force that is reaction and that is upward means it is positive in nature so I have drawn here WB by L and it is constant up to this point load and when I have to consider the shear force at this side so I am considering the right side of the section so only one force is there that is reaction that is WA by L so it is upward but our sign convention is right of the section downward is positive here upward means I have to show it negative so I am drawing here in the lower side so WA by L and it is constant up to this point load when I join this I will get a shear force diagram. This shear force diagram is for static loading condition now shear force diagram at a section for a moving unit load so when a unit load moves from A to B what will be the shear force diagram let the unit load moves from the span A to B and D be the section where we have to find out the shear force now this AD portion is having length A and DB portion is having length B when the unit load is between A and D now here the unit load is between A and D so here may be the unit load the shear force at D section so shear force at D section what it will be so I will consider the right side of the section so right side of the section only I am having one reaction that is upward reaction but right side of the section so downward is positive upward is negative so it is VB reaction so I am drawing this reaction downwards so we know that VB varies from 0 to 1 as the load moves from A to B so this I have explained the how to find out the influence line diagram for the reaction in the previous video so kindly watch that previous video for the reaction of ILD now so I have drawn here 0 to 1 in the negative side so as long as the unit load is between A and D so my shear force at D will be minus VB now when the unit load comes beyond point D when the unit load moves and it is between D and B section now I will consider the shear force to the left of this section so to the left of this section I am having only one reaction that is RA or VA so shear force at D is plus VA because now left of the section upward is positive so here again we know that VA varies from 0 to 1 as load moves from A to B so as it is positive I am drawing it upwards so this will be the ILD for reaction VA now this both the part the part of the influence line diagram between VB A and D is negative and the part of the influence line diagram for VA between D and B is positive so here both will constitute the influence line diagram for the shear force at section D now here I am drawing the section so this left side will constitute a negative and this right side will constitute the positive value for the shear force at this section D now we will determine a shear force at a section for a given system of load so now the shear force is again can be found out by multiplying intensity of load into the ordinate of the influence line diagram corresponding to the position of the load now these are the load and I have to find out the shear force at this section D so as explained earlier so we have to draw this ILD for VB in the negative direction and VA for the for in the positive direction and here I am sex drawing a section line here so at this point so from this point D of onwards it is having positive value and this is having a negative value so now I have to take this corresponding ordinate for this load for the 7 kilo Newton what will be this ordinate so I can easily find out this by using similarity triangle so this will be 1 by 6 and this will be 5 by 12 so now I have to multiply this shear force at D will be 7 into this ordinate so as it is in the negative portion I will have a negative sign here minus 1 by 6 plus 6 into this again negative ordinate minus 5 by 12 plus 5 into this ordinate it's positive that is 5 into 1 third so I will get the shear force at section D as minus 2 kilo Newton now to determine the maximum positive shear force at a section D 4 meter from left end for a 2 wheel loads any wheel load can lead the other so here the 2 wheel loads are passing which are having a fixed distance 2 meter from the distance between them so now what will be the position so when you will get the maximum positive shear force so for that I have constructed this ILD so this ILD for VB is negative ILD for VA is positive so now observing this diagram so what I understand is for that maximum positive shear force I should have both the wheel loads on the this portion that is in between D and B so now the heavier load should be just right side of the point D but for calculation we can take it exactly on the top of the point D so now I have to take this ordinate and this ordinate as both these are in the positive area so this 20 into 3 4 plus 8 into 5 8 so it will give you 20 kilo Newton positive shear force now to determine the maximum negative shear force at a section D for the same load case that is having 2 wheel loads and at a distance of 4 meter from the left end so what I have to do again you observe this influence line diagram so this portion is giving a negative value this portion is giving a positive value so for maximum negative shear force your wheel load should be in the portion AD and the maximum or the heavier load should be exactly near the point D or just left of the point D but for calculation you have to take exactly at point D so therefore maximum negative shear force at D is 20 into this negative ordinate minus 1 4 plus 8 into this ordinate minus 1 8 so that will give you minus 6 kilo Newton now to determine the shear force at a section D for a given system of load that is UDL so shear force again here we can find it by multiplying intensity of load into area of influence line diagram between the UDL so now here I have drawn the section so this part will give you positive portion this will be having negative portion so again I will take this ordinate for the starting of UDL and end of UDL so this shaded area will give you the area of influence line diagram multiplied by load intensity so here again this is a triple ZL so area you can calculate by one half sum of parallel side into height so this will give you positive value this will give negative value so solving this I will get minus point 57 kilo Newton now for the positive shear force what position should be there therefore so seeing this ILD so I should be placing the UDL on the right portion so here again I am having the right portion so this area into the intensity will give you positive shear force that is plus 8.57 kilo Newton and for maximum negative for the same case so I should place this UDL to the left of the section as this gives the negative value into this area into intensity I will get the negative minus 5.14 kilo Newton these are my references thank you thank you for watching my video