 I am Prashant Vishwanath Dinshati, Assistant Professor, Department of Civil Engineering from Vajchanth Institute of Technology, Sallapur. Today, I am here to explain you about the shear force and bending moment for a simply supported beam carrying uniformly varying load. The learning outcome of today's lecture is the students will be able to understand the behavior of simply supported beam under uniformly varying load and they can diagrammatically represent the shear force and bending moment of a simply supported beam with we will see a beam having a uniformly varying load or a triangular load. As the load is symmetric the reactions will be equal that is equal to total load divided by 2. So, the load will be now area of this loading diagram. So, the loading diagram is a triangular and the load will be one half base into height that is one half base is L into height is W. So, it is one half W into L and it is divided by 2 as we have the two supports. So, it is r A is equals to r B is W L by 4. Now, take a section at a distance of x from A find the rate of loading of this A x D that is x D. So, we can use similarity triangle. So, the triangle will be O C A. So, in this triangle to find x D we are using similarity properties of that triangle. So, O C upon A C. So, this O C upon A C is L by 2 is equals to x D upon x. So, x D upon x therefore, x D is equals to O C divided by L by 2 into x. So, but O C is having intensity W. So, putting that value W divided by L by 2 into x is equals to 2 W upon L into x. So, this ordinate or this vertical ordinate will be 2 W L upon x. Now, load on length A x of the beam is equals to area of this bending moment diagrams. So, now we will consider this load triangular load how much it is we will calculate that it is one half base into height one half base is x height is x D. So, one half into x into 2 W upon L into x. So, it is equals to W by L into x square and the load is acting at a distance of x by 3 from x. So, the resultant force acting on the left portion of the section acting upward is treated as positive. So, the shear force at this section f x is equals to r A minus load on length A x. So, f x is reaction that is W L by 4 minus W by L into x square that is equation number 1. Now, the equation shows that the shear force varies according to the parabolic law between A and C because there is a x square term. Now, we will find out the shear force putting the value of x. So, at x is equals to 0. So, the shear force will be f A. So, putting that x value in this equation number 1. So, f A is equals to W L by 4 minus W by L into 0 square. So, it is W L by 4 as it is a positive value we have to draw it above the base line W L by 4. Now, again we will find out the shear force at x is equals to L by 2 that is shear force at C that is f C is equals to W L by 4 minus W by L into x square means L by 2 bracket square. So, here W L by 4 minus W L by 4 is equals to 0. So, now we will have this diagram following the parabolic law. So, this diagram will be like this. Now, again the shear force at B. So, it will be minus R B. So, it is minus W L by 4 here. So, it is drawn below the base line and again this is connected by the parabolic curve. Now, state whether the statement is true or false the shear force diagram of a simply supported beam with uniformly varying load follows cubic law. Here pause the video and try to write answer on a paper. It is false because if you see the equation the shear force diagram varies with the parabolic law as per the equation of shear force. Now, we will find out the bending moment. So, again considering the section at x d. So, the bending moment at any section between A and C at a distance of x from end A. So, it is given by m x is equals to R A into x. So, this is the clockwise moment positive and this load will be anticlockwise. So, it is minus load on length A x into x by 3 because x by 3 is the c g of this triangular load. So, m x is equals to W L by 4 into x minus W by L into x square into x by 3. So, now the bending moment will bend in such a manner that it will have concavity at the top. So, the bending moment is positive. So, to find the bending moment at various section putting the value of x at x is equals to 0 bending moment at A is equals to W L by 4 into 0 and minus W by L into 0 will get 0. And at B x is equals to L also bending moment is 0 because we know that bending moment at simply support and at free end of the cantilever is always 0. At C at x is equals to L by 2 bending moment at C is equals to W L by 4 into this putting this value L by 2 that L by 2 minus W by L into L by 2 bracket square into this x by 3 means L by 2 into 3 that is L by 6. So, solving this we will get W L square by 12. So, we have to plot this ordinate at the top of the baseline as it is positive and we have to join all these three points following the cubic law of the curve. The maximum bending moment occurs at the center of the beam where shear force changes the sign that is at C and that maximum bending moment is also W L square by 12. Now, we will see one more beam which is having a triangular load that is 0 intensity at one end and W intensity at the other end. So, here now the load is unsymmetric. So, we have to find out the reaction by taking moment about A. So, taking moment about A this R B into L is equals to because anticlockwise moment is equals to clockwise moment. So, R B into L is equals to this triangular load is one half base into height that is L into W and it is acting at two-third of L. So, from A it will be two-third of L. So, therefore R B will be W L by 3. Now, R A is equals to total load on beam minus R B. So, total load on beam is again triangular load one half base into height that is W L by 2 minus this reaction is W L by 3. So, solving this we will get R A is equals to W L by 6. Now, again we will consider any section at a distance of x from A. So, the ordinate is again calculated by using similar to triangle O B and A. So, here W upon L. So, W upon L is equals to this x c upon x x c upon x hence x c is equals to W by L into x. So, this ordinate is W by L into x. We will consider the left portion of the section F A is equals to R A minus load on length A x. So, R A is W L by 6 minus x by 2 into W x upon L. So, F x is W L by 6 minus W x square upon 2 L. So, the above equation shows that the shear force varies according to parabolic law. Now, F x is equals to W L by 6 minus W x square upon 2 L this is equation number 1. So, we have to find out shear force at various points at x is equals to 0 then F A is equals to putting x value 0 here it is W L by 6 at x is equals to L F B is equals to W L by 6 minus W L square upon 2 L is equals to. So, solving this we will get F B is equals to minus W L by 3. So, taking this W L by 6 above the baseline as it is positive and W L by 3 as it is negative taking below the baseline and we have to join this by a parabolic curve. Now, we have to find out where the shear force is 0 or where shear force changes the sign to have the maximum bending moment. So, to find this value where shear force is 0 putting F x is equals to 0 in equation 1. So, 0 is equals to W L by 6 minus W x square upon 2 L. So, taking this negative term on the left hand side and solving we will get x square is equals to L square upon 3 therefore, x is equals to L by root 3. So, it is equals to 0.577 L. So, this is L by root 3. So, bending moment diagram. So, again we have to find out the bending moment. So, the bending moment equation is R A into x minus load on length A x into it is C g that is x by 3. So, M x is W L by 6 into x minus W x square upon 2 L into x by 3. So, this is your M x equation. Now, this M x equation shows that it follows the cubic law and the maximum bending moment occurs where shear force become 0 after. So, we will find out the bending moment at various point. So, we know the equation M x is equals to W L by 6 into x minus W x cube upon 6 L. So, putting the value at x is equals to L by root 3 to find the maximum bending moment. So, W L by 6 into L by root 3 minus W upon 6 L into L upon root 3 bracket cube. So, solving this we will get M x that is maximum bending moment where shear force changes the sign is W L square upon 9 root 3. So, these are the references which I have referred. Thank you very much for watching my video.