 Myself, Mr. A. N. Suvadeh, Assistant Professor, Mechanical Engineering Department, Walton Institute of Technology, SolarPort, today we will learn shear force and bending moment diagram. At the end of this session, students will be able to draw shear force and bending moment diagram for given load condition. For the simply supported beam, carry is UDL over its entire span. So first of all consider a beam supported at A and B having length L carries uniform distributed load over its entire length of rate of load W Newton per meter. Before we proceed we will consider or we will discuss the conventions for the shear force and bending moment. If you are referring the left side of the section, all upward forces are considered to be positive, downward forces considered to be negative. When you refer the right side of the section, all upward forces are considered to be negative, all downward forces are considered to be positive. Whereas in case of bending moment, bending moment at the section xx, if moment acting to the left side of the section is in the clockwise direction and to the right side of the section in anticlockwise direction considered to be positive and it is called as sagging moment. Whereas bending moment at section considered to be negative, if the moment to the left side of the section is in anticlockwise direction and to the right side of the section clockwise considered to be negative and it is called as hogging moment. So this is a simply supported beam carries UDL over its entire length. So consider reaction at A is rA and reaction at B is rB. Therefore some of the forces acting in the upward direction rA plus rB is equal to total load acting in the downward direction. So in case of uniform distributed load, the total load is nothing but rate of loading into distance on which it acts. Therefore total load will be acting total load acting in the downward direction that will be W into L. Now taking the moment about point A, what we will get rB into L, this moment is in anticlockwise direction which is equal to total load acting in the downward direction W into L and this load will act as a point load at a distance L by 2 and therefore reaction at B is equal to WL divided by 2. Therefore reaction at A can be obtained by putting the value of rB in the above equation and by putting that value reaction at A is WL by 2. Therefore reaction at A and B WL by 2 WL by 2. Now consider section XX at a distance x from the left end of the beam and calculate the shear force at section XX to its left side. What are the different forces and reaction acting to the left side of the section? There is a reaction rA acting vertically upward consider to be positive and there is a UDL rate of load W spread over distance X. So total load W into X it will act as a point load in the downward direction to the left side of the section therefore it is considered to be negative. So from this equation 1 we get that shear force varies according to the straight line law and therefore shear force at point A can be obtained by putting X equal to 0. Therefore shear force at A WL by 2. When X equal to L by 2 that is at the center of the beam that is at point C shear force at point C that will be equal to 0 and when you put X equal to L you will get shear force at point B which is equal to minus WL by 2 and your shear force diagram this is a baseline all positive values of shear forces are to be drawn above the baseline and negative value below the baseline. So shear force at A positive WL by 2 hence whenever there is a point load or reaction at that point the shear force value will suddenly increase or decrease by vertical straight line. So as the shear force at point A is positive WL by 2 the shear force value will increases from 0 to WL by vertical straight line. Whereas at point B the shear force is minus WL by 2 hence it decreases from 0 to minus WL by 2 by vertical straight line. So join these two points this is your shear force diagram. Now regarding the bending moment so consider same section XX at a distance X from the left end and calculate the bending moment at section XX referring the left side. So bending moment is nothing but moment due to the forces and reaction acting to the left of the section or right of the section. So bending moment to the left of the section due to the reaction RA the moment is RA into its distance from the section XX that is RA into X and this moment is in the clockwise direction consider to be positive. Whereas there is a uniform distributed load whose rate of loading is W spread over distance X this is total load W into X and it will act as a point load at a distance X by 2. The moment is in the anticlockwise direction to the left side so consider to be negative. So the bending moment at the section is RA into X minus WX square divided by 2 this is equation number 2. So from this equation we know the bending moment between point A to B varies according to the second degree curve or parabolic curve. Therefore when X equal to 0 you will get the bending moment at point A which is equal to 0. When you put X equal to L you will get the bending moment at point B equal to 0 and when you put X equal to L by 2 bending moment at point C that is WL square upon 8. Therefore the bending moment diagram this is baseline at point A and at point B the bending moment is 0 and from 0 the bending moment goes on increasing and it is maximum at the mid that is at point C and from its maximum value it goes on decreasing to 0 at B. So maximum bending moment will occur which is equal to WL square by 8 at point C where the shear force diagram changes it sign from positive to negative or negative to positive. Hence whenever shear force diagram changes its sign from positive to negative it crosses the baseline at that point shear force is 0 and corresponding to that point the bending moment is maximum. For further details my friends refer the textbook of strength of material by R.K. Bansal. Thank you.