 Hello friends, Myself Mr. A. N. Suvade, Assistant Professor from Mechanical Department. Today we will discuss about the shear force and bending moment diagram. At the end of this session, student will be able to draw shear force and bending moment diagram for given load conditions. So we will discuss today, draw shear force and bending moment diagram for a beam of length L subjected to uniform distributed load over its entire span. So consider a cantilever beam whose end A is fixed and free end B the length of the beam is L and this beam is subjected to uniform distributed load where the rate of loading is W Newton per meter. For this we have to draw shear force and bending moment diagram. Before that we will see the sign conventions for the shear force calculation and bending moment calculation. So if we consider the section xx and if we refer the left side of the section the resultant of the forces and reaction acting in the upward direction is considered to be positive where as the forces resultant of the forces and reaction acting to the right side of the section acting in the downward direction considered to be positive where the resultant of the forces and reaction acting to the left side of the section in the downward direction are considered to be negative where as to the right side of the section upward forces are considered to be negative. Similarly, for bending moment calculation the bending moment at the section is considered to be positive when the resultant of the moment due to the forces and reaction acting in clockwise direction to the left side of the section and in anti-clockwise direction to the right side of the section these moments are considered as positive and these are known as sagging moment whereas the bending moment at the section considered to be negative if the moment due to the reaction and forces acting to the left side of the section in anti-clockwise direction and to the right side of the section acting in clockwise direction such moments are considered as negative and these are called as hogging moment. So, with this we will start to solve this particular problem. First of all consider the section xx normal to the axis of the beam at a distance x from end b and calculate the shear forces to the right side of the section. Therefore, shear force at section xx to its right side is equal to so shear force is nothing but it is the forces acting to the left side of the section or to the right side of the section. So, the force which is acting to the right side of the section is the rate of load w and it spread over distance x as this is the total load w into x acting in the downward direction. So, to the right side of the section downward forces are considered to be positive. Therefore, the shear force is w into x in this case the shear force between point b to a varies according to the linear law and the shear force value at b and at a that can be obtained by putting the value x equal to 0 and x equal to l. So, when we put x equal to 0 shear force at point b is equal to 0 whereas, when we put x equal to l the shear force at point a is equal to w into l and therefore, the shear force diagram for this particular loading this is a baseline all positive values are to be drawn above the baseline all negative values of shear forces and bending moment are drawn below the baseline. So, at point b the shear force is 0 and the shear force increases from 0 to w l at a. So, the shear force varies according to the linear law and at point a the shear force is w into l as the value is positive it is drawn above the baseline. So, this is your shear force diagram. Then for bending moment calculation bending moment at section xx referring the right side of the section. So, bending moment is nothing but the resultant of the moment due to the forces and reaction acting to the left side of the section or right side of the section. So, the total force acting to the right side of the section is weight of load w spread over distance x this is your total load and this total load will act as a point load from section xx at a distance x by 2. So, this is x divided by 2. Now, the moment to the right side of the section acting in clockwise direction according to the sign convention it is considered to be negative and therefore, bending moment at section xx is minus w x square divided by 2. Now, if we see this particular bending moment equation the bending moment between point b to a varies according to the parabolic curve and it varies as the distance x varies from 0 to L. Therefore, bending moment at point b is obtained by putting x equal to 0. So, when x equal to 0 bending moment at b is 0 when you put x equal to L bending moment at point a that is equal to minus w L square upon 2. So, we have calculated the bending moment now we have to draw bending moment diagram. So, draw the baseline as the positive values are to be drawn above the baseline and negative values are to be drawn below the baseline. So, at point b the bending moment is 0 and it goes on decreasing to w L square by 2 according to the parabolic curve and at a the bending moment is w L square by 2 as the bending moment is negative it is drawn below the baseline. Hence, in this case the bending moment varies according to the parabolic curve and in case of cantilever beam the bending moment is always 0 at its free end. In reference my dear students you refer the book Strength of Material by R. K. Bansal. Thank you.