Hello friends, myself Mr. A. N. Suvre, 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 weight 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 x x and if we say set the F of the left side of the section, the resultant of the forces and reaction acting in the upper 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 down move of 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 of consider to be negative whereas to the right side of the section upward forces are consider to be negative. Similarly for bending moment calculation the bending moment at the section is consider 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 anticlockwise 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 consider to be negative if the moment due to the reaction and forces acting to the left side of the section in anticlockwise 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 x x normal to the axis of the beam at a distance x from in B and calculate the shear forces to the right side of the section. Therefore, shear force at section x x 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 weight of load W and its weight 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 have to be drawn about the baseline all all negative values of shear forces and bending moment have 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 x x referring the right side of the section. So, bending moment is nothing but the resultant of the moment due to the force and reaction acting to the left side of the section or right side of the section. Therefore, the total force acting to the right side of the section is weight off load W spread over distance x this is your total load and this total load will act as a point load from section x x at a distance x by 2. So, this is x x divided by 2. Now, the moment to the right side of the section acting in clockwise direction according to the sine convention it is considered to be negative and therefore, bending moment at section x x 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 have to be drawn above the baseline and negative values have 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 reference my dear students you refer the book Strength of Material by Arge Bansal. Thank you.