 Myself, Mr. A. N. Suvade, Assistant Professor, Mechanical Engineering Department. Today, we are going to 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. So, draw shear force and bending moment diagram for the beam supported at the ends carries a point load at its midpoint as shown in the figure. So, this is a simply supported beam AB, supported at A and B carries point load at its center. As the beam is loaded at the midpoint, the reactions at A and B are equal in magnitude and magnitude will be equal to half of the total load acting on the beam. Therefore, reaction at A is W by 2 and reaction at B is also W by 2. Now, we will see shear force calculation. So, for calculating the shear force, I will consider the section 1 1, section 2 2, section 3 3 just to the left of C, section 4 4, section 5 5 and section 6 6. So, shear force at 1 1 means shear force just to the left of point A. So, according to the sign conventions, the forces acting in the upward direction to the left side of the section considered to be positive whereas, downward forces considered to be negative. According to the sign convention, the shear force at section 1 1 or just to the left side of A is equal to 0. Shear force at 2 2. So, shear force at 2 2 to the left side of 2 2, there is a reaction R A acting vertically upward considered to be positive. Therefore, shear force at 2 2 is equal to reaction R A which is equal to W by 2. Now, shear force at 3 3 that is equal to forces acting to the left side of section 3 3, there is a reaction R A acting vertically upward considered to be positive. Therefore, shear force at 3 3 is also W by 2. Then, shear force at section 4 4 referring left side of 4 4 different forces acting to the left side of the 4 4 is reaction R A acting vertically upward considered to be positive. And there is a point load at C which is acting in the downward direction considered to be negative. And therefore, shear force at 4 4 is W by 2 minus W which is equal to minus W by 2. Shear force at 5 5 referring left side of 5 5 still the forces acting to the left R A acting vertically upward considered to be positive. And point load at C minus W according to the convention and therefore, shear force at 2 5 5 is also equal to minus W by 2. Shear force at 6 6 considering left side different forces acting to the left of 6 6 R A acting vertically upward positive. Point load at C considered to be negative and there is a reaction R B acting vertically upward considered to be positive. And therefore, shear force at 6 6 is equal to 0. So, we have calculated the shear forces at various sections. Now, we will draw the shear force diagram. So, all positive values of the shear forces are to be plotted above the baseline and negative value are to be plotted below the baseline. So, at 1 1 shear force is 0 at 2 2 shear force is W by 2. So, at A there is a reaction. So, at the section when there is a point load or reaction the shear force value will suddenly increases from 0 to positive W by 2 by vertical straight line. Now, between point A and C there is no load. Hence between the two points if there is no load then shear force value will remain constant and diagram will remain horizontal. So, we have obtained the shear force at 3 3 W by 2. Hence up to C the shear force value will remain constant and diagram will remain horizontal that is W by 2. Now, due to point load at C the shear force value suddenly drops from positive value W by 2 to minus W by 2 by vertical straight line due to the point load at C. So, it decreases to minus W by 2. Between C to B there is no load therefore, shear force value will remain constant between C to B and that will be equal to minus W by 2. And at 6 6 the shear force is 0 therefore, it goes to 0 by vertical line. So, this is your shear force diagram. So, for simply supported beam the shear force the shape of the shear force diagram is pair of rectangle. Now, calculations of bending moment. So, in case of simply supported beam at the supports the bending moment is always 0. Therefore, you can directly write bending moment at A and bending moment at B is equal to 0. Bending moment at point C. So, consider the left portion of the beam from point C and calculate the bending moment at point C referring the left side. So, what are the various moments to the left of C? There is a reaction R A its distance from point C is L by 2. Therefore, this moment to the left side of the point C acting in clockwise direction consider to be positive. And therefore, bending moment at C is W by 2 into L by 2 that is equal to W L by 4. So, bending moment at point A is 0 and bending moment at point B is 0. So, we will draw the bending moment diagram. At A bending moment is equal to 0 and it goes on increasing with linear law to W L by 4. And from W L by 4 again bending moment decreases to 0 at B. Therefore, the shape of bending moment diagram for this case is nothing, but it is a pair of isocellus triangle. That is cantilever carries point load at its free end and its length is L. So, total load acting on the beam is W acting in the downward direction. Therefore, in order to balance this force at fixed end there is a upward reaction which is R A equal to W. So, this reaction is equal to the load acting on the beam, but it is in the opposite direction. Also the moment is W into L acting in the clockwise direction. So, in order to resist that moment at fixed end there will be fixed end moment M A which is equal to W into L. And this moment will act in the anticlockwise direction. Now, we will calculate the shear force and bending moment diagram for this. So, moment at point A it is W into L it is in the anticlockwise direction and reaction at A that is R A it is W acting in the upward direction. So, consider the section xx at a distance x from end B and calculate the shear force at section xx that is equal to referring the right side of the section. The shear force is W acting in the downward direction to the right side of the section considered to be positive. Therefore, shear force diagram for cantilever carries point load at its free end. At free end there is a point load acting in the downward direction. Therefore, shear force value will suddenly increases from 0 to W by vertical straight line. Now, between point B to A there is no load. Therefore, shear force value will remain constant and diagram will remain horizontal. And at A the shear force decreases from W to 0 by vertical straight line. So, for cantilever carries point load at its free end the shape of the shear force diagram is rectangle. Similarly, we will calculate the bending moment at section xx referring the right side moment due to the load acting at point B W into x. This moment is acting in the clockwise direction to the right side considered to be negative. Therefore, bending moment between point B to A varies according to the linear law. So, bending moment at B can be obtained by putting the value x equal to 0. Therefore, bending moment at B is 0. When you put x equal to L, bending moment at point A that will be equal to minus W into L. So, bending moment diagram for this case at point B we have obtained the bending moment equal to 0. And it goes on decreasing to minus W L according to the linear law. And at A the bending moment is minus W into L. As it is negative it is drawn below the baseline. So, the shape of the bending moment diagram for cantilever carries point load at its free end is a triangle. For further details you can refer a textbook of strength of material by R K Bansal. Thank you.