 Myself, Mr. Akshay Kumar Suvde, Assistant Professor, Department of Mechanical Engineering. Today, we will discuss about shear force and bending moment diagram. At the end of the session, student will be able to determine and draw shear force and bending moment diagram for a overhang beam as well as they will determine the position of point of contour of flexure. So, draw shear force and bending moment diagram for a overhang beam and indicate the point of contour of flexure. So, overhang beam carries UDL 20 kilo Newton per meter over a span of 2 meter along with point load at D and C. So, as we know the sign conventions for the shear force and bending moment, if you consider the left side of the section, then all upward forces are considered to be positive and all downward forces are considered to be negative. If you are referring the right side of the section, all upward forces are considered to be negative and all downward forces are considered to be positive. Similarly, for the bending moment, if you consider the section xx and if you are referring the left side of the section, all clockwise movement to the left side and all anticlockwise movement to the right side are considered to be positive, whereas at the section, if the moment to the left side of the section is in anticlockwise direction and to the right side of the section is clockwise direction, then the moment are considered to be negative. So, first of all we will find out the reactions at support A and at B. Therefore, some of the forces acting in the upward direction is equal to some of the forces acting in the downward direction. Therefore, reaction at A and reaction at B is acting in the upward direction. So, r A plus r B is equal to some of the forces acting in the downward direction. So, at D 40 kilo Newton, at C 20 kilo Newton and U D L of 20 kilo Newton per meter spread over distance 2 meter and therefore, reaction of A plus R B is equal to 100 kilo Newton. Now, taking moment about point A, so we will equate clockwise movement and anticlockwise movement. Therefore, moment about A, there is a U D L 20 kilo Newton per meter spread over distance 2. So, 20 into 2 total load, it will act as a point load at half of 2. So, 2 by 2 plus 40 into 2, this movement is also in clockwise direction and at point C, there is a point load 20 and its distance from point A is 5 meter is equal to reaction at B acting in anticlockwise direction. So, r B into 4. So, by calculating above equation, we will get reaction at B is equal to 55 kilo Newton. So, putting this value in the above equation reaction at A is equal to 45 kilo Newton. Therefore, reaction at A r A 45 kilo Newton and r B 55 kilo Newton. Now, we will calculate the shear forces. So, shear force calculation, just extend the portion of the beam beyond A and beyond C, consider the section 1 1 just to the left of side A, section 2 2 just to the right side of A. Similarly, section 3 3, section 4 4, section 5 5, 6 6, 7 7 and 8 8. So, when we refer section 1 1 and consider the left side, there is no portion of the beam to the left of 1 1. Therefore, shear force at 1 1 is equal to 0. Similarly, shear force at section 2 2 referring left side of 2 2. So, shear force at 2 2 is equal to there is only reaction r A acting vertically upward according to the sign convention it is positive. And therefore, shear force at 2 2 is equal to 45 kilo Newton. Similarly, shear force just to the left of D that is shear force at section 3 3, which is equal to there is a reaction r A acting vertically upward consider to be positive. There is a U D L of 20 kilo Newton per meter spread over distance 2 meter. So, 20 into 2 this load will act in the downward direction. Therefore, it is considered to be negative. And therefore, shear force at 3 3 is 5 kilo Newton. Similarly, at section 4 4 shear force at 4 4 that is equal to referring left side r A acting upward positive U D L 20 into 2 this total load will act in the downward direction minus and at point D there is a downward load. So, it is minus 4 D. So, shear force at 4 4 is minus 35 kilo Newton. Similarly, shear force at 5 5 as between D and B there is no load. Therefore, shear force value will remain constant and diagram will remain horizontal. And therefore, shear force at 5 5 is also equal to minus 35 kilo Newton or you can calculate shear force at 5 5 referring left side of the section 5 5. Similarly, shear force at 6 6 referring left side of the section 6 6 there is a reaction r A positive minus 20 into 2 minus 4 T and there is a upward reaction at B which is considered to be positive. So, plus 55 and therefore, shear force at 6 6 is equal to 20 kilo Newton. Similarly, shear force at 7 5 is 7 7 referring left side as between B and C there is no load. So, shear force value will remain constant and therefore, shear force at 7 7 is 20 kilo Newton. Shear force at section 8 8 as there is no load to the right side of section 8 8. So, shear force value will 0. Now, we will draw shear force diagram. So, all positive values are drawn above the base line and all negative values are drawn below the base line. So, at point A there is a reaction. So, shear force will suddenly increases from 0 to 45 kilo Newton. As there is a udl between A to D, shear force will vary according to the linear law and shear force at point D is 5 kilo Newton. Due to the point load at D, shear force will suddenly drops from 5 to minus 35 by vertical straight line minus 35. Between D to B there is no load. So, shear force value will remain constant and diagram will remain horizontal. So, up to B, shear force value is minus 35. Due to the reaction at B, shear force will suddenly change by vertical straight line from minus 35 to 20. Between B to C there is no load. So, up to C, shear force will remain constant and diagram will remain horizontal and at C due to the downward load, shear force will decrease to 0 by vertical straight line. So, this is shear force diagram. Now, we will discuss about the bending moment. So, for calculation of bending moment, we will consider the section between point A to D. So, section x x at a distance x from left end A and we will calculate the bending moment at this section x x. So, bending moment at section x x is equal to r a into its distance from section x x is x moment is positive. Then, u real 20 into its distance from the section is x. So, 20 into x and it will act as a point load at distance x by 2. This moment is negative. Therefore, r a into x minus 10 x square. Therefore, as the distance x varies from 0 to 2, we can calculate the bending moment at point A and D by putting the value x equal to 0. We will get the bending moment at A is equal to 0. When you put x equal to 2, you will get the bending moment at point D which is equal to 50 kilo Newton meter. Similarly, we can calculate the bending moment at point B as well as point C. So, bending moment at point B you can refer the right side of the B and you can calculate the bending moment at B. So, bending moment at point B to the right side of B, there is a load of 20 kilo Newton and its distance from B is 1 meter. And therefore, bending moment at point B referring right side of B 20 into 1, this moment is considered to be negative. Therefore, bending moment to the right side of B is minus 20 kilo Newton meter. Bending moment at point C as there is a load acting at point C in the downward direction, but its distance from that point is equal to 0. And therefore, bending moment at point C is 0. Now, we will draw the bending moment diagram. At A, bending moment is 0. At D, the bending moment is 50 kilo Newton meter varies according to the parabolic curve. So, 50 kilo Newton meter. At B, the bending moment is minus 20. So, between two point loads, the bending moment varies according to the linear law. And at C, the bending moment is 0. So, from minus 20 it goes to 0. Now, think for a while, what is mean by point of contra flexure and how to determine the position of point of contra flexure. So, between point D to B, the bending moment diagram changes its sign from positive to negative. And therefore, the point where bending moment crosses the baseline, that particular point is called as point of contra flexure. So, in order to determine the position of point of contra flexure, we can consider these two similar triangles. And we will consider the position of point of contra flexure at a distance x from end B. And therefore, this particular distance will be 2 minus x. So, we will take the ratio of base to height of this particular triangle and the another triangle. Therefore, the ratio will be x upon 20 is equal to 2 minus x upon 50. And therefore, your distance x will be equal to 0.571 meter from end B. Therefore, the position of point of contra flexure from end B, that will be equal to 0.571 meter. The problem is referred from strength of materials by SS Bhavikati as Chan, limited. Thank you.