 Myself, Akshay Kumar and Suvade, Assistant Professor, Department of Mechanical Engineering Department. Today, we are going to discuss shear force and bending moment diagram. At the end of this session, student will be able to determine and draw shear force and bending moment diagram for a beam. So, today, we will see a problem on a cantilever AB whose end A is fixed, carries UDL over its entire length of 1 kilo Newton per meter and at mid at point C there is a couple which is acting in anticlockwise direction of magnitude 20 kilo Newton meter. So, according to the sign conventions for the shear forces, when you consider the left side of the section, all upward forces are positive, all downward forces are negative. Whereas, if you are considering the right side of the section, all upward forces are negative, all downward forces are positive. Similarly, for bending moment, if you consider the section xx, all the moments which are acting to the left side of the section in anticlockwise direction and to the right side of the section in clockwise direction, these are considered to be negative and these are called as hogging moment. Whereas, those bending moment which are acting to the left side of the section in clockwise direction and to the right side of the section in anticlockwise direction, they are considered to be positive and they are known as sagging moment. Now, for a while, think whether we can calculate the reaction and fixed end moment at the fixed support of cantilever. Reaction at point A is equal to total load acting on the beam in the downward direction. Therefore, total load acting in the downward direction, 20 kilo Newton load acting at point B and UDL 1 spread over distance 8 meter. So, total load acting in the downward direction is 28 kilo Newton and therefore, reaction at A should be acting in the upward direction having magnitude 28 kilo Newton. Similarly, moment at point A. So, in order to counter balance the clockwise moment at fixed support A, there will be anticlockwise moment. Therefore, moment at point A is equal to 20 into 8. This moment is acting in clockwise direction. So, clockwise moment we are considering positive. Then UDL 1 kilo Newton per meter spread over distance 8 meter and it will act as a point load at half of 8. This is also clockwise moment and there is anticlockwise couple acting at point C which is minus 20 kilo Newton meter and therefore, moment at point A is 172 kilo Newton meter acting in anticlockwise direction and hence fixed end moment M A is equal to 172 kilo Newton meter. Now, in order to calculate the shear forces at various points along the beam, we will consider the section xx between point C and B. So, consider section xx at a distance x from end B and therefore, shear force at section xx is equal to considering the right side of the section 20 kilo Newton force acting in the downward direction according to sign convention it is positive. There is a UDL of 1 kilo Newton per meter spread over distance x. So, 1 into x is the total load and this total load will act in the downward direction. So, consider to be positive. So, this is the shear force acting at section xx. So, this distance x varies from B to C. So, when you put x equal to 0, you are calculating the shear force at point B which is equal to 20 kilo Newton. Whereas, when you put x equal to 4, the shear force at point C is equal to 24 kilo Newton. Similarly, when we consider the shear force between point C to A, again consider the section between A to C at a distance x from end C. So, calculate the shear forces acting to the right side of the section. So, shear force at section xx when you are considering the section between A and C. Therefore, the forces acting to the right side downward force at point B 20 kilo Newton acting in the downward direction consider to be positive. There is a UDL 1 kilo Newton per meter and the total span from this section xx is 4 plus x. So, 4 plus x this is also acting in the downward direction as a point load. So, consider to be positive. Therefore, your distance x varies from point C to A. Therefore, when you put x equal to 0, you are calculating the shear force at point C again it is 24 kilo Newton. When you put x equal to 4, shear force at point A is equal to 28 kilo Newton. So, this is shear force calculation. Now, we will draw shear force diagram. So, this is your baseline all positive values are to be drawn above the baseline and negative value below the baseline. So, at point B the shear force is 20. So, due to the downward load the shear force value will increase from 0 to 20 by vertical straight line. At point C the shear force is 24 kilo Newton it varies linearly. So, from 20 it goes on increasing to 24 kilo Newton and at point A the shear force is 28. So, again from point C the shear force will increases from 24 to 28 kilo Newton. So, this is your shear force diagram. Similarly, we can calculate the bending moment also. Consider the section between C and B and calculate the bending moment. So, bending moment calculation when you consider the section between C and B. So, this is section x x considering right side. So, what are the various moments to the right side of this section? So, there is a point load 20 its distance from the section is x. So, 20 into x this moment is consider to be negative. So, it is minus. There is a UDL of magnitude 1 kilo Newton per meter spread over distance x. So, 1 into x is total load and this total load will act as a point load at x by 2 from the section. So, it is again minus. Therefore, your distance x varies from B to C. So, when you put x equal to 0 you have calculating the bending moment at point B which is equal to 0. When you put x equal to 4 you have calculating the bending moment at point C which is by putting the value x you will get minus 88 kilo Newton meter. Now, similarly consider the section between C to A at a distance x from end C and calculate the bending moment. So, bending moment at section x x when you consider the section between A and C and referring the right side of the section. So, what are the various moments to the right side of the section? This point load 20 into its distance from this section xx. So, 20 into 4 plus x this is negative. Again there is a UDL 1 spread over distance 4 plus x and it will act as a point load at a distance 4 plus x divided by 2. This moment is also minus and at point C there is anticlockwise couple. So, to the right side of the section anticlockwise moment are considered to be positive. So, while calculating the bending moment only magnitude of the couple is to be taken into account. So, it is plus 20. Therefore, bending moment at point C you can obtain by putting x equal to 0. So, bending moment at point C at x equal to 0 you will get minus 68 kilo Newton meter. Similarly, when you put x equal to 4 you will get the bending moment at point A. Which is by putting the value x equal to 4 meter you will get the bending moment at point A minus 172 kilo Newton meter. So, this is the bending moment calculation. Now, we will see the bending moment diagram. How the bending moment varies between point B to A? So, this is your base line. At point B the bending moment is 0 and at point C it goes on decreasing to minus 88 kilo Newton meter at point C. Due to the couple acting at point C there is a sudden change in the bending moment from minus 88 to minus 68 by vertical straight line. And from minus 68 again it goes on decreasing to minus 170 2 kilo Newton meter. So, from this diagram we can see at the section where couple acts there is a sudden change in the bending moment. And that change in the bending moment is equal to the magnitude of the couple. The material which is referred for this particular problem is from SS Bhavikatti strength of material S. Chan private limited fourth edition. Thank you.