 Today we will see 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. Also determine the position of point of contrast here and magnitude of maximum bending moment. So, simply supported beam AB carries point load of 20 kilo Newton per meter or a span of 3 meter along with point load of 40 kilo Newton and at point D there is a couple of 120 kilo Newton meter. Calculate shear force and bending moment diagram. So, first of all think for a while how to calculate the reactions at a simply supported beam at point A and B. So, first of all sum of the forces acting in the upward direction is equal to sum of the forces acting in the downward direction. So, reaction at A is r A and reaction at B is r B. Therefore, r A plus r B is equal to sum of the forces acting in the downward direction. So, there is a point load of 40 kilo Newton plus there is a UDL. In case of UDL total load is equal to rate of loading into distance on which it acts. So, 20 into 3 and therefore, r A plus r B is equal to 100. This is my equation number 1. Second equation taking moment about point A. So, taking moment about A 20 into 3 this total load will act as a point load at half of the distance 3 by 2 plus 40 into 3 plus the couple acting at point D in clockwise direction. So, while calculating the reaction only magnitude of couple is to be taken into account. So, 120 is equal to r B into 6 and therefore, reaction at B is equal to 55 kilo Newton. So, putting this value of r B in equation 1 you obtain reaction at point A 45 kilo Newton. So, reaction at A is 45 kilo Newton r B 55 kilo Newton. Now, for calculating the shear forces we will just extend the beam beyond point A and B. We will consider section 11 just to the left of A. Similarly, just to the right of A section 122 at C just to the left of C section 33 just to the right of C section 44 at D just to the left of D 55 just to the right of D 66 just to the left of B section 77 and just to the right of B section 88. At these various sections we will calculate the values of shear force. So, shear force calculation shear force at section 11 you can say shear force at point A of just to the left of point A. So, there is no load. So, shear force at A 11 is equal to 0. Similarly, shear force at 22. So, just to the section 22 to the left side what are the various forces acting that is there is a reaction r A acting vertically upward consider to be positive. And therefore, shear force at 22 or shear force at point A just to its right side that is equal to 45 kilo Newton. Similarly, shear force at 33 or you can say shear force at point C just to its left. So, the forces acting to the left of section 33 or just to the left of C that is reaction r A acting vertically upward and there is a union between point A and C. So, 20 into 3 is the total load and this total load will act in the downward direction. So, it is considered to be negative and therefore, the shear force at 33 is minus 15 kilo Newton. Similarly, shear force at 44 or you can say shear force at point C just to its right side. So, to the right side of C the forces acting r A 20 into 3 and there is a point load at point C 40 kilo Newton. So, it is again minus and therefore, shear force just to the right side of C is minus 55 kilo Newton. Now, at point D we have considered section 55 and 66 just to the left and to the right. So, the section where the couple acts at that section there is no change in the load. And therefore, at section 55 and at section 66 the shear force value will remain constant that is minus 55 kilo Newton. Then, shear force at 77 that is equal to r A minus 20 into 3 minus 4 D which is again equal to minus 55 kilo Newton. And shear force at 88 there is no load to the right side of section 88 therefore, the value is 0. Hence, the shear force diagram this is baseline all positive values are plotted above the baseline and negative values below the baseline. So, at point A due to the vertical reaction the shear force will suddenly increases from 0 to 45 by vertical straight line. At C at 33 the shear force value is minus 15. So, it varies according to the linear law. So, it is minus 15 and due to the point load at C the shear force suddenly drops from minus 15 to minus 55. Between C to D there is no load. So, shear force value will remain constant diagram will remain horizontal. At D the couple has no effect on the shear force value. So, value will remain constant diagram will remain horizontal. So, up to B the shear force is minus 55 and at 88 the shear force is 0. So, it decreases to 0 by vertical straight line. Now, between A to C the shear force diagram changes its sign from positive to negative of negative to positive where the shear force crosses the baseline that particular point is known as point of contour shear and at that point the bending moment is maximum. So, find out the position of point of contour shear. So, we will consider the shear force is 0 at a distance x from end A. So, at 0 and therefore, this particular distance will be 3 minus x. So, from the geometry of the shear force diagram we can take the ratio of two similar triangles that is 45 divided by x is equal to 15 divided by 3 minus x. So, solving above equation we will get x equal to 2.25 meter from A. So, at a distance x equal to 2.25 meter the bending moment will be maximum. Now, we will see how to calculate the bending moment. So, in case of simply supported beam at the support the bending moment is always 0. So, bending moment at point A and B is equal to 0. Therefore, bending moment between A to C. So, we will consider the section x x at a distance x from end A and we will calculate the bending moment at section x x. So, bending moment at section x x referring left side of the section. So, what are the various moments are A into its distance x moment is positive then udl 20 spread over distance x act as a point load at x by 2. So, it is minus. So, bending moment at this section is equal to r A into x minus 10 x square. So, bending moment between A to C varies according to the parabolic curve. So, as your distance x varies from A to C. So, when x equal to 0 bending moment at point A is equal to 0 when x equal to 3 bending moment at point C is equal to 45 kilo Newton meter. The maximum bending moment occurs at a distance of 2.25 meter from A. So, when you put x equal to 2.25 meter then maximum bending moment will be equal to 50.62 kilo Newton meter. Similarly, bending moment at point D. So, we will consider the section 55 just to the left of D and we will calculate the bending moment at phi u phi u or bending moment at point D just to its left side. So, r A into 4.5 bending moment is positive 20 into 3 into 1.5 plus 3 by 2 moment is negative minus 40 into 1.5. Therefore, bending moment at D just to the left of D that is minus 37.5 kilo Newton meter. Similarly, bending moment just to the right of D or you can say bending moment at 66. So, bending moment just to the right of D r B into 1.5 bending moment is positive. Therefore, that is equal to 82.5 kilo Newton meter. At A bending moment is 0 it goes on increasing and it will be maximum and that value is 50.62 and from maximum value it goes on decreasing to 45 with parabolic curve. From 45 it goes on decreasing to minus 37.5 and due to the couple at point D it will suddenly changes from minus 37.5 to 82.5 and from 82.5 it goes on decreasing to minus 37.5. So, this is shear force and bending moment diagram. The material is referred from the strength of material by SS Bhavi Katti. Thank you.