 Hello friends. I am Prashant Vishwanath Dinshati, assistant professor, department of civil engineering from Walchand Institute of Technology, Sulapur. Today, I am here to explain you about the shear force and bending movement diagram for a cantilever beam with point load or UDL or combination. The learning outcome of today's lecture is at the end of this session student will be able to understand the behavior of cantilever beam under different point loads and UDL and they will be able to represent diagrammatically the shear force and bending movement diagram for cantilever beam. So, what is the shear force and bending movement? The algebraic sum of vertical forces at any section of a beam to the right or left of the section is known as shear force and it is briefly written as SF. Bending movement is the reaction induced in the structural member caused by an external force acting perpendicular to the member and resulting in rotation, it is briefly written as BM. Shear force diagram is the one which shows the variation of the shear force along the length of the beam whereas bending movement diagram is the one which shows variation of the bending movement along the length of the beam. Now, there are some important points to be remembered while drawing the shear force and bending movement diagram. The shear force or bending movement are represented by ordinates that is on y-axis whereas the length of the beam is represented in on the abscissa that is x-axis. Now, for this shear force and bending movement diagram you have to remember these points so you have to consider either left or right portion of the section only. So, add the forces including the reaction normal to the beam on one of the portion. If the right portion of the section is chosen a force on the right portion acting downward is positive while a force acting upward is negative. If we consider the left portion it will be vice versa. The positive value of the shear forces and bending moments are plotted above the baseline whereas the negative values are plotted below the baseline. The shear force diagram will increase or decrease suddenly by a vertical straight line at a section where there is a vertical point load. The shear force between any two vertical load will be constant and hence the shear force diagram between two vertical load will be horizontal. The bending moment at the two supports of a simply supported beam and at the free end of the cantilever beam will be 0. Now, we will consider a cantilever beam with a point load at its free end. So, here there is a beam AB of length L which is having a point load W at its free end. So, let fx be the shear force and mx be the bending moment at any section x-axis and we will consider a distance x from the free end. So, the resultant force acting on the right portion of the section is W which is acting downward and hence it is positive. Now, as per the sign convention this W is acting downwards, but the right portion acting downward is positive. The shear force will be constant at all the section of the cantilever between A and B as there is no load. So, as there is no load between A and B it is represented horizontal. So, this is the shear force diagram for this cantilever with point load at the free end. Now, what will be the bending moment at free end for a two meter cantilever beam having two kilo Newton at its free end? Here pause the video and try to write answer on a paper. So, it will be 0 as load into its perpendicular distance. So, the load is at the free end and the distance is at the free end. So, it is a distance is 0 therefore, load into 0 it will be 0. Now, we will consider see the bending moment diagram for the same beam. So, again the section will be at x from the free end. So, the bending moment at this section will be mx is equals to minus W into x. So, this W into this perpendicular distance so, that is W into x. So, at x is equals to 0 means at point B. So, the bending moment is 0 that is W into 0 it will be 0. At x is equals to L that is at A bending moment here it will be this load into this perpendicular distance that is W into L. So, W into L so, this is negative. So, here I have drawn this on the negative side and I have to join this C to the B point to get the bending moment diagram. Now, shear force and bending moment diagram for cantilever with uniformly distributed load. So, here cantilever beam is there with uniformly distributed load of W per unit length. And I will take a section xx at a distance of x from the free end. So, again f of x will be the shear force mx will be the bending moment at section xx. Now, the resultant force acting on the right portion of the section is W into x. Now, here see the resultant force is W into x. So, this force so, it is represented W per unit length. So, W into length so, that will be W into x. And it is acting with the downward. So, it is W into x. So, the equation here above shows that it follows a straight line law. So, at B so, at B the shear force that is when x is equals to 0 putting x value here 0 f of x is 0. So, here you will be getting shear force 0. And at A when x is equals to L f of x is equals to again W into x where x is equals to L it is W L. So, this will be W L. So, after drawing this W L I will join this it will be a straight line law. So, this will be your shear force diagram for UDL. Now, bending moment again the bending moment here if I consider the resultant load so, it will be acting at this center means at x by 2. So, I have to find out the resultant load and the distance between this section xx and that point load. So, the load will be W into x and it will be acting at x by 2 from this section. So, the bending moment at a section x is given by. So, the load on the right side portion into the CG of that load from x. So, mx is equals to minus W into x and it is acting at x by 2. So, seeing this equation it shows that it follows a parabolic law as it is a square. So, to find out the bending moment at b and a we have to replace the value as x is equals to 0 and x is equals to L. So, at b when x is equals to 0 mx is 0 at a that is at x is equals to L mx is equals to minus W x square means now I will replace x with L. So, minus W L square by 2. So, as it is negative I will draw below the baseline. So, minus W L square by 2 and this follows a parabolic law. So, this is the bending moment diagram for cantilever beam with UDL. Now, shear force and bending moment diagram for cantilever with point load and uniformly distributed load. There is a point load at the end and also uniformly distributed load W 4 kilo Newton per unit length and as the point load is 5 kilo Newton. So, fx is the shear force now I will take a section x that is at a distance of x from free end. So, the resultant force acting on the right side portion of the x. So, here what it will be the point load will be 5 as it is, but UDL is 4 kilo Newton per meter. So, I have to multiply that load into this length. So, it is 4x. So, fx is equals to 5 plus 4x. So, this above equation shows that the shear force follows a straight line. So, at B when x is equals to 0 f of x is equals to 5. So, it is positive I have plot I will plot a 5 year then at x is equals to 2. So, fx is equals to 5 plus 4 into 2 that is 13. So, here I will draw this positive 13 and I will join this and I will get a shear force diagram. So, for bending moment diagram at the section xx. So, what it will be force into this perpendicular distance. So, this 5 into this distance is x. So, mx is equals to minus 5 into x then again there will be a UDL. So, this 4 into x is the total load. So, minus w into x and again it is acting at x by 2. So, into x by 2. So, therefore my equation becomes mx is equals to minus 5x minus wx square by 2. So, seeing this equation it follows a parabolic law. So, I have to find out the bending moment at x is equals to 0 and at x is equals to 2. So, at B point at x is equals to 0 mx is 0 because this 5 into 0 and this w into 0 it will be 0. Therefore, again here at x is equals to 2 that is at point A. So, mx is equals to minus 5 into 2 minus 4 into 2 square by 2 that is equals to minus 18 kilo Newton meter. So, here I have drawn minus 18 kilo Newton and this follows a parabolic law. So, these are the references which I have referred. Thank you. Thank you very much for watching my video.