 I am here to explain you about slope and deflection of cantilever beam with gradually varying load by double integration method. So, today's learning outcome is at the end of the session student will be able to understand and find the slope and deflection of cantilever beam with uniformly varying load for the following condition when uniformly varying load is acting over hole span and when it is acting for a distance A from the fixed end. So, before that we will see what is the beam. A beam is a structural element that primary resist loads applied lateral to the beam axis and the mode of deflection is by bending. So, this is the beam and it is resisting the load perpendicular or lateral to it and there is the deflection of the beam or bending of the beam. So, the beams are classified according to support conditions cantilever beam simply supported beam overhanging beam fixed beam and continuous beam. So, we are dealing with the cantilever beam. So, before that again we have to see what is slope and deflection to find the slope and deflection of the cantilever. So, every beam is frequently governed by rigidity rather than its strength. Hence, the building code specifies the limit of deflection because excessive deflection of the beam not only is visually disturbing, but may also cause damage to other parts of the building. So, therefore, many building codes limits the maximum deflection of the beam. So, we have to know what is the deflection of the beam when the load is applied. So, deflection of the beam at any point on the axis of the beam is the distance between its position before and after loading. So, initially the load beam is straight before loading when load is applied it gets deflected. So, distance between initial and final after loading. So, this distance is called a deflection. So, slope of the beam at any section in a deflected beam is defined as the angle in radiance which the tangent at the section makes with original axis of the beam. So, now this is the deflected shape of the beam. So, here I have drawn a tangent line to this deflected shape and again a horizontal line which is the shows initial position of the beam. So, this angle theta is the slope of the beam. So, cantilever beam is the beam which is fixed at one end and free at the other end. So, one end is fixed whereas the other end is free. There are various methods of finding slope and deflection for cantilevers which are mentioned below that is double integration method, maculis method, moment area method. So, we will see the double integration method. So, the equations used for double integration method how it has arrived. So, when the beam is deflected in a curvature. So, it will have some center or origin that is O and it will have some radius R. So, the radius of curvature of the deflected beam is given by the Fliccure formula that is m upon i is equals to E by R. Therefore, m upon E i is equals to 1 upon R. So, we will treat this as equation A. So, for a practical beam 1 upon R is given by d 2 y by dx square equation B. So, now equating this equation A and B we get m upon E i is equals to d 2 y d 2 y by dx square. Therefore, m is equals to E i d 2 y by dx square with this is the equation 1 which are which we are going to use for double integration method. So, now we will consider a beam which is fixed at one end at A and it is free at the other end and the load is uniformly varying. So, it is an having intensity 0 at point B that is at free end and it is having the maximum load that is w per unit length at point A. So, due to this load there is again a deflection A B dash it shows the deflected shape of the beam. So, your y B is the deflection and theta B is the slope which we have to find. So, now we will consider a section at a distance x from the fixed end. So, remaining portion will be l minus x. So, now we have to take a bending moment at this section x from the free end. So, the load is this triangular part and the value here at the section of this triangle is w l into l minus x. So, for taking this triangular moment so, we have to take the load into distance from which it is acting. So, now this m x is equals to so, now as this is a triangle it is one half base into height. So, it is minus one half. So, l minus x is the base into w by l into l minus x this is this ordinate and it is acting at one third l minus x. So, after solving this we get m x is equals to minus w upon 6 l into bracket l minus x cube this is equation 2. So, now we know the bending moment and we know from equation a moment is equals to e i d 2 y by d x square. So, putting here the moment so, we get e i d 2 y by d x square is equals to minus w upon 6 l into bracket l minus x cube. So, now integrating this above equation we get e i d y by d x is equals to minus w upon 6 l into bracket l minus x raise to 4 upon 4 into minus 1 plus c 1 where c 1 is the constant of integration. So, we will treat this as equation number 3. So, again integrating this equation number 3 we get e i y is equals to w upon 24 into l minus x raise to 5 by 5 into minus 1 plus c 1 x plus c 2 where c 2 is again a constant of integration. So, solving we get this equation e i y is equals to minus w upon 120 l into bracket l minus x raise to 5 plus c 1 x plus c 2. So, we have to find out the constant of integration c 1 and c 2 from the boundary condition of the beam. So, now what are the boundary condition for the cantilever? So, here pause the video and try to get answer. So, as the support end is fixed the values obtained for the boundary condition are at x is equals to 0, deflection y is 0 and at x is equals to 0, slope d y by d x is also 0. Now, substituting this value in equation number 4 we get 0 is equals to minus w upon 120 l into l minus now we are replacing x by 0 raise to 5 plus c 1 into x it is 0 plus c 2. So, solving this we get c 2 is equals to w l 4 by 120. Again putting the value of x is equals to 0 and slope d y by d x is equals to 0 in equation number 3 we get 0 is equals to w upon 24 l into bracket l minus 0 raise to 4 plus c 1. So, from this we get c 1 is equals to minus w l cube by 24. Now, substituting the value of c 1 minus w l cube by 24 in equation 3 we get e i d y by d x is equals to w by 24 l into bracket l minus x raise to 4 minus w l cube upon 24. This is equation number 5 which is known as slope equation. We can find out now slope at any point on a cantilever beam by substituting the value of x and let the slope is maximum at free and it will be theta b. So, putting x is equals to l and d y by d x is equals to theta b we get e i theta b is equals to w by 24 l into bracket l minus l raise to 4 minus w l cube by 24. So, l minus l will be 0 so we are getting e i theta b is equals to minus w l cube by 24. Therefore, theta b is equals to minus w l cube by 24 e i that is equation 5 a. So, negative sign shows it is a anticlockwise direction. Now, again substituting the value of c 1 is equals to minus w l cube by 24 and c 2 is equals to w l 4 by 120 in equation 4 we get e i y is equals to minus w by 120 l into bracket l minus x raise to 5 minus w l cube upon 24 into x plus w l 4 upon 120. This is equation number 6 which is known as deflection equation. So, we can find out deflection at any point on the cantilever by substituting the value of x in this above equation. So, the deflection is maximum at free and so let the deflection be y b and x is equals to l. So, putting this in equation 6 we get e i y b is equals to minus w upon 120 l into bracket l minus l raise to 5 minus w l cube by 24 into l plus w l 4 by 120. So, solving this we get minus w l 4 by 30 therefore, y b is equals to minus w l 4 by 30 e i that is equation number 6 a. So, this negative sign shows the deflection is downward. Now slope and deflection of uniformly varying load at a distance a from the fixed end. So, we will consider this from fixed end that distance a the load is acting over this triangular portion. So, now theta c is the slope at this point c and y c is the deflection at point c and y b is the deflection at point b. So, now as the load is acting over this portion in between a and c. So, the deflection of the beam is between a and c and the rest it remains straight. So, now using the same equation which we have derived in equation 5 a and 6 a instead of l we have to put the value of a because we are neglecting this means as there is a no bending moment in this portion. So, deflection is like this therefore, theta c is equals to w a 4 by 24 e i and y c is equals to w a 4 by 30 e i. So, now for finding the deflection at point b. So, as this is a straight line. So, theta c is equals to theta b that is w a 4 by 24 e i, but for y b we are have to add y c plus theta c into bracket l minus a because this is theta c and we have to solve this by triangular method. So, y b is equals to w a 4 upon 30 e i plus w a 4 upon 24 e i into bracket l minus a. So, we have replaced this theta c with w a 4 upon 24 e i. So, these are the references which I have referred. Thank you.