 Friends, myself Prashant Vishwanath Dinshati, Assistant Professor, Department of Civil Engineering from Valgen Institute of Technology, Solapur. So, today I am here to explain you about slope and deflection of a cantilever beam with point load by double integration method. So, today's learning outcome is at the end of this session students will be able to understand and find the values of slope and deflection of cantilever beams with the point load having following conditions. So, before going with slope and deflection we will see what is beam and its definition. So, beam is a structural element that primarily resist load applied laterally to the beam axis. Its mode of deflection is primarily by bending. So, this is the beam and the load applied is laterally and the beam is having a bending shape it resist by bending. So, beams are classified based on its support conditions. So, why we require slope and deflection? So, we will see. So, design of beam is frequently governed by rigidity rather than strength. So, every building course specifies the limit of deflection as well as stresses because due to excessive deflection of the beam it is not only visually disturbing but it may also damage to other parts of building. So, each building course limit the maximum deflection of beam. So, we have to see what is the deflection and slope of the beam. So, we will see the definition for the deflection of beam. The deflection at any point on the axis of the beam is the distance between its position before and after bending. So, this is a cantilever beam it is straight and when load is applied the beam bends in this way and this is the deflection. So, wherever you take the section. So, the deflection is greater at the free end and it decreases to the fixed end. Slope of the beam slope at any section in the deflected beam is defined as angle in radiance which is the tangent at the section makes with the original axis of the beam. So, this is the curvature where the beam deflects and this is the line tangent and we are having this reference horizontal axis of the beam. So, this theta b is the slope of the beam cantilever beam it is a beam which is fixed at one end and free at the other end. So, one end is fixed other end is free. So, we are seeing by double integration method. So, now what is the equation by double integration method? So, when the beam deflects this way. So, it has a curvature and here is a center point of that curvature and this is the radius of that curvature. So, the radius of curvature of the deflected beam is given by the flexural equation 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 it as equation A, but for a practically beam or mathematically it is 1 upon r is equals to d 2 y by dx square that is equation B. Equating this equation A and B we get m upon e i is equals to d 2 y by dx square. Therefore, m is equals to e i d 2 y by dx square this is the equation for the double integration method. So, this is a cantilever beam which is having length L and load W is at the free end. Due to this load there is a deflection of the beam where yB is the deflection and theta B is the slope. So, now we will consider any section in a beam that is at a distance of x from fixed end and the remaining part is L minus x. So, total length of beam is L and the W is acting at free end. So, at this section the moment mx is given by minus W into L minus x this is load into distance of the load from the section where you are going to take the bending moment. So, the minus sign is due to hogging bending moment. So, as per the equation 1 m is equals to e i d 2 y by dx square. So, putting the value of m. So, we get e i d 2 y by dx square is equals to minus W into bracket L minus x is equals to minus W L plus W x. So, integrating the above equation we get e i d y by dx is equals to minus W L into x plus W by 2 into x square plus c 1. So, this is equation number 3 where c 1 is the constant of integration. So, again integrating that equation 3 we get e i y is equals to minus W L into x square by 2 plus W by 2 into x square by 3 plus c 1 x plus c 2. So, where c 2 is the again constant of integration this is equation 4 where c 1 and c 2 are constant of integration. So, these are obtained by boundary condition. So, what are boundary conditions for cantilever? So, you are pause the video and try to get answer for the boundary condition. So, at the support is fixed the values obtained from the boundary condition are at x is equals to 0 that is at fixed end deflection y is 0 and at x is equals to 0 slope d y by dx is also 0. So, by substituting the value of x is equals to 0 y is equals to 0 in equation 4 we get c 2 is equals to 0 and by substituting x is equals to 0 and d y by dx is equals to 0 in equation 3 we get c 1 is equals to 0. So, now putting this value c 1 is equals to 0 in equation 3 we get e i d y by dx is equals to minus W L x plus W x square by 2. So, plus c 1 that is 0 is equals to minus W taking common we will get into bracket L x minus x square by 2. So, this is equation number 5 this equation is known as slope equation. So, we can find the slope at any point on the cantilever by substituting the value of x in this equation. So, we will get the theta b that is at free end where putting x is equals to L in this above equation we get e i theta b is equals to minus W into bracket L into L minus L square by 2 here I am replacing x by L. So, we will get by solving it is minus W L square by 2. So, therefore, theta b is equals to minus W L square upon 2 e i that is equation 5 a. So, this negative sign shows the tangent at b makes an angle in anticlockwise direction with line a b. Substituting the value of c 1 is equals to 0 and c 2 is equals to 0 in equation 4 we get e i y is equals to minus W L into x square by 2 plus W by 2 into x cube by 3. So, again the two terms are next two terms are 0 as c 1 and c 2 are 0. This is equals to taking minus W common minus W into bracket L x square by 2 minus x cube by 6 this is equation number 6. This equation is known as deflection equation. We can find deflection at any point on the cantilever by substituting the value of x. So, we have seen that the maximum deflection is at free end. So, we will substitute x is equals to L and the deflection at free end is y b. Therefore, e i y b is equals to minus W into bracket L into L square by 2 minus L cube by 6. Here I am replacing x by L. So, solving this we get is e i y b is equals to minus W L cube by 3. So, therefore, y b deflection is equals to minus W L cube upon 3 e i it is equation 6 a. Here again negative sign shows the deflection is downward. So, now instead of this load acting at the free end it is acting at somewhere at a distance a from the fixed end. So, this is L. So, this is a and this distance is L minus a. Now, let theta c is the slope at point c. So, now here if I draw a tangent here that will be having a theta c that is dy by dx at c. So, y c is the deflection point at c and y b is the deflection point at b. So, now the portion ac of the cantilever may be taken similar to a cantilever which is having a load at free end. So, now I will discard this. So, if I assume this only. So, now this is W is acting in a cantilever at a distance a that is at free end. Therefore, theta c as we have derived it is W a square upon 2 e i. So, here in the equation 5 a we have already derived that equation. So, instead of L I am putting it as the distance at a. So, y c is again W a cube upon 3 e i. Here I am not mentioning the negative sign because here it is anticlockwise direction and here the deflection is downward. So, I am only taking the values. So, I have not put negative sign there. So, now to find the theta b and y b. So, now beam will bend only the beam will bend only between a and c. So, here in this only portion the beam will bend as c to b there remains straight since no bending moment occurs in c b portion. So, now if no bending moments occur this will be a straight line. So, this straight line if I continue that will give a theta c or theta b will be same. So, theta c is equals to theta b is W a square upon 2 e i. So, y b is equals to y c. So, this deflection plus this deflection. So, here I am knowing this angle it is theta c. So, from that distance L minus a and theta c I will get this deflection b b dash. So, y b is equals to y c plus theta c into bracket L minus a. So, putting y c I will get y b is equals to W a cube upon 3 e i plus W a square upon 2 e i into bracket L minus a because theta c is W a square upon 2 e i. So, these are the references which I have used. Thank you.