 Hello friends, myself Prashant Vishwanath Dinshati, Assistant Professor from Department of Civil Engineering from Valjan Institute of Technology, Swalapur. So, today I am here to explain you about slope and deflection of a simply supported beam when it is subjected to a point load at center 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 a simply supported beam when it is having a point load at its center. So, what is beam? A beam is a structural element that primarily resist loads applied laterally to the beam's axis. Its mode of deflection is primarily by bending. So, this is the initial position of the beam and when load is applied it is deflected by bending. So, the beams are classified according to the support conditions that are cantilever beam, simply supported beam, overhanging beam, fixed beam, continuous beam. So, we are seeing simply supported beam. So, now why the slope and deflections are necessary to be checked? Because beam's design of beam is frequently governed by rigidity rather than its strength. So, many of the building code specifies the limits of deflection as well as stresses. So, excessive deflection of a beam not only is visually disturbing, but also causes damage to other parts of the building. Deflection of a beam, the deflection at any point on the axis of the beam is the distance between its position before and after loading. So, now this is the position of the beam before loading and this will be the position of the beam after loading. So, this difference is the deflection of the beam. So, it is maximum at the center and it decreases near the supports. Slope of a beam, slope at any section in a deflected beam is defined as an angle in radiance which the tangent at the section makes with the original axis of the beam. So, now if I draw a tangent line to this R, so this is B, this R and the angle made by this with the original axis is the slope of the beam that is theta B. Now, simply supported beam, a beam supported or resting freely on support at its both end is called as simply supported beam. So, there are various methods for finding slope and deflection for the beams subjected to various loadings. The important methods are double integration method, McColl's method and moment area method. So, today we are dealing with double integration method. Now, the equation used for double integration method. So, when a UDL or a point load acts on a beam it is get deflected like this from its original position. So, this will be having some origin point and some radius. So, the radius of curvature of the deflected beam is given by the equation m upon i is equals to E by R. Therefore, m upon E i is equals to 1 upon R. So, we will give this as a equation number A. Now, practically 1 upon R is d 2 y by dx square, this is equation B. Now, equating equation and 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. So, this is the equation number 1 which is used for double integration method. Now, slope and deflection of simply supported beam with a point load at a center. So, when a point load is acting at a center we have to find out slope and deflection of a beam. So, this A B beam shows the initial position of the beam and this A C dash B shows the beam after load is applied. So, as load is centrally it is symmetric. Therefore, reaction R A and R B are equal that is W by 2. Now, consider a section x at a distance of x from support A. So, the bending moment at this section m x is equals to this reaction into distance that is W by 2 into x. So, this is equation number 2. So, we know that from 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 W by 2 into x. So, integrating this equation above equation we get E i d y by dx is equals to W by 2 into x square by 2. So, the integration of x is x square by 2 plus C 1 where C 1 is constant of integration. So, we will treat this as equation number 3. Now, integrating this again equation number 3 we get E i y is equals to W by 4 into now x square integration is x cube by 3 plus C 1 into x plus C 2. So, it is equation number 4. So, C 1 and C 2 are the constant of integration which can be obtained from boundary condition. So, what are the boundary conditions for simply supported beam? Here pause the video for a while and try to get answers. As the beam is simply supported the values for the boundary conditions are at x is equals to 0 y is equals to 0 and at x is equals to l by 2 dy by dx is 0 that is slope is 0 as the maximum deflection at the center slope at the center is 0. So, now by substituting x is equals to l by 2 dy by dx is equals to 0 in equation number 3 we get 0 is equals to W by 2 into x square by 2 plus C 1. So, now here we are having again x is equals to l by 2 putting this value we get C 1 is equals to minus W l square by 16. Again the second boundary of condition we are putting x is equals to 0 y is equals to 0 in equation number 4 we get 0 is equals to 0 minus 0 plus C 2 therefore, C 2 is 0. Now, substituting the value of C 1 is equals to minus W l square by 16 in equation number 3 we get E i dy by dx is equals to W by 2 into x square by 2 minus W l square by 16 it is equation number 5 this equation is known as slope equation. So, we can get the slope at any point on the beam by substituting the value of x. So, the maximum slope is at the support let it be theta a. So, at a x value is 0 putting x is equals to 0 in the above equation we get E i theta a is equals to W by 2 into x square by 2 minus W l square by 16. So, solving this we get minus W l square by 16. So, now theta a is equals to theta b is equals to minus W l square by 16 E i it is equation number 5 a as the load is symmetric theta b is equals to theta a. Now, again substituting the value of C 1 is equals to minus W l square by 16 and C 2 is equals to 0 in equation number 4 we get E i y is equals to W by 4 into x cube by 3 minus W l square by 16 into x plus C 2. So, C 2 is 0. So, this is equation number 6. So, this equation is known as deflection equation. So, now again we can find the deflection at any point on a simply supported beam just by putting the value of x. The deflection is maximum at the center let it be y c. So, the x value is l by 2. So, putting x value l by 2 in the above equation we get E i y c is equals to W by 12 into l by 2 raise to cube minus W l square by 16 into l by 2 plus 0. So, solving this we get minus W l cube by 48. Therefore, y c is equals to minus W l cube by 48 E i this is equation number 6 a the negative sign shows the deflection is downward. So, these are my references which I have referred. Thank you.