 Prashant Vishwanath Dinshati, Assistant Professor, Department of Civil Engineering from Walchand Institute of Technology, Salapur. Today I am here to explain you about slope and deflection of a simply supported beam when it is subjected to uniformly distributed load by double integration method. So, today's learning outcome is the students will be able to understand and find the values of slope and deflection of a simply supported beam when subjected to UDL. Now, what is a beam element? A beam is a structural element that primarily resist the loads applied laterally to the axis. Its mode of deflection is primarily by bending. So, this is the axis of the beam and the load is applied laterally to the beam axis and this is resisted by the bending action. So, beam are classified depending upon the support conditions. They are cantilever beam, simply supported beam, overhanging beam, fixed beam and continuous beam. So, we are dealing with simply supported beam. So, why the slope and deflection are to be necessary to be measured? So, the design of beam is governed by rigidity rather than its strength. Therefore, building codes specify the limits on deflection as well as stresses. Excessive deflection of the beam not only is visually disturbing, but may also cause damage to other parts of the building. So, building codes limits the maximum deflection. So, it is necessary to measure the slope and deflection of the beam. Deflection of the 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 the initial position of the beam and this is the final position of the beam after load is applied. So, the distance between these is called as the deflection of the beam. So, as you can see for simply supported with UDL, the maximum deflection is at the center and it reduces near the support. Slope of beam. Slope at any section in a deflected beam is defined as an angle in radiance which the tangent at a section makes with the original axis of the beam. So, now if I draw a tangent to this arc, so I am getting the maximum theta at the support. So, I am draw the tangent here and this angle is the theta that is the slope of beam. Simply supported beam. A beam supported or resting freely on the support at both its end is called as simply supported beam. So, there are various method of finding slope and deflection for the beams which are subjected to various loads. So, important among them are double integration method, maculis method and moment area method. So, we are dealing with double integration method. So, the equation of double integration method how it has arrived. So, when a beam is subjected to UDL or a point load, it is deflected from its original position and it is having a curvature shape and it is having some origin and this distance to the curvature is the radius of the beam. So, the radius of curvature of a deflected beam is given by the equation m upon I z equals to E by R. Therefore, m upon E I is equals to 1 upon R. This is equation number A, but practically 1 upon R is d 2 y by d x square. This is equation number B. Therefore, equating this equation A and B we get m upon E I is equals to d 2 y by d x square. Therefore, m is equals to E I d 2 y by d x square. This is equation number 1. So, this equation is used to find slope deflection by double integration method. Now, we will see the deflection of a simply supported beam when it is subjected to UDL. So, this is the UDL which is having W per unit length intensity and A B shows this initial position of the beam before load is applied and A C dash B shows the position of the beam when load is applied. As the loading is symmetric the reaction at both ends A and B R A is equals to R B is equals to W L by 2. Now, we will consider a section x at a distance x section x at a distance x from support A. So, remaining part is L minus x total length of the beam is L. So, now bending moment at this section so, it is reaction into this distance minus this intensity of load into this system because this is clockwise moment of reaction and the UDL is anticlockwise. So, here m x is equals to reaction that is W L by 2 into x minus now, this UDL intensity is W into x square by 2. So, this is equation number 2, but the bending moment at any section is also given by equation 1 that we have seen that m is equals to E I d 2 y by dx square putting the value of m at the section x. So, we get E I d 2 y by dx square is equals to W L by 2 into x minus W x square by 2. So, integrating this above equation we get E I d y by dx is equals to W L by 2. So, integration of x is x square by 2 minus W by 2 and integration of x square is x cube by 3 plus c 1. So, c 1 is the constant of integration this we will treat as equation number 3. Now, again integrating this equation number 3 we get E I y is equals to now W L by 4 into now, x square integration is x cube by 3 minus W by 6 into x cube integration is x 4 by 4 plus c 1 into x plus c 2 this is equation number 4. Now, this c 1 and c 2 are the constants of integration which can be obtained from the boundary condition of the beam. Now, what are the boundary conditions of simply supported beam? Here pause the video for a minute and try to get answer write it on a paper as the beam is simply supported the values obtained from the boundary conditions are at x is equals to 0 y is equals to 0 and at x is equals to L y is also 0 means at both the ends deflection is 0. Now, substituting the value x is equals to 0 and y is equals to 0 in equation number 4 we get c 2 is equals to 0. Now, again substituting x is equals to L and y is equals to 0 in equation number 4 we get 0 is equals to W L by 4 into L cube by 3 minus W by 6 into L 4 by 4 plus c 1 into L plus 0. Now, this solving this W L 4 by 12 minus W L 4 by 24 plus c 1 L. So, we get from this c 1 is equals to minus W L cube by 24. Now, substituting the value of c 1 is equals to minus W L cube by 24 in equation number 3 we get E I d y by d x is equals to W L by 2 into x square by 2 minus W by 2 into x cube by 3 minus W L cube by 24. This is equation number 5 and this equation is known as slope equation. We can find a slope at any section of a beam by substituting the value of x. The slope is maximum at the support let it be theta a. So, we will put x is equals to 0 in the above equation to get the slope at support that is theta a. So, E I theta a is equals to now this term will become 0 and this term is having x value it becomes 0 minus W L cube by 24. Therefore, theta a is equals to theta b is equals to minus W L cube by 24 E I this is equation number 5 a. So, theta a is equals to theta b because the load is symmetric. Now, again putting the values of c 1 is equals to minus W L cube by 24 and c 2 is equals to 0 in equation number 4 we get E I y is equals to W L by 4 into x cube by 3 minus W by 6 into x 4 by 4 minus W L cube upon 24 into x. So, this is equation number 6 this equation is known as deflection equation. So, we can find the deflection at any point in the beam by substituting or putting the value of x. So, now for U D L the deflection is maximum at the center at x is equals to L by 2. So, we will put at x is equals to L by 2. So, that we can find the deflection y c at the center. So, putting x is equals to L by 2 in the above equation we get E I y c is equals to W L by 12 into now L by 2 cube minus W by 24 into L by 2 raise to 4 minus W L cube by 24 into L by 2. So, solving this we get now E I y c is equals to minus W L 4 by 384 E I. Therefore, y c is equals to minus W L 4 upon 384 E I this is equation number 6 A. So, the negative sign of deflection show that it the deflection is downwards. So, these are my references which I have referred