 In this problem, we need to find the deflection of this simply supported beam at x equal to 1.8 and we need to use McCollister function to find this result. So, as you can see here, we have this beam with two pin supports, A and B and we have this series of forces. We have this moment here of 3 kilometers meter applied at 1.5 meters. We have here this distributed load of 16 kilonewtons meter and we have a point load of 20 kilonewtons applied at this point. Then we need to use, as I said, McCollister function, but first of course we need to calculate what are the reaction forces at A and B. So, first of course we need to draw the free body diagram of this problem. Now, in order to calculate the reaction forces, R, A and R, B, we apply the equations of equilibrium and as usually we have the sum of forces in the vertical direction is equal to 0 and the sum of moments is also equal to 0. So, we define this positive upwards, so we have that and we have here that the sum of moments is equal to 0. We calculate this at A and we define positive clockwise. Then we have first the contribution of this moment, the contribution of this distributed load which is equal to the area of this load times the distance from this point to A, the moment that this force is creating and finally the moment that R, B is creating here at A. So, we have that. Then from the equation of moments, we find directly if we solve it that the reaction force R, B is equal to 24.36 kN and if we substitute the result into the equation of the equilibrium of forces we have that R, A is equal to 10.04 kN. Then now we can calculate what is the moment as a function of X using the McCulley-Steff function, so we have that M is equal to we define first our sine criteria, so remember that for us this internal moment is defined as positive, so for example R, A at this point for example is creating a moment like this but then the internal moment reaction is like that and it is as I said positive, so the moment created by R, A is equal to R, A times X. We continue with the moment of 3 kN, it is negative minus 3 X minus 1.5 we are applying this moment at 1.5 and the index is 0 and now the distributed force and here we have to do a trick because it is very important to remember that if we have a distributed force like this one, if we want to use McCulley-Steff function once we start having a distributed force we must have it until the end of the beam, so as I said we can do a trick so we can extend here the distributed force so now we have it until the end of the beam and we subtract the same amount here so this is equivalent to what we have before but now we can apply McCulley-Steff function then we start with the negative force and we have that is creating a negative moment so it is equal to minus the intensity of the load remember that the bracket for a distributed force is always divided by 2 and the index is 2 so now we have here X minus and this force is starting at X is equal to 1.5 so this is 1.5 now the second distributed force plus 16 divided by 2 power of 2 X minus and this is starting at 1.5 plus 0.9 2.4 ok so this is the procedure that we have to follow if we have a distributed force that starts at some point but doesn't continue until the end of the beam now we continue with the point load of 20 kN it's creating a negative moment so minus 20 the bracket has index 1 and X minus this is applied at 2.4 and what about RB in principle this is creating a moment plus RB X minus 3 but this bracket is going to be always turned off so it doesn't make sense to add this additional term so this doesn't change the problem then this is our distribution of moments using my callista function and now we can use the moment curvature relation in order to find what is the deflection of the beam we know that the moment is equal to minus EI the second derivative of V respect to X so this is equal to then if we integrate once we find that then this is the expression for the slope of the beam and we can integrate once more in order to find expression for the deflection and now we have to apply boundary conditions in order to find the values of the constants of integration A and B and now we have to use boundary conditions in order to find the values of the constants of integration A and B we know that this is a simply supported beam so then of course at A which is X is equal to 0 the deflection of the beam is equal to 0 and at B which is X is equal to 3 the deflection of the beam is also equal to 0 then from this equation at X is equal to 0 then this bracket is turned off this one as well, this one as well, this one as well so we have that minus rA divided by 6 X to the power of 3 plus AX plus B so if we particularize this for the point X is equal to 0 this is equal to B and we know that this is equal to 0 and from the second condition then if we solve this equation we find that A is equal to 12.6 then once we know the value of the constants we can use this equation here the equation for the deflection we can substitute the value of these constants so finally we have that the expression for the deflection of the beam is then finally the exercise was asking for the deflection at X equal to 1.8 then if we particularize this expression for the deflection for X equal to 1.8 finally we have that the deflection of the beam is equal to please note that in this case since X is equal to 1.8 this bracket and this bracket are turned off when we are substituting in this expression so remember that always that X is smaller than the value that is subtracting inside the bracket the whole bracket is turned off then finally this is equal to 22.4 millimeters