 In this problem we need to calculate the reaction forces for the following beam using Macaulay step functions. As you can see here, this is a simply supported beam. We have three pin supports at A, B and C. And we have here a point load W applied at a distance L-halves between A and B. Then first we can draw the free body diagram of this problem. As you can see we have three reaction forces, R, A, R, B and R, C and we can use the equilibrium equations in order to solve for them. We have that sum of forces in the vertical direction is equal to zero and sum of moments is equal to zero as well. So from this first equation we have that and we can calculate moments for example at point B. Then we have that. Then this is equation number one and this is equation number two but as you can see here we have three unknowns. So this is obviously a statically undeterminated problem and we need to look for an additional equation in order to solve it. So we need to look to the deformation of the beam and we need to find a boundary condition that provides a new relationship between these reaction forces. So the problem says that we need to use Macaulay step function so we are going to calculate first what is the distribution of moments for an arbitrary point X. And later we will calculate what are the expressions for the slope and finally what is the deformation of this beam. So if we define here this point X we have that the moment is equal to first R A times X. So this is creating here a moment like this then the internal moment distribution is like that. So it goes upwards and we define this as a positive moment. So R A times X we continue and at this point we have that the force W is creating a negative moment. So we have the bracket of Macaulay X minus this is applied at L half so L divided by 2. At this point we have the same as before but now we have also the contribution to the moments of R B and it is positive so plus R B and it is applied at X equal to L X minus L and this is the step function for this force. Then this is the internal moment distribution and now we can use the moment curvature relationship in order to find the expression for the slope and for the deflection of this beam. We know that minus M is equal to EI times the second derivative of the displacement respect to X so this is basically the negative of this. Then now in order to find the slope we can integrate once and we find that and we can integrate once more in order to find the deflection. And now we have to determine what are the constants of integration by using boundary conditions. Since we have three pin supports we know that at X equal to 0 here at point A the deflection is equal to 0 there is no displacement here. The displacement at L is also 0 and finally the displacement at point C X equal to 2L is equal to 0. So we can apply these three conditions from the first one using this equation of course this is 0 and all these brackets are 0 so we find that B is equal to 0. For the second one we have that in this case this bracket is 0 and now finally for this last condition we have that all the brackets are turned on and we have that. Then we have here two additional equations equation number three and equation number four that together with equation number one and two form a set of four equations with four unknowns. R A, R B, R C and the constant of integration A so this is just simple mathematics we can rearrange this equation and doing the proper substitutions we find that then these are the reaction forces R A, R B and R C that we were looking for. And finally just to refresh your minds we can calculate the shear force and bend the moment diagram so we can start with the shear force we have that at this point we start with a positive shear force equal to the reaction force 13 over 32 and we continue constant so this is 13 over 32W. And here we have a force applied negative so we decrease up to here so this distance is W and we continue constant until here. And now what we can do is we know that here at this last point at point D we are decreasing by 3 over 32W so that means that we know that we are finishing at zero so here we were at 3 over 32W and we can just finish the diagram. Then this distance here is equal to 11 over 16W. Now we can draw the diagram of moments. Basically what we have to know here is that dm dx is equal to this shear force. Basically the shear force is the slope of the moment. Then here we start at zero and we increase with the slope equal to 13 over 32W and then the value of the moment here is equal to 13 over 32W times this distance L over 2. And now here we drop off and the slope here is negative we have 13 over 32W minus W. So this is negative so we go here and now from this point we know that we need to reach zero here so we increase with this 3 over 32 slope. And we have that d moment here is equal to 3 over 32W times L.