 In this problem we have a cantilever beam which is supported by this spring at point B and we have a point load applied at point C with magnitude P. And we need to find what is the reaction at this spring using my call history function. So then we can start drawing the free body diagram for this problem. As you can see here we have the vertical reaction at the wall or A, the reaction moment MA and we have also the spring reaction at B here. Then we have three unknowns and from equilibrium we can find the two first equations that we need to solve this problem. We know of course that the sum of forces in the vertical direction is equal to zero and that the sum of moments is also equal to zero so from this equation we find that and if we calculate this equilibrium of moments at point A we find that then we have essentially two equations and three unknowns MA, RA and RB so we need to find one additional equation one constraint on deformation in order to solve this problem. So first since we have already two equations what we can do is we can calculate MA and RA as a function of this reaction force RB that we have to calculate because this will simplify a lot our problem later So from this first equation we have that the reaction force A is equal to B minus RB and from here we have that MA is equal to RB times L half minus PL So these are equations one and two So now using my call is the function we can calculate what is the distribution of moments for this beam so we can calculate what is the moment at any arbitrary point X So M is equal to first this moment MA plus RA times X and now this reaction force RB which is applied at L half so this is plus RB X minus L half So now the next step is we can look at the moment curvature relationship in order to find the expressions for the slope and for the deformation or deflection of this beam Then we have the moment curvature relationship We know that minus M is equal to EI the second derivative of the displacement with respect to X So this is basically the negative of this equation so this is equal to Then as you can see here the second derivative of the displacement with respect to X is a function of MA, RA and RB But remember that we know RA and MA as a function of RB So we can use here equations one and two And now we can transform this equation as an equation as a function only of RB So if we use here as I said one and two we have that this is equal to Then now we can integrate this equation once So we have that this is the expression for the slope Then we integrate once more and we have that And now we have to apply boundary conditions in order to determine what are the constants of integration A and B We know that this beam is clamped here at A so at X equal to 0 We know that the deformation is equal to 0 The deflection is equal to 0 and we know that at X equal to 0 here at the wall The angle dv dx is also 0 And we know that at X equal to 0 the angle dv dx is also equal to 0 Then from this equation we use this expression here for the deflection So this is 0, this is 0, of course this bracket is 0 and this is 0 So B is equal to 0 and for the second condition we use this equation And we have that this is 0, this is 0 and this bracket is 0 Then A is equal to 0 Then as we said before we need to find one additional equation to solve this problem Then we need one additional boundary condition And if we look at here at point B We know that the deformation of the beam at this point equals the deformation of the spring And of course the deformation of the spring is equal to the force divided by k Then this is the additional equation, the initial additional condition at X equal to l halves We have that this relation holds Then using this equation again we have that note that in this case Well remember that this is 0 and this is 0 And in this case this bracket is also 0 Then this is the third equation that we were looking for So from here from equation 3 we can calculate rb So rb is equal to Then this is the value of the reaction force at the spring And note that if this report is rigid we have that k The constant k is equal to infinity And then the reaction force rb is equal to 5p over 2