I want to find the reaction force of support A and also find an equation for deflection using Castigliano's energy method for this beam. I know that the reaction force of A is 65.625 N but my solution returns 62.5 N. I don't know if there is something that I'm doing wrong.
Assuming L=10 m and omega = 50 N
This is my solution:
Two aspects require additional attention when reviewing this solution: (1) the distributed load, which is itself changing spatially, and (2) the piecewise nature of this load.
The current solution gives "omega" as 50 N, but the units don't match, as we're talking about a distributed load that must have units of N/m.
This load increases linearly from 0 N/m to 50 N/m over 5 m, or a rate of increase of 10 N/m². It's easy to mix up these values if the units aren't correctly tracked.
You haven't defined any of the variables used in the solution, but since $w$ appears near to $x^2$ in a moment expression, let's take it to be the maximum distributed load (50 N/m in this problem).
At any point $x$ between $0$ and $L/2$, the average distributed load to the left is linearly interpolated as $0+\frac{1}{2}\frac{w-0}{L/2-0}(x-0)=\frac{wx}{L}$. The total shear to the left is therefore $Q(x)=R_A-\frac{wx^2}{L}$, where $R_A$ is the reaction load at A. (You could also write this as $Q(x)=R_A-\frac{w^\prime x^2}{2}$, where $w^\prime$ is the rate of distributed load increase.)
On the other half of the beam, the right side, at any point $x$ between $L/2$ and $L$, the shear is $Q(x)=R_A-\frac{wL}{4}$.
Integrating, we find the moment to be $M(x)=R_Ax-\frac{wx^3}{3L}$ on the left side and $M(x)=R_Ax-\frac{wLx}{4}+\frac{wL^2}{12}$ on the right side (the final term arises as a constant for continuity).
Working through the integration, we have
$$y_A=0=\frac{1}{EI}\left[\int_0^{L/2}\left( R_Ax^2-\frac{wx^4}{3L}\right)\,dx+\int_{L/2}^{L} \left(R_Ax^2-\frac{wLx^2}{4}+\frac{wL^2x}{12} \right)\,dx\right] ;$$
$$0=\left[\frac{R_Ax^3}{3}\right]^L_0+\left[-\frac{wx^5}{15L}\right]^{L/2}_0+\left[-\frac{wLx^3}{12}\right]_{L/2}^L+\left[\frac{wL^2x^2}{24}\right]_{L/2}^L;$$
$$0=\frac{R_AL^3}{3}-\frac{7wL^4}{160};$$
$$R_A=\frac{21wL}{160},$$
which gives the expected answer of $R_A=65.625\,\text{N}$.
