# Statically indeterminate beam (fixed ends) and the moments at the supports

 P: 789 Statically indeterminate beam (fixed ends) and the moments at the supports When it comes to statically indeterminate, its harder. Solving the differential equations in Mathematica, I get: $$R_a=6095/72=84.6528...$$ $$R_b=6145/72=85.3472...$$ $$M_a=2165/12=180.417...$$ $$M_b=-2215/12=-184.583...$$ The force equation is seen to be solved since $Ra+Rb-20-30-10\times12=0$ and the torque equation is seen to be solved because $Ma+Mb-3\times 20-8\times 30+Rb\times 12-10\times 12\times 6=0$. Here I use counterclockwise torques as positive, and upward forces as positive, and I calculate the torque about x=0. Since the problem is statically indeterminate, you cannot use the force and torque equations to solve for the four unknowns. Looking at the solutions in the notes, the Ra and Rb given (85) are close to the actual solutions, so I tend to think that some approximations have been made in the solution, but I don't know where. I agree with the OP, the exact solution cannot be obtained without taking into account the moment at point B.