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

by Brilliant!
Tags: beam, ends, fixed, indeterminate, moments, statically, supports
 P: 51 I'm working through my professor's solution for this problem, and I don't understand how he comes up with the reaction force at B without taking into account the moment at B. Any help would be greatly appreciated.
 P: 51 I think I've been able to reason it out: The moment at the right support is a reaction to the bending moment at that support, and the two are equal and opposite, therefore having no affect at the left end. Compare this to the moments caused by the forces acting between the supports. These forces are supported by both ends of the beam, therefore their moments affect both ends of the beam. I'm sure I could have worded that better, but I think I got the gist of it.
 P: 5,462 Well I make MA = 8820kN-m and MB = 9420kN-m So the end moments are not equal. Can you tell us more about these notes?
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.