# How to derive fixed-end moment of a beam?

Here is my attempt but I am not getting the right answer. Can someone tell me what I am doing wrong? your calculations of reaction forces $$F_A\ and\ F_B \$$ are wrong. This beam is indeterminate and you are handling it as a simply supported determinate beam.

One way of finding the fixed end reactions is to select the two fixed end moments as two degrees of redundancy and solve the beam. we will refer to this diagram at the bottom of the page.

$$R_A = \frac{Pb}{L}+ \frac{M_a}{L} -\frac{M_b}{L}\\\ RB = \frac{Pa}{L}-\frac{M_A}{L}+\frac{M_S}{L}$$

Force displacement relations:

$$(\theta A)_1 = \frac{Pab(L + b) }{6LEI}\quad$$and $$\ (\theta B)_1=\frac{Pab(L + a) }{6LEI}$$

$$(\theta A)_2=\frac{M_AL}{3EI}( \theta B )_2=\frac{M_AL }{6EI}$$

$$( \theta A )_3=\frac{M_BL}{6EI}(\theta B )_3=\frac{M_BL }{3EI}$$

Compatibilty equations:

$$\theta A = (\theta A)_1 - (\theta A)_2 - (\theta A)_3 = 0$$

$$\theta B = (\theta B)_1 - (\theta B)_2 - (\theta B)_3 = 0$$

$$\therefore \frac{M_AL}{3EI} +\frac{M_BL}{6EI}= \frac{Pab(L + b) }{6LEI}\\ \frac{M_AL}{6EI }+\frac{M_BL}{3EI }= \frac{Pab(L + a) }{6LEI}$$

Solving these equations, we obtain

$$M_A =\frac{Pab^2}{L^2} \quad M_B= \frac{Pa^2 b}{L^2}$$

and the reactions:

$$R_A =\frac{Pb^2 }{L^3}(L + 2a ) \quad R_B=\frac{Pa^2}{L^3}(L + 2b )$$