A Very Easy Way to Derive Fixed End Moments For Beams Fixed at Both Ends

In order to proceed with this problem, a zero length spans are added to the left and right hand sides

Applying the three moment theorem

M_A’ (0) + 2M_A (0+l) + M_B (l) =wl³/4

2M_A (l) + M_B (l)   = wl³/4

2M_A + M_B    = (wl²)/4 -------------------------------------------(1)

Similarly it can be shown that by adding another zero span to the right of B

M_A + 2M_B    = (wl²)/4 -------------------------------------------(2)

Equating (1) and (2)

2M_A + M_B   = M_A + 2M_B    

M_A   = M_B    

Substituting for M_A in (1)

3M_B = (wl²)/4

M_B = M_A = (wl²)/12

This is the fixed end moment for M_A and M_B