Second moment area theorem:

In the beam AB, the end B moves through a distance BB′ = Δ , such that chord rotation AB′ is positive (clockwise). Hence, MA and MB, the fixed end moments, both are anticlockwise as shown. As there is no change in slope between A and B (both tangents remain horizontal). From the first moment area theorem the area of the M/EI diagram between A and B is zero.  

∴          (M _A + M _B ) L/2EI = 0         ⇒  M _A + M _B  = 0

Also since the intercept on tangent at A, on a vertical through B is Δ , we have, by taking moment of the M/EI  diagram about B (Second moment area theorem)

(M _A + M _B /2EI)L( (2M _A + M _B/M_A+M_B). L/3) = - Δ

Here ? is taken negative as B′ is below B.

Or       2 M_A + M_B = -6E I Δ/L²

But      M_A + M_B = 0

and     ∴   M_A = - 6EI Δ/L²                                                                                                 

M_B=  6EI Δ/L²

Thus, M_A is negative (hogging) bending moment and M_B is a positive (sagging) bending moment and is equal to 6E I Δ/L² in magnitude (Figure 6(b)).

Thus, a support settlement Δ at the right hand end B of the beam AB induces a negative bending moment

M_A      = 6E I Δ/L²                                                                                                                   

on the left end A or a positive bending moment

M_B = 6E I Δ/L²                                                                                                                          

at the end B.

Both M_A and M_B are anticlockwise movements and have, therefore, negative sign. However, the cases with support settlement are beyond the scope of this course.