Fixed beam: end moments from support rotation or settlement

For a prismatic beam of span $L$ and rigidity $EI$, fixed against rotation and translation at both ends, one end is then given a rotation or a settlement. Use the slope-deflection equations (clockwise end moments positive; no load on the span, so FEM = 0):

$$M_{AB}=\frac{2EI}{L}\left(2\theta_A+\theta_B-\frac{3\Delta}{L}\right),\qquad M_{BA}=\frac{2EI}{L}\left(2\theta_B+\theta_A-\frac{3\Delta}{L}\right)$$

(a) End A rotates by $\theta$ (clockwise), no settlement

Put $\theta_A=\theta$, $\theta_B=0$, $\Delta=0$:

$$M_{AB}=\frac{4EI\theta}{L},\qquad M_{BA}=\frac{2EI\theta}{L}$$

Both are clockwise on the member ends. The moment at the far end is half of that at the rotated end (carry-over factor 1/2).

Reactions. Taking moments about B for the whole member (clockwise positive), with $V_A$ the upward force on the member at A:

$$M_{AB}+M_{BA}+V_AL=0\;\Rightarrow\;V_A=-\frac{6EI\theta}{L^2}$$

So at A the reaction is $6EI\theta/L^2$ downward and at B it is $6EI\theta/L^2$ upward (the two forces form a couple that balances the end moments).

  A: M = 4EI.theta/L (clockwise)    B: M = 2EI.theta/L (clockwise)
  A: V = 6EI.theta/L^2 down         B: V = 6EI.theta/L^2 up

(b) End B settles down by $\Delta$, no rotation

Put $\theta_A=\theta_B=0$ and $\Delta$ downward (positive chord rotation $\psi=\Delta/L$):

$$M_{AB}=M_{BA}=-\frac{6EI\Delta}{L^2}$$

Both are anticlockwise, which means hogging at A and sagging at B.

Reactions. From moments about B:

$$M_{AB}+M_{BA}+V_AL=0\;\Rightarrow\;V_A=\frac{12EI\Delta}{L^3}\ (\text{upward}),\qquad V_B=\frac{12EI\Delta}{L^3}\ (\text{downward})$$

The settlement pulls the end B down, so the support must exert a downward force at B and an upward force at A.

Summary

These are exactly the stiffness coefficients $4EI/L$, $2EI/L$, $6EI/L^2$ and $12EI/L^3$ used in the stiffness (displacement) method.