Slope deflection: B sinks, C rotates

Data and assumptions: A is fixed; AB = 1 m + 3 m = 4 m (I) with 4 t at 1 m from A; BC = 6 m ($2I$) with 2 t/m; C is fixed. Support B settles 5 mm and support C rotates 0.02 rad clockwise. The value 'EI = 20 t/mm²' gives only E (20 t/mm² $=2\times10^7$ t/m²); I is not given. The result is therefore written as: moment = (load part, independent of EI) + EI × (settlement/rotation part), and a numerical example with $I=1\times10^{-3}$ m⁴ ($EI=20\,000$ t m²) is given. Units: tonne, metre.

Step 1: Chord rotations and support rotation

• $\psi_{AB}=\dfrac{0.005}{4}=0.001250$ (B moves down, chord turns clockwise)
• $\psi_{BC}=-\dfrac{0.005}{6}=-0.000833$ (C is higher than B, chord turns anticlockwise)
• $\theta_A=0$, $\theta_C=+0.02$ rad (clockwise)

Step 2: Fixed-end moments (tm, clockwise +)

• AB ($a=1$, $b=3$): $FEM_{AB}=-\dfrac{4\times1\times3^2}{4^2}=-2.250$, $FEM_{BA}=+\dfrac{4\times1^2\times3}{4^2}=0.750$
• BC: $\mp\dfrac{2\times6^2}{12}=\mp6.000$

Step 3: Slope-deflection equations (numerical example, $EI=20\,000$ t m²)

• $M_{AB}=10000.00\left(\theta_{B}\right) - 39.75$
• $M_{BA}=10000.00\left(2\theta_{B}\right) - 36.75$
• $M_{BC}=13333.33\left(2\theta_{B} + 0.02000\right) + 27.33$
• $M_{CB}=13333.33\left(0.04000 + \theta_{B}\right) + 39.33$

Step 4: Equilibrium at B

$M_{BA}+M_{BC}=0$ gives one equation in $\theta_B$:

$$\begin{bmatrix}46666.6667\end{bmatrix}\begin{Bmatrix}\theta_{B}\end{Bmatrix}=\begin{Bmatrix}-257.250\end{Bmatrix}$$ $$\theta_{B}=-0.0055125$$

Step 5: End moments

(For any EI: $M=M_{load}+EI\times(\text{column 3})$.)

Check: $M_{BA}+M_{BC}=0.000$ ✓.

SFD and BMD (EI = 20 000 t m²)

Answer: for $EI=20\,000$ t m²: $M_{AB}=-94.875$, $M_{BA}=-147.000$, $M_{BC}=147.000$, $M_{CB}=499.167$ tm; in general $M=M_{load}+EI\times(\text{movement coefficient})$.