Slope deflection: beam with settlement of C

Assumptions (dimensions are not fully legible): A is a pin (free to rotate), AB = 3.5 m with the 2 t load 2 m from A; BC = 4 m (3I) with 3 t/m; CD = 4 m (I) with the 5 t load 3 m from C (1 m from D); D is fixed; B and C are simple supports. Units: tonne, metre. $EI=1\times10^9\ \text{t mm}^2=1000$ t m². Support C settles $\delta=4$ mm = 0.004 m.

Chord rotations (clockwise +, a member whose right end sinks rotates clockwise)

$\psi_{BC}=\dfrac{0.004}{4}=0.0010$ (C goes down relative to B), $\psi_{CD}=-\dfrac{0.004}{4}=-0.0010$ (D is higher than C). $\psi_{AB}=0$.

Fixed-end moments (tm, clockwise +)

• AB: $FEM_{AB}=-\dfrac{2\times2\times1.5^2}{3.5^2}=-0.735$, $FEM_{BA}=+\dfrac{2\times2^2\times1.5}{3.5^2}=0.980$
• BC: $\mp\dfrac{3\times4^2}{12}=\mp4.000$
• CD: $FEM_{CD}=-\dfrac{5\times3\times1^2}{4^2}=-0.938$, $FEM_{DC}=+\dfrac{5\times3^2\times1}{4^2}=2.812$

Slope-deflection equations ($M_{ij}=\tfrac{2EI}{L}(2\theta_i+\theta_j-3\psi)+FEM_{ij}$, numerical)

• $M_{AB}=571.43\left(2\theta_{A} + \theta_{B}\right) - 0.73$
• $M_{BA}=571.43\left(2\theta_{B} + \theta_{A}\right) + 0.98$
• $M_{BC}=1500.00\left(2\theta_{B} + \theta_{C}\right) - 8.50$
• $M_{CB}=1500.00\left(2\theta_{C} + \theta_{B}\right) - 0.50$
• $M_{CD}=500.00\left(2\theta_{C}\right) + 0.56$
• $M_{DC}=500.00\left(\theta_{C}\right) + 4.31$

Joint equilibrium

$M_{AB}=0$; $M_{BA}+M_{BC}=0$; $M_{CB}+M_{CD}=0$ (θ in radians)

$$\begin{bmatrix}1142.8571 & 571.4286 & 0.0000 \\ 571.4286 & 4142.8571 & 1500.0000 \\ 0.0000 & 1500.0000 & 4000.0000\end{bmatrix}\begin{Bmatrix}\theta_{A} \\ \theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}0.735 \\ 7.520 \\ -0.062\end{Bmatrix}$$ $$\theta_{A}=-0.000446259,\quad \theta_{B}=0.00217823,\quad \theta_{C}=-0.000832462$$

Final end moments (tm, clockwise on the member end +)

Checks:

• Pin A: $M_{AB}=0.00$ tm ✓
• Joint B: (3.21) + (-3.21) = 0.00 tm ✓
• Joint C: (0.27) + (-0.27) = 0.00 tm ✓

Bending moments (sagging +, tm)

Reactions (t, tm): A: $H=0.00$, $V=-0.06$, $M=0.00$; B: $H=0.00$, $V=8.80$, $M=0.00$; C: $H=0.00$, $V=5.61$, $M=0.00$; D: $H=0.00$, $V=4.66$, $M=-3.90$.

Answer: $M_{BA}=3.214$, $M_{BC}=-3.214$, $M_{CB}=0.270$, $M_{CD}=-0.270$, $M_{DC}=3.896$ tm; $M_{AB}=0$.