Slope deflection: B sinks, A rotates

Data: $E=5\times10^5$ MPa $=5\times10^8$ kN/m², $I=3\times10^7$ mm⁴ $=3\times10^{-5}$ m⁴, so $EI=15000$ kNm². AB = 5 m (1.5I) with 5 kN/m; BC = 4 m (2I) with 15 kN at mid-span; CD = 4 m (I) with 5 kN/m. A and D are fixed; support A rotates by $5^\circ=0.08727$ rad anticlockwise and B sinks 7.5 mm.

Support movements

• $\theta_A=-0.08727$ rad (anticlockwise, so negative in the clockwise-positive convention)
• $\psi_{AB}=\dfrac{0.0075}{5}=0.00150$, $\psi_{BC}=-\dfrac{0.0075}{4}=-0.00187$ (C is higher than B), $\psi_{CD}=0$

Fixed-end moments (kNm, clockwise +)

• AB: $\mp\dfrac{5\times5^2}{12}=\mp10.42$
• BC: $\mp\dfrac{15\times4}{8}=\mp7.5$
• CD: $\mp\dfrac{5\times4^2}{12}=\mp6.67$

Slope-deflection equations

With $\dfrac{2EI}{L}$: AB = 9000, BC = 15000, CD = 7500 kNm/rad.

• $M_{AB}=9000.00\left(-0.17453 + \theta_{B}\right) - 50.92$
• $M_{BA}=9000.00\left(2\theta_{B} - 0.08727\right) - 30.08$
• $M_{BC}=15000.00\left(2\theta_{B} + \theta_{C}\right) + 76.88$
• $M_{CB}=15000.00\left(2\theta_{C} + \theta_{B}\right) + 91.88$
• $M_{CD}=7500.00\left(2\theta_{C}\right) - 6.67$
• $M_{DC}=7500.00\left(\theta_{C}\right) + 6.67$ (The constants include the FEM, the effect of the rotation of A and the chord rotation.)

Equilibrium

Joint B: $M_{BA}+M_{BC}=0$; joint C: $M_{CB}+M_{CD}=0$.

$$\begin{bmatrix}48000.0000 & 15000.0000 \\ 15000.0000 & 45000.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}738.606 \\ -85.208\end{Bmatrix}$$ $$\theta_{B}=0.0178374,\quad \theta_{C}=-0.00783933$$

Final end moments (kNm)

Checks:

• Joint B: (-494.41) + (494.41) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (124.26) + (-124.26) = 0.00 kNm (applied clockwise moment 0.00) ✓

Bending moment diagram ordinates (sagging +, kNm)

Reactions: A: $H=0.00$, $V=403.62$, $M=1461.18$; B: $H=0.00$, $V=-525.78$, $M=0.00$; C: $H=0.00$, $V=216.26$, $M=0.00$; D: $H=0.00$, $V=-34.10$, $M=52.13$.

The rotation of 5° of support A is very large for a 5 m span, so the end moments are dominated by the support movements.

Answer: $M_{AB}=-1461.18$, $M_{BA}=-494.41$, $M_{BC}=494.41$, $M_{CB}=124.26$, $M_{CD}=-124.26$, $M_{DC}=-52.13$ kNm.