Slope deflection: B sinks, C rises

Data and assumptions: $EI=200\times10^6\times80\times10^{-6}=16\,000$ kNm². AB = 4 m (1.5EI) with 12 kN at mid-span; BC = 6 m (2EI) with 10 kN at mid-span; CD = 4 m (EI) with 5 kN/m; A and D are fixed. B sinks 10 mm, C rises 20 mm.

Chord rotations (clockwise +)

• $\psi_{AB}=\dfrac{0.010}{4}=0.00250$ (B goes down)
• $\psi_{BC}=\dfrac{-(0.020+0.010)}{6}=-0.00500$ (C is 30 mm higher than B, so the chord turns anticlockwise)
• $\psi_{CD}=\dfrac{0.020}{4}=0.00500$ (D is 20 mm lower than C)

Fixed-end moments (kNm, clockwise +)

• AB: $\mp\dfrac{12\times4}{8}=\mp6$
• BC: $\mp\dfrac{10\times6}{8}=\mp7.5$
• CD: $\mp\dfrac{5\times4^2}{12}=\mp6.67$

Slope-deflection equations

• $M_{AB}=12000.00\left(\theta_{B}\right) - 96.00$
• $M_{BA}=12000.00\left(2\theta_{B}\right) - 84.00$
• $M_{BC}=10666.67\left(2\theta_{B} + \theta_{C}\right) + 152.50$
• $M_{CB}=10666.67\left(2\theta_{C} + \theta_{B}\right) + 167.50$
• $M_{CD}=8000.00\left(2\theta_{C}\right) - 126.67$
• $M_{DC}=8000.00\left(\theta_{C}\right) - 113.33$

Equilibrium and solution

$M_{BA}+M_{BC}=0$ and $M_{CB}+M_{CD}=0$:

$$\begin{bmatrix}45333.3333 & 10666.6667 \\ 10666.6667 & 37333.3333\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-68.500 \\ -40.833\end{Bmatrix}$$ $$\theta_{B}=-0.00134403,\quad \theta_{C}=-0.000709741$$

Final end moments (kNm)

Checks:

• Joint B: (-116.26) + (116.26) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (138.02) + (-138.02) = 0.00 kNm (applied clockwise moment 0.00) ✓

Bending moment diagram ordinates (sagging +, kNm)

Deflected shape

• A and D: fixed (zero deflection, zero slope).
• B: moves down 10 mm; rotation $\theta_B=-0.00134$ rad (clockwise +).
• C: moves up 20 mm; rotation $\theta_C=-0.00071$ rad.
• The curve is concave upward (sagging) where the BM is positive and concave downward (hogging) where it is negative; the points of contraflexure (BM = 0) are at: AB at 1.78 m from its first end, BC at 3.09 m from its first end, CD at 1.99 m from its first end.
• The curve falls from A to B, rises across BC from the low point at B to the raised support C, and then falls again to the fixed end D. At every fixed end the tangent is horizontal.

Reactions (kN, kNm): A: $H=0.00$, $V=63.10$, $M=112.13$; B: $H=0.00$, $V=-88.48$, $M=0.00$; C: $H=0.00$, $V=121.64$, $M=0.00$; D: $H=0.00$, $V=-54.26$, $M=119.01$.

Answer: $M_{AB}=-112.13$, $M_{BA}=-116.26$, $M_{BC}=116.26$, $M_{CB}=138.02$, $M_{CD}=-138.02$, $M_{DC}=-119.01$ kNm.