Slope deflection: D rotates, C settles 4 mm

Data and assumptions: the figure is only described. A is the free end of the cantilever AB (3 m, 25 kN at A); BC = 6 m (2.5EI) carries 50 kN at 2 m and 40 kN at 4 m from B and 10 kN/m over the last 2 m; CD = 6 m (3EI) is unloaded; B and C are supports; D is fixed. Support C settles 4 mm; support D rotates $\tfrac{1}{10}$ degree $=0.001745$ rad clockwise (the unit is taken as degrees). EI is not given, so the answer is split into the load part (independent of EI) and the support-movement part (proportional to EI), and a numerical example with $EI=30\,000$ kNm² is worked out.

Step 1: Overhang AB

$M_B=25\times3=75$ kNm (hogging), known moment at B.

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

BC: 50 kN at $a=2$ ($b=4$), 40 kN at $a=4$ ($b=2$) and UDL on the last 2 m: $FEM_{BC}=-65.556$, $FEM_{CB}=70.000$ (sum of the point-load formulas $\mp Pab^2/L^2$, $\pm Pa^2b/L^2$ and the integrated UDL contribution). CD: 0.

Step 3: Support movement

• $\psi_{BC}=+\dfrac{0.004}{6}=0.000667$ (C moves down relative to B, so the chord of BC turns clockwise) and $\psi_{CD}=-\dfrac{0.004}{6}=-0.000667$ (D is higher than C, so the chord of CD turns anticlockwise); $\psi_{AB}$ does not enter (overhang).
• $\theta_D=+0.001745$ rad (clockwise) appears in $M_{CD}$ and $M_{DC}$.

Step 4: Slope-deflection equations ($EI=30\,000$ kNm², numerical)

• $M_{BC}=25000.00\left(2\theta_{B} + \theta_{C}\right) - 115.56$
• $M_{CB}=25000.00\left(2\theta_{C} + \theta_{B}\right) + 20.00$
• $M_{CD}=30000.00\left(2\theta_{C} + 0.00175\right) + 60.00$
• $M_{DC}=30000.00\left(0.00349 + \theta_{C}\right) + 60.00$

Step 5: Equilibrium and solution

Joint B: $M_{BC}=-M_{BA}$ with $M_{BA}=+75$, so $M_{BC}=-75$. Joint C: $M_{CB}+M_{CD}=0$.

$$\begin{bmatrix}50000.0000 & 25000.0000 \\ 25000.0000 & 110000.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}40.556 \\ -132.360\end{Bmatrix}$$ $$\theta_{B}=0.00159387,\quad \theta_{C}=-0.00156551$$

Step 6: Final end moments (kNm)

(The middle column multiplied by EI gives the effect of the movements for any EI: $M=M_{load}+EI\times(\text{column 3})$.)

Check: joint B: $75.00+(-75.00)=0$ ✓; joint C: $-18.43+(18.43)=0$ ✓.

Bending moment diagram (EI = 30 000)

Reactions: B: $H=0.00$, $V=90.57$, $M=0.00$; C: $H=0.00$, $V=21.73$, $M=0.00$; D: $H=0.00$, $V=22.70$, $M=-117.75$.

Answer: for $EI=30\,000$ kNm²: $M_{BA}=75.00$, $M_{BC}=-75.00$, $M_{CB}=-18.43$, $M_{CD}=18.43$, $M_{DC}=117.75$ kNm; in general $M=M_{load}+EI\times(\text{movement coefficient})$.