Slope deflection: beam, 360 kN load, fixed end

Data read from the figure: A is the free end of a 2 m cantilever AB with a 10 kN load at A; B and C are simple supports; D is fixed. BC = 10 m (3I) with 20 kN/m; CD = 6 m (2I) with a 360 kN load at mid-span (3 m from C and D). No settlement.

Step 1: Overhang AB

$M_B=10\times2=20$ kNm (hogging); it is a known moment at joint B.

Step 2: Fixed-end moments (clockwise +)

• BC: $\mp\dfrac{wL^2}{12}=\mp\dfrac{20\times10^2}{12}=\mp166.67$ kNm
• CD: $\mp\dfrac{PL}{8}=\mp\dfrac{360\times6}{8}=\mp270$ kNm

Step 3: Slope-deflection equations ($\psi=0$, $\theta_D=0$)

• $M_{BC}=\frac{2(3EI)}{10}\left(2\theta_{B} + \theta_{C}\right) - 166.67$
• $M_{CB}=\frac{2(3EI)}{10}\left(2\theta_{C} + \theta_{B}\right) + 166.67$
• $M_{CD}=\frac{2(2EI)}{6}\left(2\theta_{C}\right) - 270.00$
• $M_{DC}=\frac{2(2EI)}{6}\left(\theta_{C}\right) + 270.00$

Step 4: Joint equations

• Joint B: $M_{BA}+M_{BC}=0$, with $M_{BA}=+20$ kNm (cantilever moment, clockwise on the member end), so $M_{BC}=-20$ kNm
• Joint C: $M_{CB}+M_{CD}=0$

$$EI\begin{bmatrix}1.2000 & 0.6000 \\ 0.6000 & 2.5333\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}146.667 \\ 103.333\end{Bmatrix}$$ $$EI\theta_{B}=115.5058,\quad EI\theta_{C}=13.4328$$

Step 5: Final end moments (kNm)

Checks:

• Joint B: (20.00) + (-20.00) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (252.09) + (-252.09) = 0.00 kNm (applied clockwise moment 0.00) ✓

Bending moment diagram ordinates (sagging +, kNm)

Reactions: B: $H=0.00$, $V=86.79$, $M=0.00$; C: $H=0.00$, $V=298.73$, $M=0.00$; D: $H=0.00$, $V=184.48$, $M=-278.96$.

Answer: end moments $M_{BC}=-20.00$, $M_{CB}=252.09$, $M_{CD}=-252.09$, $M_{DC}=278.96$ kNm; BM: 20.00 (hogging) at B, 252.09 (hogging) at C, 278.96 (hogging) at D, span BC maximum 127.42 kNm (sagging), CD maximum 274.48 kNm under the 360 kN load.