--- title: "Slope deflection: beam, 360 kN load, fixed end" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:slope-deflection-beam-10kn-20knm-360kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-10kn-20knm-360kn site: BCT Engineering --- # Slope deflection: beam, 360 kN load, fixed end Theory of Structures II · Chapter 3: Displacement method - Asked in: 2075 Chaitra, 10 marks - Video: [Slope Deflection Method for Beam Analysis - Problem No 11 ( Overhanging on the Right side)](https://www.youtube.com/watch?v=APV-GnkxW8U) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-10kn-20knm-360kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Determine end moments and draw bending moment diagram by using the slope deflection method. [Figure: continuous beam; 10 kN at the left end A; AB = 2 m; BC = 10 m (3I) carrying 20 kN/m UDL; CD = 6 m (2I) with 360 kN point load, 3 m from C and 3 m from fixed end D.] ## Short answer Slope-deflection from cantilever moment 20 kNm gives M_BA=20.00, M_BC=-20.00, M_CB=252.09, M_CD=-252.09, M_DC=278.96 kNm. BC sagging maximum is 127.42 kNm at 3.84 m; under the 360 kN load in CD it is 274.48 kNm sagging. ## Answer **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) | End | $M$ (kNm) | End | $M$ (kNm) | |---|---|---|---| | AB | 0.00 | BA | 20.00 | | BC | -20.00 | CB | 252.09 | | CD | -252.09 | DC | 278.96 | 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) | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | AB | 0.00 | -20.00 | - | | BC | -20.00 | -252.09 | 127.42 at 3.84 m | | CD | -252.09 | -278.96 | 274.48 at 3.00 m | 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the moment at B from overhang AB? A. 10/2 = 5 kNm B. 10 x 2 = 20 kNm hogging C. 20 x 2 = 40 kNm D. Zero Answer: B. Load times lever arm: 10 kN x 2 m = 20 kNm, known at joint B. (Key point: Overhang moment) 2. The FEM for CD (360 kN at mid-span, 6 m) is: A. PL/4 = 540 kNm B. PL/12 = 180 kNm C. PL/8 = 270 kNm D. Pab^2/L = 360 Answer: C. PL/8 = 360 x 6 / 8 = 270 kNm. (Key point: FEM CD) 3. Why are only θB and θC unknown? A. A is fixed B. Both spans are equal C. The load is symmetric D. D is fixed so θD = 0, and there is no settlement or sway Answer: D. Fixed D has zero rotation; A is free but handled as a cantilever. (Key point: Unknowns) 4. The joint C equation is: A. MCB + MCD = 0 B. MCB = MCD C. MCB - MCD = 360 D. MCB = 0 Answer: A. No external moment at C, so the sum of the end moments is zero. 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