--- title: "Slope deflection: BMD and SFD of beam" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:slope-deflection-beam-30knm-20knm-100kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/slope-deflection-beam-30knm-20knm-100kn site: BCT Engineering --- # Slope deflection: BMD and SFD of beam Theory of Structures II · Chapter 3: Displacement method - Asked in: 2067 Asar, 15 marks - Video: [Slope Deflection Method for Beam Analysis - Problem No 8](https://www.youtube.com/watch?v=tuiMskZlEXA) — 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-30knm-20knm-100kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Use the slope deflection method to draw bending moment and shear force diagrams of the beam. [Figure: continuous beam; 30 kNm moment at the left end; spans 2 m, 6 m (2I with 20 kN/m UDL), 10 m (3I), 6 m (3I with 100 kN point load), 6 m, right end fixed.] ## Short answer Use slope-deflection with overhang AB transmitting 30 kNm to joint B and joint equilibria at B,C,D,E. Final moments in kNm: BA -30, BC 30, CB 45.29, CD -45.29, DC 16.88, DE -16.88, ED 59.46, EF -59.46, FE -29.73. Maximum sagging is 111.83 kNm under the 100 kN load. ## Answer **Assumptions (the figure is only described):** the continuous beam has a free left end A with a 30 kNm clockwise couple, a 2 m overhang AB (EI), then BC = 6 m (2I) with 20 kN/m, CD = 10 m (3I) unloaded, DE = 6 m (3I) with 100 kN at mid-span (3 m from D), and EF = 6 m (3I) unloaded with the fixed end at F. B, C, D and E are simple supports. No settlement. #### Step 1: Overhang AB The couple at the free end A is transmitted unchanged along the overhang, so the moment at B is 30 kNm (sagging). For the right-hand end B of member AB this is an anticlockwise end moment, i.e. $M_{BA}=-30.00$ kNm in the clockwise-positive convention. It is a known external moment on joint B. #### Step 2: Fixed-end moments (clockwise +) - BC: $\mp\dfrac{20\times6^2}{12}=\mp60$ kNm - DE: $\mp\dfrac{PL}{8}=\mp\dfrac{100\times6}{8}=\mp75$ kNm - CD, EF: 0 #### Step 3: Slope-deflection equations - $M_{BC}=\frac{2(2EI)}{6}\left(2\theta_{B} + \theta_{C}\right) - 60.00$ - $M_{CB}=\frac{2(2EI)}{6}\left(2\theta_{C} + \theta_{B}\right) + 60.00$ - $M_{CD}=\frac{2(3EI)}{10}\left(2\theta_{C} + \theta_{D}\right)$ - $M_{DC}=\frac{2(3EI)}{10}\left(2\theta_{D} + \theta_{C}\right)$ - $M_{DE}=\frac{2(3EI)}{6}\left(2\theta_{D} + \theta_{E}\right) - 75.00$ - $M_{ED}=\frac{2(3EI)}{6}\left(2\theta_{E} + \theta_{D}\right) + 75.00$ - $M_{EF}=\frac{2(3EI)}{6}\left(2\theta_{E}\right)$ - $M_{FE}=\frac{2(3EI)}{6}\left(\theta_{E}\right)$ #### Step 4: Joint equilibrium (B, C, D, E) Joint B: $M_{BC}+M_{BA}=0$ with $M_{BA}=-30.00$. Joints C, D, E: $\sum M=0$. $$ EI\begin{bmatrix}1.3333 & 0.6667 & 0.0000 & 0.0000 \\ 0.6667 & 2.5333 & 0.6000 & 0.0000 \\ 0.0000 & 0.6000 & 3.2000 & 1.0000 \\ 0.0000 & 0.0000 & 1.0000 & 4.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \theta_{D} \\ \theta_{E}\end{Bmatrix}=\begin{Bmatrix}90.000 \\ -60.000 \\ 75.000 \\ -75.000\end{Bmatrix} $$ $$ EI\theta_{B}=97.3532,\quad EI\theta_{C}=-59.7064,\quad EI\theta_{D}=43.9233,\quad EI\theta_{E}=-29.7308 $$ #### Step 5: Final end moments (kNm) | End | $M$ (kNm) | End | $M$ (kNm) | |---|---|---|---| | AB | 0.00 | BA | -30.00 | | BC | 30.00 | CB | 45.29 | | CD | -45.29 | DC | 16.88 | | DE | -16.88 | ED | 59.46 | | EF | -59.46 | FE | -29.73 | Checks: - Joint B: (-30.00) + (30.00) = 0.00 kNm (applied clockwise moment 0.00) ✓ - Joint C: (45.29) + (-45.29) = 0.00 kNm (applied clockwise moment 0.00) ✓ - Joint D: (16.88) + (-16.88) = 0.00 kNm (applied clockwise moment 0.00) ✓ - Joint E: (59.46) + (-59.46) = 0.00 kNm (applied clockwise moment 0.00) ✓ #### Bending moment diagram (sagging +, kNm) | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | AB | 0.00 | 30.00 | - | | BC | 30.00 | -45.29 | 86.29 at 2.37 m | | CD | -45.29 | -16.88 | - | | DE | -16.88 | -59.46 | 111.83 at 3.00 m | | EF | -59.46 | 29.73 | - | BM under the 100 kN load: 111.83 kNm. #### Shear force diagram (kN, positive when the left part is pushed upward) | Span | Shear at left end | Shear at right end | |---|---|---| | AB | 0.00 | 0.00 | | BC | 47.45 | -72.55 | | CD | 2.84 | 2.84 | | DE | 42.90 | -57.10 | | EF | 14.87 | 14.87 | The shear in BC decreases linearly by 20 kN/m, and in DE it changes by 100 kN at mid-span. Reactions: B: $H=0.00$, $V=47.45$, $M=0.00$; C: $H=0.00$, $V=75.39$, $M=0.00$; D: $H=0.00$, $V=40.06$, $M=0.00$; E: $H=0.00$, $V=71.96$, $M=0.00$; F: $H=0.00$, $V=-14.87$, $M=29.73$. **Answer: $M_{BC}=30.00$, $M_{CB}=45.29$, $M_{CD}=-45.29$, $M_{DC}=16.88$, $M_{DE}=-16.88$, $M_{ED}=59.46$, $M_{EF}=-59.46$, $M_{FE}=-29.73$ kNm (clockwise end moments +); the BMD and SFD ordinates are tabulated above.** ## 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 the 30 kNm couple at free end A? A. 30 kNm, transmitted unchanged along the overhang as a known moment B. 15 kNm C. 60 kNm D. Zero Answer: A. A couple at a free end is carried unchanged, so M_B = 30 kNm. (Key point: Overhang couple) 2. What are the FEMs of BC (20 kN/m, 6 m) and DE (100 kN at mid-span, 6 m)? A. 60 kNm and 75 kNm B. 120 and 150 C. 30 and 37.5 D. 60 and 150 Answer: A. wL^2/12 = 20x36/12 = 60 and PL/8 = 100x6/8 = 75. (Key point: FEM) 3. Which unknowns are solved? A. theta_B, theta_C, theta_D, theta_E (F is fixed) B. theta_F only C. All including theta_F D. Only theta_B Answer: A. F is fixed so its rotation is zero; four joint rotations give a 4x4 system. (Key point: Unknowns) 4. For BC (2I, 6 m) the slope-deflection coefficient is: A. 2E(2I)/6 B. 2E(3I)/6 C. 4E(2I)/6 D. E(2I)/6 Answer: A. The member stiffness 2EI/L uses its own I = 2I and L = 6. 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