--- title: "Three moment equation: beam with overhang" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:three-moment-beam-100kn-40knm-20kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/three-moment-beam-100kn-40knm-20kn site: BCT Engineering --- # Three moment equation: beam with overhang Theory of Structures II · Chapter 2: Force method - Asked in: 2072 Kartik, 7 marks - Video: [Problem No 10 on Clapeyron's Theorem of Three Moments ( With Overhanging on the right side)](https://www.youtube.com/watch?v=qMsPjvmqXEU) — 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/three-moment-beam-100kn-40knm-20kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine the support moments and draw bending moment diagram of the continuous beam shown in figure below by using the three moment equation. [Figure: beam 1-2-3, 1 fixed; 100 kN at 4 m from 1, 2 at 8 m (I); span 2-3 = 10 m (2I) with 40 kN/m UDL; overhang of 2 m with 20 kN at the end.] ## Short answer Apply three-moment equation accounting for varying I and known overhang moment. Support moments: M1 = -13.636 kNm, M2 = -272.727 kNm, M3 = -40 kNm (hogging negative). ## Answer Reading of the figure: the beam is fixed at 1; span 1-2 = 8 m with $I$ and a 100 kN load at 4 m from 1; span 2-3 = 10 m with $2I$ and 40 kN/m; the overhang 3-4 is 2 m with 20 kN at the free end. Supports 2 and 3 are on rollers. $E$ is constant. #### Known moment at 3 The overhang is statically determinate: $M_3 = -20\times2 = -40$ kNm (hogging). #### Three moment equation with different $I$ $$ M_1\frac{L_1}{I_1} + 2M_2\left(\frac{L_1}{I_1}+\frac{L_2}{I_2}\right) + M_3\frac{L_2}{I_2} = -\frac{6a_1\bar x_1}{I_1L_1} - \frac{6a_2\bar x_2}{I_2L_2} $$ Load terms: - Span 1-2: free BM peak $= \dfrac{100\times4\times4}{8} = 200$, $a_1 = \tfrac12(8)(200) = 800$, $\bar x_1 = 4$: $\dfrac{6a_1\bar x_1}{L_1} = 2400$. - Span 2-3: $\dfrac{wL_2^3}{4} = \dfrac{40(10)^3}{4} = 10000$. **Equation at 1 (fixed end):** $2M_1\dfrac{8}{I} + M_2\dfrac{8}{I} = -\dfrac{2400}{I}$ $$ 2M_1 + M_2 = -300 $$ **Equation at 2:** multiply by $I$: $8M_1 + 2M_2\left(8 + \dfrac{10}{2}\right) + M_3\dfrac{10}{2} = -2400 - \dfrac{10000}{2}$ $$ 8M_1 + 26M_2 + 5(-40) = -7400 \ \Rightarrow\ 8M_1 + 26M_2 = -7200 $$ Substituting $M_1 = \dfrac{-300 - M_2}{2}$: $-1200 - 4M_2 + 26M_2 = -7200$, so $$ M_2 = -272.727\ \text{kNm},\qquad M_1 = -13.636\ \text{kNm} $$ ($M_1$ is a small hogging moment at the fixed end.) #### Reactions With sagging-positive moments, in span 1-2: $M_2 = M_1 + 8R_1 - 100(4)$, so $$ R_1 = \frac{400 + M_2 - M_1}{8} = \frac{400 + (-272.727) - (-13.636)}{8} = 17.614\ \text{kN} $$ In span 2-3 (taking $x$ from 2): $M_3 = M_2 + 10V_{2R} - 40\frac{10^2}{2}$, so $$ V_{2R} = \frac{2000 + M_3 - M_2}{10} = \frac{2000 - 40 - (-272.727)}{10} = 223.273\ \text{kN} $$ $$ R_2 = (100 - 17.614) + 223.273 = 305.659\ \text{kN},\qquad R_3 = 400 - 223.273 + 20 = 196.727\ \text{kN} $$ Check: $R_1 + R_2 + R_3 = 520.00$ kN $= 100 + 400 + 20$. #### Bending moment (kNm) | Section | BM | |---|---| | 1 (fixed end) | -13.64 | | Under the 100 kN load | 56.82 | | 2 | -272.73 | | Maximum in span 2-3 (5.58 m from 2) | 350.41 | | 3 | -40.00 | | 4 (free end) | 0 | **Answer: $M_1 = -13.636$ kNm, $M_2 = -272.727$ kNm, $M_3 = -40$ kNm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is M3 at support 3 found and what is its value? A. From the three moment equation, 0 B. From statics of the overhang: -20 x 2 = -40 kNm (hogging) C. From the UDL: -40 x 10^2/8 D. From the fixed end: 0 Answer: B. The overhang is determinate, so the moment at support 3 is the 20 kN load times 2 m, hogging. (Key point: Overhang moment) 2. How does the equation account for span 2-3 having 2I against I? A. It is ignored B. It changes the loads C. The length terms are used as L/I, so span 2-3 uses 10/(2I) D. Only the support moments change Answer: C. The general equation uses L1/I1 and L2/I2 coefficients, so a stiffer span gets a smaller L/I. (Key point: Three moment with different I) 3. Which supports the correctness of M1 = -13.64, M2 = -272.73 kNm? A. Reactions sum to 520 kN = 100 + 400 + 20 B. M1 equals M2 C. R1 equals R3 D. Reactions sum to 100 kN Answer: A. R1 + R2 + R3 = 17.614 + 305.659 + 196.727 = 520 kN, equal to the total load. (Key point: Result and check) ## Related in Theory of Structures II - [ILD: bending moment at fixed support, 10 m beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-fixed-support-10m-1) — Theory of Structures II, chapter 4 - [Fixed beam: end moments from support rotation or settlement](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Afixed-beam-end-moments-rotation-settlement) — Theory of Structures II, chapter 3, asked 4× - [Slope deflection: beam, 360 kN load, fixed end](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-10kn-20knm-360kn) — Theory of Structures II, chapter 3 - [Three moment equation: SFD and BMD of beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Athree-moment-beam-300kn-50knm) — Theory of Structures II, chapter 2 - [Castigliano: fixed-end moment, 15 kN/m, L = 5 m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-propped-udl-15knm-5m) — Theory of Structures II, chapter 2, asked 2× - [Three moment theorem: SFD and BMD of beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Athree-moment-beam-100kn-60knm-20kn) — Theory of Structures II, chapter 2 ## Related in other subjects - [Beam thrust and bending moment diagrams with max moments](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Abeam-thrust-bmd-2kn-5kn-1knm) — Strength of Materials, chapter 1 - [Moment area: overhang beam free-end deflection](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-overhang-10kn-m-2078) — Theory of Structures I, chapter 4 - [Conjugate beam: free end of overhang, UDL and point load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-10kn-4m-3m-2071b) — Theory of Structures I, chapter 4 - [T-beam with overhang: max bending stress](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-4kn-m-10kn-overhang) — Strength of Materials, chapter 7