--- title: "Three moment equation: SFD and BMD of beam" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:three-moment-beam-300kn-50knm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/three-moment-beam-300kn-50knm site: BCT Engineering --- # Three moment equation: SFD and BMD of beam Theory of Structures II · Chapter 2: Force method - Asked in: 2078 Bhadra, 10 marks - Video: [SA60: The Three-Moment Equation for the Analysis of Continuous Beams (Part I)](https://www.youtube.com/watch?v=SHrnZtU11mc) — Dr. Structure - 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-300kn-50knm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Draw shear force and bending moment diagram of the continuous beam using the three moment equation. [Figure: beam ABC, A fixed; 300 kN point load at 4 m from A; B support at 8 m; BC = 10 m with 50 kN/m UDL; C roller.] ## Short answer Apply the three moment equation with an imaginary zero span at the fixed end. MA = -198.44 kNm, MB = -503.12 kNm hogging; reactions are RA = 111.91 kN, RB = 488.40 kN, RC = 199.69 kN; peak sagging BM in BC is 398.75 kNm. ## Answer Reading of the figure: A fixed, span AB = 8 m with a 300 kN load at 4 m from A, B an interior support, span BC = 10 m with 50 kN/m over the whole span, C a roller. EI is constant. The three moment equation is $M_A L_1 + 2M_B(L_1+L_2) + M_C L_2 = -\dfrac{6a_1\bar x_1}{L_1} - \dfrac{6a_2\bar x_2}{L_2}$ with sagging moments positive. The fixed end A is treated by adding an imaginary span of zero length to its left, which makes the slope at A zero. At the roller C the moment is $M_C = 0$. #### Load terms - Span AB: free BM is a triangle with peak $\dfrac{300\times4\times4}{8} = 600$ kNm, area $a_1 = \tfrac12(8)(600) = 2400$ kNm$^2$, $\bar x_1 = 4$ m. $\dfrac{6a_1\bar x_1}{L_1} = \dfrac{6(2400)(4)}{8} = 7200$. - Span BC: UDL, $\dfrac{wL_2^3}{4} = \dfrac{50(10)^3}{4} = 12500$. #### Equations **At A (fixed end):** $$ 2M_A(8) + M_B(8) = -7200 \quad\Rightarrow\quad 16M_A + 8M_B = -7200 $$ **At B:** $$ M_A(8) + 2M_B(8+10) + 0 = -7200 - 12500 \quad\Rightarrow\quad 8M_A + 36M_B = -19700 $$ Solving: $$ M_B = -503.125\ \text{kNm},\qquad M_A = -198.4375\ \text{kNm} $$ Both are hogging. #### Reactions With sagging-positive moments, $M(x) = M_A + R_A x - 300\langle x-4\rangle$ in AB, so $M_B = M_A + 8R_A - 1200$: $$ R_A = \frac{1200 + M_B - M_A}{8} = \frac{1200 + (-503.125) - (-198.4375)}{8} = 111.914\ \text{kN} $$ Shear just left of B $= 111.914 - 300 = -188.086$ kN. In BC, $M(x) = M_B + V_B' x - 25x^2$ with $M_C = 0$ at $x = 10$: $$ V_B' = \frac{2500 - M_B}{10} = 300.3125\ \text{kN},\qquad R_B = 188.086 + 300.312 = 488.398\ \text{kN} $$ $$ R_C = 500 - 300.3125 = 199.6875\ \text{kN} $$ Check: $R_A + R_B + R_C = 800.00$ kN $= 300 + 50(10) = 800$ kN. #### Shear force (kN) | Section | SF | |---|---| | A | +111.91 | | E (just left / right of the 300 kN load) | +111.91 / -188.09 | | B (left / right) | -188.09 / +300.31 | | C | -199.69 | SF is zero in BC at $x = 3.994$ m from C. #### Bending moment (kNm) | Section | BM | |---|---| | A | -198.44 (hogging) | | E (under the 300 kN load) | +249.22 (sagging) | | B | -503.12 (hogging) | | Maximum in BC at 3.99 m from C | +398.75 (sagging) | | C | 0 | Points of contraflexure: 1.77 m from A; 1.33 m beyond E (towards B); and 2.01 m beyond B in BC. **Answer: $M_A = -198.44$ kNm and $M_B = -503.12$ kNm (hogging); $R_A = 111.91$ kN, $R_B = 488.40$ kN, $R_C = 199.69$ kN; peak sagging BM $= 398.75$ kNm in BC.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the fixed end A treated in the three moment equation? A. A is treated as a simple support with M_A = 0 B. An imaginary span of zero length is added to the left of A, which makes the slope at A zero C. The moment at A is taken equal to the moment at B D. A is ignored and only span BC is used Answer: B. The zero-length span with the fixed end gives the equation 2 M_A L1 + M_B L1 = -6 a1 x1/L1. (Key point: Fixed end treatment) 2. For the span BC (10 m, 50 kN/m), what is the load term on the right-hand side? A. wL^3/4 = 50 x 1000/4 = 12500 B. wL^3/8 = 6250 C. wL^2/2 = 2500 D. wL^3 = 50000 Answer: A. For a UDL the term 6 a x_bar/L equals wL^3/4, giving 12500. (Key point: UDL load term) 3. Solving 16 M_A + 8 M_B = -7200 and 8 M_A + 36 M_B = -19700 gives which result? A. M_A = +198.44 kNm and M_B = +503.12 kNm, sagging B. M_A = 0 and M_B = -503.12 kNm C. M_A = -198.44 kNm and M_B = -503.12 kNm, both hogging D. M_A = -503.12 kNm and M_B = -198.44 kNm Answer: C. With sagging positive, the negative values mean the supports hog, and the fixed end carries a hogging moment. (Key point: Moments from the solution) 4. How is the maximum sagging BM in BC located? A. At mid-span of BC, 5 m from C B. At support B, where the shear is maximum C. Where the bending moment is zero D. At the section where the shear force is zero (3.994 m from C), giving 398.75 kNm Answer: D. M is maximum where dM/dx = V = 0. The shear in BC changes sign at x = 3.994 m from C. (Key point: Maximum BM in BC) ## Related in Theory of Structures II - [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 - [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 - [Slope deflection: continuous beam, SFD and BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-50kn-10knm-3ei) — Theory of Structures II, chapter 3 - [Stiffness matrix: continuous beam BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Astiffness-continuous-beam-10kn-30kn-15knm) — Theory of Structures II, chapter 3 - [Three moment equation: beam with overhang](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Athree-moment-beam-100kn-40knm-20kn) — Theory of Structures II, chapter 2 - [ILD: moment at support B, fixed-propped beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-moment-b-propped-2l-l) — Theory of Structures II, chapter 4 ## Related in other subjects - [Beam ABC with wire support: BMD and SFD](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Abeam-wire-support-sfd-bmd) — Strength of Materials, chapter 1 - [AFD, SFD, BMD of frame with UDL and triangular load](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-10knm-5kn-triangular) — Applied Mechanics, chapter 6 - [Beam AFD, SFD, BMD with hinge (80 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Abeam-afd-sfd-bmd-20knm-80kn) — Strength of Materials, chapter 1 - [AFD, SFD, BMD of frame with 40 kN/m UDL](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-40kn-frame) — Applied Mechanics, chapter 6