--- title: "Three moment equation: derive and apply to fixed beam" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:three-moment-derive-fixed-beam-udl" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/three-moment-derive-fixed-beam-udl site: BCT Engineering --- # Three moment equation: derive and apply to fixed beam Theory of Structures II · Chapter 2: Force method - Asked in: 2074 Chaitra, 6 marks - Video: [(Three moment equation) Example #1:[1/2] Fixed Ended Beam having uniformly distributed load & moment](https://www.youtube.com/watch?v=9UpkACuwDHs) — Engr Pogs - 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-derive-fixed-beam-udl - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Derive the three moment equation and use it to solve a single span fixed beam with uniform distributed load throughout the span. ## Short answer The three-moment equation is derived by equating slopes over two adjacent spans using moment-area method. For a fixed beam under UDL w over span L, adding imaginary zero-length end spans yields fixed-end moments -wL^2/12 at each end, reactions wL/2, and midspan sagging moment +wL^2/24. ## Answer #### Derivation Consider two adjacent spans AB ($L_1$) and BC ($L_2$) of a continuous beam with supports at the same level and constant $EI$. Let $M_A, M_B, M_C$ be the support moments (sagging positive). On each span the BM diagram is the free (simple beam) diagram plus a trapezoid of support moments. Let $a_1, a_2$ be the areas of the free BM diagrams, $\bar x_1$ the distance of the centroid of $a_1$ from A, and $\bar x_2$ the distance of the centroid of $a_2$ from C. **Slope at B from span AB** (moment of the $M/EI$ area about A equals the deviation of A from the tangent at B, which is $\theta_BL_1$ for level supports): $$ \theta_{B1} = \frac{1}{EI}\left[\frac{a_1\bar x_1}{L_1} + \frac{M_AL_1}{6} + \frac{M_BL_1}{3}\right] $$ **Slope at B from span BC** (taking moments about C): $$ \theta_{B2} = -\frac{1}{EI}\left[\frac{a_2\bar x_2}{L_2} + \frac{M_CL_2}{6} + \frac{M_BL_2}{3}\right] $$ The beam is continuous over B, so $\theta_{B1} = \theta_{B2}$. Multiplying by $6$ gives the **three moment equation**: $$ M_AL_1 + 2M_B(L_1+L_2) + M_CL_2 = -\frac{6a_1\bar x_1}{L_1} - \frac{6a_2\bar x_2}{L_2} $$ #### Application: fixed beam AB with a UDL $w$ over the span $L$ A fixed end is treated as a continuous span of zero length beyond the end. Add an imaginary span $A'A$ of length $L_0 = 0$ at A and $BB'$ of length $L_0 = 0$ at B; the end moments $M_A$ and $M_B$ are the unknowns. For the span AB: $a = \dfrac23\cdot L\cdot\dfrac{wL^2}{8} = \dfrac{wL^3}{12}$, $\bar x = \dfrac L2$, so $\dfrac{6a\bar x}{L} = \dfrac{wL^3}{4}$. **Equation at A** (spans $A'A$ and $AB$, the first has zero length): $$ 2M_AL + M_BL = -\frac{wL^3}{4} $$ **Equation at B** (spans $AB$ and $BB'$): $$ M_AL + 2M_BL = -\frac{wL^3}{4} $$ Subtracting gives $M_A = M_B$. Then $3M_AL = -\dfrac{wL^3}{4}$, so $$ M_A = M_B = -\frac{wL^2}{12} $$ #### Results - Reactions: $R_A = R_B = \dfrac{wL}{2}$ (symmetry). - Mid-span moment: $\dfrac{wL^2}{8} - \dfrac{wL^2}{12} = +\dfrac{wL^2}{24}$ (sagging). - Points of contraflexure at $x = \dfrac{L}{2}\left(1 \pm \dfrac{1}{\sqrt3}\right)$, that is $0.211L$ from each end. For example, with $w = 20$ kN/m and $L = 6$ m: $M_A = M_B = -60$ kNm, $R_A = R_B = 60$ kN, mid-span $+30$ kNm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On which condition is the three moment equation derived? A. Deflection at B is zero B. Moment at B is zero C. The slope at B from span AB equals that from span BC D. The reactions are equal Answer: C. The beam is continuous over support B, so the slope there is the same from both spans. (Key point: Basis of derivation) 2. How is a fixed end handled with the three moment equation? A. Moment is taken as zero B. By an imaginary span of zero length beyond the end C. By doubling the span D. By ignoring EI Answer: B. A fixed end is like a continuous span of zero length, giving 2M_A L + M_B L = -wL^3/4 at A. (Key point: Fixed end treatment) 3. What is the fixed end moment for a UDL w on span L? A. wL^2/8 B. wL^2/24 C. wL^2/2 D. wL^2/12 (hogging) Answer: D. M_A = M_B = -wL^2/12; the 6a x/L term is wL^3/4. (Key point: Fixed end moment) 4. What is the mid-span moment? A. +wL^2/24 sagging B. +wL^2/12 C. -wL^2/24 D. +wL^2/8 Answer: A. wL^2/8 - wL^2/12 = +wL^2/24 (sagging). 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