--- title: "Fixed beam: end moments from support rotation or settlement" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:fixed-beam-end-moments-rotation-settlement" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/fixed-beam-end-moments-rotation-settlement site: BCT Engineering --- # Fixed beam: end moments from support rotation or settlement Theory of Structures II · Chapter 3: Displacement method - Asked in: 2076 Chaitra, 6 marks; 2076 Asoj, 6 marks; 2075 Chaitra, 6 marks; 2069 Chaitra, 7 marks (asked 4 times) - Video: [Fixed Beam with Sinking of Supports ( Derivation For Fixed End Moments & Reactions )](https://www.youtube.com/watch?v=87Zen_ENOqo) — 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/fixed-beam-end-moments-rotation-settlement - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Determine the end moments (and reactions) of a single span fixed beam of span L and flexural rigidity EI when (a) one end rotates by an angle $\theta$, and (b) one support settles down by $\Delta$ without rotation. ## Short answer For rotation θ at one end: M_near = 4EIθ/L, M_far = 2EIθ/L, shears = 6EIθ/L². For support settlement Δ without rotation: both end moments = –6EIΔ/L² and vertical reactions are 12EIΔ/L³. ## Answer For a prismatic beam of span $L$ and rigidity $EI$, fixed against rotation and translation at both ends, one end is then given a rotation or a settlement. Use the slope-deflection equations (clockwise end moments positive; no load on the span, so FEM = 0): $$ M_{AB}=\frac{2EI}{L}\left(2\theta_A+\theta_B-\frac{3\Delta}{L}\right),\qquad M_{BA}=\frac{2EI}{L}\left(2\theta_B+\theta_A-\frac{3\Delta}{L}\right) $$ #### (a) End A rotates by $\theta$ (clockwise), no settlement Put $\theta_A=\theta$, $\theta_B=0$, $\Delta=0$: $$ M_{AB}=\frac{4EI\theta}{L},\qquad M_{BA}=\frac{2EI\theta}{L} $$ Both are clockwise on the member ends. The moment at the far end is half of that at the rotated end (carry-over factor 1/2). **Reactions.** Taking moments about B for the whole member (clockwise positive), with $V_A$ the upward force on the member at A: $$ M_{AB}+M_{BA}+V_AL=0\;\Rightarrow\;V_A=-\frac{6EI\theta}{L^2} $$ So at A the reaction is $6EI\theta/L^2$ **downward** and at B it is $6EI\theta/L^2$ **upward** (the two forces form a couple that balances the end moments). ```text A: M = 4EI.theta/L (clockwise) B: M = 2EI.theta/L (clockwise) A: V = 6EI.theta/L^2 down B: V = 6EI.theta/L^2 up ``` #### (b) End B settles down by $\Delta$, no rotation Put $\theta_A=\theta_B=0$ and $\Delta$ downward (positive chord rotation $\psi=\Delta/L$): $$ M_{AB}=M_{BA}=-\frac{6EI\Delta}{L^2} $$ Both are anticlockwise, which means hogging at A and sagging at B. **Reactions.** From moments about B: $$ M_{AB}+M_{BA}+V_AL=0\;\Rightarrow\;V_A=\frac{12EI\Delta}{L^3}\ (\text{upward}),\qquad V_B=\frac{12EI\Delta}{L^3}\ (\text{downward}) $$ The settlement pulls the end B down, so the support must exert a downward force at B and an upward force at A. #### Summary | Case | $M_{AB}$ | $M_{BA}$ | End shears | |---|---|---|---| | (a) A rotates $\theta$ | $+4EI\theta/L$ | $+2EI\theta/L$ | $6EI\theta/L^2$ | | (b) B settles $\Delta$ | $-6EI\Delta/L^2$ | $-6EI\Delta/L^2$ | $12EI\Delta/L^3$ | These are exactly the stiffness coefficients $4EI/L$, $2EI/L$, $6EI/L^2$ and $12EI/L^3$ used in the stiffness (displacement) method. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When end A of a fixed beam rotates by theta (clockwise), what are the end moments? A. M_AB = 2EI theta/L and M_BA = 4EI theta/L B. M_AB = 4EI theta/L and M_BA = 2EI theta/L C. M_AB = M_BA = 6EI theta/L D. M_AB = 4EI theta/L and M_BA = 0 Answer: B. Putting theta_A = theta, theta_B = 0 and Delta = 0 gives 4EI theta/L at A and 2EI theta/L at B, a carry-over factor of 1/2. (Key point: Rotation of one end) 2. When support B settles by Delta with no rotation, what are the end moments? A. Both equal to 6EI Delta/L^2, anticlockwise (hogging at A, sagging at B) B. M_AB = 12EI Delta/L^3 and M_BA = 6EI Delta/L^2 C. Zero at A and 6EI Delta/L^2 at B D. Opposite in sign, each 3EI Delta/L^2 Answer: A. With psi = Delta/L, M_AB = M_BA = -6EI Delta/L^2, so both end moments are equal and anticlockwise. (Key point: Settlement moments) 3. What is the end shear in the settlement case (b)? A. 6EI Delta/L^2 B. 12EI Delta/L^3 C. 4EI Delta/L^2 D. 2EI Delta/L^3 Answer: B. Moments about B give V = (M_AB + M_BA)/L magnitude = 12EI Delta/L^3, upward at A and downward at B. 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