--- title: "Moment-area theorems with proof" subject: "Strength of Materials" chapter: 8 key: "semester_4/strength_of_materials_mechanical:moment-area-theorems-statement-proof" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/moment-area-theorems-statement-proof site: BCT Engineering --- # Moment-area theorems with proof Strength of Materials · Chapter 8: Deflection of Beams by Moment- area Method - Asked in: year not given, 6 marks - Video: [Introduction to Moment Area Method | Mohr’s Theorem Derivation | Deflection Of Beam](https://www.youtube.com/watch?v=vZUUDFQ6bCM) — Tutor Jerry Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/moment-area-theorems-statement-proof - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-8 ## Question State and prove the two moment-area theorems. Explain how they are used to find the slope and deflection of a beam. ## Short answer Theorem I: change in slope between two points equals the area of the M/EI diagram between them. Theorem II: vertical deviation of a point from the tangent at another equals the moment of that area about the point. Used with a tangent of known slope (fixed end or symmetry) to find slopes and deflections. ## Answer #### Basis For small deflections $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$, so $\dfrac{d\theta}{dx} = \dfrac{M}{EI}$, where $\theta$ is the slope. The quantity $M/EI$ plotted along the beam is the **$M/EI$ diagram**. ```text tangent at A A ___.----__ \ ---.__ B theta_AB : change in slope tangent at B t_BA : deviation of B from the tangent drawn at A ``` #### Theorem I (change of slope) *The change in slope between two points A and B on the elastic curve equals the area of the $M/EI$ diagram between them.* Proof: from $d\theta = \dfrac{M}{EI}dx$, integrate from A to B: $$ \theta_{B} - \theta_{A} = \int_A^B \frac{M}{EI}\,dx = \text{Area}_{AB}\ \text{of the }M/EI\text{ diagram} $$ #### Theorem II (tangential deviation) *The vertical deviation of point B on the elastic curve from the tangent drawn at A equals the moment of the area of the $M/EI$ diagram between A and B, taken about B.* Proof: consider a small element $dx$ at distance $x$ from B. The change in slope across the element is $d\theta = \dfrac{M}{EI}dx$. The tangents at the ends of the element intercept a vertical distance $dt = x\,d\theta$ on the vertical through B. Integrating, $$ t_{B/A} = \int_A^B x\,\frac{M}{EI}\,dx = \text{Area}_{AB}\times\bar x_B $$ where $\bar x_B$ is the distance of the centroid of the $M/EI$ area from B. #### Use 1. Draw the BM diagram, divide by $EI$. 2. Choose a reference tangent where the slope is known: at the fixed end of a cantilever (zero slope) or at the centre of a symmetrically loaded beam. 3. Slope: use Theorem I from the reference point. Deflection: use Theorem II (deviation from the tangent). For a simply supported beam, find the deviation of the supports from a tangent at one support, get the slope there, then find the deflection at a section from geometry. Sign: positive area (sagging) gives an anticlockwise rotation from A to B and B lies above the tangent at A. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the first moment-area theorem state? A. The deflection at B equals the area of the M/EI diagram B. The change in slope between A and B equals the area of the M/EI diagram between them C. The change in slope equals the moment of the M/EI area about B D. The slope at B equals M/EI at B Answer: B. Integrating dθ = (M/EI)dx gives θB - θA = area of M/EI. (Key point: Theorem I) 2. What does the second theorem give? A. The change in slope between A and B B. The deviation equals the moment of M/EI area about A C. The deviation of B from the tangent at A equals the moment of the M/EI area about B D. The deviation equals the M/EI area itself Answer: C. t(B/A) = area × distance of its centroid from B. (Key point: Theorem II) 3. In the proof of Theorem II, how is the intercept dt for an element dx found? A. dt = dθ/x B. dt = dx dθ C. dt = x² dθ D. dt = x dθ, where x is the distance from B Answer: D. Small angle dθ at distance x from B subtends an intercept x dθ. (Key point: Proof of II) 4. Where is the reference tangent best chosen? A. Where the slope is known, such as the fixed end or the centre of a symmetric beam B. At the point of maximum moment C. At any point as the slope there is always zero D. At the point of zero deflection Answer: A. A known slope lets Theorem I give slopes directly. 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