--- title: "Graphical reactions of three hinged arch" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-graphical-reactions-single-load" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-graphical-reactions-single-load site: BCT Engineering --- # Graphical reactions of three hinged arch Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2075 Baisakh, 4 marks - Video: [Graphic statics. Arches.](https://www.youtube.com/watch?v=s1nvaWhORng) — Brian Storey - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-graphical-reactions-single-load - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question Explain graphical method to determine the reactions of a three hinged arch when it is subjected to a single concentrated load. ## Short answer For a single load on one half, the unloaded half acts as a two-force member, so its reaction line passes through its support and the crown hinge. This line’s intersection with the load line defines a point through which the other reaction passes. A force triangle then gives both reactions’ magnitudes and directions. ## Answer **Principle:** a three-hinged arch has the crown hinge $C$ where the bending moment is zero. If a single load $P$ acts on one half, the other half carries no load and is therefore a **two-force member**: the reaction at its support acts along the line joining the support to the hinge $C$. #### Steps (load P on the left half) 1. Draw the arch to scale and mark the line of action of $P$. 2. The right half $CB$ is unloaded, so the reaction $R_B$ acts along the line $BC$. Extend the line $BC$ until it meets the line of action of $P$ at the point $E$. 3. The three forces $P$, $R_A$ and $R_B$ are in equilibrium, so they are concurrent. Hence $R_A$ passes through $A$ and $E$: join $AE$. 4. Draw the force triangle: draw $P$ to scale (a vertical line $ab$), then from $a$ draw a line parallel to $R_A$ ($AE$) and from $b$ a line parallel to $R_B$ ($BC$). They meet at $c$. 5. Read the reactions: $R_A = ca$ in magnitude and direction, $R_B = bc$. The horizontal components are equal and opposite, which gives the thrust $H$; the vertical components give $V_A$ and $V_B$. ```text Space diagram (P on left half) Force triangle | P E *-------+--------- a /| \ |\ C / | \ P | \ R_B (parallel BC) .' | '. | \ A ' | ' B b---c R_A along AE; R_B along BC R_A (parallel AE) ``` 6. If $P$ lies on the right half, interchange the roles: $R_A$ acts along $AC$, extend it to meet $P$ at $E'$, and $R_B$ passes through $B$ and $E'$. 7. If several loads act, apply the method for the loads of each half separately and add the reactions vectorially. **Check:** the vertical components satisfy $V_A + V_B = P$ and the moment of the reactions about any point equals that of $P$; with the numerical values, $H = \dfrac{M_C}{f}$ as in the analytical method. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With P on the left half, why does R_B act along BC? A. The right half carries no load and is a two-force member B. B is a fixed support C. P is vertical D. C is a roller Answer: A. The unloaded half with hinge C at zero moment is acted on by two forces only, so they are collinear along BC. (Key point: Principle) 2. How is the direction of R_A found? A. Join A to E, where BC extended meets the line of P (three forces are concurrent) B. Draw it vertical C. Draw it parallel to BC D. Join A to C Answer: A. P, R_A and R_B are in equilibrium, hence concurrent at E, so R_A passes through A and E. (Key point: Concurrency) 3. In the force triangle, what is drawn first? A. P to scale, then lines parallel to AE and BC from its ends B. R_A C. H D. The funicular polygon Answer: A. P is known; the directions of R_A (AE) and R_B (BC) close the triangle. (Key point: Force triangle) 4. For several loads, what is done? A. Apply method for each half separately and add reactions vectorially B. Use only the largest load C. Ignore the right half D. Take their average Answer: A. By superposition, the loads of each half are treated separately and the resulting reactions are added. 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