--- title: "Graphical method for BM, shear, thrust in arch" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-graphical-method-internal-forces" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-graphical-method-internal-forces site: BCT Engineering --- # Graphical method for BM, shear, thrust in arch Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2070 Bhadra, 6 marks; 2066 Kartik, 6 marks (asked 2 times) - 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-method-internal-forces - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question Explain with necessary sketches the steps involved in determining bending moment, radial shear and normal thrust in a three hinged arch by graphical method. ## Short answer A force polygon and a funicular polygon through hinges A, C, B give the reactions and thrust line. Bending moment equals H times the vertical intercept between arch and thrust line. Resolving the section's resultant force along and perpendicular to the tangent gives normal thrust and radial shear. ## Answer The graphical method finds the reactions, then the resultant force on any section, from a force polygon and a linkage (funicular) polygon drawn to scale. #### Steps 1. **Draw the arch** to a scale (span, rise, crown hinge $C$, supports $A$ and $B$) and mark the loads $P_1, P_2, \dots$ with their lines of action. 2. **Force polygon:** draw the load line $1\text{-}2\text{-}3\ldots$ to a force scale. Choose a pole $O$ and draw the rays $O1, O2, \dots$. 3. **Reactions (using the hinge condition):** - Combine the loads on the left of $C$ into their resultant $R_1$, and those on the right into $R_2$. - For $R_1$ alone, the right part $CB$ is unloaded, so the reaction at $B$ acts along $BC$. The line of $R_1$ meets line $BC$ at $E_1$; join $AE_1$, which is the direction of the reaction at $A$ for $R_1$. Close the triangle of forces. - Do the same for $R_2$ (the reaction at $A$ acts along $AC$; join its meeting point with $R_2$ to $B$). - Add the two sets of reactions vectorially. This gives $R_A$ and $R_B$ in magnitude and direction. 4. **Linkage (thrust line) polygon:** start from $A$ with a link parallel to the ray found from $R_A$ and continue the links parallel to the rays across each load. The polygon passes through $A$, $C$ and $B$ (this fixes the pole). 5. **Bending moment at a section $D$:** measure the vertical intercept $\delta$ between the arch axis and the linkage polygon at $D$. Then $$ M_D = H\times\delta $$ where $H$ is the pole distance (horizontal thrust) scaled from the force polygon. Since $M_D = M_{beam} - H\,y = H\,(y_{thrust} - y)$, the moment is positive (sagging) where the linkage polygon lies above the arch axis and negative (hogging) where it lies below. 6. **Normal thrust and radial shear:** read from the force polygon the resultant force $R_D$ on the section (the ray, or the sum of the reaction and the loads left of $D$). Draw the tangent to the arch at $D$. Resolve $R_D$: $$ N = R_D\cos\phi \ (\text{along tangent}), \qquad V = R_D\sin\phi \ (\text{perpendicular to tangent}) $$ where $\phi$ is the angle between $R_D$ and the tangent. ```text Space diagram Force polygon P1 P2 1 *----. | | C | '. O (pole) .--+----+--. (hinge) 2 *------* A link polygon B | (through A, C, B) 3 * M = H x (vertical intercept) H = pole distance ``` 7. Check that the linkage polygon passes through all three points $A$, $C$, $B$ (the hinge condition) and that the force polygon closes. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the resultant R1 of the left-half loads, the reaction at B acts along which line? A. BC B. AB C. A to the load D. Vertical Answer: A. The right half CB is unloaded and is a two-force member, so R_B acts along BC. (Key point: Reactions using hinge) 2. How is the bending moment at D obtained? A. M = H x vertical intercept between arch axis and linkage polygon at D B. M = pole distance only C. M = R_A x span D. M = N x tangent Answer: A. The intercept delta is the ordinate of the thrust line from the arch axis; multiplying by the pole distance H gives M_D. (Key point: Bending moment by linkage polygon) 3. The linkage polygon must pass through which points? A. A, C and B B. A, D and B C. Only C D. The crown and the pole Answer: A. The hinge condition M_C = 0 means the thrust line passes through the crown hinge, plus the end hinges A and B. (Key point: Check on the linkage polygon) 4. Having the resultant R_D at angle phi to the tangent, which is right? A. N = R_D sin(phi), V = R_D cos(phi) B. N = R_D cos(phi) along the tangent, V = R_D sin(phi) perpendicular to it C. N = V D. N = R_D tan(phi) Answer: B. Normal thrust is the component along the tangent and radial shear the component perpendicular to it. (Key point: Normal thrust and radial shear) ## Related in Theory of Structures I - [Circular arch: BM, shear, thrust at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-60m-d-500kn) — Theory of Structures I, chapter 6, asked 2× - [Parabolic arch 90 m: BM, shear, thrust at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-90m-180kn-6knm-d) — Theory of Structures I, chapter 6, asked 2× - [BM, shear, thrust at any section, UDL on arch](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-general-section-udl-formula) — Theory of Structures I, chapter 6 - [Arch equation given: BM, thrust, shear at 5 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-y-equation-30knm-right-half) — Theory of Structures I, chapter 6 - [Arch: reactions, BM, thrust and shear at D by ILD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-50m-udl-50kn-ild-d-15m) — Theory of Structures I, chapter 6 - [Circular arch: BM, thrust, shear at D, BM diagram](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-d-20knm-10kn-8m) — Theory of Structures I, chapter 6 ## Related in other subjects - [Two-hinged arch: BM, thrust, radial shear, support yield](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-thrust-shear-yield-support) — Theory of Structures II, chapter 2 - [Two-hinged arch: BM at 10 m intervals](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-80kn-d-i-ic) — Theory of Structures II, chapter 2 - [Two-hinged arch: BM at section x-x](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-section-xx-60-85) — Theory of Structures II, chapter 2 - [Two-hinged arch: horizontal reaction, BM at C](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-horizontal-reaction-bm-c-50knm) — Theory of Structures II, chapter 2