--- title: "Circular arch: BM, shear, thrust at D" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-circular-60m-d-500kn" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-circular-60m-d-500kn site: BCT Engineering --- # Circular arch: BM, shear, thrust at D Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2079 Chaitra, 10 marks - Video: [Three Hinged Circular Arch - Problem No 1](https://www.youtube.com/watch?v=57BRqIgDUHg) — Stan Academy - 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-circular-60m-d-500kn - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question In three hinged circular arch shown in figure below determine bending moment, radial shear and normal thrust at section D. [Figure: three hinged circular arch ACB; span 60 m; crown hinge C with a rise of 5 m as drawn; section D on the left half, 15 m horizontally from A; UDL 25 kN/m over the left half; 500 kN vertical load on the right half, 15 m from B.] ## Short answer Vertical reactions: 687.5 kN and 562.5 kN; horizontal thrust 1875 kN from moment equilibrium about the crown. At section D (15 m from left support), bending moment = 420.6 kN m, radial shear = 4.31 kN, and normal thrust = 1900.86 kN. ## Answer **Data:** three-hinged circular arch, span $L = 60$ m, rise $f = 5$ m (crown hinge $C$ at mid-span). UDL 25 kN/m on the left half ($x = 0$ to 30 m); 500 kN vertical load on the right half, 15 m from $B$ ($x = 45$ m). Section $D$ is 15 m horizontally from $A$. #### Step 1: Equation of the arch $$ y = \sqrt{R^2-(x-30)^2} - (R-f),\quad R = \frac{L^2/4+f^2}{2f} = \frac{60^2/4+5^2}{2\times 5} = 92.5\ \text{m} $$ $$ \tan\theta = \frac{dy}{dx} = \frac{30-x}{\sqrt{R^2-(x-30)^2}} $$ At the section $x = 15$ m: $y = 3.776$ m, $\tan\theta = 0.1643$, so $\theta = 9.33^\circ$, $\sin\theta = 0.1622$, $\cos\theta = 0.9868$. #### Step 2: Vertical reactions Taking moments about $A$ for the whole arch: $$ R_B \times 60 = (25\times 30)\times 15 + 500\times 45 = 33750 \Rightarrow R_B = 562.5\ \text{kN} $$ $$ R_A = 1250 - 562.5 = 687.5\ \text{kN} $$ #### Step 3: Horizontal thrust The bending moment at the crown hinge $C$ is zero. Taking moments of the forces on the left of $C$ about $C$ (rise $f = 5$ m): $$ H \times 5 = 687.5\times 30 - (25\times 30)\times 15 = 9375 \Rightarrow H = 1875\ \text{kN} $$ #### Step 4: Forces at the section (x = 15 m) Simple-beam values at the section: $M_b = 7500$ kN m, $V_b = 312.5$ kN. $$ \begin{aligned} M &= M_b - H\,y = 7500 - 1875\times 3.776 = 420.6\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 312.5\times 0.9868 - 1875\times 0.1622 = 4.31\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 312.5\times 0.1622 + 1875\times 0.9868 = 1900.86\ \text{kN} \end{aligned} $$ **Answer: at $D$: bending moment $= 420.6$ kN m (sagging), radial shear $= 4.31$ kN, normal thrust $= 1900.86$ kN.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the radius for L = 60 m and f = 5 m? A. 92.5 m B. 60 m C. 45 m D. 100 m Answer: A. R = (900 + 25)/10 = 92.5 m. (Key point: Radius) 2. How is H found? A. Taking moments about A B. Moment at crown hinge C is zero: H x 5 = 687.5 x 30 - 750 x 15, H = 1875 kN C. Vertical equilibrium D. H = R_A Answer: B. 9375/5 = 1875 kN. (Key point: Thrust from hinge condition) 3. What are R_A and R_B? A. R_A = 687.5 kN, R_B = 562.5 kN B. R_A = 625 kN, R_B = 625 kN C. R_A = 562.5 kN, R_B = 687.5 kN D. R_A = 750 kN, R_B = 500 kN Answer: A. R_B x 60 = 750 x 15 + 500 x 45 = 33750 gives 562.5; R_A = 1250 - 562.5 = 687.5 kN. (Key point: Vertical reactions) ## Close variants set in other exams - [Parabolic arch 100 m: UDL left half and 80 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-80kn-udl-d-20m) — Theory of Structures I, chapter 6, asked 4× ## Related in Theory of Structures I - [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 - [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 - [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 - [ILD for bending moment at section D of arch](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-ild-bm-d-given-arch) — 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 ## 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: 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 - [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: 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