--- title: "Arch 18 m: forces at 3 m intervals, diagrams" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-18m-3m-moment-thrust-shear-diagrams" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-18m-3m-moment-thrust-shear-diagrams site: BCT Engineering --- # Arch 18 m: forces at 3 m intervals, diagrams Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2069 Poush, 16 marks - Video: [Structural Analysis: Three-Hinged Arch](https://www.youtube.com/watch?v=F6bKl-Msx2g) — Amanda Bao - Structural Design Teaching Channel - 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-18m-3m-moment-thrust-shear-diagrams - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question A three hinged symmetrical parabolic arch has a span of 18 m and rise of 3 m. It carries a concentrated load of 80 kN at 4.5 m from the right support and a distributed load of 5 kN/m over half portion. Determine the moment, thrust and radial shear at each 3 m interval and draw their diagrams on horizontal 'X' axis for the arch. ## Short answer At 3 m intervals from left support: moments (kN m) 0, -17.5, -17.5, 0, 57.5, 57.5, 0; normal thrusts (kN) 107.8, 101.4, 96.7, 93.8, 89.6, 114.6, 117.5; radial shears (kN) -7.3, -2.7, 2.8, 8.8, 28.9, -27.0, -7.3. ## Answer **Data and assumptions:** symmetrical parabolic three-hinged arch, $L = 18$ m, $f = 3$ m, $y = \dfrac{4\times 3\,x(18-x)}{18^2} = \dfrac{x(18-x)}{27}$. 80 kN at 4.5 m from the right support ($x = 13.5$ m); UDL 5 kN/m over half the span (taken as the left half, $x = 0$ to 9 m). #### Reactions and thrust $$ V_B\times 18 = 80\times 13.5 + (5\times 9)\times 4.5 = 1282.5 \Rightarrow V_B = 71.25\ \text{kN}, \qquad V_A = 125 - 71.25 = 53.75\ \text{kN} $$ $$ H\times 3 = V_A\times 9 - 45\times 4.5 \Rightarrow H = \frac{53.75\times 9 - 202.5}{3} = 93.75\ \text{kN} $$ #### Forces at 3 m intervals Simple-beam values: $M_b$, $V_b$ at each section; $\tan\theta = \dfrac{dy}{dx} = \dfrac{18-2x}{27}$. $$ M = M_b - H\,y, \quad V = V_b\cos\theta - H\sin\theta, \quad N = V_b\sin\theta + H\cos\theta $$ | x (m) | y (m) | M (kN m) | V (kN) | N (kN) | |---|---|---|---|---| | 0 | 0 | 0 | -7.28 | 107.82 | | 3 | 1.667 | -17.5 | -2.67 | 101.41 | | 6 | 2.667 | -17.5 | 2.85 | 96.67 | | 9 | 3 | 0 | 8.75 | 93.75 | | 12 | 2.667 | 57.5 | 28.88 | 89.62 | | 15 | 1.667 | 57.5 | -27.03 | 114.61 | | 18 | 0 | 0 | -7.28 | 117.53 | (At the supports $x=0$ and $18$, the values are those just inside the span; $M = 0$ at $A$, $C$ and $B$. The shear is the radial shear and $N$ the normal thrust.) #### Diagrams drawn on the horizontal x-axis Bending moment (+ sagging, scale not exact): ```text 0 m | 0.0 3 m | ------- -17.5 6 m | ------- -17.5 9 m | 0.0 12 m | ++++++++++++++++++++++++ 57.5 15 m | ++++++++++++++++++++++++ 57.5 18 m | 0.0 ``` Radial shear: ```text 0 m | ------ -7.3 3 m | -- -2.7 6 m | ++ 2.8 9 m | +++++++ 8.8 12 m | ++++++++++++++++++++++++ 28.9 15 m | ---------------------- -27.0 18 m | ------ -7.3 ``` Normal thrust (always compressive): ```text 0 m | ++++++++++++++++++++++ 107.8 3 m | +++++++++++++++++++++ 101.4 6 m | ++++++++++++++++++++ 96.7 9 m | +++++++++++++++++++ 93.8 12 m | ++++++++++++++++++ 89.6 15 m | +++++++++++++++++++++++ 114.6 18 m | ++++++++++++++++++++++++ 117.5 ``` Maximum sagging moment $\approx 109.7$ kN m at $x \approx 13.5$ m, and maximum hogging moment $\approx -19.7$ kN m at $x \approx 4.5$ m (in the left half, under the UDL). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 80 kN at x = 13.5 m and 45 kN (UDL 5 x 9) at 4.5 m, what is V_B? A. V_A x 18 / 2 = 62.5 kN B. (80 x 13.5 + 45 x 4.5)/18 = 71.25 kN C. 125/2 = 62.5 kN D. (80 x 4.5 + 45 x 13.5)/18 = 53.75 kN Answer: B. Moments about A give V_B x 18 = 1282.5, so V_B = 71.25 kN and V_A = 53.75 kN. (Key point: Reactions of the 18 m arch) 2. How is H = 93.75 kN obtained? A. H x 3 = V_A x 9 - 45 x 4.5, from M_C = 0 on the left part B. H = (V_A + V_B)/ 2 C. H x 3 = V_B x 9 + 80 x 4.5 on the left part D. H = total load x span / (8 x rise) Answer: A. The moment at the crown hinge is zero: 53.75 x 9 - 202.5 = 281.25, and H = 281.25/3 = 93.75 kN. (Key point: Thrust from the crown hinge) 3. At x = 9 m the table gives N = 93.75 kN, the same as H. Why? A. Because the bending moment is zero there B. Because the shear is zero there C. Because the tangent is horizontal (theta = 0), so N = H cos(0) = H D. Because the UDL ends there Answer: C. At the crown dy/dx = 0, so sin(theta) = 0 and N = V_b sin(theta) + H cos(theta) = H. (Key point: Normal thrust at the crown) 4. Which set is used to compute the internal forces at each 3 m section? A. M = M_b + H y; V = V_b sin(theta); N = H B. M = M_b - H y; V = V_b + H; N = V_b C. M = H y; V = H tan(theta); N = V_b D. M = M_b - H y; V = V_b cos(theta) - H sin(theta); N = V_b sin(theta) + H cos(theta) Answer: D. The simple-beam values are corrected by the thrust, and N, V are resolved along and across the arch axis; N is always compressive here. 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