--- title: "Arch 20 m: reactions, BM, thrust, shear, BMD" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-20m-udl-2knm-left-8m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-20m-udl-2knm-left-8m site: BCT Engineering --- # Arch 20 m: reactions, BM, thrust, shear, BMD Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2065 Chaitra, 16 marks - Video: [Structural analysis-2 3-hinge arch Numerical / Horizontal thrust / Reaction / BMD](https://www.youtube.com/watch?v=0tHG9sv5Mhk) — Yogendra Ram Meena - 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-20m-udl-2knm-left-8m - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question A three hinged parabolic arch as shown in fig-9 is loaded with udl 2 kN/m on the left 8 m length. Calculate (a) direction and magnitude of reaction at supports, (b) the bending moment, normal thrust and radial shear at 4m from left end, (c) draw bending moment diagram showing maximum positive and negative values. [Figure: three hinged parabolic arch, span 20 m, rise 4 m; 2 kN/m UDL on the left 8 m.] ## Short answer Reactions: R_A = 15.09 kN at 57.99° to horizontal, R_B = 8.62 kN at 21.8° (thrust H = 8 kN). At 4 m from left: M = 14.72 kN m, N = 9.29 kN, V = 0.87 kN. Maximum positive moment +15.06 kN m, maximum negative -8 kN m. ## Answer **Data:** three-hinged parabolic arch, span $L = 20$ m, rise $f = 4$ m ($y = 0.04\,x(20-x)$), supports at the same level; UDL 2 kN/m on the left 8 m ($x = 0$ to 8 m), total 16 kN acting at 4 m from $A$. #### (a) Reactions $$ V_B\times 20 = 16\times 4 \Rightarrow V_B = 3.2\ \text{kN}, \qquad V_A = 16 - 3.2 = 12.8\ \text{kN} $$ Crown hinge ($M_C = 0$), left part: $$ H\times 4 = 12.8\times 10 - 16\times 6 = 32 \Rightarrow H = 8\ \text{kN} $$ The reaction at each support is the resultant of its vertical component and the thrust: - At $A$: $R_A = \sqrt{12.8^2 + 8^2} = 15.09$ kN, inclined at $\tan^{-1}\dfrac{12.8}{8} = 57.99^\circ$ above the horizontal, directed up and to the right (towards the span). - At $B$: $R_B = \sqrt{3.2^2 + 8^2} = 8.62$ kN, inclined at $\tan^{-1}\dfrac{3.2}{8} = 21.8^\circ$ above the horizontal, directed up and to the left. #### (b) Forces at 4 m from the left end At $x = 4$ m: $y = 0.04\times 4\times 16 = 2.56$ m, $\tan\theta = 0.04\,(20-8) = 0.48$, $\theta = 25.64^\circ$, $\sin\theta = 0.4327$, $\cos\theta = 0.9015$. Simple-beam shear $V_b = V_A - 2\times 4 = 4.8$ kN. $$ \begin{aligned} M &= V_A\times 4 - 2\times 4\times 2 - H\,y = 12.8\times 4 - 16 - 8\times 2.56 = 14.72\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 4.8(0.9015) - 8(0.4327) = 0.87\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 4.8(0.4327) + 8(0.9015) = 9.29\ \text{kN} \end{aligned} $$ #### (c) Bending moment diagram $M = M_b - H\,y$ along the span: | x (m) | M (kN m) | |---|---| | 0 | 0 | | 2 | 10.08 | | 4 | 14.72 | | 6 | 13.92 | | 8 | 7.68 | | 10 | 0 | | 12 | -5.12 | | 14 | -7.68 | | 16 | -7.68 | | 18 | -5.12 | | 20 | 0 | The bending moment is zero at $A$, at the crown hinge $C$ ($x = 10$ m) and at $B$. The maximum positive (sagging) value is about **+15.06 kN m at $x \approx 4.7$ m** (under the UDL) and the maximum negative (hogging) value is about **-8 kN m at $x \approx 15$ m** (in the right half). ```text M (kN m) +15.06 ___ _.-' '-._ 0 ---A'---------C'-._----------B--- '-._ '--- -8 ``` **Answer: (a) $R_A = 15.09$ kN at 57.99° to the horizontal, $R_B = 8.62$ kN at 21.8° (thrust $H = 8$ kN); (b) $M = 14.72$ kN m, $N = 9.29$ kN, $V = 0.87$ kN at 4 m from the left; (c) maximum $M = +15.06$ kN m and minimum $M = -8$ kN m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 20 m arch with 2 kN/m on the left 8 m (V_A = 12.8 kN), how is H found? A. H = 16/4 = 4 kN B. H x 4 = V_A x 10 - 16 x 6 = 32, so H = 8 kN C. H x 4 = V_B x 10 + 16 x 6 D. H = wL^2/(8f) = 20 kN Answer: B. Moment at crown C is zero for the left part: 12.8 x 10 - 16 x 6 = 32 = 4H, so H = 8 kN. (Key point: Thrust from crown hinge) 2. How is the reaction at A expressed? A. Only the vertical component 12.8 kN B. The resultant of V_A and H: sqrt(12.8^2 + 8^2) = 15.09 kN at 57.99 deg to the horizontal C. H alone, 8 kN horizontal D. V_A + H = 20.8 kN Answer: B. The hinge reaction is the vector sum of its vertical and horizontal components, with angle tan^-1(12.8/8) = 57.99 deg. (Key point: Support reactions as resultants) 3. What is the bending moment at 4 m from A and how is it computed? A. 12.8 x 4 = 51.2 kN m B. 14.72 kN m = V_A x 4 - UDL moment (2 x 4 x 2) - H y with y = 2.56 m C. 16 kN m from the UDL alone D. 8 x 2.56 = 20.48 kN m Answer: B. M = 51.2 - 16 - 8 x 2.56 = 14.72 kN m. (Key point: Bending moment at 4 m) 4. Where is the bending moment zero in the BMD? A. Only at the supports B. At 4 m and 15 m C. At the crown only D. At A, the crown hinge C and B Answer: D. The moment is zero at all three hinges; it is sagging (positive, max about +15 kN m) under the UDL and hogging (about -8 kN m) in the right half. 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