--- title: "Two-hinged parabolic arch: BM at D" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:arch-bm-point-d-30-50" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-bm-point-d-30-50 site: BCT Engineering --- # Two-hinged parabolic arch: BM at D Theory of Structures II · Chapter 2: Force method - Asked in: 2079 Baishakh, 10 marks - Video: [Two Hinged Parabolic Arch (Problem No 2) With Point load acting at a distance from the Support](https://www.youtube.com/watch?v=ksywknDX3TA) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-bm-point-d-30-50 - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Find the BM at point D for the given two hinged parabolic arch. Take $I_x = I_c \sec\theta$. [Figure: two-hinged parabolic arch AB, span 60 m (dimensions 30 m, 20 m, 10 m), rise 10 m; UDL 30 kN/m on the left portion up to crown C; 50 kN point load at D, 10 m from support B.] ## Short answer For a two-hinged parabolic arch with I = I_c secθ, horizontal thrust H = ∫M₀ y dx / ∫ y² dx = 704.66 kN. At D (x=50 m), free bending moment M₀ = 2666.67 kNm and y_D = 5.556 m. Then M_D = M₀ – H y_D = –1248.10 kNm (hogging). ## Answer A two-hinged arch has one redundant, the horizontal thrust $H$. With $I_x = I_c\sec\theta$ we have $ds/I = dx/I_c$, so the thrust is $$ H = \frac{\int M_0\, y\, dx}{\int y^2\, dx} $$ where $M_0$ is the bending moment in the equivalent simply supported beam of the same span. #### Data and arch profile Span $L = 60$ m, rise $h = 10$ m, A at the left support (origin). The UDL acts on $0 \le x \le 30$ m and the 50 kN load acts at $x = 50$ m (10 m from B). $$ y = \frac{4h\,x(L-x)}{L^2} = \frac{x(60-x)}{90} $$ #### Reactions of the simple beam UDL resultant $= 30 \times 30 = 900$ kN at $x = 15$ m. $$ V_B = \frac{900(15) + 50(50)}{60} = 266.67\ \text{kN}, \qquad V_A = 950 - 266.67 = 683.33\ \text{kN} $$ #### Free bending moment $M_0$ | Range (m) | $M_0$ (kNm) | |---|---| | $0 \le x \le 30$ | $683.33x - 15x^2$ | | $30 \le x \le 50$ | $683.33x - 900(x-15)$ | | $50 \le x \le 60$ | $266.67(60-x)$ | #### Integrals $$ \begin{aligned} \int_0^{60} y^2\,dx &= \frac{8h^2L}{15} = 3200 \\ \int_0^{30} M_0 y\,dx &= 1347500, \quad \int_{30}^{50} M_0 y\,dx = 855556, \quad \int_{50}^{60} M_0 y\,dx = 51852 \\ \int_0^{60} M_0 y\,dx &= 2254907 \end{aligned} $$ $$ H = \frac{2254907}{3200} = 704.66\ \text{kN} $$ #### Bending moment at D ($x = 50$ m) $$ \begin{aligned} y_D &= \frac{50(10)}{90} = 5.556\ \text{m} \\ M_0 &= V_B \times 10 = 2666.67\ \text{kNm} \\ M_D &= M_0 - H\,y_D = 2666.67 - 704.66(5.556) = -1248.10\ \text{kNm} \end{aligned} $$ **Answer: $H = 704.66$ kN and the bending moment at D is $M_D = -1248.10$ kNm (hogging, tension on the top/extrados).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The redundant in a two-hinged arch is: A. Horizontal thrust H B. Vertical reaction C. Moment at crown D. Shear at D Answer: A. Two hinges give 4 reactions against 3 equations. (Key point: Redundant) 2. With I = Ic sec(theta), H = A. integral M0 y dx / integral y^2 dx B. integral M0 dx / integral y dx C. wL^2/8h D. M0/y Answer: A. ds/I = dx/Ic, giving this form. (Key point: Thrust formula) 3. The parabola y = 4h x(L-x)/L^2 gives: A. x(60-x)/90 B. x(60-x)/60 C. x(60-x)/45 D. x/6 Answer: A. 40/3600 = 1/90. (Key point: Profile) 4. M_D is computed as: A. M0 - H y_D, giving -1248.10 kNm B. M0 + H y_D C. H y_D only D. V_B x 10 only Answer: A. 2666.67 - 704.66(5.556) = -1248.10 kNm, hogging. 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