--- title: "Two-hinged arch: BM at section x-x" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:arch-bm-section-xx-60-85" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-bm-section-xx-60-85 site: BCT Engineering --- # Two-hinged arch: BM at section x-x Theory of Structures II · Chapter 2: Force method - Asked in: 2075 Chaitra, 6 marks - Video: [Two Hinged Arch with UDL - Horizontal Thrust Derivation and Prove BM at any section is 0](https://www.youtube.com/watch?v=CHvrnnwOCF4) — 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-section-xx-60-85 - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Find the bending moment at a given section x-x of the following loaded two hinged parabolic arch due to the given loading. Take $EI_c = 10000\ \text{kNm}^2$. [Figure: two-hinged parabolic arch with horizontal dimensions 2 m, 6 m, 8 m and 2 m as marked, rise 3 m; 60 kN/m UDL on the left part; 85 kN point load on the right part; section x-x on the left part near the support.] ## Short answer Compute horizontal thrust H = (∫ M0 y dx)/(∫ y² dx) = 369.63 kN. Bending moment at x=2 m is M = M0 – H y = 645.56 – 369.63×1.1852 = 207.48 kNm (sagging). ## Answer Reading of the figure: span $L = 18$ m, rise $h = 3$ m; A is at the left support. The dimensions 2, 6, 8 and 2 m are read as: section x-x at 2 m from A; the UDL of 60 kN/m covers the first 8 m (2 m + 6 m); the 85 kN load acts 16 m from A (8 m after the UDL, 2 m before B). The arch is taken as $I = I_c\sec\theta$ (so $EI_c = 10000$ kNm$^2$ cancels in $H$). #### Arch profile $$ y = \frac{4h\,x(L-x)}{L^2} = \frac{x(18-x)}{27} $$ #### Reactions of the equivalent simple beam UDL resultant $= 60\times8 = 480$ kN at 4 m. $$ V_B = \frac{480(4) + 85(16)}{18} = 182.222\ \text{kN},\qquad V_A = 565 - 182.222 = 382.778\ \text{kN} $$ #### Free bending moment $M_0$ (kNm) | Range | $M_0$ | |---|---| | $0 \le x \le 8$ | $382.778x - 30x^2$ | | $8 \le x \le 16$ | $382.778x - 480(x-4)$ | | $16 \le x \le 18$ | $182.222(18-x)$ | #### Horizontal thrust $$ \int_0^{18} y^2\,dx = \frac{8h^2L}{15} = 86.4 $$ $$ \int M_0\,y\,dx = 15836.2 + 15802.5 + 297.0 = 31935.6 $$ $$ H = \frac{\int M_0 y\,dx}{\int y^2\,dx} = \frac{31935.6}{86.4} = 369.63\ \text{kN} $$ #### Bending moment at section x-x ($x = 2$ m) $$ \begin{aligned} y &= \frac{2(16)}{27} = 1.1852\ \text{m} \\ M_0 &= 382.778(2) - 30(2)^2 = 645.56\ \text{kNm} \\ M_{xx} &= M_0 - Hy = 645.56 - 369.63(1.1852) = 207.48\ \text{kNm} \end{aligned} $$ **Answer: $H = 369.63$ kN and the bending moment at x-x is $M = 207.48$ kNm (sagging).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the equation of the parabolic arch axis with rise h and span L? A. y = 4h x(L-x)/L^2 B. y = h x/L C. y = h x^2/L^2 D. y = 8h x(L-x)/L^2 Answer: A. A parabola with zero ordinate at both supports and rise h at mid-span is y = 4h x(L-x)/L^2. (Key point: Arch profile) 2. How is H found for the two-hinged arch with I = Ic secθ? A. H = ∫M0 dx / L B. H = ∫M0 y dx / ∫y^2 dx C. H = total load/2 D. H = ∫y dx / ∫M0 dx Answer: B. Compatibility of zero horizontal movement gives H = ∫M0 y dx / ∫y^2 dx; EIc cancels. (Key point: Thrust formula) 3. How is the bending moment at section x-x obtained? A. M = M0 + Hy B. M = H y only C. M = M0 - Hy D. M = M0 / H Answer: C. The thrust reduces the free beam moment by H times the arch ordinate. (Key point: Moment at section) 4. What are H and M at x-x? A. H = 182.2 kN, M = 645.6 kNm B. H = 369.63 kN, M = 645.56 kNm C. H = 382.8 kN, M = 207.5 kNm D. H = 369.63 kN, M = 207.48 kNm Answer: D. H = 31935.6/86.4 = 369.63 kN and M = 645.56 - 369.63(1.1852) = 207.48 kNm. 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