--- title: "Two-hinged arch: BM at 10 m intervals" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:arch-bm-80kn-d-i-ic" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-bm-80kn-d-i-ic site: BCT Engineering --- # Two-hinged arch: BM at 10 m intervals Theory of Structures II · Chapter 2: Force method - Asked in: 2070 Asar - 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-80kn-d-i-ic - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Draw the bending moment (BM) diagram for the two hinged parabolic arch shown, $I = I_c \sec\theta$. Calculate the BM value at an interval of 10 m. [Figure: two-hinged parabolic arch AB, span 60 m (20 m + 40 m), rise 8 m at crown C; 80 kN vertical load at D, 20 m from A. Printed as the alternative (OR) to the truss question above.] ## Short answer For a two-hinged parabolic arch with I = Ic secθ, horizontal thrust H = ∫M0 y dx / ∫y² dx = 101.85 kN. Bending moment M = M0 – Hy. At 10 m intervals: 0.00, 80.66, 342.39, -14.81, -190.95, -186.01, 0.00 kNm. ## Answer A two-hinged arch is statically indeterminate to the first degree. The horizontal thrust $H$ is the redundant, found from the condition that the horizontal movement of the hinge B is zero: $$ H=\frac{\int M_0\,y\,\dfrac{ds}{EI}}{\int y^2\,\dfrac{ds}{EI}} $$ For $I=I_c\sec\theta$ we have $ds/I = dx/I_c$ ($ds=dx\sec\theta$), so $H=\int M_0 y\,dx\,/\int y^2dx$. #### Data Span $L=60$ m, rise $h=8$ m, load $P=80$ kN at $x=20$ m from A. Parabola: $$ y=\frac{4h\,x(L-x)}{L^2}=\frac{4\times8\,x(60-x)}{3600} $$ #### Reactions and simple-beam moment $M_0$ $$ V_A=\frac{80\times40}{60}=53.33\ \text{kN},\qquad V_B=80-53.33=26.67\ \text{kN} $$ $M_0 = 53.33x$ for $x\le20$ and $M_0=26.67(60-x)$ for $x\ge20$. #### Horizontal thrust $$ \int_0^{60} M_0\,y\,dx=208592.6\ \text{kN m}^3,\qquad \int_0^{60} y^2dx=\frac{8h^2L}{15}=2048.0\ \text{m}^3 $$ $$ H=\frac{208592.6}{2048.0}=101.85\ \text{kN} $$ (Check with the standard result $H=\dfrac{5PL\,\alpha(1-2\alpha^2+\alpha^3)}{8h}$ with $\alpha=1/3$: same value.) #### Bending moment $M=M_0-Hy$ at 10 m intervals | x (m) | y (m) | $M_0$ (kNm) | $Hy$ (kNm) | $M$ (kNm) | |---|---|---|---|---| | 0 | 0.00 | 0.00 | 0.00 | 0.00 | | 10 | 4.44 | 533.33 | 452.67 | 80.66 | | 20 | 7.11 | 1066.67 | 724.28 | 342.39 | | 30 | 8.00 | 800.00 | 814.81 | -14.81 | | 40 | 7.11 | 533.33 | 724.28 | -190.95 | | 50 | 4.44 | 266.67 | 452.67 | -186.01 | | 60 | 0.00 | 0.00 | 0.00 | 0.00 | Sign: positive = sagging (tension on the lower side). The BM is zero at both hinges, reaches its maximum **+342.4 kNm** under the load, changes sign at about x = 29.5 m (where $M=(60-x)(26.667-0.9053x)=0$) and is hogging beyond that, with a numerical maximum of about -211 kNm near x = 44.7 m. The values at 10 m intervals plot the BM diagram: sagging on the left of x = 29.5 m (apart from a negligible hogging moment, less than 0.3 kNm, within about 1.1 m of A) and hogging on the right. **Answer: $H=101.85$ kN; BM at 0, 10, 20, 30, 40, 50, 60 m = 0.00, 80.66, 342.39, -14.81, -190.95, -186.01, 0.00 kNm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What condition gives H? A. Vertical deflection at crown is zero B. Horizontal movement of hinge B is zero C. Moment at crown is zero D. Rotation at A is zero Answer: B. ∫M0 y ds/EI - H∫y^2 ds/EI = 0 from zero horizontal movement. (Key point: Condition for H) 2. Why does I = Ic sec θ simplify the integrals? A. ds/I = dx/Ic B. It makes y constant C. It removes the load term D. It makes the arch rigid Answer: A. ds = dx sec θ, so ds/I = dx/Ic and integration is over x. (Key point: Effect of I = Ic sec θ) 3. The BM at section x is: A. M = M0 + H y B. M = M0 - H y, positive sagging C. M = H y D. M = V_A x Answer: B. M0 is the simple-beam moment and Hy is the thrust moment; at 20 m, M = 1066.67 - 724.28 = 342.39 kNm. (Key point: BM along arch) 4. The BM diagram shows: A. Sagging everywhere B. Maximum +342.4 kNm under the load, sign change near x = 29.5 m, hogging beyond C. Hogging everywhere D. Zero throughout Answer: B. M is zero at both hinges, peaks under load and becomes hogging after about 29.5 m. 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