--- title: "Two-hinged arch: thrust with rib shortening, support yield" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:arch-60m-6m-rib-shortening" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-60m-6m-rib-shortening site: BCT Engineering --- # Two-hinged arch: thrust with rib shortening, support yield Theory of Structures II · Chapter 2: Force method - Asked in: 2078 Kartik, 8 marks - Video: [Lec 7 Unit- 1 | Effect of yielding of supports-Two Hinged Arches | SA-II](https://www.youtube.com/watch?v=Rd97iGqV1Ss) — Smart Civil Engineer - 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-60m-6m-rib-shortening - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question A two hinge parabolic arch of span 60 m and central rise of 6 m is subjected to a concentrated load of 40 kN at the crown. Allowing for rib shortening and yielding of support B by 6 mm outwards horizontally, determine the horizontal thrust. Also draw the BMD. Take $I_0 = 6\times10^{3}\ \text{cm}^4$ [?], $A_c = 1000\ \text{cm}^2$, $E = 10\ \text{kN/mm}^2$ and consider $I_x = I_0 \sec\alpha$. ## Short answer Use two-hinged parabolic arch with H as redundant, including rib shortening and 6 mm outward support yield. EI0=600 kNm², EAc=10⁶ kN. Compute numerator (∫M0y dx/EI0 – δ)=149.994, denominator (∫y²dx/EI0+∫cos²α dx/EA_c)=1.92006. H=78.12 kN. Crown BM=131.28 kNm sagging, quarter-span BM=-51.54 kNm. ## Answer A two-hinged arch has one redundant, the horizontal thrust $H$. Taking $I_x = I_0\sec\alpha$, the compatibility equation with rib shortening and yielding of a support is $$ H = \frac{\displaystyle\int \frac{M_0\,y}{EI_0}\,dx - \delta}{\displaystyle\int \frac{y^2}{EI_0}\,dx + \int \frac{\cos^2\alpha}{EA_c}\,dx} $$ where $\delta$ is the outward yield of support B. (The small axial-force term of the free structure is neglected, as is usual.) #### Data (converted to kN and m) - $L = 60$ m, $h = 6$ m, parabola $y = \dfrac{4h\,x(L-x)}{L^2} = \dfrac{x(60-x)}{150}$ - $EI_0 = 10\,\text{kN/mm}^2 \times 6\times10^3\,\text{cm}^4 = 10^7 \times 6\times10^{-5} = 600$ kNm$^2$ - $EA_c = 10^7 \times 0.1 = 10^6$ kN, $\delta = 6$ mm $= 0.006$ m - $V_A = V_B = 20$ kN; $M_0 = 20x$ for $0 \le x \le 30$ (peak 600 kNm at the crown) #### Integrals $$ \begin{aligned} \int_0^{60} M_0 y\,dx &= 2\int_0^{30} 20x\cdot\frac{x(60-x)}{150}\,dx = 90000 \\ \int_0^{60} y^2\,dx &= \frac{8h^2L}{15} = 1152 \\ \int_0^{60} \cos^2\alpha\,dx &= 57.08 \end{aligned} $$ ($\tan\alpha = \dfrac{dy}{dx} = \dfrac{4h(L-2x)}{L^2}$, so $\cos^2\alpha = \dfrac{1}{1+\tan^2\alpha}$.) $$ \begin{aligned} \frac{\int M_0 y\,dx}{EI_0} &= \frac{90000}{600} = 150.000 \\ \frac{\int y^2\,dx}{EI_0} &= \frac{1152}{600} = 1.9200 \\ \frac{\int \cos^2\alpha\,dx}{EA_c} &= \frac{57.08}{10^6} = 5.708\times10^{-5} \end{aligned} $$ #### Horizontal thrust $$ H = \frac{150.000 - 0.006}{1.9200 + 5.708\times10^{-5}} = 78.12\ \text{kN} $$ For comparison: without rib shortening and yielding, $H = \dfrac{25PL}{128h} = 78.125$ kN. With the printed $I_0$ the arch is very flexible in bending, so rib shortening and a 6 mm yield change the thrust only slightly. (A stiffer rib, say $I_0 = 6\times10^6$ cm$^4$, would give $H = 72.83$ kN.) #### Bending moment $M = M_0 - H y$ | $x$ (m) | $y$ (m) | $M_0$ (kNm) | $M$ (kNm) | |---|---|---|---| | 0.0 | 0.000 | 0.0 | 0.00 | | 7.5 | 2.625 | 150.0 | -55.06 | | 15.0 | 4.500 | 300.0 | -51.54 | | 22.5 | 5.625 | 450.0 | 10.58 | | 30.0 | 6.000 | 600.0 | 131.28 | The BMD is symmetric about the crown: zero at the hinges A and B, hogging (negative) about the quarter points, and sagging with a peak of 131.28 kNm under the load at the crown. **Answer: $H \approx 78.12$ kN; crown BM $= 131.28$ kNm (sagging), BM at $x = 15$ m $= -51.54$ kNm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the redundant in a two-hinged arch and why? A. The horizontal thrust H; with two hinges there are four reactions and only three equations B. The vertical reaction at A, since the load is at the crown C. The bending moment at the crown D. None; a two-hinged arch is determinate Answer: A. Two hinges give four unknown reactions against three equilibrium equations, so H is the single redundant. (Key point: Redundant) 2. How do outward yielding of support B and rib shortening enter the formula for H? A. Both are added to the numerator B. Yield delta is subtracted from the numerator, and the rib-shortening term cos^2(alpha)/(EA_c) is added to the denominator C. Yield is added to the denominator and rib shortening is ignored D. Both are subtracted from the denominator Answer: B. Outward movement of the support relieves thrust (reduces the numerator); axial shortening makes the arch more flexible (increases the denominator), so H falls. (Key point: Support yield and rib shortening) 3. With the printed I0 the thrust is 78.12 kN, almost the same as 25PL/(128h) = 78.125 kN. Why? A. Because the arch is very stiff in bending B. Because the axial term is exactly zero C. The bending flexibility term (1.92) is far larger than the axial term (5.7 x 10^-5), and delta = 0.006 is tiny against 150 D. Because yielding and rib shortening cancel each other Answer: C. The numerator is 150.000 - 0.006 and the denominator 1.9200 + 0.0000571, so the corrections are negligible. (Key point: Size of the effect) 4. How is BM found after H, and what is the crown value? A. M = M0 + H y = 1068.7 kNm B. M = M0 - H y = 600 - 78.12 x 6 = 131.28 kNm sagging C. M = M0 = 600 kNm, since the hinges carry no moment D. M = -H y = -468.7 kNm hogging Answer: B. The thrust acts at height y above the springing and reduces the simple-beam moment M0, leaving 131.28 kNm sagging under the load. (Key point: Bending moment at crown) ## Related in Theory of Structures II - [Two-hinged arch: BM, thrust, radial shear, support yield](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-thrust-shear-yield-support) — Theory of Structures II, chapter 2 - [Two-hinged parabolic arch: BM at D](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-point-d-30-50) — Theory of Structures II, chapter 2 - [Two-hinged parabolic arch: horizontal reaction, symmetric loads](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-secant-symmetric-loads-80m) — Theory of Structures II, chapter 2 - [Two-hinged arch: BM at 10 m intervals](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-80kn-d-i-ic) — Theory of Structures II, chapter 2 - [Compatibility equation for two-hinged arch](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-compatibility-equation) — Theory of Structures II, chapter 2 - [Two-hinged arch: horizontal reaction, BM at C](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-horizontal-reaction-bm-c-50knm) — Theory of Structures II, chapter 2 ## Related in other subjects - [Parabolic arch 80 m: inclined load, forces at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-80m-250kn-inclined-100kn) — Theory of Structures I, chapter 6, asked 2× - [Parabolic arch 100 m: UDL left half and 80 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-80kn-udl-d-20m) — Theory of Structures I, chapter 6, asked 4× - [Parabolic arch 90 m: BM, shear, thrust at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-90m-180kn-6knm-d) — Theory of Structures I, chapter 6, asked 2× - [Arch 100 m rise 15 m: rolling UDL 10 m, section 60 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-rise15-udl-10m-60m-section) — Theory of Structures I, chapter 6