--- title: "Two-hinged arch: BM, thrust, radial shear, support yield" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:arch-bm-thrust-shear-yield-support" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-bm-thrust-shear-yield-support site: BCT Engineering --- # Two-hinged arch: BM, thrust, radial shear, support yield Theory of Structures II · Chapter 2: Force method - Asked in: 2070 Chaitra (old course), 16 marks - Video: [EX 01 | ANALYSIS OF ARCH | RADIAL SHEAR | NORMAL THRUST | SLOPE | MAXIMUM MOMENTS](https://www.youtube.com/watch?v=TMdKyGVRAf0) — Structural Funda by:- KheNiSiR - 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-thrust-shear-yield-support - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question In the two hinged parabolic arch shown below, find the values of bending moment, normal thrust and radial shear at section D due to the given loading and due to yielding of support B by 10 mm. Take $EI_c = 100\times10^{6}\ \text{kNm}^2$, $I = I_c \sec\theta$. Also draw bending moment diagram. [Figure: two-hinged parabolic arch AB, span 120 m (60 m + 60 m), rise 20 m at crown C; section D at 30 m from A; 10 kN/m UDL on the right half CB.] ## Short answer For the two-hinged parabolic arch, H=(∫M0 y dx/EIc - δ)/(∫y^2 dx/EIc), giving H=410.94 kN. At D, M=-1664.06 kNm hogging, normal thrust N=437.28 kN compression, and radial shear Q=12.35 kN. ## Answer A two-hinged arch has one redundant, the horizontal thrust $H$. With $I = I_c\sec\theta$ the compatibility equation (support B yields outward by $\delta$) is $$ H = \frac{\displaystyle\frac{1}{EI_c}\int M_0\,y\,dx - \delta}{\displaystyle\frac{1}{EI_c}\int y^2\,dx} = H_{load} - \frac{\delta\,EI_c}{\int y^2\,dx} $$ #### Data Span $L = 120$ m, rise $h = 20$ m, $EI_c = 100\times10^6$ kNm$^2$, $\delta = 10$ mm $= 0.01$ m; UDL $10$ kN/m on CB (60 m to 120 m). $$ y = \frac{4h\,x(L-x)}{L^2} = \frac{x(120-x)}{180},\qquad \tan\theta = \frac{dy}{dx} = \frac{120 - 2x}{180} $$ #### Simple-beam reactions and $M_0$ UDL resultant $= 10\times60 = 600$ kN at 90 m: $V_B = \dfrac{600(90)}{120} = 450$ kN, $V_A = 150$ kN. - $0 \le x \le 60$: $M_0 = 150\,x$ - $60 \le x \le 120$: $M_0 = 150\,x - 5(x-60)^2$ #### Thrust $$ \int_0^{120} y^2\,dx = \frac{8h^2L}{15} = 25600,\qquad \int_0^{120} M_0\,y\,dx = 11520000 $$ $$ H_{load} = \frac{11520000}{25600} = 450.00\ \text{kN} $$ $$ H_{yield} = -\frac{\delta\,EI_c}{\int y^2dx} = -\frac{0.01\times100\times10^6}{25600} = -39.06\ \text{kN} $$ $$ H = 450.00 + (-39.06) = 410.94\ \text{kN} $$ #### Section D ($x = 30$ m) $$ \begin{aligned} y_D &= \frac{30(90)}{180} = 15\ \text{m},\qquad \tan\theta_D = \frac{120-60}{180} = \frac13,\ \ \theta_D = 18.43^\circ \\ M_0 &= V_A(30) = 4500\ \text{kNm},\qquad V_0 = V_A = 150\ \text{kN} \end{aligned} $$ **Bending moment** $$ M_D = M_0 - Hy_D = 4500 - 410.94(15) = -1664.06\ \text{kNm} $$ (Due to the load alone: $-2250.00$ kNm; the yield adds $+585.94$ kNm.) **Normal thrust** $$ N_D = H\cos\theta + V_0\sin\theta = 410.94(0.9487) + 150(0.3162) = 437.28\ \text{kN} $$ **Radial shear** $$ Q_D = V_0\cos\theta - H\sin\theta = 150(0.9487) - 410.94(0.3162) = 12.35\ \text{kN} $$ #### Bending moment diagram $M = M_0 - Hy$ (kNm) | $x$ (m) | $y$ (m) | $M_0$ | $M$ | |---|---|---|---| | 0 | 0.00 | 0.0 | 0.0 | | 15 | 8.75 | 2250.0 | -1345.7 | | 30 | 15.00 | 4500.0 | -1664.1 | | 45 | 18.75 | 6750.0 | -955.1 | | 60 | 20.00 | 9000.0 | 781.2 | | 75 | 18.75 | 10125.0 | 2419.9 | | 90 | 15.00 | 9000.0 | 2835.9 | | 105 | 8.75 | 5625.0 | 2029.3 | | 120 | 0.00 | 0.0 | 0.0 | The BMD is hogging for $0 < x < 54.3$ m (largest hogging about $-1683$ kNm near $x = 27$ m) and sagging for the rest, with the largest sagging moment about $2852$ kNm near $x = 88$ m, under the loaded half. It is zero at both hinges. **Answer: at D, $M = -1664.06$ kNm (hogging), $N = 437.28$ kN (compression), radial shear $Q = 12.35$ kN; $H = 410.94$ kN.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the two-hinged arch with support B yielding outward by delta, how does the yielding change the thrust? A. It increases H by delta*EIc/∫y^2dx B. It has no effect C. It reduces H by delta*EIc/∫y^2dx D. It makes H zero Answer: C. H = H_load - delta*EIc/∫y^2dx; outward yield relaxes the thrust (here -39.06 kN). (Key point: Redundant and compatibility) 2. With H_load = 450 kN and the yield effect -39.06 kN, what is the thrust used? A. 489.06 kN B. 410.94 kN C. 450 kN D. 39.06 kN Answer: B. H = 450 + (-39.06) = 410.94 kN. (Key point: Thrust value) 3. What is the formula and result for M_D at x = 30 m (y = 15 m, M0 = 4500)? A. M = M0 - H y = -1664.06 kNm (hogging) B. M = M0 + H y C. M = M0 = 4500 kNm D. M = H y = 6164 kNm Answer: A. M_D = 4500 - 410.94(15) = -1664.06 kNm. (Key point: Bending moment at D) 4. At D (tan θ = 1/3, V0 = 150 kN), how are N and Q found? A. N = H sin θ + V0 cos θ; Q = H cos θ - V0 sin θ B. N = H cos θ + V0 sin θ; Q = V0 cos θ - H sin θ C. N = H + V0; Q = 0 D. N = V0 cos θ; Q = H sin θ Answer: B. Resolving H and V0 along and normal to the arch tangent gives N = 437.28 kN and Q = 12.35 kN. 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