--- title: "Arch 40 m: ILDs for thrust, BM at 8 m, shear" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-40m-rise10-ild-thrust-bm-shear" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-40m-rise10-ild-thrust-bm-shear site: BCT Engineering --- # Arch 40 m: ILDs for thrust, BM at 8 m, shear Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2072 Asoj, 12 marks - Video: [ILD for Arches | Structural Analysis | Lecture 25 | English](https://www.youtube.com/watch?v=y1rTKKHFWQU) — Letz Prepare - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-40m-rise10-ild-thrust-bm-shear - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question A three hinged symmetrical parabolic arch has a span of 40 m and rise of 10 m. Draw ILD for: (i) horizontal thrust, (ii) BM at section 8 m from left support, (iii) ILD for normal thrust and radial shear at the same section. ## Short answer For a three-hinged parabolic arch, the ILD for horizontal thrust is triangular with peak 1.0 at the crown. The ILD for bending moment at a section 8 m from left support reaches +3.84 m at the section, zero at 15.38 m, and -2.4 m at the crown. ILDs for radial shear and normal thrust show jumps at the section due to load crossing. ## Answer **Data:** symmetrical three-hinged parabolic arch, $L = 40$ m, $f = 10$ m, $y = \dfrac{4\times 10\,x\,(40-x)}{40^2} = \dfrac{x(40-x)}{40}$. Section $D$ at $x = 8$ m: $y_D = 6.4$ m, $\tan\theta = \dfrac{40-16}{40} = 0.6$, $\theta = 30.96^\circ$, $\sin\theta = 0.5145$, $\cos\theta = 0.8575$. #### (i) ILD of horizontal thrust For a unit load at $x'$, the simple-beam moment at the crown is $\dfrac{x'}{2}$ (left half) or $\dfrac{40-x'}{2}$ (right half), so $$ H = \frac{M_C}{f} = \frac{x'}{20}\ (x'\le 20), \qquad \frac{40-x'}{20}\ (x'\ge 20) $$ This is a triangle with a peak of $1.0$ at the crown and zero at $A$ and $B$. #### (ii) ILD of bending moment at D $M_D = M_b - H\,y_D$, with $M_b = \dfrac{x'(40-8)}{40} = 0.8x'$ for $x'\le 8$ and $\dfrac{8(40-x')}{40} = 0.2\,(40-x')$ for $x'\ge 8$. - Load left of $D$: $M_D = 0.8x' - \dfrac{x'}{20}(6.4) = 0.48\,x'$ (so $+3.84$ m at $D$) - Load between $D$ and $C$: $M_D = 0.2\,(40-x') - 0.32\,x' = 8 - 0.52\,x'$ (so $-2.4$ m at the crown) - Load right of $C$: $M_D = 0.2\,(40-x') - 0.05\,(40-x')(6.4) = -0.12\,(40-x')$ The neutral point is at $s_0 = \dfrac{L^2}{3L-2a} = \dfrac{1600}{120-16} = 15.38$ m. #### (iii) ILDs of radial shear and normal thrust at D $$ V = V_b\cos\theta - H\sin\theta, \qquad N = V_b\sin\theta + H\cos\theta $$ with $V_b = -\dfrac{x'}{40}$ for a load left of $D$ and $\dfrac{40-x'}{40}$ for a load right of $D$. #### Ordinates | Load at x' (m) | H | M (m) | V (radial) | N | |---|---|---|---|---| | 0 | 0 | 0 | 0 | 0 | | 4 | 0.2 | 1.92 | -0.1886 | 0.12 | | 8- | 0.4 | 3.84 | -0.3773 | 0.2401 | | 8+ | 0.4 | 3.84 | 0.4802 | 0.7546 | | 12 | 0.6 | 1.76 | 0.2915 | 0.8746 | | 16 | 0.8 | -0.32 | 0.1029 | 0.9947 | | 20 | 1 | -2.4 | -0.0857 | 1.1147 | | 24 | 0.8 | -1.92 | -0.0686 | 0.8918 | | 28 | 0.6 | -1.44 | -0.0514 | 0.6688 | | 32 | 0.4 | -0.96 | -0.0343 | 0.4459 | | 36 | 0.2 | -0.48 | -0.0171 | 0.2229 | | 40 | 0 | 0 | 0 | 0 | (The two rows at 8 m show the jumps in $V$ ($=\cos\theta = 0.8575$) and in $N$ ($=\sin\theta = 0.5145$) as the unit load crosses the section.) #### Shapes - **H:** triangle, 0 at $A$, 1.0 at the crown, 0 at $B$. - **M at D:** 0 at $A$, rises to $+3.84$ m at $D$, falls to zero at the neutral point ($15.38$ m) and to $-2.4$ m at the crown, and returns to 0 at $B$. - **V at D:** negative from $A$ to just left of $D$ ($-0.3773$), jumps up by $\cos\theta = 0.8575$ to $+0.4802$ just right of $D$, then decreases linearly to zero at $x' = 18.18$ m; it is negative from there to $B$ ($-0.0857$ at the crown, then rising linearly to 0 at $B$). - **N at D:** positive (compressive) throughout; it rises to $0.2401$ just left of $D$, jumps by $\sin\theta = 0.5145$ to $0.7546$, rises to $1.1147$ at the crown and falls linearly to 0 at $B$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the shape of the ILD of H, and its peak ordinate here? A. Triangle, 0 at A and B, peak 1.0 at the crown B. Rectangle of height 1.0 C. Triangle with peak at the section, 0.4 D. Parabola with peak 2.0 at the crown Answer: A. H = M_C/f. A unit load at the crown gives M_C = L/4 = 10, so H = 10/10 = 1.0. (Key point: ILD of thrust) 2. What happens to the ILD of radial shear at D when the unit load crosses the section? A. It jumps by sin(theta) B. It jumps by cos(theta) = 0.8575 C. It is continuous D. It jumps by H Answer: B. The simple-beam shear jumps by 1 across the load point, so V = Vb cos(theta) - H sin(theta) jumps by cos(theta). N jumps by sin(theta). (Key point: Jump in shear ILD) 3. What is the neutral point of M at D? A. 8 m, at the section B. 20 m, at the crown C. 24 m D. 15.38 m from A, where the ILD changes sign Answer: D. s0 = L^2/(3L - 2a) = 1600/104 = 15.38 m. Left of it the M ordinates are positive, right of it negative. 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