--- title: "Parabolic arch 20 m: two loads and UDL, 5 m section" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-20m-two-loads-udl-right-half" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-20m-two-loads-udl-right-half site: BCT Engineering --- # Parabolic arch 20 m: two loads and UDL, 5 m section Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2078 Chaitra, 12 marks - Video: [3 Hinged parabolic Arch with UDL over entire span](https://www.youtube.com/watch?v=KRwpFZgltBw) — Civilholic Guy - 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-20m-two-loads-udl-right-half - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question A parabolic 3-hinged arch has a span of 20 m with a central rise of 5 m. The arch is loaded with two point loads of 20 kN and 30 kN at a distance of 3 m and 7 m respectively, from the left support. It also carries a UDL of 25 kN/m over right half of the span. Determine the bending moment, normal thrust and radial shear at 5 m from left support. ## Short answer At 5 m from the left support, bending moment = -115 kN m, normal thrust = 171.28 kN, and radial shear = 2.68 kN, using horizontal thrust H = 152 kN. ## Answer **Data:** parabolic three-hinged arch, $L = 20$ m, rise $f = 5$ m. Loads: 20 kN at 3 m and 30 kN at 7 m from $A$; UDL 25 kN/m on the right half ($x = 10$ to 20 m). Section at $x = 5$ m from $A$. #### Step 1: Equation of the arch $$ y = \frac{4f\,x\,(L-x)}{L^2} = \frac{4\times 5\,x\,(20-x)}{20^2} $$ $$ \tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2} $$ At the section $x = 5$ m: $y = 3.75$ m, $\tan\theta = 0.5$, so $\theta = 26.57^\circ$, $\sin\theta = 0.4472$, $\cos\theta = 0.8944$. #### Step 2: Vertical reactions Taking moments about $A$ for the whole arch: $$ R_B \times 20 = 20\times 3 + 30\times 7 + (25\times 10)\times 15 = 4020 \Rightarrow R_B = 201\ \text{kN} $$ $$ R_A = 300 - 201 = 99\ \text{kN} $$ #### Step 3: Horizontal thrust The bending moment at the crown hinge $C$ is zero. Taking moments of the forces on the left of $C$ about $C$ (rise $f = 5$ m): $$ H \times 5 = 99\times 10 - 20\times 7 - 30\times 3 = 760 \Rightarrow H = 152\ \text{kN} $$ #### Step 4: Forces at the section (x = 5 m) Simple-beam values at the section: $M_b = 455$ kN m, $V_b = 79$ kN. $$ \begin{aligned} M &= M_b - H\,y = 455 - 152\times 3.75 = -115\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 79\times 0.8944 - 152\times 0.4472 = 2.68\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 79\times 0.4472 + 152\times 0.8944 = 171.28\ \text{kN} \end{aligned} $$ **Answer: at 5 m from the left support: $M = -115$ kN m, normal thrust $N = 171.28$ kN, radial shear $V = 2.68$ kN.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is R_B found for the point loads and the UDL on the right half? A. Moments about C for the right half only: R_B = 125 kN B. Moments about A: R_B x 20 = 20x3 + 30x7 + (25x10)x15, giving 201 kN C. R_B = total load/2 = 150 kN D. Moments about A with the UDL taken at 10 m: R_B = 150 kN Answer: B. The UDL of 250 kN acts at its centroid 15 m from A. This gives 4020/20 = 201 kN, and R_A = 300 - 201 = 99 kN. (Key point: Vertical reaction) 2. How is H obtained? A. H = R_A = 99 kN B. H x 5 = 99 x 5, so H = 99 kN C. Moment at the crown hinge is zero: H x 5 = 99x10 - 20x7 - 30x3, so H = 152 kN D. H = wL^2/8f for the UDL = 100 kN Answer: C. Taking moments about C of the left-hand forces, both point loads are left of C. H x 5 = 990 - 140 - 90 = 760, so H = 152 kN. (Key point: Thrust from the crown condition) 3. M at x = 5 m is Mb - H y = 455 - 152 x 3.75 = -115 kN m. What does the negative sign mean? A. Hogging moment at the section because the thrust term exceeds the beam moment B. Sagging moment because the load is downward C. The section is in tension at the bottom D. The calculation is wrong, since M must be positive in an arch Answer: A. M = Mb - Hy. When H y is greater than Mb the result is negative, i.e. hogging. (Key point: Sign of the bending moment) 4. Which is the correct expression for normal thrust at the section (theta = 26.57 degrees)? A. N = Vb cos(theta) - H sin(theta) = 2.68 kN B. N = H = 152 kN C. N = Vb + H = 231 kN D. N = Vb sin(theta) + H cos(theta) = 171.28 kN Answer: D. N is the component of the forces along the tangent: 79x0.4472 + 152x0.8944 = 171.28 kN. The other expression is the radial shear. 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