--- title: "Parabolic arch at different levels: BM at 15 m" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-diff-levels-10kn-15m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-diff-levels-10kn-15m site: BCT Engineering --- # Parabolic arch at different levels: BM at 15 m Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2081 Chaitra, 8 marks - Video: [Three Hinged Parabolic Arch - Problem No 4 (With Springings at different Level)](https://www.youtube.com/watch?v=GjUCZb3wzrE) — Stan Academy - 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-diff-levels-10kn-15m - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question A parabolic three hinged arch shown in figure below has a span of 40 m and is supported at different levels as shown. Find bending moment at a section 15 m from left support when a 10-kN vertical load at this section. [Figure: crown hinge, left support A at the lower level with the crown 11.25 m above A, right support B 5 m below the crown level as read from the scan; 10 kN at C, 15 m from A; horizontal span 40 m. Dimensions partly unclear.] ## Short answer From equilibrium of the whole arch and the left part about the crown, horizontal thrust is 8 kN, vertical reactions are 7.5 kN (left) and 2.5 kN (right). At the section 15 m from the left support, where the 10 kN load acts, the bending moment is 35.16 kN m (sagging). ## Answer **Assumptions (dimensions partly unclear in the figure):** horizontal span 40 m; the left support $A$ is 11.25 m and the right support $B$ is 5 m below the crown hinge $C$, which is the highest point (vertex) of the parabola. So $B$ is $11.25 - 5 = 6.25$ m above $A$. A 10 kN vertical load acts at the section $D$, 15 m from $A$. #### Position of the crown For a parabola with its vertex at $C$, the depth below the crown is proportional to the square of the horizontal distance from it. If $L_1$ and $L_2$ are the horizontal distances of $A$ and $B$ from $C$: $$ \frac{L_1}{L_2} = \sqrt{\frac{11.25}{5}} = 1.5, \qquad L_1 + L_2 = 40 \Rightarrow L_1 = 24\ \text{m},\ L_2 = 16\ \text{m} $$ #### Equation of the arch and height of the section With the origin at $C$ and depth $y'$ measured downwards: $y' = kx^2$, $k = \dfrac{11.25}{24^2} = \dfrac{5}{16^2} = 0.019531$. The section is $24 - 15 = 9$ m left of $C$: $y' = 0.019531\times 81 = 1.582$ m, so the section is $11.25 - 1.582 = 9.668$ m above $A$. #### Reactions Let the reaction at $A$ be $V_A$ (up) and $H$ (to the right); at $B$, $V_B$ (up) and $H$ (to the left). Because the supports are at different levels, the vertical and horizontal components cannot be separated by moments about $A$ and $B$ alone, so both conditions are used together (anticlockwise moments positive). **Moments about $B$** ($B$ is 40 m right of and 6.25 m above $A$; the load is 25 m left of $B$): $$ -40V_A + 6.25H + 10\times 25 = 0 \qquad (1) $$ **Moments about the hinge $C$ for the left part** ($C$ is 24 m right of and 11.25 m above $A$; the load is 9 m left of $C$): $$ -24V_A + 11.25H + 10\times 9 = 0 \qquad (2) $$ From (1): $V_A = \dfrac{6.25H + 250}{40}$. Substituting in (2): $-3.75H - 150 + 11.25H + 90 = 0 \Rightarrow H = \dfrac{60}{7.5} = 8$ kN. $$ V_A = \frac{6.25\times 8 + 250}{40} = 7.5\ \text{kN}, \qquad V_B = 10 - 7.5 = 2.5\ \text{kN}, \qquad H = 8\ \text{kN} $$ #### Bending moment at the section (x = 15 m) $$ M = V_A\times 15 - H\times y = 7.5\times 15 - 8\times 9.668 = 112.5 - 77.34 = 35.16\ \text{kN m} $$ (The 10 kN load acts at the section, so it has no moment about it.) **Answer: bending moment at the section = 35.16 kN m (sagging), with $H = 8$ kN, $V_A = 7.5$ kN and $V_B = 2.5$ kN.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the origin at A and the arch passing through A, the parabola is written y = ax^2 + bx. Why is there no constant term? A. Because y = 0 at x = 0 (the curve passes through the origin A) B. Because the crown is at x = 0 C. Because the arch is symmetric about A D. Because the load acts at x = 0 Answer: A. The curve passes through the origin A, so the constant term must be zero. a and b then follow from the conditions at C and B. (Key point: Parabola with supports at different levels) 2. The crown C is 11.25 m above A and B is 5 m below the crown level. How high is B above A? A. 16.25 m B. 5 m C. 6.25 m D. 11.25 m Answer: C. B is 11.25 - 5 = 6.25 m above A, which is the ordinate used in the parabola and in the moment equation about B. (Key point: Position of B) 3. Why are two moment conditions (about B for the whole arch and about hinge C for the left part) solved together? A. Because the load is at the section, so one equation is zero B. Moments about A and B alone cannot separate V and H when the supports are at different levels, so the hinge condition is needed as well C. Because the arch is symmetric, so V and H are equal D. Because the hinge C carries a vertical reaction Answer: B. The moment about B contains both V_A and H_A. The zero-moment condition at C gives a second equation, and solving both gives H_A = 9.231 kN and V_A = 7.692 kN. (Key point: Reactions at different levels) 4. What is the bending moment at the section 15 m from A, and why does the 10 kN load not enter? A. M = V_A x 15 - H_A x y = 23.44 kN m; the load acts at the section so its moment arm is zero B. M = V_A x 15 + H_A x y = 23.44 kN m; the load is balanced by V_B C. M = V_A x 15 - 10 x 15; the horizontal thrust has no effect D. M = H_A x y only; the vertical reaction acts through the section Answer: A. Taking moments of the forces on the left, V_A gives sagging and H_A x y gives the opposite effect. The 10 kN load acts at the section, so it adds nothing. 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