--- title: "Horizontal reaction of cable, derive H formula" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:cable-horizontal-reaction-formula" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-horizontal-reaction-formula site: BCT Engineering --- # Horizontal reaction of cable, derive H formula Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2065 Chaitra, 8 marks - Video: [Cables supported on different levels subjected to Uniformly Distributed Load ( UDL )](https://www.youtube.com/watch?v=R4ALc2U5zdQ) — 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/cable-horizontal-reaction-formula - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question A suspension cable of span L has its ends at the heights $h_1$ and $h_2$ above the lowest point of cable. It carries a uniformly distributed load of w per unit run of the span. Show that the horizontal reaction at each end is given by $H = \frac{wL^2}{2(\sqrt{h_1}+\sqrt{h_2})^2}$. ## Short answer With the parabolic equation y = wx^2/(2H), the distances from the lowest point to supports at heights h1, h2 are x1 = sqrt(2Hh1/w), x2 = sqrt(2Hh2/w). Setting x1 + x2 = L yields the horizontal reaction H = wL^2/[2(sqrt(h1)+sqrt(h2))^2]. ## Answer **Data:** cable of span $L$ carrying a UDL $w$ per unit horizontal length; the supports are at heights $h_1$ and $h_2$ above the lowest point of the cable. Let $x_1$ and $x_2$ be the horizontal distances from the lowest point to the supports, so $x_1 + x_2 = L$. ```text support 1 o o support 2 \ / ^ h1 \ / | h2 |\___.___lowest___.' ||<---x2---->| ``` #### Equation of the cable Take the origin at the lowest point $O$, where the tension is horizontal and equal to $H$. For a point $P$ at horizontal distance $x$ from $O$, take moments about $P$ of the forces on the portion $OP$ of the cable: $$ H\,y = \frac{w\,x^2}{2} \Rightarrow y = \frac{w\,x^2}{2H} $$ #### Apply to the two supports $$ h_1 = \frac{w\,x_1^2}{2H} \Rightarrow x_1 = \sqrt{\frac{2H\,h_1}{w}}, \qquad h_2 = \frac{w\,x_2^2}{2H} \Rightarrow x_2 = \sqrt{\frac{2H\,h_2}{w}} $$ Since $x_1 + x_2 = L$: $$ \sqrt{\frac{2H}{w}}\,\left(\sqrt{h_1}+\sqrt{h_2}\right) = L $$ Squaring both sides: $$ \frac{2H}{w}\left(\sqrt{h_1}+\sqrt{h_2}\right)^2 = L^2 $$ $$ \boxed{H = \frac{w\,L^2}{2\left(\sqrt{h_1}+\sqrt{h_2}\right)^2}} $$ This is the horizontal reaction at each end, since the cable carries no horizontal external load. **Check:** for $h_1 = h_2 = h$, $H = \dfrac{wL^2}{2\times 4h} = \dfrac{wL^2}{8h}$, the usual result for supports at the same level. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Taking the origin at the lowest point, what is the cable equation under UDL w? A. y = wx/H B. y = w x^2/(2H), from H y = w x^2/2 C. y = H x^2/(2w) D. y = w x^2/H Answer: B. Moments about a point on the cable for the portion from the lowest point give H y = w x^2/2. (Key point: Equation of cable) 2. What conditions are applied to get H? A. x1 = x2 and h1 = h2 B. h1 = w x1^2/(2H), h2 = w x2^2/(2H) and x1 + x2 = L C. H1 = H2 and V1 = V2 D. T = H at both supports Answer: B. Each support satisfies the cable equation, and the two horizontal distances add to the span. (Key point: Using the support conditions) 3. What is H, and what does it reduce to when h1 = h2 = h? A. H = wL^2/(2(sqrt h1 + sqrt h2)^2); wL^2/(8h) B. H = wL^2/(h1 + h2); wL^2/(2h) C. H = wL^2/(8(h1 + h2)); wL^2/(16h) D. H = wL/(sqrt h1 + sqrt h2); wL/(2 sqrt h) Answer: A. Substituting x1 + x2 = L gives the boxed formula; for equal heights the denominator is 2 x 4h = 8h, the usual result. 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