--- title: "Cable on piers: tension, length, pier thrust (180 m)" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:cable-180m-5m-above-60-degree-backstay" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-180m-5m-above-60-degree-backstay site: BCT Engineering --- # Cable on piers: tension, length, pier thrust (180 m) Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2073 Magh, 10 marks - 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-180m-5m-above-60-degree-backstay - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question The suspension cable is suspended from two piers 180 m apart, left support being 5 m above the other. The cable carries uniformly distributed load of 15 kN/m in plan and has its lowest point 10 m below the lower support. The ends of the cables are attached to saddles on rollers on top of piers and the back stays which may be assumed straight are inclined at 60° to the vertical. Determine: (i) maximum tension in the cable, (ii) the length of the cable, (iii) maximum thrust on the pier. ## Short answer Cable length between saddles 182.31 m, maximum tension 5129.66 kN at the higher support. Maximum thrust on the pier is 4320.93 kN vertical; the roller saddle ensures zero moment on the pier. ## Answer **Set-up.** The cable carries a UDL $w$ = 15 kN/m on the horizontal span, so it is a parabola with its lowest point O as the origin. Let the left support A be higher. The lower (right) support B is 10 m above O, so A is $10+5 = 15$ m above O. Let $x_1$, $x_2$ be the horizontal distances of A and B from O. #### Position of the lowest point and $H$ For a parabola, $y = \dfrac{w x^2}{2H}$, hence $\dfrac{x_1^2}{x_2^2} = \dfrac{y_1}{y_2} = \dfrac{15}{10}$, so $x_1 = 1.2247\,x_2$. With $x_1 + x_2 = 180$ m: $$ \begin{aligned} x_2 &= \frac{180}{1+1.2247} = 80.908\ \text{m}, \qquad x_1 = 99.092\ \text{m} \\ H &= \frac{w x_2^2}{2 y_2} = \frac{15(80.908)^2}{2(10)} = 4909.60\ \text{kN} \end{aligned} $$ Check with the left side: $w x_1^2/(2\times 15) = 4909.60$ kN (same). #### Maximum tension Vertical components at the supports: $V_A = w x_1 = 1486.38$ kN, $V_B = w x_2 = 1213.62$ kN. The tension is largest at the higher support A: $$ T_{max} = \sqrt{H^2 + V_A^2} = \sqrt{4909.60^2 + 1486.38^2} = 5129.66\ \text{kN} $$ (At B: $T_B = 5057.37$ kN.) #### Length of the cable For a parabolic arc from the lowest point, with end slope $s = 2y/x$: $$ l = \frac{x}{2}\left[\sqrt{1+s^2} + \frac{\sinh^{-1}s}{s}\right] $$ | Side | x (m) | y (m) | s = 2y/x | Length (m) | |---|---|---|---|---| | A to O | 99.092 | 15 | 0.3027 | 100.585 | | O to B | 80.908 | 10 | 0.2472 | 81.725 | Total length of cable between the saddles = 182.31 m. #### Thrust on the piers The saddles are on rollers, so the horizontal pull $H$ of the main cable is balanced only by the horizontal component of the straight backstay. A backstay at 60° to the vertical has $T_b \sin 60^\circ = H$: $$ T_b = \frac{H}{\sin 60^\circ} = 5669.11\ \text{kN}, \qquad V_b = T_b\cos 60^\circ = 2834.56\ \text{kN} $$ The horizontal forces cancel at each saddle, so the pier carries only a vertical thrust (cable vertical component plus backstay vertical component) and no bending moment: - Left pier: $P_A = 1486.38 + 2834.56 = 4320.93$ kN - Right pier: $P_B = 1213.62 + 2834.56 = 4048.18$ kN **Answer:** (i) $T_{max} = 5129.66$ kN (at the higher support); (ii) cable length = 182.31 m; (iii) maximum thrust on the pier = 4320.93 kN (vertical, at the higher pier). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a parabolic cable with the lowest point O, supports are 15 m and 10 m above O. How are x1 and x2 related? A. x1/x2 = 15/10 B. x1/x2 = sqrt(15/10) = 1.2247 C. x1/x2 = (15/10)^2 D. x1 = x2 Answer: B. y = w x^2/(2H), so x1^2/x2^2 = y1/y2, giving x1 = 1.2247 x2. (Key point: Horizontal distances to lowest point) 2. Which expression gives H using the lower side? A. H = w x2^2/(2 y2) = 4909.6 kN B. H = w x2/(2 y2) C. H = w x2 y2/2 D. H = w (x1 + x2)/2 Answer: A. From y = w x^2/(2H), H = w x2^2/(2 y2) = 15(80.908)^2/20 = 4909.6 kN. (Key point: Horizontal pull) 3. Why is the tension greatest at the higher support A? A. Its vertical component w x1 is larger than w x2, while H is the same B. H is larger there C. It carries the backstay D. The cable is horizontal there Answer: A. H is constant, so T = sqrt(H^2 + V^2) is larger where V = w x is larger. That is at A, with 5129.66 kN. (Key point: Where tension is maximum) 4. With saddles on rollers and a backstay at 60 degrees to the vertical, how is the pier thrust found? A. Tb = H/cos60 and the pier takes H as bending B. Tb sin60 = H; the pier carries only vertical thrust = cable vertical component + Tb cos60 C. The pier carries H horizontally and the vertical load of the cable only D. The pier thrust equals the cable tension Answer: B. Rollers transmit no horizontal force, so the backstay horizontal component balances H. The pier gets only vertical loads, 1486.38 + 2834.56 = 4320.93 kN at A. 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