--- title: "Cable on piers: tension, length, pier thrust (200 m)" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:cable-200m-5m-above-60-degree-backstay" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/cable-200m-5m-above-60-degree-backstay site: BCT Engineering --- # Cable on piers: tension, length, pier thrust (200 m) Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2069 Poush, 8 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-200m-5m-above-60-degree-backstay - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question A suspension cable is suspended from two piers 200 m apart, left support being 5 m above the other. The cable carries a 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 backstays which may be assumed straight are inclined at 60° to the vertical. Determine: (i) the length of cable, (ii) maximum tension in cable and (iii) maximum thrust and moment on pier. ## Short answer Cable length 202.08 m, maximum tension 6282.20 kN at the higher support. Maximum vertical thrust on the pier is 5150.98 kN, and the moment is zero because the roller saddle transmits no horizontal force. ## 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 $x_1 + x_2 = 200$ m: $$ \begin{aligned} x_2 &= \frac{200}{1+1.2247} = 89.898\ \text{m}, \qquad x_1 = 110.102\ \text{m} \\ H &= \frac{w x_2^2}{2 y_2} = \frac{15(89.898)^2}{2(10)} = 6061.23\ \text{kN} \end{aligned} $$ Check with the left side: $w x_1^2/(2\times 15) = 6061.23$ kN (same). #### Maximum tension Vertical components at the supports: $V_A = w x_1 = 1651.53$ kN, $V_B = w x_2 = 1348.47$ kN. The tension is largest at the higher support A: $$ T_{max} = \sqrt{H^2 + V_A^2} = \sqrt{6061.23^2 + 1651.53^2} = 6282.20\ \text{kN} $$ (At B: $T_B = 6209.42$ 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 | 110.102 | 15 | 0.2725 | 111.450 | | O to B | 89.898 | 10 | 0.2225 | 90.634 | Total length of cable between the saddles = 202.08 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 $T_b \sin 60^\circ = H$: $$ T_b = \frac{H}{\sin 60^\circ} = 6998.91\ \text{kN}, \qquad V_b = T_b\cos 60^\circ = 3499.45\ \text{kN} $$ 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: $P_A = 1651.53 + 3499.45 = 5150.98$ kN - Right pier: $P_B = 1348.47 + 3499.45 = 4847.92$ kN **Answer:** (i) cable length = 202.08 m; (ii) $T_{max} = 6282.20$ kN at the higher support; (iii) maximum thrust on pier = 5150.98 kN (vertical), and moment on the pier = 0 because the rollers transmit no horizontal force. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With supports 15 m and 10 m above the lowest point and a 200 m span, what are x2 and x1? A. x2 = 89.9 m, x1 = 110.1 m B. x2 = 110.1 m, x1 = 89.9 m C. x2 = 100 m, x1 = 100 m D. x2 = 80 m, x1 = 120 m Answer: A. x1 = 1.2247 x2 and x1 + x2 = 200, so x2 = 200/2.2247 = 89.898 m. The lower support is nearer to the lowest point. (Key point: Position of lowest point) 2. Where is the maximum tension and what is it? A. At the lower support B, 6209.42 kN B. At the higher support A, 6282.20 kN C. At the lowest point, 6061.23 kN D. At the higher support A, 6061.23 kN Answer: B. The vertical component w x1 = 1651.53 kN is greater at A, so T = sqrt(H^2 + V^2) = 6282.20 kN there. (Key point: Maximum tension) 3. Why is the moment on the pier zero? A. The saddles are on rollers, so horizontal forces balance via the backstay and the pier gets only vertical thrust B. The pier is hinged at the base C. H is zero at the pier D. The backstay is vertical Answer: A. Rollers transmit no horizontal force; the horizontal parts of the cable and backstay cancel, so there is no bending, only vertical thrust (5150.98 kN). 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