--- title: "Suspension 200 m: max/min tension, cable length" subject: "Theory of Structures I" chapter: 7 key: "semester_4/theory_of_structures_i:suspension-200m-dip20-3x800kn" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/suspension-200m-dip20-3x800kn site: BCT Engineering --- # Suspension 200 m: max/min tension, cable length Theory of Structures I · Chapter 7: Suspension Cable Systems - Asked in: 2076 Baisakh, 12 marks - Video: [Suspension Bridges Unsymmetrical Cable Problem 1 | Tension in Cables Supported At Different Level](https://www.youtube.com/watch?v=peLiegSKNes) — Ganesh Shegar - 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/suspension-200m-dip20-3x800kn - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-7 ## Question A suspension bridge, 200 m span, has two three-hinged stiffening girders supported by two cables with a central dip of 20 m. The dead load of the bridge is 50 kN/m run and in addition, it supports three point loads of 800 kN each placed along the centre line of the roadway and dividing the span in four equal parts. Calculate the maximum tension and minimum tension with their locations in the cable and the length of the cable. ## Short answer Max tension T_max = 8885.5 kN at supports; min tension T_min = 8250 kN at mid-span. Cable length = 205.21 m. ## Answer **Data:** span $L = 200$ m, two cables and two three-hinged stiffening girders, dip $d = 20$ m. Dead load of the bridge 50 kN/m (total); three point loads of 800 kN each on the centre line at 50, 100 and 150 m. The loads are shared equally by the two cables, so for **one cable**: $w = 25$ kN/m and three loads of 400 kN. Cable: $y = \dfrac{4d\,x(L-x)}{L^2}$, $\tan\alpha_s = \dfrac{4d}{L} = 0.4$. #### Horizontal pull (one cable) **Dead load:** $H_d = \dfrac{wL^2}{8d} = \dfrac{25\times 200^2}{8\times 20} = 6250\ \text{kN}$ **Point loads:** the simple-beam reaction is $R_A = 1.5\times 400 = 600$ kN (symmetrical), and the moment at the middle hinge: $$ M_C = 600\times 100 - 400\times 50 = 40000\ \text{kN m}, \qquad H_p = \frac{M_C}{d} = \frac{40000}{20} = 2000\ \text{kN} $$ $$ H = H_d + H_p = 8250\ \text{kN} $$ #### Maximum and minimum tension in the cable - **Minimum tension:** at mid-span (the lowest point of the cable), where the cable is horizontal: $T_{min} = H = 8250$ kN. - **Maximum tension:** at the supports, where $\tan\alpha_s = 0.4$ and the vertical component is $V_s = H\tan\alpha_s = 3300$ kN: $$ T_{max} = \sqrt{H^2 + V_s^2} = H\sqrt{1+0.16} = 8885.5\ \text{kN} $$ #### Length of the cable $$ S = L\left[1 + \frac83\left(\frac{d}{L}\right)^2 - \frac{32}{5}\left(\frac{d}{L}\right)^4\right] = 200\left[1+\frac83(0.1)^2 - \frac{32}{5}(0.1)^4\right] = 205.21\ \text{m} $$ **Answer: $T_{max} = 8885.5$ kN (at the supports), $T_{min} = 8250$ kN (at mid-span), length of each cable $= 205.21$ m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With two cables sharing the loads equally, what does one cable carry? A. w = 50 kN/m and three 800 kN loads B. w = 25 kN/m and three 400 kN loads C. w = 25 kN/m and three 800 kN loads D. w = 100 kN/m and three 1600 kN loads Answer: B. Both dead load and point loads are halved: 25 kN/m and 400 kN each. (Key point: Load per cable) 2. What is the total horizontal pull H for one cable? A. 6250 kN B. 2000 kN C. 8250 kN D. 4125 kN Answer: C. Hd = 25 x 200^2/(8 x 20) = 6250 kN and Hp = 40000/20 = 2000 kN, so H = 8250 kN. (Key point: Horizontal pull components) 3. Where is the cable tension minimum, and what is its value? A. At the supports, equal to H B. At mid-span, equal to H = 8250 kN C. At the quarter points, equal to 3300 kN D. At mid-span, equal to Vs = 3300 kN Answer: B. At the lowest point the cable is horizontal, so its tension is only H = 8250 kN. (Key point: Minimum tension location) 4. Why is Tmax = 8885.5 kN at the supports? A. Because the supports have vertical component Vs = H tan(alpha) = 3300 kN added to H as a resultant B. Because the supports have the largest horizontal force C. Because Tmax = H + Vs directly D. Because the dead load acts only at the supports Answer: A. H is constant, but the slope is greatest at the supports (tan = 0.4), so T = H sqrt(1 + 0.16) = 8885.5 kN. (Key point: Maximum tension) ## Close variants set in other exams - [Suspension 400 m: max/min tension, cable length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-400m-dip30-3x300kn) — Theory of Structures I, chapter 7, asked 3× - [Suspension 100 m: max/min tension, SFD, BMD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-dip10-3x200kn-dead-40) — Theory of Structures I, chapter 7, asked 3× ## Related in Theory of Structures I - [Suspension 50 m: max tension, BM at 10 m, BMD/SFD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-50m-5m-dip-80kn) — Theory of Structures I, chapter 7 - [Suspension 100 m: UDL on right half, SF and BM at 20 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-100m-5knm2-10knm2-right-half) — Theory of Structures I, chapter 7, asked 2× - [Suspension 150 m: BM and SF at 30 m, tension, length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-dip15-180-120-dead-5) — Theory of Structures I, chapter 7, asked 2× - [Suspension 150 m: SF, BM at 30 m, tension, length](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-dip15-180-150-dead-7-5) — Theory of Structures I, chapter 7, asked 2× - [Suspension 150 m: four moving loads, BM at 40 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Asuspension-150m-four-moving-loads-200-150-300-100) — Theory of Structures I, chapter 7 - [Cable on piers: tension, length, pier thrust (200 m)](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Acable-200m-5m-above-60-degree-backstay) — Theory of Structures I, chapter 7, asked 2×