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Cable over 300 m gap: tension and length

Short answer

Horizontal tension H = wL^2/(8d) = 2.25×10^6 kN, vertical reaction V = wL/2 = 45,000 kN. Maximum tension at supports is Tmax = sqrt(H^2+V^2) ≈ 2,250,450 kN. Cable length S ≈ 300 + 8×1.5^2/(3×300) = 300.020 m.

A cable is stretched over a gap of 300 m and carries uniformly distributed load of 300 kN/m horizontally. If the central dip is 1.5 m, calculate the maximum tension in the cable. Also find the length of cable. [Load value read as 300 kN/m from the scan, partly unclear.]

Data: span L=300L = 300 m, central dip d=1.5d = 1.5 m (supports at the same level), uniformly distributed load w=300w = 300 kN/m on the horizontal span (the value is as read from the scan; the method is the same for any ww). The cable is a parabola.

Horizontal tension

H=wL28d=300×30028×1.5=2250000 kNH = \frac{wL^2}{8d} = \frac{300\times 300^2}{8\times 1.5} = 2250000\ \text{kN}

Vertical reaction at each support

V=wL2=300×3002=45000 kNV = \frac{wL}{2} = \frac{300\times 300}{2} = 45000\ \text{kN}

Maximum tension (at the supports)

Tmax=H2+V2=(2250000)2+(45000)2=2250450.0 kNT_{max} = \sqrt{H^2 + V^2} = \sqrt{(2250000)^2 + (45000)^2} = 2250450.0\ \text{kN}

Slope at the supports: tan⁡α=4dL=6300=0.02\tan\alpha = \dfrac{4d}{L} = \dfrac{6}{300} = 0.02, α=1.15∘\alpha = 1.15^\circ.

Length of the cable

S=L[1+83(dL)2−325(dL)4]=300[1+83(0.005)2−325(0.005)4]=300.020 mS = L\left[1 + \frac{8}{3}\left(\frac{d}{L}\right)^2 - \frac{32}{5}\left(\frac{d}{L}\right)^4\right] = 300\left[1 + \frac{8}{3}(0.005)^2 - \frac{32}{5}(0.005)^4\right] = 300.020\ \text{m}

Answer: Tmax=2250450.0T_{max} = 2250450.0 kN (at the supports); length of the cable ≈300.020\approx 300.020 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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