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Cable on piers: tension, length, pier thrust (180 m)

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.

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.

Set-up. The cable carries a UDL ww = 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=1510+5 = 15 m above O. Let x1x_1, x2x_2 be the horizontal distances of A and B from O.

Position of the lowest point and HH

For a parabola, y=wx22Hy = \dfrac{w x^2}{2H}, hence x12x22=y1y2=1510\dfrac{x_1^2}{x_2^2} = \dfrac{y_1}{y_2} = \dfrac{15}{10}, so x1=1.2247 x2x_1 = 1.2247\,x_2. With x1+x2=180x_1 + x_2 = 180 m:

x2=1801+1.2247=80.908 m,x1=99.092 mH=wx222y2=15(80.908)22(10)=4909.60 kN\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: wx12/(2×15)=4909.60w x_1^2/(2\times 15) = 4909.60 kN (same).

Maximum tension

Vertical components at the supports: VA=wx1=1486.38V_A = w x_1 = 1486.38 kN, VB=wx2=1213.62V_B = w x_2 = 1213.62 kN. The tension is largest at the higher support A:

Tmax=H2+VA2=4909.602+1486.382=5129.66 kNT_{max} = \sqrt{H^2 + V_A^2} = \sqrt{4909.60^2 + 1486.38^2} = 5129.66\ \text{kN}

(At B: TB=5057.37T_B = 5057.37 kN.)

Length of the cable

For a parabolic arc from the lowest point, with end slope s=2y/xs = 2y/x:

l=x2[1+s2+sinh⁡−1ss]l = \frac{x}{2}\left[\sqrt{1+s^2} + \frac{\sinh^{-1}s}{s}\right]
Sidex (m)y (m)s = 2y/xLength (m)
A to O99.092150.3027100.585
O to B80.908100.247281.725

Total length of cable between the saddles = 182.31 m.

Thrust on the piers

The saddles are on rollers, so the horizontal pull HH of the main cable is balanced only by the horizontal component of the straight backstay. A backstay at 60° to the vertical has Tbsin⁡60∘=HT_b \sin 60^\circ = H:

Tb=Hsin⁡60∘=5669.11 kN,Vb=Tbcos⁡60∘=2834.56 kNT_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: PA=1486.38+2834.56=4320.93P_A = 1486.38 + 2834.56 = 4320.93 kN
  • Right pier: PB=1213.62+2834.56=4048.18P_B = 1213.62 + 2834.56 = 4048.18 kN

Answer: (i) Tmax=5129.66T_{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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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