Cable on piers: tension, length, pier thrust (200 m)

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 $s = 2y/x$:

$$l = \frac{x}{2}\left[\sqrt{1+s^2} + \frac{\sinh^{-1}s}{s}\right]$$

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.