Cable with UDL: length, H and max tension

Set-up. Take the lowest point O of the cable as origin. A is 2 m and B is 8 m above O. For a parabolic cable under a UDL on the horizontal span, $y = \dfrac{wx^2}{2H}$.

Position of the lowest point

$$\frac{x_A^2}{x_B^2} = \frac{y_A}{y_B} = \frac{2}{8} \Rightarrow x_A = 0.5\,x_B,\qquad x_A + x_B = 45$$

So $x_B = 30.0$ m and $x_A = 15.0$ m. The lowest point is 15 m from A horizontally (30 m from B).

(b) Horizontal component of tension

$$H = \frac{w x_B^2}{2y_B} = \frac{20(30)^2}{2(8)} = 1125.0\ \text{kN}$$

Check from A: $\dfrac{20(15)^2}{2(2)} = 1125.0$ kN.

(c) Maximum tension and its position

Tension is greatest at the higher support B (30 m from the lowest point):

$$V_B = w x_B = 20(30) = 600\ \text{kN},\qquad T_{max} = \sqrt{H^2 + V_B^2} = \sqrt{1125.0^2 + 600^2} = 1275.00\ \text{kN}$$

At A: $V_A = 300$ kN and $T_A = 1164.31$ kN.

(a) Length of the cable

For each side, with end slope $s = 2y/x$: $l = \dfrac{x}{2}\left[\sqrt{1+s^2} + \dfrac{\sinh^{-1}s}{s}\right]$

Total length = 46.54 m.

Answer: (a) length = 46.54 m; (b) $H$ = 1125.0 kN; (c) $T_{max}$ = 1275.00 kN at support B (the higher end).