Cable at different levels: length and thrust

Data: span $L = 50$ m; the left support is 3 m below the right one; the lowest point of the cable is 5 m below the lower (left) support. So the heights above the lowest point are $h_1 = 5$ m (left) and $h_2 = 5 + 3 = 8$ m (right). UDL $w = 15$ kN/m on the horizontal span. The cable is a parabola with its vertex at the lowest point.

Horizontal thrust

With $y = \dfrac{w x^2}{2H}$ from the lowest point and $x_1 + x_2 = L$:

$$H = \frac{wL^2}{2\left(\sqrt{h_1}+\sqrt{h_2}\right)^2} = \frac{15\times 50^2}{2\left(\sqrt5+\sqrt8\right)^2} = \frac{37500}{2\times(5.0645)^2} = 731.0\ \text{kN}$$

Position of the lowest point

$$x_1 = \frac{L\sqrt{h_1}}{\sqrt{h_1}+\sqrt{h_2}} = 22.08\ \text{m (from the left support)}, \qquad x_2 = 50 - x_1 = 27.92\ \text{m}$$

Check: $\dfrac{w x_1^2}{2H} = 5.000$ m and $\dfrac{w x_2^2}{2H} = 8.000$ m.

Maximum tension

The tension is greatest at the higher (right) support:

$$T_{2} = \sqrt{H^2 + (w x_2)^2} = \sqrt{731.0^2 + (15\times 27.92)^2} = 842.5\ \text{kN}$$

(At the left support, $T_1 = \sqrt{H^2 + (w x_1)^2} = 802.5$ kN.)

Length of the cable

For each portion from the lowest point, with $u = \dfrac{w x}{H}$:

$$s = \frac{H}{w}\cdot\frac{u\sqrt{1+u^2} + \sinh^{-1}u}{2}$$

Left portion: $u_1 = 0.4530$, $s_1 = 22.81$ m. Right portion: $u_2 = 0.5730$, $s_2 = 29.38$ m.

$$S = s_1 + s_2 = 52.19\ \text{m}$$

(Approximation: $s \approx x\left[1 + \tfrac23\left(\tfrac{h}{x}\right)^2\right]$ gives 52.28 m.)

Answer: $H = 731.0$ kN; maximum tension $= 842.5$ kN (at the higher support); length of the cable $= 52.19$ m.