Cable 20 m: three loads, length and area

Loads 40, 30 and 20 kN are taken at 5 m, 10 m and 15 m from the left support (equal spacing on the 20 m span). The central dip (at the 30 kN load) is 0.96 m.

Reactions and horizontal pull

$$\begin{aligned} V_A &= \frac{40(15)+30(10)+20(5)}{20} = 50.0\ \text{kN},\quad V_B = 90 - 50.0 = 40.0\ \text{kN} \\ M_{10} &= 50.0(10) - 40(5) = 300.0\ \text{kN m} \\ H &= \frac{M_{10}}{0.96} = 312.50\ \text{kN} \end{aligned}$$

Cable profile and segment lengths

Depth of a load point below the supports is $y = M/H$ where $M$ is the beam moment there: $y_1 = 0.800$ m (5 m), $y_2$ = 0.960 m (10 m), $y_3 = 0.640$ m (15 m).

Length of cable required = sum of segments = 20.117 m.

Cross-sectional area

The maximum tension occurs in the end segment A-1: $T_{max} = 316.47$ kN. The permissible stress is read as 250 N/mm² (250 kN/mm² is not physically possible for a cable):

$$A = \frac{T_{max}}{\sigma} = \frac{316.47\times 10^3}{250} = 1265.9\ \text{mm}^2$$

Shape when hanging under its own weight only

With no point loads, the cable carries only its own weight, uniformly distributed along its length, and hangs in a catenary, $y = \dfrac{H}{w}\left(\cosh\dfrac{wx}{H}-1\right)$. For a flat cable (dip/span below about 1/10) this is very close to a parabola. Its length is unchanged at 20.117 m, so the dip follows from $l \approx L\left[1+\frac{8}{3}\left(\frac{d}{L}\right)^2\right]$:

$$d = \sqrt{\frac{3L(l-L)}{8}} = \sqrt{\frac{3(20)(20.117-20)}{8}} = 0.94\ \text{m}$$

The cable hangs symmetrically with its lowest point at mid-span and a dip of about 0.94 m. The point loads make it a polygon whose dip (0.96 m at the 30 kN load) is slightly larger.

Answer: length = 20.12 m; area = 1266 mm²; unloaded cable: catenary (nearly parabolic) with dip about 0.94 m.