Cable with four point loads, max tension in segments

The cable is a funicular polygon. Take A as the origin, the span as horizontal (25 m) and let $V_A$, $V_B$ be the vertical reactions and $H$ the (constant) horizontal pull. Loads: 30, 40, 25, 30 kN at $x$ = 5, 10, 15, 20 m. B is 0.5 m above A, and point 2 ($x$ = 10 m) is 2 m below the chord AB, i.e. $2 - 0.5\times\frac{10}{25} = 1.8$ m below the level of A.

Step 1: Equations for $H$ and $V_A$

Depth of the cable below A at a section $x$ is $\dfrac{V_A x - \sum P(x-x_i)}{H}$.

• At B (depth = $-0.5$ m, B is higher): $V_A(25) - [30(20)+40(15)+25(10)+30(5)] = -0.5H$, so $25V_A - 1600 = -0.5H$ ... (i)
• At point 2 (depth = 1.8 m): $V_A(10) - 30(5) = 1.8H$, so $10V_A - 150 = 1.8H$ ... (ii)

From (i), $V_A = 64 - 0.02H$. Put in (ii): $640 - 0.2H - 150 = 1.8H$, so $2H = 490$.

$$\begin{aligned} H &= 245.0\ \text{kN} \\ V_A &= 64 - 0.02(245.0) = 59.1\ \text{kN} \\ V_B &= 125 - 59.1 = 65.9\ \text{kN} \end{aligned}$$

Check: $V_A + V_B = 125$ kN = total load.

Step 2: Cable levels (depth below A)

Points 1, 2, 3, 4 are 1.206 m, 1.800 m, 1.578 m and 0.845 m below A; B is 0.5 m above A.

Step 3: Tension in each segment

Vertical component in a segment = $V_A$ minus the loads to its left; $T = \sqrt{H^2+V^2}$.

Answer: $H = 245.0$ kN; segment tensions are 252.03, 246.72, 245.24, 247.62 and 253.71 kN for segments A-1 to 4-B. The maximum tension is 253.71 kN in segment 4-B (next to the higher support B).