Cable area, BMD of girder, cable length

Taking the figure as read: span 100 m, dip 8 m, central hinge C, loads 50 kN at 20 m and 80 kN at 30 m from A, girder weight 10 kN/m. A single cable and a single girder are assumed.

Cable pull $H$

$$\begin{aligned} R_A &= \frac{50(80)+80(70)}{100} = 96.0\ \text{kN} \\ M_{0C} &= 96.0(50) - 50(30) - 80(20) = 1700.0\ \text{kN m} \\ H_p &= \frac{1700}{8} = 212.5\ \text{kN}, \qquad H_d = \frac{10(100)^2}{8(8)} = 1562.5\ \text{kN} \\ H &= 1562.5 + 212.5 = 1775.0\ \text{kN} \end{aligned}$$

Cable area

At the supports $\tan\theta = 4d/L = 0.32$, $\sec\theta = 1.0500$:

$$T_{max} = H\sec\theta = 1775.0(1.0500) = 1863.67\ \text{kN}$$ $$A = \frac{T_{max}}{\sigma} = \frac{1863.67\times 10^3}{1500} = 1242.4\ \text{mm}^2$$

BMD of the girder

The self-weight is carried by the cable (no girder moment). For the point loads, $M = M_0 - H_p y$, $y = \dfrac{4(8)x(100-x)}{100^2}$:

$M$ = 0 at both ends and at the hinge C. The BMD is sagging from A to C (peak 952.0 kN m under the 80 kN load) and hogging from C to B (peak -425.0 kN m at 75 m).

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(top = 952.0, bottom = -425.0; '-' is zero line)

Length of the cable

Using $l = L\left[1+\dfrac{8}{3}\left(\dfrac{d}{L}\right)^2-\dfrac{32}{5}\left(\dfrac{d}{L}\right)^4\right]$ with $d/L = 0.08$:

$$l = 100\left[1+\frac{8}{3}(0.08)^2-\frac{32}{5}(0.08)^4\right] = 101.68\ \text{m}$$

Answer: Cable area = 1242.4 mm²; $M_{20} = 832.0$ kN m, $M_{30} = 952.0$ kN m, $M_C = 0$, maximum hogging $-425.0$ kN m at 75 m; cable length = 101.68 m.