Suspension 120 m: SF, BM at 25 m, cable area

Load shares. The bridge has two cables and two girders, so each takes half. Dead load: $5\times 6 = 30$ kN/m, so 15 kN/m per girder. The live loads (200 kN at 20 m and 150 kN at 30 m from the left support, both on the centre line) give 100 kN and 75 kN per girder. $L$ = 120 m, $d$ = 12 m.

Cable pull $H$

$$\begin{aligned} R_A &= \frac{100(100)+75(90)}{120} = 139.58\ \text{kN} \\ M_{0C} &= 139.58(60) - 100(40) - 75(30) = 2125.0\ \text{kN m} \\ H_p &= \frac{2125}{12} = 177.08\ \text{kN}, \qquad H_d = \frac{15(120)^2}{8(12)} = 2250.0\ \text{kN} \\ H &= 2250.0 + 177.08 = 2427.08\ \text{kN} \end{aligned}$$

SF and BM at 25 m (one of the two girders)

The dead load is carried entirely by the cable and produces no moment or shear in the girder.

$$y_{25} = \frac{4(12)(25)(95)}{120^2} = 7.9167\ \text{m},\qquad \frac{dy}{dx} = \frac{4(12)(120-50)}{120^2} = 0.2333$$ $$\begin{aligned} M_{25} &= R_A(25) - 100(5) - H_p\,y_{25} = 2989.58 - 1401.91 = 1587.67\ \text{kN m} \\ S_{25} &= (R_A - 100) - H_p\,\frac{dy}{dx} = 39.58 - 41.32 = -1.74\ \text{kN} \end{aligned}$$

(If the whole roadway is taken as carried by a single girder, the girder values double: $M_{25} = 3175.3$ kN m and $S_{25} = -3.47$ kN. The cable force per cable is unchanged.)

Cross-sectional area of the cable

Maximum tension (at the supports, $\tan\theta = 4d/L = 0.40$, $\sec\theta = 1.0770$):

$$T_{max} = H\sec\theta = 2427.08(1.0770) = 2614.05\ \text{kN}$$ $$A = \frac{T_{max}}{\sigma_{allow}} = \frac{2614.05\times 10^3}{120} = 21784\ \text{mm}^2$$

Answer: $M_{25} = 1587.67$ kN m (sagging); $S_{25} = -1.74$ kN (almost zero, so 25 m is close to the section of maximum BM); required cable area = 21784 mm² for each of the two cables.