Suspension 150 m: UDL right half, SF and BM at 50 m

Data and assumptions: span $L = 150$ m, two cables with two three-hinged stiffening girders, central sag $d = 15$ m. The UDL of 15 kN/m on the right half is the total load on the bridge, shared equally by the two girders and cables, so each girder-cable system carries $w = 7.5$ kN/m on the right half ($x = 75$ to 150 m). (The road width is not needed for the force calculation.) If 15 kN/m acted on each girder, all answers below would be doubled. Section at 50 m from the right end, i.e. $x = 100$ m from the left. Cable: $y = \dfrac{4d\,x(L-x)}{L^2} = \dfrac{4\times 15\,x(150-x)}{22500}$.

Forces for one cable and girder

Total load on the right half $= 7.5\times 75 = 562.5$ kN acting at $x = 112.5$ m.

$$R_A = \frac{562.5\times 37.5}{150} = 140.625\ \text{kN}, \qquad M_C = R_A\times 75 = 10546.875\ \text{kN m}$$

The mid-span hinge gives:

$$H = \frac{M_C}{d} = \frac{10546.875}{15} = 703.125\ \text{kN}$$

SF and BM at x = 100 m

Cable ordinate: $y = \dfrac{4\times 15\times 100\times 50}{22500} = 13.333$ m; slope: $\tan\alpha = \dfrac{4\times 15\,(150-200)}{22500} = -0.1333$.

Simple-beam values: $M_b = R_A\times 100 - 7.5\times 25\times 12.5 = 11718.75$ kN m; $V_b = R_A - 7.5\times 25 = -46.875$ kN.

$$\begin{aligned} M &= M_b - H\,y = 11718.75 - 703.125\times 13.333 = 2343.75\ \text{kN m}\\ V &= V_b - H\tan\alpha = -46.875 - 703.125\times(-0.1333) = 46.88\ \text{kN} \end{aligned}$$

For the whole bridge (two girders) these values are doubled: $M = 4687.50$ kN m, $V = 93.75$ kN.

Maximum tension in the cable

The cable carries a uniform load $p = \dfrac{8Hd}{L^2} = 3.750$ kN/m (total $pL = 562.5$ kN), so it is a parabola and the vertical component at each support is $V_s = \dfrac{pL}{2} = 281.250$ kN.

$$T_{max} = \sqrt{H^2 + V_s^2} = \sqrt{703.125^2 + 281.250^2} = 757.3\ \text{kN (in each cable)}$$

Answer (per girder): $V = 46.88$ kN, $M = 2343.75$ kN m (sagging) at 50 m from the right end; maximum cable tension $= 757.3$ kN.