Suspension 100 m: UDL on right half, SF and BM at 20 m

Data and assumptions: span $L = 100$ m, two cables with two three-hinged stiffening girders, dip $d = 10$ m, roadway width 5 m. Dead load $5$ kN/m² and live load $10$ kN/m² (right half only). Load per metre run of the bridge: dead $5\times 5 = 25$ kN/m, live $10\times 5 = 50$ kN/m. Each of the two cables and girders carries half: per cable/girder, dead load $w_d = 12.5$ kN/m (whole span, taken by the cable) and live load $w_l = 25$ kN/m on the right half ($x = 50$ to 100 m). Cable: $y = 0.004\,x(100-x)$.

Horizontal pull (per cable)

$$H_d = \frac{w_dL^2}{8d} = \frac{12.5\times 10^4}{80} = 1562.5\ \text{kN}$$

Live load: total $25\times 50 = 1250$ kN acting at $x = 75$ m.

$$R_A = \frac{1250\times 25}{100} = 312.5\ \text{kN}, \qquad M_C = R_A\times 50 = 15625.0\ \text{kN m}, \qquad H_l = \frac{M_C}{d} = 1562.5\ \text{kN}$$ $$H = H_d + H_l = 3125.0\ \text{kN}$$

SF and BM in the girder at 20 m from the left support

The dead load is carried by the cable, so only the live load acts on the girder. At $x = 20$ m: $y = 0.004\times 20\times 80 = 6.40$ m, $\tan\alpha = 0.004\,(100-40) = 0.24$. For the section (left of the load): $M_b = R_A\times 20 = 6250.0$ kN m, $V_b = R_A = 312.5$ kN.

$$\begin{aligned} M &= M_b - H_l\,y = 6250.0 - 1562.5\times 6.40 = -3750.0\ \text{kN m}\\ V &= V_b - H_l\tan\alpha = 312.5 - 1562.5\times 0.24 = -62.5\ \text{kN} \end{aligned}$$

(These values are per girder; the negative moment is hogging.)

Maximum tension in the cable

At the supports, $\tan\alpha_s = \dfrac{4d}{L} = 0.4$, $V_s = H\tan\alpha_s = 1250.0$ kN:

$$T_{max} = \sqrt{H^2 + V_s^2} = \sqrt{3125.0^2 + 1250.0^2} = 3365.7\ \text{kN (each cable)}$$

Answer (per girder): $V = -62.5$ kN, $M = -3750.0$ kN m at 20 m from the left support; maximum tension in each cable $= 3365.7$ kN.