Suspension 150 m: SF, BM at 30 m, tension, length

Data: three-hinged stiffening girder, span $L = 150$ m, cable dip $d = 15$ m. Point loads 180 kN at 50 m and 150 kN at 100 m from the left support. Dead load of the girder $w = 7.5$ kN/m, taken wholly by the cable (so the girder carries no dead-load moment). Cable: $y = \dfrac{4d\,x(L-x)}{L^2} = \dfrac{x(150-x)}{375}$.

Horizontal pull

Dead load: $H_d = \dfrac{wL^2}{8d} = \dfrac{7.5\times 150^2}{8\times 15} = 1406.25\ \text{kN}$

Point loads (hinge at mid-span):

$$R_A = \frac{180\times 100 + 150\times 50}{150} = 170.0\ \text{kN}, \qquad M_C = 170\times 75 - 180\times 25 = 8250\ \text{kN m}$$ $$H_p = \frac{M_C}{d} = \frac{8250}{15} = 550\ \text{kN}$$

Total: $H = 1406.25 + 550 = 1956.25$ kN.

SF and BM in the girder at 30 m from the left

Cable ordinate: $y = \dfrac{30\times 120}{375} = 9.60$ m; slope $\tan\alpha = \dfrac{4\times 15\,(150-60)}{22500} = 0.24$. Only the point loads act on the girder (the dead load is carried by the cable):

$$\begin{aligned} M &= R_A\times 30 - H_p\,y = 170\times 30 - 550\times 9.60 = -180.0\ \text{kN m}\\ V &= R_A - H_p\tan\alpha = 170 - 550\times 0.24 = 38.0\ \text{kN} \end{aligned}$$

Maximum tension in the cable

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

$$V_s = H\tan\alpha_s = 782.50\ \text{kN}, \qquad T_{max} = \sqrt{H^2 + V_s^2} = 2106.9\ \text{kN}$$

Length of the cable

$$S = L\left[1 + \frac83\left(\frac{d}{L}\right)^2 - \frac{32}{5}\left(\frac{d}{L}\right)^4\right] = 150\left[1 + \frac83(0.1)^2 - \frac{32}{5}(0.1)^4\right] = 153.90\ \text{m}$$

Answer: at 30 m from the left support, $V = 38.0$ kN and $M = -180.0$ kN m; maximum cable tension $= 2106.9$ kN; length of cable $= 153.90$ m.