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

Single cable and three-hinged stiffening girder: span $L$ = 150 m, dip $d$ = 15 m, dead load $w$ = 5 kN/m, loads 180 kN at 50 m from A and 120 kN at 30 m from B (that is, 120 m from A).

Cable pull $H$

$$\begin{aligned} R_A &= \frac{180(100)+120(30)}{150} = 144.0\ \text{kN} \\ M_{0C} &= 144.0(75) - 180(25) = 6300.0\ \text{kN m} \\ H_p &= \frac{M_{0C}}{d} = 420.0\ \text{kN}, \qquad H_d = \frac{wL^2}{8d} = \frac{5(150)^2}{8(15)} = 937.5\ \text{kN} \\ H &= H_d + H_p = 1357.5\ \text{kN} \end{aligned}$$

BM and SF at 30 m from the left support

The dead load is carried entirely by the cable (parabolic), so it gives no moment or shear in the girder; only the point loads act on it.

$$y_{30} = \frac{4(15)(30)(120)}{150^2} = 9.60\ \text{m},\qquad \frac{dy}{dx} = \frac{4d(L-2x)}{L^2} = \frac{4(15)(90)}{150^2} = 0.24$$ $$\begin{aligned} M_{30} &= R_A(30) - H_p\,y_{30} = 4320.0 - 420(9.60) = 288.0\ \text{kN m} \\ S_{30} &= R_A - H_p\,\frac{dy}{dx} = 144.0 - 420(0.24) = 43.2\ \text{kN} \end{aligned}$$

Maximum tension in the cable

At the supports $\tan\theta = 4d/L = 0.40$, so $\sec\theta = 1.0770$:

$$T_{max} = H\sec\theta = 1357.5(1.0770) = 1462.07\ \text{kN}$$

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 $d/L = 0.1$:

$$l = 150\left[1+\frac{8}{3}(0.1)^2-\frac{32}{5}(0.1)^4\right] = 153.90\ \text{m}$$

Answer: $M_{30} = 288.0$ kN m (sagging), $S_{30} = 43.2$ kN; $T_{max} = 1462.07$ kN; cable length = 153.90 m.