Suspension 200 m: max/min tension, cable length

Data: span $L = 200$ m, two cables and two three-hinged stiffening girders, dip $d = 20$ m. Dead load of the bridge 50 kN/m (total); three point loads of 800 kN each on the centre line at 50, 100 and 150 m. The loads are shared equally by the two cables, so for one cable: $w = 25$ kN/m and three loads of 400 kN. Cable: $y = \dfrac{4d\,x(L-x)}{L^2}$, $\tan\alpha_s = \dfrac{4d}{L} = 0.4$.

Horizontal pull (one cable)

Dead load: $H_d = \dfrac{wL^2}{8d} = \dfrac{25\times 200^2}{8\times 20} = 6250\ \text{kN}$

Point loads: the simple-beam reaction is $R_A = 1.5\times 400 = 600$ kN (symmetrical), and the moment at the middle hinge:

$$M_C = 600\times 100 - 400\times 50 = 40000\ \text{kN m}, \qquad H_p = \frac{M_C}{d} = \frac{40000}{20} = 2000\ \text{kN}$$ $$H = H_d + H_p = 8250\ \text{kN}$$

Maximum and minimum tension in the cable

• Minimum tension: at mid-span (the lowest point of the cable), where the cable is horizontal: $T_{min} = H = 8250$ kN.
• Maximum tension: at the supports, where $\tan\alpha_s = 0.4$ and the vertical component is $V_s = H\tan\alpha_s = 3300$ kN:

$$T_{max} = \sqrt{H^2 + V_s^2} = H\sqrt{1+0.16} = 8885.5\ \text{kN}$$

Length of the cable

$$S = L\left[1 + \frac83\left(\frac{d}{L}\right)^2 - \frac{32}{5}\left(\frac{d}{L}\right)^4\right] = 200\left[1+\frac83(0.1)^2 - \frac{32}{5}(0.1)^4\right] = 205.21\ \text{m}$$

Answer: $T_{max} = 8885.5$ kN (at the supports), $T_{min} = 8250$ kN (at mid-span), length of each cable $= 205.21$ m.