Suspension 400 m: max/min tension, cable length

Assumptions. The data are for the whole bridge; two girders and two cables share the load equally, so per cable $w$ = 30/2 = 15 kN/m and $P$ = 300/2 = 150 kN. (If a single cable is intended, all forces below double: $H$ = 22000 kN, $T_{max}$ = 22968.7 kN. The length is unchanged.) Span 400 m, dip 30 m, point loads at 100, 200 and 300 m.

Horizontal pull $H$

• Dead load: $H_d = \dfrac{wL^2}{8d} = \dfrac{15(400)^2}{8(30)} = 10000.0$ kN
• Point loads: $R_A = 1.5P = 225.0$ kN, $M_{0C} = 225.0(200) - 150(100) = 30000.0$ kN m, so $H_p = \dfrac{30000}{30} = 1000.0$ kN

$$H = 10000.0 + 1000.0 = 11000.0\ \text{kN}$$

Maximum and minimum tension

$\tan\theta = 4d/L = 0.300$ at the supports, $\sec\theta = 1.04403$.

• Maximum (at the two supports): $T_{max} = H\sec\theta = 11000.0(1.04403) = 11484.34$ kN
• Minimum (at mid-span, where the cable is horizontal): $T_{min} = H = 11000.0$ 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 $d/L = 0.075$:

$$l = 400\left[1+\frac{8}{3}(0.075)^2-\frac{32}{5}(0.075)^4\right] = 405.92\ \text{m}$$

Answer: $T_{max} = 11484.34$ kN at the supports; $T_{min} = 11000.0$ kN at mid-span; length of cable = 405.92 m (per cable, loads shared equally by the two cables).