Suspension 120 m: UDL on left half, SF and BM at 30 m

Loads per girder. The roadway is 6 m wide and is carried by two girders and two cables, each taking half:

• Dead load: $5\times 6 = 30$ kN/m on the bridge, so $w_d$ = 15 kN/m per girder
• Live load on the left half (0 to 60 m): $10\times 6 = 60$ kN/m, so $w_l$ = 30 kN/m per girder

Span $L$ = 120 m, dip $d$ = 12 m.

Cable pull $H$

• Dead load: $H_d = \dfrac{w_d L^2}{8d} = \dfrac{15(120)^2}{8(12)} = 2250.0$ kN
• Live load on the left half: $R_A = \dfrac{30(60)(90)}{120} = 1350.0$ kN, then $M_{0C} = 1350.0(60) - 30(60)(30) = 27000.0$ kN m
• $H_l = \dfrac{M_{0C}}{d} = \dfrac{27000}{12} = 2250.0$ kN

$$H = H_d + H_l = 2250.0 + 2250.0 = 4500.0\ \text{kN}$$

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

The dead load causes no girder BM or SF (the cable carries it). For the live load:

$$y_{30} = \frac{4(12)(30)(90)}{120^2} = 9.00\ \text{m},\qquad \frac{dy}{dx} = \frac{4(12)(120-60)}{120^2} = 0.20$$ $$\begin{aligned} M_{0,30} &= 1350(30) - 30(30)(15) = 27000.0\ \text{kN m} \\ M_{30} &= M_{0,30} - H_l\,y_{30} = 27000.0 - 2250.0(9.00) = 6750.0\ \text{kN m} \\ S_{30} &= \left[R_A - 30(30)\right] - H_l\,\frac{dy}{dx} = 450.0 - 2250.0(0.20) = 0.0\ \text{kN} \end{aligned}$$

Maximum tension in the cable for this loading

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

$$T_{max} = H\sec\theta = 4500.0(1.0770) = 4846.65\ \text{kN}$$

Answer: At 30 m from the left end, $M = 6750.0$ kN m (sagging) and $S = 0.0$ kN (the shear is zero, so this is the section of maximum BM in the loaded half); maximum cable tension = 4846.65 kN in each cable.