Suspension 100 m: SF and BM at 30 m, tension

Data: three-hinged stiffening girder (hinges at both ends and at mid-span), span $L = 100$ m, loads 200 kN at 25 m and 300 kN at 50 m from the left end; cable dip $d = 10$ m. The cable is parabolic: $y = \dfrac{4d\,x(L-x)}{L^2} = 0.004\,x\,(100-x)$, slope $\tan\alpha = \dfrac{4d\,(L-2x)}{L^2}$.

Horizontal pull in the cable

Reaction of the simple beam (taking moments about the right end):

$$R_A = \frac{200\times 75 + 300\times 50}{100} = 300.0\ \text{kN}$$

Moment at the mid-span hinge $C$ of the simple beam:

$$M_C = 300\times 50 - 200\times 25 = 10000\ \text{kN m}$$

The girder has a hinge at $C$, so the cable thrust makes the net moment there zero:

$$H\,d = M_C \Rightarrow H = \frac{10000}{10} = 1000\ \text{kN}$$

Shear force and bending moment at 30 m from the left

At $x = 30$ m: $y = 0.004\times 30\times 70 = 8.40$ m; $\tan\alpha = 0.004\,(100-60) = 0.16$.

$$M_b = 300\times 30 - 200\times 5 = 8000\ \text{kN m}, \qquad V_b = 300 - 200 = 100\ \text{kN}$$ $$\begin{aligned} M &= M_b - H\,y = 8000 - 1000\times 8.40 = -400\ \text{kN m}\\ V &= V_b - H\tan\alpha = 100 - 1000\times 0.16 = -60\ \text{kN} \end{aligned}$$

Maximum tension in the cable and its slope

The tension is greatest at the supports, where the slope is $\tan\alpha_s = \dfrac{4d}{L} = 0.4$:

$$\alpha_s = \tan^{-1}0.4 = 21.80^\circ, \qquad V_s = H\tan\alpha_s = 400\ \text{kN}$$ $$T_{max} = \sqrt{H^2 + V_s^2} = \sqrt{1000^2 + 400^2} = 1077.0\ \text{kN}$$

Answer: at 30 m from the left, $V = -60$ kN and $M = -400$ kN m (hogging); maximum cable tension $= 1077.0$ kN at the supports, inclined at 21.80° to the horizontal.