Suspension 120 m: girder with different support levels

Assumptions. The girder is horizontal and the hangers are vertical and closely spaced, so the cable is parabolic. The "maximum dip of 12 m from the upper end" is taken as the depth of the lowest point of the cable below the higher end $A$; the lower end $B$ is then $12 - 6 = 6$ m above the lowest point. The self-weight of the girder is not given, so only the two point loads act (a uniform dead load would be carried wholly by the cable and would not change the girder forces).

Dip of the cable below the chord at mid-span

With heights $h_1 = 12$ m and $h_2 = 6$ m above the lowest point, $H = \dfrac{wL^2}{2(\sqrt{h_1}+\sqrt{h_2})^2}$, and the dip below the chord $AB$ at mid-span is

$$d = \frac{wL^2}{8H} = \frac{\left(\sqrt{h_1}+\sqrt{h_2}\right)^2}{4} = \frac{(3.464+2.449)^2}{4} = 8.743\ \text{m}$$

Cable pull $H$

Simple-beam reaction and moment at the centre (span 120 m, loads measured from the higher end):

$$\begin{aligned} R_A &= \frac{200(90)+100(65)}{120} = 204.17\ \text{kN} \\ M_{0C} &= 204.17(60) - 200(30) - 100(5) = 5750.0\ \text{kN m} \end{aligned}$$

Since the girder has a hinge at C ($M_C = 0$): $M_{0C} - H\,d = 0$

$$H = \frac{M_{0C}}{d} = \frac{5750.0}{8.743} = 657.7\ \text{kN}$$

(If the 12 m is instead read as the cable depth below $A$ at mid-span, then $d = 12 - 3 = 9$ m and $H = 638.9$ kN. The girder SF and BM below are the same in either case, because $H\,y = M_{0C}\,\dfrac{4x(L-x)}{L^2}$ does not depend on $d$.)

Girder moments and shears

The parabolic cable below the chord is $y = \dfrac{4d\,x(L-x)}{L^2}$. For the girder:

$$M(x) = M_0(x) - H\,y(x), \qquad S(x) = S_0(x) - H\,\frac{dy}{dx}, \qquad \frac{dy}{dx} = \frac{4d(L-2x)}{L^2}$$

so that $H\,y = 1.5972\,x(120-x)$ and $H\,\dfrac{dy}{dx} = 1.5972\,(120-2x)$.

Shear force (kN): at A = 12.50; at 30 m: 108.33 (left), -91.67 (right); at 55 m: -11.81 (left), -111.81 (right); at C: -95.83; zero at 90 m; just left of B: 95.83. The shear varies linearly between these points because $dy/dx$ is linear in $x$.

Girder end reactions: $R_A = 12.50$ kN (up) and $R_B = -95.83$ kN (a downward anchor force at B). Check: $12.5 - 95.83 + 383.33\ (\text{total hanger pull } 8Hd/L) = 300$ kN, equal to the applied load.

Diagrams (left end A at the top; $x$ from 0 to 120 m)

Shear force diagram (kN), linear between the listed points (+ above, - below the zero line):

  x =   0      | +            12.5
  x =  30 left | ++++++++++  108.3
  x =  30 right| ---------   -91.7
  x =  55 left | -           -11.8
  x =  55 right| ----------  -111.8
  x =  60 (C)  | ---------   -95.8
  x =  90      |               0
  x = 120      | +++++++++    95.8

Bending moment diagram (kN m, + sagging, - hogging; parabolic curves between points):

  x =   0      |  0
  x =  30      | ++++++++++++++++++  1812.5
  x =  55      | +++++                519.1
  x =  60 (C)  |  0
  x =  90      | --------------     -1437.5
  x = 120      |  0

Answer: $H = 657.7$ kN (638.9 kN if the 12 m is taken at mid-span); $M_{max}^{+} = 1812.5$ kN m under the 200 kN load (30 m); $M$ at 55 m $= 519.1$ kN m; $M_C = 0$; $M^{-}_{max} = -1437.5$ kN m at 90 m. The maximum shear force is $-111.8$ kN just right of the 100 kN load (55 m); the largest positive shear is $108.3$ kN just left of the 200 kN load.