UDL transferred to main cable from point load

In a suspension bridge with a three-hinged stiffening girder (hinges at the two ends and at mid-span $C$), the hangers carry a point load to the cable as a uniform load.

Assumptions

The hangers are closely spaced and inextensible, so the cable transmits an upward uniform load $p$ per unit length to the girder. The cable is a parabola of dip $d$.

Derivation

Let a point load $W$ act on the girder (span $L$) at a distance $x$ from the left end, $x\le L/2$.

        W
   A    |        C(hinge)       B
   o----+---------o-------------o
   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^   p per unit length (from the cable)

The bending moment at the middle hinge is zero. Treating the girder as a simple beam under $W$ and the upward uniform load $p$:

$$M_C = \frac{W x}{2} - \frac{p L^2}{8} = 0 \Rightarrow p = \frac{4Wx}{L^2}$$

Here $\dfrac{Wx}{2}$ is the mid-span moment of the simple beam under $W$ (the reaction $R_B = Wx/L$ times the lever arm $L/2$). For a load on the other half, put $L - x$ in place of $x$:

$$p = \frac{4W\,(L-x)}{L^2} \quad (x\ge L/2)$$

Result

The load transferred to the main cable is a uniformly distributed load of intensity

$$p = \frac{4Wx}{L^2}\ \ (x\le L/2), \qquad p = \frac{4W(L-x)}{L^2}\ \ (x\ge L/2)$$

The total load on the cable is $pL = \dfrac{4Wx}{L}$, which is $4\times$ the simple-beam reaction $R_B = Wx/L$. It is largest ($p = \dfrac{2W}{L}$) when $W$ is at mid-span. The horizontal pull in the cable is

$$H = \frac{pL^2}{8d} = \frac{Wx}{2d}$$

Example: $W = 100$ kN at $x = 10$ m on a girder of span $L = 50$ m: $p = \dfrac{4\times 100\times 10}{2500} = 1.6$ kN/m.