# Equation of the parabolic suspension bridge cable when the deck mass is known

I'm reading the derivation of the equation of the hanging catenary from Wikipedia:

https://en.wikipedia.org/wiki/Catenary

In the same article the parabolic equation governing the shape of the main cable of a suspension bridge is discussed, and its equation is obtained as following:

$$y={\frac {w}{2T_{0}}}x^{2}+\beta$$

Here $w$ is the density of the deck and $T_{0}$ is the horizontal tension on the cable at lowest point.

Let's say I'm designing a suspension bridge with the length and density of the deck and its other features are known. I would be interested in knowing the equation of the main cable to calculate the length of the secondary cables (the vertical ones). But how do I get $T_{0}$?

• Are you sure that parabolic isn't just a first-order approximation to the catenary? – Carl Witthoft Aug 22 '18 at 18:02
• @CarlWitthoft If the deck weight is far larger than the weight of the cable, then the parabola is a better fit; I explore the transition here. – Chemomechanics Aug 22 '18 at 18:17
• @Chemomechanics interesting.... I would have guessed that the deck weight could be treated as a series of masses attached to the support cable at specified locations. Does this mean a non-uniform chain fails to fit to a catenary? – Carl Witthoft Aug 22 '18 at 18:19
• @CarlWitthoft You're exactly right, but those specific locations for the case of a massive deck are per unit horizontal distance rather than per unit distance along the cable; the difference is important unless the sag is very small. Yes, the general non-uniform chain assumes a more complex shape than a catenary. – Chemomechanics Aug 22 '18 at 18:30

Since $x=0$ marks the midpoint of the parabola, let $y(0)=0$ at this point to eliminate $\beta$. Then, we only need to know the sag $s$ of the cable (which is typically set during the design process). The sag is equivalent to the value of $y$ at one of the symmetric towers (where $x=L/2$, where $L$ is the span or horizontal distance between the towers).
Thus, if we know the sag $s$ and span $L$ (and the distributed weight $w$), then we can immediately obtain $T_0$ from $y=\frac{w}{2T_0}x^2$.
If we know the length $l$ of the cable but not the sag, then we can use the formula for the length of a parabola $$l=\sqrt{\left(\frac{L}{2}\right)^2+4s^2}+\frac{L^2}{8s}\sinh^{-1}\left(\frac{4s}{L}\right)$$ to obtain the sag $s$. (We'll probably have to solve this equation numerically.)
• Ah, now I understand. So to design a bridge, I plug in the density of the deck, I choose some acceptable tension on the cable (within its limits) and then solve for the height of the tower? Apparently the cable is in higher tension when the tower is lower? Also a final question: Are the different parts of the cable in different tension? For some reason I initially thought the tension would be the same at every point.. Is the tension highest in the lowest point, $T_0$? – S. Rotos Aug 23 '18 at 6:51