# Catenary differential equation (suspension bridge)

• Jul 13th 2011, 06:31 PM
cbjohn1
Catenary differential equation (suspension bridge)
So I am getting stuck trying to solve this. pgA and Tx are constants
d^2h/dx^2 = (pgA/Tx)*sqrt(1+(dh/dx)^2)

I substitute dh/dx = sinh(u), d^2h/dx^2 = coshu(du/dx)

then you get (pgA/Tx)*cosh(u) = cosh(u)(du/dx)

... and I'm unsure where to go after this. I tried a couple more things but I'm not sure if it's the right track:

(pgA/Tx) * cosh(u) = cosh(u)(du/dx) ... integrate and solve for u = (pgAx/Tx)+C
and then substitute back in for u??? Sorry this isn't in Latex but I couldn't figure it out.
• Jul 13th 2011, 11:42 PM
CaptainBlack
Re: Catenary differential equation (suspension bridge)
Quote:

Originally Posted by cbjohn1
So I am getting stuck trying to solve this. pgA and Tx are constants
d^2h/dx^2 = (pgA/Tx)*sqrt(1+(dh/dx)^2)

I substitute dh/dx = sinh(u), d^2h/dx^2 = coshu(du/dx)

then you get (pgA/Tx)*cosh(u) = cosh(u)(du/dx)

... and I'm unsure where to go after this. I tried a couple more things but I'm not sure if it's the right track:

(pgA/Tx) * cosh(u) = cosh(u)(du/dx) ... integrate and solve for u = (pgAx/Tx)+C
and then substitute back in for u??? Sorry this isn't in Latex but I couldn't figure it out.

1. Lump the constants, so then you have:

$\frac{d^2h}{dx^2}=K\left(1+\left(\frac{dh}{dx} \right) ^2\right)^{1/2}$

where $K=pgA/Tx$.

Now consider reducing the order by putting $u=dh/dx$

CB