Show that the total length $l$ of the stretched cord is given by: $l = l_0(1 + \frac{\rho_0gl_0}{2E})$

Question

Apparently, for bungie jumpers it is more entertaining if the cord is long enough that the jumper comes close to hitting the ground. Getting the right length cord is not a simple matter of just knowing the weigth of the jumper and the height of the jump. The reason is that the weigth of the cord will cause it to extend. To determine this suppose the cord is hung with the upper end $X = 0$ attached and the lower end is free. Assume gravity is the only force present, and the cord starts off with a constant density $ρ_0$ and length $l_0$. Given that the cord is in steady-state we have :

$0 = \rho_0B + \frac{dP}{dX}$, with $B = g$ where g is the gravitational acceleration. As for the boundary conditions, the lower end is free and this means the stress is zero there: $P(l_0) = 0$.

I have the following three questions which I have to answer:

1. Solve the above ode for the given boundary condition:

For the boundary condition $P(l_0) = 0$ I found that $P(X) = -\rho_0gX + C$, where C is a constant. Solve C with the given boundary condition gives: $P(X) = -\rho_0gX + \rho_0gl_0$.

2. To determine the displacement $U(X)$ of the cord we need to specify the constitutive law for the stress. Here, we will use $P = EU_X$:

By integration, I found that $U(X) = \frac{PX}{E} + C$, where C is a constant. Using the boundary $U(0) = 0$ gives that $C=0$. Hence, $U(X) = \frac{PX}{E}$.

3. Show that the total length $l$ of the stretched cord is given by: $l = l_0(1 + \frac{\rho_0gl_0}{2E})$

I could use some help with the last question. I have not figured out how to solve that question. Thanks in advance.

Answer 1

Based on your workings -

$P(X) = -\rho_0gX + \rho_0gl_0$

Displacement $U(X) = \frac{PX}{E}$

Total displacement over length $\displaystyle l_0 = \frac{1}{E} \int_0^{l_0} P(X)dX$

$\displaystyle = \frac{1}{E}\int_0^{l_0} (-\rho_0gX + \rho_0gl_0) \, dX$

$\displaystyle = \frac{\rho_0g}{E} \int_0^{l_0} (l_0 - X) \, dX$

$\displaystyle = \frac{\rho_0g}{E} |(l_0X - \frac{1}{2}X^2)|_0^{l_0}$

$\displaystyle = \frac{\rho_0gl_0^2}{2E}$

So, the length of the stretched cord = $l_0 + $ displacement $= l_0 (1 + \frac{\rho_0gl_0}{2E})$