Finding the spring constant of a torsion spring.

[Silvershield]

Hi,

How would I go about finding the spring constant of a bundle of nylon rope acting as a torsion spring?

I know the length. I don't know the width of the bundle. When twisted I wanted roughly 6cm diameter (is that a wrong method)?

I ll be twisting it by 1080 degrees.


When twisted as above ^ -

how can I calculate how much force/energy it will project if an arm in the middle of the torsion spring is pulled against it by a certain length/angle.

Im guessing that the longer the arm, the more it can pull back.

Like a ballista.

Thank you.

 

[IsaacOh]

I'm sorry, could you clarify your question? But here are some stuff I know about torsion.


Torque = -$\kappa$$\theta$
Where Kappa is the torsional coefficient and theta the angle rotated.

Period = 2$\pi$$\sqrt{I/\kappa}$

 

[Silvershield]

Yes, the question is a bit over the place.

Im trying to find the spring constant (k) of a bundle of nylon-6 rope.

The total diameter of the rope is 6cm roughly.

I've twisted the rope bundle 3 full revolutions and have fixed it at equilibrium.

The length of this rope bundle spring is 36cm

What is the working out/steps/equation to find the spring constant?

Furthermore, If I fix an arm in the middle of the spring, I wish to find the force needed to pull back 50 degrees (length 50cm of arm), and the forces the arm will project when it returns to equilibrium.

I know how to find the spring constant of a metal spring coil, however.
k = ( E*d ^3) / (8 * D^4 * n)
n = number of coils
d= wire diameter, D = diameter of spring, E = young's modulus