Torsion constant of a bent spring

[tex]K_t= \frac{T}{\theta}= \frac{Fr}{\frac{x}{r}}= \frac{F}{x}r^2= Kr^2[/tex]

Where [itex]K[/itex] is the linear stiffness of the spring and [itex]r[/itex] is the distance between the spring and the center of rotation.

 

Sorry about that.

I'm going to let other chimes in, but I don't think it change anything for the torsional stiffness; Just like the bent doesn't change anything to linear stiffness (for example, the way I thought you were using the spring). The torsional stiffness is:



and none of these variables should change if the spring is bent. The exception would be if the bent deformed the spring so much that it doesn't allow the spring to freely rotate. For example, if the bent is so pronounced that it looks like 2 springs side by side joint by a wire, you will probably end up with 2 independent springs with half the active coils of the initial spring (i.e. twice the initial spring torsional stiffness).

However, because there is an initial deformation, the torsional stress limit that it can withstand will be lower.