Aluminum 6061-T6 has a density of 2.7 g/cm^3, a Young's modulus of 69 GPa, and an ultimate tensile strength of 310 MPa.

If a "spring" were to be made by pulling a 10cm long cylinder of material in tension with a desired stiffness of 5000kN/m, what should the diameter (in millimeters) be for an Aluminum "spring"?

I dont want the answer for this question but the FORMULA for working it out as i am totally unsure how to derive the diameter from the info given..


With a little bit of searching, a good theoretical value of the spring rate/stiffness $k$ (i.e. ratio of force to extension) can be calculated using:

$$k=\frac{G d^4}{8D^3 N}$$

Where $G$ is the shear modulus of the material, $d$ is the wire diameter of the spring, $D$ is the diameter of the turns of the spring, and $N$ is the effective number of turns.

This can be calculated from scratch by equating the work done by the tensile/compressive force acting on the spring with the elastic torsional energy stored inside the wire of the spring. The formula assumes that the length of the wire of the spring is approximately equal to the number of turns multiplied by the circumference of a circle of diameter $D$, i.e. the turns of the spring are rather flat.


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