This question is similar to a question I am stuck on, but I have changed it to get an understanding how to works.

A spring is being made by pulling a 12cm long cylinder of material in tension with a desired stiffness of 6200kN/m.

It has a density of 2.9 g/cm^3, an ultimate tensile strength of 375 MPa and a Young's modulus of 70 GPa.

What should the diameter (in mm) be for the metal "spring"?

So, my question is how can I go about solving this question?

I have looked at the Young's Modulus and Hooke Law's formulas. But keep getting stuck with material displacement. Do I need to know the amount of length the object changes to work out this solution, or am I on the wrong track?

Can I assume: $\Delta L = 1$ if I used the following formula to find the $A_0$, which I can then calculate the diameter from: $$ F = \frac{E A_0 \Delta L}{L_0} $$

Any help will be most appreciated.

  • 1
    $\begingroup$ Well you know the stiffness. What is the formula for the axial stiffness of a cylinder? $\endgroup$
    – atom44
    Commented Feb 26, 2017 at 14:49
  • $\begingroup$ The formula is Force = Young's Modulus * Area. (F=EA). $\endgroup$
    – Jan
    Commented Feb 26, 2017 at 16:29
  • $\begingroup$ axial stiffness = (Young's Modulus * Area) / Length. I think I was over-complicating the question. Thanks for the tip. $\endgroup$
    – Jan
    Commented Feb 26, 2017 at 16:38

1 Answer 1


The axial stiffness of an isotropic material with a uniform cross-section is a fundamental part of most engineering mechanics of materials concepts. We can derive it as follows:

The force-displacement relation of a spring is described by Hooke's Law,

$$F = k \Delta L$$

where $F$ is the force exerted on the spring, $\Delta L$ is the change in length or displacement, and $k$ is the stiffness or spring constant.

We can rearrange to get an expression of the stiffness

$$k = \frac{F}{\Delta L}$$

Now we want to express $k$ exclusively in terms of the geometry and material properties. For that, we need to make use of the engineering stress $\sigma = F/A_0$ where $A_0$ is the cross-sectional area of the material before deformation. We can rearrange and substitute this back into the previous equation

$$k = \frac{\sigma A_0}{\Delta L}$$

We still need to get rid of $\sigma$, and to do that we can use $\sigma = E \epsilon$ where $E$ is the Young's modulus of the material and $\epsilon$ is the engineering strain. Substituting that and using the definition of engineering strain $\epsilon = \Delta L/L_0$ where $L_0$ is the length of the material before deformation and we get

$$ k = \frac{E(\Delta L/L_0)A_0}{\Delta L}$$

$$ k = \frac{E A_0}{L_0}$$

Note that $\Delta L$ cancels, so we don't need to know the change in length.

Now if you want to figure out the necessary dimensions for a block of material to have a given stiffness, just substitute the appropriate equation for $A_0$ and rearrange. For example, a cylinder has cross-sectional area $A_0 = \frac{\pi}{4} D_0^2$, sub that in and you get

$$ D_0 = \sqrt{\frac{4 k L_0}{\pi E}}$$

  • $\begingroup$ Thank you so much BarbalatsDilemma. That cleared a lot of questions I had. I really appreciate the explanation, it all makes sense now. $\endgroup$
    – Jan
    Commented Feb 26, 2017 at 20:15

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