Is the maximum tensile strength a property of a material ? If it is so then suppose I am given the maximum tensile strength of two flat spiral springs to be $30MN/m^2$ and I have to find the dimensions (length per unit thickness $l/t$) of the spring I need to have to make some spring controlled system in an indicating instrument.

Do I need to divide the maximum tensile strength by two, multiply it by two or leave it as it is ?

I know I can calculate the dimensions of the spring by $$f_{max} = \frac{Et\theta}{2l}$$ where

$E$ is the modulus of elasticity (given to me)

$\theta $ is the maximum angle of deflection $\pi/2$

  • $\begingroup$ you have not described the geometry of your spring, but your formula is clearly wrong as the right side expression does not have dimensions of force. In any case there is no way to answer this without a clear description of the geometry of the spring. $\endgroup$
    – agentp
    Sep 17, 2017 at 14:39
  • $\begingroup$ @agentp it is not force on the right side, it is maximum tensile strength $\endgroup$ Sep 20, 2017 at 17:54

1 Answer 1


You have a few questions there:

Is the maximum tensile strength a property of a material ?

Yes. There is such a thing as maximum tensile strength, or ultimate tensile strength. This is a property of the material, and therefore the shape does not matter. If any section of the spring experiences a stress equal to or above the ultimate tensile strength of the material, it will break!

Geometry and loading conditions can cause different sections of a part to experience more or less stress, but the ultimate tensile strength determines if it will break or not. (This makes sense when you think of two identically shaped springs which are of different material: why does the plastic one break before a metal one, even if they are loaded the same?)

It is also worth knowing that there is the yield strength, which is the stress limit between permanent deformation and elastic deformation. It is a generally good idea that you should stay below the yield strength of a material to keep parts the same size. Working (well) below a materials yield strength is good design.

Do I need to divide the maximum tensile strength by two, multiply it by two, or leave it as it is?

Leave it. No dividing, multiplying, or anything to that number.

For a spring, a quick approximation would be to figure out the maximum force the spring will support (including its own weight), and divide that by its cross sectional area. If that force/area goes above the ultimate tensile strength, it will break for sure. (If that force/area goes above the yield strength, the spring will be permanently deformed, but not always broken.)

Of course, you may introduce a factor of safety which may alter the numbers you are using. Maybe you want the spring to handle 2, 3, or more times the loading you actually think it will experience. In this case, I would multiply the maximum expected stress by this safety factor, which would allow you to easily compare materials you wish to use.

  • $\begingroup$ note that you would never design anywhere near the ultimate strength. I don't know where he got the factor of two but it might be some ad hoc factor of safety. $\endgroup$
    – agentp
    Sep 17, 2017 at 14:41
  • $\begingroup$ @agentp good point about the factor of safety. I think I'll add it. $\endgroup$
    – PipperChip
    Sep 17, 2017 at 17:02
  • $\begingroup$ Yup, i do have that safety factor in mind, thankyou :) $\endgroup$ Sep 20, 2017 at 17:56

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