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Assuming a steel wire with a Young's modulus of 400 GPa, which has a 1 mm. square cross section and 10 meters in length, what would be it's spring constant?

It is understood that Young's modulus is the special case spring constant where the cross section and length are one unit each.

I'm simpler words, the question could rephrased as: how to derive the spring constant from Young's modulus for a homogenous wire of arbitrary dimensions.

Ashley
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  • Why does your steel have such a high Young's Modulus. – JMac Mar 20 '17 at 09:42
  • @JMac Sorry, my bad. The value for steel apparently is around 200 GPa. Anyways, the actual value is relatively unimportant to the question than the method or formula used to deduce the answer. Thank you. – Ashley Mar 20 '17 at 10:27

3 Answers3

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For a straight wire, the stiffness is given by $$k = \frac{EA}{L}$$ where $k$ is the stiffness (Newtons/meter), $E$ is Young's modulus (Pascals), $A$ is the cross section area (square meters), and $L$ is the length (meters).

Be careful with the units! Note, Young's Modulus (Pascals, or Newtons / meter squared) does not have the same units as the spring constant (Newtons / meter).

alephzero
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  • Thanks for the answer. From the equation you've provided, is it correct to say that a 1 meter long wire will have a spring constant that is 10 times that of a 10 metres long wire? so, then the amount of potential energy stored in 10 1 meter long wires will be lesser than the amount of potential energy stored in 1 10 meter long wire, since the energy is proportional to displacement squared, and the longer wire would naturally have a displacement 10 times that of the smaller wire. – Ashley Mar 20 '17 at 17:30
  • But, the volume of material in both the cases is the same (10 1 meter long wires and 1 10 meter long wire), and by extension, the manner of arrangement of the material should have no impact on the amount of potential energy being stored. How to resolve this seeming contradiction, or am I missing something. – Ashley Mar 20 '17 at 17:38
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    Are you sure you're calculating and summing the potential energies correctly? The spring constant of each short wire is larger by a factor of 10. The elongation is smaller by a factor of 10. Therefore, the potential energy is smaller by a factor of 10. Add the potential energy (or elongation) from 10 short wires in series and you get a potential energy (or elongation) identical to that of the long wire. – Chemomechanics Mar 21 '17 at 18:42
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Young's modulus $E = \dfrac{\text{tensile stress}}{\text{tensile strain}} = \dfrac{\left(\frac F A \right )}{\left (\frac lL\right )} \Rightarrow F = \dfrac {EA}{L}\,l \Rightarrow \Delta F = \dfrac {EA}{L}\, \Delta l$

This gives the spring constant $k = \dfrac{\Delta F}{\Delta A} = \dfrac {EA}{L}$

If $A= 1\,\rm m^2$ amd $L = 1\,\rm m$ then the numerical value of the Young's modulus $E$ is equal to numerical value of the spring constant.

Farcher
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  • Thanks for confirming the analogy of Young's modulus as a special case spring constant. However, would you be able to clear my apparent misconception that I've raised in comments to alephzero's answer? It seems only logical to me that the elastic energy stored per unit volume of a material must be intrinsic and not dependent on the particular shape of the structure. – Ashley Mar 21 '17 at 14:56
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$\Delta E = \frac 1 2 k \Delta L^2 = \frac 1 2 \frac {EA} L \Delta L^2$

Volume $V = AL$

$\Delta E = \frac 1 2 \frac {EA} L \Delta L^2 = \frac 1 2 (EV/L^2) \Delta L^2 = \frac {EV} 2 \left( \frac { ΔL } L \right)^2$ is proportional to volume only for same strain $\Delta L / L$

warlock
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David Tomanek
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Sep 13 '22 at 19:20