What is the spring constant of a steel wire? 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.
 A: 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).
A: 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.
A: $\Delta E = \frac 1 2 k \Delta L^2 = \frac 1 2 \frac {EA} L \Delta L^2$
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 $\Delta L / L$
