TIL about the elasticity of steel
(sketch is self-made)
So, many solid materials have a so-called "Young modulus E" which describes the mechanical stress that develops inside a material when it is deformed. Basically, when a material is deformed by a deformation ε, then the stress σ is σ = E · ε.
ε is measured as Δx/x, i.e. when you take 1m of steel and deform (stretch) it by 1mm, then ε = 1mm / 1m = 10^-3. The value of E is set for any material, i.e. E(steel) = 200 GPa, or 200 kN/mm². So if you stretch it by a deformation of ε = 10^-3, you get a stress σ = 10⁻³ · 200 GPa = 200 MPa.
A practical application of this would be if you have an elastic wire (a wire made from steel, in this case), and you want to calculate how much force you need to stretch it.
This means that if the steel bar has a cross-section area of A = 1mm², and a length of 1 m, and you stretch it by an additional 1mm, then ε = 10^-3 and to calculate the force F, you multiply σ by A and get F = σ · A = 200 MPa · 1 mm² = 200 N/mm² · 1 mm² = 200 N. This is the force that you have to exert to stretch the wire to this length.
In a follow-up article i intend to show how you calculate how much energy you need to stretch the wire to this length, which is surprisingly simple to do if you know some calculus and the basic rule that dW = F · dx. To sum it up, the result would be W = ∫ F · dx = ½ · F · Δx. For the above example this would be W = ½ · F · Δx = ½ · 200 N · 1 mm = 0.1 J. Which is surprisingly little energy, in my opinion.
Did i calculate this correctly? Can somebody check please?
9 replies
200 MPa = 200 N/m2, not 200 N/mm2.
Edit: I was corrected below.
1 Pa = 1 N/m² (says wikipedia). So 200 N/m² would be 200 Pa.
Yep, you're right, thanks.
Once you get into the 300 level classes in an engineering degree you learn that everything is a spring, the difference is only in the degree of springiness (Young's modulus)
Yes, your calculations are correct. The force of 200 N corresponds to the weight of a mass of 20 kg.
Pa is N/m2, not N/mm2. The answer is off by a factor of 1e6, right?
Edit: nm, I was corrected elsewhere.
Sounds plausible. In order to store more energy you'd need a longer wire. To compactify your setup you can wind the wire to a spiral and get a spring ( maths changing a bit due to Shear Modulus however). This way you can store more energy in the etall wire
It may sound contradictory, but a "softer" (more compliant) spring (in this example corresponding to a longer wire) stores more energy than a stiff one when the same load (force) is applied.
With the caveat that the Young's Modulus of a material can vary with temperature.
In many cases, that's not important. But sometimes it is.