I was thinking about the relative costs of materials, and it occurred to me that there are three principal figures of merit on which you might consider using a material for mechanical engineering:
Of course, for non-mechanical and even manufacturing purposes, many other properties are relevant: transparency, color, smell, machinability, magnetic permeability, conductivity, dielectric constant, thermal conductivity, dielectric strength, melting point, electronegativity, heat capacity, chemical resistance, UV resistance, viscosity when molten, resistance to galling, lubricity, fatigue resistance, and so on. And even for mechanical purposes, ratios of weight or volume to tensile support, stiffness, and impact energy are of secondary importance; and there are other kinds of stiffness and strength other than tensile: shear and compressive, normally. Additionally, these fundamental properties give rise to emergent properties like Vickers hardness.
But mostly those three are the ones we care about.
Cost is an extensive quantity — twice as many grams of the same alloy costs twice as much — so, to take these ratios, we need to convert tensile strength, stiffness (Young’s modulus), and impact resistance into extensive quantities too.
Consider ASTM A36 structural steel, with a yield stress of 250 MPa. A square-millimeter fiber of it can thus support 250 N, about 25 kg weight. Suppose it costs US$0.60 per pound ($1.32/kg) and has a density of 7.8 g/cc; then it costs about 10.3 μ$ per cubic millimeter, which means that it costs 10.3 μ$ per millimeter to support 250 N.
Normalizing, this works out to about 41 μ$ per newton-meter. If you need to hang 1000 N one meter below a support, or 10kN 100 mm below, this steel will do it for you for 4.1¢.
This is just reinterpreting 250 MPa from 250 N / m² to 250 N m / m³, then dividing by the density to get 32 kN m / kg. Dividing the cost by that gets you to the 41 μ$ / (N m):
$ 0.60 / pound / (250 MPa / (7.8 g/cc))
The Young’s modulus E has, like tensile strength, units of pressure; it’s the stress that would stretch a cable to twice its natural length if it stayed Hookean over that range, which nothing does. A less fanciful, but more arbitrary, way of looking at it is that it’s a million times the stress per microstrain.
If we consider the square-millimeter cable of A36 steel again, whose Young’s modulus is about 200 GPa, it will elongate by a microstrain under a force of 0.2 N. But we have a somewhat different situation. While the force needed to break the cable is independent of its length, that microstrain is a micron if it’s a meter long, or ten microns if it’s ten meters long. But if we make the cable ten times thicker instead of ten times longer, it becomes ten times stiffer rather than ten times less stiff.
How do we derive an extensive quantity from this, so that we can divide the cost per pound by it? I have no idea. I have to think about this.
When our aforementioned square-millimeter cable of A36 steel is stretched, it stores energy in its elastic deformation. If deformed too much, beyond its tensile strength, it will deform plastically or break (sometimes the difference between these is important, but I’ll gloss over it here). The ratio between its tensile strength (yield stress S in this case) and its Young’s modulus is its elongation at break (or at yield), about 0.125%†, which is an intensive quantity; but in itself that doesn’t tell us much.
† this differs by more than two orders of magnitude from the elongation at break in MatWeb, and ultimate tensile strength isn’t that much higher than yield stress, so I’m probably wrong.
It turns out, though, that the energy it can store is also an intensive quantity — which ought not to be surprising in retrospect. If we integrate the force over the deformation distance up to the yield stress times the wire length, we get the energy that can be stored in the cable before breaking. This energy is proportional to the force, which is proportional to the cross-section of the cable and doesn’t change with its length; it’s also proportional to the distance, which is proportional to the length of the cable and doesn’t change with its cross-section. The force F = EAx/L is also proportional to the Young’s modulus E: it’s the cross-section A times the Young’s modulus E times the distance x, divided by the length to get the strain. So we have ∫EAx/L dx from x=0 to x=LS/E, so the total energy is ½E A/L (LS/E)² = ½ALS²/E. The AL part of this is just the volume of the wire, so ½S²/E is the energy that can be stored per unit volume.
In the case of A36 steel, this gives us an impact energy (before plastic deformation) of about 156kJ/m³, or 20 J/kg, which works out to 6.6¢/J of elastic deformation at the US$0.60/pound price above:
$ 0.60 / pound / (1/(7.8 g/cc) * (250 MPa)^2/200GPa/2)