# How do I calculate the strength to weight ratio of a material

I want to manufacture something using metal 3d printing. And I'm now choosing the material. Stainless steel is stronger but much heavier than aluminum, so I want to calculate which material can achieve a lighter weight with the same performance. I know titanium or other super alloys such as scalmalloy is better but they're just too expensive.

The best strength to weight ratio depends on the type of loading. I.e. if the loading is in tension of flexure the strength to weight ratio changes.

A unified way to access materials is the concept of material indices, which help creation of Ashby charts. For a given loading you can access different materials.

i.e. the optimal material for strength to ratio

• for tension is the one with the highest $$\frac{\sigma}{\rho}$$
• for pure bending is the one with the highest $$\frac{\sqrt{\sigma}}{\rho}$$
• etc

figure: Ashby diagram

You need to know what kind of loading and geometry you have. You also need to know if you are volume limited or mass limited. You can't just have a part of identical geometry and compare. You need to compensate the geometry for the material used.

For example, aluminum is three times less dense than steel and has three times lower tensile strength (measured in force per cross sectional area). But tensile strength is not the only strength. In bending, one side is under tension while the other side is under compression, and often the primary mode of failure in bending is on the compression side where it buckles. This is highly dependent on volume of material so it helps if the material is less dense so you can use more volume of it without increasing mass.

For example, compared to a steel rope of the same cross sectional area, an aluminum rope will be one third the mass and one third the tensile strength. But if you designed that aluminum rope to be three times the cross sectional area as the steel rope, then it would be about the same mass and tensile strength.

However, it would have much higher bending stiffness (triple the cross sectional area works out to be $$\sqrt 3$$ increase in radius and I believe it is raised to the power of three so 5 times stiffer) and usable bending strength (i.e. before it bends or yields unacceptably) because the larger cross sectional area means there is much more material away from the axis of bending. This would matter if instead of a rope, you were designing a pillar or beam. This metric of how much material is positioned away from the bending axis is called the second moment of inertia and is used in with the strength of materials numbers in the strength calculation.

It's why a hollow tube is not as strong or as stiff as a solid rod of the same diameter and material, but is almost as stiff and strong but weighs much, much less. It's also why I-beams have the flat horizontal caps to position space apart the opposite sides of the beam that are in tension and compression.

It's why a board of plywood is stiffer and has higher bending strength (which is when one side is under tension and the other side is in compression) than the sheet of steel if they both have the same mass. The plywood board is going to be thicker since it is less dense. But the sheet of steel might still have higher tensile strength the plywood board. Two different aspects of strength.