What do tensile strength values mean and why are they reported in units of pressure?

Question

How does one interpret the numbers when reading data about tensile strength, yield strength, and the likes?

Say for example reinforcing bars. Grade 40 Rebars are rated at 70,000 PSI for its ultimate tensile strength and 40,000 PSI for its yield strength. I know a material's tensile strength is the point in the stress-strain curve at which the material will ultimately fail while the yield strength is the point on the curve at which the material will permanently deform (correct me if I'm wrong).

What I don't understand though is where this stress (pressure?) is supposed to be applied on the object for it to break. What does the "per square inch" (or whatever unit area) point to, the cross section? How should I interpret this data if I plan on hanging some load directly on the center of a rebar whose ends are securely mounted by the ends? What if the load is evenly distributed throughout the bar? What if I use a pipe whose center is hollow?

I need to understand this in a practical sense and not just numerically since I'm planning on building some equipment where structural integrity is a must. I'm pretty sure interpreting this should be fairly simple, but I can't seem to get a grasp on the 'real' and 'practical' side of this, regardless on how much I use Google and Wiki for this.

Answer 1

Q1: How does one interpret tensile strength, yield strength, etc.?

The answer is to interpret them as the result of a test that tells you what the material can withstand in an engineering application.

The type of machine used to measure tensile strength is popularly called an Instron machine (the most famous manufacturer is Instron; kind of like how tissue paper is called Kleenex). Here's a picture from Instron's website of what it looks like:

A specimen of the material to be tested is placed between two clamps which move apart at a predetermined speed, e.g. 10 millimeters per minute. The machine provides whatever force is necessary to keep that speed constant as the material deforms. The record of the force provided and the distance the clamps have stretched the material provides the stress-strain curve for that material:

^{Image source: KeyToMetals.com}

You read the values of "tensile strength" and "yield strength" off of this curve, as you stated in your question. The units are units of pressure, e.g. $psi$ or $N/m^2$.

Q2: Why units of pressure?

The area does indeed refer to the cross-sectional area. Different applications call for vastly different diameters of steel bars. Most simply, the answer is because you will need a bigger force to rip apart a thicker bar of steel.

Example: Bar A and Bar B are made of the same steel, but Bar A has a cross-sectional area twice that of Bar B. Bar A will require twice the force to rip apart. However, they are both made of steel and so both should be the same "strength," right? If you divide by the cross-sectional area, they do have the same strength. The reason for reporting the strength of materials as pressure is practical: it is a way to compare the strength of materials ignoring the thickness of the bar.

Answer 2

A measure like 70,000 PSI is found by making a dent in a bar, of reduced area, and then stressing it. The ultimate load a beam can carry, for example, is governed by things like cracks in the bar etc, which reduces the minimum cross-section of the bar.

Other area-reducing elements are holes drilled for fasteners (which is why welding is being also used).

Some attempts to alleviate this can be done with laminated metals and the such, which arrest the spread of cracks across the whole beam.

But really, if you want to build something of structural integrity at the level of hundreds of PSI, i would really reccomend getting a real engineer in to look at the proposed structure, and not try to crowd-source an answer to a hypothetical.

Answer 3

There are different types of deformations that result in different stresses. There are axial, bending, shearing, twisting and buckling modes that may need to be considered. If you are dealing with beams, I suggest read up on the theory of beams, and look into a Engineer's Handbook, or even better the Roark formulas for stress and strain book.

Find an engineer friend of yours to give you a crash course on some basics, because this is not something to learn from Wikipedia or [Physics.SE].