(Steel for now; wood later)
E stands for Modulus of Elasticity, or Elastic Modulus, also Young’s Modulus. It is also abbreviated MOE (`Em – Oh – Ee’). By definition it is the amount of stress (tensile or compression) required to produce a unit deformation (elongation or shortening) of a material. Structural steel has an MOE of 29,000,000 psi (per inch per inch). This means that a stress of 29 MILLION psi (pounds per square inch) will stretch (or compress) a 1-inch (long) piece of steel … 1 inch. Obviously, the piece of steel would have long before been destroyed. (You can’t stretch a 1-inch long piece of steel an additional inch before it breaks apart.) A more reasonable look might be 29 THOUSAND psi; a stress of 29,000 psi will stretch (or compress) a 1-inch long piece of steel … 1000th of an inch. Most structural steel can `handle’ this kind of stress. And when we relax the steel (take away the stress), it will return to its original length. And that’s what we mean by `elastic’. We can push or pull on it, compressing or elongating it, and as long as we are in the `elastic’ range (of stress) … when we `let go’ it returns to it’s original condition (original length). When a material is `linear elastic’, the relationship between stress and deformation is `linear’ (straight line!); and we can say that deformation is proportional to stress (in the linear elastic range).
Let’s consider a stress of 29,000 psi, applied to piece (bar) of steel 100 inches long (about 8 feet). It will stretch … 1/1000th of an inch per inch … and (but) it’s 100 inches long, so, 1/1000th of an inch per inch, times 100 inches, equals … one-tenth of an inch! Small, but we could probably actually see it! It’s real! The beauty of steel is that most steels can handle that amount of stress, and, while it does deform, it’s not very much!
Here are the equations:
E = σ / ε …
E = Young’s modulus, or modulus of elasticity, or elastic modulus, E, MOE
σ = stress (force per area), and
ε = unit deformation, a.k.a., `strain’.
Strain, if you would like an equation, is …
ε = δ / L
δ = amount of deformation, and
L = length (original) of material being deformed.
Stress, if you want an equation, is …
σ = F / A
σ = stress,
F = force, either tension or compression
A = cross section area.
Now you can use the equations to confirm the number juggling above.
Let’s say we have a steel bar that is 1 inch x 1 inch (square) cross section, and 100 inches long.
If we apply a tension force of 29,000 pounds (a lot!) to both ends,
σ = F / A = 29,000 pounds / (1 inch x 1 inch) = 29,000 psi.
The strain will be … (rearranging the equation above)
ε = σ / E = 29,000 psi / 29,000,000 psi per inch per inch = 1/1000 inch per inch.
Then, (and rearranging),
δ = ε x L = 1/1000 inch per inch x 100 inches = 0.1 inch.
The 8-ft bar will elongate a bit less than 1/8th of an inch supposing the force is tensile. If the force is compressive, the bar will, likewise, shorten 0.1 inch, but only if we can keep it from buckling (like a noodle, like pushing on a rope).
A few more words on `Elastic’ …
Let’s take, for example, Grade 50 ksi steel. First, ksi stands for `thousands’ of psi. And the 50 is the amount (of ksi) at which the steel yields (yield stress). Up to the yield stress (0 ksi to 50 ksi) the steel essentially elongates and compresses under the action of stress in accordance with `E’ (Young’s modulus). The steel deforms, but when the stress goes away, so does the deformation. Above the yield stress two important things happen. First, small amounts of additional stress cause (relatively) large amounts of additional deformation. We could also say that the relationship between stress and deformation is no longer linear (no longer governed by `E’). Second, when we take the stress away, the steel does not return to its original state (length). It gets a certain amount of permanent deformation. The steel behaves in a plastic fashion (no longer elastic). You can learn more about this via any (good) steel or engineering mechanics textbook.
But let me say a few more things, about steel. I have cast the conversation above in terms of a bar of steel (1 inch x 1 inch x 100 inches long), subject to a (tensile) force. If we apply a force of 1 pound (lb), we get a stress of 1 lb / (1 inch x 1 inch) or 1 psi; we get a strain of 1 psi /29,000,000 psi per inch per inch, and an elongation of 1/29,000,000 inch per inch times 100 inches or 1/290,000 of an inch (pretty small). A force of 100 lb would give us 100 times as much, and so on. Deformation goes with force (up to the yield stress, or elastic limit).
But now let’s think in terms of taking the bar and stretching it determined amounts of deformation (instead of applying force or stress). A tiny deformation will require a tiny force; a deformation ten times as much will require a force of ten times as much, and so on. Up to the yield stress. The deformation associated with a stress of 50 ksi is …
ε = σ / E = 50,000 psi / 29,000,000 psi per inch per inch = 0.0017 inch per inch …
and for the 100-inch-long rod …
100 inches x 0.0017 inch per inch = 0.17 inch (about 3/16 inch).
If we keep elongating the rod … up to 1/4 inch, 3/8 inch, and so on, we’ll continue to elongate the bar, but it will take on only a bit more stress (force). No matter how much more we stretch the bar, it will not take on (`carry’) much more than 50,000 psi, or, in this case, 50,000 pounds. Eventually the steel will … BREAK APART! The stress at which it (ultimately) breaks is called the `ultimate stress’. The deformation at which the steel ultimately breaks may be multiple times that associated with the yield stress. The ability of steel to take on a lot of deformation, before finally `coming apart’, lends to it being a `safe’ construction material, as well as strong. The steel will take on gigantic deformations, but still hold (things) together.