Consider a case where a column supporting a building is bearing a heavy load. We can speak of the stress and the strain on such a column. The stress denoted with a Greek sigma (like our 's') is the force applied per cross sectional area, or [math]\sigma = F/A.[/math] When the column is under stress the consequence is strain, which is the fact that it gets shorter. The strain is denoted with a Greek epsilon (like our 'e'), and is [math]\epsilon = \Delta L/L[/math] where L is the unstressed length of the column. Take note that the strain is a dimensionless quantity.

Young's modulus quantifies the stiffness of a solid. It is often called the elastic modulus. It is a material-specific quantity and does not depend on the shape of the object made with the material. In other words, you can look up Young's modulus for some alloy of steel or of titanium, but not for a given spring made of that steel. For springs, we instead use the elastic constant k, which will be specified by the manufacturer of the spring. It will depend on both the material and its geometry, unlike Young's modulus which only depends on the material. Young's modulus applies in cases where objects are being stretched (like a wire under tension) or compressed (like a column). These are called [b]tensile[/b] and [b]compressive stresses, [/b]respectively.[br][br]Young's modulus is defined as the ratio of the stress to the strain:[br][br][center][math]E=\frac{\sigma}{\epsilon}.[/math][/center]If we plug in physical dimensions for a column, for instance, we can see that it is really Hooke's law in disguise. Let's do that. Suppose the column has length L and cross sectional area A, and is subject to a compressive force F. That means we can write [math]E=\frac{F/A}{\Delta L/L}.[/math] Let's try to rearrange that to look like F=kx (noting that we're not worried about the minus sign that belongs). There is already an F in the equation, and the [math]\Delta L[/math] clearly plays the role of x (the stretch or compression distance). That means we can write [math]F=\frac{EA}{L}\Delta L.[/math] The group of constants in the fraction must represent the elastic constant, so [math]k=\frac{EA}{L}.[/math][br][br]This should give us insight about stiffness of materials. Since E depends on the material, certainly the stiffness of a given object will depend on the material from which it's made. But we also see that the thicker it is in cross sectional area A, the stiffer it gets. While that is probably intuitive, we can also see that longer objects are less stiff. In the case of the column, a longer column of equal cross section will be compressed more easily. Likewise a longer spring will be easier to stretch than a shorter one.[br][br]While that discussion can give us insight about the behavior of springs as well, they technically are under a different kind of stress called shear stress, and so Young's modulus doesn't technically apply. We can nonetheless safely conclude that longer springs will be less stiff and that thicker ones will be stiffer.

The diagram below shows the idea of shear stress. Conceptually it is very similar to tensile or compressive stress except that the two surfaces are stressed differently. A material has a corresponding shear modulus to indicate its stiffness when stressed in this fashion.[br][br]The shear stress is denoted by a greek tau. It is the applied force F in the diagram below divided by the coplanar area A, or [math]\tau=F/A.[/math] The strain in this context is [math]\epsilon = \Delta x/l.[/math] The shear modulus G is just the stress over the strain as before, or [br][center][math]G=\frac{F/A}{\Delta x/l}.[/math][/center]

The bulk modulus is a measure of the stiffness of a material when under pressure from all sides such as occurs when submerged in a fluid. The squeezing causes a proportional decrease in volume. The higher the bulk modulus the harder it is to compress a material. For a given change in ambient pressure [math]\Delta P[/math] we can write the bulk modulus K for a given volume V of a material as:[br][center][math]K=-V\frac{\Delta P}{\Delta V}.[/math][/center]This is often defined as a derivative rather than using deltas, but it's easier for first time students to see what it means this way. For small deformations it doesn't matter anyway. And while for whatever reason the bulk modulus is the one of these three moduli that is most often written as a derivative, in principle all of them should be. They should all have the following replacement [math]\frac{F}{\Delta L}\rightarrow \frac{dF}{dL}.[/math] Yet for small deformations it isn't too important to worry about it.[br][br]The math is simple to do in order to see why it doesn't matter. The approximation is good to about a stress of 10% which most materials won't tolerate anyway. Try making the above substitution and integrating both sides by separation of variables. If you do the math correctly you'll see that the difference comes down to the difference between [math]y=\frac{x}{a}[/math] and [math]y=\ln(\frac{x+a}{a}).[/math] While those look different on paper, the plots are very similar for small values of deformation x.