The practical goal of elasticity theory is often to calculate the deformation of an elastic body under an applied stress. We want to know how much force can be applied to the wing of an airplane before it breaks, or how much force it would take to crumple a car body in a collision, or how much weight a bridge can carry.
The situations I just described represent large deformations and the eventual failure, or rupture, of the object. But elasticity theory usually applies only to small deformations (say, 3% or less). Often, however, it only takes a small deformation to cause trouble. For calculations involving the strength of materials, one wants to know how much stress can be applied before the material yields, i.e. is no longer elastic.
An elastic material is one which has a "natural" state, that it tries to return to when an external force is applied, and will return to when that force is released. This statement applies to each point in the body. (A material that does not return to its initial state, but remains deformed, is said to be "plastic.")
Consider a point in an elastic body which has no applied stress - no force is being exerted on its surface. We think of this point as being attached to the material, so that if the body is deformed in some way by an applied stress, this "material point" may move, relative to a coordinate system that is fixed in space.
What we may want from the theory is a way to calculate the displacement, the distance and direction that any material point in the body moves due to an applied force. Alternatively, we may want to know the distribution of forces within the body, given a certain deformation.
To get started, imagine that you are holding an exercise band - or any length of rubber tubing with a handle at each end. Put one handle on the floor and hold it there with your foot, and use one hand to pull the other handle up away from the floor, as you might to exercise your bicep. We refer to the end of the band attached to the handle on the floor as "the foot end," and the other end as "the hand end."
Imagine a cross sectional area through the tubing, and the total force acting over that area on the part of tubing that is above that cross section - between it and the hand end. The total force, the integral of the stress vector over that area, must equal the force you applying with your hand, since you are hold the band very still.
Since this is true for any cross section through the band, it must be that the force of the material acting on itself (the integral of the stress tensor over the surface) is constant along the band. Because the rubber is uniform, it seems reasonable that the stress tensor is uniform over the cross section, and is therefore itself constant along the length of the band.
Suppose the band has the length l when no force is applied. Consider a one dimensional coordinate system, with the foot end of the band at x=0, and the hand end at x=l. Now when the band is stretched to some length l+dl, every point material point fixed in the band moves away from the foot end. The hand end moves the distance dl, but the foot end doesn't move at all.
Similarly, a material point in the band that was at the position x in the natural state will move to a new position x.' The displacement of the band due to the stretching is u=x'-x. We see that u(0)=0, and u(l) =dl.* Since we are considering a uniform, or homogeneous piece of rubber, whose elastic properties are the same throughout, it seems reasonable to expect the displacement to be a linear function of position. That is, u is a constant times x, and from the boundary conditions just given, u=(dl/l)x.
You can try this, by marking the band with a piece of chalk 1" from one end. Hold the band with both hands on a flat surface, alongside a tape measure, with the chalk mark near the left end. Suppose the band is 20 inches long. Hold the left end fixed at the zero end of the tape measure. Now move the right end 5" to the right, so that it has moved 25% of the initial ("natural") distance from the left end. You will see that the chalk mark has moved 1/4," which also represents a 25% stretch from its initial position.
It seems plausible that the fractional movement of each point along the band under an applied force will be the same, as long as the elastic properties of the band are uniform. On the other hand, if there were a weak region in the material, it would stretch more there.
We already know how to calculate the stress as a function of the applied force. In fact, in this simple case, the stress is constant, and equal and opposite to the applied force per unit area, from the discussion above. But how can we determine the displacement from the applied force, i.e., from the stress?
We see that the stress cannot be an explicit function of the displacement, since the displacement is not constant, but increases linearly from the foot end, along the band. However, the derivative of the displacement is constant, since du/dx=dl/l. This derivative is called the strain.
So we will postulate that for this piece of rubber, the stress is equal to the strain times a proportionality constant, which is a property of the material. (This postulate is known as Hooke's Law, originally developed for a spring). We write the equation below, where sigma is the normal stress on a surface cut through the band normal to x (the y-z plane), dl/l is the strain (which is dimensionless), and E is the Young's Modulus of elasticity. Note that E has the same units as the stress, i.e. force per unit area
Since the stress is constant and equal to the applied force per unit area, say P, this completes the solution to this simple problem. The strain is constant, and equal to the applied force per unit area, divided by the Young's Modulus. The strain is a constant fraction of the total length of the band. For rubber this fraction can be fairly large without causing permanent damage - 20 % or more. For other materials, like metals, the strain for elastic behavior must remain quite small.
From the equation above, the applied force must behave correspondingly - i.e., not exceed around 20% of the Young's Modulus for the rubber material used to make the exercise tubing. (Actually, you can get them with different wall thicknesses, and or elastic properties). In my case, I didn't have to pull very hard to stretch the band 5", probably with no more than 25 pounds of force. The tubing is 3/8" diameter, with a wall thickness of 1/16", so the cross-sectional area is a little more than 0.07 sq in. The force per unit area, which is the stress in the material, is therefore about 350 psi. So from the equation above, the elastic modulus for this particular piece of rubber is about 1750 psi.
According to the literature (Table 6-1, Barrett, Nix and Tetelman, The Principles of Engineering Materials, Prentice-Hall, 1973, P. 197), the elastic modulus for natural rubber varies from 1,000 to 10,000 psi.
Other materials have much larger moduli - Aluminum's, for example is 10 million psi. If I applied the same force to a piece of aluminum tubing with the same length and cross-section, its length would increase by 1750 x 10E-7, which is less than 0.2 mil, or 0.0002".
From the above discussion, we now have a basis for understanding Cauchy's definition of an elastic material: The stress is determined by the strain only, and that strain is not taken with respect to an arbitrary state but with respect to one particular state, the preferred or 'natural state of the material.'" (Truesdell, The Principles of Continuum Mechanics).
Since stress is a tensor, it seems likely that strain must be a tensor as well, so we'll leave that discussion for the next blog.
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