# The Phyisics Philes, lesson 67: I-Beam, You-Beam, We-All-Beam for…Physics!

So far in our discussion of equilibrium and elasticity we’ve focused solely on rigid bodies. But real life isn’t composed of only rigid bodies. There are squishy things and stretchy things. How on Earth are we supposed to deal with those? We need to talk about stress and strain.

Stress and strain are familiar words, but, as I learned while I was preparing for this post, they have specific definitions that are different than how they are colloquially used. Stress is a value that characterizes the strength of the forces causing deformation. Strain describes that deformation. We need to keep these definitions in mind as we discuss squishy objects.

When the stress and strain on an object are small enough, the two are directly proportional. This proportionality is called the elastic modulus, and it’s calculated by dividing the stress by the strain. It’s also known as Hooke’s law, which isn’t actually a law, per se. It’s an experimental finding that valid over a limited range. Basically, the harder you squeeze, the more the body compresses. Conversely, the harder you pull on a body, the more it stretches. Pretty intuitive, really.

But this definition isn’t really helpful unless we know how to calculate stress and strain. So I guess we should learn how to do that, huh?

First, let’s talk about tensile stress and strain. The easiest way to think about this is to imagine that you’re pulling each end of a rod or bar. When you do that, the bar is in tension.

OK, now imagine this. The rod you’re pulling on has a uniform cross-sectional area A and an initial length. When you apply equal and opposite forces on the bar – like, say, by pulling on each side – you put the object in tension. You can find an actual numerical value for tensile stress by finding the ratio of the force to the area A. That gives us the strength of the forces causing the object to be deformed.

But what about tensile strain? Remember that strain describes the deformation caused by the stress. To find the tensile strain of an object, we need to figure out how much it’s deformed. Simply put, tensile strain is the ratio of the change in length to the original length. Since tensile strain is a ratio of two lengths, it is a dimensionless number with no units. This is different from tensile stress, whose SI unit is the pascal, which is equal to 1 N/m^2.

For a sufficiently small tensile stress, experiment shows that the stress and strain are proportional. The ratio of tensile stress to tensile strain is known as Young’s modulus (denoted by Y). The bigger the value of Young’s modulus is, the less stretchable it is. That is, the larger the value for Y, the more stress is required for a given strain.

That’s all well and good, but what about when our hypothetical rod is squeezed together? This is called compression. We calculate compressive stress the same way we calculate tensile stress: force divided by the area. Compressive strain is calculated in the same way tensile strain is calculated, except that the change in the length is in the opposite direction, i.e. the rod gets squeezed shorter rather than stretched longer. Hooke’s law is also valid for compression as long as that compressive stress isn’t too great, and Young’s modulus is the same value for tensile and compressive stress for many materials.

These concepts of tensile and compressive stress are why I-beams are shaped the way they are. A horizontal beam will sag under its own weight. As the beam sags, compressive stress acts on the top half of the beam, and tensile stress acts on the bottom half. The middle is in neither compression nor tension. In order to minimize the stress, the top and bottom halves are given bigger cross-sectional areas. This isn’t needed in the middle, hence the characteristic I shape of the I-beam.

Dude! It’s physics!

This is just the beginning. In the coming weeks we’ll learn about all kinds of different types of stress and strain. (OK, two more types. But still!)

*Featured image credit: slgckgc*

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