### Video Transcript

In this video, we’re going to learn about stress, strain, and elastic modulus. We’ll learn what these terms mean, how they relate to one another, and how they describe deformable solids.

To start out, imagine that you are training to learn how to walk on a tightrope. The rope is made of fairly elastic material. And because you expect to fall fairly often before you get the hang of it, you want to be able to fasten the rope as low to the ground as possible so the fall is as small as possible.

You know that as you stand on the rope, your weight will cause it to stretch a bit. And you like to figure out how low you can tie the ends of the rope onto the supporting posts while still staying off the ground when you’re at the midpoint of the line. To figure this out, it will be helpful to know something about stress, strain, and elastic modulus. All three of these terms relate to the deformation or shape change of solid objects.

Stress is pressure with units of force per unit area applied to a cross section. When we walk along with a hiking stick and press down on it, when we press a key on our computer keyboard, or when we pull on a rope in a game of tug of war, we exert stress. From an engineering or a mathematical perspective, we symbolize stress using the Greek letter 𝜎.

Though we sometimes use the terms interchangeably in our everyday language, strain is different from stress. Strain is a measure of how much a solid deforms or changes shape due to stress. For example, when we’re walking along with our hiking stick, the stress that we put on the hiking stick actually causes it to compress a tiny amount. That tiny compression of the stick along its length would be its strain. Or in the case of our tug-of-war game, the rope would extend a bit due to the stress of our pulling on it from either side. And that extension would be the strain.

In engineering terms then, stress causes strain. We can write strain as an equation using the Greek letter 𝜀 to represent it. Strain is equal to the change in an object’s length divided by its original length, 𝐿 sub zero. So a stress is a pressure applied to a cross section of a material. And strain is how much that material changes shape because of the stress.

We can combine these two terms into a third term called elastic modulus. And this is a material property. The elastic modulus, also called Young’s modulus, is a measure of the ratio of the stress applied to an object and the resulting strain or deformation of that object. Using symbols, we can write it as 𝜎, stress, divided by 𝜀, strain.

As we mentioned, the elastic modulus is a property of materials. So aluminum has an elastic modulus, so does nylon, so does Teflon, and so on. The elastic modulus of a material is very high if it takes a lot of stress or pressure in order to change the shape or deform that object a small amount. On the other hand, if very little applied stress creates a significant change of shape or length in a substance, we expect its elastic modulus to be relatively low.

Knowing that, if we were to compare the elastic modulus of steel with the elastic modulus of rubber, which one do you think will be greater? In the case of steel, it would take quite a large stress or pressure in order to create a very small shape change, whereas rubber might require a significantly smaller amount of stress to create the same amount of strain. Just going off our intuition then, we would expect the elastic modulus of steel to be greater than that of rubber.

One last thing about elastic modulus, if we look at the strain equation, which is a change in length over a length, we realize that strain will be unitless. This means that elastic modulus 𝐸 will take on the units of stress 𝜎 since 𝜀, the strain, has no units in and of itself. So Young’s modulus or the elastic modulus of a material is given to us in terms of pressure, force per unit area or pascals.

Let’s get some practice with these new ideas through a couple of examples. A uniform rope of cross-sectional area 0.600 centimeters squared breaks when the tensile stress in it reaches 7.2 times 10 to the sixth newtons per meter squared. What is the maximum load that can be lifted slowly at a constant speed by the rope? What is the maximum load that can be lifted by the rope with an acceleration of 4.0 meters per second squared?

In this two-part problem, we wanna solve first for the maximum load the rope can lift at a constant speed. We’ll call that 𝐿 sub 𝑠. And then we wanna solve for the maximum load the rope can lift when that load is accelerating upward at 4.0 meters per second squared. We’ll call this load 𝐿 sub 𝑎.

In this scenario, we have a weight we’ve called 𝑊, which at first is being raised at a steady speed by a rope. Knowing the maximum tensile stress the rope can handle before breaking, we want to solve for the maximum load 𝐿 sub 𝑠. If we choose the upward direction to be positive and consider the forces on the weight 𝑊, we can write that the tension force created by the rope is equal to the maximum load, 𝐿 sub 𝑠, that can be lifted at a steady speed. That tension force 𝑇 is equal to the maximum stress the rope can handle multiplied by the cross-sectional area of the rope.

When we substitute in these two values, we’re careful to write our cross-sectional area 𝐴 in units of square meters. That’s to agree with the units of our stress, newtons per meter squared. We see when we multiply these numbers, we’ll get a result in units of newtons. And, to two significant figures, 𝐿 sub 𝑠 is 430 newtons. That’s the maximum load this rope can lift at a steady speed without breaking.

Next, we imagine that our scenario changes a bit in that, instead of our weight being lifted at a constant speed, it’s now accelerating upward at 4.0 meters per second squared. Under these new conditions, we wanna solve for the maximum load that this rope can handle. When we write out the forces on 𝑊 now, we again have the tension force from the rope 𝑇. And subtracted from that is the weight of our load, 𝐿 sub 𝑎, whereas before, that difference was equal to zero.

Now because we have a nonzero acceleration by the second law of motion, it’s equal to 𝑚 times 𝑎, where 𝑚 is the mass of our load, 𝐿 sub 𝑎. Speaking of the mass of this weight, as we think about it, 𝐿 sub 𝑎 is the mass of this weight times 𝑔, the acceleration due to gravity, where 𝑔 we’ll assume is exactly 9.8 meters per second squared.

If we add 𝑚 times 𝑔 to both sides of this equation and we rewrite the tension force in terms of the maximum stress, 𝜎 max, times the area 𝐴, then we now have an expression that says 𝜎 max times 𝐴 is equal to 𝑚 times the quantity 𝑔 plus 𝑎. It might seem like we’re moving backwards because nowhere in this expression do we see a value for 𝐿 sub 𝑎, what we want to solve for. But 𝐿𝑎 is equal to 𝑚 times 𝑔. And we know 𝑔. So if we can solve for 𝑚, then we’ll have solved effectively for 𝐿𝑎.

So back to our force balance equation, rearranging this equation for the mass 𝑚, we know 𝜎 sub max, 𝐴, 𝑔, and the acceleration 𝑎. And when we plug these values in, again expression our cross-sectional area 𝐴 in units of meters squared, and calculate this term, we find it’s roughly equal to 31.3 kilograms. Going back to our equation for 𝐿𝑎, when we plug this value in for 𝑚 and then 9.8 meters per second squared for 𝑔, our result, to two significant figures, is 310 newtons. That’s the maximum weight that this rope could sustain while accelerating upward at 4.0 meters per second squared.

Now let’s look at an example involving elastic modulus. The lead in pencils is a graphite composition with a Young’s modulus of 1.0 times 10 to the ninth newtons per meter squared. Calculate the change in length of the lead in an automatic pencil when it is tapped straight into a pencil with a force of 4.0 newtons. The lead is 0.50 millimeters in diameter and 60 millimeters long.

We want to calculate the change in the length of the lead, which we can call Δ𝐿. And we’re told the Young’s modulus of this material, which we’ll label 𝑌 sub 𝑚. The force with which the lead is tapped into the pencil of 4.0 newtons we’ll call 𝐹. And we’re told the lead’s diameter as well as its original length. We’ll label these 𝐷 and 𝐿, respectively.

We can start on our solution by recalling that the Young’s modulus, 𝑌 sub 𝑚, is equal to stress divided by strain, where stress is equal to a force divided by an area and the strain 𝜀 is equal to the change in length Δ𝐿 divided by the original length. Combining all three of these relationships, we can say that the Young’s modulus, 𝑌 sub 𝑚, is equal to force divided by area, all divided by Δ𝐿, the change in length, over 𝐿, the original length.

We want to solve of course for Δ𝐿. So we rearrange this equation. Δ𝐿 is equal to 𝐹 times 𝐿 divided by 𝑌 sub 𝑚 times 𝐴. The area 𝐴 is the cross section of our circular shaft of graphite and therefore equals 𝜋 over four times 𝐷 squared. Altogether then, we have an expression now for Δ𝐿 in terms of variables we’ve been given in the problem statement. Plugging in all these values, being sure to convert our length to units of meters as well as our diameter 𝐷, we find that Δ𝐿, to two significant figures, is 1.2 millimeters. That’s the change in the length of this shaft of graphite in response to the applied stress.

Let’s summarize what we’ve learned so far about stress, strain, and elastic modulus. We’ve seen that stress is pressure in units of newtons per square meter applied to a cross section. The symbol for stress is 𝜎. Strain, on the other hand, represented by the Greek symbol 𝜀, is how much a solid deforms due to an applied stress. Written as an equation, strain is equal to the change in an object’s length divided by its original length.

And finally, we saw that elastic or Young’s modulus ties stress and strain together. As a material property, the elastic modulus is equal to stress divided by strain, or 𝜎 over 𝜀. Overall, stress, strain, and elastic modulus help us understand forces that are acting on solid objects and how those solid objects change shape as a result of those forces.