What affects how much an object can be stretched? So we have an object here, and we want to stretch it, and so it's this long instead. We're going to call this extension here delta L. So what affects how big this extension will be?
In order to stretch something, first of all, we need a force, don't we? A bigger force will increase it. What else do we have? We have cross-sectional area. If you've got a bigger area then it actually makes it harder to stretch something because it's thicker so it's the opposite.
Its original length will have a big effect as well. The bigger the original length the more it will stretch. And then we've got one more thing as well.
What material it is. There's got to be some property of a material that means that it's harder or easier to stretch. And we're going to call this the Young's modulus.
We're going to give that the capital letter E. Sadly we have a capital E here that doesn't mean energy, so be careful regarding what E you're actually using. Now we're going to say that if we have a bigger Young's modulus it's going to be harder to stretch. kind of like spring constant of a spring.
Let's figure out an equation that incorporates all these things here. So, how far something stretches, that is dependent on force, because a bigger force will increase the extension. It's also dependent on the original length, but it's inversely proportional to the area and this Young's modulus. What we're going to do is rearrange this for the Young's modulus E. We can measure the Young's modulus of material by measuring the force applied, its original length, the cross-sectional area and how far it's extended as well.
Now a couple of other definitions. When we talk about materials we've got something called stress and we give it the symbol sigma. Stress is very similar to pressure and it has the units, newtons per meter squared. That's also known as pascals.
Very similar to pressure. Generally, we use stress when we're pulling something and pressure when we're pushing something. And we have this as well, strain. Give that the symbol epsilon, not emf.
That's the ratio of how far a material has stretched compared to its original length. It doesn't have a unit because they're both meters. So it turns out that the Young's modulus of material can be calculated using these two things here, stress over strain.
So that's equals to the stress divided by the strain Putting these two fractions together we end up with our equation that we have above. So here's our equation again. Young's modulus equals stress over strain. And that's equals to F L over A.
Delta L. So in order to find out this value for a material, we can do an experiment and this is actually a required practical. What we can do is hang two of these wires from a beam. One is just going to have a weight on to give it some tension and the other one we're going to be adding on some masses onto that to make it stretch further.
Now what we do is compare these two wires and we have a vernier scale. So this is what our setup looks like. I have my two wires coming down here, I have my reference wire and then I have my extending wire.
So this is what my Vernier scale looks like when it's set up to begin with. My two bits of my Vernier scale are lined up zero to zero at the top, it's calibrated. As I add masses to this wire here it's going to move this half of the Vernier scale down past the other one. First thing we need to do is see how many millimetres this zero point on this half of the Vernier scale has gone down.
down. So to read a vernier scale, step one, how many millimetres has zero on scale moved? So let's say that my vernier scale has moved to here, I can see that my zero has moved one, two, three, it's gone past three millimetres there.
But the vernier scales are clever because this half of the scale is different to the scale here. What we need to do then is have a look at all of our lines on here. There should be 10 lines.
I've drawn way too many on here, but you get the idea. There should be 10 lines on here. What do I do? I find out which one of these lines matches up with one of these lines the best. So does the one line up here?
No, it doesn't. I'm going to say that it's my number two that lines up best. So step two is find mark that lines up best with the other side of the scale. In this case, it's two. So that actually gives me 0.2 millimeters.
That means that my extension is now three millimeters plus that two tenths of a millimeter. That's what my extension is. And this is how any Vernier scale works. They could have different resolutions.
Usually the scale goes to millimeters. This is usually tenth of millimeters. Sometimes it goes to 20th of millimeters. So you can go to the nearest 0.05 millimeter.
So just to recap. Find out how many millimetres the zero has gone past for your millimetres and then find out which line lines up best with the other side on here to find out how many tenths of millimetres, to find out how many tenths of millimetres on top it's gone. By the way having this reference wire here is useful because if there are any changes to temperature and that results in expansion then it's going to happen to both of the wires. So what could we do?
Well we could draw a graph of stress against strain. And that would give us a nice straight line and the gradient of that is going to be our Young's modulus. We can see that from the equation. But when we're doing this experiment, we're just changing the mass and we're measuring the extension.
So it's much better. to have a graph of mass against extension. That's still going to give us a nice straight line.
If it starts levelling off like that, that's when we stop taking our recordings. We'll talk about what's going on there in a moment. How do we find out...
what's going on here? Well we know that the gradient here is equal to the mass divided by the extension. So let's go back to our full Young's modulus equation.
We have FL divided by A delta L but we know that the force force in this case is going to be mg. So that ends up being mgl divided by a delta l. What part of this equation gives us our gradient? It's this part here, m divided by delta l. So that means all we have to do is take our gradient times by gl.
Divided by a and that will give us our young's modulus Gb 9.8 lb in the original length of the wire a be in the cross-sectional area if this is a wire And this is probably going to be pi r squared or pi d squared over 4. I'd recommend measuring three times to get the mean radius or the mean diameter of the wire. I would recommend measuring the diameter three times along the wire before you start and get the mean which gives you a far more reliable value for the cross-sectional area. So we said that if you do this experiment and you plot stress against strain you should get a nice straight line.
So that means that if we... put more force, that's more stress, on our wire, then we're getting more strain. In other words, it's stretching further. But eventually, it's going to start levelling off like that.
Now, we do see some weird things happening here as well, past that point. This is called a yield point. This point up here is called the ultimate tensile strength.
Then here is the breaking point. But we're only concerned about these points here. We care about this straight line and this point up here. Now, we can see that...
while we've got a straight line, stress and strain are proportional. When it starts levelling off like that, it's no longer proportional. So that's why we call this the limit of proportionality.
This here is called the elastic limit. We call that the point of no return brackets into... Original length.
Even though we've gone slightly past the limited proportionality, even up until that point, if we take the load off the wire, it will actually return to its original length again. Past this point, in other words, we say that it's still active. elastically. Past this point it's no longer elastic but plastic, it's being plastically deformed. In other words if we stretch it past this point it's not going to return to its original length.
Incidentally, very similarly to what happens with a spring, if we've got force against extension then the area under the graph is going to be the energy stored in the wire. Now then, let's say that we have a stress stone graph for a couple of materials. This material here. goes up and then breaks there. This material goes up like that, plastically deforms like so.
This is what we would call a very brittle material, this is what we call a ductile material. Brittle can't be plastically deformed. In other words, once it reaches its limit of proportionality then it just breaks.
So that's going to be something like glass or ceramics. Something that's ductile, that could be copper or something, that can be stretched and it can be deformed plastically. And it makes sense that brittle materials have a very high gradient, therefore a very high Young's modulus for their elastic deformation. Ductile materials, they can be stretched more easily and they can be stretched plastically as well.
The last thing, loading and unloading curves. Now we said that if something goes up to its limit of proportionality, go past and it reaches its elastic limit, if we take off the force it will actually... go back part way to its original length.
It's been permanently stretched. Now talking about area, we know that to stretch whatever this material is up to here, the area under the graph up to that point is the energy put in when it's being loaded. When it's being unloaded however, again we know that the area under the graph gives you the energy but this time it's the energy given out.
So that must mean that whatever the area is of this here, that's going to be the energy left in the wire or the material. So we've put more energy in extending the spring then we've got out to let it go back to nearly its original length so some of that energy has been lost. Now generally that's going to be lost due to heat.
in the process of stretching it past its elastic limit. Now these curves can be all kinds of shapes. They don't have to be straight going up. They can be curved going up and curved coming back down. But it's important that you understand that the area between the loading curve and the unloading curve, that's going to be the energy lost.
So that's stress, strain and Young's modulus. Hope you found that useful. If you did, please leave a like. If you think I've missed anything or have any questions, please leave a comment down below and I'll see you next time.