 Alright, so let us take a couple of numericals. We can't just keep on doing theory. So I'll project a numerical. Which book in school you guys follow? Oh, um, the, like the, the official Cambridge one. I don't know exactly who it's by. Okay. Can you send me the name of that book? Yeah, I'll send it. I think I might have it already. Just there is Cambridge. This one, this one. No, I don't have this one. I think I maybe what I have is a later version of this. I'll check and send. Start doing this one. See, in this chapter is like thermodynamics where you exactly know how to solve, but people make a lot of calculation error. So just curtail that and get the answer correctly because if you get it wrong, then you have wasted a question. Okay, so it is all about calculation. Do it properly. Young's modulus is given as 2 into 10 is for 11. It's very high Newton per meter square for steel. Yeah, done. Done? Yeah. Yeah, I got doing the last part. Okay, yeah, done. Okay, what is the answer for stress? For 10, I got 10 to the power of 9 by pi. Yeah, 10 to the power of 9 by pi. Okay, you need to simplify that. Okay, you can't leave it like this. It is 3.18 into transfer rate. Yeah. What is the elongation? Delta L? Yeah, I got 0.5 into 10 to the power of minus 2 pi. Arushi, what do you got? Wait, I'm not getting that. Okay, what did you say? 0.5 into 10 to the power of minus 2 pi. Wait, give me a minute. You have to simplify that. You can't let me answer it in terms of pi. Do that quickly and tell me. Yeah, okay, wait. I think I got something. I got 1.6 into 10 to the power of minus 3. Yeah, I got 1.57 into 10 to the power of minus 2. Oh. It is minus 3. 1.59 into 10 to the power of minus 3. You might as well hear that. It's a factor of 10 somewhere. Okay. Yeah, that's it. And strain, how much? Strain, I got... Wait, I'm sorry. Yeah, so strain, I got the same thing. Yeah, me too. Same, 1.6 into 10 to the power of minus 2. Yeah. 1.6, what? In 10 to the power of minus 3. Minus 3. But when you write strain, you also write in terms of percentage. What is the percentage of deformation? That into 100, which is 0.159 percentage. Oh, okay. Okay. Is that clear? Yeah. Okay, next one. Steel Young's Modellers, you know, that is from the previous question, it is steel and the copper Young's Modellers is 1.1 into 10 to the power of 11. You know, you can substitute the values at the end. Take L1, L2, A1, A2, D1, D2, whatever it is, then Y1, Y2, find out in terms of these variables and substitute at the end the values when you get an expression. Are you stuck? No, I'm stuck. I'm on the side. I think I am stuck. Yeah, I think so too. Okay. So, there is wire like this. Fine, let me move. There is this copper wire, all right, whose Young's Modellers is Y, let's say Y1, length is L1, and area processing is same for both, right? Yeah. A, this is, let's say Young's Modellers Y2, length L2, and area A. They're stretched by a load, let's say the load is F, it has to be stretched from both sides, one end will be fixed, other end is stretched, or both sides, same force is applied. Okay, so this is the situation and the net elongation is given, let's say delta L is a net elongation, which is 0.7, okay? So, can I say that total elongation delta L is equal to elongation in the first one plus elongation in the second one? Yeah. Yeah. Okay, and the formula I have for Young's Modellers and any other thing, it is with respect to a single material, but the problem here is that you have two different materials connected together. Yeah. I cannot use directly the formula, but if I somehow break this, if I consider the copper separately, and if I find out what is the stress in the copper, then I can use a formula for the copper, right? So, it is applied here, copper is also stationary only, so F will be here, so F will be below also, right? It is not going anywhere, and the second one, the steel, this F is applied by the steel on the steel upside over here, it will be applied, Newton's third law equal and opposite, like this can see that the force is same for steel and copper, and since area is same, even the stress is same. So, that is why the Young's Modellers, Y1 is F L by A delta L, this is the formula, right? So, delta 1 and F is same. Yeah. Friend, so basically I am getting delta L1 as F L1 divided by AY1, similar delta L2 will be F L2 by AY2. Yeah. Total extension will be equal to F L1 by AY1 plus F L2 by AY2, right? Yeah. I know everything now, I know delta L, I know L1, L2, AY1, AY2, so I can get the value of F, I can take F common and divide delta L by the rest of the thing. Oh, nice, nothing wrong. Okay. This is how you solve these kind of problems. Now, something about thermal stress, you guys live in a flat? I live in a flat. So, the top floor, top floor of the flat, top floor of the building? I live in some of the middle floors. So, you might not have experienced that the top floor, the crack comes the most. Okay. So, when you move in a freshly prepared building, the top floor, initially it will not have any cracks, but slowly the cracks will start coming in. Why? Because it receives heat due to sun and because it receives heat, its temperature goes up to thermal expansion happens. Okay. If it doesn't happen uniformly, then cracks will appear, okay? Okay. But if you are somewhere in the middle, then you get compressed from the top, so that get compensated. But anyways, what I'm trying to say is that there can be strain because of the heat as well. Strain is what? Strain is deformation? Correct. And the length will change because of change in temperature also, thermal expansion we have seen. Okay. And we have seen that according to a thermal expansion formula, delta L should be equal to what? L, alpha delta T? Yeah. Okay. So delta L by L is what? Alpha delta T? Yeah, which is true. Yeah. Which is not strain by the way, it is misleading. See what, when I say length L, this length L is the original length. And what do I mean by that? Original length add a particular temperature. So basically, if I increase the temperature, length will become L plus delta L, which is L naught, one plus alpha delta T. Okay. So if I let it expand, if the length, I do not stop it to expand, there will not be any stress. Because the original length is itself has increased. Original length is no longer L naught. Okay. You compare the deformation with respect to what should be its length, right? Its length should be something else at a different temperature. Okay. But if you do not let it expand, if you do not let it expand, then there will be deformation. Okay. Yeah, then there will be deformation because it is no longer L naught, one plus alpha delta T, you are not letting it reach there. So whatever the difference between this and what you are allowing, that will create a strain and because of that stress will be there. Okay. One such example is this. Suppose you have in between the two walls, you're clamping a rod. All right. The rod has Young's modulus Y area of cross section A and length L. Okay. And this is the original values at a temperature T. Now you are increasing your temperature by delta T. Okay. So if you increase your temperature by delta T, what will happen to the length? Anyone? Length will change or not? It will. Yeah. You're not letting it change. Oh yeah. So then it will deform. It will get deformed. Apparently, when you look it from your eyes, there is zero deformation. Ever was earlier, same thing built, you know, after also, but the reality is that this rod will behave as if it is compressed now. Oh, letting it expand. All right. And how much you're not letting it expand? You're not letting it expand by the value of entire delta. So this should be your strain now because you're not letting it expand. Okay. Okay. So when it comes to longitudinal compression or extension, both are similar. So this is the strain and you'll get the stress as well. So once modulus is stress divided by alpha delta T, the stress corresponding to that strain is why alpha delta T. Okay. In reality, this denominator should have been this original length, but we say that it is okay because almost the same. Okay. Yes. And suppose this is not the case. Originally, it is like this and the wall was not touching this end. Okay. Let's say you leave, let's say a very, very small distance of delta X. Then you heat it by delta T. Now, can you tell me what is the value of stress developed? So it'll expand then. Do it, do it. Solve it like a numerical. We have to calculate the stress, right? Hmm. Would it just be y dx by L? Yeah. Okay. See, its change in length should be how much L alpha delta T? Yeah. But you are, if you're making it expand by delta X only. It wants to expand by L alpha delta T, you're making it expand by delta L itself. Delta X, yeah. So the compression delta L dash will be equal to L alpha delta T minus delta X. Okay. Oh, okay. The strain will be this divided by L. So L alpha delta T minus delta X by L. Oh. Got it. Okay. Okay. The denominator is this. This one. And then you can get the value of sigma. Okay. So there are a few questions on thermal stresses that you will see.