 see now let me tell you something as here suppose you have a plate this is a plate fine all right so l is equal to l naught alpha delta l is equal to l naught alpha delta t delta b is b naught alpha delta t ok what about this if I draw a randomly a line like this this line expand if I give need to this substance will this line expand it will expand it will expand fine so if original length is x then delta x will be what x alpha into delta t ok very very simple so all the linear dimensions all the linear dimension in a substance will follow linear expansion formula getting it so if the unit is meters centimeter or m m like unit of length is there dimension is length then that will expand as linear expansion getting it so it need not be only external dimension of a substance fine is it clear now similarly if I draw a square over here whose area is a naught this will follow up area expansion fine similarly if it is a block inside the block there is a small volume that will follow volumetric expansion fine now let me give you another thing tell me what will happen here you have a plate and this plate has a hole in it if you heat it temperature increases delta t what happens to the diameter of the hole increase or decrease if I heat this the hole will increase in size or decrease in size so my question is is the material try to occupy this m d space or it will simply move out which one it will increase why 2 pi r will increase because that is a perimeter of the circle that has a linear dimension that will increase and automatically the diameter increases now diameter will follow what expansion law this diameter linear expansion fine d is equal to d naught into 1 plus alpha delta t or delta t is d naught alpha delta t fine have you learned about mechanical properties of solids already you have done Young's modulus and all that right now tell me one thing I am just talking about scenarios ok I am not getting into the numerical of it right now so if you have a rod like this length is a fine it is clamped between the two walls fine now you are increasing the temperature by delta t this rod is not allowed to expand what will happen this rod is not allowed to expand remains the same length it remains the same length but actually what this rod wants to do is increase its length of it by probably this much how much that would have been if you are allowing it to increase this delta l would have been equal to l alpha delta t but you are not allowing it to increase so in a way it is in compressed state getting a point and what is a strain in the rod delta l by l which is what strain is delta l by l which is equal to alpha delta t this is called thermal strain because the natural length of the rod changes at different different temperature at 20 degrees Celsius length should be this much to have zero strain but if you do not allow it to expand there will be a strain from what it should have been getting it so that is the reason why you will see that there is a small gap in the railway tracks those rails there is a provision for it to expand if its temperature increases it will expand little bit but if you do not allow it to expand it will become with a curve sort of thing and it will anyway change its length right so like for example this is the two rails you are not allowing it to expand you are clamping it so if temperature increases it is just you know randomly it bulges out from here and there so this is thermal strain and there are a lot of nice numerical that are made out of just thermal strain there comes the Young's modulus also you know how much stress will be developed we know that sigma divided by epsilon should be equal to Young's modulus the sigma should be equal to Young's modulus into strain so y alpha delta t will be stress and if a def procession is a force will be equal to sigma into a fine so you should be very comfortable with playing around all these things but in now since we are you know looking at for the first time probably right now we can just focus on the NCIT numerical in this chapter you will see that many times you know exactly how to solve a numerical but still you will not be getting the answers probably because you are not very careful with the calculation okay make sure that you are good at calculation and do not use calculator because if you are a habit of using calculator you know the problem will be that the difficult chapters probably some difficult question come will not be able to probably you know some hard question not able to solve it and these question it will be pretty that you know exactly how to solve it still you are not able to solve okay so be good at calculation fine so here is the first numerical you know how how the bullet cards wheel is made you have been to village you are like cosmopolitan you never seen a bullet card television okay have you seen the wheel how it is there's an iron ring on it if iron ring is not there to just wear out right so that iron ring is placed very tightly on the wooden wheel otherwise you come out have you ever wondered how it is done exactly you you heat that iron ring expands then you put it on the wooden wheel and then let it cool down so it will contract and hold that wheel okay so the question is on that only write down a blacksmith fixes you don't have to write the entire version okay just write down the data which is required to solve the question a blacksmith fixes iron ring on the ring of the wooden wheel of a bullet card the diameter of the rim the rim diameter is 5.243 meters okay diameter of iron ring is 5.231 diameter of iron ring is 5.231 meters okay both these dimensions are given at 27 degrees Celsius you need to find out what temperature should the ring be heated so that it fits on the wooden rim okay the coefficient of linear expansion for iron alpha is given as 1.2 into 10 raise to power minus 5 what do you think the unit should be look at delta l is equal to l alpha delta t so alpha is delta l by l into 1 by delta t meter meter cancels or degree Celsius this is called degree Celsius get the value of temperature up to which the iron ring should be heated so that it gets fitted into the wooden ring all you do is heat from 27 to 33 and expand something wrong how much should be change in diameter difference right delta t should be around 46 that's why i'm saying you know exactly how still not how much should be just this minus that which is 0.012 meters right this should be equal to diameter of the iron ring into alpha to delta t okay so delta t will be what 0.012 divided by diameter of iron ring that is 5.231 into alpha which is 1.2 into 10 raise to power minus 5 this is delta t all of you getting this as delta t oh so calculation this goes in the numerator and it becomes 1200 divided by 5.23 and just simplifying it this this is delta t how much this comes out to be no idea see and let me do the calculation in front of you 6 4 10 3 2 5 so it is 6.27 okay so this is 1200 divided by 6.27 right so i'll take it as 6.3 just a simplification so whatever i get the answer will be slightly more than that so this if it would have been 6 it would have been 200 but this is 6.3 okay so this you can roughly say around 180 and you can check whether it is reading it okay delta t is this the final temperature is what plus this 180 plus 227 and if you want to get it exactly just divide this you can divide 63 from this okay like that divide you get the answer 200 12 with approximation i'm getting 207 let's do a numerical on this thermal stress also write down a rod of length 2 meter is at a temperature of 20 degrees Celsius so a rod of length 2 meter is at a temperature of 20 degrees Celsius okay you need to find out the free expansion of the rod free expansion as in you are not restricting its expansion find the free expansion of the rod if the temperature is increased to 50 degrees Celsius so from t1 the temperature goes to 50 degrees Celsius you need to find the free expansion this is the part 1 okay the value of alpha is 15 into 10 is power minus 6 please solve this then i'll tell you just find the expansion how much it expand delta l 0.9 mm 9 into 20 minus 4 right that's correct so delta l is l alpha delta t okay now you need to find out the part if you prevent it to expand you're not allowing it to expand okay you need to find what is a stress that is developed how you know young's modulus is not given you got the strain stress young's modulus is given as 2 into 10 is power 11 Newton per meter find out what is the stress if you're preventing it completely same as what we have done earlier so strain should be equal to alpha delta t okay the stress should be what y alpha delta t in and out no c part it is permitted to expand only by 0.4 mm if you're allowing it to expand only 0.4 mm now what is the stress okay so free expansion is what 0.9 mm you're allowing it to expand 0.4 mm how much is the restriction 0.5 so strain corresponding to 0.5 mm will be there getting it when you're completely restricting it strain corresponding to 0.9 mm was there when you are allowing it to expand 0.4 0.5 is a restriction any doubts so strain should be equal to what 0.5 mm which is 0.5 to 10 iso minus 3 divided by 2 meters this is strain okay this is strain this into young's modulus is the stress any doubts okay so that's how we deal with the expansion property of the substances okay so still there are some varieties of numericals that you have to do it as homework and ask me doubts in the next class where to find the strain stress stress strain is this okay so we'll take next property after the break