Preparation session. Today we're going to be discussing measurements. So we're going to look at more questions relating to measurements, where we calculate the perimeter of a shape, the area, and the volume, and also how to convert from one unit to the other. So without wasting any time, we can start. So with measurements, you need to know the formulas, right? Because they are standard formulas, especially to calculate regular shapes. So for example, how we calculate the perimeter, which means the outside of a shape. We just add the sides, with an exception to a circle where we need to find the circumference or the radius of this, where we need to define the diameter or the radius of a circle in order to use it in the calculation. The other thing when you calculate the perimeter or any other length of a circle, always use the pi function of your calculator. Do not use 3,14, but use the pi function. There is a pi function on all the calculators. You need to look for it on your specific calculator, so that you can identify which function to use. Also, calculating an area, an area of a shape, it is the inside from all to all of that shape. It's the inside of the shape. However, you need to use the formula. So there is no way that you can just add all the values that they have given you. You need to apply the formula. So you need to know the formula to calculate the square, the formula to calculate the cycle, the triangle, and a rectangle. The other thing you also need to remember as well is calculating the volume. And please always remember to keep your mic muted all the time when you join the session. To calculate the volume, so the volume is what can fit inside the shape or whatever the three-dimensional figure of a circle, a rectangle, or a square. For example, a box or a bottle, which is, we call them a cylinder or a cube or a pyramid. What can we fit into that? That is the volume. So also you need to be able to calculate the volume of any shape, and especially with the volume to always remember what the formula should be. With the volume is the area times the height, because we just elevate that area with a height. So always remember that if you're not multiplying with a height, you're not calculating the volume. And you also need to remember the units for each and every type of measurement that you are calculating. That if it is a perimeter, it is at that same unit level. So if it's meters, it will just be the answer will be in meters. If it's an area because it's multiplication, therefore it will be in square. So it will be meter squared, millimeter square, centimeter squared. If it's a volume, because there are three things that you are multiplying together, the area times also the height, which then means for the volume, the answer will be in centimeters cubed, meter cubed. But also because it's a volume and the volume is always calculated in liters, because we calculate the thing that goes inside how much of that can go in. It's the volume, it's always calculated either in liters or milliliters or kiloliters. So you need to be able to also be able to change from cubic to liters as well. And also if they give you the the length or the the the sides length and the width with different measurement units, you need to be able to know how to move from one unit to the other unit. So those are the things that we need to be aware of today as we go along and do the the measurements. So let's look at a couple of questions so that we can refresh our mind in terms of the type of questions that you get in the exam. So we um maybe I should share this again this week with everyone. I'm gonna check if I'm able to share in the chat function, in the file chat function. You will let me know if you are able to see the papers right, especially for those who just joined today, so that everyone can have the same documents that we're going to use. You just see and put in the course of the session, you can download them and you can also use them for preparation on your own as well. Okay so let's look at possible questions and sometimes some of the questions are not as clear as possible, the way we would like them to be, but let's see if we can make sense of what we are looking at and because it's not clear which area is shaded because here they're asking I'm going to abandon this question because the first question that we have now it's not clear. Consider the figure below the semi-cycle with the center M and the diameter of 18 centimeter perfectly into the rectangle as indicated in the sketch. We can see the sketch and calculate the perimeter of the shaded figure, so which one is the shaded figure? Is it just this area that is shaded? I can make sense of that, but let's assume that the shaded area is this part of a rectangle which is cut off by a triangle that is eating up out of the area and I'm going to assume that the semi-cycle is also not shaded. I'm not sure if the semi-cycle is shaded, it's very difficult going to be challenging. It's going to be very challenging because I don't know whether the semi-cycle is also shaded. Okay let's assume that the shaded area is the whole semi-cycle plus the triangle. Let's we're going to assume that if we need to calculate the the perimeter of the shaded area which is the semi-cycle plus this rectangle shape which is cut off out by the rectangle by the triangle. We then can consider this as a what we call a composite shape so then it has the half a cycle and it has a rectangle shape. So what we can do because what we know is the perimeter means eating up all the sites and because for the site of a the site of a cycle we use the formula and to calculate the perimeter of a cycle we need to know what the formula is. The circumference of a cycle is calculated by 2 pi r but now because this is a half a cycle therefore it means we need to divide that cycle by half and we need to know what our radius is. We were told what the diameter is and our diameter is two so therefore diameter is made up of two radiuses so it means in order for us to find the radius we're going to say the radius we divide the diameter by two so our diameter is 18 divided by two will give us half of that and which our radius is so then we can come to our half a cycle calculation which will be pi r therefore we just need to calculate pi times nine and that will give us remember to use your pi function from your calculator and let me know if you don't know what your pi function looks like a way to find it so if you're using a case here you'll press second function and you will press the button that relates to the pi next it's the times 10 to the power raised to the power of x button next to the answer and multiply that with nine and that gives us 28 28.27 so we know what this circumference is now we need to add this side and that side and this side and this side and this side they didn't give us because this side since they didn't give it to us we can assume that is the same as that side it's 12 centimeters so this is 12 centimeter so we go into it this circumference of this whole figure which is the shaded part will be 28.27 you can start adding up plus 12 plus 7.5 plus 14.8 plus 12 and it will the units will be in centimeters because all of them are centimeters so what do you get it might be correct it might not be correct because I'm making assumptions on a picture that is not clear so let's see if we are right plus 12 plus 7.5 plus 14.8 74.57 which is option number two that is how you will find the circumference of a shaded area any questions and remember circumference is adding all the sites and since we are adding the sites you can ignore the triangle that is eating up the rectangle within the volume that creates the volume of a rectangle sorry ma'am hello it's Michelle ma'am yes can you please explain again how to calculate the circumference on the calculator because I didn't get that we explained what using the like the steps on using the the pi symbol so I know how to get it like the shift and then obviously we to get the pi symbol but like you explained like how to use your calculator to calculate the pi using the pi function on your calculators so depending on which calculator you have I'm going to show you on the case you and then I'll show you on the sharp calculator especially for those who have a financial calculator so to calculate the pi r because it's just pi multiplied by the value of r which is nine as I wrote it this way I can put it the same way on the calculator but instead of using bracket I will have to use multiply I'm going to stay away from using brackets so shift pi they use my pi and I'm going to say multiply by nine or not 99 nine which is equals two as you can see straightforward you just need to use the multiplication I'm not sure if I use the bracket it will work so let's see shift pi open bracket nine close bracket equal and it should work because it's brackets are multiplication on your case your calculate or sharp calculators you can also do the same just need to put back my calculator to normal mode so on your financial calculator go and find your pi function it should be somewhere on the I think on the dot those who have a financial calculator on my calculator it is a pi function it's written already on the button so I just press that pi function and then press the bracket then put nine and close bracket then that's that's it that's how you will find the part you use that pi function always thank you lazy thank you so let's go to the next question just want to see they have more questions see yes they do refer to this sketch in 16 that is the sketch that we just used and now they want us to calculate the area of a shaded region now with the area it's going to work differently so now we know that for a parameter we were adding the sides now we are not adding the sides anymore but we have I just want to I just want to redraw this just give me a second so that I can write all these values okay so let's go down here I'm gonna redraw it here so that is a bit clean it might not be 100% the way you saw it there but I will try this thing there to take that and you have and this is six centimeter I'm just gonna write the six you will understand what is that this was 14.8 this is 12 this is 12 and this is 7.5 and what do we have here was the the midpoint something there in the circle at the top and this was 18 okay so right now we know that the shaded area is this part right this is our shaded area so because it's the area we need to treat this as true composite shape so we go into first talk about three things here so there's not only two shapes there are three shapes but they are made up of four things that we need to take into consideration when we we do all this so now there is the cycle a half a cycle there is a rectangle I'm going to treat the whole thing as a rectangle and there is a triangle with a right angle with a height so we can assume that this is the base and this is the height so we have a triangle so we need to calculate the area of each and every one of them and we need to add the area so we'll have to add the area of a cycle and a rectangle but we also need to subtract the area of a triangle because it's heating up some portion of the rectangle so we can subtract the area of a triangle from the rectangle and add them together so since I've already laid it out there let's do the calculation what is so you the other thing you need to always remember is what are the formulas to calculate each one of them so to calculate the area of a cycle the area of a cycle it's pi r squared but because we dealing with a half we need to take it a half of it right so this will be pi r squared divided by two plus into brackets we also need the area of a this is the area of a cycle the area of a rectangle is length times breadth so you also need to remember length times breadth so I can also use this as my breath but that's the thing so I'm going to have to do length times breadth of the rectangle but I need to also subtract the area of a triangle which is half base times height our base for this triangle is 18 because it's the same length as the diameter so we need to take that into consideration and our height of this triangle they gave us because they gave us a 90 degree angle somewhere there there is a 90 degree angle which is not clearly visible a lot but it is there which tells you the height so which makes things easier so we know what the height of this is six so we need to use half base times height okay now we can substitute the values and start calculating you can also do this separate one by one by calculating the area of a cycle calculate the area of a rectangle and then come in and calculate the substituting through the formula so our area of a cycle of a half a cycle we know that our radius we did calculate it previously is nine so you will say pi times nine squared divided by two plus open bracket open bracket our length doesn't matter which value I use times 18 minus half times breadth will be 18 times height will be six and that's how you go through there let's calculate so what is pi to the power of no sorry pi times nine to the power of two divided by by two will be nine times nine is 18 18 times pi divided by two so let's do the calculation you tell me the values since you are able to calculate the pi as well what is the answer for pi times nine to the power of two divided by two are you guys still here are you guys calculating are you waiting for me to give you the answer nobody I don't know if it's right but mine is I get 127.23 127 point and remember while we still in the problem mode we need to write all the values so right okay so it's two three two three four five and this is zero and it's zero two five zero two five oh I thought you were using a right I'll calculate okay so now let's go into the bracket 12 times 18 216 216 minus half which is 0.5 times 18 times 6 54 is 54 what must rule says like what is inside the bracket well you can use your calculate to do everything so inside the bracket it's 162 sorry ma'am I'm writing also and what is the answer 127 223 4 5 0 25 it's it's equal to 289.2 so it's number option three 289.2 centimeters which is option three and that's how you will calculate we all wish I did let's look at more questions unless if you have any question look at question 18 question 18 says there is a square fish pond oh a square fish pond has a solid fixed circular pillar or an island at the center so it's displayed on this figure the measurement of the pond are indicated on the diagram so the point the pond is a rectangular figure because the sides are not equal if it was a cube or a square it would have been oh no sorry it is a square the height so because they told us that it's a square so therefore it means all the sides are 4444 and the heights will be 222 222 okay so they did tell us that and they also told us that there is a circular thing which makes it easy and they say the circular island has a diameter of one meter so it means from this end of the circular to that end it is one meter so we can calculate the radius because the radius will be half of that diameter the question is determine the volume of water that would fill the pond to its brim and give the answer to your nearest liters now because you need to calculate the volume of the brim inside the brim there is a circular thing inside a there is a solid pillar there is a a pillar inside of this box we need to also take that into consideration when we calculate the volume because there the volume of the square where that circular is needs to be subtracted because it will not be included in the amount of water that will be filled in because it's just an island right so the same we take this as a composite problem because there are two figures that we need to take into consideration since we're talking about volume so you need to remember this is a cube what is the volume the formula to calculate the volume of a cube length times breadth which is the area times you need to always use the height because it's a it's a cube or we could have just said it's a cube or something like that but because it's not a cube because they gave us the height of this is and this side and this side times that side that's what we need to be calculating when we calculate using the volume as well so our length is four times four times our height of two but we have calculated the volume we calculate the volume of the the fish pot which is how much thirty two thirty two and I must put the meter cube you must always remember that because it you must put the the units as well so now let's calculate the volume of this circular thing the volume that we need to subtract from the volume of this circular of the point so because this is a cycle it will be pi r squared times height because it's a cycle plus the height and the height of this it's similar to the height of the the box so what is our radius so we can calculate r is diameter over two and we know that one over two it's our diameter because our diameter sorry one is our diameter so there will be pi times half square times the height of two calculate what do you get I get one point five seven oh seven nine oh seven nine six seven yes I get the same seven seven oh meter cubed meter cubed we need to take 32 minus we need to subtract those two together to get the volume the actual volume of this fish pot of the water that will go into the fish pot to the brain so 82 minus so you need to subtract and the answer is thirty point two nine two zero three six seven three six seven we're not done because this is in meter cubed so you need to make sure that you know how to convert from meter cube to liters so how many liters are in meter cube how many meter cubes are in liters cubic meters two meters so there is one one cubic meter in 1000 so it means we need to multiply by a thousand one oh not cubic or one cubic meters it's 1000 liters am I right what is cubic meters how do we write cubic meters is it meter cubed yes ma'am I'm looking in our in our study guide now ma'am that says here obviously they have like centimeters now here but they're saying that one liter for example one liter is equal to 10 centimeters times 10 centimeters times 10 centimeters it gives you a hundred cubic centimeters 1000 cubic centimeters that's for the centimeters I don't see one for cubic meters oh yeah ma'am they say it's you all right ma'am it's right so it's the first one because it says here in our textbook one cubic meter is equal to 1000 liters yes that I know so if we know that right if we know that one meter cube or cubic meter is equals to 1000 liters therefore it means 30 cubic meters will be equals to 30 times 1000 yes just multiply this by a thousand and the answer will be option one option one it will be 30,429 liters if we get it to the nearest liters because it will be point two zero yeah so 30,429 okay so that's measurements let's see if we have more questions on this paper should be it so let's move to the next question paper measurements yeah it is our winner consider the diagram below the measurements are indicated on the diagram and this looks like a triangle with a triangle within calculate the shaded area in the diagram what oh so you need to calculate the area of a shaded area in the diagram because the answer I in meter millimeter squared so which is a little bit it makes it a little bit tricky right because this triangle is a triangle within another triangle any ideas let's think through this ma'am when when I look at this right the first thing I think is that we should convert the the centimeters either yeah the centimeters to millimeters because all the other the other measurements are in millimeters so that is the first thing that I think so let's convert millimeters or centimeters to millimeters so we can start there yes let's start there how many centimeters are in a millimeter so there is one centimeter one centimeter in ten meters right so if we know that one centimeter mix up ten millimeters two times by ten then you multiply by ten so four one comma can I just say something centimeter will be the same as yes um you're working out centimeters but I think it's actually a mistake I think the 41 should be millimeters no it's not a mistake you need to convert like at the other line at the top it's 14 millimeters yes it doesn't matter the line outside what it's whether it's in in millimeters or centimeters what they want you to do is convert it's not a mistake they want you to convert they want you to use the conversion there will be 412 millimeters yes and the 4.4 will be 44 44 millimeters so but that does not solve your problem because we need to calculate the area of that shaded area which um if I think of it to this way this is a right angle so I can take any of the sides maybe the 40 millimeter is my height and this is the base then if that is the base and this is my height 10 plus 10 plus 10 is 30 millimeter that's my base and my height or 40 is my height it doesn't really matter which one is your base or which one is your height because at the end it will be half times 40 times 30 but that gives you the area of the entire rectangle or triangle and because the shaded area they didn't give us that which one is your height which one is the base there is no right angle to the excuse me ma'am um the formula that we're using for this is it half base times height um because the the area of a triangle it's half base times height if you have a right angle triangle right okay now the shaded part is not a right angle triangle unless if we're going to assume that if the bigger triangle is a right angle triangle then the inner triangle will automatically become right angle triangle then we can apply half base times right if we use that same assumption and if we say it's a right angle triangle there therefore then it means this is our base this is our height if we put the right angle triangle this side this will be our base this will be our height that's how i'm going to assume this whole thing so what is going to be our formula we're going to use here that's the thing i want us to think through because in your module they never gave you any other yes the formula they only have provided you with with that so this is more about maths thinking outside of your basic numeracy scope because if i don't have uh the height there are there are other formulas that we can use because if our like let's assume that our triangle it's a scaling triangle then we can use the square root of your side times your side minus your because if i have three sides of a triangle right a b c it will be a you will calculate the perimeter of that and the perimeter will always be the area of all the sides that you have right so you will say a plus b plus c but now because we will be calculating lezzie i have a question the 41.2 don't you think it should have been millimeters already it looks like it's millimeters they on top and i think the first lady also asked the question because when you do that it comes to the answer number two what number two yes option two yes it's not yes i think it shouldn't be centimeters no no no wait wait wait wait what answer number two what are you calculating to get to answer number two if you take the half base times height like say for example no if you say which half times base times height like say for example you only calculate the shaded area right and you say half um 10 right times by the height and you take 41.2 then you get 206 no okay you can't because this is not a right angle triangle so alternatively in order for us to calculate this shape because this triangle is almost it's not a right angle triangle in a way so because they didn't tell us that if they would have given us a right angle yeah we can use that so now i'm gonna give you another option to do let's calculate the circumference or the yes the perimeter of this which is we need to convert this because it's centimeters is not millimeters it regardless of whether you think that it is millimeters or centimeters at this point but they gave you the unit there as centimeters unless if this is millimeter in in disguise but you can see that there is a centimeter so 412 plus 4.4 to 4 plus 10 and we're going to find that it is how much 466 5 466 466 yes okay who just want to do this because i think this might be when we calculate this 466 right we need to take half of this so that we calculate a semi because it's not a so divide this by two and that gives you 466 divide by two will give you 233 233 we're still working with millimeters the next step is to calculate the area of this triangle which is large we use the formula your semi perimeter of this and you're going to say your times your semi perimeter minus one of the sides so we're going to start first with there you can start with with yeah it's going to be a negative but anyway let's see uh let's not use that and let's use 44 times what do we get a square root of your 2.33 and i'll take the square root first 2.33 not 2.33 it's 283 times minus 44 bracket 233 it's not but it's not gonna work no no no it's not gonna work it's not gonna work it's not gonna work so sorry does you have a question the other thing that i would think is that obviously i can be wrong but i would think that like you in the beginning when you started like don't we calculate the entire triangle right and then calculate the shaded area and then subtract the shaded area from the entire triangle yes yes but the challenge is with you and you're with the shaded area triangle we need to find the formula to use to calculate that shaded area because it is not a right angle triangle do you understand we know that this the big triangle it's a right angle triangle so that one it's easy because if it's a right angle triangle then one line the the other line of the way it makes a right angle we will take it as the height whether it's the 40 or the 30 will be one of them will be the height which makes it easy to calculate because then it will be your height your half times the height times the breadth which will give you 600 right for the entire shaded area oh sorry the big triangle the big t will give you half times 40 times 30 which is equals to 600 the inner one one not given to us as a right angle triangle we can unless if we say if we take it from here today if we only include this we can find a right angle triangle okay now based on how I just highlighted it right now right so if we take it not including this part of the triangle but only this small portion right will give us because then we will need it to subtract those two other triangles because we can calculate them separately because they divided it in this way to help us know what is this length and what is this length and what is this length so which makes it easy so instead of calculating the entire big triangle let's calculate this half of it so we're going to use half 40 times 20 what do we get 400 we get 400 now we can also subtract this part because if we subtract this part then it means we are left with only that shaded part right so let's subtract that part with that part is half 40 times 10 which is how much 200 which is 200 200 subtract 200 or 400 subtract 200 is 200 eta squared which is option one which is option one and because it's a composite triangle made up of small little triangle within but all those other small triangles are not the right angle triangle but they also gave us which one is the right angle triangle so even if we did it the other way right if we said like um even if we said like the 40 times 30 like for example like you have it on the screen right even if we did it like that we'd still get to 200 right which way or or rather my question would rather be like when we see something like like as an example says like a different a different shape for example right how how would we like like you like you the the the light went on now for you right you go on how to calculate how would we know how to identify that we should do it this way might you have to uh write different limitations to get it right right you can't just rely on the first instant because the other thing if we use the big triangle it means we needed to subtract this side and that side but we cannot subtract this side because it's not a right angle triangle this one we are not told if they could have put here a right angle triangle then it would have made it easier for us to identify that this is our height or breadth and this is our height but they didn't give us that so we cannot use that information so based on the information that we know we can do other things so because they also split this to tell us this from this distance to that distance from this distance to that this distance and from year to year if they could have given us the whole entire line and say this is 30 centimeter we couldn't we can't get this area by not knowing how much that area is and that area is so we need to also think about all those other things that they give you to help you okay so so so Lucy in an instance like this you would say that the key thing is to look first of all if there's a right angle triangle which helps which would obviously help us to see way to what areas to calculate in order to get to our answer yes for a triangle it's always useful to have a right angle triangle to calculate the area especially for the triangle because you need the height if you don't know the height the right angle creates you that height gives you the opportunity to have that height whereas if it doesn't have the height they need to tell you what the height is of that triangle you cannot assume that this line is your height yeah so they will give it they will either give you the height of the triangle or they will give it to you in a right angle and you will know that one of the sides of the right angle creates the height of your triangle oh sorry you still learn the same question paper so let's look at more complex questions and here is another question so at least this one is clear it's shaded so consider the diagram below the measurements are indicated on the diagram the semicircle fits perfectly on the shorter side of this figure so now as you can see here they also did a because then it means this side as well it's a right angle because on a tangent of a right angle where it creates a right angle it creates a 90 degree angle on that tangent and then even if the square is on this side we must also know that this side is also a square because at the the angle at this point is 180 because it's 90 plus 90 it creates a and these are just simple maths things that's why I'm saying some of these questions they give it to you thinking that you know about maths but this should be a basic numeracy question okay so this is also we can consider this a composite figure because there are different figures within the figure there is a this figure if I draw a line here that join this side to the side we can say almost mostly this is like a rectangle possibly but not not really but we can draw that and hence you have this dotted line that moves from this point to that half of our cutting through the length of the other's opposite side right so this line the dotted line is to help you know what this great diameter is so the diameter of this circle it's four centimeters four centimeters based on that line is not based on this skewed line that creates this triangle here it's based on the dotted line so you need to also think of that okay so you are given a semi-cycle that eats up part of this shape and you are given a almost similar rectangle circular shape but not through that which has the sides at it because then they tell you that from this side to this side the six centimeters the same from that side to that side the only difference is the side it's extended by two centimeter hence this line creates a additional point eight centimeter from the normal four centimeter line of the square or the rectangle that they have here so they want you to calculate the perimeter of this shaded area so it means you need to calculate adding this side this side this side and that side remembering that this is a half a cycle or a half a cycle circumference is two pi r divided by two because two pi r it's a circumference of a cycle divide that by two because you need to have that cycle it's not a full cycle it's half a cycle so you will do the same as what we did previously so the perimeter which is the circumference will be two I can just say it's pi r because two and two will cancel out and you will be left with pi r plus six plus six plus two plus four point eight and that should give you the answer that you are looking for so you need to first calculate the radius because you are given the diameter of four centimeters radius is two so just go and substitute into this formula two times pi plus six plus six plus two plus four point eight is it right what is the answer the answer is 25.08 so if you add two times pi times oh plus six plus six plus two plus four point eight will give you that much okay so moving on to the next question consider the diagram with the measurement indicated on it below the rectangle with a diagonal of five centimeter so because of that diagonal of five centimeter you need to also look at it because it touches from this side of the cycle to the other side and what do we call that diameter so it means they gave you a diameter and this figure that we have what can we call this it is a composite figure how many figures do we have here two right you have a cycle and within a cycle there is a wide area which is cut off by the it's a rectangle because the side of this rectangle it's four centimeter if you don't see this is four I hope it's four centimeter and this is three centimeter so that side is three and this side is four centimeter so your question is calculate the area of the shaded part what do we need to do here do you know um don't we have to do the same like calculate the the circle and then the um right angle and minus thing from each other that's correct you need to subtract the area of a rectangle from the cycle the area of a cycle the area of a cycle remember it's a full cycle it's pi r squared minus the area of a rectangle it's length times breadth right substitute the values calculate let's see if you can do this without my help at least ma'am I get option two seven point six four option two precision option one okay so what is our radius two point five five times two point five squared minus our length times breadth four times three four times three and what is the answer that you got seven point six three four nine seven point six three four nine six three four nine yes and if we round it off at seven point six four huh is it six three nine four or six three four nine four nine then it cannot be if you round it off to be seven point six four it can ma'am because the nine makes a four or five and the five makes the three oh yeah ma'am we do basic numeracy hey gosh yes then the five becomes five and it's bigger than then it's four hey oh you guys are teaching me now things that I told you yo gosh some of these things that's good ma'am it shows that we're learning hi yep that's it does so at least it's good that I know that you guys are getting it and you understanding oh it makes my heart sink and it makes me wanna dance okay so here is another question so looking at this I hope I will be able to see what these values are they don't look visible clearly visible let's see if they have mentioned them on their thing refer to the sketch below this is a cylindrical still rod which is 30 millimeter long okay so that is the rod that is inside has a radius oh this is the rod the whole of it has a radius of six centimeter so I'm trying to see oh this is the six centimeter so this is the radius so they gave you the radius of six centimeters so this is six centimeter uh and has a square hole right through it so there is a hole right through from this side to the other side the hole is in the center of this rod so it's in the middle the sides of the square are five millimeters so this is five millimeters so this side and that side and this side and this side individually there are five millimeters oh this is millimeter not centimeter think of all this so the only visible number here actually was 30 millimeter so it was it makes it easy because this is the height you can just point it out to that way so the question is calculating the volume of the metal used for this rod with the square in a hole so we need to calculate the volume and we need to take away the volume of the square thing in the middle so what is the volume of a cylinder a cylinder is made up of a circle and the height right so think about the area of a circle and times that with the height so the area of a circle it's pi r squared times the height now we need to think about the cube that is inside which is the rod you can see that it goes right through so it's got the height as well so we need to subtract the volume of this cube thing as well so what is the volume of a of a cube also remember it's the surface of a square which means we we normally use l squared for the area of a of a square right but you can always you can it's up to you what you want to do l squared times height because we have the height of that so you need to subtract that from the volume of the rod so what is our radius it was easy i times our radius is six centimeter they told us you don't have to cut divide that by two it's not the diameter is the radius so six squared times thirty minus five squared times thirty oh we could have just said length times breadth times height which is five times five times thirty which is the same thing as five squared because they are two sides let me not confuse you calculate and get the answer and remember now the answer will be in millimeters and on this question they are they've got millimeters cubed and centimeter cubed and they didn't tell you how the answer should look but it means if it's not millimeter cube then it means we need to do some conversions yeah okay so as you calculated yes ma'am the answer is option three let's first find get the answer and then think about how do we convert from that one to the answer that you want so what is the answer um it's two two thousand six hundred and forty two two thousand six hundred and forty two two point nine two point nine two zero zero oh i'm just gonna keep it that's fine millimeter cube millimeter cubed and if you look at this it's not one it's not two so it means we need to do some conversion so how many millimeter cube are in centimeter cube one thousand so if there are one thousand therefore it means we're going to have to divide by a thousand because we're moving from a smaller value to a bigger value so we divide previously we used to multiply now we are dividing by a one thousand so two six four two divided by a thousand would give us two point six four centimeter cube one when when we were growing up and studying our learning methods in the olden days when we were using matchsticks and all that the other way that we were taught how to divide especially when it's thousand hundreds so if i have two thousand six hundred and forty two point nine two and i'm dividing by a thousand and they say thousand are three zero so it means i must move three spaces one two three spaces so it will be two point six four two and i still remember that and this i'm talking about long time ago in the 1980 something not to give my age anyway but yeah it was that long ago nowadays we have calculate this previously we used to use matchsticks and water caps and all those things all right enough with history lesson let me move on to the next question are there any questions no questions we have 30 minutes so let's open this last session paper we'll have nice questions and we get the most the most difficult question because it's got so many lines within it consider the sketch below a rectangle piece of a material with the sides of 150 centimeter and 100 centimeter is used to make a circular table cloth the diameter of the completed cloth of must be zero point nine meter and five centimeter okay so there are two diameters and i'm working with two different units so you have meters and you have centimeters is provided for the hem which is how much the material is wasted in terms of this so we need to calculate wastage okay so how do we go about on this we know that we have this rectangular material that is given so he is 150 and 100 centimeter so the length comes the breadth of this rectangle and we are told that the hem of this cloth has a diameter so it needs to be the cloth needs to be cut into a circular so into this rectangular cloth needs to be cut into a circular cloth to create a table cloth with the radius for the hem to be zero point nine meters and five centimeters so let's convert meters to centimeters yeah the diameter yes let's convert the meters to centimeters so how many what is zero point nine meters it's 90 centimeters it will be 90 centimeters so now we're working with the same units 90 centimeter so the diameter so i'm going to assume that the diameter from here to the outer side circle that those two this is 90 centimeter and the diameter for this one inside to that side is five centimeter like we need to calculate the wastage how much will be how much of the material will be wasted so let's calculate first the area of a rectangle which is length times breadth which is 150 times 100 and let's calculate the area of a circle now i'm tempted to use only one of them not two of them because the circle that will be completed because it's a circular thing so the whole circle diameter will be 90 centimeter but they need to create the hem which is the folding so it will just be of five centimeters and a little bit of it it's not it's not wasted then it means that five centimeters also included in the 90 if we calculate for the bigger circle the outside the outer one or is it something that they cut off i'm not sure let's see let's calculate the area of a cycle for the bigger cycle so which is high r squared all right so if we use the bigger diameter then which is the outer one then it will be pi times what is our radius 90 divided by 2 which is 45 square so let's what is 150 times 100 15 000 15 000 and on the other side 6361 6361 0.725 124 yeah that's right okay so this is in centimeters right so let's subtract that from that so that we can get the area of what are we calculating wait let's make sense of this before i make i make you subtract things from one another so oh we calculated the area of a wastage sorry my bad oh gosh wastage it will be 15 000 minus the hand the the tablecloth what is remaining what is it that is remaining 8 638.275 and this is in centimeters wait right let's go to our answer is in meter cube so we need to convert centimeter square to meter cube how many sorry centimeter square to meter square how many centimeters squared are in meters 10 000 10 000 yes man yeah so 10 000 it means there are four zeros so if i move four zeros one two three four it will be zero comma six eight three four yes man so the answer will be option one you just need to divide by 10 000 get to meter right so it takes 8000 638 divide by 10 000 it will give you zero comma six eight zero comma eight six four now here is another question there is our right angle given make it easy right because we need to know whether the triangle is the right angle or not if they didn't give us the right angle but they would have given you this line then you would assume that that is your height it gives you the height of this triangle but now it's a right angle triangle so and we are told what the height is is the 10 centimeters so let's read a semicircle is drawn inside the triangle as indicated on the sketch the side length of the triangle are 30 40 and 50 centimeters respectively calculate the perimeter of the shaded part composite as well you have the bigger part is the triangle but remember as well you are calculating what are we calculating the parameter of this because the parameter is adding up all the sides right but now what they didn't give you is this portion and that portion which makes it difficult to calculate if we just adding all the sides so in order for us not to worry too much we can calculate the circumference of a cycle or let's first calculate the circumference of a triangle which is the whole thing we calculate it regardless of whether there is a missing piece we're going to subtract the circumference of a semicircle we call it a half a cycle so let's do that so we know that the circumference of a triangle is just adding all the sides so you just add all the sides 50 plus 30 plus 40 minus pi r because it's half a cycle we know that circumference of a cycle is 2 pi r you always need to remember that night so because it's half 2 and 2 will cancel you will have to tie up now what you do not know you still have a problem right we still have a problem a huge problem actually because one we are not being told what is the distance between this and that unless if you guys have some what you call those you are able to read the numbers here ma'am there is numbers I think you just have to zoom in maybe I think like there is a number there at the bottom yes it's five I think it's five it's five centimeters yes ma'am and this side five as well five yes ma'am yeah so if we know those then we didn't even have to do this because we could have 50 plus 40 plus five plus five plus half a cycle we can do that but anyway we can do that as another as another activity you can check double check so if we know that this is five and this is five therefore this is 10 right therefore the remainder yeah will be five plus five is 10 then it means this is 20 so if our diameter is 20 then 20 divided by 2 is 10 so we're going to have minus pi times 10 50 plus 30 is 80 plus 40 it's 120 minus 120 minus okay which is equals to 80 and it's none of those and they set the perimeter of this she's 88 88.58 centimeters and it's none of them probably maybe this answers they calculated them wrong but let's double check something so we can also calculate it by saying 50 plus 40 plus five plus five plus pi r right which is 90 100 100 plus five times 20 20 is 10 so it all probably has to yes mammy's option too if we do it that way which is 100 131.415 yes so we need to do it this way 131.442 hence this why this one will not work it's because we take in this but we were subtracting this area we also need to subtract the 20 again let's see because I'm adding it twice in this I didn't take it out I didn't take out this 20 that I added so it should be the same now 120 minus so you need to also take away the additional 20 minus no it's not right 10 it's still not working minus 20 no okay so not that way you will not do it that way because it's the circumference we just add all the other areas it only works when we do the the area it's not going to work with the circumference we have six minutes let's see if we do have another question we do calculate the volume of a container shown in the figure below it is a rectangular box with a keft lid that is a cylinder sliced down in the middle okay so this is also a composite figure because we've got a box and the cylinder half a cylinder as well so we are asked to calculate the volume of this container so we need to take into consideration two things because we'll have to add them so we need to calculate the volume of the rectangle or this rectangle piece which will be the length times the breadth times the height which is all the values you see and you need the area the volume of the cylinder piece but it is half of it it's not the full of it so but the volume of a cylinder is because it's a circle it will be i r squared times the height and I'm going to divide that by two because it's split into two it's half half of it is not a full cylinder so let's substitute the value so first we need to find the radius the radius because this length from here to here it's our diameter which is eight it's the same as that length so it will be eight so our radius will be four so let's substitute the values ten times eight times six plus five times four squared times the height of this cylinder will be the tenth it will be the same as the breadth of this so it will be ten the height of this cylinder thing this cylinder thing this height will be the same as that which is ten divide a thing by two the radius is the same as the length of this because the radius is on the square thing on this so it will be the same and the height of the cylinder will be the same as the breadth of the sorry man the box the height of this cylinder will be the same as the length from here the ten this long length will be similar to the height of because the cylinder so we have to like turn it basically yes because if you cannot say this is the height of the cylinder it's not the height of the cylinder the cylinder is faced down it's if it was like that you will say the height is this but now your cylinder is like this i get option two let's get all the values yes option two uh okay so ten times eight times six is four hundred and eighty and i times four squared times ten divided by two is two hundred and fifty one points three three two five four eight five three two oh three sorry three adding them together will give you seven so you can also look for other questions let's see in this other paper i've shared them in the chat right you did get them let's see on this one if it's difficult no it also there are so many numbers i cannot make up some of the numbers that they wrote here but i think there is a 12 so this is also a composite because it's a figure that has a rectangle and a semicircle at the top and also a triangle now the challenge with this is because they didn't tell you that this triangle is the right angle but it doesn't matter because you need to be calculated the parameter the parameter of the figure so it means eddie hi lozi are you there i think she lost connection okay he's there um last week's um session can i get it some way um i think she's gone guys yeah the recordings are um supposed to be on myonisa and our module i think um there is a section where they have all the recordings of the previous classes so i think maybe just check there um we can actually check now while we on the course would be the saturday recordings are not there i already checked and additional resources there's nothing it's mainly the weekly sessions it was on thursdays that is there but the saturday classes are not there um why did you cover last saturday which modules um last saturday i'll tell you now we did um let's check here so we did like simplifying um yeah we did like expressions and equations like simplifying and then like making extra subject that kind of stuff we did last week which is also really nice um it was a really nice session as well there's also a lady here right i don't know if she's on the call now but it doesn't look like she's here but she's also like like an edmund kind of thing because i thought we could just maybe ask her about the the recordings but otherwise the other option is to um maybe email the lecturer um for this module and then ask him like where can we find these recordings yeah because she said it is recorded so it may have to be some way look on teams no um i i checked there there's nothing there so especially for me i can't see anything that's of the first semester for you or the second semester for you guys as i was part of the first semester team and they don't give me the notifications or when the classes is i had like a average 85 percent i just the day before we had our final exam i had a C-section so i didn't write the exam on the day seeing that i was in hospital yes sorry guys i got cut off i don't know why hello leslie hello i'm yeah leslie we was on the Saturday sessions where can we find it because i didn't attend class yesterday all the sessions are on on the unisa platform uh let me get the link and somebody shared in the chat so now one of the classes so you can get that one from the last week's class yes but but but leslie i think maybe like the link could be very helpful as well for those who are maybe not on here and maybe might ask us in the groups they became you shared with him uh yes i will share that um the way you can get all the sessions not just the schedule i don't want the schedule only the schedule i want all the sessions let me see the recordings i want the recordings not the schedule sorry just give me a second i need to get the right link okay so all the sessions including also the sessions that we used to have during the week you can catch them up under the numeracy center um on the so they are all recorded so let me post it in there oh someone someone just posted the the link to the recording not the not the that's for last week's recording yeah in the chat they um thank you leslie share the link to all the other recordings including the recordings that we have been doing for the past the past months um on mondays so you can go there let me just stop sharing that for a minute and share what i'm referring to so that when you get to that platform you know what exactly you need to be looking out for when you click on that link it will take you to a thing that looks like this depending whether you're clicking on it on the phone or on on your laptop so there are tutorial classes that have been happening some of the recordings have been recorded since last semester so you can also look at other modules that you have that have classes and there are also the writing center where they do the english and um all that and then we have the numeracy center where i am so under the basic numeracy there's all the sessions that we had on on every monday for the past three months or so or four months and then the exam preparation you will find all the recordings for the exams you will find them there if you're doing qmi you will find them under that but those are the two places where you will find the recording so if you click on the basic numeracy one it will take you to this will you will be able to see all the recorded sessions that we had on a weekly basis that you know is take you know to respond on a weekly weekly basis until i think the 19 of september that was the last recording and if you want the notes relating to that you can just click on open notes and it should take you to where the notes are in order for you to follow through what we have been doing in those sessions as well so the notes are also uploaded but uploaded in here we do have we had sessions in the first semester but there are no recordings for that if you want the recordings for that you can follow my youtube channel but you will have to see through a whole lot of videos to get to um but it's easy you just search under my youtube channel you search for your right um you can type there be a new they will all all the recordings for being you will pop up and you can follow those um yeah so those are the things i was part of the semester one group and it was really nice i i'm actually just recapping yeah so because they didn't keep the semester one the recordings but i do have have them uh let's see so maybe probably under the some people said they can't access the the chat you can find your um the papers here i will post the other two i think i didn't include them so i'll also include the other two on here then you will have some of the exam papers anyway i need to get to go eat and get ready for my next session with the qmi students at 12 o'clock so i will stop it right here and stop the recording and i will see you next week as we do our last um session on financial uh financial meds so see you next week bye thanks thanks lizzie have an awesome weekend further everyone


Let's warm up with a problem much like the type we were looking at when we finished on Friday. So imagine a tubular member of some kind rigidly fixed on the back wall and inside of it with a much smaller diameter than even the inside diameter of the outer tube is yet another one. So I have two tubes like that, concentric or a tube with a concentric rod in it and there's some kind of rigid cap over those, so I'll draw that as if it's out here, but it's really bonded to the top of those so the cap itself is rigid so there will be no deformation in that. That's sort of always been one of the idealizations we've used and both of the structural member, both parts of that are intimately made and inlay adhered to that rigid cap that's on the outside and then there's some kind of torsion applied to that. So if you need a side view to help, we have this tubular member now cut open in cross-section, another material, remember down the inside and then that cap over the outside to which we've applied some kind of torsion. So if we use our right hand rule, we can also draw that like that then. All right, pretty simple setup, a little more details to it. Let's say that it's 500 millimeters in length, that the inside material is steel and the outside material is aluminum. Maybe there's room for coolant flow between the two or something, for some reason we've got a setup like that. Outside diameter is 76 meters, inside diameter of the rod on the inside is 50 and then the walls of the tube are 8 millimeters. See that well enough? Is that all right? Got the picture, a steel solid shaft inside a tube, all capped off of a rigid cap to which we apply some kind of torsion. All right, some of the material limits with which we must stay below, we have an allowable shear stress on the tube and we need to stay under both of those, 120 megapascals for the steel, 70 for the aluminum, other material properties, modulus of rigidity, 77 megapascals for the steel and 27 for the aluminum. No gigapascals, we struggle with that elastic, I know you'd rather I let you get halfway through before I make those corrections, but let me do them now, okay, I think that's all the pieces we've got. All right, so we need to determine the maximum applied torque, there's one torque that will drive one of the materials or the other to its limit in terms of the shear stress and that then would be the design of maximum on the applied torque. Got the picture, take a second to get the whole drawing down. This is one of those that's statically indeterminate because we certainly know that if we look just at the rigid cap that it's got some applied torque in one direction but because of its immaculate adhesion to the other two materials, they've got some amount of torsion that they're withstanding or applying to the cap and all we know is that total torsion has got equal some load taken by the two inner structures of this machine component of some kind. Static alone is not going to help us, this is statically indeterminate but that's okay, we're not worried because we've got extra tools this term that allows us to finish these problems that we couldn't have finished last fall. That's all the farther we could have gotten is to here but what we need to do is decide what of the stuff from this term is going to be helpful to us. You have to look at the problem to decide what of the tools from this term are going to be of use to us. That's there, not bad. And what are you writing? You're doing something, you want to share it? So two unknowns, how much torsion goes into each, actually this is three unknowns I guess since we don't know any one of those we just know that the applied torques to the individual members have got equal to total torque applied to the entire thing so what other things can we bring to our aid from this term? We're looking at deformation so it's going to have to be something of that. Any ideas? Are you mad at me? Is that what happened? Did I ruin your weekend? You're going to make me pay for it? Chris you're going to say something? Well yeah, but not really much change of length expected here. The angle of twist is going to be the same on the two however much one twist the other is going to twist just as much. That again is on the assumption that whatever this cap is rigid. So we can also use the fact that whatever angle of twist we see in one we'll see in the other. So we started to set that up then depends on a couple things. First the load experienced by that number and the overall length. What else? Which all those last ones independent of the load. G of course is a material. Property and L and GA are properties of the geometry of the problem as set up. So those two must be equal. Trouble is that we're still short because there's three unknowns. This is an independent equation doesn't introduce any more unknowns but we're still short one equation. The equation bank. Now this is one where since we have two materials and we know they're not going to be, well I hope we know, I hope we have a feel that even though they have limits on the stress they're not going to be at the same stress level at the same time and there's going to be one situation that dominates these two. So we have to assume one of them is at the limit and then use that to calculate what the total load is then check that for the other one and if it's over the limit then we know our assumption is wrong and we have to redo it the other way. So we can add that to it by bringing into it the stress limits just for a little bit of help here. I'll give you the two moments, polar moments so you don't have to bother with calculating those the deal is 2.03 x 10 to the minus 6.003 10 to the minus 6 inch millimeters to the fourth that'd be meters meters to the fourth and the aluminum is 0.064 All right, all this stuff here is constants except for the two torsional loads that we're looking at so those can all be calculated and you can get this down to just a simple relationship between the two it's just a proportionality constant between the two that just slimlines things a little bit so then what we need to do is assume one or the other is critical take that through the calculation until we find a load on the other one and just determine whether or not it has exceeded the limit so for example maybe assume the word at the limit on the aluminum assume that's the shear stress we're at the limit on the shear stress of the aluminum remember these are allowable limits so assume we're at the shear stress on the aluminum use that then to find the load on the aluminum so assume one or the other is critical you don't know which, you have to take a guess assume that we're at this critical limit on the aluminum 70 megapascals everything else is known so that we can then find the torsion expected on the aluminum that will make it critical we can then go back and then test that if we're at the limit then on the steel as well because once we find the aluminum torsion we can then find the steel torsion that goes with that we can then check to see if that's at the limit of the steel as well so there's the bare bones plan sir one thing, is it possible you've got the j-aluminum and j-steel switch it's possible j of the aluminum yeah why don't you check these things thanks David trouble was I changed my subscripts from my notes to the board so now they're all messed up alright everybody understand a bit of the game plan then solving this problem Joey make a little sense assume the aluminum stress is critical that's this one put that in here then we can solve for the torsion on the aluminum with that we can then find the torsion on the steel with that in the same way we can find out the shear stress on the steel then and see if we're over this limit if we're over this limit then our assumption is wrong and we have to redo this very same calculation with the opposite material being under the stress limit at the stress limit don't forget that C is the outer radius yeah for this part right here all these values in here are for the aluminum if you're looking we're using the value for the stress limit on the aluminum that works Travis go anywhere that's what you're finding okay now then that gives you a value in the stress and the steel but then you have to also determine then for the steel what the shear stress is in the steel and compare that to this and if we're over that then this assumption was wrong yeah the this assumption for the most for essential purposes is our second equation our third equation we're assuming a value for one of the shear stresses that serves as a second equation a third equation that's the torsion in the aluminum if the aluminum is at the stress limit what are the numbers I have yeah that's what I have yeah check your numbers John show if you're confirming mine well you have 36.9 hang on hang on where's what's 74 not 74, 70 just put it in an actualistic case yeah so you know I appreciate it you're mimicking my style at the board where did I get that from? from Joey it's just way way back there as far away as they can be there's too much flying and chalk dust up here to go close to it alright most people are getting something like 36.90 meters on that with the John and then you can use that to find what it is in the steel then once we've got the torsion in the steel you can find the shear stress on the steel and see if we're over the limit for the steel once you get a torsion in the steel well 4 yeah you've got a decimal point on the of the 76 find a torsion in the steel then you can in the same way find shear stress on the steel now that we've got that compare that number to the limit on the steel find that bar and yeah so the assumption is wrong it's a working bill alright now we've got an estimated load in the steel then the dimensions for the steel and that should give you then a shear stress based upon the assumption that we were already at the limit in the aluminum but if you wasn't wrong by that much we're all on traps and we can find other ones okay yeah and that's passed so good and that tells you the assumption was wrong huh well this would be the steel inside 2 so this number is 25 that would be yours and the j thanks today because we got corrected Bill you got something? not yet? first word you get well that's a little different oh that's that's for the aluminum yeah so now you have to find it for the steel using equation 2 which is just the torsion and the 2 related by a bunch of constants so you have to say this number use it to find this number which is the torsion and the steel then that's the one used in here because it's for the steel yeah did you use j steel to get the 3694 no that's aluminum 2 yeah 2.0 I used the 2.0 I used the right numbers in here and the c was the outside diameter which is half of the 76 so that comes to be about 131 John 131 3 give or take a little bit it doesn't really matter because that's greater than our our allow of 120 which indicates that the assumption is wrong because that's the torsional load in the aluminum you use equation 2 to find the torsional load in the steel and then that should be about this that you use then to find the shear stress on the steel and that we find is over the limit a sine limit of 120 where that comes from we're not saying that's one of the givens in the problem agree David with those numbers coming up over there then you have to assume now that the steel is critical redo essentially the same calculation you did there and we know that it will come below the shear stress for the aluminum because we assume that was critical the first one and went over so we have to come down getting those numbers down Chris that's the torsion in the aluminum we're going to use equation 2 we have the torsion in the aluminum all these others are just geometry things so now we can get this so once you have that which should be that number over there then you can figure out the shear stress in the aluminum you got this okay this conclusion Chris fill you up to it then it's a matter of going through the same numbers you should get then the torsion in the steel you have that Tom? yeah 2950 2.9 if you use kilonewtons kilometers or 2950 doesn't matter, wait as long as you're consistent then back into equation 2 to find the load in the aluminum and then determine the shear stress in the aluminum and we already know it will be below this because we reduced the entire load since we first found the steel was the limiting number of the torsion and the aluminum then is 3375 which we know is less because we have been at 3690 with her original assumption it's not quite the answer the answer was what's the total load and so it's a matter of summing all those up because that's what we've done now is found the division between the two and then adding it together to give us a total Chris you getting those numbers now a little better Joe okay getting those just we had to pick one, we picked the wrong one of course I did that by design so you can see how to recognize that your assumption is wrong so I'm going to go back and calculate the same steps we did here the same steps we did here you don't need to actually calculate the stress on the aluminum because we already know it's below that was these were the stresses at the aluminum limit and we're already below those so we don't need to calculate the stress on the aluminum that doesn't mean you don't want to know that number but we don't need to check it to see if we're alright, we already know we're alright I'm sorry 29 47 3371 oh yeah that's fun that's just round off because if you want three significant figures we agree just fine and don't forget we're not saying where these came from we don't know if there's a factor of safety on those yet we're just going through the calculations in the real design there's more details that need to be brought out yeah so you're right here you know that that's over the limit for the steel so now we go back and then assume this value with the limit value for the steel and essentially we do the population only each of the different pieces swap the roles so now that we know that the steel is the limit calculate the load and the steel using this same equation only for the steel so now we've got that as the limit on the steel j for the steel c for the steel we'll give you the load on the steel then in equation two again we've got the load on the steel we can find the load on the aluminum and we already know that's going to be below the stress limit for the aluminum because we are way below those not way below quite a bit below enough below that we know we're not at the limit of that aluminum Chris okay so you should give us a total load of 663 25 Travis you got that Joe you okay do we have to check that and like we did with the aluminum no because being at the aluminum limit pushes over the steel limit so we drop down on the steel which is automatically going to take all the loads off to a lower level which means exactly below the load on the aluminum anyway so you don't need to check it you might need to know it for some reason but you don't need to check it okay John press those numbers down a little bit later then because we've got a little bit more we've got to hit here you sit and just make sure you're not crossing aluminum values with steel values alright to wrap up our look at torsion typically in drive shafts and the like and there's several problems of doing just that notice that we can also do the very same kind of thing we did with the the surfaces in axial loading that looks kind of terrible let's just draw it on side because we're it's a lot easier to draw a drive shaft on the side we can have these changes in thickness where we get stress concentrations so depending upon what the load is and the geometry of each of these it's no surprise I don't think it depends upon the major and minor diameters of the piece because whatever stress is felt in the large part that same amount of stress is going to have to flow into the smaller one causing then stress concentrations and it also depends upon the radius of the fillet and we define it just like we did just like we did for the axial loading where we have the stress concentration factor some kind of allowable or maximum limit compared to the average values we have been calculating and that's what all of these are remember that assumes that we're not at any limit on the stresses but at a an average if we look at the load levels in the piece the real profile is something like this it's actually quite severe where we've been assuming that we have average stress levels the distribution of the shear stress torsionally loaded shafts was at a maximum of the outside and decreased linearly to the inside but these are all average values so our average distribution there looks something like this it's linearly increasing from the inside to the outside and that's what we've been using whereas the maximum might be there and if that's the value we need to keep below the allowable limit so the stress concentration factor allows us to not have to think that these places of greater concern are possible and the chart we use it looks very, very much like that well it's in your book depending on which edition you have it's 536 or 532 I'm not sure which for some reason the device is not open see if that's necessary is it necessary oh yeah that was nice there we go the table is very, very much like the one we saw for the the axially loaded members our distribution is very much different these tend to be kind of severe so let's try a couple problems so imagine we have a step shaft like that with a major diameter of seven and a half inches minor of 3.75 inches and a fillet radius of 9 sixteenths good old American values the shaft rotates at 900 rpm and the allowable stress that's the one that's on the upper part of the definition of the spring stress concentration factor is 8 ksi so we want to find then the maximum power output so there's a little bit extra in here that we need to work through just to make sure we've got the parts to it the power from a rotating loaded shaft of this kind is 2pi f where f is the frequency of rotation the engineering frequency not the angular frequency so it's essentially this number in per seconds rather than per minute and then whatever the torsional load is this is the very same t we've been using as we go along here for a while so we can take this value then and using the allowable limit and the stress concentration factor find then a maximum load we can put on the piece just to help you out a little bit whether you can do it so that's the engineering frequency the revolution per second or hertz if you will and then that's the number we use in here so we need to find a maximum torsion from the allowable once we find that then we can calculate the maximum allowable power any power load over that is going to over stress the shaft all is calculated on the smaller diameter because that's the part that's under the greatest stress the torsional load is the same the stress is going to be a lot greater and we need to calculate what effect the radius has on the stress we're going to figure out a stress concentration factor use that to find the torsion that keeps us at or below that stress limit that we can find the power and finding K is just like it was before with the other axially loaded members all these are torsionally loaded it just comes right off the chart you can imagine in real design situations there's a lot of these graph spectra probably all embedded in software now so find the K use the allowable limit to find an average stress use that to find the torsional load use that to find the power we need R over D and D over D for this particular problem and then those things can be adjusted as needed as you do with time iteration both in inches of course these little indices values are so we're at about 0.15 and we're on the two line which is this second one here so where those two hit looks like about 1.3 looks pretty close to it easy one to read this is near the end of chapter 5 now we can use that to find the torsion which allow us to find the average load no you put it into horsepower when you go down to A's hardware you're looking for a motor to connect to this shaft you're going to want to know horsepower that's how they sell it down there Earl's a bright guy but he wants horsepower 550 foot-pounds per horsepower intermediate values but you're going to have to move that horsepower that's probably one unit that's going to be with us for a very long time well no European vehicles are sold over here without mention of horsepower what they actually have in their catalog and how they go about the design the torsion the T before you went to the power there should be able to get a T from that and then once you get a T from that you can get a power up there and we're getting 63.7 per pinches do you want to underestimate K or overestimate K overestimate I'm not sure how they come up with those I don't know if it says in the book or not how they actually come up with these charts should get something like 900 horsepower 910 911 is also close there's 550 foot-pounds per horsepower very easily make a change in the radius of the fillet if that's not enough if 900 horsepower isn't enough you can easily change the radius would you go bigger or smaller increase the available power output would you go to a bigger radius or smaller radius on the fillet so size matters you know without even looking at the charts that'll just confirm it but if you have a very very sharp corner in there that's an extreme stress concentration those areas tend to break very easily part of what's good about this too is that when you put these shafts together this fillet might well be just a matter of the welding of the two shafts together well if done well enough conserve that as a fillet just make sure you have the intermediate value 6 times 10 is 6 inch-pounds is what I have to have per second that's a power unit Chris are you alright Joel did you get that you're tired too how do you get J oh back here how do you get J one-half pi radius to the fourth but which radius the smaller one these calculations are all done on the smaller that's going to be under the greatest load this 550 foot-pounds oh foot-pounds per second we have 63.7 kits over here yeah and that goes in here so that becomes units of something like the band where your decimal place is inch-pounds per sec because the frequency is in per seconds but 2 pi here is the revolution but it's actually the radius this is actually the angular flossing fill alright did no same numbers Samantha make sure that all these calculations in here are on the smaller diameter same thing that we did for the actually loaded steps on the quality of the weld and the material used in the weld so yeah it could make it weaker but it could also make it stronger also possible of course the machine needs that one piece material the the science of welding there's you can get a master's degree on welding alone non-destructive testing techniques how to check welds without taking the piece apart which then doesn't help you any because that really weakens it when you saw through the thing me? no I don't want to play with fire alright increasing the fillet radius to just a little bit more to 15-16 raises the power by about 11% so then it's an engineering design decision then if that's worth it drops your k-value yeah if you increase the radius you'll appear the k-value drops and you get more power out if you can put in more torque alright let's do one more and then we're done alright the drive shaft with a diameter change but multiple loads on it so 700 pound inches back here further along so all of these will be in pound inches 300 up here in the opposite direction the only one in the opposite direction 1500 pounds there all of those foot pounds and diameters 1.25 inches fillet radius 0.05 inches so find the maximum expected shear stress that's the same thing as the allowable on the definition of k-factor stress-construction factor we can make that a get out of class question that'll incentivize you so we have a shaft of different loads on it somehow gears are placed there maybe even belts boys what about the length of this we only needed the length when we were looking at the angle of twist other than that the length of any of this doesn't matter how far apart those loads are other than that they're not right over each other we need to find the maximum shear stress maybe label the different regions the shear stress changes in every one of those regions because the load changes and the radius change so you're going to need to look at it for each of these regions A to B, B to C C to D and D to E figure out the maximum stress in each of those determine which is the worst case the design limit you then decide if you have material that can take that or you're going to have to change some of the design specifications each of the sections for example A to B we know that there's a load of 700 pound inches so there must be an internal torsion then of all the very same so you need to do that for each of the sections that the internal torsion loads are and then figure out where the area is to worry about this doesn't seem like too much of a worry because there's other places with greater load and smaller diameter so the suspicion would be A to B is not going to be the limiting region some other region will be that's a pretty low load not a great concern we actually want to take the diameter no you have to look at what's happening in each section that will be a C a little bit longer we've got the 700 and 500 so you know the internal load there is 1200 pound inches and we only do B up to C because that's one diameter there's a transition at C that could be the critical point actually this the 1200 will take you all the way to D we have the 700 there 500 there not quite up at the D load yet so that's not on there so there must be an internal load of 1200 yeah because in the transition there's no increase in load the load doesn't come until down here farther so when we're below D not quite up at D yet remember this section B to D we're going until just until we get the load at D which is out here farther so you have to look at a tau max in here well that wouldn't be a concern because here we're at even smaller diameter but then we also have to take into account the stress concentration then it's probably easier to go from the other end so D to E we have the 1500 there not concerned about the direction we would be if we were looking at the angle and twist on this entire shaft but we're not we're just looking to make sure other than the stress limits so it looks like two areas of big concern this is a large load and a small diameter so A to B is much better off much slower load but a much greater diameter so we're not even concerned with A to B B to C is no real trouble but we are worried about what happens at the transition and then we're worried about what happens at this this region there when we're down in the smaller end of the shaft out here between B and E the areas of concern that area where we have the diameter change and the stress concentration and this area where we have the large T and the small D both of those areas for the areas of concern we know the other areas are not as much of concern the loads are less and the diameters are greater well, you should because it's the largest load near the end of that shaft and it's the smallest diameter so we want to calculate here the expected shear stress with those loads direction doesn't matter the only time direction matters if we're looking at the angle of twist where twist can be one direction and the other this all that matters is this magnitude this is the internal load in that end of the shaft which is the biggest load of any and the smallest diameter so it's going to be this is a critical area that we have to look at then so is this one where you have the area change because that can with the stress concentration factor make things even worse in fact this value will be the average value to use there we'll have that one or we can use that same number just with the diameter in instead of this radius we'll have this one PSI yep that's actually the lower number on this one because remember the calculation actually wouldn't be because we'd have 1200 in here instead we're at the slightly lower load but we have the stress concentration factor so we have to calculate a tau average calculate a from that which is greater than 1 that will be greater concern do you have this intermediate number with 1200 in here using the smaller diameter remember that's that's the average and not taking into account the stress concentration factor which we can squeeze in here and we're done our over D is 0.05 D over D 5 so very small radius on the fillet and there's a 1.25 line it's this one so we've got a pretty high stress concentration factor it looks like 1.58 or so if you take a little bit that gives you a multiply of 1.58 times this average shear stress in the little piece before the 300 foot pounds is added so this comes out to be about 96.60 which is quite a bit worse than the 76.40 so the area of concern is at the step change so I want to investigate increasing the fillet radius if it's machined in that's usually not too difficult a change to make that would increase this number bringing it down to a lower K factor and can significantly drop this number then Phil, okay Travis you got those same numbers Chris David 1232 so I was using problem D to E you got 1232 on here oh yeah at the fillet there's not a load of 1500 there's a load of 1200 at the fillet you wanted us to look at the fillet well you have to look at both this had the greatest load on the smallest diameter so we had to look at it but then we also had to look back at the fillet area even though the load was not as great oh so the area of concern is going to be right at these corners because that's where the stress concentration factor is yes because that's the whole point of the stress concentration factor is these area changes okay that's why I circled that region of its own to compare those two numbers