WEBVTT
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So in this question we're asked to estimate the stress
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in your femur when you're standing still and upright
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with your weight distributed evenly on two feet.
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And so, we're just going to idealise the femur
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as a cylinder and it's going to have a force
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acting downwards and putting it in compression
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and that's going to be reacted by a stress
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that we're being asked to calculate
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which is going to be acting over
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the cross section area A.
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So, I haven't given you very much information here
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so you can find your own mass
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in order to find the force here
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and so I looked up the average woman in the UK
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has a mass of m = 70.2kg
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according to our National Office of Statistics;
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and the same average person has a femur diameter
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at the neck, so that's the narrowest point
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and we'll call that d, and that's equal to 25mm,
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and so their typical dimensions and of course
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you can estimate the size of your own femur
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and weigh yourself to get your own mass
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and then use that data and so you will get a slightly
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different answers to the ones I going to get.
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And so, we can say now that stress,
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sigma will be equal to force divided by the cross-section area
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and our force is going to be mg,
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mass times the gravitational constant,
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divided by the cross-section area of the cylinder here,
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and so that's going to be Pi d squared upon four.
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Now we have two legs so there's two of these A's,
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and I'm going to put a little n in here,
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where n is the number of legs that you have,
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because in the later part of the question we're going
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to use the same formula, but for four legged animals.
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And so now we can substitute some numbers into here:
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so, the mass for the average woman in the UK
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is 70.2 times 9.81 for the gravitational acceleration
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divided by 2 into Pi and then multiply by
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the average diameter 0.025, to put it into metres,
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Squared, divided by 4, and so
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with a calculator you can work out that comes to 701,465 Pascals
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or Newtons per metre squared whichever you prefer.
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So we can say that's approximately equal to 0.7 Mega Pascals
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Ok, and so then in the next part of the question
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we're asked to repeat this exercise for an African elephant.
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And so for African elephants, so I've done it for a cow elephant,
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you could do it for a bull elephant if you want,
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you'd get a slightly different answer,
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and I looked up some data, I used Google scholar to find it,
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and I've put the references in the transcript,
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so you can have a look at the data there,
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so an African cow elephant typically weighs 3500kg
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and has a diameter of their femur of 140mm,
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and there's a nice website you can look up
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called www.elephant.se that has a lot of information on it.
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So, we can now say sigma will be equal,
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using this formula again here, 3500 x 9.81
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divided by, now this time n is, elephants got 4 legs,
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so I have a 4 in there, times Pi times 0.14 squared
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upon 4 and that comes out at 557386 Pascals
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or approximately equal to 0.6 Pascals, sorry Mega Pascals.
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And then finally, it asks us to do it for a mouse;
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and so, we can say and for a mouse,
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so again I used Google scholar,
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and I found a typical value mass for a mouse is 35g,
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and d is equal to 0.7mm;
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and hence, using the same formula, I found that my sigma is 0.2 MPa.
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So, you can see that there's not a lot of difference
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between a typical UK woman and
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an African elephant stress in their femurs,
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but there's a big drop down to what the average mouse
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might expect to have in terms of stress in it's femurs.
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Now, these are just estimates because
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we're not taken account of dynamic loading
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or the effect of muscles and so on,
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but they give us an idea of the scale of stress
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in our femurs and so because the stress levels
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in a mouse is so much smaller,
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their much less susceptible to breaking their legs,
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and that's why they can do big jumps and so on,
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without worrying about fracturing their femurs,
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where as larger mammals have to be a lot more careful.