Worked Example: Femoral Stresses

So in this question we’re asked to estimate the stress in your femur when you’re standing still and upright with your weight distributed evenly on two feet. And so, we’re just going to idealise the femur as a cylinder and it’s going to have a force acting downwards and putting it in compression and that’s going to be reacted by a stress that we’re being asked to calculate which is going to be acting over the cross section area A.

So, I haven’t given you very much information here so you can find your own mass in order to find the force here and so I looked up the average woman in the UK has a mass of m = 70.2kg according to our National Office of Statistics; and the same average person has a femur diameter at the neck, so that’s the narrowest point and we’ll call that d, and that’s equal to 25mm, and so their typical dimensions and of course you can estimate the size of your own femur and weigh yourself to get your own mass and then use that data and so you will get a slightly different answers to the ones I going to get.

And so, we can say now that stress, sigma will be equal to force divided by the cross-section area and our force is going to be mg, mass times the gravitational constant, divided by the cross-section area of the cylinder here, and so that’s going to be Pi d squared upon four. Now we have two legs so there’s two of these A’s, and I’m going to put a little n in here, where n is the number of legs that you have, because in the later part of the question we’re going to use the same formula, but for four legged animals.

And so now we can substitute some numbers into here: so, the mass for the average woman in the UK is 70.2 times 9.81 for the gravitational acceleration divided by 2 into Pi and then multiply by the average diameter 0.025, to put it into metres, Squared, divided by 4, and so with a calculator you can work out that comes to 701,465 Pascals or Newtons per metre squared whichever you prefer. So we can say that’s approximately equal to 0.7 Mega Pascals Ok, and so then in the next part of the question we’re asked to repeat this exercise for an African elephant.

And so for African elephants, so I’ve done it for a cow elephant, you could do it for a bull elephant if you want, you’d get a slightly different answer, and I looked up some data, I used Google scholar to find it, and I’ve put the references in the transcript, so you can have a look at the data there, so an African cow elephant typically weighs 3500kg and has a diameter of their femur of 140mm, and there’s a nice website you can look up called www.elephant.se that has a lot of information on it.

So, we can now say sigma will be equal, using this formula again here, 3500 x 9.81 divided by, now this time n is, elephants got 4 legs, so I have a 4 in there, times Pi times 0.14 squared upon 4 and that comes out at 557386 Pascals or approximately equal to 0.6 Pascals, sorry Mega Pascals. And then finally, it asks us to do it for a mouse; and so, we can say and for a mouse, so again I used Google scholar, and I found a typical value mass for a mouse is 35g, and d is equal to 0.7mm; and hence, using the same formula, I found that my sigma is 0.2 MPa.

So, you can see that there’s not a lot of difference between a typical UK woman and an African elephant stress in their femurs, but there’s a big drop down to what the average mouse might expect to have in terms of stress in it’s femurs. Now, these are just estimates because we’re not taken account of dynamic loading or the effect of muscles and so on, but they give us an idea of the scale of stress in our femurs and so because the stress levels in a mouse is so much smaller, their much less susceptible to breaking their legs, and that’s why they can do big jumps and so on, without worrying about fracturing their femurs, where as larger mammals have to be a lot more careful.