# Young's modulus for bone

Young's modulus for bone is 1.5x10^10 N/m^2 and that bone will fracture if more than 1.5X10^8 N/m^2 is exerted.

What is the max force that can be exerted on the femur if the effective diameter is 2.5cm?

$$Y=F/A$$
$$1.50x10^8N/m^2 =\frac{F}{\pi (0.0125m^2)^2}$$
$$F=73631N$$

is that correct?

if this force is applied compressively, by how much does the 25cm long bone shorten?

$$\Delta L=\frac{FL}{AY}$$
$$\Delta L=\frac{73631N*0.25m}{\pi (0.0125m^2)^2 * 1.5x10^{10} N/m^2}$$
$$\Delta L=0.0025m$$

I think I did some of this wrong but I'm not sure how to approch these problems. Any guidance?

Last edited:

Doc Al
Mentor
UrbanXrisis said:
$$Y=F/A$$
$$1.50x10^8N/m^2 =\frac{F}{\pi (0.0125m^2)^2}$$
$$F=73631N$$

is that correct?
Yes. But realize that 1.5x10^8 N/m^2 is not Y (which stands for Young's modulus) but is the maximum stress the bone can support. (Stress = F/A). Also: round off to a sensible number of significant figures.

if this force is applied compressively, by how much does the 25cm long bone shorten?

$$\Delta L=\frac{FL}{AY}$$
$$\Delta L=\frac{73631N*0.25m}{\pi (0.0125m^2)^2 * 1.5x10^{10} N/m^2}$$
$$\Delta L=0.0025m$$
Looks good. But realize you could have saved a bit of arithmetic by starting with the maximum stress (F/A) instead of the force.

so what was the point in giving me $$1.5X10^8 N/m^2$$? Was that to throw me off?

Doc Al
Mentor
I don't understand your question. You were given two numbers: max stress and Young's modulus. You used them both.