Young's modulus for bone

  • #1
1,197
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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?

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

is that correct?

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

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

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

Answers and Replies

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

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?

[tex]\Delta L=\frac{FL}{AY}[/tex]
[tex]\Delta L=\frac{73631N*0.25m}{\pi (0.0125m^2)^2 * 1.5x10^{10} N/m^2}[/tex]
[tex]\Delta L=0.0025m[/tex]
Looks good. But realize you could have saved a bit of arithmetic by starting with the maximum stress (F/A) instead of the force.
 
  • #3
1,197
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so what was the point in giving me [tex]1.5X10^8 N/m^2[/tex]? Was that to throw me off?
 
  • #4
Doc Al
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45,204
1,540
I don't understand your question. You were given two numbers: max stress and Young's modulus. You used them both.
 

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