# Drilling a hole into a bone and measuring the force required to break it

I'm currently planning an experiment (high school biology) where I will drill a hole into a bone, then load weights onto it until it breaks. I will be repeating this with holes of different diameters.

So my question is- what is the name of the property I'm measuring? I thought this might be the 'fracture toughness', or 'lateral shear' as a physicist suggested, but this is similar to how a three-point flexural test works, so it could be 'flexural strength' too?

Any help would be greatly appreciated and many apologies if this is the wrong place to ask • You need to add a diagram (or a good description) how how you will load the drilled bone with weights. What stresses the bone will experience depend on the mode of loading. – Gert Aug 22 '15 at 16:40
• Added a rather rough diagram. The two ends of the bone will be clamped and the weights will be loaded onto a piece of material around the hole (thought this might spread the weight more evenly). Thank you. – sonder Aug 22 '15 at 17:26 Your test will not be measuring a single property of the test specimens, so it won't have a specific name.

But we can try and draw some inferences about what you'll be measuring with these experiments, using the idealised schematic above. In it the bone has been replaced with a uniform bar, with a hole drilled through it at the halfway point. The bar is suspended on two frictionless supports.

The green horizontal line is the so called neutral line. With the loading force $F$ in place as shown, the area below the neutral line and between the hole and the horizontal edge of the bar is under tension. Similarly, area above the neutral line and between the hole and the horizontal edge of the bar is under compression.

The force $F_b$ at which the bone breaks is likely to be correlated with the so called stress at break of the bone, $\sigma_b$. We can expect:

$F_b \propto \sigma_b$.

Increasing the diameter $D$ of the drilled hole reduces the area of the zone under tension, so that for comparable values of $F$, the stress at break of the bone will be reached earlier. We can expect:

$F_b \propto \frac{1}{D}$.

Combined, we can expect:

$F_b \propto \frac{\sigma_b}{D}$.

• Just out of interest, what would be the forces acting on the bone if the hole was perpendicular to the length of bone and would an inverse relationship still be expected? Thank you very much for your help. – sonder Aug 24 '15 at 3:13
• Yes, the relationship $F_b \propto \frac{\sigma_b}{D}$ would still hold but all things being equal the values of $F_b$ would be higher. Thank you. – Gert Aug 24 '15 at 11:03