## Friday, November 23, 2012

### The Physics of Tearing an LCL

By Jessica Mondon
I am very accident prone, and have sustained many knee injuries over my years. A few years ago I tore my lateral collateral ligament (LCL). For those who don’t know the  LCL runs along the outside of the knee.

There is some elasticity in ligaments, but when they stretch too much they will snap. In my case, I tore my left LCL when I was playing softball. I was sliding and the catcher went to tag me and forcefully pushed my tibia to the right, while the top half of my leg and my femur stayed still. The force from this was more than the ligament could endure so it tore. This can be examined by looking at elasticity in solids. I would want to look at the equation with Shear’s modulus, but unfortunately I could only find Young’s modulus for the LCL. So instead of finding the force that would need to be applied when it was hit in order to break, I will be finding the force of tension in the ligament before it snaps. I found that the ultimate tensile strength (UTS) of the LCL is 14.9 ± 3.9 MPa (I used 1.49 x 107 Pa), and that Young’s modulus for LCL is 16.9 ± 4.07 MPa (1.69 x 107 Pa). I also found that the length of the ligament is about 9 mm (0.009 m). The UTS represents the maximum stress that the ligament can endure before it snaps. We can assume it is cylindrical with radius 2 mm, so cross sectional area would be Π(.002m)2=  1.26 x 10-5 m2.
Stress = F/A (Eq. 1)
F = Stress x A (Eq. 2)
We also know that:
F = E (ΔL/Lo)A (Eq. 3)
We can plug equation 2 into equation 3 and get:
Stress x A = E (ΔL/Lo) A
We can cancel out the A on both sides and get:
Stress = E (ΔL/Lo)
UTS = E (ΔL/Lo)
We know UTS, Young’s modulus (E), and the length of the ligament, so we can solve for ΔL to see how far the ligament stretches when maximum stress is exerted.
1.49 x 107 Pa = 1.69 x 107 Pa (ΔL/0.009m)
ΔL= 0.0079m= 7.9 mm

We can then find the force that would be exerted when there was maximum stress using the equation:
Stress = F/A
F = Stress x A
F = 1.49 x 107 Pa x 1.26 x 10-5 m2
F= 190 N
The force would have to exceed 190 N in order for the ligament to tear.

Works Referenced: