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 10

^{7}Pa), and that Young’s modulus for LCL is 16.9 ± 4.07 MPa (1.69 x 10^{7}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}m^{2}.
Stress = F/A (Eq. 1)

F = Stress x A (Eq.
2)

We also know that:

F = E (ΔL/L

_{o})A (Eq. 3)
We can plug equation 2 into equation 3 and get:

Stress x A = E (ΔL/L

_{o}) A
We can cancel out the A on both sides and get:

Stress = E (ΔL/L

_{o})
UTS = E (ΔL/L

_{o})
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 10

^{7}Pa = 1.69 x 10^{7}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 10

^{7}Pa x 1.26 x 10^{-5}m^{2}**F= 190 N**

The force would have to exceed 190 N in order for the
ligament to tear.

__Works Referenced__:

http://www.plosone.org/article/info%3Adoi%2F10.1371%2Fjournal.pone.0026178

http://www.wheelessonline.com/ortho/lateral_collateral_ligament

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