An Analytical Model for Knee Ligaments

Abstract

In this work, an analytical model of porcine knee ligaments, statically loaded, is presented. Although introductory, this elastic analytical model provided an estimative of the main mechanical variables results, as longitudinal forces and displacements, and its respective longitudinal stresses and strains, in each of four knee ligaments: lateral collateral ligament (LCL), anterior cruciate ligament (ACL), posterior cruciate ligament (PCL) and medial collateral ligament (MCL). An interesting conclusion of the application of this analytical model was that the cruciate ligaments were submitted to greater forces than the collateral ones.

Appendices

Appendix

The 1D Model in Coronal Plane

Figure 4 shows the geometric representation of the ligaments of a knee of the 1D model in coronal plane, adapted from Silva et al. [13]. The proposed analytic model uses mechanics of solids to estimate the load share between four porcine ligaments, namely: LCL, ACL, PCL, MCL. The model considers a 1D representation of the knee of porcine ligaments, in the coronal plane.

It is supposed that ligaments force components in the axis z direction are in a parallel arrangement, being submitted to the same vertical displacement and sharing forces in function of its own stiffnesses and angles. They are renamed as follows: LCL = 1, ACL = 2, PCL = 3 and MCL = 4.

The model uses (1) as vertical equilibrium condition and (2) as compatibility condition, adapted from Silva et al. [13].

Fig. 4

A Simplified 1D analytical model in the coronal plane

$$F_{1} + F_{2} \cdot \sin \left( {\theta_{2} } \right) + F_{3} \cdot \sin \left( {\theta_{3} } \right) + F_{4} = P$$

(8)

$$\delta_{1} = \delta_{2} \cdot \sin \left( {\theta_{2} } \right) = \delta_{3} \cdot \sin \left( {\theta_{3} } \right) = \delta_{4}$$

(9)

(linear displacements in the unit vector $\vec{k}$direction)

Note that $\delta_{n} = \frac{{F_{n} L_{n} }}{{A_{n} E_{n} }} = \frac{{F_{n} }}{{K_{n} }}$ where $K_{n} = \frac{{A_{n} E_{n} }}{{L_{n} }}$ for n = 1, 2, 3 or 4. So, (9) can be re-written as:

$$\frac{{F_{1} }}{{K_{1} }} = \frac{{F_{2} \cdot {\text{sin}}\left( {\theta_{2} } \right)}}{{K_{2} }} = \frac{{F_{3} \cdot {\text{sin}}\left( {\theta_{3} } \right)}}{{K_{3} }} = \frac{{F_{4} }}{{K_{4} }}$$

(10)

So, the longitudinal forces for the LCL, ACL, PCL and MCL, can be calculated. For the LCL and MCL, respectively:

$$F_{1} = \frac{P}{{\left( {\frac{K1 + K2 + K3 + K4}{{K1}}} \right)}}\quad F_{4} = \frac{P}{{\left( {\frac{K1 + K2 + K3 + K4}{{K4}}} \right)}}$$

(11)

For the ACL and PCL, respectively:

$$F_{2} = \frac{P}{{{\text{sin}}\left( {\theta_{2} } \right)\left( {\frac{K1 + K2 + K3 + K4}{{K2}}} \right)}}\quad F_{3} = \frac{P}{{{\text{sin}}\left( {\theta_{3} } \right)\left( {\frac{K1 + K2 + K3 + K4}{{K3}}} \right)}}$$

(12)

where, F_n are ligaments longitudinal forces, P is the vertical loading force, θ₂ and θ₃ are ligament angles (note that ligaments 1 and 4 are supposed to be in vertical positions), δ_n are the ligaments longitudinal displacements, A_n are the ligaments transversal areas, E_n are the ligaments elasticity modulus, L_n are ligament longitudinal lengths, K_n are the ligaments stiffnesses. Also, n varies from 1 to 4.