# Biomechanics of thr

Páginas: 3 (513 palabras) Publicado: 12 de diciembre de 2011
Biomechanics of THR

Basic understanding of the factors influencing the forces acting around the hip joint is fundamental to the success of total hip arthroplasty (THA). In a properly performed THAthis forces should be similar to those acting upon a natural hip joint.
When standing on boths legs, the body weight is distributed equally on both hips and the center of gravity lies in the midsagital plane. In a static balanced position, minimal muscular forces are required, and for practical purposes, these can be disregarded. The force exerted upon both hips represents the total body weightminus the weight of the lower limbs. Since this load is transmited simetrically to both hips, each femoral head supports half of this load as shown in figure 1, where:
X1=X2 (distance between centreof natural femoral head / prosthetic femoral head and mid-line)
COG: centre ogf gravity
A=B=1/3 total body weight

However in a single leg stance, the center of gravity is translated distally andaway from the supporting leg because the non weight-bearing leg is part of the body weight acting upon the supporting hip. Since the bearing axis is eccentric to the axis of the centre of gravity, arotating force takes place around the femoral head. This rotating force must be counterbalanced by the muscle action (abductor mechanism) inserted into the greater trochanter (vector of force M infigure 2).

Since the effective lever arm of force M (X2)is shorter than the lever arm of body weight acting through the center of gravity (X1), the combined force of the abductor mechanism must be amultiple of body weight. The vectors of force W and force M produces a resultant compressive load (vector of force R)on the femoral head.

Free body diagram analysis (figure 2) when X1 = 2.5X2:In equilibrium, the sum of moments in the Y axis is cero:

M = 0
2.5W – My = 0
My = 2.5W

In equilibrium, the sum of forces in the Y axis is cero:

Fy = 0
Ry – W – My = 0
Ry = W + My...

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