Biomechanics of the Hip

Activity

Typical peak force (BW)

Walking, slow

1.6 ± 4.1

Walking, normal

2.1 ± 3.3

Walking, fast

1.8 ± 4.3

Jogging/running

4.3 ± 5.0

Ascending stairs

1.5 ± 5.5

Descending stairs

1.6 ± 5.1

Standing up

1.8 ± 2.2

Sitting down

1.5 ± 2.0

Standing/2-1-2 legs

2.2 ± 3.7

Knee bend

1.2 ± 1.8

Stumbling

7.2 ± 8.7

Table adapted from “The Adult Hip, Volume 1” Table 5-1, page 84. Callaghan, John J; Rosenberg, Aaron G; Rubash, Harry E (eds)

Fig. 2.1

During static load of the hip joint in single leg stance, the lever arm between the center of rotation of the hip and the body center line (o-b) leads to a downward force towards the non-supporting leg. The hip abductors counteract this torque with the respective lever arm (c-o) and horizontally stabilize the pelvis. The sum of all forces acting on the joint is equal to zero. In normal conditions, the resultant force (R) passes through the center of the femoral head and forms an angle of 16° to the vertical. The respective magnitude of the total load (R) is dependent on the body weight (K) and the length ratio of the lever arms and results in roughly four times the body weight (original figures from Pauwels)

In a stable pelvis, the sum of all moment arms is 0 (∑M = 0). Assuming c-o is 1 and o-b is 3, then we can come up with the following equation

$-{M}_y+3\ K=0$

where M _yis the vertical component of the abductor muscle moment arm and K represents the body weight moment arm (minus the weight of the ipsilateral leg that corresponds to roughly 1/6 of the body weight). This equation then yields

${M}_y=3\ K$

Assuming the sum of all forces around the hip is also 0 (∑F _y= 0), then we can say that

$-{M}_y-K+{R}_y=0$

which is to say that the vertical force created by the abductor muscle pull (M _y), the body weight (K), and the vertical component of the joint reaction force (R _y), must add up to 0. Substituting what we have already established is the relationship between M _yand K, we are left with

${R}_y={M}_y+K=4\ K$

$R=\frac{R_y}{\cos\ {16}^{{}^{\circ}}}$

$\mathrm{R}=4.2\ K$

Since K represents the force created by approximately 5/6 body weight (since the ipsilateral limb weight is not included), then the joint reaction force across a stable hip during single leg stance is approximately 3.5 times the body weight. Dynamic forces generated during walking additionally increase the load of the femoral head by 50%. Thus, the maximum applied pressure on the femoral head when walking is about 4.5 times the total body weight.

During Trendelenburg gait , the body weight is shifted over the affected hip, thereby decreasing the moment arm K. If the weight is shifted such that c-o remains 1, but o-b is reduced to ½, then the joint reactive force decreases to 1.7 K, or approximately 1½ times the body weight (Fig. 2.2).

Fig. 2.2

In the case of Trendelenburg gait , the patient inclines the upper body over the affected hip, thus shifting the center of gravity (S) closer to the center of rotation of the hip. The lever arm of the body weight is thereby shortened, and the necessary counteracting forces of the abductors are reduced. This significantly reduces the compressive stress of the hip joint to 1.5 times the body weight (original figures from Pauwels)

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