DR. DIBYENDUNARAYAN BID [PT]T H E S A R VA J A N I K C O L L E G E O F P H Y S I O T H E R A P Y,

R A M P U R A , S U R AT

Biomechanics of the

Hip Complex: 3

Hip Joint Forces and

MuscleFunction in Stance

Bilateral StanceIn erect bilateral stance, both hips are in neutral

or slight hyperextension, and weight is evenly distributed between both legs.

The LoG falls just posterior to axis for flexion/extension of the hip joint.

The posterior location of the LoG creates an extension moment of force around the hip that tends to posteriorly tilt the pelvis on the femoral heads.

The gravitational extension moment is largely checked by passive tension in the hip joint capsuloligamentous structures, although slight or intermittent activity in the iliopsoas muscles in relaxed standing may assist the passive structures.

In the frontal plane during bilateral stance, the superincumbent body weight is transmitted through the sacroiliac joints and pelvis to the right and left femoral heads.

Hypothetically, the weight of the HAT (two thirds of body weight) should be distributed so that each femoral head receives approximately half of the superincumbent weight.

As shown in Figure 10-30, the joint axis of each hip lies at an equal distance from the LoG of HAT; that is, the gravitational MAs for the right hip (DR) and the left hip (DL) are equal.

Because the body weight (W) on each femoral head is the same (WR = WL), the magnitude of the gravitational torques around each hip must be identical (WR x DR = WL x DL).

The gravitational torques on the right and left hips, however, occur in opposite directions.

The weight of the body acting around the right hip tends to drop the pelvis down on the left (right adduction moment), whereas the weight acting around the left hip tends to drop the pelvis down on the right (left adduction moment).

These two opposing gravitational moments of equal magnitude balance each other, and the pelvis is maintained in equilibrium in the frontal plane without the assistance of active muscles.

Assuming that muscular forces are not required to maintain either sagittal or frontal plane stability at the hip joint in bilateral stance, the compression across each hip joint in bilateral stance should simply be half the superimposed body weight (or one half of HAT to each hip).

Example 10-4Calculating Hip Joint Compression in Bilateral Stance

Using a hypothetical case of someone weighing 825 N (~185 lb), the weight of HAT (2/3 body weight) will be 550 N (~124 lb).

Of that 550 N, half will presumably be distributed through each hip.

Because we are assuming no additional compressive force produced by hip muscle activity, the total hip joint compression at each hip in bilateral stance is estimated to be 275 N (~50 lb); that is, total hip joint compression through each hip in bilateral stance is one third of body weight.

# 825N/3=275N

The rationale presented in Example 10-4 for assuming that each hip received one third of body weight in bilateral stance is reasonable.

However, Bergmann and colleagues showed in several subjects with an instrumented pressure-sensitive hip prosthesis that the joint compression across each hip in bilateral stance was 80% to 100% of body weight, rather than one third (33%) of body weight, as commonly proposed.

When they added a symmetrically distributed load to the subject’s trunk, the forces at both hip joints increased by the full weight of the load, rather than by half of the superimposed load as might be expected.

Although the mechanics of someone standing who has a prosthetic hip may not fully represent normal hip joint forces, the findings of Bergmann and colleagues call into question the simplistic view of hip joint forces in bilateral stance.

The slight activity in the iliopsoas muscle may account for more joint compression than previously thought.

Alternatively, capsuloligamentous tension may contribute to joint compression.

When bilateral stance is not symmetrical, frontal plane muscle activity will be necessary to either control the side-to-side motion or to return the hips to symmetrical stance.

In Figure 10-22, the pelvis is shifted to the right, resulting in relative adduction of the right hip and abduction of the left hip. To return to neutral position, an active contraction of the right hip abductors would be expected.

However, a contraction of the left hip adductors would accomplish the same goal. [In bilateral stance, the contralateral abductors and adductors may function as synergists to control the frontal plane motion of the pelvis.]

Under the condition that both extremities bear at least some of the superimposed body weight, the adductors may assist the abductors in control of the pelvis against the force of gravity or the GRF.

In unilateral stance, activity of the adductors either in the weight-bearing or non–weight-bearing hip cannot contribute to stability of the stance limb.

Hip joint stability in unilateral stance is the sole domain of the hip joint abductors. In the absence of adequate hip abductor function, the adductors can contribute to stability—but only in bilateral stance.

Unilateral Stance

In Figure 10-31, the left leg has been lifted from the ground and the full superimposed body weight is being supported by the right hip joint.

Rather than sharing the compressive force of the superimposed body weight with the left limb, the right hip joint must now carry the full burden.

In addition, the weight of the non–weight-bearing left limb that is hanging on the left side of the pelvis must be supported along with the weight of HAT.

Of the one-third portion of the body weight found in the lower extremities, the nonsupporting limb must account for half of that, or one sixth of the full body weight.

The magnitude of body weight (W) com-pressing the right hip joint in right unilateral stance, therefore, is:

Right hip joint compression body weight =[2/3 x W] + [1/6 x W]

Right hip joint compression body weight = 5/6 x W

In our hypothetical subject from Example 10-4 who weighs 825 N, HAT in this individual weighs 550 N.

One lower extremity weighs one sixth of body weight, or 137.5 N.

Therefore, when this individual lifts one leg off the ground, the supporting hip joint will undergo 687.5 N (or five sixths of body weight) of compression from body weight alone.

Although we have accounted for the increase in hip joint compression from body weight as a person moves from double-limb support (bilateral stance) to single-limb support, the problem is more complex.

Not only is the hip joint in unilateral stance being compressed by body weight (gravity), but also that body weight is concomitantly creating a torque around the hip joint.

The force of gravity acting on HAT and the non–weight-bearing left lower limb (HATLL) will create an adduction torque around the supporting hip joint; that is, gravity will attempt to drop the pelvis around the right weight-bearing hip joint axis.

The abduction countertorque will have to be supplied by the hip abductor musculature. The result will be joint compression or a joint reaction force that is a combination of both body weight and abductor muscular compression.

The total joint compression can be calculated for our hypothetical 825-N subject.

The LoG of HATLL can be estimated to lie 10 cm (0.1 m, or ~4 inches) from the right hip joint axis (that is, MA 0.1 m), although the actual distance will vary among individuals.

The 10-cm estimate of the gravitational MA is for symmetrical stance.

The actual MA is likely to be slightly greater because the weight of the hanging left leg will pull the center of gravity of the superimposed weight slightly to the left, although the LoG will simultaneously be shifted slightly right to get the LoG within the single foot base of support.

Example 10-5Calculating Hip Joint Compression in Unilateral Stance

For simplicity, the possible increase in the MA of HATLL from the non–weight-bearing limb will be ignored, but our torque calculation here is likely to under-estimate the actual gravitational torque.

In our simplified hypothetical example, the magnitude of the gravitational adduction torque at the right hip will be as follows:

HATLL torque adduction: 687.5 N x 0.1 m = 68.75 Nm

To maintain the single-limb support position, there must be a countertorque (abduction moment) of equivalent magnitude.

The countertorque must be produced by the force of the hip abductors (gluteus medius, minimus, and tensor fascia lata muscles) acting on the pelvis.

Assuming that the abductor muscles act through a typical MA of 5 cm (0.05 m, or 2 inches) and knowing that the muscles must generate an abduction torque equivalent to the adduction torque of gravity (68.75 Nm), we can solve for the magnitude of muscle contraction (Fms) needed to maintain equilibrium in our hypothetical example.

Torque abduction: 68.75 Nm= Fms x 0.05 mFms = 68.75 Nm ÷ 0.05 m= 1375 N

***Assuming that all the abductor

muscular force is transmitted through the acetabulum to the femoral head, the 1375-N abductor muscular compressive force is now added to the 687.5 N of compression caused by body weight passing through the supporting hip.

***

Thus, the total hip joint compression, or joint reaction force, at the stance hip joint in unilateral support can be estimated for our hypothetical subject at:

1375 N abductor joint compression+ 687.5 N body weight (HATLL) compression------------------------------------------------------=2062.5 N total joint compression

Total hip joint compression or joint reaction forces are generally considered to be: 2.5 to 3 times body weight in static unilateral stance.

Investigators have calculated or measured forces of four and seven times body weight in, respectively, the beginning and end of the stance phase of gait and seven times the body weight in activities such as stair climbing.

Although weight loss can reduce the hip joint reaction force, the larger component of the joint reaction force is generated by the contraction of muscles, primarily presumed to be the hip abductors.

The magnitude of hip abductor force required can be affected by individual differences in the angle of pull of the abductor muscles in particular, although peak pressures may not vary within the normal variations in abductor angle of pull.

Presuming that muscle force requirements are unchanged, substantial changes in the angle of inclination of the femoral head or in the angle of femoral torsion can affect the contact areas within the hip joint, thereby resulting in the potential for increased pressures in certain segments of the femoral head or acetabulum.

Next Episode………………

End of Part - 3