BPT notes

1. DR. DIBYENDUNARAYAN BID [PT]
T H E S A R V A 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 A T
Biomechanics
of the
Hip Complex: 3

2. Hip Joint Forces
and
Muscle
Function in Stance

3. Bilateral Stance
 In 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.

4.  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.

5.  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.

6.  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.

8.  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.

9.  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).

11. Example 10-4
Calculating 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

12.  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.

13.  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.

14.  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.

16.  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.

17.  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.

18. 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.

19.  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

21.  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.

22.  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.

23.  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.

24.  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.

25.  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.

26.  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.

27. Example 10-5
Calculating 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

28.  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.

29.  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 m
Fms = 68.75 Nm ÷ 0.05 m= 1375 N

30. ***
 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.

31. ***
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

32.  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.

33.  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.

34.  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.

35.  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.

36.  Next Episode………………

37. End of Part - 3