Bones - Biomechanics of the Human Body - page 122

The region of the column that is subject to the greatest forces in bending over to
pick up a heavy object from the ground is the lumbar, as seen through the calcula-
tion of Example 6.5. Moreover, the intensity of this force increases very much for
incorrect postures.
Measurements of pressure in the intervertebral discs of human beings have been
made by A. Nachemson and published as Disc Pressure Measurements in Spine
6 93, 1981. For an adult of 70 kg, standing erect, the pressure on a disc between the
third and the fourth lumbar vertebra is 5.5 atm 10 5 Pa. The force applied

there results from the weight force of the set trunk/neck/head/upper arms/fore
arms/hands equal to 460 N. Using these data we can find the cross-sectional area
A of a disc:
A

¼ 5.6

10 5 10 4 m 2

¼ weight force/pressure ¼ 460 N/(5.6 Pa) ¼ 8.2 ¼

8.2 cm 2 .
From these data, we can calculate the pressure on the same disc assuming that
this subject now holds in each hand a 10 kg mass, hence a total mass of 20 kg. In this
case the total weight will be ¼ 660 N, and we obtain for the pressure, p ¼ 7.9 atm.

There was a 44 % increase in the pressure.
Nachemson also measured the pressure on the same disc while picking up an
object of 200 N weight with both hands and bending correctly the knees. He
obtained an average value of approximately 13.0 atm. When the knees are not
bent, the pressure reached 35.0 atm during a very small interval of time, which is
due to a compressive force of 2,907 N.
In Example 6.5 in which a person lifts a weight of 200 N without bending his
knees, we have obtained, using a simple model, for the contact force applied by the
sacrum on the last lumbar disc the value of 3,381 N. As the area of this disc is
slightly larger than that used above, we can consider it as being 9.0 cm 2 . With these
data the pressure can be obtained, giving 37.1 atm, which is coherent with the value
measured by Nachemson.
The compressive strength, that is, the compressive breaking stress, for any
intervertebral disc is, according to Table 7.3 , 1.10

10 7 Pa which is equal to

11 N/mm 2 ¼ 1,100 N/cm 2 . In the case of Example 6.5, in which the applied force

on the last lumbar disc is 3,381 N, we obtain for the pressure the value 3,381 N/
9.0 cm 2 375.7 N/cm 2 . That means that there is a safety factor of 2.9 times to

cause a fracture, if the force is uniformly applied on the total area of 9.0 cm 2 .
Passing from the cervical part to the thoracic and then to the lumbar, the weight
force that the column must support increases, as already discussed in Chap. 1 . Since
the compressive strength of any intervertebral disc is practically constant with a
value of 1,100 N/cm 2 , as the weight force increases, the area of the disc must also
increase.
This question becomes important, when a person who practices yoga decides to
do a posture of sirsha-asana, the headstand posture in which this subject stays
upside-down, with the top of his head on the floor. In this posture, almost all of the
body weight falls on the cervical vertebrae. For an adult of 70 kg, the whole body
mass, less that of the head, weighs 65.2 kg. Hence, the first cervical disc must
support a weight of about 652 N. If this posture is achieved with care, it is still

¼