Kinesiology 201 Solutions
Tissue Mechanics
Tony Leyland
School of Kinesiology
Simon Fraser University
Connective Tissue
1 a) Tensile stress = 2500/2 = 1250 N/cm2
b) The area has been reduced by 10% and
therefore is now 1.8 cm2
Tensile stress = 3000/1.8 = 1667 N/cm2
c) The ligament may fail completely. As can be
seen from a typical ligamentous tissue stressstrain curve opposite, once the collagen fibres
begin to fail, the ligament will tend to fail
completely without much further increase in
stress (indeed the stress may even reduce
but the ligament still will fail).
Point of rapid failure
Stress
2. A visco-elastic tissue is able to resist deformation
with a greater force when loaded quickly. Many
fractures occur due to acute loading (except stress
fractures) so this is a beneficial property. As can
be seen from the graph opposite the bone not only
offers greater resistance when loaded quickly it
can also withstand greater deformation before
failing. [3]
3. Strain is defined as change in length divided by
original length. Therefore:
0
Load
5
Strain (percent)
10
Fracture
Quick
Slow
Fracture
Deformation
Δlength = 0.02 m and original length = 0.1 m. Therefore: strain = 0.2 (or 20%) [1]
As strain is a ratio of lengths it has no units. [1]
4. (i) Spreads load. Cartilage can reduce maximum contact stresses by 50% or more.
(ii) Reduces friction during movement.
5. a) The table shows that the total mass is 74.4 kg and that 4.4 kg is located in
shank and foot (lower = 3.4 kg, foot = 1 kg). So the total mass located above
knees is 65.6 kg [74.4 - (4.4 x 2)] representing a force of -644 N. Therefore
compressive force on each knee is -322 N.
b) If he stands on one foot then the entire body except the mass of one lower
and one foot is above support by that knee joint. Therefore, the mass above
knee is 70 kg (74.4 - 4.4) representing a force of -687 N.
the
the
the
leg
the
1
c) Obviously the contact forces would be higher still. If you factored in the muscle
forces needed to accelerate and decelerate in activities like running you can
imagine how high the loads are on the lower joints (approximately three times
body weight in distance running and higher still in sprinting).
6. Total relative mass above L5/S1 level = M (we must account for two arms, forearms
and hands)
M = 0.073 + 0.507 + (2 x (0.026 + 0.016 + 0.007)) = 0.678
Weight on this disk
=
=
total body mass x relative mass
(90 x 0.678) x 9.81 = 598.40 N = 594 N
Pressure = Force/area = 598.40/25 = 23.94 N/cm2
(or 239442.48 Pascal [N/m2])
(or 239.44 KiloPascal [kPa])
7. Ligament Crimp [2]
8. Bone is composed of water (approximately 25-30%) [1], protein (about 5-15%
primarily in the form of collagen fibres) [1] and minerals (60-70% primarily calcium
phosphate) [1]. Note that giving percentages is more information then was requested
in the question. It is termed a two-phase material as it is structured to effectively
resist both compression and tension. [1]
Anisotropic Characteristics
9. Anisotropic material is a material that does not
have the same strength when loaded in different
ways. The graph opposite shows the quantitative
strength of bone under three loading conditions. [2]
Compression is loading that is directed along the
long axis of the specimen and tends to crush the
specimen. [1] tension is directed along the long
axis and tends to pull the specimen apart. [1] and
shear is force directed at right angles to the long
axis and tends to bend or torque the specimen. [1]
Stress to Fracture
Compression
Tension
Shear
2