HKIN 298 Biomechanics Assignment 2
Biomechanical Properties of Tissues Questions (21.5 marks)
*Please type your answers into the MS Word doc in BOLD
Due in Class February 9th at 9:30am, 2023
Name: ________Ava Krepp_____________
1. What is the difference between stress and strain? (1 mark)
Stress is the external force acting to deform a material. For a uniformly distributed load,
stress material is equal to the magnitude of the of the applied load divided by the cross-sectional
area upon which it acts. Strain is the resulting magnitude of deformation as a result of the applied
stress. Strain is the percentage of the change in length due to the applied stress.
2. Although we didn’t cover it in class, the concept of coefficient of restitution is related to stress-strain, in
particular, the elastic components of tissues. Review the concept of the coefficient of restitution in McLester
pp.125-126
Review this short video: https://www.youtube.com/watch?v=YOfJ2as3Pp4
Sports like volleyball and basketball use balls inflated to a standardized pressure. In light of this integrate
your understanding of the coefficient of restitution and explain why this is an important factor for sporting
performance in these sports (2 marks)
The coefficient of restitution is a parameter observed after reformation that indicates the
ability of an object to return to its original shape after deformation. In sports, it can be calculated by
dropping a ball from a known height and observing the rebound. It can also be calculated by using
the change in velocities of the objects involved in the collision. For example, hitting a volleyball or
dribbling a basketball, the collision would be with a particular sports implement, such as the hitter’s
or dribblers hand. The hand and the ball each have velocities before impact. After impact the
velocities will have decreased because of the energy lost on collision. The higher the coefficient of
restitution, the less energy lost at impact. Furthermore, the higher the coefficient of restitution, the
greater the energy remaining after impact in the form of either bounce height or velocity. That being
said, if volleyballs or basketballs were not inflated to a standardized pressure, it could create an
unfair advantage. For example, if a ball were pumped with a higher standardized pressure giving it a
higher coefficient of restitution, it would return to its original shape very quickly after deformation,
loose less energy on impact and have a higher energy remaining after impact.
3. Tendons and ligaments are made of similar materials yet tendons handle tensile stresses better than
ligaments. Why is this? (1 mark)
Tendons handle tensile stresses better than ligaments because tendons contain a higher
percentage of collagen and a lower percentage of ground substance than ligaments. Furthermore,
the collagen in tendons is more highly oriented than in ligaments, with fibers near all oriented along
the long axis of the tendon. Collagen is also the strongest fibrous protein allowing the tendon to
handle more stress.
4. What is the stress if a 450N force is applied to an area of 0.42m2? Show your work (1 mark)
𝝈=
𝑭
𝑨
= 450N/ 0.42m2
= 1071.43 N/m2
5. Calculate the strain if the original length of a tissue is 3.9 cm and the final length is 4.2 cm. Show your
work (1 mark)
∑=
𝚫𝐥
𝒍𝒊
= (lfinal – linitial) / li
= (4.2cm - 3.9cm) / 3.9cm
= 0.0769
= 7.7%
6. A tendon has a tensile stress applied to it (30N over 0.4 m2). The initial length of the tendon is 2.4 cm
and the final length is 2.6 cm). Calculate the following (show your work):
a. The resulting strain (1 mark)
∑=
𝚫𝐥
𝒍𝒊
= (lfinal – linitial) / li
= (2.6cm – 2.4cm) / 2.4cm
= 0.083
= 8.3%
b. Young’s modulus (1 mark)
𝑭
𝝈=
𝑨
= 30N/0.4m2
= 75N/m2
𝚬=
𝝈
𝜺
= (75N/m2) / (0.083)
= 903.6N/m2
7. Review the definition and description of Young’s modulus (also known as the elastic modulus). Now
consider the following tissues: Tissue A has a Young’s modulus of 3.0 and Tissue B has a Young’s
modulus of 6.2. Which of the tissues would be considered more stiff? Give a reason for your answer (1.5
mark)
Tissue B would be considered stiffer than Tissue A. Young’s modulus is the linear slope of
the stress/strain (elastic) graph. As such, the steeper the slope (the higher value of Young’s
modulus) the greater the stiffness of the tissue and the shallower the slope (the lower the value of
Young’s modulus) the greater the ductility of the material.
8. Young’s modulus of a specific tissue remains constant at 1.7 for a few seconds, but at a certain point
Young’s modulus decreases to 0.4 and remains at this level over time. Explain what you think happened
here, based on your knowledge of stress-strain relationships. (1 mark)
Young’s modulus is the slope of the stress/strain relationship. In this case, the initial slope
value is 1.7 meaning the stress would equal 0.425 and the stress would be 0.25 because
0.425/0.25=1.7. Therefore, a decreased value for young’s modulus (from 1.7 to 0.4) would mean that
the strain would have increased. Namely, if the slope value is now 0.4 then the strain would have
increased to 1.0625, while the stress stays constant at 0.425 because 0.425/1.0625=0.4.
9. Silly Putty and Plasticine: rates of loading and tissue responses
To answer the following questions refer to the following short videos and documents posted:



Video: Rapid Compression – Silly Putty and Plasticine
Video: Silly Putty – Different rates of tensile loading
Document: ‘Timecourses for Silly Putty and Plasticine under gravitational loading’
(a.) Which of the tissues is more elastic – silly putty or plasticine? Explain. (1 mark)
The silly putty is more elastic because it remained in the elastic region whereas the
plasticine made it into the plastic region when the same rapid, strong compressive force was
applied to it. When the silly puddy was slammed with the textbook, it very quickly returned to its
original shape unlike the plasticine which changed dramatically and remained deformed even after
the compressive stress was removed.
(b.) Which of the tissues is more viscous – silly putty or plasticine? Explain. (1 mark)
Silly putty is also more viscous than plasticine. A viscous property resists shear force.
When the silly putty and the plasticine were exposed to the same rapid, high magnitude
compressive force the silly putty was able to resist the force and maintain its original formation,
whereas the plasticine was unable to resist the force and became deformed after the removal of the
stress.
(c.) Describe how fast and slow rates of loading of both Silly Putty and Plasticine affect their response
characteristics. Recall that loading can mean both compression and tensile (2 marks)
The fast rate of high magnitude compressive loading affected the plasticine, much more
than it did the silly putty in that the plasticine was permanently deformed after the removal of the
stress, and the silly putty changed its shape but quickly ‘recoiled’ back to its original shape. Thus,
when plasticine had a fast rate of compressive loading it didn’t take a lot of strain and loading for it
to become deformed.
When the fast rate tensile stress was applied to the silly putty it very quickly passed its
yield point and was permanently deformed. Yet, when a slow rate tensile stress was applied to the
silly putty it took longer to reach the yield point and deform it after the stress was removed.
Therefore, under a compressive force the silly putty was very tough and the plasticine very ductile
and under a slow tensile force the silly putty had a greater slope and thus greater stiffness, and
with a fast tensile force the silly putty had a shallower slope and thus greater ductility.
10. Examine the following two graphs (hint: check the numerical data). Each of which involves a tissue
undergoing tensile stress. Which of the following tissues would return more strain energy to the moving
system? Explain your answer. (2 marks)
Tissue B Stress-Strain graph
30
0,5
25
0,4
Stress N/m2
STRESS (N/m2)
Tissue A Stress-Strain graph
20
15
10
0,3
0,2
0,1
5
0
0
0
1
2
3
STRAIN (change in length/initial Length)
0
2
4
6
8
10
Strain (Change in length/initial length)
a
Tissue A:
Tissue B:
Strain = 2.5
Stress = 27 N/m2
Strain = 9
Stress = 0.45 N/m2
Energy absorbed = area under the graph
= (27 x 2.5) / 2
= 33.75
Energy absorbed = area under the graph
= (0.45 x 9)/2
= 2.025
Strain energy of the amount of energy absorbed by the tissue as a result of the deformation
and it can be calculated by determining the area underneath the graph. However, when simply
observing the graphs, they appear to have a similar slope and area underneath the graph. As such,
you must calculate the area using the numerical data. Upon doing so, it shows that Tissue A had a
higher area under the graph (33.75) and thus absorbed much more strain energy than Tissue B,
which had a lower value area underneath the graph (2.025).
11. Open the Excel Data file: Stress-Strain Student.
These data are for Silly Putty that has undergone both a slow loading (low force) (Silly Putty Tensile 1) and
Silly Putty that has undergone rapid loading (high force) (Silly Putty Tensile 2). You can access each of
these datasets by clicking on the tabs in the lower left hand corner. The default will be Silly Putty Tensile 1.

For each dataset you will see columns showing Time, Length, Strain and Stress. Note also the
original length of the tissue at 4.0 cm.

You will need to calculate the Strain of the tissue. To do this enter the formula into the cell C4:
=(B4-4)/4. After you have done this, click on cell C4, and copy and paste down the entire column
(see screenshots).

Review the video on Basic Excel Graphing and create Stress-Strain graphs for both Silly Putty
Tensile 1 and Silly Putty Tensile 2 data. Copy and attach these graphs at the end of your
assignment (ensure they are labeled) – 2 marks for graphs
Answer the following questions based on these graphs:
(a.) Describe the mechanical properties of Silly Putty under each condition. Refer to terms like stiff, ductile,
etc. (2 marks)
When comparing the stress-strain graphs, the graph for the slow loading (low force) Silly
Putty Tensile 1 had a much steeper slope than the rapid loading (high force) Silly Putty Tensile 2
graph. Thus, because the curve of the Silly Putty Tensile 1 graph had a steeper slope, we know that
it is stiffer, more resistant to change with a lower ductility, higher strength, and lower toughness.
We also know that because the Silly Putty Tensile 2 graph, had a gradual slope, it has a greater
ductility, or relents more easily to loading, with a lower strength and moderate toughness.
(b.) At what level of strain do you see the yield point for each scenario? (1 mark)
Silly Putty Tensile 1 yield point: stress – 47 N/m2 , strain – 0.20
Silly Putty Tensile 2 yield point: stress – 55 N/m2 , strain – 0.0010
Attach Silly Putty Graphs here:
Silly Putty Tensile 1
60
55
50
45
Stress (N/cm2)
40
35
30
25
20
15
10
5
0
0,00
0,05
0,10
0,15
0,20
0,25
0,30
0,35
0,40
0,45
0,50
0,55
0,60
Strain
Silly Putty Tensile 2
60
55
50
Stress (N/cm2)
45
40
35
30
25
20
15
10
5
0,002
0,003
0,004
0,005
0,006
0,007
Strain
0,008
0,009
0,010
0,011
0,012