BMES 345 CH02 Problem Set v20220329(1)
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BMES 345 PROBLEM SET – CHAPTER 2. INTRODUCTION TO STRESS AND STRAIN
Contents
Problem 2.1: Normal Stress and Strain
.......................................................................................................................
2
Problem 2.2: Normal Stress and Strain
.......................................................................................................................
3
Problem 2.3: Normal Stress and Strain
.......................................................................................................................
4
Problem 2.4: Shear Stress and Strain
.........................................................................................................................
5
Problem 2.5: Shear Stress and Strain
.........................................................................................................................
6
Problem 2.6: Normal Stress and Strain
.......................................................................................................................
7
[Solution] Problem 2.1
....................................................................................................................................................
8
[Solution] Problem 2.2
....................................................................................................................................................
9
[Solution] Problem 2.3
...................................................................................................................................................
10
[Solution] Problem 2.4
...................................................................................................................................................
12
[Solution] Problem 2.5
...................................................................................................................................................
13
[Solution] Problem 2.6
...................................................................................................................................................
14
1
Problem 2.1: Normal Stress and Strain
As part of an experiment to study osteoarthritis, the meniscus was removed (meniscectomy) from the knees of dogs (hereafter referred to as “treated”). After 12 weeks, articular cartilage samples from the femoral condyles of the treated dogs were harvested, along with cartilage from control dogs with normal menisci. The articular cartilage samples were cut into a traditional “dog bone” shape for tensile testing, as shown on the left. The cross-section of the samples (not pictured) was rectangular, with a width of 2 mm and a thickness of 0.4 mm. The samples of knee articular cartilage were subjected to tensile testing and pulled to specific lengths, as noted in the table. The resulting forces were measured and recorded in the table below. Using the data provided, complete the columns for stress and strain. Please show your work. L (mm)
Strain
Force (N)
Control
Stress (MPa)
Control
Force (N)
Treated
Stress (MPa)
Treated
5.00
0
0.0
0
0.0
0
5.50
0.1
1.8
2.25
1.0
1.25
5.55
0.11
2.2
2.75
1.2
1.5
5.60
0.12
2.5
3.125
1.4
1.75
5.65
0.13
2.9
3.625
1.6
2
5.70
0.14
3.2
4
1.8
2.25
5.75
0.15
3.6
4.5
2.1
2.625
We know:
Strain
ℇ
=
∆L
L
0
Stress
σ
=
F
A
where F = force
where
L = change in length (final -initial)
A = area L
0
= original length
Therefore: Area = 2 mm * 0.4 mm = 0.8 mm
@ 5.50 mm
Strain
ℇ
=
∆L
L
0
= 5.50
mm
−
5.00
mm
5.00
mm
=
0.5
mm
5.00
mm
=
0.01
Control @ 1.8 N
Stress
σ
=
F
A
=
1.8
N
0.8
mm
=
2.25
MPa
Treated @ 1.0 N
Stress
σ
=
F
A
=
1.0
N
0.8
mm
=
1.25
MPa
2
3
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Problem 2.2: Normal Stress and Strain
Consider the cross-sections of the humeri from an adult tennis player (adapted from Jones et al., Journal of Bone and Joint Surgery
1977), assuming the cross-sections are annular (i.e., the humerus is a hollow circular cylinder):
Dimension
Non-playing arm
Playing arm
Outer diameter (cm)
2.195
2.45
Inner diameter (cm)
1.10
0.975
a)
For an applied axial force of 750 N, what is the normal stress in each humerus?
b)
How much more force
(in terms of percent) can the humerus in the playing arm bear while still experiencing the same stress
as the non-playing arm?
c)
Using your basic understanding of normal stress, can you explain the difference in cross-sectional dimensions?
4
Problem 2.3: Normal Stress and Strain
Consider the forearm in 90
o
flexion and holding a ball, as shown.
The mass of the forearm is 1.1 kg, and the diameter of the biceps tendon is 3 mm. a)
Determine the stress in the biceps tendon as a function of the weight of the ball. State your assumptions. b)
What is the stress generated in the biceps tendon when holding a 2 kg ball?
c)
If the biceps tendon elongates by 4 mm while lifting the ball, and the resulting strain in the tendon is 0.04, what is the original length of the tendon? What is the stretch ratio of the tendon under these conditions? 5
Problem 2.4: Shear Stress and Strain
A sample of porcine brain tissue (used as an analogue for human brain) is subjected to pure shear testing, as pictured below. A slab of tissue is sandwiched between two rigid plates. The bottom plate remains stationary, while the top plate moves when the tissue is loaded by a shear force V, and the displacement of the top plate is measured. For a shear force V = 513 µN, the top plate displaces 45 µm. Answer the following questions: a)
What is the shear strain of the brain tissue sample?
b)
What is the shear stress of the brain tissue sample?
6
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Problem 2.5: Shear Stress and Strain
The lining of blood vessels, called the endothelium, is made up of cells called endothelial cells. These cells are
constantly subjected to shear forces generated by the fluid flow in the circulatory system. Consider an endothelial cell in a coronary artery, as pictured (side view):
The cell has a surface area of 440 µm
2
, a height of 2 µm and a length of 30 µm. The shear force applied to the
cell surface by blood flow is V = 5.28 nN (1 nN = 10
-9
N). Using microscopy, we are able to measure that the apical (upper) surface of the endothelial cell displaces 0.045 µm in the direction of blood flow. Answer the following questions (state all assumptions):
a)
What is the applied shear stress?
shear stress = τ
=
V
A
where V = shear force
A = cross-sectional area
surface area = 440 µm
2
440 * 10
-12
m
2
τ
=
V
A
=
5.28
nN
440
μm
2
=
5.28
∗
10
−
9
N
440
∗
10
−
12
m
2
=
0.012
∗
10
−
3
=
12
Pa
b)
What is the resulting shear strain?
Assumption(s):
Small deformations
We know:
shear strain = γ
=
tan
θ
=
x
h
=
0.045
μm
2
μm
=
0.0225
*Recall: tangent
opposite/adjacent
7
Problem 2.6: Normal Stress and Strain
In a 2008 study (Couppé et al, J Appl Physiol
), differences in patellar tendon cross-sectional area (CSA) of the lead and nonlead legs were quantified in 22 badminton and fencing athletes. In both sports, athletes frequently lunge forward and do so with a dominant leg. Here are some of the results:
Lead Leg
Nonlead Leg
Force (N)
6,886 ± 883
5,832 ± 631
Cross-sectional area (mm
2
)
106 ± 7
83 ± 4
Normal stress (MPa)
Please answer the following questions:
a)
What are the average normal stresses in the patellar tendon of the lead and nonlead legs?
Stress
σ
=
F
A
Lead:
Nonlead:
σ
=
F
A
=
6886
N
106
m m
2
=
64.96
MPa
σ
=
F
A
=
5832
N
83
mm
2
=
70.27
MPa
b)
What assumptions did you make to calculate the normal stresses in part a?
1.
Uniaxial loading
2.
Homogenous material
3.
Prismatic bar
constant cross-sectional area of leg c)
Why do you think the cross-sectional area of the patellar tendon is greater in the lead leg?
The cross-sectional area of the lead patellar tendon is greater because the lead would need to use the patellar tendon more to adjust to the more force placed on it.
8
[Solution] Problem 2.1
As part of an experiment to study osteoarthritis, the meniscus was removed (meniscectomy) from the knees of dogs (hereafter referred to as “treated”). After 12 weeks, articular cartilage samples from the femoral condyles of the treated dogs were harvested, along with cartilage from control dogs with normal menisci. The articular cartilage samples were cut into a traditional “dog bone” shape for tensile testing, as shown on the left. The cross-section of the samples (not pictured) was rectangular, with a width of 2 mm and a thickness of 0.4 mm. The samples of knee articular cartilage were subjected to tensile testing and pulled to specific lengths, as noted in the table. The resulting forces were measured and recorded in the table below. Using the data provided, complete the columns for stress and strain. Please show your work. L (mm)
Strain
Force (N)
Control
Stress (MPa)
Control
Force (N)
Treated
Stress (MPa)
Treated
5.00
0
0.0
0
0.0
0
5.50
0.10
1.8
2.25
1.0
1.2
5.55
0.11
2.2
2.75
1.2
1.5
5.60
0.12
2.5
3.125
1.4
1.75
5.65
0.13
2.9
3.625
1.6
2
5.70
0.14
3.2
4
1.8
2.25
5.75
0.15
3.6
4.5
2.1
2.625
First we assume that the cartilage is homogeneous and the sample is a prismatic bar undergoing uniaxial tension.
To complete this table, you need to calculate the cross-sectional area A, and use our equations for normal stress and normal strain:
A
=
(
2
mm
) (
0.4
mm
)
=
0.8
mm
2
σ
=
F
A
=
F
0.8
mm
2
=
? MPa
ε
=
δ
L
=
L
−
5.00
mm
5.00
mm
=
?
9
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[Solution] Problem 2.2
Consider the cross-sections of the humeri from an adult tennis player (adapted from Jones et al., Journal of Bone and Joint Surgery
1977), assuming the cross-sections are annular (i.e., the humerus is a hollow circular cylinder):
Dimension
Non-playing arm
Playing arm
Outer diameter (cm)
2.195
2.45
Inner diameter (cm)
1.10
0.975
a)
For an applied axial force of 750 N, what is the normal stress in each humerus?
First we assume the humerus is a prismatic bar with an annular cross-section, that it undergoes uniaxial loading, and that it is made of a homogeneous material.
σ
non
−
playing
=
F
non
−
playing
A
non
−
playing
=
750
N
π
(
(
0.02195
m
2
)
2
−
(
0.011
m
2
)
2
)
=
2.65
MPa
σ
playing
=
F
playing
A
playing
=
750
N
π
(
(
0.0245
m
2
)
2
−
(
0.00975
m
2
)
2
)
=
1.89
MPa
b)
How much more force
can the humerus in the playing arm bear while still experiencing the same stress
as the non-playing arm?
σ
=
F
non
−
playing
A
non
−
playing
=
F
playing
A
playing
F
playing
F
non
−
playing
=
A
playing
A
non
−
playing
=
π
(
(
0.0245
m
2
)
2
−
(
0.00975
m
2
)
2
)
π
(
(
0.02195
m
2
)
2
−
(
0.011
m
2
)
2
)
=
1.4
The playing arm can bear 40% more force
than the non-playing arm for the same stress.
c)
Using your basic understanding of stress, can you explain the difference in cross-sectional dimensions?
Since normal stress is axial force divided by cross-sectional area, the humerus remodels and increases its cross-sectional area in the playing arm. Thus, while the forces experienced in the playing
arm are typically higher than in the non-playing arm, the increased cross-sectional area maintains a similar stress
. 10
[Solution] Problem 2.3
Consider the forearm in 90
o
flexion and holding a ball, as shown.
The mass of the forearm is 1.1 kg, and the diameter of the biceps tendon is 3 mm. a)
Determine the stress in the biceps tendon as a function of the weight of the ball. State your assumptions. Assume that the arm is in equilibrium, and that the biceps tendon inserts perpendicular to the forearm.
∑
M
=
0
=
0.04
F
biceps
−
0.15
(
1.1
kg
)
(
9.81
m s
−
2
)
−
0.35
F
ball
F
biceps
=
1.62
Nm
−
0.35
F
ball
0.04
m
σ
biceps
=
F
biceps
A
=
1.62
Nm
+
0.35
F
ball
(
0.04
m
)
π
(
0.003
m
2
)
2
=
5729578
Pa
+
(
1237872
m
−
2
)
F
ball
b)
What is the stress generated in the biceps tendon when holding a 2 kg ball?
σ
biceps
=
5729578
Pa
+
(
1237872
m
−
2
)
(
2
kg
)
(
9.81
m s
−
2
)
=
30
MPa
11
0.35 m
0.04 m
0.15 m
F
ball
W
arm
c)
If the biceps tendon elongates by 4 mm while lifting the ball, and the resulting strain in the tendon is 0.04, what is the original length of the tendon? What is the stretch ratio of the tendon under these conditions? ε
=
δ
L
→L
=
δ
ε
=
4
mm
0.04
=
100
mm
=
10
cm
λ
=
1
+
ε
=
1
+
0.04
=
1.04
12
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[Solution] Problem 2.4
A sample of porcine brain tissue (used as an analogue for human brain) is subjected to pure shear testing, as pictured below. A slab of tissue is sandwiched between two rigid plates. The bottom plate remains stationary, while the top plate moves when the tissue is loaded by a shear force V, and the displacement of the top plate is measured. For a shear force V = 513 µN, the top plate displaces 45 µm. Answer the following questions: a)
What is the shear strain of the brain tissue sample?
The situation depicted is a case of simple shear. Since the displacement of the top plate is relatively small, we are going to analyze it as if the entire block is experiencing pure shear. The shear strain is the change in angle between two adjacent sides, which can be calculated using the geometry shown in the side view (right image):
tan
γ
=
x
L
=
45
μm
1
mm
=
0.045
mm
1
mm
=
0.045
γ
=
tan
−
1
(
0.045
)
=
0.04497
≈
0.045
Alternatively, we can observe that the deformation is sufficiently small to assume small strains / deformations
, so we can approximate our shear strain:
γ
≅
tan
γ
=
0.045
b)
What is the shear stress of the brain tissue sample?
We will assume that the brain tissue is homogeneous. The average shear stress can be calculated using the applied shear force V over the area that it is acting upon, which is the top surface of the plate:
τ
average
=
V
A
=
513
μN
(
3
mm
)(
5
mm
)
=
513
x
10
−
6
N
(
0.003
m
)(
0.005
m
)
=
34.2
Pa
Note that this is an approximation; the shear stress will not be completely uniform throughout the block of brain tissue in this configuration.
13
[Solution] Problem 2.5
The lining of blood vessels, called the endothelium, is made up of cells called endothelial cells. These cells are
constantly subjected to shear forces generated by the fluid flow in the circulatory system. Consider an endothelial cell in a coronary artery, as pictured (side view):
The cell has a surface area of 440 µm
2
, a height of 2 µm and a length of 30 µm. The shear force applied to the
cell surface by blood flow is V = 5.28 nN (1 nN = 10
-9
N). Using microscopy, we are able to measure that the apical (upper) surface of the endothelial cell displaces 0.045 µm in the direction of blood flow. Answer the following questions (state all assumptions):
a)
What is the applied shear stress?
τ
=
V
A
=
5.28
x
10
−
9
N
440
x
10
−
12
m
=
12
Pa
b)
What is the resulting shear strain?
Since the upper surface displaces 0.045 µm, we can calculate the shear strain using the tangent of the resulting angle (assuming small strains and therefore small angles):
γ ≈
tan
γ
=
x
h
=
0.045
μm
2
μm
=
0.0225
14
[Solution] Problem 2.6
In a 2008 study (Couppé et al, J Appl Physiol
), differences in patellar tendon cross-sectional area (CSA) of the lead and nonlead legs were quantified in 22 badminton and fencing athletes. In both sports, athletes frequently lunge forward and do so with a dominant leg. Here are some of the results:
Lead Leg
Nonlead Leg
Force (N)
6,886 ± 883
5,832 ± 631
Cross-sectional area (mm
2
)
106 ± 7
83 ± 4
Normal stress (MPa)
Please answer the following questions:
a)
What are the average normal stresses in the patellar tendon of the lead and nonlead legs?
We can calculate the normal stress using the equation:
σ
=
F
A
Using the mean values for force and cross-sectional area provided, we find that:
σ
lead
=
6886
N
106
mm
2
=
65.0
N
σ
nonlead
=
5832
N
83
mm
2
=
70.3
N
b)
What assumptions did you make to calculate the normal stresses in part a?
To compute the normal stresses using the simple equation in part a, we need to make several key assumptions:
The patellar tendon has a constant cross-sectional area, a.k.a. it is a prismatic bar
Force is applied normal to the cross-sectional area, a.k.a. uniaxial force
Homogeneous material
Uniformly applied force
c)
Why do you think the cross-sectional area of the patellar tendon is greater in the lead leg?
Tendons and ligaments have been shown to adapt to differences in loading by remodeling their size, i.e., increasing their cross-sectional area in response to higher than expected loading and decreasing their cross-sectional area in response to lower than expected loading. 15
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- Write the definitions for engineering stress, true stress,engineering strain, and true strain for loading along a singleaxis.arrow_forwardHow is the nominal or engineering strain found directly from the strain gage?arrow_forwardTrue stress (MPa) 8-7 Figure 8-21 is a plot of true stress versus true strain for a metal. For total imposed strains of 0.1, 0.2, 0.3 and 0.4, determine the elastic and plastic components of the strain. The modulus of elasticity of the metal is 100 GPa. 600 500 400 300 200 100 0.1 0.2 0.3 0.4 0.5 True strain Figure 8-21 A true stress versus true strain curve for a metal (for Problem 8-7).arrow_forward
- Which one of the following is the correct definition of ultimate tensile strength, as derived from the results of a tensile test on a metal specimen: O the stress encountered when the stress strain curve transforms from elastic to plastic behavior the maximum load divided by the final area of the specimen the maximum load divided by the original area of the specimen O the stress observed when the specimen finally failsarrow_forwardName the 7 different points on a stress-strain diagram?arrow_forwardQuestion 4 You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.1 plastic strain, ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= ✓ MPa. Hence, we can predict that we need a prior plastic strain of approximately ✓to obtain a hgth of 120 MPa True Stress (MPa) 250 200 100 85 95 50 105 115 125 135 145 300 350 0.12 0.14 150 0.16 0.18 155 165 175 200 250 0.1 True Stress (MPa) Ep = 0.1 Zoomed version of left plotarrow_forward
- Question 4 Save Answer You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is ✓ MPa for 0.1 plastic strain, proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= MPa. Hence, we can predict that we need a prior plastic strain of approximately True Stress (MPa) True Stress (MPa) True Stress (MPa) 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 Ep = 0.1 0.01 Ep=0.2 0.01 Ep=0.3 0.01 0.02 True Strain 0.02 0.03 0.02 0.03 True Strain 0.03 0.04 0.04 0.04 True Stress (MPa) 0.05 0.000 0.001 True Stress (MPa) Ep=0.1 True…arrow_forwardDistinguish between the Normal and Shear Stress Components?arrow_forwardUnder which of the conditions below is the generalized Hooke’s law applicable for use in analysis? a. When the material distorts due to any of the normal stresses b. When the material has a negative value for Poisson’s ratio c. When the material exceeds its proportional limit but still remains elastic d. When the material’s cross-section changes along one direction Which of the following statements is false about the Poisson’s ratio? a. It is a dimensionless quantity b. It relates the modulus of elasticity E and the modulus of rigidity G c. It has a single value for isotropic and homogeneous materials d. Its magnitude is equal to the negative ratio of the longitudinal strain to the lateral strain.arrow_forward
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