BMES 345 CH05 Problem Set 20220927
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BMES 345 PROBLEM SET – CHAPTER 5. AXIAL LOADING
Contents
Problem 5.1: Axial Loading
............................................................................................................................................
3
Problem 5.2: Non-Uniform Axial Loading
...................................................................................................................
4
Problem 5.3: Non-Uniform Axial Loading
...................................................................................................................
5
Problem 5.4: Non-Uniform Axial Loading
...................................................................................................................
6
Problem 5.5: Non-Uniform Axial Loading
...................................................................................................................
7
Problem 5.6: Non-Uniform Axial Loading
...................................................................................................................
8
Problem 5.7: Non-Uniform Axial Loading
...................................................................................................................
9
Problem 5.8: Non-Uniform Axial Loading
.................................................................................................................
10
Problem 5.9: Non-Uniform Axial Loading
.................................................................................................................
11
Problem 5.10: Statically Indeterminate Axially Loaded Systems
.......................................................................
12
Problem 5.11: Statically Indeterminate Axially Loaded Systems
.......................................................................
13
Problem 5.12: Statically Indeterminate Axially Loaded Systems
.......................................................................
14
Problem 5.13: Statically Indeterminate Axially Loaded Systems
.......................................................................
15
Problem 5.14: Statically Indeterminate Axially Loaded Systems
.......................................................................
16
Problem 5.15: Axial Loading
........................................................................................................................................
17
Problem 5.16: Non-Uniform Axial Loading
..............................................................................................................
18
Problem 5.17: Non-Uniform Axial Loading
..............................................................................................................
19
Problem 5.18: Non-Uniform Axial Loading
..............................................................................................................
20
Problem 5.19: Non-Uniform Axial Loading
..............................................................................................................
21
[Solution] Problem 5.1
...................................................................................................................................................
22
[Solution] Problem 5.2
...................................................................................................................................................
23
[Solution] Problem 5.3
...................................................................................................................................................
25
[Solution] Problem 5.4
...................................................................................................................................................
27
[Solution] Problem 5.5
...................................................................................................................................................
28
[Solution] Problem 5.6
...................................................................................................................................................
30
[Solution] Problem 5.7
...................................................................................................................................................
32
[Solution] Problem 5.8
...................................................................................................................................................
33
[Solution] Problem 5.9
...................................................................................................................................................
35
[Solution] Problem 5.10
................................................................................................................................................
37
[Solution] Problem 5.11
................................................................................................................................................
40
1
[Solution] Problem 5.12
................................................................................................................................................
42
[Solution] Problem 5.13
................................................................................................................................................
45
[Solution] Problem 5.14
................................................................................................................................................
48
[Solution] Problem 5.15
................................................................................................................................................
50
[Solution] Problem 5.16
................................................................................................................................................
53
[Solution] Problem 5.17
................................................................................................................................................
55
[Solution] Problem 5.18
................................................................................................................................................
56
[Solution] Problem 5.19
................................................................................................................................................
58
2
Problem 5.1: Axial Loading
The human Achilles tendon has an average length of 150 mm, an average cross-sectional area of 45 mm
2
, and an effective elastic modulus of 800 MPa. Stating your assumptions, determine the following:
1.
Prismatic bar
2.
Uniaxial force
3.
Homogenous, Linear elastic material
a)
Stiffness of the Achilles tendon
We know: stress
=
σ
=
F
A
=
Eε
F
A
=
E
δ
L
0
F
=
EA
L
0
δ
k
=
EA
L
0
=
(
800
MPa
)(
45
mm
2
)
150
mm
=
240
N mm
−
1
b)
Axial rigidity of the Achilles tendon
EA
=
(
800
MPa
)
(
45
mm
2
)
=
36,000
MPamm
2
=
36,000
N
c)
Force required to cause 6 mm of elongation
F
=
EA
L
0
δ
F
=
(
800
MPa
)(
45
mm
2
)
(
150
mm
)
(
6
mm
)
=
(
240
N mm
−
1
)
(
6
mm
)
=
1,440
N
3
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Problem 5.2: Non-Uniform Axial Loading
Consider the portion of the cervical spine known as C3-
C6, consisting of four cervical vertebrae and three
interposing intervertebral discs (figure on the right).
Assume you can model each vertebral body and
intervertebral disc as prismatic bars made of linear elastic
materials. Each vertebral body is made of cortical bone
(
E = 15 GPa) and has a height of 15 mm. Each
intervertebral disc has a height of 9 mm and E = 980 kPa.
Both vertebrae and intervertebral discs have cross-
sectional areas of 750 mm
2
. Under a compressive load F
= 50 N (roughly equivalent to the weight of a human
head), what is the total deformation (shortening) of the
C3-C6 segment of the cervical spine?
4
Problem 5.3: Non-Uniform Axial Loading
Synthetic biomaterial scaffolds are being developed to induce bone regeneration in critical size defects (defects that are too large for normal healing to occur). In this example, the scaffold fills a 1.5
cm gap in the tibia. In the diagram, in addition to the axial compressive loads, there is an intermediate tensile load applied by the patellar ligament (transmitting the quadriceps muscle force): Assume the bone has a circular cross-sectional area of A
= 480 mm
2
. The elastic modulus of the tibia
is E = 15 GPa, and the elastic modulus of the space filler is E = 3.5 GPa. The force applied by the ankle joint is 1.5 kN. The force applied by the knee is 2.2 kN. Determine the total change in length of
the tibia.
Assumptions:
1.
isotropic, homogeneous, linear elastic materials
2.
segment experiences uniaxial loading
3.
segments are in prismatic bars with same cross-sectional area
F
ligament
=
F
knee
−
F
ankle
=
2.2
kN
−
1.5
kN
=
700
N
N
1
=
700
N
+
1500
N
=
2200
N
N
2
=
N
3
=
N
4
=
1500
N
δ
=
∑
i
=
1
4
F
i
L
i
E
i
A
=
(
(
2200
N
)
(
40
mm
)
(
15,000
MPa
)(
480
mm
2
)
+
(
1500
N
)
(
165
mm
)
(
15,000
MPa
)(
480
mm
2
)
+
(
1500
N
)
(
15
mm
)
(
3,500
MPa
)(
480
m m
2
)
+
(
1500
N
)
(
180
mm
)
(
15,000
MPa
)(
480
mm
2
)
)
The overall system is under compressive forces therefore the tibia’s length will experience a decrease
in overall length by 0.0975 mm (-0.0975 mm).
5
Problem 5.4: Non-Uniform Axial Loading
After fracture of the tibia, a callus forms around the fracture site, consisting of a mixture of woven bone and cartilage. Over time, the callus is replaced with normal, organized bone tissue (called lamellar bone) that is much stronger than the temporary callus. Let us consider a tibia at the early stages of the callus. The callus has a larger radius than the normal tibia, but is composed of weaker woven bone:
Let us assume that both the cortical bone of the normal tibia and the woven bone of the callus are linear elastic materials with E
cortical = 18 GPa and E
woven
= 0.5 GPa. The proximal and distal tibia are prismatic bars with cross-sectional areas A = 350 mm
2
. The callus is also a prismatic bar, with cross-
sectional area A = 720 mm
2
. The entire tibia is subjected to end-applied compressive forces F
= 500 N, as shown above. Based on these assumptions and parameters, answer the following questions: a)
What is the internal force in the callus?
b)
What is the change in length of the entire tibia?
6
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Problem 5.5: Non-Uniform Axial Loading
In total knee replacements (TKRs), the tibial plateau is replaced with a tibial component, consisting of a polyethylene (PE) insert on a titanium alloy tray. We can idealize a tibia with a TKR tibial component as a segmented bar under compressive loading:
If the entire tibia (including the plastic and metal tibial component) experiences a change in length of δ = -0.1 mm, what is the magnitude of the compressive force F
? The elastic modulus for the PE, titanium alloy, and bone are E
PE
= 0.9 GPa, E
Ti
= 90 GPa, and E
bone
= 15 GPa, respectively. Remember to show your work and state all relevant assumptions.
7
Problem 5.6: Non-Uniform Axial Loading
Ruptures of the Achilles tendon are typically repaired by suturing
together the two ruptured ends. We can model the repaired
tissue as a segmented bar under uniaxial tensile loading (pictured
on the right). Both the Achilles tendon and the sutures joining the two portions
of the Achilles tendon are modeled as prismatic bars made of
linear elastic materials, with the (undeformed) lengths given
above. The cross-sectional area of the tendon (for both portions)
is A = 55 mm
2
and the elastic modulus is E = 1177 MPa. The
total cross-sectional area of the sutures is A
= 0.46 mm
2
and the
elastic modulus is E = 3700 MPa. Even with limited activity after
surgery, the repaired Achilles tendon can be subjected to
significant loads. Under a tensile force F
, the entire surgically
repaired Achilles tendon stretches by δ =
0.59 mm. Given this
information, answer the following questions (remember to show your work and state assumptions):
a)
Based on the observed change in length, what is the uniaxial tensile force F applied to the repaired Achilles tendon?
b)
Draw a stress element representing the state of stress in the sutures. What are the principal stresses and corresponding principal angles for this state of stress?
8
Problem 5.7: Non-Uniform Axial Loading
Ruptures of the Achilles tendon are typically repaired by suturing together the two ruptured ends. We
can model the repaired tissue as a segmented bar under uniaxial tensile loading:
Both the Achilles tendon and the sutures joining the two portions of the Achilles tendon are modeled as prismatic bars made of linear elastic materials. The cross sectional area of the tendon (for both portions) is A = 58 mm
2
and the Young’s modulus is E = 820 MPa. The cross-sectional area of the sutures (made of polyester) is A
= 20 mm
2
and the Young’s modulus is E = 240 MPa.
Even with limited activity after surgical repair, the Achilles tendon can be subjected to significant loads. Under a tensile force of 1800 N, what is the change in length of the repaired Achilles tendon?
9
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Problem 5.8: Non-Uniform Axial Loading
Consider a bone fixation plate made of titanium alloy (
E = 90 GPa), subjected to intermediate axial loads as pictured:
The cross-sectional area of the plate is 16 mm
2 at the ends, and 8 mm
2
in the center portion (where the cut-out is). Based on the illustration provided, answer the following questions (remember to show
your work and state additional assumptions):
a)
What is the total change in length of the fixation plate under these loads?
b)
What is the maximum shear stress in the fixation plate and where does it occur?
10
Problem 5.9: Non-Uniform Axial Loading
Consider the following bone fixation plate made of Ti-6V-4Al (
E = 90 GPa), with a cross-sectional area of 60 mm
2
:
The plate is subjected to 7000 N loads (as pictured) at each location where a screw is inserted. Answer the following questions:
a)
What are the internal forces in each portion of the fixation plate?
b)
What is the deformed length of the fixation plate under the pictured loads?
11
Problem 5.10: Statically Indeterminate Axially Loaded Systems
A number of polymer biomaterials have been considered for
orthopedic applications, but the strength of these polymers is often
a concern. To address this deficiency, biomaterials researchers
have reinforced these polymers with much stronger materials,
such as carbon fibers. Consider the polymer polyether ether
ketone (PEEK) reinforced with carbon fibers. The carbon fibers
are aligned in the direction of loading, and make up 70% of the
total volume of the material (and therefore we can approximate
that carbon fibers are 70% of the cross-sectional area). PEEK has
a Young’s modulus E
= 5 GPa, while the carbon fibers have a
Young’s modulus E
= 150 GPa. Let us model a hip implant stem
made of carbon fiber-reinforced PEEK as a solid circular cylinder
with length L = 12 cm and a radius of 8 mm. During preliminary
mechanical testing, the stem is subjected to a uniaxial
compressive force of 2000 N. Answer the following questions:
Assumptions
1.
Isotropic, Homogenous, Linear elastic materials
2.
Uniaxial compression
3.
The two different components (PEEK, carbon fiber) can be approximated as prismatic bars
a)
What the internal forces in the PEEK and the carbon fibers?
Force equations:
N
PEEK
+
N
fibers
=
2000
N
There are more variables than equations, therefore the system is statically indeterminate and since the two systems are in direct contact then:
δ
PEEK
=
δ
fibers
N
PEEK
L
E
PEEK
(
0.3
A
)
=
N
fibers
L
E
fibers
(
0.7
A
)
N
fibers
=
(
0.7
) (
150
GPa
)
(
0.3
) (
5
GPa
)
N
PEEK
=
70
N
PEEK
N
PEEK
+
70
N
PEEK
=
2000
N
N
PEEK
=
28.17
N
N
fibers
=
1971.9
N
12
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b)
What are the axial normal strains in the PEEK and the carbon fibers?
δ
=
FL
EA
A
=
π r
2
=
π
∗(
0.008
)
2
=
0.000201
m
2
δ
PEEK
=
(
28.17
N
)(
0.12
m
)
(
5
x
10
9
Pa
)(
0.3
)(
0.000201
m
2
)
=
1.121
x
10
−
5
m
δ
fibers
=
(
1971.9
N
)(
0.12
m
)
(
150
x
10
9
Pa
)(
0.7
)(
0.000201
m
2
)
=
1.121
x
10
−
5
m
Normal strain (identical for the PEEK and the carbon fibers because δ
is the same):
ε
=
δ
L
=
1.121
x
10
−
5
m
0.12
m
=
9.342
x
10
−
5
13
Problem 5.11: Statically Indeterminate Axially Loaded Systems
Stress shielding is a phenomenon where bone mineral density decreases due to the presence of an implant that shifts the loading away from the bone. This is usually because the implant is made of a material with a significantly higher elastic modulus than bone tissue. Let us consider a simplified situation where the stem of a hip implant sits inside the proximal femur. The bone-implant hybrid consists of an outer shell of cortical bone (
E = 18 GPa, A = 530 mm
2
) and an
inner core of Ti-13Nb-13Zr (
E = 75 GPa, A = 430 mm
2
) that is press-fit inside the bone. Determine the relative amount of force that the cortical bone and titanium alloy stem will experience during uniaxial compressive loading. Compare this to the amount of loading that the cortical bone would experience if the implant stem was not present.
Assumptions:
1.
Homogeneous, Isotropic, Linear elastic materials
2.
The two compartments can be approximated as prismatic bars
3.
Uniaxial compression
4.
Well-bonded
Force Equations:
F
=
N
cortical
+
N
Ti
13
Nb
13
Zr
There are more unknowns than equations, therefore the system is statically indeterminate and since the two systems are in direct contact:
δ
cortical
=
δ
Ti
N
cortical
L
E
cortical
A
cortical
=
N
Ti
L
E
Ti
A
Ti
N
cortical
=
E
cortical
A
cortical
E
Ti
A
Ti
N
Ti
=
(
18
GPa
)
(
530
mm
2
)
(
75
GPa
)
(
430
mm
2
)
N
Ti
N
cortical
=
0.296
N
Ti
F
=
N
Ti
+
0.296
N
Ti
=
1.296
N
Ti
N
Ti
=
0.772
F
(77.2% of force is born by Ti)
N
cortical
=
0.228
F
(22.8% is born by cortical)
The cortical bone is experiencing about 23% of the load than normal.
14
Problem 5.12: Statically Indeterminate Axially Loaded Systems
Stress shielding in bone due to metallic orthopedic implants is not limited to joint replacements. The presence of an internal fixation plate on the femur can also cause significantly reduced loading on the
bone. Consider the simplified example below:
In this model of a fixation plate and a segment of the femur undergoing compressive axial loading (
F = 750 N), both the femur and the fixation plate are assumed to be prismatic bars, with cross-sectional
areas A
femur
= 350 mm
2
and A
plate
= 75 mm
2
. The length of the femur-fixation plate system is L
= 10 cm. The femur is made of cortical bone (
E = 15 GPa) and the plate is made of 316L stainless steel (
E
= 200 GPa). In our simplified model, we will not consider the effects of the screws used to keep the fixation plate in place. Based on these initial assumptions and the schematic provided, answer the following questions (remember to show your work and state additional assumptions):
a)
Show that this system is statically indeterminate. Write an appropriate equation of compatibility for this system.
b)
What are the internal forces (acting on the cross-sections) in the femur and the plate under these loading conditions?
c)
What is the maximum shear stress in the fixation plate and where does it occur?
d)
To prevent stress shielding, the femur should experiencing an axial compressive strain = 0.0001 under these loading conditions. Does at least that amount of strain occur?
15
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Problem 5.13: Statically Indeterminate Axially Loaded Systems
Consider a 15 cm long segment of the normal femur, and then the same femur where the cancellous bone has been removed and replaced with a titanium alloy hip implant, as illustrated (assume the femur is a prismatic bar with a circular cross-section of outer diameter 2.8 cm):
Assume that the dimensions between situation A and B are identical; the only change is the complete
replacement of the cancellous bone interior of the femur with the hip implant stem (made of Ti-6Al-
4V). Under an axial compressive load of 3000 N, answer the following questions.
a)
What are the normal stresses and normal strains in the cortical bone and cancellous bone in situation A?
b)
What are the normal stresses and normal strains in the cortical bone and Ti-6Al-4V implant stem in situation B?
16
Problem 5.14: Statically Indeterminate Axially Loaded Systems
Consider the following system, which consists of a bar with constant diameter of 80 mm, but made of two different materials. The bar is fixed between two immovable walls at A and C, and a load P
= 100
kN is applied at D. Source: Introductory Mechanics of Materials (via YouTube) (2017)
Answer the following questions:
a)
Show that the system is statically indeterminate.
A
x
+ C
x
- 100 kN = 0 Since there are more unknowns than equations, the system is statically indeterminate.
b)
Write an appropriate equation of compatibility that would allow you to solve this system.
The wall and bars are immovable, therefore:
δ
total
=
0
=
δ
AB
+
δ
BD
+
δ
DC
c)
What are the internal forces in the two bars (AB and BC)?
17
Problem 5.15: Axial Loading
The vertebral bodies in the lumbar spine are subjected to axial compressive forces during a variety of everyday activities. Consider a simple model of a vertebral body consisting of the following:
Two endplates of cortical bone (
E = 15 GPa), with cross-sectional area A = 283 mm
2
and length L
= 1 mm.
A centrum of trabecular bone (
E = 700 MPa) with cross-sectional area A = 226 mm
2
, surrounded by a cortical bone shell of cross-sectional area A = 57 mm
2
. The centrum and cortical shell have a length L = 26 mm. If the entire vertebral body is subjected to a uniaxial compressive force of 600 N, determine the following (remember to show your work and state assumptions): a)
Internal normal forces in the two endplates, the centrum, and the cortical shell (surrounding the
centrum)
b)
Total change in length of the vertebral body 18
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Problem 5.16: Non-Uniform Axial Loading
Fixation plates are sometimes used to keep bone fractures in union during healing. Consider a situation where a healing humerus is subjected to tension (e.g., when someone lifts something heavy). The fixation plate is modeled by the intermediate axial loads:
The normal portions of the humerus have E = 18 GPa. The callus has E = 5 GPa. The cross-
sectional area over the entire humerus is a relatively constant A = 710 mm
2
. Answer the following questions (remember to show your work):
Assumptions: -
System (bone and callus) are made of homogenous, isotropic, linear elastic materials
-
Each segment is a prismatic bar
-
Each segment experiences uniaxial loading
a)
What is the total change in length of the humerus?
b)
To keep the healing fracture in union, the central segment of the humerus must remain under a
small amount of compression. Is the healing fracture under compression in this case?
19
a)
δ
total
=
∑
(
FL
EA
)
i
The system can be broken into 5 segments based on the different lengths, however, segments 1 and 5 are the same, segments 2 and 4 are the same. Based on the pictures drawn above, N
1
= N
5
= 200 N (in tension), N
2
= N
4
= 40 N and N
3
= 40 N (both in compression).
δ
total
=
2
(
200
N
)(
105
mm
)
(
18,000
MPa
)(
710
mm
2
)
−
2
(
40
N
) (
40
mm
)
(
18,000
MPa
)
(
710
mm
2
)
−
(
40
N
) (
30
mm
)
(
5,000
MPa
)
(
710
mm
2
)
δ
total
=
2
(
0.001643
mm
)
−
2
(
0.000125
mm
)
−
0.000338
mm
=
0.002698
mm
**Because segments 1 = 5 and 2
= 4, we can just multiply 2
a)
To keep the healing fracture in union, the central segment of the humerus must remain under a
small amount of compression. Is the healing fracture under compression in this case?
The callus segment which is the central segment of the humerus is under 40 N of compression, therefore the fracture is healing. 20
Problem 5.17: Non-Uniform Axial Loading
Most hip implant stem components have a gradual taper. Let us consider a titanium alloy (
E = 100 GPa) stem component with L = 128 mm with a tapering rectangular cross-section. The width of the cross-section is a constant w = 7 mm, but the thickness t
tapers from 20 mm (at x = 0 mm) to 6 mm (at x = 128 mm). We can describe the cross-sectional area A(x) as a function of x
(position along the length of the stem):
A
(
x
)
=
w∙t
(
x
)
=−
0.76563
x
+
140
In this case, A(x)
is in units of mm
2
if x
is inputted in units of mm. If a uniaxial compressive
force F = 500 N is applied to the tapered stem, determine its total change in length. 21
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Problem 5.18: Non-Uniform Axial Loading
For large bone defects (e.g., after removal of a bone tumor), surgeons sometimes use fibular grafts to
fill the gap. The surgeon removes a large segment of the fibula (which has relatively limited weight bearing function in our lower leg), and transplants that segment into the bone needing repair.
Consider a situation where a fibular graft is used to fill a large defect in the humerus (upper arm bone). A simplified diagram of the system look as follows:
You know the following information:
The cross-sectional area of the humerus (for both humeral segments) is
A
humerus
= 315 mm
2
The cross-sectional area of the fibular graft is A
fibula
= 200 mm
2
F
biceps
= 400 N
F
elbow
= 1500 N
E
bone
= 20 GPa
The lengths are as shown on the diagram
Given this information, answer the following questions (remember to show your work):
a)
State any relevant assumptions.
b)
What is F
shoulder
?
c)
How many segments are there?
d)
What are the internal normal forces in each segment?
e)
What is the total change in length, in millimeters, of the entire system (make sure you use the correct sign)? 22
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Problem 5.19: Non-Uniform Axial Loading
Consider the following simplified model of a dental implant, which consists of a solid titanium cap on top of an exterior of porous tantalum surrounding a core of solid titanium:
In addition to the lengths shown on the diagram, you have the following information:
Total change in length δ
= -0.0015 mm
Elastic modulus of titanium E
Titanium
= 100 GPa
Elastic modulus of porous tantalum E
Tantalum
= 10 GPa
Cross-sectional area of the titanium cap = 12 mm
2
Cross-sectional area of the porous tantalum exterior = 7 mm
2
Cross-sectional area of the titanium core = 5 mm
2
Answer the following questions and remember to show your work:
a)
State your assumptions.
b)
Write an equation of equilibrium that shows that a portion of this system is statically indeterminate.
c)
Write an appropriate equation of compatibility for the statically indeterminate portion of the system.
d)
Determine the relationship between the internal normal forces in the porous tantalum exterior and the titanium core.
e)
Determine the applied force F necessary to cause the indicated total change in length δ = -
0.0015 mm.
23
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[Solution] Problem 5.1
The human Achilles tendon has an average length of 150 mm, an average cross-sectional area of 45 mm
2
, and an effective elastic modulus of 800 MPa. Stating your assumptions, determine the following:
For this problem, we will make the following assumptions
:
4.
Prismatic bar
5.
Uniaxial force
6.
Homogeneous material
7.
Linear elastic material
d)
Stiffness of the Achilles tendon
The stiffness, k,
is the proportionality constant relating force to displacement. If we use 1-D Hooke’s law, we can start with:
σ
=
Eε
Then we can substitute equations for stress and strain and eventually end up with:
F
A
=
E
δ
L
0
F
=
EA
L
0
δ
k
=
EA
L
0
=
(
800
MPa
)(
45
mm
2
)
150
mm
=
240
N mm
−
1
e)
Axial rigidity of the Achilles tendon
Axial rigidity is defined as the quantity EA:
EA
=
(
800
MPa
)
(
45
mm
2
)
=
36,000
MPamm
2
=
36,000
N
f)
Force required to cause 6 mm of elongation
Going back to the equations above, we can state:
F
=
EA
L
0
δ
F
=
(
800
MPa
)(
45
mm
2
)
(
150
mm
)
(
6
mm
)
=
(
240
N mm
−
1
)
(
6
mm
)
=
1,440
N
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[Solution] Problem 5.2
Consider the portion of the cervical spine known as C3-
C6, consisting of four cervical vertebrae and three
interposing intervertebral discs (figure on the right).
Assume you can model each vertebral body and
intervertebral disc as prismatic bars made of linear elastic
materials. Each vertebral body is made of cortical bone
(
E = 15 GPa) and has a height of 15 mm. Each
intervertebral disc has a height of 9 mm and E = 980 kPa.
Both vertebrae and intervertebral discs have cross-
sectional areas of 750 mm
2
. Under a compressive load F
= 50 N (roughly equivalent to the weight of a human
head), what is the total deformation (shortening) of the
C3-C6 segment of the cervical spine?
Assumptions
1.
Vertebral bodies and intervertebral discs are prismatic bars
2.
Linear elastic materials
3.
Isotropic materials
4.
Homogeneous materials
5.
Uniaxial compression – this is a significant assumption, since we know that the spine curves, and therefore the load is not perfectly uniaxial (it is actually more like combined compression and bending)
Based on the assumptions we made, we can use the equation for determining the deformation
of an axial bar that can be represented by multiple axial segments (in this case 7 different segments):
δ
=
∑
i
=
1
7
N
i
L
i
E
i
A
i
At this point we make two observations. First, since the cross-section area of the vertebral bodies and intervertebral discs are assumed to be identical, A
i
= A = 750 mm
2
. Second, since the only axial loads are applied at the end, we can see that the internal forces N
i
will be equal to F in every single segment. Thus, our equation becomes:
δ
=
F
A
∑
i
=
1
7
L
i
E
i
Since we have 4 identical vertebral bodies with the same L and E, and the 3 identical intervertebral discs, we can quickly determine the total shortening of the C3-C6 spine:
δ
=
50
N
750
mm
2
(
−
4
(
15
mm
)
15000
MPa
−
3
(
9
mm
)
0.98
MPa
)
=−
1.84
mm
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Note: Since this system is under compression, and each individual segment is also under compression, the seven terms of the sum are all negative and result in a negative change in length.
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[Solution] Problem 5.3
Synthetic biomaterial scaffolds are being developed to induce bone regeneration in critical size defects (defects that are too large for normal healing to occur). In this example, the scaffold fills a 1.5
cm gap in the tibia. In the diagram, in addition to the axial compressive loads, there is an intermediate tensile load applied by the patellar ligament (transmitting the quadriceps muscle force): Assume the bone has a circular cross-sectional area of A
= 480 mm
2
. The elastic modulus of the tibia
is E = 15 GPa, and the elastic modulus of the space filler is E = 3.5 GPa. The force applied by the ankle joint is 1.5 kN. The force applied by the knee is 2.2 kN. Determine the total change in length of
the tibia.
Assumptions:
4.
Linear elastic materials
5.
Isotropic materials
6.
Homogeneous materials
7.
Forces are all axial
8.
All segments are prismatic bars with same cross-sectional area
First we need to determine the intermediate axial force applied by the patellar ligament. From the free body diagram above, we can write the equation of equilibrium for forces in the x-
direction:
∑
F
x
=
0
=
F
knee
−
F
ligament
−
F
ankle
F
ligament
=
F
knee
−
F
ankle
=
2.2
kN
−
1.5
kN
=
700
N
To solve for the length of the tibia, we need to determine the total deformation of the tibia. To do this, we analyze the tibia by segments, starting with a free body diagram to determine the internal forces of each segment:
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We can solve for the internal forces for each segment:
N
1
=
700
N
+
1500
N
=
2200
N
N
2
=
N
3
=
N
4
=
1500
N
Now we use our equation for determining the deformation of a segmented bar under axial load. Remember that L
i
is the length of each individual segment:
δ
=
∑
i
=
1
4
F
i
L
i
E
i
A
=
1
480
mm
2
(
(
2200
N
)
(
40
mm
)
15,000
MPa
+
(
1500
N
)
(
165
mm
)
15,000
MPa
+
(
1500
N
)
(
15
mm
)
3,500
MPa
+
(
1500
N
)
(
180
mm
)
15,000
MPa
)
=
¿
0.0975
mm
¿
Since the tibia is under compressive load, and each segment is also under compression, both the individual segments and the whole tibia experience a negative change in length = -0.0975 mm
.
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[Solution] Problem 5.4
After fracture of the tibia, a callus forms around the fracture site, consisting of a mixture of woven bone and cartilage. Over time, the callus is replaced with normal, organized bone tissue (called lamellar bone) that is much stronger than the temporary callus. Let us consider a tibia at the early stages of the callus. The callus has a larger radius than the normal tibia, but is composed of weaker woven bone:
Let us assume that both the cortical bone of the normal tibia and the woven bone of the callus are linear elastic materials with E
cortical = 18 GPa and E
woven
= 0.5 GPa. The proximal and distal tibia are prismatic bars with cross-sectional areas A = 350 mm
2
. The callus is also a prismatic bar, with cross-
sectional area A = 720 mm
2
. The entire tibia is subjected to end-applied compressive forces F
= 500 N, as shown above. Based on these assumptions and parameters, answer the following questions: a)
What is the internal force in the callus?
Since there are no intermediate axial loads applied to the tibia in this case, the internal force throughout the tibia is equal to the externally applied load, i.e., F
callus = 500 N.
b)
What is the change in length of the entire tibia?
Assumptions:
1.
Linear elastic materials
2.
Isotropic materials
3.
Homogeneous materials
4.
Segments are prismatic bars
5.
Uniaxial compression
Since this is a case of non-uniform axial loading, we need to divide our tibia into three segments: proximal tibia, callus, and distal tibia. That allows us to use the following formula:
δ
=
∑
i
=
1
n
N
i
L
i
E
i
A
i
Important note: since the end-applied loads compress the tibia, all internal forces N
i
must be negative when entered in this equation.
δ
=
(−
500
N
)(
0.18
m
)
(
18
x
10
9
Pa
)(
0.00035
m
2
)
+
(−
500
N
)(
0.06
m
)
(
0.5
x
10
9
Pa
)(
0.00072
m
2
)
+
(−
500
N
)(
0.12
m
)
(
18
x
10
9
Pa
)(
0.00035
m
2
)
29
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δ
=−
0.000107
m
=−
0.107
mm
=−
107
μm
[Solution] Problem 5.5
In total knee replacements (TKRs), the tibial plateau is replaced with a tibial component, consisting of a polyethylene (PE) insert on a titanium alloy tray. We can idealize a tibia with a TKR tibial component as a segmented bar under compressive loading:
If the entire tibia (including the plastic and metal tibial component) experiences a change in length of δ = -0.1 mm, what is the magnitude of the compressive force F
? The elastic modulus for the PE, titanium alloy, and bone are E
PE
= 0.9 GPa, E
Ti
= 90 GPa, and E
bone
= 15 GPa, respectively. Remember to show your work and state all relevant assumptions.
Key assumptions:
1.
All materials are isotropic, homogeneous, and linear elastic
2.
Each segment is a prismatic bar
3.
Each segment experiences uniaxial loading
This allows us to use our equation for the behavior of a system that can be segmented in prismatic bars undergoing uniaxial loads in each segment:
δ
=
∑
i
=
1
n
N
i
L
i
E
i
A
i
In this case, we already know the total change in length. We also observe, from the formulation of the problem, that there are no intermediate axial loads. This means that the internal force in each segment, N
i
, is equal the applied force F
(alternatively, you could also show that this is true via a series of free body diagrams to find each internal force). We also observe that the tibia can be segmented into 5 segments: PE insert, metal tray, proximal tibia, tibial shaft, and distal tibia. We can now set up our equation:
30
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δ
=−
0.1
mm
=
∑
i
=
1
5
N
i
L
i
E
i
A
i
=
F L
1
E
PE
A
1
+
F L
2
E
Ti
A
2
+
F L
3
E
bone
A
3
+
F L
4
E
bone
A
4
+
F L
5
E
bone
A
5
−
0.1
mm
=
F
[
−
20
mm
(
900
MPa
)
(
3600
mm
2
)
−
6
mm
(
90,000
MPa
)
(
3600
mm
2
)
−
18
mm
(
15,000
MPa
)
(
3600
mm
2
)
−
288
mm
(
15,000
MPa
)
(
600
mm
2
)
−
(
1
F
=
2,462
N
31
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[Solution] Problem 5.6
Ruptures of the Achilles tendon are typically repaired by suturing
together the two ruptured ends. We can model the repaired tissue
as a segmented bar under uniaxial tensile loading (pictured on the
right). Both the Achilles tendon and the sutures joining the two portions
of the Achilles tendon are modeled as prismatic bars made of
linear elastic materials, with the (undeformed) lengths given
above. The cross-sectional area of the tendon (for both portions)
is A = 55 mm
2
and the elastic modulus is E = 1177 MPa. The total
cross-sectional area of the sutures is A
= 0.46 mm
2
and the elastic
modulus is E = 3700 MPa. Even with limited activity after surgery,
the repaired Achilles tendon can be subjected to significant loads. Under a tensile force F
, the entire surgically repaired Achilles
tendon stretches by δ =
0.59 mm. Given this information, answer
the following questions (remember to show your work and state
assumptions):
a)
Based on the observed change in length, what is the uniaxial tensile force F applied to the repaired Achilles tendon?
We identify that this system, based on changes in material and cross-sectional area, must be divided into 3 segments. Then we draw free body diagrams to determine the internal normal force in each of these segments:
Based on the FBDs, we can see the internal normal force is a constant value equal to F
(in tension) for all segments. Since we are given the total change in length, we can use our equation for the total change in length of a segmented bar to find the applied force F
:
δ
Total
=
∑
i
=
1
3
δ
i
=
∑
i
=
1
3
N
i
L
i
E
i
A
i
=
F
[
40
mm
(
1177
MPa
)(
55
mm
2
)
+
3
mm
(
3700
MPa
)(
0.46
mm
2
)
+
30
mm
(
1177
MPa
)(
55
mm
2
)
]
32
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δ
Total
=
0.59
mm
=
F
[
40
mm
(
1177
MPa
)(
55
mm
2
)
+
3
mm
(
3700
MPa
)(
0.46
mm
2
)
+
30
mm
(
1177
MPa
)(
55
mm
2
)
]
0.59
mm
=
F
(
0.002844
mm
N
)
→ F
=
207.5
N
b)
Draw a stress element representing the state of stress in the sutures. What are the principal stresses and corresponding principal angles for this state of stress?
Since we assume that the tendon is undergoing uniaxial tension, the state of stress in the sutures will be uniaxial normal stress. All we have to do is define the direction of loading (I choose x here, but it is arbitrary – x or y is appropriate) and then calculate the normal stress in the sutures:
σ
x
=
N
2
A
=
F
A
=
207.5
N
0.46
mm
2
=
451
MPa
Since this is uniaxial normal stress, we know that the principal stresses correspond to these reference configuration stress components (you could also go through the process of solving for the principal stresses and angles):
σ
1
=
451
MPa@
0
°
σ
2
=
0
MPa@
90
°
(
alternatively@
−
90
°
)
Note that if you defined the y-direction as the direction of loading, the only thing that would change are the principal angles (they would switch).
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[Solution] Problem 5.7
Ruptures of the Achilles tendon are typically repaired by suturing together the two ruptured ends. We
can model the repaired tissue as a segmented bar under uniaxial tensile loading:
Both the Achilles tendon and the sutures joining the two portions of the Achilles tendon are modeled as prismatic bars made of linear elastic materials. The cross sectional area of the tendon (for both portions) is A = 58 mm
2
and the Young’s modulus is E = 820 MPa. The cross-sectional area of the sutures (made of polyester) is A
= 20 mm
2
and the Young’s modulus is E = 240 MPa.
Even with limited activity after surgical repair, the Achilles tendon can be subjected to significant loads. Under a tensile force of 1800 N, what is the change in length of the repaired Achilles tendon?
Because the repaired Achilles tendon has sutures with different cross-sectional area and properties, we immediately recognize that this is a case of non-uniform axial loading.
First we need to determine how many segments to divide our repaired tendon into. By inspection, we see that force, cross-sectional area, and/or properties change twice, dividing the system into three segments: the upper portion of the tendon, the sutures, and the lower portion of the tendon. Next we determine the internal forces in each segment. We can do so using free body diagrams that cut through each segment (to reveal the internal force), or we can show by inspection that, since the only applied loads are at the ends, the internal forces in all segments are equal to 1800 N (tensile). Now we can use our summation equation to find the change in length of the repaired tendon:
δ
=
∑
i
=
1
3
N
i
L
i
E
i
A
i
=
(
1800
N
)(
36
mm
)
(
820
MPa
)(
58
mm
2
)
+
(
1800
N
)(
1
mm
)
(
240
MPa
)(
20
mm
2
)
+
(
1800
N
)(
30
mm
)
(
820
MPa
)(
58
mm
2
)
=
2.87
mm
34
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[Solution] Problem 5.8
Consider a bone fixation plate made of titanium alloy (
E = 90 GPa), subjected to intermediate axial loads as pictured:
The cross-sectional area of the plate is 16 mm
2 at the ends, and 8 mm
2
in the center portion (where the cut-out is). Based on the illustration provided, answer the following questions (remember to show
your work and state additional assumptions):
a)
What is the total change in length of the fixation plate under these loads?
This is clearly a system that we need to segment into individual segments of prismatic bars undergoing uniaxial loading. We first assume that the plate is made of a linear elastic material
(titanium alloy), and we observe that the changes in loading and cross-section require us to break it into 5 different segments. We draw free body diagrams that allow us to determine the internal forces in each of these five segments:
N
1
−
300
N
=
0
→N
1
=
300
N
N
2
−
300
N
−
300
N
=
0
→N
2
=
600
N
N
3
−
300
N
−
300
N
=
0
→N
3
=
600
N
35
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N
4
−
300
N
−
300
N
=
0
→ N
4
=
600
N
N
5
+
300
N
−
300
N
−
300
N
=
0
→N
5
=
300
N
Note that all the forces are tensile. Now we can apply our equation for total change in length of a non-uniform axially loaded system:
δ
=
∑
i
=
1
n
N
i
L
i
E
i
A
i
δ
=
(
300
N
)(
10
mm
)
(
90,000
MPa
)(
16
mm
2
)
+
(
600
N
)(
5
mm
)
(
90,000
MPa
)(
16
mm
2
)
+
(
600
N
)(
50
mm
)
(
90,000
MPa
)(
8
mm
2
)
+
(
600
N
)(
5
mm
)
(
90,000
MPa
)(
16
mm
2
)
+
(
300
N
)(
10
m
(
90,000
MPa
)(
1
Alternatively, you can lump some segments together:
δ
=(
2
)
(
300
N
)(
10
mm
)
(
90,000
MPa
)(
16
mm
2
)
+(
2
)
(
600
N
)(
5
mm
)
(
90,000
MPa
)(
16
mm
2
)
+
(
600
N
)(
50
mm
)
(
90,000
MPa
)(
8
mm
2
)
Either way, you evaluate your equation, and find the total change in length:
δ
=
0.05
mm
b)
What is the maximum shear stress in the fixation plate and where does it occur?
In an axially loaded system, the maximum shear stress occurs wherever the maximum normal stress occurs, but on a 45
o
inclined section (instead of a cross-section). The highest normal stress occurs in the middle segment of the plate (where the cut out is located and the cross-
sectional area is smallest).
σ
x
=
F
A
=
600
N
8
mm
2
=
75
MPa
Since the maximum shear stress is always one-half the maximum normal stress, we can quickly determine it:
τ
max
=
σ
x
2
=
37.5
MPa
36
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[Solution] Problem 5.9
Consider the following bone fixation plate made of Ti-6V-4Al (
E = 90 GPa), with a cross-sectional area of 60 mm
2
:
The plate is subjected to 7000 N loads (as pictured) at each location where a screw is inserted. Answer the following questions:
a)
What are the internal forces in each portion of the
fixation plate?
This problem requires using our segmented bar under
uniaxial loading approach. First we need to solve for
the internal forces in each of the 5 segments of the bar,
keeping in mind that while the elastic modulus and
cross-sectional area do not change, the internal force
will vary because of the intermediate loads.
∑
F
x
=
0
−
N
1
−
7000
N
−
7000
N
+
7000
N
+
7000
N
=
0
→ N
1
=
0
−
N
2
−
7000
N
+
7000
N
+
7000
N
=
0
→ N
2
=
7000
N
−
N
3
+
7000
N
+
7000
N
=
0
→N
3
=
14000
N
−
N
4
+
7000
N
=
0
→N
4
=
7000
N
−
N
5
=
0
b)
What is the deformed length of the fixation plate under the pictured loads?
Using the values for the internal forces found in part (b) and our equation for the change in length of an axially loaded prismatic bar, we can find the new length of the fixation plate:
δ
=
∑
i
=
1
5
N
i
L
i
E
i
A
i
=
1
EA
∑
i
=
1
5
N
i
L
i
37
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δ
=
1
(
90000
MPa
)(
60
mm
2
)
[
2
(
0
) (
20
mm
)
+
2
(
7000
N
) (
10
mm
)
+
(
14000
N
) (
60
mm
)
]
=
0.181
mm
L
=
L
0
+
δ
=
120
mm
+
0.181
mm
=
120.181
mm
=
12.0181
cm
38
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[Solution] Problem 5.10
A number of polymer biomaterials have been considered for
orthopedic applications, but the strength of these polymers is often
a concern. To address this deficiency, biomaterials researchers
have reinforced these polymers with much stronger materials,
such as carbon fibers. Consider the polymer polyether ether
ketone (PEEK) reinforced with carbon fibers. The carbon fibers
are aligned in the direction of loading, and make up 70% of the
total volume of the material (and therefore we can approximate
that carbon fibers are 70% of the cross-sectional area). PEEK has
a Young’s modulus E
= 5 GPa, while the carbon fibers have a
Young’s modulus E
= 150 GPa. Let us model a hip implant stem
made of carbon fiber-reinforced PEEK as a solid circular cylinder
with length L = 12 cm and a radius of 8 mm. During preliminary
mechanical testing, the stem is subjected to a uniaxial
compressive force of 2000 N. Answer the following questions:
a)
What the internal forces in the PEEK and the carbon
fibers?
Assumptions
4.
Linear elastic materials
5.
Isotropic materials
6.
Homogeneous materials
7.
Uniaxial compression
8.
The two different components (PEEK, carbon fiber) can be approximated as prismatic bars
If we write the force balance equations for this system, we quickly ascertain that it is a statically indeterminate system:
2000
N
=
N
PEEK
+
N
fibers
To solve a statically indeterminate system, we need to first define an equation of compatibility:
δ
PEEK
=
δ
fibers
Then we assume the PEEK and the carbon fibers behave as prismatic bars made of linear elastic materials under a uniaxial load, and in doing so we can substitute for the change in lengths:
N
PEEK
L
E
PEEK
(
0.3
A
)
=
N
fibers
L
E
fibers
(
0.7
A
)
Now we can solve for the internal force in the carbon fibers as a function of the internal force in the PEEK (or vice versa):
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N
fibers
=
0.7
E
fibers
0.3
E
PEEK
N
PEEK
=
(
0.7
) (
150
GPa
)
(
0.3
) (
5
GPa
)
N
PEEK
=
70
N
PEEK
Now we can solve our force balance equation:
2000
N
=
N
PEEK
+
70
N
PEEK
N
PEEK
=
28.17
N
N
fibers
=
1971.9
N
b)
What are the axial normal strains in the PEEK and the carbon fibers?
There are two typical approaches to answering this question.
Approach #1:
You can treat the carbon fibers and the PEEK portion of the stem each as prismatic bars made
of linear elastic materials under a uniaxial load. Using the information provided and the forces
computed in part (a), you can determine the change in length of EITHER the carbon fibers or PEEK (since we showed in part (a) that they must be equal):
δ
=
FL
EA
A
=
π r
2
=
0.000201
m
2
δ
PEEK
=
(
28.17
N
)(
0.12
m
)
(
5
x
10
9
Pa
)(
0.3
)(
0.000201
m
2
)
=
1.121
x
10
−
5
m
δ
fibers
=
(
1971.9
N
)(
0.12
m
)
(
150
x
10
9
Pa
)(
0.7
)(
0.000201
m
2
)
=
1.121
x
10
−
5
m
Then we compute the normal strain (identical for the PEEK and the carbon fibers):
ε
=
δ
L
=
1.121
x
10
−
5
m
0.12
m
=
9.342
x
10
−
5
Approach #2:
You compute the stress in either the PEEK or carbon fiber portion of the stem, and then use Hooke’s law to find the normal strain. Since the two materials are undergoing the same change length, and have the same overall length, their normal strain should be identical.
σ
PEEK
=
28.17
N
(
0.3
)(
0.000201
m
2
)
=
467,164
Pa
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σ
fibers
=
1971.9
N
(
0.7
)(
0.000201
m
2
)
=
14,014,925
Pa
Then we substitute either value into Hooke’s law to find the axial strain:
ε
=
σ
E
=
467,164
Pa
5
x
10
9
Pa
=
9.343
x
10
−
5
ε
=
σ
E
=
14,014,925
Pa
150
x
10
9
Pa
=
9.343
x
10
−
5
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[Solution] Problem 5.11
Stress shielding is a phenomenon where bone mineral density decreases due to the presence of an implant that shifts the loading away from the bone. This is usually because the implant is made of a material with a significantly higher elastic modulus than bone tissue. Let us consider a simplified situation where the stem of a hip implant sits inside the proximal femur. The bone-implant hybrid consists of an outer shell of cortical bone (
E = 18 GPa, A = 530 mm
2
) and an
inner core of Ti-13Nb-13Zr (
E = 75 GPa, A = 430 mm
2
) that is press-fit inside the bone. Determine the relative amount of force that the cortical bone and titanium alloy stem will experience during uniaxial compressive loading. Compare this to the amount of loading that the cortical bone would experience if the implant stem was not present.
We first realize that this is a statically indeterminate system, since we only have 1 equation of equilibrium with two unknown internal forces:
F
=
N
cortical
+
N
Ti
13
Nb
13
Zr
We can write a simple equation of compatibility to help us analyze this system:
δ
cortical
=
δ
Ti
Assumptions:
5.
Linear elastic materials
6.
Isotropic materials
7.
Homogeneous materials
8.
The two compartments can be approximated as prismatic bars
9.
Uniaxial compression
Given our assumptions, we can substitute expressions for the changes in length:
N
cortical
L
E
cortical
A
cortical
=
N
Ti
L
E
Ti
A
Ti
Rearranging and substituting in the values, we can solve for the force in the cortical bone:
N
cortical
=
E
cortical
A
cortical
E
Ti
A
Ti
N
Ti
=
(
18
GPa
)
(
530
mm
2
)
(
75
GPa
)
(
430
mm
2
)
N
Ti
N
cortical
=
0.296
N
Ti
F
=
N
cortical
+
N
Ti
=
1.296
N
Ti
N
Ti
=
0.772
F
N
cortical
=
0.228
F
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Recall that, in the normal case, the interior of the femur mostly contains yellow marrow, which
contributes very little to the overall load-bearing of the femoral shaft. If we assume that the cortical bone would normally carry ~99% of the applied force, then the change from the healthy to the implanted case is approximately 0.99/0.228, or a 4.3-fold decrease in loading (or in other words, the cortical bone is experiencing about 23% of the load it would normally expect to carry).
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[Solution] Problem 5.12
Stress shielding in bone due to metallic orthopedic implants is not limited to joint replacements. The presence of an internal fixation plate on the femur can also cause significantly reduced loading on the
bone. Consider the simplified example below:
In this model of a fixation plate and a segment of the femur undergoing compressive axial loading (
F = 750 N), both the femur and the fixation plate are assumed to be prismatic bars, with cross-sectional
areas A
femur
= 350 mm
2
and A
plate
= 75 mm
2
. The length of the femur-fixation plate system is L
= 10 cm. The femur is made of cortical bone (
E = 15 GPa) and the plate is made of 316L stainless steel (
E
= 200 GPa). In our simplified model, we will not consider the effects of the screws used to keep the fixation plate in place. Based on these initial assumptions and the schematic provided, answer the following questions (remember to show your work and state additional assumptions):
a)
Show that this system is statically indeterminate. Write an appropriate equation of compatibility for this system.
To show that the system is statically indeterminate, you need to show that you have more unknowns than equations when applying force or moment balance equations. In this case, you have 1 equation and 2 unknowns:
∑
F
y
=
0
→ N
femur
+
N
plate
=
750
N
Now that we know it is a statically indeterminate system, we begin to solve the system by writing an equation of compatibility. In this case, the equation should be:
δ
femur
=
δ
plate
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b)
What are the internal forces (acting on the cross-sections) in the femur and the plate under these loading conditions?
To solve for the internal forces, we need to assume that cortical bone and stainless steel are linear elastic
. Using that assumption, the fact that the plate and femur are prismatic bars, and are subjected to uniaxial loads, we can use our equation of compatibility to write the following
equality:
δ
femur
=
δ
plate
N
femur
L
E
femur
A
femur
=
N
plate
L
E
plate
A
plate
We can then solve for 1 of the internal forces in terms of the other:
N
femur
=
(
EA
)
femur
(
EA
)
plate
N
plate
=
(
15
GPa
)(
350
mm
2
)
(
200
GPa
)(
75
mm
2
)
N
plate
N
femur
=
0.35
N
plate
Now we substitute into our force balance equation and solve for the unknown internal forces:
N
femur
+
N
plate
=
750
N
1.35
N
plate
=
750
N → N
plate
=
555.6
N
N
femur
=
194.4
N
c)
What is the maximum shear stress in the fixation plate and where does it occur?
The maximum shear stress will occur on an inclined section. We know that the magnitude of the maximum shear stress is ½ the maximum normal stress, and those shear stress values occur at ±45
o
. You can either state this, or use the inclined section stress equations to directly calculate the max shear stress value. To find the stress value, we need to know the max normal stress:
σ
x
=
N
plate
A
plate
=
555.6
N
75
mm
2
=
7.4
MPa
τ
max
=
σ
x
2
=
3.7
MPa
Alternatively:
τ
max
=
−
σ
x
2
sin 2
θ
=
3.7
MPa
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d)
To prevent stress shielding, the femur should experiencing an axial compressive strain = 0.0001 under these loading conditions. Does at least that amount of strain occur?
We need to calculate the strain in the femur. We can either do this directly, using the stress in
the femur and Hooke’s law, or observe that since the femur and plate have the same length and undergo the same length change, they experience the same strain. Therefore, we can also calculate the strain in the plate and use that, since it is equal to the strain in the femur. Either approach is valid. I opt for the latter, since we have already calculated the stress in the plate in part c.
ε
femur
=
ε
plate
=
σ
plate
E
plate
=
7.4
MPa
200,000
MPa
=
0.000037
Therefore, the answer is no, the femur does not experience normal strain ≥ 0.0001 under these
conditions
.
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[Solution] Problem 5.13
Consider a 15 cm long segment of the normal femur, and then the same femur where the cancellous bone has been removed and replaced with a titanium alloy hip implant, as illustrated (assume the femur is a prismatic bar with a circular cross-section of outer diameter 2.8 cm):
Assume that the dimensions between situation A and B are identical; the only change is the complete
replacement of the cancellous bone interior of the femur with the hip implant stem (made of Ti-6Al-
4V). Under an axial compressive load of 3000 N, answer the following questions.
a)
What are the normal stresses and normal strains in the cortical bone and cancellous bone in situation A?
This is a statically indeterminate problem—we cannot determine the internal forces within the cortical and cancellous bone from our equation of equilibrium alone. Thus, we will need to identify an equation of compatibility and use force-displacement relationships (assuming linear elastic materials) to solve for the forces, and thus the stresses and strains in the bone tissues.
The equation of equilibrium is thus:
∑
F
y
=
0
=
3
kN
−
F
cortical
−
F
cancellous
The equation of compatibility comes from the observation that the deformation of the cortical bone and cancellous bone, since they are one structure, must be equal:
δ
cortical
=
δ
cancellous
=
δ
Finally, we can assume that the two types of bone behave as linear elastic materials. Since
this is an axial loading situation, we can invoke our force-displacement relationships for an
axially loaded prismatic bar:
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δ
cortical
=
F
cortical
L
E
cortical
A
cortical
;δ
cancellous
=
F
cancellous
L
E
cancellous
A
cancellous
Assuming the cross-section of the bone is circular, the cross-sectional areas of the cortical
bone shell and the cancellous bone core are as follows:
A
cortical
=
π
[
(
14
mm
)
2
−(
8
mm
)
2
]
=
414.69
mm
2
A
cancellous
=
π
(
8
mm
)
2
=
201.06
mm
2
We can now solve for the force in the cortical bone in terms of the force in the cancellous bone:
F
cortical
L
E
cortical
A
cortical
=
F
cancellous
L
E
cancellous
A
cancellous
F
cortical
=
E
cortical
A
cortical
E
cancellous
A
cancellous
F
cancellous
=
(
17
GPa
)
(
414.69
mm
2
)
(
3
GPa
)
(
201.06
mm
2
)
F
cancellous
F
cortical
=
11.69
F
cancellous
Substituting back into our equation of equilibrium, we find the forces:
3
kN
−
F
cortical
−
F
cancellous
=
0
11.69
F
cancellous
+
F
cancellous
=
3000
N
F
cancellous
=
236.4
N ; F
cortical
=
2763.6
N
Then using our relationships for stress and strain, we obtain those values as well (note that the strain is equal in both cortical and cancellous bone, because the deformation is equal and the original lengths are the same):
σ
cortical
=
F
cortical
A
cortical
=
2763.6
N
414.69
mm
2
=
6.664
MPa
σ
cancellous
=
F
cancellous
A
cancellous
=
236.4
N
201.06
mm
2
=
1.176
MPa
ε
=
σ
cancellous
E
cancellous
=
1.176
MPa
3,000
MPa
=
0.000392
=
0.0392%
strain
(
compressive
)
b)
What are the normal stresses and normal strains in the cortical bone and Ti-6Al-4V implant stem in situation B?
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In principle, we would follow the same procedure as in part (a), except that we would replace the cancellous bone with Ti-6Al-4V. Effectively, we just need to re-run our numbers
using the elastic modulus of the alloy (E = 113 GPa) to obtain the stresses and strains.
F
cortical
=
E
cortical
A
cortical
E
Ti
6
Al
4
V
A
Ti
6
Al
4
V
F
Ti
6
Al
4
V
=
(
17
GPa
)
(
414.69
mm
2
)
(
113
GPa
)
(
201.06
mm
2
)
F
Ti
6
Al
4
V
=
0.3103
F
Ti
6
Al
4
V
0.3103
F
Ti
6
Al
4
V
+
F
Ti
6
Al
4
V
=
3000
N
F
Ti
6
Al
4
V
=
2289.6
N ;F
cortical
=
710.4
N
σ
cortical
=
F
cortical
A
cortical
=
710.4
N
414.69
mm
2
=
1.713
MPa
σ
Ti
6
Al
4
V
=
F
Ti
6
Al
4
V
A
Ti
6
Al
4
V
=
2289.6
N
201.06
mm
2
=
11.38
MPa
ε
=
σ
Ti
6
Al
4
V
E
Ti
6
Al
4
V
=
11.38
MPa
113,000
MPa
=
0.000101
=
0.0101%
strain
(
compressive
)
Note that the stress born by the cortical bone decreases from 6.6 MPa to 1.7 MPa, and the strain in the bone drops from 0.039% to 0.01%.
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[Solution] Problem 5.14
Consider the following system, which consists of a bar with constant diameter of 80 mm, but made of two different materials. The bar is fixed between two immovable walls at A and C, and a load P
= 100
kN is applied at D. Source: Introductory Mechanics of Materials (via YouTube) (2017)
Answer the following questions:
a)
Show that the system is statically indeterminate.
If we do a force balance in the axial direction (let’s call it the x-direction), we have 3 forces: the
reaction force at A, the reaction force at C, and the intermediate axial load P = 100 kN:
∑
F
x
=
0
→ A
x
+
C
x
−
100
kN
=
0
Since this is the only equation of equilibrium we can write for the system, it is statically indeterminate.
b)
Write an appropriate equation of compatibility that would allow you to solve this system.
Since the walls at A and C are fixed and immovable, the total change in length of the entire system must be 0. We observe that this statically indeterminate system is also one that needs
to be segmented, because of the change in material at B and the load P
. Therefore, we can write our equation of equilibrium as:
δ
total
=
0
=
δ
AB
+
δ
BD
+
δ
DC
Since the walls at A and C are fixed and immovable, the total change in length of the entire system must be 0. We observe that this statically indeterminate system is also one that needs
to be segmented, because of the change in material at B and the load P
. c)
What are the internal forces in the two bars (AB and BC)?
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The key to this problem is two-fold: (1) the reaction forces @ A and C are both acting to the right, in opposition of the force P, and (2) the reaction forces are related to the internal forces in the 3 segments:
A
x
+
C
x
−
100,000
N
=
0
C
x
=
100,000
N
−
A
x
A
x
=
N
AB
A
x
=
N
BD
C
x
=
N
DC
So now that we have segmented the system and equated the internal forces to the reaction forces @ the wall, we can use our equation of compatibility and, assuming the segments are linear elastic, prismatic bars subjected to uniaxial load
, to solve for A
x
and C
x
(and therefore the internal forces in each segment):
δ
AB
+
δ
BD
+
δ
DC
=
0
N
AB
L
AB
E
AB
A
+
N
BD
L
BD
E
BC
A
+
N
DC
L
DC
E
BC
A
=
0
−
A
x
(
600
mm
)
101,000
MPa
−
A
x
(
200
mm
)
200,000
MPa
+
(
100,000
N
−
A
¿¿
x
)(
200
mm
)
200,000
MPa
=
0
¿
A
x
=
N
AB
=
N
BD
=
12,600
N
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C
x
=
N
DC
=
87,400
N
[Solution] Problem 5.15
The vertebral bodies in the lumbar spine are subjected to axial compressive forces during a variety of everyday activities. Consider a simple model of a vertebral body consisting of the following:
Two endplates of cortical bone (
E = 15 GPa), with cross-sectional area A = 283 mm
2
and length L
= 1 mm.
A centrum of trabecular bone (
E = 700 MPa) with cross-sectional area A = 226 mm
2
, surrounded by a cortical bone shell of cross-sectional area A = 57 mm
2
. The centrum and cortical shell have a length L = 26 mm. If the entire vertebral body is subjected to a uniaxial compressive force of 600 N, determine the following (remember to show your work and state assumptions): a)
Internal normal forces in the two endplates, the centrum, and the cortical shell (surrounding the
centrum)
This is another segmented system, where one of the segments is actually a statically indeterminate system. We note that there are 3 segments due to material changes: the two endplates and the centrum / cortical shell combo:
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From the FBDs, we see that the internal normal forces in the endplates equal the applied 600 N
compressive force
: N
1
=
N
3
=
600
N
However, for the 2
nd
segment (the centrum / cortical shell combo), we have two unknown internal forces that must oppose the 600 N applied load; however we have no other equations to solve for the forces at the moment:
N
centrum
+
N
shell
=
600
N
Thus it is a statically indeterminate system. To solve this conundrum, we first state an equation of compatibility:
δ
=
δ
centrum
=
δ
shell
Now we assume linear elasticity
, which allows us to turn the equation of compatibility into a second equation incorporating our two unknown internal forces:
δ
centrum
=
δ
shell
N
centrum
L
E
trab
A
centrum
=
N
shell
L
E
cort
A
shell
We solve for one of the internal forces in terms of the other, then substitute back into our original balance of forces equation for solve for both internal forces:
N
centrum
=
(
700
MPa
)
(
226
mm
2
)
(
15,000
MPa
)
(
57
mm
2
)
N
shell
=
0.185
N
shell
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N
centrum
+
N
shell
=
1.185
N
shell
=
600
N
N
shell
=
506.3
N
N
centrum
=
93.7
N
As expected, most of the compressive force is resisted by the cortical bone shell. b)
Total change in length of the vertebral body To find the total change in length of the vertebral body, we use our equation for the change in length of a segmented bar:
δ
Total
=
∑
i
=
1
3
δ
i
=
∑
i
=
1
3
N
i
L
i
E
i
A
i
Importantly, for the 2
nd
segment (the centrum / cortical shell combo), we can use the values for
either the shell or the centrum, since both will deform the same amount (as we stated in our equation of compatibility in the previous section). We also note that all the internal forces are compressive (from our FBDs), and therefore are negative:
δ
Total
=
(
−
600
N
) (
1
mm
)
(
15,000
MPa
)
(
283
mm
2
)
+
(
−
93.7
N
) (
26
mm
)
(
700
MPa
)
(
226
mm
2
)
+
(
−
600
N
) (
1
mm
)
(
15,000
MPa
)
(
283
mm
2
)
δ
Total
=−
0.015
mm
54
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[Solution] Problem 5.16
Fixation plates are sometimes used to keep bone fractures in union during healing. Consider a situation where a healing humerus is subjected to tension (e.g., when someone lifts something heavy). The fixation plate is modeled by the intermediate axial loads:
The normal portions of the humerus have E = 18 GPa. The callus has E = 5 GPa. The cross-
sectional area over the entire humerus is a relatively constant A = 710 mm
2
. Answer the following questions (remember to show your work):
b)
What is the total change in length of the humerus?
Assumptions:
Bone and callus are linear elastic materials
Bone and callus are homogeneous materials
Bone and callus are isotropic materials
Each segment is a prismatic bar
Each segment is experiencing uniaxial force
We observe that the system can be broken up into 5 segments: the two regions between the end applied and intermediate loads, the two regions between the intermediate loads and the callus, and the callus itself. Through free body analysis, we can show that the 1
st
and 5
th
segments experience N = 200 N (tension), while the central 3 segments experience N
= 40 N (compression).
Using this information, we can compute the total change in length of this system:
δ
total
=
∑
i
=
1
5
δ
i
=
∑
i
=
1
5
N
i
L
i
E
i
A
i
55
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Since the 1
st
and 5
th
segments, as well as the 2
nd
and 4
th
segments, behave identically, I will lump together those segments when computing the change in length:
δ
total
=
2
(
200
N
)(
105
mm
)
(
18,000
MPa
)(
710
mm
2
)
−
2
(
40
N
) (
40
mm
)
(
18,000
MPa
)
(
710
mm
2
)
−
(
40
N
) (
30
mm
)
(
5,000
MPa
)
(
710
mm
2
)
δ
total
=
2
(
0.001643
mm
)
−
2
(
0.000125
mm
)
−
0.000338
mm
=
0.002698
mm
c)
To keep the healing fracture in union, the central segment of the humerus must remain under a
small amount of compression. Is the healing fracture under compression in this case?
Since the callus is experience 40 N of compressive force (resulting in δ = -0.000338 mm), we can state that yes, the healing fracture is under compression.
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[Solution] Problem 5.17
Most hip implant stem components have a gradual taper. Let us consider a titanium alloy (
E = 100 GPa) stem component with L = 128 mm with a tapering rectangular cross-section. The width of the cross-section is a constant w = 7 mm, but the thickness t
tapers from 20 mm (at x = 0 mm) to 6 mm (at x = 128 mm). We can describe the cross-sectional area A(x) as a function of x
(position along the length of the stem):
A
(
x
)
=
w∙t
(
x
)
=−
0.76563
x
+
140
In this case, A(x)
is in units of mm
2
if x
is inputted in units of mm. If a uniaxial compressive
force F = 500 N is applied to the tapered stem, determine its total change in length. Assumptions
Uniaxial force
Gradual change in cross-sectional area
Linear elastic material
Isotropic material
Homogeneous material
If the cross-sectional area changes in a sufficiently gradual way, we can use an integral equation to compute the change in length:
δ
=
∫
0
128
mm
F
EA
(
x
)
dx
Since F and E
are constants, we can move them outside the integral:
δ
=
−
500
N
100,000
MPa
∫
0
128
mm
dx
−
0.76563
x
+
140
I put this into a symbolic integrator (e.g., Wolfram Alpha, MATLAB) to evaluate the integral and find the
total change in length:
δ
=
(
−
500
N
100,000
MPa
)
(
1.57255
mm
−
1
)
=−
0.00786
mm
Note that the change in length must be negative because the applied force is compressive.
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[Solution] Problem 5.18
For large bone defects (e.g., after removal of a bone tumor), surgeons sometimes use fibular grafts to
fill the gap. The surgeon removes a large segment of the fibula (which has relatively limited weight bearing function in our lower leg), and transplants that segment into the bone needing repair.
Consider a situation where a fibular graft is used to fill a large defect in the humerus (upper arm bone). A simplified diagram of the system look as follows:
You know the following information:
The cross-sectional area of the humerus (for both humeral segments) is
A
humerus
= 315 mm
2
The cross-sectional area of the fibular graft is A
fibula
= 200 mm
2
F
biceps
= 400 N
F
elbow
= 1500 N
E
bone
= 20 GPa
The lengths are as shown on the diagram
Given this information, answer the following questions (remember to show your work):
a)
State any relevant assumptions.
Assumptions
Bone is a linear elastic material
Bone is a homogeneous material
Each segment is a prismatic bar
Each segment is subjected to uniaxial load
b)
What is F
shoulder
?
We can do a simple force balance to find F
shoulder
:
F
biceps
+
F
shoulder
−
F
elbow
=
0
400
N
+
F
shoulder
−
1500
N
=
0
F
shoulder
=
1100
N
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c)
How many segments are there?
Because of the intermediate axial load applied the biceps and the change in cross-section of the fibular graft, there are four (4)
segments.
d)
What are the internal normal forces in each segment?
Here we use a series of free body diagrams to isolate and solve for the internal forces N
in each segment:
N
1
=
F
shoulder
=
F
elbow
−
F
biceps
=
1100
N
(
compression
)
N
2
=
F
biceps
+
F
shoulder
=
F
elbow
=
1500
N
(
compression
)
N
3
=
F
biceps
+
F
shoulder
=
F
elbow
=
1500
N
(
compression
)
N
4
=
F
biceps
+
F
shoulder
=
F
elbow
=
1500
N
(
compression
)
e)
What is the total change in length, in millimeters, of the entire system (make sure you use the correct sign)? Since the system is segmented, the total change in length is equal to the sum of the individual
changes in length:
δ
total
=
∑
i
=
1
4
δ
i
=
∑
i
=
1
4
N
i
L
i
E
i
A
i
Remember to account for whether a segment is undergoing compression or tension – in this case, all segments are under compression. Therefore, we need to account for this when summing together the individual changes in length. You also need to get the units correct (e.g., if you change everything to N, mm, and MPa, the result will be in mm):
δ
=
−
(
1100
N
) (
10
mm
)
(
20
e
3
MPa
)
(
315
mm
2
)
−
(
1500
N
) (
60
mm
)
(
20
e
3
MPa
)
(
315
mm
2
)
−
(
1500
N
) (
120
mm
)
(
20
e
3
MPa
)
(
200
mm
2
)
−
(
1500
N
) (
130
mm
)
(
20
e
3
MPa
)
(
315
mm
2
)
=−
0.092
mm
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[Solution] Problem 5.19
Consider the following simplified model of a dental implant, which consists of a solid titanium cap on top of an exterior of porous tantalum surrounding a core of solid titanium:
In addition to the lengths shown on the diagram, you have the following information:
Total change in length δ
= -0.0015 mm
Elastic modulus of titanium E
Titanium
= 100 GPa
Elastic modulus of porous tantalum E
Tantalum
= 10 GPa
Cross-sectional area of the titanium cap = 12 mm
2
Cross-sectional area of the porous tantalum exterior = 7 mm
2
Cross-sectional area of the titanium core = 5 mm
2
Answer the following questions and remember to show your work:
a)
State your assumptions.
Assumptions:
1.
Uniaxial loading
2.
Each component / segment is a prismatic bar
3.
Titanium and porous tantalum are linear elastic materials
4.
Titanium and porous tantalum are homogeneous materials
5.
Titanium and porous tantalum are perfectly bonded to each other (see part c)
b)
Write an equation of equilibrium that shows that a portion of this system is statically indeterminate.
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We can write an equation of equilibrium for the portion of the implant that has a porous tantalum outside and a solid titanium core:
F
=
N
Ti
+
N
Ta
c)
Write an appropriate equation of compatibility for the statically indeterminate portion of the system.
Since the second segment of the implant is statically indeterminate, we need an equation of compatibility to solve for the internal forces in the tantalum and titanium compartments. If we assume perfect bonding (a.k.a. well bonded or something about it being well attached in some
way)
, we can write the following equation of compatibility:
δ
Ti
=
δ
Ta
d)
Determine the relationship between the internal normal forces in the porous tantalum exterior and the titanium core.
Using our assumptions from part (a), we can substitute relationships for the change in length of each compartment, giving us the following:
N
Ti
L
E
Ti
A
Ti
=
N
Ta
L
E
Ta
A
Ta
Simplifying and solving for one force in terms of the other (I solved for the internal force in the
titanium core, but it does not matter which you choose), we get:
N
Ti
=
E
Ti
A
Ti
E
Ta
A
Ta
N
Ta
N
Ti
=
E
Ti
A
Ti
E
Ta
A
Ta
N
Ta
=
(
100
GPa
)(
5
mm
2
)
(
10
GPa
)(
7
mm
2
)
N
Ta
=
7.143
N
Ta
e)
Determine the applied force F necessary to cause the indicated total change in length δ = -
0.0015 mm.
Now that we have found the relationship between the internal forces in part d, we can figure the value of the uniaxial compressive force F
. First we note that this is a segmented system:
δ
total
=
δ
cap
+
δ
core
∨
exterior
For the second segment, since each part deforms the same as the other, we can either use the
change in length of the titanium core OR the tantalum exterior – BUT not both. Since we are given the total change in length, we can use that to solve for the force F.
Our first step is to figure out the internal force in the second (statically indeterminate) segment as a fraction of F
:
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F
=
N
Ti
+
N
Ta
=
8.143
N
Ta
N
Ta
=
0.1228
F
Now we can use our segmented system approach to find the total change in length (note the negative signs because the force and resultant deformation is compressive):
δ
total
=
δ
cap
+
δ
Taext
=
N
1
L
1
E
Ti
A
cap
+
N
Ta
L
2
E
Ta
A
Ta
δ
total
=−
0.0015
mm
=
−
F
(
1.5
mm
)
(
100,000
MPa
)
(
12
mm
2
)
−
0.1228
F
(
8.5
mm
)
(
10,000
MPa
)
(
7
mm
2
)
F
=
92.8
N
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