Biomaterials Science: An Introduction to Materials in
Medicine, Fourth Edition
(Instructor's Solution Manual) by William R Wagner
,Chapter Questions G=
E
2 ( 1 + ν)
1.
A stainless-steel rod with a circular cross-section has
a length of 100 mm and a diameter of 2.0 mm. It is we have
deformed elastically in compression by a force of 500 N
applied parallel to its length. The material has a Young’s E 200 (GPa)
ν= −1= − 1 = 0.299
modulus of 200 GPa, a shear modulus of 77 GPa, and 2G 2 × 77 (GPa)
a yield strength of 0.3 GPa. Calculate (A) the amount
by which the rod will decrease in length; and (B) the From the definition of the Poisson’s ratio
amount by which the rod will increase in diameter. εtransverse
ν=−
εlongitudinal
Solution:
and with the longitudinal strain determined earlier, we cal-
(A) Before we solve the problems, it is helpful to convert culate
all units to the SI units. Doing so, we have the length εtransverse = − νεlongitudinal =
of the rod L = 0.1 m and diameter d = 0.002 m. By the − 0.299 × ( − 7.962) × 10 − 4 = 2.378 × 10 − 4
definition of stress, σ = F/A, we calculate the compres-
sive stress as With this, the change in diameter can be deter-
mined as
F − 500 (N)
σ= = π 2 = − 0.159 (GPa) δ = dεtransverse = 0.002 (m) × 2.378 × 10 − 4
A (0.002 m)
4 = 4.757 × 10 − 7 (m)
Since the induced compressive stress is below the yield The positive value suggests that the diameter of
strength of 0.3 GPa, the rod is deforming elastically. More- the rod is becoming wider due to a compressive force.
over, because we are dealing with a rod under uniaxial
loading, we can use the simplified Hooke’s Law to capture 2. Use a stress-strain curve to depict the determination of
the linear relationship between stress and strain with: the followings: (A) strain energy, (B) resilience, and (C)
toughness.
σ = Eε
and find the induced strain in the rod
σ − 0.159 (GPa) Solution:
ε= = = − 7.962 × 10 − 4
E 200 (GPa)
Using the stress-strain curve of a typical metallic material,
Knowing the strain value, the change in length of we can sketch the followings.
the rod can be determined by (A) Strain energy is typically obtained by the area under the
stress-strain curve at a given stress which is below the
δ = Lε = 0.1 (m) × ( − 7.962) × 10 − 4 = elastic yield point.
−7.962 × 10 − 5 (m) = − 7.962 × 10 − 2 (mm) (B) Resilience is determined by the area under the stress-
strain curve up to the point where stress reaches the
The negative value suggests that the length of the
elastic yield point.
rod is getting shorter due to a compressive force.
(C) Toughness is determined by the area under the stress-
(B) To determine the change in diameter, we will need to know
strain curve up to the point where stress reaches the
the Poisson’s ratio. From the known relationship between
breaking point.
Yong’s modulus (E) and shear modulus (G), namely,
52.e1
,52.e2 Chapter Questions
3. Referring to the generalized Hooke’s Law relation and
the [c] matric given in Fig. 1.2.3.5A for an isotropic
material, derive the reduced stress-strain relation for a
2D plane-strain and plane-stress situation.
Solution:
With the [c] matrix for an isotropic material given in Fig. Referring to Fig. 1.2.3.8A, we have the three zero stress
1.2.3.5A, we can write its stress and strain relations accord- components: σz = τyz = τxz for a plane-stress situation. Plug-
ing to the generalized Hooke’s Law as, ging in these zero components into this equation we have
With simplification, we obtain the reduced strain-stress
relation for a 2D plane-stress situation:
By substituting the given expressions in this chapter for
c11 and c12, we have
or its inverse stress-strain relation
For a plane-strain situation, by referring to Fig. 1.2.3.8B
ε =γ =γ .
the following three strain components are zero: z yz xz
Plugging in these zero components into the above matrix
equation we have
along with
4. For an isotropic elastic material having Young’s modulus
of E =210 GPa and Poisson’s ratio of ν =0.33, find the
[c] matrix for 2D simplified plane-stress and plane-strain
With simplification, we obtain the following reduced situations.
stress-strain relation for a 2D plane-strain situation:
Solution:
For a plane-strain situation, by plugging the given values
for E and ν into the reduced stress-strain relations found in
Question 3, we have
along with
For a plane-stress situation, we first express the above
generalized stress-strain relations in strain-stress relations by For a plane-stress situation, by plugging the given values
taking the inverse of the generalized Hooke’s Law matrix for E and ν into the reduced stress-strain relations found in
equation as Question 3, we get
, Chapter Questions 52.e3
6. The image given below shows a helical fracture in a femur
bone. Explain your observation.
Apparently, the two [c] matrices are not the same. The [c]
matrix of the plane-strain situation is stiffened as compared
with that of the plane-stress situation. The two terms in the
first row are 3.110×1011 GPa and 1.530×1011 GPa for the
plane-strain situation, which are approximately 1.32 and
1.97 times higher than the respective terms in the plane-
stress situation (i.e, 2.360×1011 GPa and 0.778×1011 GPa).
This suggests that treating a plane-stress problem by mistake
as a plane-strain problem would lead to stiffening the mate-
rial significantly. On the contrary, treating a plane-strain
problem as a plane-stress one would lead to softening the
material.
5. The constitutive relations between strains and stresses for
a 2D plane-stress simplified problem can be expressed in
a reduced 3×3 matrix equation as follows
Solution:
A helical fracture line is surely the result of torsional
Show that when the situation can be further simplified loading. Referring to the Mohr’s circle and stress ele-
to a 1D problem, a simple relationship between stress and ments shown in Fig. 1.2.3.6, it is known that under a
strain as torsional load, the femur could break either in a shear
failure mode or a tensile failure mode. If in a shear-failure
σx = Eεx
mode, the fracture surface should be in a cross-cutting
Will govern the constitutive relation. plane perpendicular to the shaft of the femur because the
this plane is where the maximum shear stress lies, and
if in tensile-failure mode, the fracture surface will fol-
Solution: low a helical line due to the existence of the maximum
For a 1D situation, we can further assume σy = τxy = 0. By tensile stress following a helical pattern. From the frac-
plugging these two zero terms into the given relations, we ture shown, it is believed that the bone fractured under a
have torsional load. However, the load was applied very likely
at a relatively high speed such that the bone behaves
more like a brittle material. In this situation, the tensile
strength of the bone becomes the weakest link because
it will not be able to resist the highest tensile stress gen-
erated in the bone, thereby leading to a tensile-stress
induced helical fracture due to the high-speed torsional
With simplification, we obtain load. This example highlights the fact that regardless the
νσ external loading situations, the actual fracture mode in a
σx = Eεx along with εy = − x and γxy = 0 material is always dictated by the internal stress state and
E
The first equation is the 1D Hooke’s Law governed con- the nature of material, as well as the loading speed in the
stitutive relation, and the first two equations will lead to the case of materials of viscoelastic nature.
Poisson’s ratio equation 7. Of the two images given below, one showing the stress
trajectories obtained from a finite element analysis of a
εy − νσx 2D femoral-head-like structure, and the other showing
− =− σE =ν the X-ray microarchitectures in a femoral head. Explain
εx x
E your observations.
Medicine, Fourth Edition
(Instructor's Solution Manual) by William R Wagner
,Chapter Questions G=
E
2 ( 1 + ν)
1.
A stainless-steel rod with a circular cross-section has
a length of 100 mm and a diameter of 2.0 mm. It is we have
deformed elastically in compression by a force of 500 N
applied parallel to its length. The material has a Young’s E 200 (GPa)
ν= −1= − 1 = 0.299
modulus of 200 GPa, a shear modulus of 77 GPa, and 2G 2 × 77 (GPa)
a yield strength of 0.3 GPa. Calculate (A) the amount
by which the rod will decrease in length; and (B) the From the definition of the Poisson’s ratio
amount by which the rod will increase in diameter. εtransverse
ν=−
εlongitudinal
Solution:
and with the longitudinal strain determined earlier, we cal-
(A) Before we solve the problems, it is helpful to convert culate
all units to the SI units. Doing so, we have the length εtransverse = − νεlongitudinal =
of the rod L = 0.1 m and diameter d = 0.002 m. By the − 0.299 × ( − 7.962) × 10 − 4 = 2.378 × 10 − 4
definition of stress, σ = F/A, we calculate the compres-
sive stress as With this, the change in diameter can be deter-
mined as
F − 500 (N)
σ= = π 2 = − 0.159 (GPa) δ = dεtransverse = 0.002 (m) × 2.378 × 10 − 4
A (0.002 m)
4 = 4.757 × 10 − 7 (m)
Since the induced compressive stress is below the yield The positive value suggests that the diameter of
strength of 0.3 GPa, the rod is deforming elastically. More- the rod is becoming wider due to a compressive force.
over, because we are dealing with a rod under uniaxial
loading, we can use the simplified Hooke’s Law to capture 2. Use a stress-strain curve to depict the determination of
the linear relationship between stress and strain with: the followings: (A) strain energy, (B) resilience, and (C)
toughness.
σ = Eε
and find the induced strain in the rod
σ − 0.159 (GPa) Solution:
ε= = = − 7.962 × 10 − 4
E 200 (GPa)
Using the stress-strain curve of a typical metallic material,
Knowing the strain value, the change in length of we can sketch the followings.
the rod can be determined by (A) Strain energy is typically obtained by the area under the
stress-strain curve at a given stress which is below the
δ = Lε = 0.1 (m) × ( − 7.962) × 10 − 4 = elastic yield point.
−7.962 × 10 − 5 (m) = − 7.962 × 10 − 2 (mm) (B) Resilience is determined by the area under the stress-
strain curve up to the point where stress reaches the
The negative value suggests that the length of the
elastic yield point.
rod is getting shorter due to a compressive force.
(C) Toughness is determined by the area under the stress-
(B) To determine the change in diameter, we will need to know
strain curve up to the point where stress reaches the
the Poisson’s ratio. From the known relationship between
breaking point.
Yong’s modulus (E) and shear modulus (G), namely,
52.e1
,52.e2 Chapter Questions
3. Referring to the generalized Hooke’s Law relation and
the [c] matric given in Fig. 1.2.3.5A for an isotropic
material, derive the reduced stress-strain relation for a
2D plane-strain and plane-stress situation.
Solution:
With the [c] matrix for an isotropic material given in Fig. Referring to Fig. 1.2.3.8A, we have the three zero stress
1.2.3.5A, we can write its stress and strain relations accord- components: σz = τyz = τxz for a plane-stress situation. Plug-
ing to the generalized Hooke’s Law as, ging in these zero components into this equation we have
With simplification, we obtain the reduced strain-stress
relation for a 2D plane-stress situation:
By substituting the given expressions in this chapter for
c11 and c12, we have
or its inverse stress-strain relation
For a plane-strain situation, by referring to Fig. 1.2.3.8B
ε =γ =γ .
the following three strain components are zero: z yz xz
Plugging in these zero components into the above matrix
equation we have
along with
4. For an isotropic elastic material having Young’s modulus
of E =210 GPa and Poisson’s ratio of ν =0.33, find the
[c] matrix for 2D simplified plane-stress and plane-strain
With simplification, we obtain the following reduced situations.
stress-strain relation for a 2D plane-strain situation:
Solution:
For a plane-strain situation, by plugging the given values
for E and ν into the reduced stress-strain relations found in
Question 3, we have
along with
For a plane-stress situation, we first express the above
generalized stress-strain relations in strain-stress relations by For a plane-stress situation, by plugging the given values
taking the inverse of the generalized Hooke’s Law matrix for E and ν into the reduced stress-strain relations found in
equation as Question 3, we get
, Chapter Questions 52.e3
6. The image given below shows a helical fracture in a femur
bone. Explain your observation.
Apparently, the two [c] matrices are not the same. The [c]
matrix of the plane-strain situation is stiffened as compared
with that of the plane-stress situation. The two terms in the
first row are 3.110×1011 GPa and 1.530×1011 GPa for the
plane-strain situation, which are approximately 1.32 and
1.97 times higher than the respective terms in the plane-
stress situation (i.e, 2.360×1011 GPa and 0.778×1011 GPa).
This suggests that treating a plane-stress problem by mistake
as a plane-strain problem would lead to stiffening the mate-
rial significantly. On the contrary, treating a plane-strain
problem as a plane-stress one would lead to softening the
material.
5. The constitutive relations between strains and stresses for
a 2D plane-stress simplified problem can be expressed in
a reduced 3×3 matrix equation as follows
Solution:
A helical fracture line is surely the result of torsional
Show that when the situation can be further simplified loading. Referring to the Mohr’s circle and stress ele-
to a 1D problem, a simple relationship between stress and ments shown in Fig. 1.2.3.6, it is known that under a
strain as torsional load, the femur could break either in a shear
failure mode or a tensile failure mode. If in a shear-failure
σx = Eεx
mode, the fracture surface should be in a cross-cutting
Will govern the constitutive relation. plane perpendicular to the shaft of the femur because the
this plane is where the maximum shear stress lies, and
if in tensile-failure mode, the fracture surface will fol-
Solution: low a helical line due to the existence of the maximum
For a 1D situation, we can further assume σy = τxy = 0. By tensile stress following a helical pattern. From the frac-
plugging these two zero terms into the given relations, we ture shown, it is believed that the bone fractured under a
have torsional load. However, the load was applied very likely
at a relatively high speed such that the bone behaves
more like a brittle material. In this situation, the tensile
strength of the bone becomes the weakest link because
it will not be able to resist the highest tensile stress gen-
erated in the bone, thereby leading to a tensile-stress
induced helical fracture due to the high-speed torsional
With simplification, we obtain load. This example highlights the fact that regardless the
νσ external loading situations, the actual fracture mode in a
σx = Eεx along with εy = − x and γxy = 0 material is always dictated by the internal stress state and
E
The first equation is the 1D Hooke’s Law governed con- the nature of material, as well as the loading speed in the
stitutive relation, and the first two equations will lead to the case of materials of viscoelastic nature.
Poisson’s ratio equation 7. Of the two images given below, one showing the stress
trajectories obtained from a finite element analysis of a
εy − νσx 2D femoral-head-like structure, and the other showing
− =− σE =ν the X-ray microarchitectures in a femoral head. Explain
εx x
E your observations.
