Solutions Manual for Biomaterials Science - An Introduction to Materials in Medicine 4th Edition by Wagner, William R.; Zhang, Guigen; Sakiyama-Elbert, Shelly E.; Yaszemski, Michael J.

Solutions Manual for Biomaterials Science - An Introduction to
Materials in Medicine 4th Edition by Wagner, William R.; Zhang,
Guigen; Sakiyama-Elbert, Shelly E.; Yaszemski, Michael J.




Solutions Manual for Biomaterials Science - An Introduction to
Materials in Medicine 4th Edition by Wagner, William R.; Zhang,
Guigen; Sakiyama-Elbert, Shelly E.; Yaszemski, Michael J.

,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. ε
ν = − transverse
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-
F − 500 (N) mined as
σ= = = − 0.159 (GPa)
A π (0.002 m)2 δ = dεtransverse = 0.002 (m) × 2.378 × 10 − 4
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
ε = = – 0.159 (GPa) = − 7.962 × 10 − 4 Solution:
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

, Solutions Manual for Biomaterials Science - An Introduction to Materials in Medicine 4th Edition by Wagner,
William R.; Zhang, Guigen; Sakiyama-Elbert, Shelly E.; Yaszemski, Michael J.


52.e2 Questions



3. Referring to the generalized Hooke’s Law relation and

the [c] matric given in Fig. 1.2.3.5A for an isotropic along with


For a plane-stress situation, we first express the above
generalized stress-strain relations in strain-stress relations by
material, derive the reduced stress-strain relation for a taking the inverse of the generalized Hooke’s Law matrix
2D plane-strain and plane-stress situation. equation as

Solution:




With the [c] matrix for an isotropic material given in Fig.
1.2.3.5A, we can write its stress and strain relations accord-
ing to the generalized Hooke’s Law as,



By substituting the given expressions in this chapter for
c11 and c12, we have




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




With simplification, we obtain the following reduced
stress-strain relation for a 2D plane-strain situation:
Solutions Manual for Biomaterials Science - An Introduction to Materials in Medicine 4th Edition by Wagner,
William R.; Zhang, Guigen; Sakiyama-Elbert, Shelly E.; Yaszemski, Michael J.

, Referring to Fig. 1.2.3.8A, we have the three zero stress
components: σz = τyz = τxz for a plane-stress situation.
Plug- 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:


or its inverse stress-strain relation



along with


4. For an isotropic elastic material having Young’s modulus
of E =210 GPa and Poisson’s ratio of =0.33, find
the
[c] matrix for 2D simplified plane-stress and plane-strain
situations.

Solution:
For a plane-strain situation, by plugging the given values
for E and into the reduced stress-strain relations found in
Question 3, we have



For a plane-stress situation, by plugging the given values
for E and into the reduced stress-strain relations found in
Question 3, we get