SOLUTIONS MANUAL for Biomaterials Science - An Introduction to Materials in Medicine 4th Edition by

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Chapter Questions

G=

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 deformed elastically in compression by a force of 500 N applied parallel to its length. The material has a Young’s modulus of 200 GPa, a shear modulus of 77 GPa, and a yield strength of 0.3 GPa. Calculate (A) the amount by which the rod will decrease in length; and (B) the amount by which the rod will increase in diameter.

Solution: (A) Before we solve the problems, it is helpful to convert all units to the SI units. Doing so, we have the length of the rod L = 0.1 m and diameter d = 0.002 m. By the definition of stress, σ = F/A, we calculate the compressive stress as σ=

− 500 (N) = π 2 = − 0.159 (GPa) A (0.002 m) 4 F

Since the induced compressive stress is below the yield strength of 0.3 GPa, the rod is deforming elastically. Moreover, because we are dealing with a rod under uniaxial loading, we can use the simplified Hooke’s Law to capture the linear relationship between stress and strain with: σ = Eε

E 2 ( 1 + ν)

we have ν=

E 2G

−1=

200 (GPa) 2 × 77 (GPa)

− 1 = 0.299

From the definition of the Poisson’s ratio εtransverse ν=− εlongitudinal and with the longitudinal strain determined earlier, we calculate εtransverse = − νεlongitudinal = − 0.299 × ( − 7.962) × 10 − 4 = 2.378 × 10 − 4 With this, the change in diameter can be determined as δ = dεtransverse = 0.002 (m) × 2.378 × 10 − 4 = 4.757 × 10 − 7 (m)

The positive value suggests that the diameter of the rod is becoming wider due to a compressive force.

2. Use a stress-strain curve to depict the determination of the followings: (A) strain energy, (B) resilience, and (C) toughness.

and find the induced strain in the rod ε=

σ E

=

− 0.159 (GPa) 200 (GPa)

= − 7.962 × 10 − 4

Knowing the strain value, the change in length of the rod can be determined by δ = Lε = 0.1 (m) × ( − 7.962) × 10 − 4 = −7.962 × 10 − 5 (m) = − 7.962 × 10 − 2 (mm)

The negative value suggests that the length of the rod is getting shorter due to a compressive force.

(B) To determine the change in diameter, we will need to know the Poisson’s ratio. From the known relationship between Yong’s modulus (E) and shear modulus (G), namely,

Solution: Using the stress-strain curve of a typical metallic material, we can sketch the followings.

(A) Strain energy is typically obtained by the area under the stress-strain curve at a given stress which is below the elastic yield point.

(B) Resilience is determined by the area under the stressstrain curve up to the point where stress reaches the elastic yield point.

(C) Toughness is determined by the area under the stressstrain curve up to the point where stress reaches the breaking point.

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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. 1.2.3.5A, we can write its stress and strain relations according to the generalized Hooke’s Law as,

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

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

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 situations.

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 taking the inverse of the generalized Hooke’s Law matrix equation as

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

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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 planestress 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 material 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: Show that when the situation can be further simplified to a 1D problem, a simple relationship between stress and strain as σx = Eεx

Will govern the constitutive relation.

Solution: For a 1D situation, we can further assume σy = τxy = 0. By plugging these two zero terms into the given relations, we have

With simplification, we obtain νσ σx = Eεx along with εy = − x and γxy = 0 E The first equation is the 1D Hooke’s Law governed constitutive relation, and the first two equations will lead to the Poisson’s ratio equation − νσx εy − =− σE =ν x εx E

A helical fracture line is surely the result of torsional loading. Referring to the Mohr’s circle and stress elements shown in Fig. 1.2.3.6, it is known that under a torsional load, the femur could break either in a shear failure mode or a tensile failure mode. If in a shear-failure mode, the fracture surface should be in a cross-cutting 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 follow a helical line due to the existence of the maximum tensile stress following a helical pattern. From the fracture shown, it is believed that the bone fractured under a 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 generated in the bone, thereby leading to a tensile-stress induced helical fracture due to the high-speed torsional load. This example highlights the fact that regardless the external loading situations, the actual fracture mode in a material is always dictated by the internal stress state and the nature of material, as well as the loading speed in the case of materials of viscoelastic nature.

7. Of the two images given below, one showing the stress trajectories obtained from a finite element analysis of a 2D femoral-head-like structure, and the other showing the X-ray microarchitectures in a femoral head. Explain your observations.

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52.e4

Chapter Questions

dεd dt

=

σ η

Taking the time derivative of the first two equations, we obtain dε 1 dσ dε dεs dεd and s = = + dt E dt dt dt dt With the three derivative expressions, we can eliminate the εs and εd terms, and arrive at the following governing differential equation in time domain for the viscoelastic material: dε dt

Solution: Inside the thin layer of strong cortical (or compact) bone of a femoral head is coarsely structured trabecular (or cancellous, spongy) bony structure. The microarchitecture of the spongy bony structure is not random, instead it appears to follow the stress trajectories inside the femoral head. According to Wolff’s Law, living bones will remodel in adaptation to the external loads they experience. The microarchitecture of the spongy bone is likely the result of the bone remodeling process in meeting the strengthening needs of human/animal growth and daily activities, guided by the stress trajectories dictated by the internal stress state.

8. A viscoelastic material represented by a spring (with modulus of E) and a dashpot (with viscosity of η) connected in series is illustrated below. The spring is regarded as governed by Hooke’s Law and the dashpot by Newton’s Law. Derive the governing differential equation for the material and find the time-dependent stress function in response to a step function of strain.

=

1 dσ E dt

+

σ η

To find the time-dependent stress function in response to a step function of strain, we first multiply the above differential equation by E and rearrange it to dσ

E dε + σ=E dt η dt

Then by taking the Laplace transformation of the above equation, we arrive at the following frequency-domain differential equation ( ) E σ ( s) s + = Eε (s) s η in which σ (s) and ε (s) are stress and strain in frequency domain, corresponding to the σ (t) and ε (t) in time domain, respectively, and s is the Laplace variable.

Let a step function of strain take the following form ε , t ≥ 0 ε ( t) = 0 0, t<0 then its Laplace transformed function will be ε0 ε ( s) = s Substituting this expression into the frequency-domain differential equation with some rearrangement, we obtain the stress response function in frequency-domain as

Solution: Since the spring and dashpot are connected in series, the stress in the material (σ ) will be experienced by the spring (σs = σ ) and the dashpot (σd = σ ), and the strain in the material will be shared by the spring and dashpot in the following relationship: ε = εs + εd

By applying Hooke’s Law to the spring, we have: σ εs = E and by applying the Newton’s Law to the dashpot, we have:

σ ( s) = (

Eε0 ) s+ E η

To find the time-dependent stress function, we will just take the inverse Laplace transformation. Do so leads us to the following stress response function in time domain: ( ) E σ (t) = E ε0 exp − t η Clearly, the time-dependent stress function in response to a step function of strain is an exponential delaying function. This behavior is commonly referred to as the stress

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