Solution Manual for The Mechanics of Biological Materials, 1E By Manuel Elices, Gustavo Guinea, John

Solution Manual for The Mechanics of Biological Materials, 1E By Manuel Elices, Gustavo Guinea, John Morton

EXERCISE 1(BF) A useful and simple approach to the behaviour of bundles of fibres like tendons and ligaments is that shown in figure A, which is part of a more complex model proposed by Lanir1. The model considers the effect of the nonuniform waviness of the fibres that, upon stretching, are gradually recruited due to their differences in their extended ("free") lengths, giving rise to the characteristic J-shaped macroscopic stress-strain curve of these bundles of fibres.

The stretching of the bundle is characterised by its elongation Λ = L/L0, where L and L0, are respectively the actual (stretched) and initial (unloaded, extended) lengths. A single fibre (i) in the bundle is also characterised by its elongation λi = L/L0i, where L0i is now the extended (or free) length of the unloaded fibre. When the elongation of the bundle reaches the value of the straightening elongation θ i = L0i/L0, the fibre (i) will begin to carry load (si>0). The model assumes that each fibre is thin and perfectly flexible, has no compressive strength and buckles under zero load if compressed.

This way, the fibre (i) will be unloaded until an elongation of the bundle Λ = θi and its stress will be determined by its elongation λi , that can be expressed as λi = L/L0i = (L/L0) (L0/L0i) = Λ / θi .

To account for the waviness of the bundle, an undulation density distribution function Φ(θ) is defined such that Φ(θ) dθ is the joint cross-sectional area dA of the fibres with straightening elongation between θ and θ + dθ, as shown in figure B. The value θm corresponds to the elongation at which all the fibers in the bundle will have been stretched.

For a bundle of linear elastic fibres each with modulus of elasticity E and elongation at break λR, derive an expression for the stress (S) - elongation (Λ) relationship and compute the modulus of elasticity E* of the bundle for the region with linear behaviour in the following cases: a) constant undulation density function Φ=const. b) positive linear undulation density function Φ= const. (θ –1) c) negative linear undulation density function Φ= const. (θm – θ)

Figure A

Figure B

1

Lanir Y., Constitutive equations for fibrous connective tissues. J Biomech. 1983;16(1):1-12. doi: 10.1016/00219290(83)90041-6

1

Solution: The force in the bundle will be the sum of the contribution of all the fibres: ∫

∫

∫

where we have made use of the definition of the undulation density distribution and the relationship between the elongations in the bundle and the fibre.

Considering that in a fibre = (λ –1) E >0 for 1< λ < λR, we can obtain:

and the limits of the integral can be made explicit as [

]

∫ [

]

a) In this case Φ= Φ 0 = A/(θm – 1) , since the total area of the bundle is A = Φ 0 (θm – 1) .

The equation to calculate the force is: [

]

∫

( [

[ [

,

)

]

] ]

[

]

[

]-

and the stress-elongation curve for the bundle results: [ [

,

] ]

[

]

[

]-

Linear bundle behaviour is reached when all fibres have been recruited (Λ > θm) and none have broken (Λ < λR). In this case min[Λ,θm]= θm and max[1,Λ/λR]=1. The S - Λ relationship for this region is: {

}

and the slope is:

b) now the undulation density function is: And the total force on the bundle: [

]

∫

( [

{ *

)

]

+

[

]}

2

and the stress-elongation relationship for the bundle is given by: { *

[

+

]}

The linear bundle behaviour (θm < Λ < λR ; min[Λ,θm]= θm , max[1,Λ/λR]=1) is: { [

]

[

]}

and the slope is:

c) finally, in a similar manner, the undulation density function is: and the total force on the bundle: [

]

∫

( [

)

]

{ [

]

[

]

[

]}

with the stress-elongation relationship for the bundle being: { [

[

]

]

[

]}

Again, for linear bundle behaviour (θm < Λ < λR ; min[Λ,θm]= θm , max[1,Λ/λR]=1) we have: { [

]

[

]}

and the slope is:

It can be shown that for any θm > 1 is The figure shows the three curves S - Λ for a bundle of fibres with E=300MPa, λR =1.19 and θm =1.05. The respective values of E* are given by: E*a /E = 0.9758

E*b /E = 0.9679

E*c /E = 0.9837

and which lie within 4% of E. The experimental values of Huang et al.2 for the human palmarius longus tendon are also shown in the figure.

2

Huang et al. Biomechanical Properties of Knee Medial Collateral Ligament Compared to Palmaris Longus for Ulnar Collateral Ligament Reconstruction. Ann Biomed Eng. 2023 Aug;51(8):1795-1801. doi: 10.1007/s10439-023-03188-z

3

EXERCISE 2(BF) For the models described in EXERCISE 1 prove that for any θm >1

Solution:

To demonstrate the inequalities, let us consider the differences,

which will be greater that zero as long as

.

To prove this last inequality, consider the function,

whose derivative is

This last function confirms that,

and consequently,

Similar reasoning leads us to see that if

, then,

4

for any θm >1, and the inequalities are proven.

EXERCISE 3(BF) To reinforce an injured ligament (I) a parallel grafting with another ligament (G) has been performed. The two ligaments can be considered as two separate bundles of collagen fibres, whose cross-sectional area and free (unloaded) extended length are provided in the table.

If the repaired ligament has an initial resting length (unloaded) L0=15 mm and the modulus of the elasticity of all the collagen fibres is E=933 MPa, calculate the following: a) The stress-elongation (S – Λ) curve for the repaired ligament. b) The modulus of elasticity E* of the repaired ligament when all the fibers are subjected to tension.

bundle

cross-sectional area

free length

1.2 2.5

15 19

I G

Solution: The force in the repaired ligament will be the sum of the force of the two bundles,

where are the cross-sectional areas and their respective elongations. We have made use the linear elastic relationship between stress and elongations. We also assume that fibres have no compressive strength, and the load is zero for λ <1.

When the repaired ligament has an elongation Λ = L/ L0, the elongations in the bundles are and with θI and θG being the elongations when the fibres begin to carry load. From the data we obtain: θI =15/15= 1, and θG = 19/15 = 1.267.

The stress-elongation relationship becomes,

(

[

)

Inserting values [

5

The stress-elongation (S – Λ) and force-length (F– L) curves are shown in the figure

b) For Λ ≥ θG, all the fibres are loaded and (

)

which gives E*= 800.3MPa .

Note that if θI = θG = 1, E* = E = 933MPa

6

EXERCISE 4(BF) Consider the general behaviour of elastomer fibres under constant length, as observed by Meyer and Ferry, given by equation chap.2-22, a) show that the general force-length relationship is given by b) show that, under a constant force, fibres contract with an increase in temperature c) derive an expression to calculate the coefficient of thermal expansion α. d) calculate the value of α for an incompressible Gaussian fibre.

Solution: a) The constitutive equation for a fibre of length at temperature T is given by the relationship . Under constant length, , the force-temperature relationship will be and for elastomers given by equation chap.2-22 we have:

and consequently,

Note that for a stretched fibre, F and T are positive and so is function of .

, that is an increasing

b) If we consider a fibre under constant force at two different temperatures T and T* we have

with

being the corresponding lengths. If we suppose that

since

we can obtain

, and then

so that . Consequently, the length of the fibre length reduces from to when temperature increases from to . This is the general behaviour for all fibres satisfying equation chap.2-22, rooted in the entropic origin of their forces (equation chap.2-25). c) The coefficient of thermal expansion is defined as the relative variation of length upon a change in temperature under constant force

From equation chap.2-22 we have

with

⁄

and then

which has a negative value since both

and

are positive.

7

d) The force-length relationship for a Gaussian incompressible fibre is given by equation chap.2-38

where

(

)

(

)

⁄ . We note that,

and then (

with

⁄

⁄ ⁄

)

.

Values of (β T) as a function of the elongation are shown in the figure. For large values of elongation β becomes –1/T.

8

EXERCISE 5(BF) Obtain the constitutive equation following experimental behaviour:

for an incompressible fibre meeting the

– under constant temperature is – under constant force is where is a positive function of temperature and derive the stress strain-curve.

is a constant. Using the result,

Solution: From the constant temperature condition, we obtain by integration in :

Now, we differentiate this equation to obtain.

taking d

and substituting

from the second condition (

which only can be achieved for any

)

if both, and

From these two equations we obtain, and being

.

Imposing as reference results

and

, the constitutive equation for the fibre (

)

The engineering stress – elongation curve will be (

)

and is represented in the figure.

9

The true stress – elongation curve is obtained recalling equation chap.2-7 for incompressible fibres ( and is also shown in the figure. Note that For small

)

when

.

we have [

] [

]

10

EXERCISE 6(BF) Show that the viscoelastic standard linear solid model described in section 2.5.3 can be also represented by the mechanical model shown in the figure.

Obtain the relationships between the two sets of spring and damper parameters.

Solution: Consider the stress-strain relations in the components of the model: Hookean spring E1: Damper spring E2: Damper: with

and

̇

being the stresses on the springs and the damper, respectively.

Equilibrium requires that

, and compatibility of strains leads to

.

Combining the above equations, we obtain: ̇ from which ̇

̇

̇

̇

This can be rearranged as ̇

̇

or ̇

̇

An equation with the same functional dependence on σ and ε as equation chap.2-45.

Identifying the terms in the variables and their time derivatives between these two equations, we obtain:

11

from which we can draw the equivalences:

(

)

12

EXERCISE 7(BF) Use the viscoelastic standard linear solid model to obtain the tensile stress-strain curve of a viscoelastic fibre under a loading-unloading cycle at constant rate α, as shown in the figure.

Assume. that the fibre has never been subjected to previous stretching.

Solution: a) The constitutive equation of viscoelastic standard linear model is given section 2.5.3, equation chap.2-45:

Τhe strain input is defined piecewise as: { For the first interval (loading), the constitutive equation is:

This is a first-order differential equation. The homogeneous solution is:

The particular solution for the linear term is:

Thus, the total solution is: The constant A can be determined using the initial condition, σ1(0) =0, giving:

For the second interval (unloading), the constitutive equation now becomes:

13

This is also a first-order linear differential equation, having the particular and homogeneous solutions:

The constant A´ is determined by ensuring continuity from the previous interval: and then

from which the stress solution for interval 2 is obtained:

The complete solution for the cycle is , The figure plots the stress solution for a standard linear solid with EH = 35.5kPa, EV = 22kPa; τR = 31s and a loading-unloading cycle with ε0 = 0.5 and T = 560s (values derived for skeletal muscle fibres from Böll3.

It is worth noting that the value of stress at the end of the cycle is always: [

]

since cosh(x) ≥ 1. This value is only possible is the fibre does not buckle, which is unrealistic. A fibre subjected to the prescribed cycle will buckle before reaching ε = 0, as show in the stress-strain figure.

3

Böll et al. Mechano-geometrical skeletal muscle fibre characterization under cyclic and relaxation loading, https://doi.org/10.1016/j.jmbbm.2020.104001

14

EXERCISE 8(BF) Use the viscoelastic standard linear solid model to obtain the tensile stress-strain curve of a viscoelastic fibre under a loading-unloading cycle at constant rate α as shown in the figure.

Assume that the fibre has been previously subjected to many similar loading-unloading cycles until reaching a steady state.

FIGURE A

Solution: Consider the steady state cycle 0 – T shown in figure B.

FIGURE B Due to the internal residual strains, the stress in fibre drops to 0 at time t 2 < T during unloading (see EXERCISE 7). From this point until a time T+t1 in the next cycle the fibre will remain unloaded. In general, we have that and for any sufficiently large integer n.

To obtain the values for t1, t2 consider the stress solution for the loading and unloading phases of the cycle. The constitutive equation of the viscoelastic standard linear model is given by equation chap.2-45:

Τhe strain input can be defined piecewise as: { For the first (loading) interval, the constitutive equation is:

15

This is a first-order differential equation whose general solution is (see EXERCISE 7):

For the second (unloading) interval, the constitutive equation is:

whose general solution is:

The following conditions apply: i) at t =T/2, stress continuity leads to: or:

ii) at t1 and t2 stress is zero:

iii) considering that from t2 until a time T+t1 the fibre is completely unloaded we can write and then

The differential strain on the fibre in the interval (t2,T+t1) is

since

Solving for dε, and taking into account that dε = dεH and εH = σH/EH= –σV/EH

this equation can be integrated in the interval (t2, T+t1) to obtain,

Note that since T+t1–t2 >0, must be ε2 > ε1 . Finally, substituting ε1=α t1 and ε2= α(T–t2) we arrive at

The above four equations from i), ii) and iii) can be solved numerically for t1, t2, A and A´ to obtain the steady stress/strain solution.

16

As an example, the steady state has been obtained for a standard linear solid with EH = 35.5kPa, EV = 24kPa; τR = 31s and a loading-unloading cycle with εmax = 0.5 and T = 560s (values derived from Böll4.

The set of equation to solve are the following:

with A, A´in kPa and t1, t2 in s. The numerical solution is A= – 2.939kPa A´= + 22234kPa t1 = 11.27s t2= 539.07s The strains at times t1, t2 are ε1 = α t1 =0.02013 ε2=α (T– t2) =0.03738 Figure C shows the steady state cycle and compares it with the first loading path.

Differences are of minor.

FIGURE C

4

Böll et al. Mechano-geometrical skeletal muscle fibre characterization under cyclic and relaxation loading, https://doi.org/10.1016/j.jmbbm.2020.104001

17

EXERCISE 9(BF) Figure A shows the average of relaxation tests performed by Böll 5. They first applied an increasing deformation at high rate until reaching a stretching elongation of λ=1.5, and then held this deformation constant. a) Use the viscoelastic standard linear solid model to fit this relaxation tests and obtain its parameters. b) Apply the results to model the average tensile behaviour of the fibres shown in figure B.

Figure A

Figure B

5

Böll et al. Mechano-geometrical skeletal muscle fibre characterization under cyclic and relaxation loading, https://doi.org/10.1016/j.jmbbm.2020.104001

18

Solution: a.1) The viscoelastic standard linear model is described in section 2.5.3. The solution for the relaxation test is given by equation chap.2-46: ⁄

A fit to the stress relaxation curve –neglecting the initial loading part– gives: ⁄

with ε0=λ-1=0.5, and σ in kPa and t in s. The curve is shown as fit 1 in figure C.

FIGURE C Note that if t' is considered, so that t'=0 at the point of maximum stress (t=5.78s), a similar fit is obtained (fit 2 in figure C): ⁄

with ε0=0.5, and σ in kPa and t´ in s. In both cases, the fit is reasonable (R=0.97). a.2) A more exact approach can be obtained if the whole tensile process is modelled.

In this case, the strain input is defined piecewise as: { with ε0 = αT. The constitutive equation of viscoelastic standard linear model is given Section I Biological Fibres 2.5.3, equation chap.2-45:

For the first (loading) interval, this equation is:

The homogeneous solution of this first-order differential equation is:

and the particular solution for the linear term is:

19

The total solution is then: The constant A is determined using the initial condition, σ1(0) = 0, and the stress results:

For the second (relaxation) interval, the constitutive equation now is:

This is again a first-order linear differential equation, having the particular and homogeneous solutions:

The constant A´ is determined by ensuring continuity from the first interval: and then

from which the stress solution for interval 2 is obtained:

The complete solution is then: , Note that when the term with equation chap.2-46.

The fit of experimental results for

,and the solution for

coincides

=5.78s to the above equation results in: ⁄

⁄

with ε0 =0.5, α = ε0 / T, and σ in kPa and t in s. The values obtained and the coefficient of regression R are practically identical to the previously obtained. The fit is show in figure C as fit 3. The table summarises the results.

EH (kPa) EV (kPa) τR (s) R

fit 1 35.54 26.41 30.93 0.97

fit 2 35.54 21.90 30.97 0.97

fit 3 35.54 24.01 30.95 0.97

20

b.1) Noval curve.

EXERCISE 7obtained the solution for the tensile stress-strain when the fibre had not been previously loaded - we call this the noval curve of a viscoelastic fibre - under a loadingunloading cycle at constant rate α shown in figure D.

FIGURE D The solution for the cycle is: , From figures B and D, we can set:

with these values and EH = 35.54kPa, EV = 24.01kPa and τR = 30.95s the stress-strain curve shown in figure E is obtained. b.2) Steady state.

EXERCISE 8 derived the solution for the steady-state tensile stress-strain curve of a viscoelastic fibre under a loading-unloading cycle at constant rate α. The curve is shown in figure E. Final comments: The fit in both cases, noval and steady-state, is poor, mainly due to the fact that in the standard linear solid the elastic behaviour is linear, in contrast to all soft tissues that show a characteristic J-shaped stress-strain curve.

Furthermore, using a relaxation curve to fit the standard linear solid model is quite complicated since, to obtain a set of parameters which is good enough to model tests with different strain rates, the whole relaxation curve must be well fitted. Otherwise, the ability to model high frequencies may be lost if the relaxation curve is a poorly fit for small t values, and conversely, low frequencies will be not well represented if the relaxation curve is poorly fitted for large t values.

In our case, if we perform a fit of the relaxation curve in figure C for the time interval (100 to 400), the parameters obtained are: EH= 34.4kPa, EV= 80.7kPa, τR= 34.3s, with R= 0.83.

Though the coefficient of regression is very low, this new fit better approximates the maximum load and unloading behaviour as shown in Figure E.

21

FIGURE E

22

EXERCISE 10(BF) Analyse the effect of specimen size on the tensile strength of fibres of diameter D and length L using Weibull theory for bulk flaws without threshold. Determine the dependence of the mean failure stress on specimen size in the following three cases: (a) D and L vary proportionally (b) D varies while L is kept constant (c) D is kept constant while L varies.

Solution: a) Let and two groups of fibres of different size, with constant of proportionality. The volumes of the two groups will be given by

being the

The mean failure stress for the Weibull model without threshold is given by equation chap.2-67: ( ) and then, the mean failure stress for the two groups will be. ( ) (

)

from which we obtain ( b) When the two groups are

)

( and

)

( )

. Their volumes are given by,

and the ratio of the mean failure stresses for the two groups becomes, ( c) Similarly, with the two groups

) and

(

) the volumes are

and the ratio becomes, (

)

( )

23

The figure shows the fit of the average failure stresses data from Neves 6 obtained on bamboo fibres with a constant length L0 = 60mm and diameters in the ranges 0.1-0.2mm, 0.2-0.3mm, 0.3-0.4mm, 0.4-0.5mm, 0.5-0.6mm and 0.6-0.7mm.

The group of fibres with D0 = 0.15mm mean diameter was taken as reference. Error bars show the standard error (n=20).

6

Neves et al. Weibull analysis of the tensile strength dependence with fiber diameter of giant bamboo; J. Materials Research and Technology, 6, 4, October–December 2017, pp. 317-322, http://dx.doi.org/10.1016/j.jmrt.2017.07.001

24

EXERCISE 11(BM) Surface tension arises at the interface between two phases as the result of cohesive forces between molecules at the interface. The surface tension coefficient  is defined as the energy required to increase the interfacial area by one unit. a) Show that the normal force per unit length acting along the boundary ∂A of the interface is . b) Show that, at equilibrium, an interface satisfies Laplace’s law (

)

where ∆p is the pressure difference between the inside and outside of the interface, and R1 and R2 are the radii of curvature in two perpendicular directions along the interface. Apply the results above to obtain the internal pressure of a soap bubble that has a radius of 1cm and 0.8µm thickness. Assume a surface tension  =0.03 N/m c) The figure depicts a simple model of a lung alveolus. The alveolar wall is covered by an internal film primarily composed of water and is surrounded by pleural fluid. Obtain the alveolar radius at rest R1, and after normal tidal volume inspiration R2, and compute the circumferential elongation of the alveolar wall.

Apply the following assumptions: - the alveolar wall is unloaded before inspiration - alveoli are spherical and identical, with a total volume of 2500ml before inspiration - the tidal volume is 500ml - intrapleural gauge pressure is constant and equal to –4mmHg. - the surface tension coefficient for the internal fluid film is  =0.03 N/m.

Solution: a) Consider a segment d𝓁 along the boundary ∂A as shown in the figure. df represents the normal force acting on it. If the segment moves a distance dx, the work done by the force will be dW = df ∙ dx = df dx ∙ n, and the area generated is dA= d𝓁 dx ∙ n .

The work per unit surface is then

𝓁

25

b) The pressure difference between the inside and outside of a membrane was given by equation chap.5-27 (

)

where h is the thickness of the membrane and σ11 and σ22 are the normal stresses acting in directions 1, 2. Noting that the product h σii represents the normal force per unit length acting along direction i, we can express this equation as: (

)

(

)

(

)

which is the required equation. Note that this result is applicable only to a single interface.

The figure depicts the situation of a soap bubble where we have two interfaces. Laplace’s law applied to them gives

(

)

from which

with the values given we obtain ∆p =pin–pout = 12Pa. c) At rest, the alveolar wall is unstressed, and according to Laplace's equation for membranes (equation chap.5-27) we have

where σ1 is the circumferential stress on the wall, and pfluid,1 and ppleura the pressures on both sides of the alveolar wall. From this equation we obtain pfluid,1= ppleura.= – 4mmHg = – 533.3Pa. Now, using Laplace's law for the interface fluid-air we obtain

26

from which we obtain

since pair,1= 0 is the atmospheric pressure.

The total volume of the alveoli before inspiration is then

where N is the number of alveoli. From this equation we obtain N=419.09 106.

In the same way, after completing the inspiration, the new volume will be

and then R2=119.56µm.

The circumferential elongation can be calculated as

27

EXERCISE 12(BM) Use the model presented in section 5.2.4, for the simultaneous extension and inflation of a homogeneous thin-walled tube made from an isotropic material, to obtain an estimate of the circumferential stress-elongation curve of the arterial wall from pressure-diameter tests.

Apply the result to the pressure-diameter curves shown in figure chap.5-17 for a Yeoh incompressible material and compare the results to the uniaxial tensile curve shown in figure chap.5-17 for that same material.

Solution: According to the model in section 5.2.4, the circumferential stress σθθ produced by the extension and inflation of the tube is given by equation chap.5-64:

where p is the pressure. R and H are the initial radius and thickness, and λθ and λ3 the elongation in circumferential and axial direction, respectively.

The value of λθ can be obtained from the variation of the diameter according to chap.5-54:

In which d=2r, is the current diameter, and D=2R the reference, initial, diameter. The required relationships are then: and

and the values of σθθ and λθ can be directly derived from p-D curves.

Application to Yeoh's p-D curves The left figure shows the computed p-D curves for a Yeoh's incompressible material with parameters α = 40 kPa, β = – 142 kPa and γ = 1200 kPa. In this case p-D curves are all referred to a same initial unstressed (D=3mm, H=0.5mm) configuration. Consequently, to obtain a reasonable σθθ – λθ estimate from curves other than that λ3 =1 we take, as reference for circumferential elongation, the initial diameter of each p-D test, to have λθ =1 as the starting elongation in all cases. With this correction, and using the above equations, we obtain the σθθ – λθ curves shown in the figure on the right.

28

300 Cauchy stress s (kPa)

Uniaxial s - l curve 250

Estimates from p-D curves with l3= 1

200

1.1 1.2

150

1.3 1.4

100 50 0 0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

elongation, l

We can observe that the estimates for σθθ – λθ are in reasonable agreement with the uniaxial behaviour even when they are obtained from p-D curves with high axial elongation λ3. This is usually the case for isotropic membranes. However, the greater the anisotropic behaviour, the lower the agreement between the estimates and the actual behaviour, as shown in the data below for coronary arteries – data from Claes7 – where predictions even from curves with λ3 =1 are not satisfactory.

250

150 coronary artery thickness, H=1.0mm, l =1 3

Cauchy stress, s(kPa)

pressure (mmHg)

200

150

100

100

estimate sqq - lqq

50

50

0 6.2

6.4

6.6

6.8

7.0

diameter (mm)

7.2

7.4

0 1

1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 elongation, l

7

Claes, Els (2010). Estudio mecánico de las arterias coronarias humanas y sus sustitutos vasculares. , UPM, https://doi.org/10.20868/UPM.thesis.3859

29

EXERCISE 13(BM) a) Show that the velocity divergence is zero (div preserves the area (det F = 1)

= 0) for any motion in a membrane that

b) Use the result in a) to demonstrate that in such a motion any hydrostatic stress σ = q 1 (q = constant) defined over an area of the membrane does not produce work on any arbitrary enclosure defined on the membrane.

Solution: a) The deformation gradient tensor Fij is defined as (equation chap.2-5a)

where x, X are the vector positions of a particle in the deformed and reference configurations, respectively. The velocity of any particle is obtained by taking the time derivative at constant X , i.e. at a fixed point (material derivative)

The divergence can be calculated as (

)

with , being the components of the velocity vector. We have used the Einstein summation convention, where when an index variable appears twice it implies the sum of all possible values (1,2) of that index. Now, noting that derivate in time (t) and particle (X) commute we can write (

)

since

is the inverse deformation gradient tensor. The values of F–1 can be derived from F as [

]

with det F = F11 F22 F12 F21 . The divergence is then

This is easily seen to be equal to zero since det F = 1.

30

b) The work done by the hydrostatic stress σ = q 1 on a segment ds on the boundary of an enclosure in a time interval dt is (see the figure) ̂ ̂ is the stress vector acting on the surface (h ds). The total where is the velocity and work done on an enclosure B on the membrane will be obtained by integration at the boundary ∂B ∫

̂

∫

̂

∫

where we have made use of Gauss's theorem (or divergence theorem) to transform the contour integral into an integral over the area enclosed. Since in a) we have shown that div = 0 when det F = 1 we conclude that the hydrostatic stress σ = q 1 does not produce work on any enclosure of the membrane.

The two results above can be generalized for a 3D situation. In general it can be shown 8 that

and div v = 0 when detF = 1. (*).

8

P. Chadwick, Continuum Mechanics: Concise Theory and problems. Dover Publications, 1999

31

EXERCISE 14(BM) Determine the Cauchy stress tensor and the Green-Lagrange strain tensor for a biaxial test performed on a membrane with initial dimensions L1 =100, L2 =100, H =5 mm. Assume that the material is homogeneous, isotropic and incompressible, with a neo-Hookean constitutive equation with strain energy density per unit volume W = 2α tr( ), where is the Green-Lagrange tensor and α = 100MJ/m3.

Calculate the stresses and strains: a) for f1 = 50kN and f2 = 100kN b) for f1 = 50kN and f2 = 0 c) for f1 = 0 and f2 = 100kN

Solution: Recalling the analysis shown in section 5.2.3 for a homogeneous biaxial test, we can write the equations of the displacement as

and with λ1 λ2 λ3 = 1 as required by the conservation of volume. The deformation gradient tensor is [

]

and the Green-Lagrange strain tensor

(

) [

]

The Cauchy stress tensor is obtained as (section 5.2.3, Biological Membranes)

[

]

is a constant hydrostatic stress that can be obtained by equlibrium. The force balance at the faces of the membrane leads to

32

[

]

and consequently

from which we obtain

(

)

(

)

a) solving the above equations for f1 = 50kN and f2 = 100kN, with α = 100MJ/m3, L1 =100mm, and H =5mm we obtain λ1= 1.01113, λ2 = 1.37569, λ3 = 0.7189. The stress and strain tensors are [

]

[

]

b) the solution for f1 = 50kN and f2 =0 is λ1= 1.19743, λ2 = 0.91385 and λ3 = 0.91385. The stress and strain tensors are [

]

[

]

c) the solution for f1 =0 and f2 = 100kN is λ1= 0.826031, λ2 = 1.465571 and λ3 = 0.826031.

The stress and strain tensors are [

]

33

[ Note that σa ≠ σb + σc and

a ≠

b+

]

c

34

EXERCISE 15(BM) Determine the Cauchy stress tensor and the Green-Lagrange strain tensor for the biaxial test of EXERCISE 14 assuming that the membrane preserves its area.

Solution: The equations of the displacement are

and

with the conditions

as required by the conservation of volume and the area of the membrane. This conditions can be rewritten as

The 2D deformation gradient tensor

is [

]

and the Green-Lagrange strain tensor

(

) [

]

The Cauchy stress is (section 5.2.3)

*

+ [

]

that must balance the forces on the faces of the membrane. The resulting equations are

That, together with λ1 λ2=1, can be solved for λ1, λ2 and q

35

a) solving for f1 = 50kN and f2 = 100kN, with α = 100MJ/m3, L1 = L2 =100mm and H =5mm, we obtain λ1= 0.821618 , λ2 = 1.21711, and q = –52.85MPa. The stress and strain tensors are then *

+

*

+

b) the solution for f1 = 50kN and f2 =0 is λ1= 1.15278; , λ2 = 0.86747, q = –150.5MPa. The stress and strain tensors are *

+

*

+

c) the solution for f1 =0 and f2 = 100kN is λ1= 0.72449; , λ2 = 1.38028, q = –105.0MPa. The stress and strain tensors are * * Note that again σa ≠ σb + σc and

a ≠

b+

+ +

c

36

EXERCISE 16(BM) Determine the residual circumferential deformations in the wall of a blood vessel as a function of the angle of aperture α (see section 4.3.3). Make use of Navier’s hypothesis that through-thickness plane sections of the wall remain planar after cutting the circular section of the vessel.

Solution: According to Navier’s hypothesis, the deformation of a small section of the wall ds after the cutting produces a rotation angle dφ (see figure below)

where λi and λo are the inner and outer residual elongations, respectively, and h is the wall thickness. Consequently, the curvature of the deformed section is

where the negative sign indicates that the centre of curvature lies outside the vessel. The total curvature, after the cutting, is obtained by adding the initial value 1/ρ0

Now, the perimeter of the cut ring is From these two last equations we can solve for the elongations (

)

(

)

(

)

(

)

If we neglect any change in the perimeter, πρ0 = (π-α)ρ and, then, the elongations are

37

EXERCISE 17(BM) Determine the equations governing the radius of curvature of a membrane made up of several layers, each with different length, thickness, and modulus of elasticity, which are joined together, as indicated in the figure.

Assume that the layers are linearly elastic, homogeneous, and isotropic and that the deformations are small. Use Navier's hypothesis that plane sections through the thickness remain plane after deformation.

Apply the result to predict the residual curvature of an axial section of arterial wall, as measured by Holzapfel9, if it is formed of three layers (intima, media and adventitia) with the characteristics given in the table for the separated, unstressed configuration.

layer intima media adventitia

1 2 3

thickness, hi (mm) 0.23 0.57 0.40

length, Li (mm) 21.7 19.6 18.8

modulus of elasticity10, Ei (kPa) 120 93 51

Solution: According to Navier’s hypothesis, and due to the linear-elastic behaviour, the stress state in a layer can be represented by a uniform extensional stress σm and a bending stress distribution with maximum value σb, as depicted in the figure.

Mechanical equilibrium for the membrane on any cross-section requires: a) Equilibrium of axial forces 9

Holzapfel et al., Layer-specific 3D residual deformations of human aortas with non-atherosclerotic intimal thickening. Ann Biomed Eng. 2007;35(4):530-45. doi: 10.1007/s10439-006-9252-z 10 Holzapfel GA.; Determination of material models for arterial walls from uniaxial extension tests and histological structure. J Theor Biol. 2006;238(2):290-302. doi: 10.1016/j.jtbi.2005.05.0067

38

∑

∑

with Pi = σmi hi being the axial tensile force per unit width in each layer. Note that bending distributions do not generate net axial force. b) Equilibrium of moments about the bottom of the cross-section ∑

∑

∑

(

)

Mi is the bending moment per unit width on each layer, given by

which is related with the radius of curvature ρi at the neutral axis of the layer through equation chap.8-20

with ρi being ∑

c) Finally, to ensure that the lengths at the interface match for every pair of adjacent layers (i,i+1) we have:

where εi,t and εi,b are the axial strains at the top and at the bottom surfaces of layer i, respectively. These strains are due to the uniform axial extension in the layer

plus the bending strain

Taking into account the different signs of the bending strain in the top (compression, negative) and bottom (tension, positive) surfaces of the layer, we can write:

and

and the compatibility equations become: (

)

(

)

The above equations can be expressed in terms of Pi and ρ

39

∑

∑

(

∑

(

∑

(

∑

)

) )

∑

(

)

(

∑

(

) )

and solved for obtaining Pi and ρ .

In the case of a membrane made of three layers, the system of equations becomes:

(

(

)

(

)

(

)

)

Solving the system for: h1=0.23mm h2=0.57mm h3=0.40mm

L1=21.7mm L2=19.6mm L3=18.8mm

E1=120kPa E2=93kPa E3=51kPa

we obtain layer intima-1 media-2 adventitia-3

Pi (N/m) Mi (10–3 N/m2) –0.825 0.986 –0.161

0.0158 0.1961 0.0398

top, εi,t −0.0448 −0.0203 −0.0372

strain bottom, εi,b −0.0150 0.0576 0.0214

stress (kPa) top, σi,t bottom, σi,b –5.375 –1.799 –1.891 5.353 –1.897 1.089

The computed radius of curvature is then ρ=6.633mm. This value is consistent with the values measured by Holzapfel, which had an average of 6.9mm.

The interfacial lengths are 20.7mm between the intima and the media, and 19.2mm between the media and the adventitia. The average length of the membrane is 19.7mm.

40

EXERCISE 18(BM) Determine the minimum adhesion energy necessary to prevent an adherent film from peeling off a substrate under tension, as depicted in the figure. Assume that both the film and substrate exhibit linear elastic behaviour.

Apply this result to the detachment of a plaster (band-aid) from the skin. For this calculation, use the thicknesses and elastic moduli provided in the table.

band-aid human skin11

thickness, h (mm) 0.5 1.0

Elastic modulus, E (MPa) 400 0.08

Solution: Consider an interfacial crack which increases in length by da, as shown in figure (A) below.

Due to symmetry, we consider only half of the assembly. The energy balance when work Fdx is done is:

where Λ is the elastic energy of the system, and RD is the (specific) adhesion energy.

The value of x can be determined by taking into account that, to a first approximation (see figure B), both the substrate – over a length a – and the film-substrate assembly – over a length (L – a) –are subjected to uniaxial traction. Then,

in which εs and εm are the values of the deformation in the substrate and in the filmsubstrate assembly, respectively. These values are

11

Feng et al., In vivo stiffness measurement of epidermis, dermis, and hypodermis using broadband Rayleigh-wave optical coherence elastography, Acta Biomaterialia, 2022, https://doi.org/10.1016/j.actbio.2022.04.030

41

and where Em is the modulus of elasticity of the composite (film+substrate), which can be computed from equation chap.8-77 in terms of the volume fractions

The final value of x is then

The elastic energy is

Differentiating x and Λ and substituting into the energy balance equation we obtain

from which

which is independent of the crack propagation length.

Applying the above equation to the detachment of a band-aid from the skin, and assuming a maximum value of εs =0.2, the threshold value of RD to propagate the crack is

This value is small compared to common adhesives, which range from 100 to 200 J/m2 ( e.g. Magic Tape).

In the above analysis, a single substrate strip of width B, equal to the width of the film, has been considered. Figure (C) represents a more realistic situation, in which the substrate can be considered to have a width much greater than that of the film.

42

In this case we can assume that beyond a width B* the effect of the film adhered to the sustrate is negligible. Consequently, the analisys of the deformation of film and substrate can be made with a width B for the film and a width B* for the sustrate.

This assumption changes the value of the deformation in the substrate, that now is

In addition, the modulus of elasticity of the composite (film + substrate) Em now is computed as

and the deformation εm is

where we have defined,

The final result, following the same steps as above, is

Since the elasticity modulus of the skin is several orders of magnitude lower than that of the plaster (band-aid) material, the correction made is not relevant for the calculated value of RD . For the same εs =0.2 and a value B* = 10B we obtain ⁄ ⁄

43

EXERCISE 19(BM) Figure a) illustrates the results of a tear test (also known as the Elmendorf or trouser test) conducted by Purslow12 on the aortic wall of a pig. Figure b) shows the results of a uniaxial tensile test on the same material. Both tests were carried out with the loading applied in the circumferential direction. Calculate the specific tear resistance R of the aortic wall in circumferential direction.

Solution: Assuming nonlinear elastic behaviour, the specific tear resistance can be obtained by equation chap.5-91

where the subscript R indicates values at rupture. If strains far from the tear are negligible compared to 1 (i.e. λR ≈ 1) and, consequently, the elastic strain energy density WR is also small, the tear resistance is approximated by

12

Purslow PP. Positional variations in fracture toughness, stiffness and strength of descending thoracic pig aorta. J Biomech.

1983;16(11):947-53. doi: 10.1016/0021-9290(83)90058-1

44

The average tearing force from the graph provided is FR=1.18N and, with B=1.5mm, the resulting value is R=1.57kJ/m2.

Alternatively, we can account for the effect of the elasticity through equation chap.5-91, which can be expressed as ∫ since F = s B W, where s is the engineering stress and e the engineering strain.

During crack propagation, the average stress in each leg of the sample is

and from the stress-strain curve (figure b) we obtain eR= 0.297, from which λR = 1.297. The value of the integral is the area under the s–e curve between 0 and eR, which gives 14.13 kJ/m3. With these values we obtain R=1.869 kJ/m2. This value is 19% greater than the previous one.

Finally, we can compute R more accurately through the work done during the test. The work will be the area A enclosed by the force-displacement curve. Its value is A=35.32Nmm. The crack propagation ∆a=26mm, is estimated from the displacements at the beginning (2mm) and end (28mm) of the test. Then, the specific tear resistance can be computed from equation chap.5-90 as:

where the factor 2 accounts for the work done by the two forces. This result aligns closely with the value derived using equation chap.5-91.

45

EXERCISE 20(BM) Derive an expression for the suture retention strength (SRS) of a biological material from its specific tear strength, R. The SRS evaluates the maximum force that a suture thread can withstand before tearing. This is determined using a standardised test13, as illustrated in the accompanying figure.

Assume that the membrane is subjected to a state of uniaxial tension and that its behaviour is linear-elastic. Apply the result to the determination of SRS of pig aorta (B=1.5mm, R=1.6 kJ/m2), bovine pericardium (B=0.26 mm, R=34kJ/ m2) and rat skin (B=0.65mm, R=20kJ/ m2).

Solution: The SRS test is shown schematically in the figure. Under a load F, the membrane will be elongated by an amount (L0+a) = (1+e) (L0+a).

The deformation

is calculated using Hooke's Law

where E is the modulus of elasticity. If a tear of length da occurs in the membrane, the work done by the external force will be [

] (

)

During tearing, the elastic energy stored in the membrane will undergo a change of 13

Cardiovascular implants-Tubular vascular prostheses. Association for the Advancement of Medical Instrumentation, ANSI/AAMI/ISO 7198:1998/2001(R) 2004

46

(

)

(

)

where W = s e /2 is the energy density, and s=F/(BW), the engineering stress.

The energy balance is [

]

from which we deduce

since the term F2/(2BWE) is typically negligible compared to F. The value of the SRS is then.

The table compares the values of FSRS for the required tissues. In addition, some, values for human vessels have also been included

pig aorta calf pericardium rat skin internal mammary artery14 saphenous vein15

B(mm) 1.5 0.26 0.65 0.30 0.25

R(kJ/m2) 1.6 34 20 6.4 5.4

FSRS(N) 2.4 8.84 13.0 1.92 1.35

Note that the value of the FSRS is highly dependent on the thickness. In vascular surgery, the minimum FSRS value for a tissue to be successfully used** is between 0.5 and 0.7N.

14

Konig et al., Mechanical properties of completely autologous human tissue engineered blood vessels compared to human saphenous vein and mammary artery; Biomaterials 30 (2009); doi: 10.1016/j.biomaterials.2008.11.011 15 L’Heureux et al. Human tissue-engineered blood vessels for adult arterial revascularization. Nat Med. 2006;12(3). doi: 10.1038/nm1364

47

EXERCISE 21(BBm) A sample of articular cartilage with a fixed charge density of 0.2M is submerged in a saline buffer solution containing NaCl and Na₂HPO₄. At equilibrium, the solution is observed to contain 150 mmol of Na⁺ per litre and has an osmolarity of 280 mmol/litre.

If [ ]ₑ is the concentration in the external solution and [ ]ᵢ is the concentration in the interior of the cartilage, determine: a) the ion concentrations outside and inside the cartilage b) the osmotic pressure inside the cartilage, assuming the temperature is 37°C. Assume that both NaCl and Na₂HPO₄ dissociate completely.

Solution: a) From the dissociation reactions we obtain the ions involved in the electro-chemical equilibrium:

The electrical equilibrium condition is derived from the fact that the net charge on both sides must balance: [ ] [ ] [ ] [ ] [ ] [ ] where FCD is the fixed charge density (negative) attached to the proteoglycans of the cartilage. The Gibbs-Donnan equation (see section 9.4.3) now reads as: (

[ [

] ) ]

(

[ [

] ) ]

[ ( [

] ]

)

since the charge is Z=+1 and the charges of and are Z= –1 and Z=–2 , respectively. Additionally, the osmolarity C of the external solution is: [ ] [ ] [ ] The above five equations, together with C =280 mmol/litre, [ ] * + * + * = 0.2mol/litre, can be solved for [ + [ [ [ [ [

] =150mmol/litre and FCD *

+ to give:

] ] ] ] ]

b) The osmotic pressure inside the cartilage can be calculated as ] [ ] [ [] [] ([ ] [

]

[

]

[

] )

where [ ] [ ] is the difference in the ionic concentration between the inside and outside of the cartilage, T the absolute temperature and R the Gas Constant. Substituting the values of concentrations and T = 300K and R =8.31 J/mol/K we obtain: MPa

48

EXERCISE 22(BBm) Muscle contraction happens through the sliding of actin (thin) and myosin (thick) filaments, as depicted in the figure. Myosin heads bind to actin, pivoting and pulling in a “power stroke" action, shortening the sarcomere unit it and generating contraction. The figure also shows a micrograph of the cross-sectional view through the dotted line.

Given that the muscle can generate a stress of 200 kPa, determine the average number of myosin heads if each can sustain a load of 0.7pN16.

Solution: Assuming an equilateral triangular arrangement, as illustrated in the figure, we can calculate the force exerted by myosin fibres. The area of the equilateral triangle shown is: √

With a stress of 200 kPa, the corresponding force is:

This force is exerted by a total of 1/6 +1/6 +1/6 =1/2 myosin fibres. If N is the number of heads per fibre, we have:

Solving for N, we find that N=619 heads per myosin fibre. The number of myosin heads varies depending on muscle type and species, with vertebrate skeletal muscle fibres typically containing 600-800 heads17. .

16

Tyska et al.; Two heads of myosin are better than one for generating force and motion Proc. Natl. Acad. Sci. USA Vol. 96, pp. 4402–4407, April 1999; DOI:10.1073/pnas.96.8.4402 17 Huxley, H. E. ,The double array of filaments in cross-striated muscle. J Biophys Biochem Cytol. 1957 Sep 25;3(5):631-48. doi: 10.1083/jcb.3.5.631

49

EXERCISE 23(BBm) A homogeneous ceramic rod of length L and diameter D is dropped from rest in a horizontal position from a height H. Calculate the maximum possible value of H that guarantees that the rod does not break transversely.

Apply the result to two rods of aragonite and mother-of-pearl, each of 5 cm length.

Data: Bulk aragonite Mother-of-pearl

Density (g/cm3) 2.9 2.8

Fracture energy (J/m2) 50 600

Solution: a) A lower bound to the energy required to produce a transverse fracture of the bar is

where GC is the fracture energy, and A is its cross-sectional area. Any other process will consume more energy, either due to the presence of damage and/or deformation or due to a larger fracture surface.

If we assume –as a limiting case– that the potential energy of the bar is entirely converted into energy available for the fracture, we have

from which we obtain the minimum height required for the fracture:

Substituting the given values:

50

EXERCISE 24(BBm) The red abalone (Haliotis rufescens) shell exhibits anisotropic mechanical properties due to its unique microstructure. This structure, shown in the figure, consists of microscopic calcium carbonate (aragonite) tiles layered with organic proteins. Under small deformations the shell can be considered to be a transversely isotropic material with the axis of symmetry perpendicular to the layers, and with uniform properties in directions parallel to them.

Using the compression and shear test date obtained by Menig18 shown below, derive the elastic stress-strain relationship for a compression test in which the lamellae are oriented at 45º relative to the loading direction, and apply it to the experiments performed by Wang 19 also depicted in the figure.

100

Stress (MPa)

80

60

40

20

0 0.00

0.01

0.02

0.02

0.03

0.04

Strain

Wang et al.

16 700

14

Shear stress (MPa)

Stress (MPa)

600 500 400 300 200

10 8 6 4

100 0 0.00

12

2 0.01

0.02 Strain

Menig et al.

0.03

0.04

0 0.00

0.03

0.05

0.08

0.10

Shear strain

Menig et al.

18

Menig et al. Quasi-static and dynamic mechanical response of Haliotis rufescens (abalone) shells; Acta Materialia, 48, Issue 9, 29 May 2000, pp 2383-2398; https://doi.org/10.1016/S1359-6454(99)00443-7 19 Wang et al., Deformation mechanisms in nacre, J. Mater. Res., 16 (2001), pp. 2485-2493; DOI 10.1557/JMR.2001.0340

51

Solution: Determination of elastic constants Referring to transversely isotropic materials (section 8.3.2 and exercise 65), we can express the stress-strain relationship using the flexibility tensor:

[

]

[

[

]

]

Here, axes 1 and 2 are parallel to the lamellae, and axis 3 is perpendicular. From the compression tests performed by Menig, we identify the following relationships: for compression parallel to the lamellae, and

for the perpendicular direction. A fit to the initial linear part of the stress-strain curves gives and .

For the shear test:

A linear fit yields

.

Analysis of the Oblique Compression Test For the oblique compression tests performed by Wang, we derive expressions for the stresses and strains after a rotation of axes by an angle θ, as shown in the accompanying figure.

Stresses From equilibrium of forces along the rotated axes (1, 3), we derive:

52

and

from which

Note that in our problem is σzz = σxz = σyz = 0 and σ22 = σ12 = σ23 = 0 .

Strains

For an infinitesimal strain ε11 (see figure above):

⁄ ⁄ ( since

)

(

⁄

and

⁄

)

.

For a strain ε33 is now (see figure)

⁄

53

⁄ ( since

)

(

⁄

⁄

)

.

And for a shear strain ε13

since

and

.

Similarly, we can obtain, The shear strain εxy is Combining the effect of strains ε11, ε33 and ε13, we obtain

With the results above we can compute the stresses on the axes of the laminate (1, 2, 3) for a uniaxial loading at θ=45º with σyy , σxx = σxy = σzz = σxz = σyz = 0 Substituting θ = 45º in the equations obtained for the stresses, we obtain

jointly with σ22 = σ12 = σ23 = 0.

54

Then, through the flexibility (or compliance) tensor we can compute the strains to be

(

)

with ε23 = ε12 = 0.

Finally, the strains become, (

)

(

)

(

)

The fit of a linear relationship to the initial part of the stress-strain curve obtained by Wang et al. gives , that together with the values previously obtained from the experiments of Meng et al. ( , ) lead to an unrealistic value of the Poisson's coefficient .

On the other hand, if we make use of the values obtained by Yamamoto20 of predicted relationship between and is (

the

)

This predicts a shear modulus of almost an order or magnitude lower than measured by Wang et al. In addition to the variability of natural samples, the reason for this discrepancy might lie in the low value of obtained from the shear tests, in which it is very difficult to have complete control of the boundary conditions.

20

Yamamoto et al. ; Determination of transverse isotropic elastic constants of nacre and constituent tablets based on geneticalgorithm-assisted resonant ultrasound spectroscopy; Results in Materials, 15 (2022) 100312, https://doi.org/10.1016/j.rinma.2022.100312

55

EXERCISE 25(BBm) Using beam theory analysis, determine the relationship between the apparent modulus of elasticity (E*/ Es) and the density ratio (d*/ds) for the flat (plate-like) trabeculae shown in the figure when loaded, a) in the direction parallel ( ) to the planes of the trabeculae b) in the plane perpendicular ( ) to the planes of the trabeculae

Solution: a) In this case, the apparent stress and strain ( , ) are related through

where is the total load, and the apparent modulus of elasticity, and the apparent area perpendicular to the load. Since the load is only supported by the actual section of trabeculae, strain can be computed as

with Es and As being the modulus of elasticity and cross-sectional area of the trabeculae, respectively. From these equations we can develop the relationship

and then

since

the

mass

of

the

representative

prismatic

cell

is

.

56

b) Let us now consider the deformation of the representative prismatic cell in the plane perpendicular ( ) to the planes of the trabeculae. If we model the trabeculae as elastic beams of length 𝓁 and depth (see the figure), the displacement under an applied load must be 𝓁

𝓁

where is the second moment of area of the beam section. The strain in the cell will be proportional to the relative deflection of the beams, 𝓁 , so that 𝓁 𝓁

since the apparent area is now

𝓁

. The apparent strain is

and then 𝓁

Finally, taking into account the fact that the mass of the representative prismatic cell can be expressed as 𝓁 𝓁 and also 𝓁 , then 𝓁 r, resulting in ( )

Note the anisotropy in the behaviour of this trabeculae.

57

EXERCISE 26(BBm) Using the results of simple compression tests on trabecular bone performed by Hayes and Carter21 on subchondral trabecular bone shown in Figure 3.3.3, determine the relationship between the apparent modulus of elasticity (E*) and the density ratio (d*/ds) for this material, and fit an expression of the type (E*/Es) = C (d*/ds).

From this fit, indicate which trabecular model (see section 2.3.4, bulk biological materials) would be most suitable for the material: flat ( =1, 3, see EXERCISE 25), columnar ( =2, equation chap.8-104) or other type.

Solution: Figure A show the compression tests and the fit of the initial slope. The values of the modulus of elasticity as a function of apparent density are given in the table. d*/ds 0.31 0.39 0.53

Figure A

E* (MPa) 193 848 1150

Figure B

The fit of initial elastic modulus to apparent density is shown in Figure B. The coefficient of correlation is 0.90. As can be noticed, E* is best fit by a columnar trabeculae model, although the error is significative.

21

Hayes W.C., Carter D.R. (1976) Post yield behavior of subchondrial trabecular bone, J Biomed Mat Res Symp 7, 537-544

58

EXERCISE 27(BBm) A trabecular bone implant is needed to repair a lesion in a lumbar vertebra. The graft is sourced from the proximal tibia of a donor.

Given that the maximum working compressive stress on the vertebra is 0.7 MPa: a) Calculate the safety factor for the implant. b) Calculate the safety factor for the healthy vertebra.

Data: - Porosities: trabeculae in lumbar vertebra: Φvertebra=91% trabeculae in proximal tibia: Φtibia=84% - Yield strength dependency of trabecular bone (σys, s and ds are the yield stress and density of cortical bone, respectively): σ*ys / σys, s = (d*/ds)2 Assume that the maximum possible density of trabecular bone matches that of cortical bone (1800 kg/m3) and that σys, s =193 MPa. Solution: a) The yield stress for the implant is estimated from the dependence on density: (

)

where we have used the relationship between densities and porosity (Φ):

with V* being the apparent volume and Vs the volume of solid.

The safety factor for the implant is then obtained as

b) Similar reasoning gives, for the healthy vertebra:

59

EXERCISE 28(BBm) Bones serve as the primary reservoir of calcium in the body. The human skeletal system undergoes continuous remodelling, with bone mass turnover involving the resorption and deposition of calcium. Approximately 70% of bone mass is composed of hydroxyapatite (Ca5 (PO4)3 (OH)), a mineral in which calcium is the main component. For an average healthy adult weighing 70 kg (14% bone) the daily calcium turnover is approximately 0.5 grams22. a) Calculate the daily turnover of bone mass, assuming that the entire hydroxyapatite structure, not just the calcium component, is replaced at this rate. b) If the lost bone is not recovered, calculate the time required to reduce the bone mass by half.

Solution: a) The molecular weight of Hydroxyapatite (HA) is Calcium (Ca) 40.08 g/mol x 5 =200.4 g/mol Phosphorus (P) 30.97 g/mol x 3=92.9 g/mol Oxygen (O) 16.00 g/mol ×13=208.0 g/mol Hydrogen (H) 1.01 g/mol Total = 200.4 + 92.9 + 208.0 +1.01 =502.31 g/mol The mass of HA corresponding to a calcium turnover of 0.5 g/day will be: and the daily turnover of bone mass, assuming the entire hydroxyapatite structure is replaced, is approximately:

that represents a 100 x 1.8/(70,000 x 0.14) = 0.02 % / day or ~ 7% /yr.

In humans, the turnover rate depends on bone type and skeletal site, and can range from 5% /year for cortical bone to 20 % /year for trabecular23 b) The mass variation of bone is

Integration of this equation yields ∫

∫

and then ∆t = ln2/β = 3466 days ~ 9,5 years.

Alternatively, if the rate of bone loss is considered constant (and not relative to the current mass) and now ∆t = (m0/2)/1.8g/day = (0.14 x 3.5 104/2)/1.8 = 2722 days ~7.5 years.

22

Hall, J. E. (2015). Guyton and Hall Textbook of Medical Physiology (13th ed.). W B Saunders Parfitt, A.M., 2002. Misconceptions (2): turnover is always higher in cancellous than in cortical bone. Bone 30, 807–809. https://doi.org/10.1016/S8756-3282(02)00735-4 23

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EXERCISE 29(BBm) According to Frost’s mechanostat theory for bone growth24, a minimal effective strain of between 0.005% and 0.02% is required to prevent net bone mass loss. Within the range up to 0.1%–0.3% strain, bone remodelling remains in a steady state, supporting normal maintenance and adaptation.

The figure illustrates a simplified model for trabecular bone inspired by Frost's concepts. In this model, the growth rate of the solid fraction of bone, dƒ/dt, is influenced by compression deformation. Outside the strain range defined by εlow and εhigh strains, where bone mass remains stable, the growth rate maintains constant positive (+α) or negative (-α) values.

To validate the model, experiments were conducted on bone samples with initial mass, m0, and porosity Φ0. The samples were subjected to constant uniaxial compression at nominal (apparent) stress , while mass changes were tracked over time until they reached a constant value mc at time tc.

Derive an expression for εhigh and α based on the experimental data. To simplify the analysis, assume a homogeneous strain (Voigt) model with an elastic modulus Es and density for the solid bone.

Solution: a) In the Voigt model (see section 8.3.3) the loading is parallel to the laminas of solid bone (see figure). Under an external load the strain will be

where ƒ = Vs / V* = As / A* is the volume fraction of the solid bone, , are the strain and stress on the solid bone, and ,( ) denote the cross-sectional area and volume of the whole sample and the solid bone, respectively.

While

the fraction of solid bone will increase as ∫

∫

where ƒ(0)=ƒ0 =1– Φ0. Similarly, will have Frost, H.M. (2003). Bone’s Mechanostat: A 2003 Update. The Anatomical Record Part A: Discoveries in Molecular, Cellular, and Evolutionary Biology, 275A(2), 1081–1101. doi:10.1002/ar.a.10119 24

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∫

∫

and the mass ∫

∫

(

)

∫

According to the model, the growth of the bone will cease at time t such that

Note that the larger the stress applied

, the longer the time required to reach

.

Substituting the experimental values in the above equations we obtain

(

)

from which (

)

(

)

The figure shows the mass changes over time

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EXERCISE 30(BBm) Evaluate the probability of failure for structures subjected to nonhomogeneous uniaxial stresses. Assume an elastic-brittle material for which Weibull’s analysis is applicable and a simplified Weibull distribution function (refer to equation chap.2-64), setting the threshold stress to zero (σth=0).

Use this approach to obtain: a) the mean flexural strength of a prismatic beam under three-point bending –ignoring shear effects. b) the relationship between the flexural and the tensile strengths. Apply the result to the experimental values obtained by Currey25 for bone shown in the figure and give an estimation of the average Weibull's modulus of bone.

Solution: For structures under a uniform stress (such as fibres under tension), the probability of failure Pf according to Weibull's theory is given by (equation chap.2-62)

with being the Weibull defect concentration function for a zero threshold. Here A, m are constants, with m being the Weibull modulus.

If we consider a structure subjected to different uniaxial stresses σi acting in discrete volumes ∆Vi , the joint probability of survival will be ∑

∏ and for a continuous distribution of stresses, we have, ∫

∫

25

Currey JD. What determines the bending strength of compact bone? J Exp Biol. 1999 Sep;202(Pt 18):2495-503. doi: 10.1242/jeb.202.18.2495

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a) If we now consider a prismatic beam under three-point bending, such as that shown in the figure, the stress in the fibre at position y is (see the flexure test, section 8.2.1) | | with being the maximum stress, occurring at the lowest fibre (y = D/2). The probability of failure will be, (

∫

)

∫

| |

∫

since and , which is the flexural strength under three-point bending.

Differentiation of Pf with respect to σ3P yields the probability density function

and the mean flexural strength can be obtained as ̅​̅​̅​̅​̅

∫

(

)

where we have used the gamma function ∫ b) If a prism of the same material is subjected to a uniform uniaxial stress (i.e a tensile test), the mean failure stress σu will be given by (equation chap.2-67)

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with being the prism dimensions. The ratio of flexural (3P) and tensile mean strength values predicted by the Weibull model will be *

+

Data from Currey (figure) show that σƒ/σu ~ 350/200 = 1.75 and since the volume ratio of the specimens used is 0.2 , we have

from which m= 1.23.

The Weibull modulus for cortical bone has been reported26 to lie between 5 and 9, so the value obtained here does not align with direct experimental measurements. This discrepancy is likely due to the fact that, unlike pure brittle materials, which fail as soon as a microscopic flaw reaches a critical size, bone fails only after significant, stable growth of crack zones. This process leads to stress redistributions and the release of stored energy, which ultimately dominates over any potential statistical effects.

26

Bigley et al., Volume effects on yield strength of equine cortical bone. J Mech Behav Biomed Mater. 2008 Oct;1(4):295-302. doi: 10.1016/j.jmbbm.2007.11.001; Pithioux et al.,Comparison of compact bone failure under two different loading rates: Experimental and modelling approaches; Medical Engineering & Physics, 26 (2004), pp. 647-653

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