The Materials of the Living

Abstract

This chapter delves into the mathematical toolbox of continuum mechanics, necessary to describe the mechanical deformation of biological materials. After being relegated for a long time to little more than the (useful, but quite uninspiring) role of providing healing and implants for broken body parts, the domain of biomechanics has been recently promoted to the highest level of attention. The modern view of biomaterials relies on connecting structures over multiple length-scales. Functional organisation is found at all levels, from the molecules, to fibrils and coils, to the carefully networked microstructures, up to the macroscopic scale. At odds with the variety of materials available in nature, just a small set of recurring molecule types, arranged in ever different structures, can give rise to such diverse tissues as skin, cartilage and bone, green stems, rose buds or wood. The secret is the hierarchical multi-level structure of biomaterials.

Appendices

Appendix H: Materials Elasticity Theory for Dummies

9.1.1 Stress

Consider a body with a generic shape, and apply a force oriented along a generic direction, at a point of its surface. If we define the perpendicular to the surface through that point by some unit vector $\mathbf n =(n_x,n_y,n_z$), and the force as another vector $\mathbf f =(f_x,f_y,f_z$), the stress can be defined by combining the dependence on both vectors, as shown in Fig. 9.22(left). This is a mathematical quantity with two subscript indices, $\underline{\sigma }=\sigma _{ab}$, with $a,b=(x,y,z)$, describing the possible combinations of the Cartesian components of the two vectors, called a tensor:

$$\begin{aligned} \sigma _{ab} = \frac{1}{A} \frac{\partial f_a}{\partial n_b} \end{aligned}$$

(9.21)

The $3 \times 3$ components of the tensor can be written in the form of a matrix. If the force is acting only along one or more of the three directions, and there is no coupling between the forces in different directions, the stress tensor matrix will have non zero components only on its diagonal. If moreover the three components of the force are identical, we have $\sigma _{xx}= \sigma _{xx} =\sigma _{xx} = \sigma _0$, and it can be proved that the pressure p is the trace of the stress tensor:

$$\begin{aligned} p = \frac{1}{3} \text {Tr}[\underline{\sigma }] = \frac{\sigma _{xx}+\sigma _{yy}+\sigma _{zz}}{3} = \sigma _0 \end{aligned}$$

(9.22)

Fig. 9.22

Schematic representation of the geometrical interpretation of the stress (left) and strain (right) tensors. Left A force f is applied at a point on the surface of a body with arbitrary shape. The vector n indicates the perpendicular direction to the element of surface dA (little grey circle). The stress components $\underline{\sigma }=(\sigma _{xx},\sigma _{xy},...,\sigma _{zz})$ represent the variation of each of the components $f_x$, $f_y$, $f_z$ of the force vector, according to the components $n_x$, $n_y$ or $n_z$ of the perpendicular vector. Right A point located at the vector position r in the undeformed orange body, is displaced at $\mathbf r ^{\prime }$ in the deformed body. The strain components $\underline{\varepsilon }=(\varepsilon _{xx},\varepsilon _{xy},...,\varepsilon _{zz})$ represent the variation of each of the components $u_x$, $u_y$, $u_z$ of the displacement vector $\mathbf u =\mathbf r -\mathbf r ^\prime$, with respect to the components of the undeformed position vector $r_x$, $r_y$, $r_z$

If the stress tensor is given, the force across a surface element of the body can be defined as a mechanical tension:

$$\begin{aligned} t_{a} = \frac{1}{A} \sum _{b} \int _{A} (\underline{\sigma } \otimes \mathbf n ) dA = \frac{1}{A} \sum _{j} \int _{A} (\sigma _{ij} \cdot n_j) dA \end{aligned}$$

(9.23)

where the symbol ‘$\otimes$’ indicates that the product between a tensor and a vector follows the special rules of matrix multiplication.

9.1.2 Strain

Similar to stress, the strain is mathematically defined as well as a tensor, $\underline{\varvec{\varepsilon }}$ with indices describing the deformation $\mathbf u =\mathbf{r }-\mathbf{r }^{\prime }$ of a vector $\mathbf{r }^{\prime }$ to any point along each Cartesian direction, with respect to the original position vector r, in that same, or another Cartesian direction, as shown in Fig. 9.22(right). The general definition of the strain tensor, symmetric in the Cartesian components, is:

$$\begin{aligned} \varepsilon _{ij} = \frac{1}{2} \left( \frac{\partial u_i}{\partial r_j} + \frac{\partial u_j}{\partial r_i} \right) \end{aligned}$$

(9.24)

The rigorous definition of the displacement vector $\mathbf u$ should also include local material rotation, $\mathbf u =\mathbf r -\mathbf{r }^{\prime }+\omega$; however, the rotational component of deformations will not be considered in this book, except in special cases.

It can be noted that the symmetry of the Euclidean space implies that the off-diagonal components of the stress and strain tensors are symmetrical, i.e. $a_{ij}=a_{ji}$ for $i \ne j$. Therefore the nine components of each tensor are reduced to six.

Any tensor a admits a unique and additive decomposition into a diagonal (or trace) component, $a_{kk}$, and a deviatoric (or traceless) component $\overline{a}_{ij}$:

$$\begin{aligned} a_{ij} = \tfrac{1}{3}a_{kk}\delta _{ij}+\overline{a}_{ij} \end{aligned}$$

(9.25)

(the Kronecker symbol $\delta _{ij}$ is always zero except for $i=j$).

To make a bit less cumbersome the notation in the following Sections, the so-called Voigt convention for numbering the tensor components can be followed, namely: $xx=1$, $yy=2$, $zz=3$, $xy=yx=4$, $yz=zy=5$, $xz=zx=6$. In this way, tensors have only one index running from 1 to 6.

9.1.3 Elastic Constants and Compliances

In the regime of small deformations for which the linear approximation can be applied, stress and strain are proportional to each other via the matrix of elastic constants, C, and elastic compliances, S :

$$\begin{aligned} \sigma _i = \sum _j C_{ij} \varepsilon _j \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, i,j=1,...,6 \end{aligned}$$

(9.26)

$$\begin{aligned} \varepsilon _i = \sum _j S_{ij} \sigma _j \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, i,j=1,...,6 \end{aligned}$$

(9.27)

where the Voigt notation of the indices was used. Note that the C have dimension of an energy density (energy/volume), and the S are just their inverse (also in the more mathematical sense of inverse matrix).

In principle, the matrices C or S have $6 \times 6$ independent components, relating each of the stress components to a different strain component, and vice versa. The same symmetry concept applies, however, and the 36 components are reduced to the diagonal plus one full triangle of the matrix, $6(6+1)/2=21$ components:

$$\begin{aligned} \begin{pmatrix} \sigma _1 \\ \sigma _2 \\ \sigma _3 \\ \sigma _4 \\ \sigma _5 \\ \sigma _6 \end{pmatrix} = \begin{pmatrix} C_{11} &{} C_{12} &{} C_{13} &{} C_{14} &{} C_{15} &{} C_{16} \\ &{} C_{22} &{} C_{23} &{} C_{24} &{} C_{25} &{} C_{26} \\ &{} &{} C_{33} &{} C_{34} &{} C_{35} &{} C_{36} \\ &{} &{} &{} C_{44} &{} C_{45} &{} C_{46} \\ &{} &{} &{} &{} C_{55} &{} C_{56} \\ &{} &{} &{} &{} &{} C_{66} \end{pmatrix} \begin{pmatrix} \varepsilon _1 \\ \varepsilon _2 \\ \varepsilon _3 \\ \varepsilon _4 \\ \varepsilon _5 \\ \varepsilon _6 \end{pmatrix} \end{aligned}$$

(9.28)

Moreover, further symmetry considerations relative to the particular internal arrangement of the atoms and molecules of the material can further reduce the number of independent “Hooke-like” relations between the different components of stress and strain.

For a material with perfectly isotropic response to the applied forces, the three Cartesian directions are equivalent and only two coefficients are independent: $C_{11}$ (also equal to $C_{22}$ and $C_{33}$), and $C_{12}$ (also equal to all the combinations $C_{ab}$ with $a,b=1$, 2, 3); the coefficients $C_{aa}$ for $a=4$, 5, 6 are all equal to $(C_{11}-C_{12})/2$, and all other coefficients are zero. Examples of isotropic materials are any liquid, or amorphous solid, like a glass; most mixtures of polymers and plastic materials are practically isotropic.

9.1.4 Elastic Moduli for Solid Materials

For an isotropic material only two elastic constants are needed to specify its response to an applied stress, in the linear regime of deformation. As we saw above, these are $C_{11}$ and $C_{12}$. Just by looking at their indices ‘11’ and ‘12’, it is not immediate to understand what kind of deformation these coefficients relate to. On the other hand, their value is not directly accessible to a simple experiment, and a more convenient way is to deduce them from experiments in which some combination of their values occurs. It turns out that in this way, also the interpretation of their physical meaning becomes more transparent. The combinations of elastic constants are called elastic moduli, and each one of them corresponds to an experimentally realizable mode of deformation.

Elastic moduli for isotropic materials were identified already in the first half of the XIX century, and called Lamé coefficients. In terms of the elastic constants, they would be written as $\lambda =C_{12}$ and $\mu =(C_{11}-C_{12})/2$. The two coefficients are still derived mathematically, by looking at the symmetries of deformation.

Experimental quantities related to the $C_{ij}$ are the bulk modulus, B, the shear modulus, G (equal to $\mu$), the Young’s modulus, E, or the Poisson’s ratio, $\nu$. Obviously, the same isotropic material will be fully described by any pair of these, but some may be more convenient than others.

By decomposing the stress tensor according to the strain components (the overbar denoting the deviatoric components), we can write:

$$\begin{aligned} \sigma _{ii}&= 3B\varepsilon _{ii} \end{aligned}$$

(9.29)

$$\begin{aligned} \overline{\sigma }_{ij}&= 2G{\overline{\varepsilon }_{ij}} \end{aligned}$$

(9.30)

as well as:

$$\begin{aligned} \sigma _{ij} - \tfrac{1}{3} \sigma _{kk}\delta _{ij} = \sigma _{ij} - B\varepsilon _{kk}\delta _{ij}=2G \left[ \varepsilon _{ij} - \tfrac{1}{3}\varepsilon _{kk}\delta _{ij} \right] \end{aligned}$$

(9.31)

From the last two identities, it also follows that:

$$\begin{aligned} \sigma _{ij} = 2\mu \varepsilon _{ij}+\lambda \varepsilon _{kk}\delta _{ij} \end{aligned}$$

(9.32)

with $\lambda =B-2G/3$.

Alternatively, one can write the strain tensor according to the stress components, as:

$$\begin{aligned} \varepsilon _{ij} - \tfrac{1}{3} \varepsilon _{kk}\delta _{ij} = \varepsilon _{ij} - \frac{1}{9B}\sigma _{kk}\delta _{ij}=\frac{1}{2G} \left[ \sigma _{ij} - \tfrac{1}{3}\sigma _{kk}\delta _{ij} \right] \end{aligned}$$

(9.33)

from which it follows the expression:

$$\begin{aligned} \varepsilon _{ij}= \frac{1}{2G} \sigma _{ij} - \left( \frac{1}{6G} - \frac{1}{9B} \right) \sigma _{kk}\delta _{ij} = \frac{1+\nu }{E} \sigma _{ij} - \frac{\nu }{E}\sigma _{kk}\delta _{ij} \end{aligned}$$

(9.34)

Accordingly, the elastic constants and compliances matrices are written in terms of E and $\nu$ as:

$$\begin{aligned} C_{ij} = \frac{E}{(1+\nu )(1-2\nu )} \begin{pmatrix} {1-\nu } &{} {\nu } &{} {\nu } &{} {0} &{} {0} &{} {0} \\ &{} 1 - \nu &{} \nu &{} 0 &{} 0 &{} 0 \\ &{} &{} 1 - \nu &{} 0 &{} 0 &{} 0 \\ &{} &{} &{} 1 - 2\nu &{} 0 &{} 0 \\ &{} &{} &{} &{} 1 - 2\nu &{} 0 \\ &{} &{} &{} &{} &{} 1 - 2\nu \end{pmatrix} \end{aligned}$$

(9.35)

$$\begin{aligned} S_{ij} = \frac{1}{E} \begin{pmatrix} {1} &{} {-\nu } &{} {-\nu } &{} {0} &{} {0} &{} {0} \\ &{} 1 &{} -\nu &{} 0 &{} 0 &{} 0 \\ &{} &{} 1 &{} 0 &{} 0 &{} 0 \\ &{} &{} &{} 4(1+\nu ) &{} 0 &{} 0 \\ &{} &{} &{} &{} 4(1+\nu ) &{} 0 \\ &{} &{} &{} &{} &{} 4(1+\nu ) \end{pmatrix} \end{aligned}$$

(9.36)

9.1.5 Bulk modulus

This parameter defines the relative variation of the volume induced by an isotropic compression/dilation of the solid, typically a variation of hydrostatic pressure corresponding to a stress tensor $(\sigma _0, \sigma _0, \sigma _0, 0, 0, 0)$, under the assumption that the material is homogeneous at the scale of the applied deformation (see Fig. 9.23, top):

$$\begin{aligned} \varDelta P = B \left( \frac{\varDelta V}{V} \right) \end{aligned}$$

(9.37)

In terms of the Lamé parameters it is $B=\lambda +(2/3)\mu$, and in terms of the independent elastic constants, $B=(C_{11}+2 C_{12})/3$. We already encountered this expression, in Eq. (9.48) above, to define the elastic energy of hydrostatic compression/dilation under a strain $(\varepsilon _0, \varepsilon _0, \varepsilon _0, 0, 0, 0)$. In fact, we can also write:

$$\begin{aligned} E_{el} = \frac{1}{2} B \varepsilon ^2 \end{aligned}$$

(9.38)

which suggests a harmonic character of the deformation (energy proportional to a squared variation). Also, the $\varepsilon ^2$-dependence makes the energy symmetric for compression ($\varepsilon <0$) and dilation ($\varepsilon >0$). In fact, this is characteristic of any perturbation in the linear regime, in which the force is proportional to the perturbation, and the energy is correspondingly proportional to its square. We will find other similar relationships also later on.

Since the strain is dimensionless, B has the same units of energy/density, or force/surface. Some typical values of B for various materials are given in Fig. 9.23.

Fig. 9.23

The simplest homogeneous deformation modes (left column), corresponding to: hydrostatic compression/dilation (top); shear (middle); uniaxial tension/compression (bottom). On the right column, some values (in units of 10$^9$ J/m$^3$, or GPa) of the corresponding elastic moduli for typical materials and biological tissues: bulk modulus (top), shear modulus (middle), and Young’s modulus (bottom)

9.1.6 Shear Modulus

In this type of deformation, also homogeneous within the test body, the stress is applied parallel to the surface (Fig. 9.23, middle); if we take the xy axis in the plane of the figure, the applied stress tensor would be $(0, 0, 0, \sigma _0, 0, 0)$. If we imagine the body cut in slices parallel to the direction of the force, the corresponding deformation tends to slide all slices with respect to each other. The overall deformation is d/D, occurring symmetrically with xy and yx components. This mode of deformation is called shearing, and the corresponding parameter is the shear modulus. The relation between the applied stress and the resulting deformation is:

$$\begin{aligned} \sigma _0 = G \left( \frac{d}{D} \right) \end{aligned}$$

(9.39)

In terms of the Lamé parameters it is $G=\mu$, and in terms of the independent elastic constants, $G=(C_{11}- C_{12})/2$. For many solids, the shear modulus is of the same order of magnitude of the bulk modulus. On the other hand, it is zero by definition for any fluid, in fact the very definition of a fluid is that of a material that does not support shear stress. It may be worth noting, however, that in the context of fluid mechanics, the Lamé parameter $\mu$ is often identified with the dynamic viscosity of the medium.

9.1.7 Young’s Modulus

Named after the same English physician Thomas Young that we mentioned in Chap. 2 for introducing the term ‘energy’, this elastic modulus is the most appropriate to describe an experiment of traction or compression of a body along an axis, e.g. with a stress tensor $(\sigma _0, 0, 0, 0, 0, 0)$ if deforming along x. In this case, the deformation is again homogeneous, but in a more subtle way. It is quite common to observe that a material undergoing axial compression will contract along the compression direction, while it will dilate in the perpendicular plane, and vice-versa in extension. With the stress specified above, this results in a strain tensor $(\varepsilon , \varepsilon _{tr}, \varepsilon _{tr}, 0, 0, 0)$, with $\varepsilon _{tr}$ the relative deformation in the perpendicular directions y and z. The negative of the ratio between the two resulting values of deformation is called the Poisson’s coefficient of the material:

$$\begin{aligned} \nu = - \frac{\varepsilon _{tr}}{\varepsilon } \end{aligned}$$

(9.40)

It is easily seen that only for $\nu =0.5$ the volume of the body is conserved during the deformation, for example for a traction along x of a body with lengths $L_x, L_y, L_z$ and volume $V=L_x L_y L_z$:

$$\begin{aligned} \varDelta V = L_x(1+\varepsilon )L_y(1-\varepsilon _{tr})L_z(1-\varepsilon _{tr}) - L_x L_y L_z = \nonumber \\ = V (\varepsilon - 2 \varepsilon _{tr}) + O(\varepsilon ^2) \simeq V \varepsilon (1 - 2\nu ) \end{aligned}$$

(9.41)

(the terms in $\varepsilon ^2$ and $\varepsilon ^3$ going to zero for small deformation). On the other hand, for most materials the Poisson’s ratio is different from 0.5, and the deformation changes also the volume of the object (as it is done also by the hydrostatic compression and dilation). In terms of the elastic constants or Lamé coefficients, it is $\nu =C_{12}/(C_{11}+C_{12})$, or $\nu =\lambda /2(\lambda +\mu )$.

As shown in the Fig. 9.23, bottom, we imagine to apply a force F at the extremities of a rod with cross section A and initial length L. The stress-strain relation is:

$$\begin{aligned} \sigma = E \varepsilon \,\,\,\, \rightarrow \,\,\,\, \frac{F}{A} = E \frac{\varDelta L}{L} \end{aligned}$$

(9.42)

with the cross section A being taken at its reference, undeformed value. During the uniaxial deformation, the cross section changes as $A/(1\pm \varepsilon )^{2\nu }$, the ‘+’ and ‘–’ sign being for traction or compression, respectively.

In terms of the Lamé parameters it is $E=2\,\mu +\lambda$, and in terms of the independent elastic constants we have the more complicate expression $E=\tfrac{C_{11}-C_{12}}{C_{11}+C_{12}} (C_{11}+2C_{12})$. It will be noticed that the traction/compression experiment is geometrically similar to Robert Hooke’s experiments on his springs. Indeed, starting from the stress-strain relation (9.42) we can determine an effective “spring” constant for the material, which will depend on its geometrical shape. By multiplying both sides of Eq. (9.42) by A, we get:

$$\begin{aligned} F = \left( \frac{EA}{L} \right) \varDelta L \end{aligned}$$

(9.43)

from which an effective ’spring constant’ of the body can be defined as $k=EA/L$.

A related elastic modulus that is of utility in biological materials is the aggregate modulus, H, obtained in an experiment in which only the uniaxial deformation is allowed, while no transverse strain occurs. In terms of E, this modulus is formally defined:

$$\begin{aligned} H = \frac{E(1-\nu )}{(1+\nu )(1-2\nu )} \end{aligned}$$

(9.44)

but when looking at its expression in terms of the independent elastic constants, it is just $H=C_{11}$. To have a zero component of the transverse deformation it is necessary to apply a stress in the plane that would contrast the natural tendency of the material to contract or dilate, according to the sign of its Poisson’s ratio, therefore the stress tensor in this case looks like $(\sigma , \pm \sigma _{tr}, \pm \sigma _{tr}, 0, 0, 0)$.

For the deformation in a plane upon a force applied along two directions, an equivalent expression to that of uniaxial deformation can be obtained, by using Eq. (9.35) or (9.36). For example, we may have $\{\sigma _1, \sigma _2, \sigma _4\} \ne 0$, if two forces $f_x$ and $f_y$ are applied in the xy plane without allowing to deform along z. The matrix product $\varepsilon _i=\sum _j S_{ij} \sigma _j$ gives:

$$\begin{aligned} {\left\{ \begin{array}{ll} E \varepsilon _1 &{} = \sigma _1 - \nu \sigma _2 \\ E \varepsilon _2 &{} = \sigma _2 - \nu \sigma _1 \\ \mu \varepsilon _4 &{} = 2 \sigma _4 \end{array}\right. } \end{aligned}$$

(9.45)

Fig. 9.24

Example of a stress-strain curve for a material exhibiting a mixed mechanical response to a uniaxial pulling force f. The red curve describes the “apparent” stress, $\sigma =f/S_0$, the black curve the “real” stress, $\sigma =f/S$, where $S_0$ and S are, respectively, the initial and instantaneous cross section area

9.1.8 Elastic Deformation Energy

In solid mechanics, it is common practice to follow the response of a material to an applied load (in the form of an imposed stress, or deformation) by tracing its stress-strain diagram. Such a representation contains much information about the mechanical behaviour of our material. A typical example of such a diagram is given in Fig. 9.24, for a uniaxial force f pulling a material sample along one direction; the initial area of the cross section transverse to the direction of f is $S_0$, and it evolves into S(f) as far as the force stretches the material. The point 1 is the elastic limit: up to this point, the relation between stress and strain is linear, such as in a Hookean spring; beyond this point, the material could break if it is fragile (such as glass), or continue to deform if it is ductile (such as steel); the corresponding value of stress is called the yield stress $\sigma _y$ of the material. The point 1$^{\prime }$ is a generic value of strain beyond the elastic limit: if the force is released at this point, the material will go back to zero stress but retaining a residual deformation $\varepsilon _r$. Upon increasing the stress, the material can continue to deform, up to the point 3 where final rupture occurs; the corresponding values of $\sigma _f$ and $\varepsilon _f$ are called fracture stress and fracture strain. The maximum value of stress supported during all the long deformation, marked as 2, is the ultimate strength of the material. The region 4 under the curve is called the “strain hardening” region, since the material responds in a complex way however the stress keeps increasing (the red and black curve run approximately parallel to each other). The region 5 is the “necking” region, in which the cross section rapidly decreases (note the widening difference in this region, between the red and black curve): this is also the region of maximum plastic deformation. The integral of the area under the stress-strain plot from zero to $\varepsilon _f$, $\tau _0=\int _0^{\varepsilon _f}\sigma d\varepsilon$, is the toughness .

When a material body is deformed by an external force, this force performs a work equal to $dW=\sigma d\varepsilon$, to be integrated over the entire volume V of the deformed body. The work done by the external force is stored in the material in the form of a deformation energy, which is given back to the environment when the force is removed.

Always remaining in the limit of small deformations, so that the continuum linear elasticity theory can be applied, the elastic deformation energy in the volume V is written:

$$\begin{aligned} E_{el} = \int _V dW = \int _V \underline{\sigma } \otimes d\underline{\varepsilon } = \sum _{ij} \int _V ( \varepsilon _i^T C_{ij} ) d\varepsilon _j \end{aligned}$$

(9.46)

by using the formal stress-strain relation Eq. (9.26), and the explicit matrix components of stress and strain with the matrix multiplication rules (note that the writing “$\varepsilon _i^T C_{ij}$” indicates the product between the line-vector $\varepsilon ^T$, transposed of the column-vector $\varepsilon$, and the matrix $\underline{C}$).

Carrying out the formal integration, in a homogeneous volume (in which the elastic constants are constant) and homogeneously deformed (i.e., every point of the volume is deformed in the same way), the elastic energy density is:

$$\begin{aligned} E_{el} = \frac{1}{2} \sum _{ij} \varepsilon _i C_{ij} \varepsilon _j \end{aligned}$$

(9.47)

For example, the compression by equal amounts $\varepsilon _0$ along x, y, z (or “hydrostatic” deformation) of an isotropic material (with only $C_{11}$ and $C_{12}$ non-zero, and $C_{44}=C_{55}=C_{66}=(C_{11}-C_{12})/2$), would result in a strain tensor $\underline{\varepsilon }=(\varepsilon _0, \varepsilon _0, \varepsilon _0, 0, 0, 0$) (in Voigt notation). By using the matrix-vector multiplication rules, the elastic energy for this case is:

$$\begin{aligned} E_{el} = \frac{1}{2} \left( \frac{C_{11}+2C_{12}}{3} \right) \varepsilon _0^2 \end{aligned}$$

(9.48)

Note that this is an energy density, i.e. it must be multiplied by the volume element V, to obtain a proper energy value. Furthermore, if a body can be decomposed into pieces of elements, each with different elastic constants and with different local deformations, the overall elastic energy is the sum $E=\sum _i E_{el,i}V_i$, calculated for each elementary volume $V_i$.

Fig. 9.25

Example of a stress-strain curve for a resilient material exhibiting a hysteresis loop. The grey area comprised between the loading and unloading ramps represents the amount of work (stored elastic energy) lost during the cyclic loading

If a non-linear material is loaded cyclically between two values of strain, it may display hysteresis (Fig. 9.25), i.e. the two ramps of the cyclic curve (loading/unloading) are not equal. This means that a fraction of the work expended to perform the tensile deformation is not recovered when the deforming force is removed. The lost energy (likely in the form of heat) is given by the area (grey shaded in the figure) comprised within the hysteresis loop, and its inverse is called the resilience of the material. A material that loses a large fraction of the elastic energy stored in the cyclic loading is said to have a low resilience, and the opposite is true if the hysteresis loop is more narrow. A resilient material is one that can efficiently cycle back and forth in a repeated and sustained deformation, such as a tendon stretched during the walk or run.

Problems

9.1

Average elastic modulus

Using the known volume fractions from the text, (a) calculate the volume fractions of mineral and collagen in dry cortical bone. (b) For values of Young’s modulus $E=54$ GPa for mineral, and 1.25 GPa for collagen, calculate the resulting Young’s modulus of wet and dry bone.

9.2

Skin stretching

The plot below reproduces the stress-strain curve of abdominal skin, stretched along the direction perpendicular and parallel to the body height. (a) What stress is developed for a stretching of 35 %? (b) What strain is developed for stretching to 5 MPa? (c) What is the elastic modulus E in the two principal directions? (d) What is the toughness?

9.3

Arterial stress relaxation

In a mechanical test, a stress of 1 MPa is applied to a 2-cm aorta strip, which as a result is stretched to 2.3 cm. The strain is held constant for an hour, and the stress in the strip drops to 0.75 MPa. Assume that the mechanical properties of the tissue do not change during the experiment. (a) Use the Maxwell model of a viscoelastic material, to obtain the relaxation time of the biomaterial. (b) Calculate the stress in the tissue, if the experiment is continued up to a time of 3 h. (c) The same experiment is performed in a different way, by holding the stress constant at 1 MPa for the same time of 1 h, after which the stress is released. Use the Kelvin-Vogt model to obtain the strain relaxation time, if 1 h 25 min after the release, the strip length is back to 2.2 cm.

9.4

Stretch the leg

Compare the charge on the tendon and the bone in the calf of a man walking in the street. Take E(bone) $=$ 20 GPa, E(tendon) $=$ 1.5 GPa, ${\text {diameter}} \times {\text {length}}$ equal to $1.5\times 8$ cm for the tendon, and $4\times 35$ cm for the tibia.

9.5

Jumping cat

A cat of mass $M=4.5$ kg jumps on the ground from a height $h=3$ m. For simplicity, assume that the leg has two equal muscles in each of the upper and lower half, attached to the leg bones as shown in the figure. Each muscle is simulated by a cylinder of average diameter 4 cm and length 12 cm, and their attachment point (enthesis) is $a=1$ cm off-axis; take the Young’s modulus of striated muscle $E=20$ kPa. Leg bones are represented as two thinner, straight cylinders of infinite rigidity, hinged at the knee. Calculate the bending angle of the legs. What does this calculation demonstrates? Compare with your answer to the previous question.

9.6

Muscles and temperature

We measure the relative elongation of an insect’s muscle with the help of a dynamometer (a simple instrument that allows to impose a constant load to a structure and measure the elongation). The muscle can be represented as a homogeneous cylinder of length L and radius R. From a measurement performed at the temperature $T=10\,^{\circ }$C, we obtain a relative elongation $\varDelta L/L=+2.\%$ and a contraction in the radial direction $\varDelta R/R= -0.25\,\%$; a second measurement at the same load, performed at $T=15\,^{\circ }$C, gives an elongation $\varDelta L/L=+4.\%$ and a contraction $\varDelta R/R= -0.5\,\%$. Again, by measuring at $T=20\,^{\circ }$C we find a $\varDelta L/L=+6.\%$ and a $\varDelta R/R= -0.75\,\%$.

(a) Which elastic moduli are of interest in this kind of experiment?

(b) Find how the ratio $\varDelta V/V$ varies as a function of the temperature.

(c) Predict the values of the relevant elastic moduli and the elongation at $T=23.5\,{^{\circ }}$C.

9.7

Implant materials

Some biocompatible materials for implants are made by a relatively soft matrix, in which a fraction h of harder fibres are dispersed for reinforcement. The overall elastic modulus is calculated on the basis of a “shear-lag” model, which considers that the fibres are too short to be in contact, and cannot share the stress on the material:

$$\begin{aligned} E=hE_{f}\left( 1-\frac{\tanh {n_s}}{n_s} \right) + (1-h)E_{m} \,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,\,\,\,\,\, n_s\simeq \sqrt{\frac{2E_{m}}{E_{f} \ln (1/h)}} \end{aligned}$$

(the $\simeq$ sign in $n_s$ comes from the assumption of a Poisson’s ratio $\nu \sim 0$ for the artificial polymer matrix). Discuss the behaviour of the resulting modulus as a function of h. How does h relate to the ratio $E_{f}/E_{m}$?

9.8

Bend, break or twist

Compare the elastic and strength moduli of the materials in the following table, from each of which a hypothetical stick in form of a full cylinder of diameter 1 cm and length 50 cm is fabricated. Which sticks will bend, break, or twist, under the loads shown in the accompanying drawing?

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