Keywords

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

5.1 Linear Biphasic Model

In this section we introduce the linear biphasic theory, which models articular cartilage as a binary mixture of an intrinsically incompressible elastic matrix (skeleton) and an inviscid (i.e., dissipationless) incompressible fluid. We also present and discuss the formulation of the governing differential equations along with the different types of boundary conditions.

5.1.1 Linear Biphasic Theory

According to the biphasic theory of Mow et al. [60], articular cartilage is modeled as a biphasic mixture consisting of a solid phase (representing collagen, proteoglycans, chrondrocytes, and other quantitatively minor glycoproteins) and a fluid phase (representing mobile interstitial fluid and dissolved electrolytes). The fluid phase typically ranges between 65 and 90 % of the articular cartilage tissue by weight [8].

Note also that various biphasic and poroelastic models were used to describe the deformation behavior of bone [20], skin [62], polymeric and silk hydrogels [16, 42], and arterial walls [40]. An overview of computational models for the mechanical behavior of articular cartilage was given in [27, 49, 76].

Let the fluid volume fraction (porosity) be denoted by \(\phi _\mathrm{f}=V_\mathrm{f}/V\), and the solid volume fraction be \(\phi _\mathrm{s}=V_\mathrm{s}/V\), where \(V_\mathrm{f}+V_\mathrm{s}=V\), so that

$$\begin{aligned} \phi _\mathrm{f}+\phi _\mathrm{s}=1. \end{aligned}$$
(5.1)

The continuity equation for a biphasic medium is

$$\begin{aligned} \nabla \cdot (\phi _\mathrm{f}\mathbf{v}^\mathrm{f}+\phi _\mathrm{s}\mathbf{v}^\mathrm{s})=0, \end{aligned}$$
(5.2)

where \(\mathbf{v}^\mathrm{f}\) and \(\mathbf{v}^\mathrm{s}\) are solid and fluid velocities, \(\nabla \) is the gradient operator.

Under quasi-static conditions, and in the absence of body forces, the momentum equations for each phase are given by

$$\begin{aligned} \begin{array}{r} \nabla \cdot \varvec{\sigma }^\mathrm{s}-\varvec{\pi }^\mathrm{f}=\mathbf{0}, \\ \nabla \cdot \varvec{\sigma }^\mathrm{f}+\varvec{\pi }^\mathrm{f}=\mathbf{0}, \end{array}\end{aligned}$$
(5.3)

where \(\varvec{\pi }^\mathrm{f}\) is the momentum exchange between the phases due to frictional drag of relative fluid flow through the porous-permeable solid matrix. In articular cartilage, it has been shown [35] that this momentum exchange term creates a frictional drag several orders of magnitude greater than the viscous shear stress within the interstitial fluid due to the viscosity of the fluid. The internal fluid viscosity can usually be neglected except for very small layers of very permeable materials [11].

Thus, neglecting the frictional dissipation between the fluid particles, the interstitial water is assumed to be inviscid, and the fluid phase stress is given by

$$\begin{aligned} \varvec{\sigma }^\mathrm{f}=-\phi _\mathrm{f}p\mathbf{I}. \end{aligned}$$
(5.4)

Here, p is the fluid pressure, \(\mathbf{I}\) is the identity tensor.

The single pore-fluid flow is governed by the local interaction force per unit volume defined as follows [12, 55, 65]:

$$\begin{aligned} \varvec{\pi }^\mathrm{f}=-\mathbf{K}\cdot (\mathbf{v}^\mathrm{f}-\mathbf{v}^\mathrm{s}). \end{aligned}$$
(5.5)

Here, \(\mathbf{K}\) represents a hydraulic resistivity (or inverse permeability) tensor, which is related to the permeability tensor, \(\mathbf{k}\) , through

$$\begin{aligned} \mathbf{K}=\phi _\mathrm{f}^2 \mathbf{k}^{-1}, \end{aligned}$$
(5.6)

and apparently depends on the deformable pore structure and the interstitial fluid properties [58, 59, 67]. Note also that the permeability of the tissue decreases when the pore volume decreases [24, 33].

Generally, \(\mathbf{k}\) is a positive definite and symmetric tensor. For a transversely isotropic skeleton, if \(x_3=0\) is the plane of isotropy, the matrix of the permeability tensor takes the form

$$ \mathbf{k}=\left( \begin{array}{ccc} k_1 &{} 0 &{} 0 \\ 0 &{} k_1 &{} 0 \\ 0 &{} 0 &{} k_3 \end{array} \right) . $$

As was shown in [25], the transverse isotropy of permeability in articular cartilage is caused by its microstructural anisotropy. In particular, the permeability is greater in the direction parallel to the collagen fibres than the orthogonal.

In the isotropic case [44], \(\mathbf{K}=K\mathbf{I}\), where K is the diffusion coefficient and is related to the permeability coefficient of the solid matrix, k, by \(k=\phi _\mathrm{f}^2/K\).

The stress-strain relation for the solid matrix is assumed to have the form

$$\begin{aligned} \varvec{\sigma }^\mathrm{s}=-\phi _\mathrm{s}p\mathbf{I}+\varvec{\sigma }^\mathrm{e}, \end{aligned}$$
(5.7)

where \(\varvec{\sigma }^\mathrm{e}\) is the effective (or elastic) stress of the solid matrix. Note that the concept of effective stress was originally formulated by Terzaghi [74] in a geotechnical consolidation problem, presuming that the effective soil stress is determined by the total stress minus the excess pore pressure.

Thus, under these assumptions the total stress in the biphasic material, which is defined as the sum

$$\begin{aligned} \varvec{\sigma }=\varvec{\sigma }^\mathrm{s}+\varvec{\sigma }^\mathrm{f}, \end{aligned}$$
(5.8)

in light of (5.4) and (5.7) is given by

$$\begin{aligned} \varvec{\sigma }=-p\mathbf{I}+\varvec{\sigma }^\mathrm{e}, \end{aligned}$$
(5.9)

while from (5.3) it follows that

$$\begin{aligned} \nabla \cdot \varvec{\sigma }=\mathbf{0}. \end{aligned}$$
(5.10)

For an anisotropic linearly elastic material, the effective stress \({\varvec{\sigma }^\mathrm{e}}\) is related to the infinitesimal strain tensor of the solid matrix, \({\varvec{\varepsilon }}\), by Hooke’s law

$$\begin{aligned}{\varvec{\sigma }^\mathrm{e}=\mathbf{C}:\varvec{\varepsilon },} \end{aligned}$$

where \(\mathbf{C}\) is a fourth-order stiffness tensor, and the strain tensor is given by

$$\begin{aligned} {\varvec{\varepsilon }=\frac{1}{2}\bigl (\nabla \mathbf{u}+\nabla \mathbf{u}^\mathsf{{T}}\bigr ),} \end{aligned}$$
(5.11)

where \(\mathbf{u}\) is the displacement vector of the solid phase. Note also that the solid velocity is

$$\begin{aligned} \mathbf{v}^\mathrm{s}=\frac{\partial \mathbf{u}}{\partial t}, \end{aligned}$$
(5.12)

where t is a time variable.

Following Cohen et al. [19], we assume that the solid-phase material is transversely isotropic, so that

$$\begin{aligned} \left( \begin{array}{c} \sigma _{11}^\mathrm{e} \\ \sigma _{22}^\mathrm{e} \\ \sigma _{33}^\mathrm{e} \\ \sigma _{23}^\mathrm{e} \\ \sigma _{13}^\mathrm{e} \\ \sigma _{12}^\mathrm{e} \end{array} \right) = \left[ \begin{array}{cccccc} A_{11}^\mathrm{s} &{} A_{12}^\mathrm{s} &{} A_{13}^\mathrm{s} &{} 0 &{} 0 &{} 0 \\ A_{12}^\mathrm{s} &{} A_{11}^\mathrm{s} &{} A_{13}^\mathrm{s} &{} 0 &{} 0 &{} 0 \\ A_{13}^\mathrm{s} &{} A_{13}^\mathrm{s} &{} A_{33}^\mathrm{s} &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 2A_{44}^\mathrm{s} &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 2A_{44}^\mathrm{s} &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2A_{66}^\mathrm{s} \end{array} \right] \left( \begin{array}{c} \varepsilon _{11} \\ \varepsilon _{22} \\ \varepsilon _{33} \\ \varepsilon _{23} \\ \varepsilon _{13} \\ \varepsilon _{12} \end{array} \right) , \end{aligned}$$
(5.13)

where \(A_{11}^\mathrm{s}\), \(A_{12}^\mathrm{s}\), \(A_{13}^\mathrm{s}\), \(A_{33}^\mathrm{s}\), and \(A_{44}^\mathrm{s}\) are five independent elastic constants of the solid skeleton.

The substitution of (5.4), (5.5), and (5.7) into Eq. (5.3) yields

$$\begin{aligned} {-\phi _\mathrm{s}\nabla p+\nabla \cdot \varvec{\sigma }^\mathrm{e} +\mathbf{K}\cdot (\mathbf{v}^\mathrm{f}-\mathbf{v}^\mathrm{s})=\mathbf{0},} \end{aligned}$$
(5.14)
$$\begin{aligned} {-}\phi _\mathrm{f}\nabla p-\mathbf{K}\cdot (\mathbf{v}^\mathrm{f}-\mathbf{v}^\mathrm{s})=\mathbf{0}, \end{aligned}$$
(5.15)

whereas the substitution of (5.9) into Eq. (5.10) gives

$$\begin{aligned} {-\nabla p+\nabla \cdot \varvec{\sigma }^\mathrm{e}=\mathbf{0}.} \end{aligned}$$
(5.16)

Observe [55] that in the equations above, the vector \(\mathbf{v}^\mathrm{f}-\mathbf{v}^\mathrm{s}\) represents the seepage velocity, describing the fluid motion relative to the deforming solid matrix. Moreover, relating this vector only to the fluid part of the mixture, the so-called relative fluid flux (or filter velocity), \(\mathbf{w}^\mathrm{f}\) , can be defined by the formula

$$\begin{aligned} \mathbf{w}^\mathrm{f}=\phi _\mathrm{f}(\mathbf{v}^\mathrm{f}-\mathbf{v}^\mathrm{s}). \end{aligned}$$
(5.17)

Then, introducing the relative fluid flux into Eq. (5.15) according to its definition (5.17) and taking into account the relation (5.6) between the hydraulic resistivity tensor \(\mathbf{K}\) and the permeability tensor \(\mathbf{k}\), we arrive at the equation

$$\begin{aligned} \mathbf{w}^\mathrm{f}=-\mathbf{k}\cdot \nabla p, \end{aligned}$$
(5.18)

which can be interpreted [10] as Darcy’s law relative to the motion of the solid matrix.

As a result of (5.1) and (5.17), the continuity equation (5.2) can be recast as

$$ \nabla \cdot (\mathbf{v}^\mathrm{s}+\mathbf{w}^\mathrm{f})=0, $$

which after the substitution of (5.12) and (5.18) is reduced to the equation

$$\begin{aligned} \frac{\partial }{\partial t}\nabla \cdot \mathbf{u}=\nabla \cdot (\mathbf{k}\cdot \nabla p), \end{aligned}$$
(5.19)

where \(\nabla \cdot \mathbf{u}\) is the dilatation of the solid matrix. It is important to note that Eqs. (5.16) and (5.19) can be solved independently of Eq. (5.18).

Observe [19] that the linear transversely isotropic biphasic model requires altogether seven constitutional parameters: five elastic constants (Young’s moduli and Poisson’s ratios in the transverse plane and out-of-plane, \(E_{1}^\mathrm{s}\), \(\nu _{12}^\mathrm{s}\) and \(E_{3}^\mathrm{s}\), \(\nu _{31}^\mathrm{s}\), respectively, and the out-of-plane shear modulus, \(G_{13}^\mathrm{s}\)) and two permeability coefficients \(k_{1}\) and \(k_{3}\), which are called the axial (in-plane) and transverse (out-of-plane) permeability coefficients, respectively.

Finally, it should be emphasized [57] that in addition to its mechanical response, articular cartilage also exhibits complex electrochemical phenomena due to the charged nature of its solid phase and the electrolytes dissolved in the interstitial water. A number of constitutive theories [31, 37, 45] for charged-hydrated soft tissues like articular cartilage have been developed using the multiphasic approach (see comprehensive review by Mow and Guo [57]), and a generalized correspondence principle for the equilibrium deformational behavior in the framework of the triphasic model was introduced [52, 53].

5.1.2 Boundary and Initial Conditions

Following Barry and Holmes [10], we consider the most common boundary conditions applicable to thin fluid-saturated porous layers. We assume that a biphasic layer is firmly attached to a rigid impermeable substrate, on the bottom of the layer, \(x_3=h\), in which case the boundary conditions become

$$\begin{aligned} \mathbf{u}\bigr \vert _{x_3=h}=\mathbf{0}, \end{aligned}$$
(5.20)
$$\begin{aligned} \frac{\partial p}{\partial x_3}\biggr \vert _{x_3=h}=0. \end{aligned}$$
(5.21)

On the upper surface, \(x_3=0\), a number of different boundary conditions may be formulated depending on the problem setting. If the porous layer is in contact with a porous filter, then the boundary condition

$$\begin{aligned} p\bigr \vert _{x_3=0}=0 \end{aligned}$$
(5.22)

is imposed on the top surface.

If the layer is pressed against an impermeable punch, then

$$\begin{aligned} \frac{\partial p}{\partial x_3}\biggr \vert _{x_3=0}=0. \end{aligned}$$
(5.23)

Further, the normal stress balance under a rigid punch \(\sigma _{33}\bigr \vert _{x_3=0}=-q\) gives

$$\begin{aligned} {-}p+A_{13}^\mathrm{s}\frac{\partial u_1}{\partial x_1} +A_{13}^\mathrm{s}\frac{\partial u_2}{\partial x_2} +A_{33}^\mathrm{s}\frac{\partial u_3}{\partial x_3}\biggr \vert _{x_3=0}=-q, \end{aligned}$$
(5.24)

where q is the load distribution on the top surface transferred by the punch. Note that the left-hand side of Eq. (5.24) represents the normal total stress \(\sigma _{33}\).

For the frictionless contact, the tangential stresses \(\sigma _{31}\) and \(\sigma _{32}\) are zero, so that

$$\begin{aligned} \frac{\partial u_1}{\partial x_3}+\frac{\partial u_3}{\partial x_1}\biggr \vert _{x_3=0}=0,\quad \frac{\partial u_2}{\partial x_3}+\frac{\partial u_3}{\partial x_2}\biggr \vert _{x_3=0}=0. \end{aligned}$$
(5.25)

In an idealized situation, the normal load q would be a known function of variables \(t,x_1,x_2\). However, if the surface load is transferred from a punch, then within the contact area, \(\omega \), the contact condition is formulated as

$$\begin{aligned} u_3\bigr \vert _{x_3=0}=\delta _0(t)-\varphi (x_1,x_2),\quad (x_1,x_2)\in \omega , \end{aligned}$$
(5.26)

where \(\delta _0(t)\) is the normal displacement of the punch, \(\varphi (x_1,x_2)\) is the punch shape function (defining the initial gap between the contacting surfaces).

Before continuing, we observe that the punch equilibrium implies that

$$\begin{aligned} \iint \limits _{\omega } q(t,\mathbf{y})\,d\mathbf{y}=F(t), \end{aligned}$$
(5.27)

where F(t) is an applied external force.

In the case of frictionless contact between two biphasic layers, based on the results of Hou et al. [35] for porous layers saturated with inviscid interstitial fluids, Ateshian et al. [6] formulated the following interface boundary conditions:

$$\begin{aligned}{}[\![p]\!]=0,\quad [\![\mathbf{u}\cdot \mathbf{n}]\!]=\mathbf{0},\quad [\![\mathbf{w}^\mathrm{f}\cdot \mathbf{n}]\!]=\mathbf{0},\quad [\![\sigma _N^{(n)}]\!]=0, \end{aligned}$$
(5.28)
$$\begin{aligned} {\varvec{\sigma }^{(n)}_T=0.} \end{aligned}$$
(5.29)

Here, \(\mathbf{n}\) is the normal unit vector on the contact interface, \([\![\,\cdot \,]\!]\) denotes the jump across the interface of the quantity within the brackets (e.g., \([\![p]\!]=p_+-p_-\), where \(p_+\) and \(p_-\) are the limit values of p at the two opposite sides of the interface), \(\sigma _N^{(n)}\) and \({\varvec{\sigma }^{(n)}_T}\) are the normal and tangential components of the total stress vector \({\varvec{\sigma }^{(n)}=\varvec{\sigma }\cdot \mathbf{n}}\), such that \({\sigma _N^{(n)}=\varvec{\sigma }^{(n)}\cdot \mathbf{n}}\) and \({\varvec{\sigma }^{(n)}_T=\varvec{\sigma }^{(n)} -\sigma _N^{(n)}\mathbf{n}}\).

Note that the interface boundary conditions (5.28) simply state that the fluid pressure, p, the normal component of the solid displacement vector, \(\mathbf{u}\cdot \mathbf{n}\), the normal component of the relative fluid flow, \(\mathbf{w}^\mathrm{f}\cdot \mathbf{n}\), and the normal component of the total stress vector, \(\sigma _N^{(n)}\), must be continuous across the interface.

Finally, we consider the usual initial conditions that the displacement vector, \(\mathbf{u}\), the relative fluid flux, \(\mathbf{w}^\mathrm{f}\), and the fluid pressure, p, are zero before the external load is applied, i.e.,

$$\begin{aligned} \mathbf{u}=\mathbf{0},\quad \mathbf{w}^\mathrm{f}=\mathbf{0},\quad p=0,\quad -\infty <t<0, \end{aligned}$$
(5.30)

throughout the biphasic medium.

5.1.3 Equivalent Elastic Material Properties of a Transversely Isotropic Biphasic Material for the Instantaneous Response

It is known [23] that during short-duration impact events, articular cartilage can be considered as an elastic material. Moreover, it has been shown that the instantaneous response of a biphasic material is equivalent to that of an incompressible elastic material [57]. Following Garcia et al. [29], we introduce the elastic properties of the equivalent transversely isotropic incompressible elastic material, which can be used to model its instantaneous response.

In the biphasic model (see Eq. (5.8)), the total stresses in a biphasic material are defined as

$$\begin{aligned} \sigma _{ij}=\sigma _{ij}^\mathrm{s}+\sigma _{ij}^\mathrm{f}, \end{aligned}$$
(5.31)

where \(\sigma _{ij}^\mathrm{s}\) are the stresses in the solid matrix, \(\sigma _{ij}^\mathrm{f}\) are the stresses in the fluid phase.

The stresses in the fluid phase are equal to the pressure in the fluid, p, averaged over the whole volume, i.e.,

$$\begin{aligned} \sigma _{ij}^\mathrm{f}=-\phi _\mathrm{f}p\delta _{ij}, \end{aligned}$$
(5.32)

where \(\phi _\mathrm{f}\) is the volume fraction of the fluid phase, \(\delta _{ij}\) is the Kroneker delta.

At the same time, the stresses in the solid matrix are determined in terms of the effective stresses, \(\sigma _{ij}^\mathrm{e}\), as

$$\begin{aligned} \sigma _{ij}^\mathrm{s}=-\phi _\mathrm{s}p\delta _{ij}+\sigma _{ij}^\mathrm{e}, \end{aligned}$$
(5.33)

where \(\phi _\mathrm{s}=1-\phi _\mathrm{f}\) is the volume fraction of the solid phase.

From (5.31)–(5.33), it follows that the total stresses in the biphasic tissue can be decomposed as

$$\begin{aligned} \sigma _{ij}=-p\delta _{ij}+\sigma _{ij}^\mathrm{e}. \end{aligned}$$
(5.34)

It should be emphasized that, since both the fluid and the material forming the skeleton are assumed to be incompressible, the strains in the biphasic tissue are due to the effective stresses. Thus, the strains in the tissue are given by

$$\begin{aligned} \begin{array}{rclrcl} \varepsilon _{11} &{} = &{} \displaystyle \frac{1}{E_1^\mathrm{s}}(\sigma _{11}^\mathrm{e}-\nu _{12}^\mathrm{s}\sigma _{22}^\mathrm{e})-\frac{\nu _{31}^\mathrm{s}}{E_3^\mathrm{s}}\sigma _{33}^\mathrm{e}, &{} \varepsilon _{23} &{} = &{} \displaystyle \frac{1}{2G_{13}^\mathrm{s}}\sigma _{23}^\mathrm{e}, \\ \varepsilon _{22} &{} = &{} \displaystyle \frac{1}{E_1^\mathrm{s}}(-\nu _{12}^\mathrm{s}\sigma _{11}^\mathrm{e}+\sigma _{22}^\mathrm{e})-\frac{\nu _{31}^\mathrm{s}}{E_3^\mathrm{s}}\sigma _{33}^\mathrm{e},\quad &{} \varepsilon _{13} &{} = &{} \displaystyle \frac{1}{2G_{13}^\mathrm{s}}\sigma _{13}^\mathrm{e}, \\ \varepsilon _{33} &{} = &{} \displaystyle -\frac{\nu _{31}^\mathrm{s}}{E_3^\mathrm{s}}(\sigma _{11}^\mathrm{e}+\sigma _{22}^\mathrm{e})+\frac{1}{E_3^\mathrm{s}}\sigma _{33}^\mathrm{e}, &{} \varepsilon _{12} &{} = &{} \displaystyle \frac{1}{2G_{12}^\mathrm{s}}\sigma _{12}^\mathrm{e}, \end{array} \end{aligned}$$
(5.35)

where \(E_1^\mathrm{s}\) and \(E_3^\mathrm{s}\) are Young’s moduli of the solid matrix in the plane of transverse isotropy and in the orthogonal direction, respectively, \(\nu _{12}^\mathrm{s}\) and \(\nu _{31}^\mathrm{s}\) are Poisson’s ratios characterizing the lateral strain response in the plane of transverse isotropy to a stress acting parallel or normal to it, respectively, \(G_{13}^\mathrm{s}\) is the shear modulus in planes normal to the plane of transverse isotropy, and \(G_{12}^\mathrm{s}=E_1^\mathrm{s}/[2(1+\nu _{12}^\mathrm{s})]\) is the in-plane shear modulus.

For a transversely isotropic material, the incompressibility condition is attained if its Poisson’s ratios are as follows [29, 39]:

$$\begin{aligned} \nu _{31}=\frac{1}{2},\quad \nu _{12}=1-\frac{E_1}{2E_3}. \end{aligned}$$
(5.36)

Let the three independent constants of the equivalent incompressible elastic material be denoted by \(E_1\), \(E_3\), and \(G_{13}\), while its Poisson’s ratios are given by (5.36).

Observe [7] that upon sudden loading of a biphasic tissue, the interstitial fluid does not have sufficient time to leave the tissue, except at permeable boundaries. At time \(t=0^+\), the matrix pores change shape but not volume. Thus, it is assumed [29] that the general stress field in a transversely isotropic biphasic material at time zero can be decomposed into the pressure in the fluid and the effective stresses in the solid skeleton according to Eq. (5.34), in such a way that the fluid pressure at time zero is

$$\begin{aligned} p=-\alpha \sigma _{kk}. \end{aligned}$$
(5.37)

Here, \(\sigma _{kk}=\sigma _{11}+\sigma _{22}+\sigma _{33}\) is the trace of the total stress tensor, and \(\alpha \) is a dimensionless parameter, to be determined.

Thus, on one hand, the deformation of a biphasic material at time zero is given by Eq. (5.35), where in light of the hypothesis (5.37) we have

$$\begin{aligned} \sigma _{ij}^\mathrm{e}=\sigma _{ij}-\alpha \sigma _{kk}\delta _{ij}. \end{aligned}$$
(5.38)

On the other hand, the same deformation must be equal to the deformation of the equivalent incompressible tissue under the total stress field \(\sigma _{ij}\). This means, first, that the shear moduli \(G_{12}^\mathrm{s}\) and \(G_{13}^\mathrm{s}\) should be the same for the solid skeleton and the equivalent incompressible elastic material, and in particular,

$$\begin{aligned} G_{13}=G_{13}^\mathrm{s}, \end{aligned}$$
(5.39)

since the inviscid interstitial fluid sustains only pressure.

Therefore, applying the decomposition (5.38) to normal stresses, one arrives at a homogeneous system of linear algebraic equations with respect to \(\sigma _{11}\), \(\sigma _{22}\), and \(\sigma _{33}\), with the coefficients depending on \(E_1\), \(E_3\), and \(\alpha \). As was shown in [29], this system is satisfied for any combination of the normal stresses, if the equivalent elastic moduli are given by

$$\begin{aligned} \begin{array}{l} \displaystyle E_1=\frac{E_1^\mathrm{s}\bigl [1-4\nu _{31}^\mathrm{s}+2(1-\nu _{12}^\mathrm{s})(E_3^\mathrm{s}/E_1^\mathrm{s})\bigr ]}{ 1+\bigl (1-\nu _{12}^{\mathrm{s}2}\bigr )(E_3^\mathrm{s}/E_1^\mathrm{s})-2\nu _{31}^\mathrm{s}(1+\nu _{12}^\mathrm{s}) -\nu _{31}^{\mathrm{s}2}(E_1^\mathrm{s}/E_3^\mathrm{s})}, \\ \displaystyle E_3=\frac{E_3^\mathrm{s}\bigl [1-4\nu _{31}^\mathrm{s}+2(1-\nu _{12}^\mathrm{s})(E_3^\mathrm{s}/E_1^\mathrm{s})\bigr ]}{ 2(1-\nu _{12}^\mathrm{s})(E_3^\mathrm{s}/E_1^\mathrm{s})-4\nu _{31}^{\mathrm{s}2}}. \end{array} \end{aligned}$$
(5.40)

Note that the corresponding decomposition parameter \(\alpha \) depends on the applied stresses \(\sigma _{11}\), \(\sigma _{22}\), and \(\sigma _{33}\).

Thus, the instantaneous response of any transversely isotropic biphasic tissue is equivalent to that of an incompressible transversely isotropic elastic material with the material constants \(\nu _{12}\), \(\nu _{31}\), \(G_{13}\), \(E_1\), and \(E_3\), given by (5.36), (5.39), and (5.40).

We note that in the isotropic case, the equivalent elastic material will also be isotropic, with Poisson’s ratio \(\nu =0{.}5\) and the shear modulus \(G=G_\mathrm{s}\), where \(G_\mathrm{s}\) is the shear modulus of the elastic skeleton. As a consequence of the relations \(E_1^\mathrm{s}=E_3^\mathrm{s}=2(1+\nu _\mathrm{s})G_\mathrm{s}\) and \(\nu _{12}^\mathrm{s}=\nu _{31}^\mathrm{s}=\nu _\mathrm{s}\), Eq. (5.40) yield the same result \(E_1=E_3=3G_\mathrm{s}\), irrespective of the value of the Poisson’s ratio \(\nu _\mathrm{s}\) of the skeleton.

Note also that if the elastic constants of the elastic skeleton satisfy the incompressibility condition (5.36), then Eq. (5.40) give \(E_1=E_1^\mathrm{s}\) and \(E_3=E_3^\mathrm{s}\), respectively, when successively taking the limits \(\nu _{31}^\mathrm{s}\rightarrow 0{.}5\) (with \(\nu _{12}^\mathrm{s}\) fixed) and after that as \(\nu _{12}^\mathrm{s}\rightarrow 1-E_1^\mathrm{s}/(2E_3^\mathrm{s})\), and vice-versa.

5.1.4 Axisymmetric Biphasic Model

Let us consider a biphasic material with the axis of symmetry oriented along the z axis of an \((r,\theta ,z)\) cylindrical coordinate system. The constitutive equations (5.13) for the effective stresses of the solid matrix are as follows:

$$\begin{aligned} \begin{array}{rcl} \sigma _{rr}^\mathrm{e} &{} = &{} A_{11}^\mathrm{s}\varepsilon _{rr}+A_{12}^\mathrm{s}\varepsilon _{\theta \theta } +A_{13}^\mathrm{s}\varepsilon _{zz}, \\ \sigma _{\theta \theta }^\mathrm{e} &{} = &{} A_{12}^\mathrm{s}\varepsilon _{rr}+A_{11}^\mathrm{s}\varepsilon _{\theta \theta } +A_{13}^\mathrm{s}\varepsilon _{zz}, \\ \sigma _{zz}^\mathrm{e} &{} = &{} A_{13}^\mathrm{s}\varepsilon _{rr}+A_{13}^\mathrm{s}\varepsilon _{\theta \theta } +A_{33}^\mathrm{s}\varepsilon _{zz}, \\ \sigma _{rz}^\mathrm{e} &{} = &{} 2A_{44}^\mathrm{s}\varepsilon _{rz}. \end{array} \end{aligned}$$
(5.41)

The strain-displacement relations (5.11) become

$$\begin{aligned} \varepsilon _{rr}=\frac{\partial u_r}{\partial r},\quad \varepsilon _{\theta \theta }=\frac{u_r}{r},\quad \varepsilon _{zz}=\frac{\partial u_z}{\partial z},\quad \varepsilon _{rz}=\frac{1}{2}\Bigl (\frac{\partial u_r}{\partial z} +\frac{\partial u_z}{\partial r}\Bigr ), \end{aligned}$$
(5.42)

where \(u_r\) and \(u_z\) denote the radial and axial displacements.

The equilibrium equations (5.10), which are written with respect to the total stress

$$\begin{aligned}{\varvec{\sigma }=-p\mathbf{I}+\varvec{\sigma }^\mathrm{e},} \end{aligned}$$

now reduce to the following equations of equilibrium:

$$\begin{aligned} \begin{array}{rcl} \displaystyle -\frac{\partial p}{\partial r}+\frac{1}{r}\frac{\partial (r\sigma _{rr}^\mathrm{e})}{\partial r} -\frac{\sigma _{\theta \theta }^\mathrm{e}}{r} +\frac{\partial \sigma _{rz}^\mathrm{e}}{\partial z} &{} = &{} 0, \\ \displaystyle -\frac{\partial p}{\partial z}+\frac{1}{r}\frac{\partial (r\sigma _{rz}^\mathrm{e})}{\partial r} +\frac{\partial \sigma _{zz}^\mathrm{e}}{\partial z} &{} = &{} 0. \end{array} \end{aligned}$$
(5.43)

Correspondingly, Eq. (5.19) takes the form

$$\begin{aligned} \frac{\partial }{\partial t}\biggl ( \frac{\partial u_r}{\partial r}+\frac{u_r}{r}+\frac{\partial u_z}{\partial z}\biggr ) =\frac{1}{r}\frac{\partial }{\partial r}\Bigl (r k_1\frac{\partial p}{\partial r}\Bigr ) +\frac{\partial }{\partial z}\Bigl (k_3\frac{\partial p}{\partial z}\Bigr ), \end{aligned}$$
(5.44)

where \(k_1\) and \(k_3\) are the in-plane and out-of-plane permeability coefficients, while the relative fluid flux can be expressed as

$$\begin{aligned} \mathbf{w}^\mathrm{f}=-k_1\frac{\partial p}{\partial r}\mathbf{e}_r -k_3\frac{\partial p}{\partial z}\mathbf{e}_z, \end{aligned}$$
(5.45)

with the radial and axial unit coordinate vectors \(\mathbf{e}_r\) and \(\mathbf{e}_z\), respectively.

5.2 Confined Compression of a Biphasic Material

In this section, we outline a linear confined compression biphasic model. In particular, the biphasic stress relaxation and the biphasic creep tests in confined compression are considered.

5.2.1 Confined Compression Problem

In the confined compression test, a cylindrical plug of biphasic material is constrained in a confining chamber with impermeable rigid walls, and is subjected to a compressive load, F(t), via a porous loading plate (see Fig. 5.1). Observe that the non-linear confined compression problem has been considered in a number of publications [8, 9, 62].

Fig. 5.1
figure 1

Schematic of the confined compression configuration [52]

In the cylindrical coordinate system, the boundary conditions on the lateral surface are

$$\begin{aligned} w_r^\mathrm{f}\bigr \vert _{r=a}=0,\quad \sigma _{rz}\bigr \vert _{r=a}=\sigma _{r\theta }\bigr \vert _{r=a}=0,\quad u_r\bigr \vert _{r=a}=0, \end{aligned}$$
(5.46)

where \(w_r^\mathrm{f}\) is the transverse (in-plane) relative fluid flux, \(\sigma _{rz}\) and \(\sigma _{r\theta }\) are out-of-plane and in-plane total shear stresses, and \(u_r\) is the radial displacement of the solid matrix.

The boundary conditions at the bottom surface, \(z=h\), are as follows:

$$\begin{aligned} w_z^\mathrm{f}\bigr \vert _{z=h}=0,\quad \sigma _{zr}\bigr \vert _{z=h}=\sigma _{z\theta }\bigr \vert _{z=h}=0.\quad u_z\bigr \vert _{z=h}=0, \end{aligned}$$
(5.47)

Here, \(w_z^\mathrm{f}\) and \(u_z\) are the vertical relative fluid flux and the vertical displacement of the solid matrix, respectively.

Meanwhile, the boundary conditions at the top surface, \(z=0\), have the form

$$\begin{aligned} p\bigr \vert _{z=0}=0,\quad \sigma _{zr}\bigr \vert _{z=0}=\sigma _{z\theta }\bigr \vert _{z=0}=0, \end{aligned}$$
(5.48)

where p is the interstitial fluid pressure.

Observe that the boundary condition (5.48)\(_1\) describes the free-draining porous interface, where no resistance to fluid movement is assumed at the interface between the porous loading plate and the sample surface.

For the stress relaxation test, the additional boundary condition at the top is

$$\begin{aligned} u_z\bigr \vert _{z=0}=w(t), \end{aligned}$$
(5.49)

where w(t) is a specified displacement of the loading plate.

For example, the ramp displacement is defined as

$$\begin{aligned} w(t)=\left\{ \begin{array}{l} V_0 t,\quad 0\le t\le t_0, \\ V_0 t_0,\quad t_0\le t, \end{array} \right. \end{aligned}$$
(5.50)

and \(V_0\), \(t_0\) are given constants.

On the other hand, for creep, the boundary condition is

$$\begin{aligned} \sigma _{zz}\bigr \vert _{z=0}=-\frac{F(t)}{A}, \end{aligned}$$
(5.51)

where F(t) is a specified external load acting on the porous loading plate, and \(A=\pi a^2\) is the sample cross-sectional area.

For example, if the load is applied instantaneously, then

$$\begin{aligned} F(t)=F_0\fancyscript{H}(t), \end{aligned}$$
(5.52)

where \(\fancyscript{H}(t)\) is the Heaviside step function.

The experimental setup of the confined compression test for articular cartilage represents a one-dimensional problem in the axial direction, so that

$$\begin{aligned} u_r=u_\theta =0,\quad \varepsilon _{rr}=\varepsilon _{\theta \theta }=0,\quad w_r^\mathrm{f}=w_\theta ^\mathrm{f}=0, \end{aligned}$$
(5.53)

while all non-trivial variables are dependent on t and z only.

Note that the assumptions (5.53)\(_1\) and (5.53)\(_2\) dictate that the rigid confining chamber prevents any lateral deformation. Therefore, the axial total stress is

$$\begin{aligned} \sigma _{zz}=-p+H_A\varepsilon _{zz}, \end{aligned}$$
(5.54)

where \(H_A\) is the confined compression equilibrium modulus (aggregate elastic modulus) of the solid matrix given by

$$\begin{aligned} H_A=A_{33}^\mathrm{s}. \end{aligned}$$
(5.55)

Hence, taking into account the free-draining condition (5.48)\(_1\), the traction boundary condition (5.51) can be reduced to the following:

$$\begin{aligned} H_A\frac{\partial u_z}{\partial z}\Bigr \vert _{z=0}=-f(t). \end{aligned}$$
(5.56)

Here, f(t) is the applied compressive stress, i.e.,

$$\begin{aligned} f(t)=\frac{F(t)}{A}. \end{aligned}$$
(5.57)

Finally, to complete the confined compression problem formulation, we assume the usual initial conditions (5.30).

5.2.2 Governing Equation of the Confined Compression Model

Under the assumptions made in the confined compression model, the only equilibrium differential equation (5.43)\(_2\) takes the form

$$\begin{aligned} -\frac{\partial p}{\partial z}+\frac{\partial \sigma _{zz}^\mathrm{e}}{\partial z}=0, \end{aligned}$$
(5.58)

where, in light of (5.53) and (5.54), we have

$$\begin{aligned} \sigma _{zz}^\mathrm{e}=H_A\varepsilon _{zz}. \end{aligned}$$
(5.59)

Integrating Eq. (5.59) with respect to the coordinate z, we arrive at the equation

$$\begin{aligned} {-}p+\sigma _{zz}^\mathrm{e}=\sigma _{zz}(t). \end{aligned}$$
(5.60)

Here, \(\sigma _{zz}(t)\) is the integration constant (being a function of the time variable t only).

Now, comparing Eqs. (5.54) and (5.60), we find that the total normal stress, \(\sigma _{zz}(t)\), is uniform through the depth of the biphasic sample.

Further, since all non-trivial variables are dependent on t and z only, the equilibrium equation (5.44) for the fluid phase after integration with respect to the coordinate z reduces to

$$\begin{aligned} \frac{\partial u_z}{\partial t}=k_3\frac{\partial p}{\partial z}, \end{aligned}$$
(5.61)

where \(k_3\) is the axial permeability coefficient.

Now, collecting Eqs. (5.59)–(5.61), we arrive at the governing equation

$$\begin{aligned} \frac{\partial u_z}{\partial t}=k_3\frac{\partial }{\partial z}\Bigl ( H_A\frac{\partial u_z}{\partial z}\Bigr ), \end{aligned}$$
(5.62)

which for a homogeneous biphasic material simplifies to a Fourier equation

$$\begin{aligned} \frac{\partial u_z}{\partial t}=k_3H_A\frac{\partial ^2 u_z}{\partial z^2}. \end{aligned}$$
(5.63)

Equation (5.63) is supplemented by the homogeneous initial condition

$$ u_z(0,z)=0 $$

and the boundary condition at the bottom surface

$$ u_z(t,h)=0. $$

For stress relaxation, the general boundary condition at the top surface is

$$\begin{aligned} u_z(t,0)=w(t), \end{aligned}$$
(5.64)

where w(t) is the prescribed surface displacement as a function of time, while for creep, in light of (5.56), we have

$$\begin{aligned} \frac{\partial u_z}{\partial z}(t,0)=-\frac{f(t)}{H_A}, \end{aligned}$$
(5.65)

where f(t) is the prescribed compressive stress as a function of time.

Note that the second-order parabolic partial differential equation (5.62) was solved in [26] using a semi-analytical approach based on the finite difference and Laplace transform methods. The effect of the depth-dependent aggregate modulus on articular cartilage stress-relaxation in confined compression was studied in [75].

5.2.3 Biphasic Stress Relaxation in Confined Compression

The following formula [64] gives the general solution to the problem (5.63) and (5.64):

$$\begin{aligned} u_z(t,z)=w(t)\Bigl (1-\frac{z}{h}\Bigr ) -\sum _{n=1}^\infty \frac{2}{\pi n}\sin \Bigl (\pi n\frac{z}{h}\Bigr ) \bigl (\fancyscript{K}_n w\bigr )(t). \end{aligned}$$
(5.66)

Here we have introduced the notation

$$\begin{aligned} \bigl (\fancyscript{K}_n w\bigr )(t)=\int \limits _{0^-}^t \exp \biggl (-\frac{n^2(t-\tau )}{\tau _R^\prime } \biggr )\dot{w}(\tau )\,d\tau , \end{aligned}$$
(5.67)

where \(\tau _R^\prime \) is the characteristic relaxation time in confined compression defined by

$$\begin{aligned} \tau _R^\prime =\frac{h^2}{\pi ^2 k_3 H_A}. \end{aligned}$$
(5.68)

Recall that the lower integration limit \(0^-\) in the integral operator above indicates that the integration in (5.67) starts at infinitesimally negative time so as to include the displacement discontinuity at time zero.

The strain of the solid matrix can be simply obtained from the relationship

$$\varepsilon _{zz}=\frac{\partial u_z}{\partial z}.$$

Then according to Eq. (5.60), we have

$$\begin{aligned} p=H_A\varepsilon _{zz}-\sigma _{zz}(t), \end{aligned}$$
(5.69)

from which we can determine the total normal stress, in light of the free-draining boundary condition (5.48)\(_1\), as follows:

$$ \sigma _{zz}(t)=H_A\varepsilon _{zz}\bigr \vert _{z=0}. $$

The interstitial hydrostatic pressure can be expressed in the form

$$\begin{aligned} p(t,z)=\frac{2H_A}{h}\sum _{n=1}^\infty \Bigl [1-\cos \Bigl (\pi n\frac{z}{h}\Bigr )\Bigr ] \bigl (\fancyscript{K}_n w\bigr )(t). \end{aligned}$$
(5.70)

The corresponding stress relaxation response in confined compression is then given by

$$\begin{aligned} \sigma _{zz}(t)=-H_A\frac{w(t)}{h} -\frac{2H_A}{h}\sum _{n=1}^\infty \bigl (\fancyscript{K}_n w\bigr )(t). \end{aligned}$$
(5.71)

In the case of constant strain rate compression in the loading phase \(t\in (0,t_0)\), followed by the hold period \(t\in (t_0,+\infty )\), formula (5.50) yields

$$ \dot{w}(t)=\left\{ \begin{array}{l} V_0,\quad 0\le t\le t_0, \\ 0,\quad t_0\le t. \end{array} \right. $$

Therefore, in the case of ramp displacement (5.50), the integral (5.67), which appears on the right-hand sides of (5.66), (5.70), and (5.71), is evaluated as

$$\begin{aligned} \bigl (\fancyscript{K}_n w\bigr )(t)= & {} \frac{V_0\tau _R^\prime }{n^2}\biggl (1-\exp \Bigl (-\frac{n^2 t}{\tau _R^\prime }\Bigr )\biggr ),\quad 0\le t <t_0, \nonumber \\ \bigl (\fancyscript{K}_n w\bigr )(t)= & {} \frac{V_0\tau _R^\prime }{n^2}\exp \Bigl (-\frac{n^2 t}{\tau _R^\prime }\Bigr ) \biggl (\exp \Bigl (\frac{n^2 t_0}{\tau _R^\prime }\Bigr )-1\biggr ),\quad t_0\le t. \nonumber \end{aligned}$$

Now, taking into account the above formulas and the identity

$$ \sum _{n=1}^\infty \frac{1}{n^2}=\frac{\pi ^2}{6}, $$

we arrive at the following formulas [60]:

$$\begin{aligned} \sigma _{zz}(t)=-H_A\frac{V_0 t}{h}-\frac{V_0 h}{3 k_3} +\frac{2V_0 h}{\pi ^2 k_3}\sum _{n=1}^\infty \frac{1}{n^2}\exp \Bigl (-\frac{n^2 t}{\tau _R^\prime }\Bigr ) \end{aligned}$$
(5.72)

for \(0\le t <t_0\), and

$$\begin{aligned} \sigma _{zz}(t)=-H_A\frac{V_0 t_0}{h} +\frac{2V_0 h}{\pi ^2 k_3}\sum _{n=1}^\infty \frac{1}{n^2} \biggl \{\exp \Bigl (-\frac{n^2 t}{\tau _R^\prime }\Bigr ) -\exp \biggl (-\frac{n^2 (t-t_0)}{\tau _R^\prime }\biggr )\biggr \} \end{aligned}$$
(5.73)

for \(t\ge t_0\).

Fig. 5.2
figure 2

The effect of \(t_0/\tau _R^\prime \) on the stress-relaxation time history in response to a ramped displacement. (The values taken by \(t_0/\tau _R^\prime \) are indicated on the figure.) Note the limit value 1 as \(t\rightarrow \infty \) in all cases due to the normalization

According to Eqs. (5.72) and (5.73), the effect of \(t_0/\tau _R^\prime \) on the dimensionless stress-relaxation time history \(-\sigma _{zz}(t)/[H_A\dot{\varepsilon }_0 t_0]\), where \(\dot{\varepsilon }_0=V_0/h\), is shown in Fig. 5.2.

Equations (5.72) and (5.73) can be used for determining the aggregate modulus, \(H_A\), and the constant axial permeability coefficient, \(k_3\), from the stress relaxation experiment by fitting the theoretical solution on to the experimental curve for the measured total normal stress [69].

5.2.4 Biphasic Creep in Confined Compression

The following formula [64] gives the general solution to the problem (5.63) and (5.65):

$$\begin{aligned} u_z(t,z)=\frac{2k_3}{h}\sum _{n=1}^\infty \cos \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr ) \bigl (\fancyscript{N}_n f\bigr )(t). \end{aligned}$$
(5.74)

Here we have introduced the notation

$$\begin{aligned} \bigl (\fancyscript{N}_n f\bigr )(t)=\int \limits _0^t \exp \biggl (-(2n-1)^2\frac{(t-\tau )}{\tau _R^{\prime \prime }} \biggr )f(\tau )\,d\tau , \end{aligned}$$
(5.75)

where \(\tau _R^\prime \) is the characteristic time having the meaning of a retardation time in confined compression defined as

$$\begin{aligned} \tau _R^{\prime \prime }=\frac{4h^2}{\pi ^2 k_3 H_A}. \end{aligned}$$
(5.76)

Thus, the nominal sample-average strain (surface-to-surface strain), which is calculated from the distance \(h-u_z(t,0)\) between the loading platens, is

$$\begin{aligned} \frac{u_z(t,0)}{h}=\frac{2k_3}{h^2}\sum _{n=1}^\infty \bigl (\fancyscript{N}_n f\bigr )(t). \end{aligned}$$
(5.77)

Taking into account the constitutive equation (5.60) and the boundary conditions (5.48)\(_1\) and (5.56), we find that the total normal stress in the biphasic sample is

$$ \sigma _{zz}(t)=-f(t). $$

Hence, Eq. (5.69) allows evaluation of the interstitial hydrostatic pressure via

$$\begin{aligned} p=H_A\varepsilon _{zz}+f(t). \end{aligned}$$
(5.78)

Evaluating the strain of the solid matrix according to formula (5.74) and substituting the obtained result into Eq. (5.78), we find

$$\begin{aligned} p(t,z)=f(t)-\frac{\pi k_3 H_A}{h^2}\sum _{n=1}^\infty (2n-1)\sin \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr ) \bigl (\fancyscript{N}_n f\bigr )(t). \end{aligned}$$
(5.79)

For a creep experiment, where a constant external load, \(F_0\), is applied instantaneously, the following loading law holds true (see Eq. (5.57)):

$$\begin{aligned} f(t)=f_0\fancyscript{H}(t). \end{aligned}$$
(5.80)

Here, \(\fancyscript{H}(t)\) is the Heaviside step function, \(f_0=F_0/A\) is the constant compressive stress, and A is the cross-sectional area of the biphasic sample.

Correspondingly, formula (5.77) yields the following result [14]:

$$\begin{aligned} \frac{u_z(t,0)}{h}=\frac{f_0}{H_A}\biggl \{1-\frac{8}{\pi ^2} \sum _{n=1}^\infty \frac{1}{(2n-1)^2} \exp \biggl (-(2n-1)^2\frac{t}{\tau _R^{\prime \prime }}\biggr )\biggr \}. \end{aligned}$$
(5.81)

Note that in writing (5.81), we used the identity

$$ \sum _{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{\pi ^2}{8}. $$

In the same way, in the case (5.80), formula (5.79) is rearranged to obtain

$$\begin{aligned} p(t,z)=\frac{4 f_0}{\pi }\sum _{n=1}^\infty \frac{1}{2n-1}\sin \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr ) \exp \biggl (-(2n-1)^2\frac{t}{\tau _R^{\prime \prime }}\biggr ). \end{aligned}$$
(5.82)

In writing (5.82), the following identity was used (see, e.g., [30], formula (1.442.1)):

$$\begin{aligned} \sum _{n=1}^\infty \frac{\sin (2n-1)\zeta }{2n-1}=\frac{\pi }{4},\quad 0<\zeta <\pi . \end{aligned}$$
(5.83)

By substituting \(z=h\) into Eq. (5.82), we readily obtain

$$\begin{aligned} p(t,h)=\frac{4 f_0}{\pi }\sum _{k=0}^\infty \frac{(-1)^k}{2k+1}\exp \biggl (-(2k+1)^2\frac{t}{\tau _R^{\prime \prime }}\biggr ). \end{aligned}$$
(5.84)

Note also [69] that the interstitial fluid pressure at the impermeable interface, \(z=h\), can be represented by the formula

$$ p(t,h)=H_A\Bigl (\varepsilon _{zz}\bigr \vert _{z=h}-\varepsilon _{zz}\bigr \vert _{z=0}\Bigr ), $$

where \(\varepsilon _{zz}\) is the strain of the solid matrix, \(\varepsilon _{zz}=\partial u_z/\partial z\), while the displacement \(u_z\) is taken from (5.74) for creep and from (5.66) for stress relaxation.

Correspondingly, the so-called fluid load support

$$ \frac{W^p}{W}=\frac{p(t,h)}{f(t)} $$

is evaluated as follows [63]:

$$ \frac{W^p}{W}=\biggl (\frac{\partial u}{\partial z}\Bigr \vert _{z=0} -\frac{\partial u}{\partial z}\Bigr \vert _{z=h}\biggr )\biggr / \frac{\partial u}{\partial z}\Bigr \vert _{z=0}. $$

In the stepwise creep test, according to (5.80) and (5.84), we have

$$ \frac{W^p}{W}=\frac{4}{\pi }\sum _{k=0}^\infty \frac{(-1)^k}{2k+1}\exp \biggl (-(2k+1)^2\frac{t}{\tau _R^{\prime \prime }}\biggr ). $$

Therefore, as a consequence of (5.83), the above formula yields the maximum value

$$ \frac{W^p}{W}\Bigr \vert _{t=0}=1. $$

It should be noted [63] that in real creep testing configurations, there exists a delay in pressurization due to the impedance of the pressure transducer in measuring the pressure p(th). Namely, the greater the compliance of the pressure transducer, the greater the delay in the interstitial fluid pressurization achieving the peak value of \({W^p{/}W}\).

Equation (5.81) is commonly used in determining the aggregate modulus, \(H_A\), and the constant axial permeability coefficient, \(k_3\), from the confined compression creep experiment by fitting the theoretical solution on to the experimental curve for the nominal sample-average strain. It should be mentioned that for improved mechanical characterization of articular cartilage, testing experiments may involve multiple-step ramp loading [50] as well as an alternating sequence of stress relaxation and creep transients [17].

Finally, observe [13] that the presence of a gap between the loading plate and the confining chamber walls, which is necessary to guarantee a correct plate movement, allows flow exudation and tissue extrusion around the plate, thus leading to underestimation of \(H_A\) and overestimation of \(k_3\).

5.2.5 Dynamic Behavior of a Biphasic Material Under Cyclic Compressive Loading in Confined Compression

Let us now consider the following loading history for cyclic compressive confined compression [70, 73]:

$$\begin{aligned} f(t)=f_0(1-\cos \omega t)\fancyscript{H}(t). \end{aligned}$$
(5.85)

Here, \(f_0\) is the median magnitude of the applied cyclic stress, and \(\omega \) is the loading angular frequency. Recall that \(\omega /(2\pi )\) is the loading frequency measured in hertz.

First of all, using integration by parts and the identity (see, e.g., [30], formula (1.444.6))

$$ \sum _{n=1}^\infty \frac{\cos (2n-1)\zeta }{(2n-1)^2}=\frac{\pi }{4}\Bigl (\frac{\pi }{2}-\zeta \Bigr ),\quad 0<\zeta <\pi , $$

we transform formulas (5.74) and (5.79) as follows:

$$\begin{aligned} u_z(t,z)= & {} \frac{h}{H_A}\biggl \{ f(t)\Bigl (1-\frac{z}{h}\Bigr ) \nonumber \\ { }{} & {} {}-\frac{8}{\pi ^2}\sum _{n=1}^\infty \frac{1}{(2n-1)^2} \cos \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr ) \bigl (\fancyscript{M}_n f\bigr )(t)\biggr \}, \end{aligned}$$
(5.86)
$$\begin{aligned} p(t,z)= & {} \frac{4}{\pi }\sum _{n=1}^\infty \frac{1}{2n-1}\sin \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr ) \bigl (\fancyscript{M}_n f\bigr )(t). \end{aligned}$$
(5.87)

Here we have introduced the notation

$$\begin{aligned} \bigl (\fancyscript{M}_n f\bigr )(t)=\int \limits _{0^-}^t \exp \biggl (-(2n-1)^2\frac{(t-\tau )}{\tau _R^{\prime \prime }} \biggr )\dot{f}(\tau )\,d\tau . \end{aligned}$$
(5.88)

As usual, the notation \(0^-\) in the lower limit of the integral above means that

$$ \int \limits _{0^-}^t e^{-a_n(t-\tau )}\dot{f}(\tau )\,d\tau =f(0^+)e^{-a_n t}+\int \limits _0^t e^{-a_n(t-\tau )}\dot{f}(\tau )\,d\tau , $$

where \(f(0^+)\) is the limit of the function f(t) when the independent variable t approaches 0 from the right.

Since \(f(0)=0\) for the function defined by formula (5.85), we have

$$ \dot{f}(\tau )=\omega f_0\fancyscript{H}(t)\sin \omega t, $$

so that Eqs. (5.86) and (5.87) yield

$$\begin{aligned} u_z(t,z)= & {} \frac{f_0 h}{H_A}\biggl \{ (1-\cos \omega t)\Bigl (1-\frac{z}{h}\Bigr ) \nonumber \\ { }{} & {} {}-\frac{8}{\pi ^2}\sin \omega t \sum _{n=1}^\infty \frac{\omega a_n}{(2n-1)^2(a_n^2+\omega ^2)} \cos \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr ) \nonumber \\ { }{} & {} {}+\frac{8}{\pi ^2}\cos \omega t \sum _{n=1}^\infty \frac{\omega ^2}{(2n-1)^2(a_n^2+\omega ^2)} \cos \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr ) \nonumber \\ { }{} & {} {}-\frac{8}{\pi ^2}\sum _{n=1}^\infty \frac{\omega ^2}{(2n-1)^2(a_n^2+\omega ^2)} \exp (-a_n t)\cos \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr )\biggr \} \quad \qquad \end{aligned}$$
(5.89)

for the vertical displacement of the solid matrix and

$$\begin{aligned} p(t,z)= & {} \frac{4f_0}{\pi }\biggl \{ \sin \omega t \sum _{n=1}^\infty \frac{\omega a_n}{(2n-1)(a_n^2+\omega ^2)} \sin \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr ) \nonumber \\ { }{} & {} {}-\cos \omega t \sum _{n=1}^\infty \frac{\omega ^2}{(2n-1)(a_n^2+\omega ^2)} \sin \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr ) \nonumber \\ { }{} & {} {}+\sum _{n=1}^\infty \frac{\omega ^2}{(2n-1)(a_n^2+\omega ^2)} \exp (-a_n t)\sin \Bigl (\frac{\pi (2n-1)z}{2h}\Bigr )\biggr \} \end{aligned}$$
(5.90)

for the interstitial fluid pressure, where we have introduced the notation

$$ a_n=\frac{(2n-1)^2}{\tau _R^{\prime \prime }}. $$

It can be checked that Eqs. (5.89) and (5.90) coincide with the corresponding results obtained by Suh et al. [73], apart from notation.

5.3 Unconfined Compression of a Biphasic Material

In this section, the unconfined compression biphasic model is developed. In particular, the biphasic stress relaxation and the biphasic creep tests in unconfined compression are studied.

5.3.1 Unconfined Compression Problem

In the axisymmetric unconfined compression test, a thin cylindrical disk of biphasic material is compressed between two smooth (frictionless) and impermeable rigid platens (see Fig. 5.3). Therefore, the material is free to expand radially, and free fluid flow is enabled across the lateral cylindrical surface.

Fig. 5.3
figure 3

Schematic of the unconfined compression configuration [52]. The articular cartilage sample has to be stripped off from the subchondral bone and cut into a perfect cylinder

In the cylindrical coordinate system, the boundary conditions on the lateral surface are

$$\begin{aligned} p\bigr \vert _{r=a}=0,\quad \sigma _{rz}\bigr \vert _{r=a}=\sigma _{r\theta }\bigr \vert _{r=a}=\sigma _{rr}\bigr \vert _{r=a}=0. \end{aligned}$$
(5.91)

Here, p is the interstitial fluid pressure, \(\sigma _{rz}\) and \(\sigma _{r\theta }\) are shear stresses, \(\sigma _{rr}\) is the radial normal stress.

The boundary conditions at the bottom surface, \(z=h\), are

$$\begin{aligned} w_z^\mathrm{f}\bigr \vert _{z=h}=0,\quad \sigma _{zr}\bigr \vert _{z=h}=\sigma _{z\theta }\bigr \vert _{z=h}=0,\quad u_z\bigr \vert _{z=h}=0, \end{aligned}$$
(5.92)

where \(w_z^\mathrm{f}\) and \(u_z\) are the vertical relative fluid flux and the vertical displacement of the solid matrix, respectively.

At the top surface, \(z=0\), we have the following boundary conditions:

$$\begin{aligned} w_z^\mathrm{f}\bigr \vert _{z=0}=0,\quad \sigma _{zr}\bigr \vert _{z=0}=\sigma _{z\theta }\bigr \vert _{z=0}=0,\quad u_z\bigr \vert _{z=0}=w(t). \end{aligned}$$
(5.93)

Here, w(t) is the vertical displacement of the upper platen.

Again, we can consider either the creep test (load-controlled) or the stress-relaxation test (displacement-controlled) in unconfined compression. For the stress-relaxation experiment, w(t) is a prescribed function of time.

Following Armstrong et al. [5] and Cohen et al. [19], we assume the radial displacement of the solid skeleton, \(u_r\), the fluid relative radial velocity, \(w_r^\mathrm{f}\), and the fluid pressure, p, to be of the form

$$\begin{aligned} u_r=u_r(t,r),\quad w_r^\mathrm{f}=w_r^\mathrm{f}(t,r), \quad p=p(t,r). \end{aligned}$$
(5.94)

Moreover, the axial strain, \(\varepsilon _{zz}\), is assumed to be uniform throughout the sample, i.e.,

$$\begin{aligned} \varepsilon _{zz}=\varepsilon (t), \end{aligned}$$
(5.95)

where \(\varepsilon (t)\) is a time-dependent function.

Hence, in light of the assumption (5.95), the integration of the equation

$$\varepsilon _{zz}=\frac{\partial u_z}{\partial z}$$

with the boundary conditions (5.92)\(_3\) and (5.93)\(_3\) taken into account yields

$$\begin{aligned} \varepsilon (t)=-\frac{w(t)}{h}, \end{aligned}$$
(5.96)
$$\begin{aligned} u_z(t,z)=-\varepsilon (t)(h-z). \end{aligned}$$
(5.97)

According to (5.94), (5.95), and (5.97), the only nonzero strain components are

$$ \varepsilon _{rr}=\frac{\partial u_r}{\partial r},\quad \varepsilon _{\theta \theta }=\frac{u_r}{r},\quad \varepsilon _{zz}=\varepsilon , $$

and, correspondingly, Eq. (5.41) yield the following nonzero effective stresses:

$$\begin{aligned} \begin{array}{rcl} \displaystyle \sigma _{rr}^\mathrm{e} &{} = &{} \displaystyle A_{11}^\mathrm{s}\frac{\partial u_r}{\partial r}+A_{12}^\mathrm{s}\frac{u_r}{r} +A_{13}^\mathrm{s}\varepsilon , \\ \displaystyle \sigma _{\theta \theta }^\mathrm{e} &{} = &{} \displaystyle A_{12}^\mathrm{s}\frac{\partial u_r}{\partial r}+A_{11}^\mathrm{s}\frac{u_r}{r} +A_{13}^\mathrm{s}\varepsilon , \\ \displaystyle \sigma _{zz}^\mathrm{e} &{} = &{} \displaystyle A_{13}^\mathrm{s}\frac{\partial u_r}{\partial r}+A_{13}^\mathrm{s}\frac{u_r}{r} +A_{33}^\mathrm{s}\varepsilon . \end{array} \end{aligned}$$
(5.98)

The substitution of (5.98) into the equilibrium equations for the solid matrix (5.43) results in the differential equation

$$\begin{aligned} {-}\frac{\partial p}{\partial r}+A_{11}^\mathrm{s}\biggl ( \frac{\partial ^2 u_r}{\partial r^2}+\frac{1}{r}\frac{\partial u_r}{\partial r}-\frac{u_r}{r^2}\biggr )=0, \end{aligned}$$
(5.99)

while Eq. (5.44) takes the form

$$\begin{aligned} \frac{\partial }{\partial t}\biggl ( \frac{\partial u_r}{\partial r}+\frac{u_r}{r}+\varepsilon \biggr ) =\frac{1}{r}\frac{\partial }{\partial r}\Bigl (r k_1\frac{\partial p}{\partial r}\Bigr ), \end{aligned}$$
(5.100)

where \(k_1\) is the transverse (in-plane) permeability coefficient.

Finally, the relative fluid flux (5.45), in light of (5.94), is given by

$$\begin{aligned} \mathbf{w}^\mathrm{f}=-k_1\frac{\partial p}{\partial r}\mathbf{e}_r, \end{aligned}$$
(5.101)

where \(\mathbf{e}_r\) is the radial unit vector.

Equations (5.99)–(5.101), with the boundary conditions (5.91)–(5.93) and the zero initial conditions, constitute the unconfined compression problem.

It has been established [57] that the incorporation of a transverse isotropy for material properties into the linear biphasic theory improves its predictive power in unconfined compression analysis [19].

5.3.2 Solution of the Unconfined Compression Problem

Let us first turn to Eq. (5.100). Taking into account that

$$ \frac{\partial u_r}{\partial r}+\frac{u_r}{r}=\frac{1}{r}\frac{\partial }{\partial r}(r u_r), $$

we can integrate Eq. (5.100) once with respect to the radial coordinate to get

$$\begin{aligned} \frac{\partial p}{\partial r}=\frac{1}{k_1}\frac{\partial }{\partial t}\Bigl ( u_r+\frac{\varepsilon }{2}r\Bigr ), \end{aligned}$$
(5.102)

where the integration constant vanishes due to the regularity condition at the center of the sample, \(r=0\).

Therefore, Eqs. (5.99) and (5.102) yield the following equation [19]:

$$\begin{aligned} \frac{\partial ^2 u_r}{\partial r^2}+\frac{1}{r}\frac{\partial u_r}{\partial r}-\frac{u_r}{r^2} =\frac{1}{A_{11}^\mathrm{s}k_1}\frac{\partial }{\partial t}\Bigl ( u_r+\frac{\varepsilon }{2}r\Bigr ). \end{aligned}$$
(5.103)

Recall that the boundary conditions (5.91)\(_2\) are formulated in terms of the components of the total stress tensor

$$\begin{aligned} {\varvec{\sigma }=-p\mathbf{I}+\varvec{\sigma }^\mathrm{e}.} \end{aligned}$$
(5.104)

Hence, as a consequence of the constitutive equation (5.98), the boundary conditions (5.91) become

$$\begin{aligned} p\bigr \vert _{r=a}=0,\quad A_{11}^\mathrm{s}\frac{\partial u_r}{\partial r}+A_{12}^\mathrm{s}\frac{u_r}{r}+A_{13}^\mathrm{s}\varepsilon \,\Bigr \vert _{r=a}=0. \end{aligned}$$
(5.105)

To simplify our treatment of the problem, we introduce dimensionless variables

$$\begin{aligned} \rho =\frac{r}{a},\quad U=\frac{u_r}{a},\quad \tau =\frac{A_{11}^\mathrm{s}k_1}{a^2}t,\quad P=\frac{p}{A_{11}^\mathrm{s}}. \end{aligned}$$
(5.106)

Then, Eqs. (5.102)–(5.105) take the form

$$\begin{aligned} \frac{\partial P}{\partial \rho }=\frac{\partial }{\partial \tau }\Bigl ( U+\frac{\varepsilon }{2}\rho \Bigr ), \end{aligned}$$
(5.107)
$$\begin{aligned} \frac{\partial ^2 U}{\partial \rho ^2}+\frac{1}{\rho }\frac{\partial U}{\partial \rho }-\frac{U}{\rho ^2} =\frac{\partial }{\partial \tau }\Bigl ( U+\frac{\varepsilon }{2}\rho \Bigr ), \end{aligned}$$
(5.108)
$$\begin{aligned} P\bigr \vert _{\rho =1}=0,\quad \frac{\partial U}{\partial \rho }+\alpha _{12}\frac{U}{\rho }+\alpha _{13}\varepsilon \,\biggr \vert _{\rho =1}=0, \end{aligned}$$
(5.109)

where we have introduced the notation

$$\begin{aligned} \alpha _{12}=\frac{A_{12}^\mathrm{s}}{A_{11}^\mathrm{s}},\quad \alpha _{13}=\frac{A_{13}^\mathrm{s}}{A_{11}^\mathrm{s}}. \end{aligned}$$
(5.110)

Now, let \(\tilde{P}(s)\), \(\tilde{U}(s)\), and \(\tilde{\varepsilon }(s)\) denote the Laplace transforms with respect to the dimensionless time \(\tau \). Taking into account the zero initial conditions, the Laplace transformation of Eqs. (5.107)–(5.109) leads to the system

$$\begin{aligned} \frac{\partial \tilde{P}}{\partial \rho }=s\Bigl ( \tilde{U}+\frac{\tilde{\varepsilon }}{2}\rho \Bigr ), \end{aligned}$$
(5.111)
$$\begin{aligned} \frac{\partial ^2 \tilde{U}}{\partial \rho ^2}+\frac{1}{\rho }\frac{\partial \tilde{U}}{\partial \rho }-\frac{\tilde{U}}{\rho ^2} =s\Bigl ( \tilde{U}+\frac{\tilde{\varepsilon }}{2}\rho \Bigr ), \end{aligned}$$
(5.112)
$$\begin{aligned} \tilde{P}\bigr \vert _{\rho =1}=0,\quad \frac{\partial \tilde{U}}{\partial \rho }+\alpha _{12}\frac{\tilde{U}}{\rho } +\alpha _{13}\tilde{\varepsilon }\,\biggr \vert _{\rho =1}=0. \end{aligned}$$
(5.113)

The general solution of Eq. (5.112) can be represented in the form

$$\begin{aligned} \tilde{U}=-\frac{\tilde{\varepsilon }}{2}\rho +\tilde{U}_0, \end{aligned}$$
(5.114)

where \(\tilde{U}_0\) is the general solution of the homogeneous equation corresponding to Eq. (5.112), i.e.,

$$\begin{aligned} \frac{\partial ^2 \tilde{U}_0}{\partial \rho ^2}+\frac{1}{\rho }\frac{\partial \tilde{U}_0}{\partial \rho } -\Bigl (s+\frac{1}{\rho ^2}\Bigr )\tilde{U}_0=0. \end{aligned}$$
(5.115)

Making use of the change of the independent variable \(\rho =\rho ^\prime /\sqrt{s}\), Eq. (5.115) can be reduced to the modified Bessel’s equation. In this way, taking into account the regularity condition at \(\rho =0\), we obtain

$$\begin{aligned} \tilde{U}_0=C_0 I_1(\sqrt{s}\rho ), \end{aligned}$$
(5.116)

where \(C_0\) is an arbitrary function of the Laplace transform parameter s. This integration constant (with respect to the variable \(\rho \)) should be determined from the boundary condition (5.113)\(_2\), which in light of (5.114) becomes

$$ \frac{\partial \tilde{U}_0}{\partial \rho }+\alpha _{12}\frac{\tilde{U}_0}{\rho }\,\biggr \vert _{\rho =1} =\frac{\tilde{\varepsilon }}{2}(1+\alpha _{12}-2\alpha _{13}). $$

By substituting the expression (5.116) into the above equation and taking into account the identity \(I_1^\prime (x)=I_0(x)-(1/x)I_1(x)\) for the modified Bessel functions of the first kind, we find

$$\begin{aligned} C_0=\frac{(1+\alpha _{12}-2\alpha _{13})\tilde{\varepsilon }(s)}{ 2\bigl [ \sqrt{s}I_0(\sqrt{s})-(1-\alpha _{12})I_1(\sqrt{s})\bigr ]}. \end{aligned}$$
(5.117)

Collecting formulas (5.114)–(5.117), we obtain

$$\begin{aligned} \tilde{U}=-\frac{\tilde{\varepsilon }}{2}\rho \left( 1-\frac{(1+\alpha _{12}-2\alpha _{13})I_1(\sqrt{s}\rho )}{\displaystyle \sqrt{s}\rho \biggl ( I_0(\sqrt{s})-(1-\alpha _{12})\frac{I_1(\sqrt{s})}{\sqrt{s}} \biggr )}\right) . \end{aligned}$$
(5.118)

We now calculate the Laplace transform of the dimensionalized pressure, \(\tilde{P}\), from Eq. (5.111), which as a result of (5.114) can be rewritten as

$$ \frac{\partial \tilde{P}}{\partial \rho }=s\tilde{U}_0. $$

The integration of the above equation with respect to \(\rho \), in light of the identity \(I_0^\prime (x)=I_1(x)\), yields

$$ \tilde{P}=C_0\sqrt{s} I_0(\sqrt{s}\rho )+C_1, $$

where \(C_1\) is an arbitrary function of s. By satisfying the boundary condition (5.113)\(_1\), we immediately get \(C_1=-C_0 \sqrt{s} I_0(\sqrt{s})\) and

$$\begin{aligned} \tilde{P}=C_0\sqrt{s}\bigl (I_0(\sqrt{s}\rho )-I_0(\sqrt{s})\bigr ), \end{aligned}$$
(5.119)

where \(C_0\) is given by (5.117).

Finally, we consider the force response of the biphasic sample

$$\begin{aligned} F(t)=-2\pi \int \limits _0^a\sigma _{zz}(t,r)r\,dr, \end{aligned}$$
(5.120)

where, in light of (5.98) and (5.104), the out-of-plane normal total stress is given by

$$\begin{aligned} \sigma _{zz}=-p+A_{13}^\mathrm{s}\frac{\partial u_r}{\partial r}+A_{13}^\mathrm{s}\frac{u_r}{r}+A_{33}^\mathrm{s}\varepsilon . \end{aligned}$$
(5.121)

In terms of the dimensionless variables (5.106), Eq. (5.120) takes the form

$$\begin{aligned} \fancyscript{F}(\tau )=-2\int \limits _0^1 \biggl (-P+\alpha _{13}\Bigl ( \frac{\partial U}{\partial \rho }+\frac{U}{\rho }\Bigr )+\alpha _{33}\varepsilon \biggr )\rho \,d\rho , \end{aligned}$$
(5.122)

where we have introduced the notation

$$\begin{aligned} \fancyscript{F}(\tau )=\frac{F(t)}{\pi a^2 A_{11}^\mathrm{s}}, \end{aligned}$$
(5.123)
$$\begin{aligned} \alpha _{33}=\frac{A_{33}^\mathrm{s}}{A_{11}^\mathrm{s}}. \end{aligned}$$
(5.124)

After application of the Laplace transform, Eq. (5.122) becomes

$$\begin{aligned} \tilde{\fancyscript{F}}(s)=-2\int \limits _0^1 \biggl (-\tilde{P}+\alpha _{13}\Bigl ( \frac{\partial \tilde{U}}{\partial \rho }+\frac{\tilde{U}}{\rho }\Bigr )+\alpha _{33}\tilde{\varepsilon }\biggr )\rho \,d\rho . \end{aligned}$$
(5.125)

Now, taking into account Eqs. (5.114), (5.117), (5.119) and formulas

$$I_0^\prime (x)=I_1(x),\quad xI_1^\prime (x)=xI_0(x)-I_1(x),$$

the integral (5.125) becomes

$$\begin{aligned} \tilde{\fancyscript{F}}(s)=-\frac{\displaystyle \gamma _1 I_0(\sqrt{s})-\gamma _2 \frac{I_1(\sqrt{s})}{\sqrt{s}}}{\displaystyle I_0(\sqrt{s})-\gamma _0 \frac{I_1(\sqrt{s})}{\sqrt{s}}}\tilde{\varepsilon }(s), \end{aligned}$$
(5.126)

where we have also introduced the notation

$$\begin{aligned} \gamma _0= & {} 1-\alpha _{12}, \nonumber \\ \gamma _1= & {} \frac{1}{2}(1+\alpha _{12}+2\alpha _{33}-4\alpha _{13}), \\ \gamma _2= & {} \alpha _{33}(1-\alpha _{12})+1+\alpha _{12}-4\alpha _{13}+2\alpha _{13}^2. \nonumber \end{aligned}$$
(5.127)

Formulas (5.126) and (5.127) coincide with the corresponding results given by Cohen et al. [19], up to the notation. In the isotropic case, we have

$$ \alpha _{12}=\alpha _{13}=\frac{\lambda _\mathrm{s}}{H_A},\quad \alpha _{33}=1,\quad \gamma _0=\frac{2\mu _\mathrm{s}}{H_A},\quad \gamma _1=\frac{3\mu _\mathrm{s}}{H_A},\quad \gamma _2=\frac{8\mu _\mathrm{s}^2}{H_A^2}, $$

where \(\lambda _\mathrm{s}\), \(\mu _\mathrm{s}\), and \(H_A\) are the Lamé elastic constants and the aggregate elastic modulus of the solid skeleton, respectively, and the original results of Armstrong et al. [5] are immediately recovered.

5.3.3 Unconfined Compression Model

Following [3], we rewrite Eq. (5.126) in the form

$$\begin{aligned} \tilde{\fancyscript{F}}(s)=-s\tilde{\varepsilon }(s)\tilde{\fancyscript{K}}(s), \end{aligned}$$
(5.128)

where we have introduced the notation

$$\begin{aligned} \tilde{\fancyscript{K}}(s)=\frac{\displaystyle \gamma _1 I_0(\sqrt{s})-\gamma _2 \frac{I_1(\sqrt{s})}{\sqrt{s}}}{\displaystyle s\biggl ( I_0(\sqrt{s})-\gamma _0 \frac{I_1(\sqrt{s})}{\sqrt{s}}\biggr )}. \end{aligned}$$
(5.129)

By applying the convolution theorem to Eq. (5.128), we obtain

$$\begin{aligned} \fancyscript{F}(\tau )=-\int \limits _{0^-}^\tau \frac{d\varepsilon (\tau ^\prime )}{d\tau ^\prime } \fancyscript{K}(\tau -\tau ^\prime )\,d\tau ^\prime , \end{aligned}$$
(5.130)

where \(\fancyscript{K}(\tau )=\fancyscript{L}^{-1}\{\tilde{\fancyscript{K}}(s)\}\) is the original function of \(\tilde{\fancyscript{K}}(s)\), and \(\tau =0^-\) is the dimensionless time moment just preceding the initial moment of loading \(\tau =0\). In deriving Eq. (5.130), we have used the formula

$$ \fancyscript{L}^{-1}\{s\tilde{\varepsilon }(s)\}=\frac{d\varepsilon (\tau )}{d\tau }+\varepsilon (0^+)\delta (\tau ), $$

where \(\delta (\tau )\) is the Dirac function.

Following [5], we calculate the inverse Laplace transform by using the residue theorem (see, e.g., [46, 48]) to find

$$\begin{aligned} \fancyscript{K}(\tau )=\frac{2\gamma _1-\gamma _2}{2-\gamma _0} +\sum _{n=1}^\infty \frac{2(\gamma _2-\gamma _0\gamma _1)}{\alpha _n^2-\gamma _0(2-\gamma _0)} e^{-\alpha _n^2\tau }, \end{aligned}$$
(5.131)

where \(\alpha _n\) are the roots of the transcendental equation

$$\begin{aligned} J_0(x)-\gamma _0 \frac{J_1(x)}{x}=0, \end{aligned}$$
(5.132)

in which \(J_0(x)\) and \(J_1(x)\) are Bessel functions of the first kind.

The inverse relation for Eq. (5.130) can be represented by

$$\begin{aligned} \varepsilon (\tau )=-\int \limits _{0^-}^\tau \frac{d\fancyscript{F}(\tau ^\prime )}{d\tau ^\prime } \fancyscript{M}(\tau -\tau ^\prime )\,d\tau ^\prime , \end{aligned}$$
(5.133)

where \(\fancyscript{M}(\tau )=\fancyscript{L}^{-1}\{\tilde{\fancyscript{M}}(s)\}\), and \(\tilde{\fancyscript{M}}(s)\) is defined by the formula

$$\begin{aligned} \tilde{\fancyscript{M}}(s)=\frac{\displaystyle I_0(\sqrt{s})-\gamma _0 \frac{I_1(\sqrt{s})}{\sqrt{s}}}{\displaystyle s\biggl (\gamma _1 I_0(\sqrt{s})-\gamma _2 \frac{I_1(\sqrt{s})}{\sqrt{s}}\biggr )}. \end{aligned}$$
(5.134)

Again making use of the residue theorem, we obtain

$$\begin{aligned} \fancyscript{M}(\tau )=\frac{2-\gamma _0}{2\gamma _1-\gamma _2} -\sum _{n=1}^\infty \frac{2(\gamma _2-\gamma _0\gamma _1)}{ \gamma _1^2\beta _n^2+\gamma _2(\gamma _2-2\gamma _1)} e^{-\beta _n^2\tau }, \end{aligned}$$
(5.135)

where \(\beta _n\) are the roots of the transcendental equation

$$\begin{aligned} J_0(x)-\frac{\gamma _2}{\gamma _1}\frac{J_1(x)}{x}=0. \end{aligned}$$
(5.136)

The short-time asymptotic approximation for the kernel \(\fancyscript{K}(\tau )\), as can also be found for \(\fancyscript{M}(\tau )\), can be obtained by evaluating the Laplace inverse of \(\tilde{\fancyscript{K}}(s)\) as \(s\rightarrow \infty \). For this purpose, we apply the following well known asymptotic expansion (see, e.g., [30], formula (8.451.5)):

$$ I_n(z)=\frac{e^z}{\sqrt{2\pi z}}\Bigl \{1+\frac{(1-4n^2)}{8z}+O(z^{-2})\Bigr \}. $$

By making use of the above asymptotic formula, we expand the right-hand sides of (5.129) and (5.134) in terms of \(1/\sqrt{s}\). As a result, we arrive at the following asymptotic expansions:

$$\begin{aligned} \fancyscript{K}(\tau )=\gamma _1-\frac{2}{\sqrt{\pi }}(\gamma _2-\gamma _0\gamma _1)\sqrt{\tau }+O(\tau ),\quad \tau \rightarrow 0^+, \end{aligned}$$
(5.137)
$$\begin{aligned} \fancyscript{M}(\tau )=\frac{1}{\gamma _1}+\frac{2}{\sqrt{\pi }}\frac{(\gamma _2-\gamma _0\gamma _1)}{\gamma _1^2}\sqrt{\tau }+O(\tau ),\quad \tau \rightarrow 0^+. \end{aligned}$$
(5.138)

The asymptotic approximations (5.137) and (5.138) can be used in evaluating unconfined impact compression tests where the impact duration is relatively small compared to the so-called [5] gel diffusion time for the biphasic material \(t_\mathrm{g}=a^2/(H_A k_1)\), which is the time taken for a cylindrical biphasic sample of radius a to reach equilibrium in unconfined stepwise compression.

Further, let us introduce the notation

$$ \fancyscript{K}_0=\fancyscript{K}(0),\quad \fancyscript{M}_0=\fancyscript{M}(0). $$

In light of (5.137) and (5.138), we have

$$\begin{aligned} \fancyscript{K}_0=\gamma _1,\quad \fancyscript{M}_0=\frac{1}{\gamma _1}. \end{aligned}$$
(5.139)

Hence, the following identities hold true:

$$\begin{aligned} \begin{array}{rcl} \displaystyle \frac{2\gamma _1-\gamma _2}{2-\gamma _0} +\sum _{n=1}^\infty \frac{2(\gamma _2-\gamma _0\gamma _1)}{\alpha _n^2-\gamma _0(2-\gamma _0)} &{} = &{} \gamma _1, \\ \displaystyle \frac{2-\gamma _0}{2\gamma _1-\gamma _2} -\sum _{n=1}^\infty \frac{2(\gamma _2-\gamma _0\gamma _1)}{ \gamma _1^2\beta _n^2+\gamma _2(\gamma _2-2\gamma _1)} &{} = &{} \displaystyle \frac{1}{\gamma _1}. \end{array} \end{aligned}$$
(5.140)

Using Eq. (5.140), we can rewrite the kernel functions (5.131) and (5.135) as

$$\begin{aligned} \begin{array}{rcl} \displaystyle \fancyscript{K}(\tau ) &{} = &{} \displaystyle \fancyscript{K}_0 -\sum _{n=1}^\infty \frac{2(\gamma _2-\gamma _0\gamma _1)}{\alpha _n^2-\gamma _0(2-\gamma _0)} \bigl (1-e^{-\alpha _n^2\tau }\bigr ), \\ \displaystyle \fancyscript{M}(\tau ) &{} = &{} \displaystyle \fancyscript{M}_0 +\sum _{n=1}^\infty \frac{2(\gamma _2-\gamma _0\gamma _1)}{ \gamma _1^2\beta _n^2+\gamma _2(\gamma _2-2\gamma _1)} \bigl (1-e^{-\beta _n^2\tau }\bigr ), \end{array} \end{aligned}$$
(5.141)

and then introduce the normalized kernel functions

$$\begin{aligned} K(t)=\frac{1}{\fancyscript{K}_0}\fancyscript{K}\Bigl (\frac{A_{11}^\mathrm{s}k_1}{a^2}t\Bigr ),\quad M(t)=\frac{1}{\fancyscript{M}_0}\fancyscript{M}\Bigl (\frac{A_{11}^\mathrm{s}k_1}{a^2}t\Bigr ), \end{aligned}$$
(5.142)

so that, in light of (5.106) and (5.139)–(5.142), we have

$$\begin{aligned} K(t)= & {} 1-\sum _{n=1}^\infty \fancyscript{A}_n\biggl (1-\exp \Bigl (-\frac{t}{\rho _n}\Bigr )\biggr ), \end{aligned}$$
(5.143)
$$\begin{aligned} M(t)= & {} 1+\sum _{n=1}^\infty \fancyscript{B}_n\biggl (1-\exp \Bigl (-\frac{t}{\tau _n}\Bigr )\biggr ), \end{aligned}$$
(5.144)

where we have introduced the notation

$$\begin{aligned} \fancyscript{A}_n=\frac{2(\gamma _2-\gamma _0\gamma _1)}{ \gamma _1\bigl [\alpha _n^2-\gamma _0(2-\gamma _0)\bigr ]},\quad \fancyscript{B}_n=\frac{2\gamma _1(\gamma _2-\gamma _0\gamma _1)}{ \gamma _1^2\beta _n^2+\gamma _2(\gamma _2-2\gamma _1)}, \end{aligned}$$
(5.145)
$$\begin{aligned} \rho _n=\frac{a^2}{\alpha _n^2 A_{11}^\mathrm{s}k_1},\quad \tau _n=\frac{a^2}{\beta _n^2 A_{11}^\mathrm{s}k_1}. \end{aligned}$$
(5.146)

Note that, by analogy with the viscoelastic model, the functions K(t) and M(t) may be called the normalized biphasic relaxation function and the normalized biphasic creep function for unconfined compression, respectively. Note also that the sequences \(\rho _1>\rho _2>\ldots \) and \(\tau _1>\tau _2>\ldots \), which are defined by formulas (5.146), represent the discrete relaxation and retardation spectra, respectively.

In the dimensional form, Eqs. (5.130) and (5.133) can be recast as follows (see formulas (5.96), (5.106), and (5.123)):

$$\begin{aligned} F(t)=\frac{\pi a^2 E_3}{h}\int \limits _{0^-}^t K(t-t^\prime )\dot{w}(t^\prime )\,dt^\prime , \end{aligned}$$
(5.147)
$$\begin{aligned} w(t)=\frac{h}{\pi a^2 E_3}\int \limits _{0^-}^t M(t-t^\prime )\dot{F}(t^\prime )\,dt^\prime . \end{aligned}$$
(5.148)

Here we have introduced the notation

$$E_3=A_{11}^\mathrm{s}\fancyscript{K}_0=\frac{A_{11}^\mathrm{s}}{\fancyscript{M}_0},$$

which according to Eqs. (5.110), (5.124), (5.127), and (5.139), has the form

$$\begin{aligned} E_3=\frac{1}{2}\bigl (A_{11}^\mathrm{s}+A_{12}^\mathrm{s}+2A_{33}^\mathrm{s}-4A_{13}^\mathrm{s}\bigr ). \end{aligned}$$
(5.149)

Note that the elastic constant \(E_3\) defined by formula (5.149) coincides with the out-of-plane Young’s modulus of the equivalent (for instantaneous response) transversely isotropic incompressible elastic material given by formula (5.40)\(_2\).

In terms of the technical elastic constants, formulas (5.127) yield

$$\begin{aligned} \gamma _0=\frac{1-\nu _{12}^\mathrm{s}-2\nu _{31}^\mathrm{s 2}n_1}{ 1-\nu _{31}^\mathrm{s 2}n_1}, \end{aligned}$$
(5.150)
$$\begin{aligned} \frac{\gamma _2}{\gamma _1}=\frac{2\bigl \{ \bigl [1-4\nu _{31}^\mathrm{s}(1-\nu _{12}^\mathrm{s}\nu _{31}^\mathrm{s})\bigr ]n_1 +(1-\nu _{12}^\mathrm{s})^2-\nu _{31}^\mathrm{s 2}(1-4\nu _{31}^\mathrm{s})n_1^2\bigr \}}{ \bigl (1-n_1\nu _{31}^\mathrm{s 2}\bigr )\bigl [(1-4\nu _{31}^\mathrm{s})n_1+2(1-\nu _{12}^\mathrm{s}) \bigr ]}, \end{aligned}$$
(5.151)

where we have introduced the notation

$$ n_1=\frac{E_1^\mathrm{s}}{E_3^\mathrm{s}}. $$

Finally, we note that in light of (5.150), Eq. (5.132) coincides with the corresponding equation derived in [19].

5.3.4 Biphasic Stress Relaxation in Unconfined Compression

For an imposed step displacement, i.e.,

$$w(t)=w_0\fancyscript{H}(t),$$

where \(\fancyscript{H}(t)\) is the Heaviside step function, by formula (5.147) we have

$$\begin{aligned} F(t)=\pi a^2 E_3\frac{w_0}{h}K(t), \end{aligned}$$
(5.152)

for K(t) and \(E_3\) given by (5.143) and (5.149).

Following Cohen et al. [19], we consider the ratio of the peak load intensity (\(F_\mathrm{peak}\), at \(t\rightarrow 0^+\)) to the one at equilibrium (\(F_\mathrm{eq}\), at \(t\rightarrow +\infty \)), which in light of the relations \(K(0)=1\) and \(K(+\infty )=1-\sum _{n=1}^\infty \fancyscript{A}_n\) takes the form

$$\begin{aligned} \frac{F_\mathrm{peak}}{F_\mathrm{eq}}=\frac{\gamma _1(2-\gamma _0)}{2\gamma _1-\gamma _2}. \end{aligned}$$
(5.153)

We note that in writing the above equation we have used the first identity (5.140).

In the isotropic case, the right-hand side of (5.153) reduces to \(3/[2(1+\nu _\mathrm{s})]\), which is a strictly decreasing function of Poisson’s ratio \(\nu _\mathrm{s}\) and for positive \(\nu _\mathrm{s}\) attains a maximum value of 1.5 at \(\nu _\mathrm{s}=0\), as shown by Armstrong et al. [5].

Taking into account formulas (5.110), (5.124), and (5.127), we rewrite Eq. (5.153) in the following form [19]:

$$\begin{aligned} \frac{F_\mathrm{peak}}{F_\mathrm{eq}}=\frac{2(1-\nu _{12}^\mathrm{s})+(1-4\nu _{31}^\mathrm{s})E_1^\mathrm{s}/E_3^\mathrm{s}}{ 2\bigl (1-\nu _{12}^\mathrm{s}-2\nu _{31}^\mathrm{s 2}E_1^\mathrm{s}/E_3^\mathrm{s}\bigr )}. \end{aligned}$$
(5.154)
Fig. 5.4
figure 4

The effect of \(\nu _{12}^\mathrm{s}\) and \(E_1^\mathrm{s}/E_3^\mathrm{s}\) on the peak to equilibrium ratio of the load intensity in the stress-relaxation response to a step displacement. (The values taken by \(E_1^\mathrm{s}/E_3^\mathrm{s}\) are indicated on the figure)

For the particular case of \(\nu _{31}^\mathrm{s}=0\), the maximum values of the load intensity ratio are depicted in Fig. 5.4 for different values of \(E_1^\mathrm{s}/E_3^\mathrm{s}\) and \(\nu _{12}^\mathrm{s}\in (0,1{.}0)\). Note the high values that this ratio can attain (much greater than the maximum value 1.5 for the ratio \(F_\mathrm{peak}/F_\mathrm{eq}=3/[2(1+\nu )]\) in the isotropic case for \(\nu =0\)).

For an imposed ramp displacement, i.e.,

$$ w(t)=\left\{ \begin{array}{l} V_0 t,\quad 0\le t\le t_0, \\ V_0 t_0,\quad t_0\le t, \end{array} \right. $$

when a constant strain rate \(-V_0/h\) is maintained until time \(t_0\), the general solution (3.49) yields

$$\begin{aligned} F(t)=\pi a^2 E_3\frac{V_0}{h}\biggl \{ \frac{(2\gamma _1-\gamma _2)}{\gamma _1(2-\gamma _0)}t +\sum _{n=1}^\infty \fancyscript{A}_n\rho _n\bigl (1-e^{-t/\rho _n}\bigr )\biggr \} \end{aligned}$$
(5.155)

for \(0\le t\le t_0\), and for \(t\ge t_0\) gives

$$\begin{aligned} F(t)=\pi a^2 E_3\frac{V_0}{h}\biggl \{ \frac{(2\gamma _1-\gamma _2)}{\gamma _1(2-\gamma _0)}t_0 +\sum _{n=1}^\infty \fancyscript{A}_n\rho _n\bigl ( e^{-(t-t_0)/\rho _n}-e^{-t/\rho _n} \Bigr )\bigr )\biggr \}. \end{aligned}$$
(5.156)

Note that at equilibrium (as \(t\rightarrow \infty \)), the load intensity will be

$$ F_\mathrm{eq}=\pi a^2 E_3\frac{(2\gamma _1-\gamma _2)}{\gamma _1(2-\gamma _0)}\frac{V_0 t_0}{h}, $$

which in light of (5.110), (5.124), and (5.127), reduces to

$$\begin{aligned} F_\mathrm{eq}=\pi a^2 E_3^\mathrm{s}\frac{V_0 t_0}{h}. \end{aligned}$$
(5.157)
Fig. 5.5
figure 5

The effect of \(\nu _{12}^\mathrm{s}\) on the stress-relaxation time history in response to a ramped displacement, when \(E_1^\mathrm{s}/E_3^\mathrm{s}=5\) and \(t_0/t_\mathrm{g}=1\)

The characteristic relaxation time in unconfined compression can be defined by

$$\begin{aligned} \tau _R^\prime =\frac{a^2}{\alpha _1^2 A_{11}^\mathrm{s} k_1}, \end{aligned}$$
(5.158)

where \(\alpha _1\) is the first root of the transcendental equation (5.132).

For the special case in which \(\nu _{31}^\mathrm{s}=0\), \(E_1^\mathrm{s}/E_3^\mathrm{s}=5\), and the ratio of the ramp time to the gel diffusion time, \(t_\mathrm{g}=a^2/(A_{11}^\mathrm{s}k_1)\) (cf. formula (5.158)), is \(t_0/t_\mathrm{g}=1\), the effect of \(\nu _{12}^\mathrm{s}\) on the dimensionless stress-relaxation time history \(f(t)/[E_3^\mathrm{s}\dot{\varepsilon }_0 t_0]\), where \(f(t)=F(t)/[\pi a^2]\) and \(\dot{\varepsilon }_0=V_0/h\), is shown in Fig. 5.5. Also, according to Cohen et al. [19], Fig. 5.6 illustrates the effect of the ratio \(t_0/t_\mathrm{g}\) on the stress relaxation time history for the special case of \(\nu _{12}^\mathrm{s}=0{.}3\), \(\nu _{31}^\mathrm{s}=0\), and \(E_1^\mathrm{s}/E_3^\mathrm{s}=5\).

Fig. 5.6
figure 6

The effect of \(t_0/t_\mathrm{g}\) on the stress-relaxation time history in response to a ramped displacement, when \(\nu _{12}^\mathrm{s}=0{.}3\), \(\nu _{31}^\mathrm{s}=0{.}0\), and \(E_1^\mathrm{s}/E_3^\mathrm{s}=5\). (The values taken by \(t_0/t_\mathrm{g}\) are indicated on the figure)

We also note an alternative representation for F(t) in the loading stage, similar to the one obtained in [19], which follows from the direct inverse Laplace transform of Eq. (5.128) for \(\tilde{\varepsilon }(s)=-(V_0t_\mathrm{g}/h)s^{-2}\). Then, for \(0\le t\le t_0\) we have

$$ F(t)=\pi a^2 E_3\frac{V_0}{h}\biggl \{ \frac{(2\gamma _1-\gamma _2)}{\gamma _1(2-\gamma _0)}t +\frac{\gamma _2-\gamma _0\gamma _1}{4\gamma _1(2-\gamma _0)^2} -\sum _{n=1}^\infty \fancyscript{A}_n\rho _n e^{-t/\rho _n}\biggr \}. $$

Equations (5.155) and (5.156) can be used in determining the material properties from the unconfined stress relaxation experiment, by fitting the theoretical solution on to the experimental curve for the total normal stress [19, 32].

5.3.5 Biphasic Creep in Unconfined Compression

For an imposed step loading, \(F(t)=F_0\fancyscript{H}(t)\), we obtain from formula (5.148) that

$$\begin{aligned} w(t)=\frac{hF_0}{\pi a^2 E_3}M(t), \end{aligned}$$
(5.159)

where M(t) and \(E_3\) are given by (5.144) and (5.149).

Taking into account Eqs. (5.140)\(_2\) and (5.145)\(_2\), we find

$$\begin{aligned} M(t)=\frac{\gamma _2+2\gamma _1^2-\gamma _1(\gamma _0+\gamma _2)}{ \gamma _1(2\gamma _1-\gamma _2)} -\sum _{n=1}^\infty \fancyscript{B}_n\exp \Bigl (-\frac{t}{\tau _n}\Bigr ), \end{aligned}$$
(5.160)

where \(\gamma _0\), \(\gamma _1\), \(\gamma _2\) and \(\fancyscript{B}_n\), \(\tau _n\) are given by (5.127) and (5.145)\(_2\), (5.146)\(_2\), respectively.

The characteristic retardation time in unconfined compression can be defined by

$$\begin{aligned} \tau _R^{\prime \prime }=\frac{a^2}{\beta _1^2 A_{11}^\mathrm{s} k_1}, \nonumber \end{aligned}$$

where \(\beta _1\) is the first root of the transcendental equation (5.136).

In the isotropic case, formulas (5.159) and (5.160) reduce to the solution originally obtained by Armstrong et al. [5].

Equations (5.159) and (5.160) are used in determining the biphasic materials properties from the unconfined compression creep experiment by fitting the theoretical solution with the experimental curve for the nominal strain [5, 47].

Observe that the deformation response is characterized by an instantaneous jump

$$ w(0^+)=\frac{hF_0}{\pi a^2 E_3} $$

followed by a decreasing slope until equilibrium is reached, where

$$ w(+\infty )=\frac{hF_0}{\pi a^2 E_3^\mathrm{s}}. $$

Note that the identity

$$ E_3=\frac{\gamma _1(2\gamma _1-\gamma _2)}{\gamma _2+2\gamma _1^2-\gamma _1(\gamma _0+\gamma _2)}E_3^\mathrm{s} $$

can be directly proved using the expressions (5.110), (5.124), (5.127), and (5.149).

In the isotropic case, when the deformation behavior of the solid phase is described by two elastic constants \(E_\mathrm{s}\) and \(\nu _\mathrm{s}\), the equilibrium response of biphasic material in unconfined and confined compression allows us to evaluate the Young’s modulus \(E_\mathrm{s}\) and the aggregate modulus \(H_A\). Thus, taking into account that

$$ H_A=\frac{1-\nu _\mathrm{s}}{(1+\nu _\mathrm{s})(1-2\nu _\mathrm{s})}E_\mathrm{s}, $$

the following formula for Poisson’s ratio can be derived [43]:

$$ \nu _\mathrm{s}=\frac{1}{4}\biggl ( \frac{E_\mathrm{s}}{H_A}-1+\sqrt{\Bigl (\frac{E_\mathrm{s}}{H_A}-1\Bigr ) \Bigl (\frac{E_\mathrm{s}}{H_A}-9\Bigr )}\biggr ). $$

Finally, note that platen/specimen friction influences the mechanical response of articular cartilage in unconfined compression [5, 71, 77]. In particular, the peak reaction forces in unconfined stress-relaxation experiments exceed the corresponding maximum values predicted analytically. Consequently, the frictional effect becomes more significant for specimens with large aspect (diameter/height) ratios.

5.3.6 Cyclic Compressive Loading in Unconfined Compression

It is well known that the long-term creep and relaxation tests, typically used for determining viscoelastic and biphasic/poroelastic properties, are not appropriate for rapidly assessing the dynamic biomechanical properties of biological tissues like articular cartilage. For in vivo measurements of tissue viability, Appleyard et al. [1] developed a dynamic indentation instrument, which employs a single-frequency (20 Hz) sinusoidal oscillatory waveform superimposed on a carrier load.

Following Li et al. [51], we assume that a biphasic tissue sample is subjected to a cyclic displacement input

$$\begin{aligned} w(t)=[w_0(1-\cos \omega t)+w_1]\fancyscript{H}(t), \end{aligned}$$
(5.161)

where \(w_1/h\) is the prestrain resulting from the initial deformation applied to the sample to create the desired preload, \(w_0\) is the displacement amplitude, i.e., \(w_0/h\) is equal to one-half the peak-to-peak cyclic strain input superimposed on the prestrain, and \(\omega =2\pi f\) is the angular frequency, f being the loading frequency.

Differentiating (5.161), we obtain

$$\begin{aligned} \frac{dw(t)}{dt}=\fancyscript{H}(t)w_0\omega \sin \omega t+w_1\delta (t). \end{aligned}$$
(5.162)

Substituting expression (5.162) into Eq. (5.147), we arrive, after some algebra, at the following stress output:

$$\begin{aligned} \frac{F(t)}{\pi a^2}= & {} \frac{E_3}{h}\Biggl \{ w_1 K(t) +w_0\biggl ( \frac{2\gamma _1-\gamma _2}{\gamma _1(2-\gamma _0)} +\sum _{n=1}^\infty \frac{\rho _n^2\omega ^2 \fancyscript{A}_n}{1+\rho _n^2\omega ^2}\exp \Bigl (-\frac{t}{\rho _n}\Bigr ) \biggr ) \nonumber \\ {}{} & {} {}-w_0\bigl [ K_1(\omega )\cos \omega t-K_2(\omega )\sin \omega t \bigr ]\Biggr \}. \end{aligned}$$
(5.163)

Here we have introduced the notation

$$\begin{aligned} K_1(\omega )=1-\sum _{n=1}^\infty \frac{\fancyscript{A}_n}{1+\rho _n^2\omega ^2}, \end{aligned}$$
(5.164)
$$\begin{aligned} K_2(\omega )=\sum _{n=1}^\infty \frac{\rho _n\omega \fancyscript{A}_n}{1+\rho _n^2\omega ^2}. \end{aligned}$$
(5.165)

To assign a physical meaning to the introduced functions \(K_1(\omega )\) and \(K_2(\omega )\), let us compare the oscillating part of the input strain, that is \(-(w_0/h)\cos \omega t\), with the corresponding oscillating part of the compressive stress, which is equal to \(-E_3(w_0/h)\bigl [K_1(\omega )\cos \omega t-K_2(\omega )\sin \omega t\bigr ]\). By analogy with the viscoelastic model, we obtain that \(K_1(\omega )\) and \(K_2(\omega )\) represent, respectively, the apparent relative storage and loss moduli. Correspondingly, the apparent loss angle, \(\delta (\omega )\), can be introduced by the formula

$$\begin{aligned} \tan \delta (\omega )=\frac{K_2(\omega )}{K_1(\omega )}. \end{aligned}$$
(5.166)

The apparent loss angle \(\delta (\omega )\) describes the phase difference between the displacement input and force output.

In the case of load-controlled compression, following Suh et al. [73], we will assume that the tissue sample is subjected to a cyclic compressive loading

$$\begin{aligned} F(t)=[F_0(1-\cos \omega t)+F_1]\fancyscript{H}(t), \end{aligned}$$
(5.167)

where \(F_0\) is the force amplitude, and \(F_1\) is the initial preload.

After substitution of the expression (5.167) into Eq. (5.148), we finally obtain the following resulting strain output:

$$\begin{aligned} \frac{w(t)}{h}= & {} \frac{1}{\pi a^2 E_3}\Biggl \{ F_1 M(t) +F_0\biggl ( \frac{\gamma _1(2-\gamma _0)}{2\gamma _1-\gamma _2} -\sum _{n=1}^\infty \frac{\tau _n^2\omega ^2 \fancyscript{B}_n}{1+\tau _n^2\omega ^2}\exp \Bigl (-\frac{t}{\tau _n}\Bigr ) \biggr ) \nonumber \\ {}{} & {} {}-F_0\bigl [ M_1(\omega )\cos \omega t+M_2(\omega )\sin \omega t \bigr ]\Biggr \}, \end{aligned}$$
(5.168)

where we have introduced the notation

$$\begin{aligned} M_1(\omega )=1+\sum _{n=1}^\infty \frac{\fancyscript{B}_n}{1+\tau _n^2\omega ^2}, \end{aligned}$$
(5.169)
$$\begin{aligned} M_2(\omega )=\sum _{n=1}^\infty \frac{\tau _n\omega \fancyscript{B}_n}{1+\tau _n^2\omega ^2}. \end{aligned}$$
(5.170)

Note that \(M_1(\omega )\) and \(M_2(\omega )\) have physical meanings of the apparent relative storage and loss compliances, respectively.

5.3.7 Displacement-Controlled Unconfined Compression Test

Following Argatov [3], we consider an unconfined compression test with the upper plate displacement specified according to the equation

$$\begin{aligned} w(t)=w_0\sin \omega t, \quad t\in (0,\pi /\omega ). \end{aligned}$$
(5.171)

The maximum displacement, \(w_0\), will be achieved at the time moment \(t_m=\pi /(2\omega )\). The moment of time \(t=\tilde{t}_M^\prime \), when the contact force F(t) vanishes, determines the duration of the contact. The contact force itself can be evaluated according to Eqs. (5.147) and (5.171) as follows:

$$\begin{aligned} F(t)=\frac{\pi a^2 E_3}{h}w_0\omega \int \limits _0^t K(t-\tau )\cos \omega \tau \,d\tau . \end{aligned}$$
(5.172)

According to Eq. (5.172), the contact force at the moment of maximum displacement is given by

$$\begin{aligned} F(t_m)=\frac{\pi a^2 E_3}{h}w_0\tilde{K}_1(\omega ), \end{aligned}$$
(5.173)

where we have introduced the notation

$$\begin{aligned} \tilde{K}_1(\omega )=\omega \int \limits _0^{\pi /(2\omega )} K(\tau )\sin \omega \tau \,d\tau . \end{aligned}$$
(5.174)

By analogy with the viscoelastic case [2, 4], the quantity \(\tilde{K}_1(\omega )\) will be called the reduced incomplete apparent storage modulus.

Substituting the expression (5.143) into the right-hand side of Eq. (5.174), we obtain

$$\begin{aligned} \tilde{K}_1(\omega )=1-\sum _{n=1}^\infty \frac{\fancyscript{A}_n}{1+\rho _n^2\omega ^2} -\sum _{n=1}^\infty \frac{\rho _n\omega \fancyscript{A}_n}{1+\rho _n^2\omega ^2} \exp \Bigl (-\frac{\pi }{2\omega \rho _n}\Bigr ). \end{aligned}$$
(5.175)

Now, taking into consideration Eqs. (5.164) and (5.175), we may conclude that the difference between the reduced apparent storage modulus \(K_1(\omega )\) and the reduced incomplete apparent storage modulus \(\tilde{K}_1(\omega )\) is relatively small at low frequencies. To be more precise, the difference \(K_1(\omega )-\tilde{K}_1(\omega )\) is positive and of order \(O\bigl (\omega \rho _1\exp (-\pi /(2\omega \rho _1))\bigr )\) as \(\omega \rightarrow 0\), where \(\rho _1\) is the maximum relaxation time.

In the high frequency limit, the upper limit of the integral (5.174) tends to zero as \(\omega \) increases. Thus, the behavior of \(\tilde{K}_1(\omega )\) as \(\omega \rightarrow +\infty \) will depend on the behavior of K(t) as \(t\rightarrow 0\). According to (5.137), as \(\omega \rightarrow \infty \), we have

$$\begin{aligned} \tilde{K}_1(\omega )=1-\frac{2s_{1/2}}{\sqrt{\pi }} \frac{(\gamma _2-\gamma _0\gamma _1)}{\gamma _1} \sqrt{\frac{A_{11}^\mathrm{s} k_1}{a^2}} \frac{1}{\sqrt{\omega }}+O(\omega ^{-1}), \end{aligned}$$
(5.176)

where \(s_{1/2}=\int _0^{\pi /2} \sqrt{x}\sin x\,dx\).

On the other hand, due to the asymptotic formula (5.176), the following limit relation holds true: \(\lim K_1(\omega )=1\) as \(\omega \rightarrow \infty \). Thus, we conclude that \(\tilde{K}_1(\omega )\simeq K_1(\omega )\) for \(\omega \rightarrow \infty \) as well as \(\tilde{K}_1(\omega )\simeq K_1(\omega )\) for \(\omega \rightarrow 0\). In other words, the reduced incomplete apparent storage modulus \(\tilde{K}_1(\omega )\) obeys both asymptotic behaviors of the reduced apparent storage modulus \(K_1(\omega )\).

5.3.8 Force-Controlled Unconfined Compression Test

Consider now an unconfined compression test where the external force is as specified by the equation

$$\begin{aligned} F(t)=F_0\sin \omega t, \quad t\in (0,\pi /\omega ). \end{aligned}$$
(5.177)

The maximum contact force, \(F_0\), will be achieved at the time moment \(t_M=\pi /(2\omega )\). The moment of time \(t_M^\prime =\pi /\omega \), when the contact force F(t) vanishes, determines the duration of the compression test. According to Eqs. (5.148) and (5.177), the upper plate displacement can be evaluated as follows:

$$\begin{aligned} w(t)=\frac{h}{\pi a^2 E_3}F_0\omega \int \limits _0^t M(t-\tau )\cos \omega \tau \,d\tau . \end{aligned}$$
(5.178)

Due to Eq. (5.178), the displacement at the moment of maximum contact force is given by

$$\begin{aligned} w(t_M)=\frac{h}{\pi a^2 E_3}F_0 \tilde{M}_1(\omega ), \end{aligned}$$
(5.179)

where we have introduced the notation

$$\begin{aligned} \tilde{M}_1(\omega )=\omega \int \limits _0^{\pi /(2\omega )} M(\tau )\sin \omega \tau \,d\tau . \end{aligned}$$
(5.180)

By analogy with the viscoelastic case [2, 4], the quantity \(\tilde{M}_1(\omega )\) will be called the reduced incomplete apparent storage compliance.

Substituting the expression (5.138) into the right-hand side of Eq. (5.180), we obtain

$$\begin{aligned} \tilde{M}_1(\omega )=1+ \sum _{n=1}^\infty \frac{\fancyscript{B}_n}{1+\tau _n^2\omega ^2} +\sum _{n=1}^\infty \frac{\tau _n\omega \fancyscript{B}_n}{1+\tau _n^2\omega ^2} \exp \Bigl (-\frac{\pi }{2\omega \tau _n}\Bigr ). \end{aligned}$$
(5.181)

Using the same method as for \(\tilde{K}_1(\omega )\), it can be shown that the incomplete apparent storage compliance \(\tilde{M}_1(\omega )\) obeys both asymptotic behaviors of the apparent storage compliance \(M_1(\omega )\), that is \(\tilde{M}_1(\omega )\simeq M_1(\omega )\) for \(\omega \rightarrow 0\) along with \(\tilde{M}_1(\omega )\simeq M_1(\omega )\) for \(\omega \rightarrow \infty \).

5.4 Biphasic Poroviscoelastic (BPVE) Model

In this section, the biphasic poroviscoelastic model is briefly outlined. The confined and unconfined compression tests as well as the torsion test are considered.

5.4.1 Linear Biphasic Poroviscoelastic Theory

The biphasic theory was extended by Mak [54] to account for the inherent viscoelasticity of the solid matrix, by replacing the effective stresses (5.13) in the constitutive equation (5.7) with viscoelastic constitutive relations in the hereditary integral form. Therefore, for a transversely isotropic material, we have

$$\begin{aligned} \begin{array}{rclrcl} \sigma _{11}^\mathrm{VE} &{} = &{} B_{11}^\mathrm{s}\!*\varepsilon _{11}+B_{12}^\mathrm{s}\!*\varepsilon _{22} +B_{13}^\mathrm{s}\!*\varepsilon _{33}, &{} \sigma _{23}^\mathrm{VE} &{} = &{} 2B_{44}^\mathrm{s}\!*\varepsilon _{23}, \\ \sigma _{22}^\mathrm{VE} &{} = &{} B_{12}^\mathrm{s}\!*\varepsilon _{11}+B_{11}^\mathrm{s}\!*\varepsilon _{22} +B_{13}^\mathrm{s}\!*\varepsilon _{33}, &{} \sigma _{13}^\mathrm{VE} &{} = &{} 2B_{44}^\mathrm{s}\!*\varepsilon _{13}, \\ \sigma _{33}^\mathrm{VE} &{} = &{} B_{13}^\mathrm{s}\!*\varepsilon _{11}+B_{13}^\mathrm{s}\!*\varepsilon _{22} +B_{33}^\mathrm{s}\!*\varepsilon _{33}, &{} \sigma _{12}^\mathrm{VE} &{} = &{} 2B_{66}^\mathrm{s}\!*\varepsilon _{12}, \end{array} \end{aligned}$$
(5.182)

where the \(*\) sign denotes the Stieltjes integral, i.e.,

$$\begin{aligned} B_{kl}^\mathrm{s}\!*\varepsilon _{ij}=\int \limits _{-\infty }^t B_{kl}^\mathrm{s}(t-\tau )\,d\varepsilon _{ij}(\tau ). \end{aligned}$$
(5.183)

For simplicity’s sake, following [68], we assume that the deformation behavior of the solid phase is governed by a single reduced stress-relaxation function, \(\psi (t)\) , which is usually assumed to be in the following form proposed by Fung [28]:

$$\begin{aligned} \psi (t)=1+\int \limits _0^\infty S(\tau )e^{-t/\tau }d\tau , \end{aligned}$$
(5.184)

where

$$\begin{aligned} S(\tau )=\left\{ \begin{array}{l} \displaystyle \frac{c}{\tau },\quad \tau _1\le \tau \le \tau _2, \\ \displaystyle 0,\quad \tau < \tau _1,\quad \tau > \tau _2. \end{array} \right. \end{aligned}$$
(5.185)

We note (see, e.g., [38, 49]) that the relaxation spectrum (5.185 ) with constant amplitude over a range of frequencies \(\tau \in (\tau _1,\tau _2)\), which was originally introduced by Neubert [61], has least sensitivity to strain rate, which has been believed to be the case for some biological tissues [28].

The function \(S(\tau )\) defines a continuous relaxation spectrum, where the parameter c is a proportionality constant for the amplitude of the spectrum \(S(\tau )\). The width of the spectrum is defined by the time constants \(\tau _1\) and \(\tau _2\), which govern the fast and slow relaxation phenomena, respectively.

Note that at initial times after loading and at equilibrium, respectively, we have

$$\begin{aligned} \psi (0)=1+c\ln \frac{\tau _2}{\tau _1},\quad \psi (+\infty )=1. \end{aligned}$$
(5.186)

Therefore, under the above assumptions, the relaxation functions \(B_{kl}^\mathrm{s}(t)\) can be represented as

$$\begin{aligned} B_{kl}^\mathrm{s}(t)=B_{kl}^\mathrm{s\infty }\psi (t), \end{aligned}$$
(5.187)

where \(B_{kl}^\mathrm{s\infty }=B_{kl}^\mathrm{s}(+\infty )\) are the equilibrium elastic moduli, while the instantaneous elastic moduli, \(B_{kl}^\mathrm{s0}=B_{kl}^\mathrm{s}(0)\) , are given by

$$\begin{aligned} B_{kl}^\mathrm{s0}=B_{kl}^\mathrm{s\infty }\Bigl (1+c\ln \frac{\tau _2}{\tau _1}\Bigr ). \end{aligned}$$
(5.188)

The viscoelastic parameters c, \(\tau _1\), \(\tau _2\) are material properties of the solid skeleton that need to be determined from experimental data. The confined and unconfined compression problems for a BPVE material were considered in [34, 36, 54, 68].

Thus, the constitutive equations for the solid matrix in the biphasic poroviscoelastic (BPVE) theory have the form

$$\begin{aligned} {\varvec{\sigma }^\mathrm{s}=-\phi _\mathrm{s}p\mathbf{I}+\varvec{\sigma }^\mathrm{VE},} \end{aligned}$$
(5.189)

where p is the pressure of the fluid phase, \(\mathbf{I}\) is the identity tensor, and the components of the stress tensor \({\varvec{\sigma }^\mathrm{VE}}\) are given by (5.182).

The reduced stress-relaxation function (5.184) and (5.185) can be represented by

$$\begin{aligned} \psi (t)=1+c\Bigl [E_1\Bigl (\frac{t}{\tau _2}\Bigr )-E_1\Bigl (\frac{t}{\tau _1}\Bigr )\Bigr ], \end{aligned}$$
(5.190)

where \(E_1(x)\) is the exponential integral function, i.e.,

$$ E_1(x)=\int \limits _0^1 \frac{1}{\xi }\exp \Bigl (-\frac{x}{\xi }\Bigr )d\xi . $$

Observe [68] that, if the width of the relaxation spectrum reduces to zero, i.e., \(\tau _1\rightarrow \tau _2\), the reduced relaxation function (5.190) becomes \(\psi (t)=1\). It is also readily seen that the intrinsic viscoelastic effect diminishes as \(c\rightarrow 0\). Thus, for the limiting cases \(c\rightarrow 0\) or \(\tau _1\rightarrow \tau _2\), and the BPVE theory reduces to the linear biphasic theory.

Note also that for the sake of numerical efficiency the discrete form of the relaxation function

$$ \psi (t)=1+\sum _{i}C_i\exp \Bigl (-\frac{t}{\tau _i}\Bigr ) $$

has also been used for articular cartilage [21, 72]. Multiple discrete spectrums (different sets of \(C_i\) and \(\tau _i\)) can be used to fit the experimental data for the short-term, mid-term and long-term responses [49].

We finally note [57] that the inclusion of intrinsic matrix viscoelastic properties for the solid matrix in the biphasic theory [54] improved the prediction in the unconfined compression case [68] as well as the material property determination [36, 66].

5.4.2 Confined Compression of a Biphasic Poroviscoelastic Material

Under the idealized conditions of the confined compression experiment described in Sect. 5.2.1, the displacement of the solid matrix and the fluid movement occur only in the axial direction, and the governing differential equation (5.63) of the biphasic model should be replaced with the following [54, 68]:

$$\begin{aligned} \int \limits _{-\infty }^t B_{33}^\mathrm{s}(t-\tau )\frac{\partial }{\partial \tau } \Bigl (\frac{\partial ^2 u_z}{\partial z^2}\Bigr )d\tau = \frac{1}{k_3}\frac{\partial u_z}{\partial t}. \end{aligned}$$
(5.191)

Here, \(u_z(t,z)\) is the axial displacement of the solid phase, \(k_3\) is the axial permeability, and \(B_{33}^\mathrm{s}(t)\) is the axial aggregate relaxation modulus of the solid phase.

For creep, the initial and boundary conditions are

$$\begin{aligned} u_z(t,z)=0,\quad -\infty <t<0, \end{aligned}$$
(5.192)
$$\begin{aligned} \int \limits _{-\infty }^t B_{33}^\mathrm{s}(t-\tau )\frac{\partial }{\partial \tau } \Bigl (\frac{\partial u_z}{\partial z}\Bigr )d\tau \Biggr \vert _{z=0}=-\sigma (t)\fancyscript{H}(t), \end{aligned}$$
(5.193)
$$\begin{aligned} u_z\bigr \vert _{z=h}=0, \end{aligned}$$
(5.194)

where \(\sigma (t)\) is the applied compressive stress (see Eq. (5.63)).

Following [54, 68], we put

$$\begin{aligned} B_{33}^\mathrm{s}(t)=H_A \psi (t), \end{aligned}$$
(5.195)

where \(H_A=B_{33}^\mathrm{s\infty }\) is the equilibrium aggregate elastic modulus, and \(\psi (t)\) is the reduced stress-relaxation function given by (5.184) and (5.185).

To solve the problem (5.191)–(5.194), we introduce dimensionless quantities

$$\begin{aligned} \zeta =\frac{z}{h},\quad \tau =\alpha _1 t,\quad \tau _i^\prime =\alpha _1\tau _i,\quad i=1,2,\quad \alpha _1=\frac{H_A k_3}{h^2} \end{aligned}$$
(5.196)

and apply the Laplace transform to Eqs. (5.191), (5.193), and (5.194) with respect to the dimensionless time variable \(\tau \). In this way, remembering that the Laplace transform of the function \(\psi (\tau /\alpha _1)\) (see Eqs. (5.184) and (5.185)) is given by

$$\begin{aligned} \tilde{\psi }(s)=\frac{1}{s}\biggl ( 1+c\ln \frac{1+s\tau _2^\prime }{1+s\tau _1^\prime }\biggr ), \end{aligned}$$
(5.197)

we arrive at the problem

$$\begin{aligned} \begin{array}{c} \displaystyle \frac{\partial ^2 \tilde{u}_z}{\partial \zeta ^2}-f(s)\tilde{u}_z=0, \quad \zeta \in (0,1), \\ \displaystyle \frac{\partial \tilde{u}_z}{\partial \zeta }\Bigr \vert _{\zeta =0}= -\frac{h}{H_A}\frac{f(s)\tilde{\sigma }(s)}{s},\quad \tilde{u}_z\bigr \vert _{\zeta =1}=0, \end{array} \end{aligned}$$
(5.198)

where we have introduced the notation

$$\begin{aligned} f(s)=\frac{1}{\tilde{\psi }(s)}. \end{aligned}$$
(5.199)

From (5.198), we readily obtain

$$\begin{aligned} \tilde{u}_z=\frac{h\tilde{\sigma }(s)\sqrt{f(s)}}{H_A s} \frac{\sinh \bigl [\sqrt{f(s)}(1-\zeta )\bigr ]}{ \cosh \sqrt{f(s)}}. \end{aligned}$$
(5.200)

In the case where a constant external load, \(F_0\), is applied instantaneously, we have \(\sigma (t)=\sigma _0\fancyscript{H}(t)\), where \(\sigma _0=F_0/A\) with A being the sample cross-sectional area, and \(\tilde{\sigma }(s)=\sigma _0/s\), so that formula (5.200) reduces to the fallowing [54, 68]:

$$ \tilde{u}_z=\frac{h\sigma _0}{H_A} \frac{\sqrt{f(s)}\sinh \bigl [\sqrt{f(s)}(1-\zeta )\bigr ]}{ s^2\cosh \sqrt{f(s)}}. $$

An asymptotic approximation of the surface displacement \(u_z\vert _{z=0}\) for small times after loading can be obtained by evaluating the inverse Laplace transform as \(s\rightarrow \infty \), when \(f(s)\sim s/\alpha _2\), for a constant \(\alpha _2\). Taking into account (5.197), (5.199) and the relation \(\tau _2^\prime /\tau _1^\prime =\tau _2/\tau _1\) (see Eq. (5.196)\(_3\)), we get

$$ \alpha _2=1+c\ln \frac{\tau _2}{\tau _1}. $$

Thus, the short-time asymptotic approximation for the nominal sample-average strain is given by the following formula [54, 68]:

$$ \frac{u_z(t,0)}{h}\simeq \frac{2\sigma _0}{H_A}\sqrt{\frac{\alpha _1}{\pi \alpha _2}t}. $$

Note that \(\alpha _2=\psi (0)\) (see Eqs. (5.184) and (5.185)). This dimensionless parameter characterizes the intrinsic solid matrix viscoelastic effects. Observe also [68] that larger values of \(\alpha _2\), reflecting increased effects of matrix viscoelasticity, or lower values of \(\alpha _1\), and reflecting a low value of the solid matrix permeability, will have the same effect in reducing the early creep response.

5.4.3 Unconfined Compression of a BPVE Material

In the framework of the BPVE theory, the unconfined compression problem (considered previously in Sect. 5.3.1) differs in essence via the constitutive equations (5.98), which now take the form

$$\begin{aligned} \begin{array}{rcl} \sigma _{rr}^\mathrm{VE} &{} = &{} \displaystyle B_{11}^\mathrm{s}\!*\frac{\partial u_r}{\partial r} +B_{12}^\mathrm{s}\!*\frac{u_r}{r} +B_{13}^\mathrm{s}\!*\varepsilon , \\ \sigma _{\theta \theta }^\mathrm{VE} &{} = &{} \displaystyle B_{12}^\mathrm{s}\!*\frac{\partial u_r}{\partial r}+B_{11}^\mathrm{s}\!*\frac{u_r}{r} +B_{13}^\mathrm{s}\!*\varepsilon , \\ \sigma _{zz}^\mathrm{VE} &{} = &{} \displaystyle B_{13}^\mathrm{s}\!*\frac{\partial u_r}{\partial r}+B_{13}^\mathrm{s}\!*\frac{u_r}{r} +B_{33}^\mathrm{s}\!*\varepsilon . \end{array} \end{aligned}$$
(5.201)

The equilibrium equation of the solid matrix (5.99) now has the form

$$ -\frac{\partial p}{\partial r}+B_{11}^\mathrm{s}\!*\biggl ( \frac{\partial ^2 u_r}{\partial r^2}+\frac{1}{r}\frac{\partial u_r}{\partial r}-\frac{u_r}{r^2}\biggr )=0. $$

We recall that the boundary conditions (5.91)–(5.93) are formulated in terms of the components of the total stress tensor

$$\begin{aligned} {\varvec{\sigma }=-p\mathbf{I}+\varvec{\sigma }^\mathrm{VE}.} \end{aligned}$$
(5.202)

To solve this problem, let us first introduce the dimensionless variables

$$ \rho =\frac{r}{a},\quad U=\frac{u_r}{a}, $$

where we have refrained from using the variables t and p in the non-dimensionalization. Secondly, we apply the Laplace transform with respect to the time variable t, denoting by a tilde the transformed quantities. We then introduce auxiliary notation

$$\begin{aligned} \tilde{P}=\frac{\tilde{p}}{\bar{B}_{11}^\mathrm{s}(s)},\quad \alpha _{12}=\frac{\bar{B}_{12}^\mathrm{s}(s)}{\bar{B}_{11}^\mathrm{s}(s)},\quad \alpha _{13}=\frac{\bar{B}_{13}^\mathrm{s}(s)}{\bar{B}_{11}^\mathrm{s}(s)},\quad \alpha _{33}=\frac{\bar{B}_{33}^\mathrm{s}(s)}{\bar{B}_{11}^\mathrm{s}(s)}, \end{aligned}$$
(5.203)
$$\begin{aligned} f(s)=\frac{a^2 s}{k_1\bar{B}_{11}^\mathrm{s}(s)},\quad \bar{B}_{kl}^\mathrm{s}(s)=s\tilde{B}_{kl}^\mathrm{s}(s). \end{aligned}$$
(5.204)

In this case, the biphasic unconfined compression problem (5.111)–(5.113) is replaced with the following:

$$ \frac{\partial \tilde{P}}{\partial \rho }=f(s)\Bigl ( \tilde{U}+\frac{\tilde{\varepsilon }}{2}\rho \Bigr ), $$
$$ \frac{\partial ^2 \tilde{U}}{\partial \rho ^2}+\frac{1}{\rho }\frac{\partial \tilde{U}}{\partial \rho }-\frac{\tilde{U}}{\rho ^2} =f(s)\Bigl ( \tilde{U}+\frac{\tilde{\varepsilon }}{2}\rho \Bigr ), $$
$$ \tilde{P}\bigr \vert _{\rho =1}=0,\quad \frac{\partial \tilde{U}}{\partial \rho }+\alpha _{12}(s)\frac{\tilde{U}}{\rho } +\alpha _{13}(s)\tilde{\varepsilon }\,\biggr \vert _{\rho =1}=0. $$

In the same way as was done in Sect. 5.3.2, we find

$$\begin{aligned} \tilde{U}=-\frac{\tilde{\varepsilon }(s)}{2}\rho \left( \displaystyle 1-\frac{\displaystyle (1+\alpha _{12}(s)-2\alpha _{13}(s))\frac{I_1(\sqrt{f(s)}\rho )}{\sqrt{f(s)}\rho }}{\displaystyle I_0(\sqrt{f(s)})-(1-\alpha _{12}(s))\frac{I_1(\sqrt{f(s)})}{\sqrt{f(s)}}}\right) , \end{aligned}$$
(5.205)
$$\begin{aligned} \tilde{p}=\frac{\displaystyle (1+\alpha _{12}(s)-2\alpha _{13}(s))\tilde{\varepsilon }(s) \bigl [I_0(\sqrt{f(s)}\rho )-I_0(\sqrt{f(s)})\bigr ]}{\displaystyle 2\biggl ( I_0(\sqrt{f(s)})-(1-\alpha _{12}(s))\frac{I_1(\sqrt{f(s)})}{\sqrt{f(s)}}\biggr )}. \end{aligned}$$
(5.206)

We now consider the force response of the BPVE sample

$$\begin{aligned} F(t)=-2\pi \int \limits _0^a\sigma _{zz}(t,r)r\,dr, \end{aligned}$$
(5.207)

where, as a result of (5.201) and (5.202), the normal total stress is given by

$$ \sigma _{zz}(t,r)=-p+B_{13}^\mathrm{s}\!*\Bigl (\frac{\partial u_r}{\partial r}+\frac{u_r}{r}\Bigr ) +B_{33}^\mathrm{s}\!*\varepsilon . $$

Upon application of the Laplace transform to Eq. (5.207), rewritten in terms of the dimensionless variables (5.203), we obtain

$$ \frac{\tilde{F}(s)}{\pi a^2\bar{B}_{11}^\mathrm{s}(s)}=-2\int \limits _0^1 \biggl (-\tilde{P}+\alpha _{13}(s)\Bigl ( \frac{\partial \tilde{U}}{\partial \rho }+\frac{\tilde{U}}{\rho }\Bigr ) +\alpha _{33}(s)\tilde{\varepsilon }\biggr )\rho \,d\rho . $$

Taking into account formulas (5.205) and (5.206), and performing the integration in the above equation, we arrive at formula (5.126), where the coefficients \(\gamma _0\), \(\gamma _1\), and \(\gamma _2\) are evaluated by Eq. (5.127), where \(\alpha _{12}\), \(\alpha _{13}\), and \(\alpha _{33}\) are as given by (5.203).

Finally, note that the instantaneous axial modulus \(E_3\), in light of (5.149), is

$$ E_3=\frac{1}{2}\bigl (B_{11}^\mathrm{s0}+B_{12}^\mathrm{s0}+2B_{33}^\mathrm{s0}-4B_{13}^\mathrm{s0}\bigr ), $$

where \(B_{kl}^\mathrm{s0}=B_{kl}^\mathrm{s}(0)\) are the instantaneous elastic moduli of the solid matrix.

We can thus hypothesize that the instantaneous response of a transversely isotropic biphasic poroviscoelastic tissue is equivalent to that of an incompressible transversely isotropic elastic material with the material constants given by formulas (5.36), (5.39), and (5.40), where \(\nu _{12}^\mathrm{s}\), \(\nu _{31}^\mathrm{s}\), \(G_{13}^\mathrm{s}\), \(E_1^\mathrm{s}\), and \(E_3^\mathrm{s}\) are regarded as instantaneous elastic material properties of the poroviscoelastic matrix.

Observe that the deformation response of a biphasic or BPVE sample depends on how it is tested. In particular, while only the aggregate relaxation modulus \(B_{33}^\mathrm{s}(t)\) governs the behavior of a sample tested in confined compression, all four relaxation moduli \(B_{11}^\mathrm{s}(t)\), \(B_{12}^\mathrm{s}(t)\), \(B_{13}^\mathrm{s}(t)\), and \(B_{33}^\mathrm{s}(t)\) have significant influence on the deformation behavior of a BPVE sample tested in unconfined compression.

Finally, we note [34] that the failure to account for either anisotropy or viscoelasticity of the articular cartilage matrix could result in flawed predictions of the tissue deformation under general external loading.

5.4.4 Torsion of a Biphasic Poroviscoelastic Material

We consider a cylindrical sample of a BPVE material of radius a and height h subjected to a torque T (see Fig. 5.7). Let \(\theta \) denotes the angle of torsional displacement imposed on the upper surface of the sample to achieve a specified shear strain, \(\gamma \). Between these geometrical parameters, the following relation takes place:

$$\begin{aligned} \gamma =\vartheta a. \end{aligned}$$
(5.208)

Here, \(\vartheta =\theta /h\) is the so-called twist, defined as the angle of rotation per unit length along the axis of the sample.

Fig. 5.7
figure 7

Schematic of the pure torsional shear testing configuration

The components of the in-plane displacement vector are \(u_r=0\) and \(u_\theta =\vartheta rz\), and so the only nonzero component of strain is

$$\begin{aligned} \varepsilon _{z\theta }=\frac{1}{2}\vartheta r. \end{aligned}$$
(5.209)

The torque is given by

$$\begin{aligned} T=2\pi \int \limits _0^a r^2\sigma _{z\theta }\,dr, \end{aligned}$$
(5.210)

where according to (5.182) and (5.202) the total shear stress \(\sigma _{z\theta }\) is related to the shear strain component \(\varepsilon _{z\theta }\) as follows:

$$\begin{aligned} \sigma _{z\theta }= 2\int \limits _{-\infty }^t B_{44}^\mathrm{s}(t-\tau ) \frac{\partial \varepsilon _{z\theta }}{\partial \tau }(\tau )\,d\tau . \end{aligned}$$
(5.211)

Therefore, the substitution of (5.209) and (5.211) into Eq. (5.210) yields

$$\begin{aligned} T(t)= I_\mathrm{p} \int \limits _{-\infty }^t B_{44}^\mathrm{s}(t-\tau ) \frac{\partial \vartheta }{\partial \tau }(\tau )\,d\tau , \end{aligned}$$
(5.212)

where \(I_\mathrm{p}=\pi a^4/2\) is the polar moment of inertia of the sample’s cross section area.

Furthermore, introducing the so-called shear stress

$$ \tau =\frac{aT}{I_\mathrm{p}}, $$

which is defined as maximum shear stress in the sample, and taking into account (5.208), we can rewrite Eq. (5.212) in terms of the shear strain as follows:

$$\begin{aligned} \tau (t)= \int \limits _{-\infty }^t B_{44}^\mathrm{s}(t-t^\prime ) \frac{\partial \gamma }{\partial t^\prime }(t^\prime )\,d t^\prime . \end{aligned}$$
(5.213)

Now, assuming that the employed herein viscoelastic material of the solid phase is described by the Fung model [28], we represent the above equation in the form

$$\begin{aligned} \tau (t)= B_{44}^\mathrm{s\infty }\int \limits _{-\infty }^t \psi (t-t^\prime ) \frac{\partial \gamma }{\partial t^\prime }(t^\prime )\,d t^\prime , \end{aligned}$$
(5.214)

where \(B_{44}^\mathrm{s\infty }\) is the equilibrium out-of-plane shear modulus, and \(\psi (t)\) is the reduced relaxation function given by (5.184) and (5.185), i.e.,

$$\begin{aligned} \psi (t)=1+c\int \limits _{\tau _1}^{\tau _2} \frac{1}{\tau }\exp \Bigl (-\frac{t}{\tau }\Bigr )d\tau . \end{aligned}$$
(5.215)

Following Iatridis et al. [38], we now outline the main shear testing protocols.

Stress-relaxation behavior. In the stress-relaxation experiments, the sample is subjected to a ramping phase, where the strain increases linearly at a constant strain rate, followed by a relaxation phase, where the shear strain is held constant, i.e.,

$$ \gamma (t)=\left\{ \begin{array}{l}\displaystyle \frac{\gamma _0}{t_0} t,\quad 0\le t\le t_0, \\ \gamma _0,\quad t_0\le t, \end{array} \right. $$

where \(\gamma _0\) and \(t_0\) are given constants.

For the ramping phase (\(0\le t\le t_0\)), we have

$$ \tau (t)= \frac{B_{44}^\mathrm{s\infty }\gamma _0}{t_0}\bigl [ t+c_1 F_1(t,\tau _1,\tau _2)\bigr ], $$

where (with \(E_\mathrm{i}(x)\) being the exponential integral)

$$ F_1(t,\tau _1,\tau _2)=\tau _2\bigl (1-e^{-t/\tau _2}\bigr )-\tau _1\bigl (1-e^{-t/\tau _1}\bigr ) -t\bigl [E_\mathrm{i}(t/\tau _1)-E_\mathrm{i}(t/\tau _2)\bigr ], $$
$$ E_\mathrm{i}(x)=\int \limits _x^\infty \frac{\exp (-\xi )}{\xi }\,d\xi . $$

For the stress-relaxation phase, the solution is given in terms of the shifted time parameter \(\hat{t}=t-t_0\) in the following form [38]:

$$ \tau (t)= \frac{B_{44}^\mathrm{s\infty }\gamma _0}{t_0}\bigl [ t_0+c_1 G_1(t,\hat{t},\tau _1,\tau _2)\bigr ], $$

where

$$\begin{aligned} G_1(t,\hat{t},\tau _1,\tau _2)= & {} \tau _2\bigl (e^{-\hat{t}/\tau _2}-e^{-t/\tau _2}\bigr ) -\tau _1\bigl (e^{-\hat{t}/\tau _1}-e^{-t/\tau _1}\bigr ) \nonumber \\ { }{} & {} {}-\bigl \{ \hat{t}\bigl [E_\mathrm{i}(\hat{t}/\tau _2)-E_\mathrm{i}(\hat{t}/\tau _1)\bigr ] -t\bigl [E_\mathrm{i}(t/\tau _2)-E_\mathrm{i}(t/\tau _1)\bigr ]\bigr \}. \nonumber \end{aligned}$$

Creep behavior. Let us introduce the out-of-plane creep compliance in shear of the solid matrix, \(J_{44}^\mathrm{s}(t)\) , which governs the deformation response of the solid phase under application of a step out-of-plane shear stress of unit magnitude. Hence, the inverse relation for (5.213) is given by

$$\begin{aligned} \gamma (t)=\int \limits _{0^-}^t J_{44}^\mathrm{s}(t-t^\prime )\frac{\partial \tau }{\partial t^\prime }(t^\prime )\,dt^\prime . \end{aligned}$$
(5.216)

For a given relaxation modulus \(B_{44}^\mathrm{s}(t)\), the corresponding creep compliance can be evaluated via its Laplace transform

$$\begin{aligned} \tilde{J}_{44}^\mathrm{s}(s)=\frac{1}{s^2\tilde{B}_{44}^\mathrm{s}(s)}. \end{aligned}$$
(5.217)

Let us now introduce the reduced creep function, \(\varphi (t)\), by the formula

$$\begin{aligned} J_{44}^\mathrm{s}(t)=J_{44}^\mathrm{s\infty }\varphi (t), \end{aligned}$$
(5.218)

where \(J_{44}^\mathrm{s\infty }=J_{44}^\mathrm{s}(+\infty )\) is the equilibrium compliance. Since the reduced stress-relaxation function is defined by \(\psi (t)=B_{44}^\mathrm{s}(t)/B_{44}^\mathrm{s\infty }\), where \(B_{44}^\mathrm{s\infty }\) is the equilibrium modulus such that \(B_{44}^\mathrm{s\infty }=1/J_{44}^\mathrm{s\infty }\), the following normalization conditions hold:

$$ \psi (+\infty )=1,\quad \varphi (+\infty )=1. $$

The Fung reduced creep function \(\varphi (t)\) corresponding to the reduced relaxation function \(\psi (t)\) given by Eq. (5.218) can be obtained by employment of the Laplace transform and Eq. (5.217), that is

$$ \tilde{\psi }(s)\tilde{\varphi }(s)=\frac{1}{s^2}, $$

where the Laplace transform \(\tilde{\psi }(s)\) is given by formula (5.197).

According to Dortmans et al. [22], the following formula holds:

$$\begin{aligned} \varphi (t)= & {} 1-\frac{(\tau _c-\tau _2)(\tau _c-\tau _1)}{c\tau _c(\tau _2-\tau _1)} e^{-t/\tau _c} \nonumber \\ { }{} & {} {}-c\int \limits _{\tau _1}^{\tau _2}e^{-t/\tau }\frac{1}{\tau } \frac{1}{\displaystyle \Bigl (1+c\ln \frac{\tau _2-\tau }{\tau -\tau _1} \Bigr )^2+\pi ^2 c^2}\,d\tau , \end{aligned}$$
(5.219)

where we have used the notation

$$ \tau _c=\frac{\tau _2 e^{1/c}-\tau _1}{e^{1/c}-1}. $$

In the creep experiment, a constant torque, \(T_0\), is applied instantaneously, i.e., \(T(t)=T_0\fancyscript{H}(t)\) and \(\tau (t)=\tau _0\fancyscript{H}(t)\), where \(\tau _0=aT_0/I_\mathrm{p}\). Therefore, by formulas (5.216) and (5.218), we obtain

$$ \gamma (t)=\tau _0 J_{44}^\mathrm{s\infty }\varphi (t), $$

where \(\varphi (t)\) is given by (5.219).

Steady sinusoidal behavior. If the sample is subjected to a dynamic frequency sweep, the sinusoidal shear strain input is given by

$$ \gamma =\gamma _0 e^{\mathrm{i}\omega t}, $$

where \(\gamma _0\) (rad) is the peak shear strain, \(\omega \) (rad\(/\mathrm{s}\)) is the angular frequency, and \(\mathrm{i}\) is the imaginary unit equal to the square root of \(-1\).

In the absence of inertial forces, the corresponding shear stress output will be

$$ \tau =\tau _0 e^{\mathrm{i}\omega t}, $$

where \(\tau _0\) is a complex quantity.

The ratio of the amplitudes \(\tau _0\) and \(\gamma _0\) determines the reduced complex elastic shear modulus, \(\psi ^*\), such that

$$ \tau _0=B_{44}^\mathrm{s\infty }\psi ^*\gamma _0. $$

The reduced complex modulus \(\psi ^*\) is comprised of real and imaginary parts, i.e.,

$$\psi ^*=\psi _1+\mathrm{i} \psi _2,$$

which defines as the reduced storage, \(\psi _1\), and loss, \(\psi _2\), modulus, respectively.

The reduced storage and loss moduli as functions of angular frequency are evaluated as follows [38]:

$$ \psi _1(\omega )=1+\frac{c}{2}\ln \frac{1+(\omega \tau _2)^2}{1+(\omega \tau _1)^2}, $$
$$ \psi _2(\omega )=c\bigl \{\tan ^{-1}(\omega \tau _2)-\tan ^{-1}(\omega \tau _1)\bigr \}. $$

The torsional shear configuration shown in Fig. 5.7 has been used to study equilibrium and dynamic shear moduli as well as the characterization of the stress-relaxation behavior of articular cartilage. Note also [15] that other types of shear testing (for instance, single-lap test) may be important in determining the capacity of cartilage to repair.

It is also noteworthy that under a small shear strain no volumetric changes or pressure gradients occur in a cylindrical sample of BPVE material, and therefore no interstitial fluid flow is induced. Thus, shear tests under infinitesimal strain enable evaluation of the intrinsic viscoelastic, flow-independent properties of the collagen-proteoglycan solid matrix [18, 41].

Finally, we note that a finite element formulation for describing the large deformation response of biphasic materials in torsion was presented in [56], with a specific focus on the consideration of nonlinear coupling between torsional deformation and fluid pressurization in articular cartilage.