4.3 Hyperelastic materials and strain energy

The deformation work per unit time or stress power for short, is given in in the previous section by equation (4.71), where $ T_{ij} $ and $ D_{ij} $, represent the stress tensor and is the rate of deformation tensor, respectively. The stress power per unit volume $ \omega $, may then naturally be introduced as: $$ \begin{equation} \tag{4.73} \omega = T_{ij} D_{ij} \end{equation} $$ which readily allows for the introduction of the stress work per unit volume $ W $ as: $$ \begin{equation} w = \int_{t_0}^t \omega \, dt %= \int_{E_0}^E T_{ij} \, dE_{ij} \tag{4.74} \end{equation} $$

For small deformations we have the relation: $$ \begin{equation} \tag{4.75} \boldsymbol{D} = \dot{\boldsymbol{E}} \end{equation} $$ and then the stress work per unit volume may be computed as: $$ \begin{equation} \tag{4.76} w = \int_{t_0}^t \omega \, dt = \int_{t_0}^t T_{ij} \, \dot{E}_{ij} \, dt = \int_{\boldsymbol{E}_0}^{\boldsymbol{E}} T_{ij} \, dE_{ij} % = \int_{E_0}^E T_{ij} \, dE_{ij} \end{equation} $$ i.e., the relation in equation (4.75) allows for a variable substitution on the computation of the stress work per unit volume such that the integral in equation (4.76) may be computed as an integral between the tqo strain states $ \mathbf{E}_0 $ and $ \mathbf{E} $.

Figure 31: Stress work per unit volume $ w $ and elastic energy $ \phi $ for a uniaxial stress situation.

Now, to simplify, consider a case of uniaxial stress (see figure 31), where the principal stress $ \sigma $ is co-axial with the principal strain $ \varepsilon $, and they are both functions of time. The stress power per unit volume is the given as $ \omega = \sigma \cdot \dot{\varepsilon} $. The stress work per unit volume can then be computed as: $$ \begin{equation} w = \int_{t_0}^t \omega \, dt = \int_{t_0}^t \sigma \dot{\epsilon} \, dt = \int_{\epsilon_0}^\epsilon \sigma \, d\epsilon \tag{4.77} \end{equation} $$ A material is classified as hyperelastic if $ \omega $ and $ w $ may derived from a scalar valued potential $ \phi(\boldsymbol{E}) $ such that: $$ \begin{equation} \tag{4.78} \omega = \dot{\phi} = \partd{\phi}{E_{ij}} \, \dot{E_{ij}} \qquad \text {and} \qquad \phi=\phi(\boldsymbol{E}) \end{equation} $$ Mathematically, this has the beneficial consequense that the stress work per unit volume for a hyperealstic material may be computed simply as: $$ \begin{equation} w = \int_{t_0}^t \omega \, dt = \left . \phi \right |_{t_0}^t = \phi(\boldsymbol{E}) - \phi(\boldsymbol{E}_0) \tag{4.79} \end{equation} $$ i.e., as the difference of the potential $ \phi $ evaluated at the current and reference strain state. The potential $ \phi(\boldsymbol{E}) $ is commonly denoted the elastic energy or strain energy per unit volume.

Now, from the definition of the stress power per unit volume given in equation (4.73) and from the definition of the hyperelastic potential in equation (4.78) we have to expressions for the stress power per unit volme: $$ \begin{equation} \tag{4.80} \omega = T_{ij} \dot{E}_{ij} = \partd{\phi}{E_{ij}} \, \dot{E}_{ij} \end{equation} $$ As the hyperelastic potential per definition is a function of strain only (i.e., $ \phi(\boldsymbol{E}) $) and not the strain rate $ \dot{E}_{ij} $, we get: $$ \begin{equation} \tag{4.81} T_{ij} = \partd{\phi}{E_{ij}} \end{equation} $$ Therefore, is the hyperelastic potential $ \phi $ is is given for a material, one may compute all the stress components from equation (4.81) for given strains $ E_{ij} $. From equations (4.78) and (4.81) we deduce that $ \boldsymbol{T} = \boldsymbol{T(E)} $, and according to equation (4.1), the assumption of a hyperelastic material implies that the material is elastic too (i.e., hyperelasticity $ \Rightarrow $ elasticity). Note that the converse is not necessarily true.

4.3.1 Hyperelasticity for large derformations

For large deformations it is customary to introduce the elastic energy (or hyperelastic potential) per mass unit $ \psi(\boldsymbol{C}) $ rather than per volume unit $ \phi(\boldsymbol{E}) $, and further to express the energy as a function of the Green's deformations tensor $ \boldsymbol{C} $ rather than the Green strain tensor (see previous the section 4.3 Hyperelastic materials and strain energy for small deformations) . The stress power $ \omega \, dV $ for a differential volume $ dV $ may be derived from the potential $ \psi \, \rho \, dV $ in the following manner: $$ \begin{equation} \tag{4.82} \omega \, dV = \frac{d}{dt} \left ( \psi \rho \, dV \right ) = \dot{\psi} \, \rho dV \end{equation} $$ Consequently, the stress power per unit volume an the elastic energy per unit mass are related by: $$ \begin{equation} \tag{4.83} \omega = \rho \dot{\psi} = \rho \partd{\psi}{C_ {ij}} \dot{C}_{ij} \qquad \text{and} \qquad \psi=\psi(\boldsymbol{C}) \end{equation} $$

In order to relate the potential to the stress as given in equation (4.73), we need to establish a relation between $ \dot{\boldsymbol{C}} $ and $ \boldsymbol{D} $. From the definition of Green's deformation tensor in equation (3.15) we get: $$ \begin{equation} \tag{4.84} \dot{\boldsymbol{C}} = \dot{\boldsymbol{F}}^T \cdot \boldsymbol{F} + \boldsymbol{F}^T \cdot \dot{\boldsymbol{F}} \end{equation} $$

Further, from equations (3.11), (3.78), and (3.79), the rate of deformation gradient tensor, the velocity gradient tensor, and the strain rate tensor are given by: $$ \begin{equation} \tag{4.85} F_{ik} = \partd{x_i}{X_k} \quad \mathrm{and} \quad L_{il} = v_{i,l} \quad \mathrm{and} \quad D_{ik} = \frac{1}{2} (v_{i,k} + v_{k,i}) \end{equation} $$ Now, the expression for $ \dot{\boldsymbol{F}} $ may be computed from equation (4.85) as: $$ \begin{equation*} \tag{4.86} \dot{F_{ik}} =\frac{\partial^2 x_i}{\partial t \partial X_k} =\partd{}{X_k} \left (\partd{x_i}{t} \right ) = \partd{}{X_k} v_i=\partd{v_i}{x_l} \partd{x_l}{X_k} \nonumber \end{equation*} $$ which yields in tensor notation has the representtion: $$ \begin{equation} \tag{4.87} \dot{\boldsymbol{F}} = \boldsymbol{L \cdot F} \quad \mathrm{and} \quad \dot{\boldsymbol{F}^T} = \boldsymbol{F}^T \cdot \boldsymbol{L}^T \end{equation} $$ And finally, from equation (4.87), (4.85), and (4.84) we get the wanted relation $ \dot{\boldsymbol{C}} $ and $ \boldsymbol{D} $: $$ \begin{align} \dot{\boldsymbol{C}} &= \dot{\boldsymbol{F}}^T \cdot \boldsymbol{F} + \boldsymbol{F}^T \cdot \dot{\boldsymbol{F}} \nonumber \\ &= 2 \, \boldsymbol{F}^T \cdot \boldsymbol{D \cdot F} \tag{4.88} \end{align} $$

By substitution of the expression for $ \dot{C}_{ij} $ in equation (4.88) in to equation (4.83) we get: $$ \begin{equation} \tag{4.89} \dot{\psi} = \frac{\psi}{C_{ij}} \dot{C_{ij}} = \frac{\psi}{C_{ij}} \left ( 2 F_{ki} D_{kl} F_{lj} \right ) = 2 \, \left ( \boldsymbol{F \partd{\psi}{C} F}^T \right ): \boldsymbol{D} \end{equation} $$

Having proceeded in the same way as in derivation of hyperelasticity for small deformations (see equation (4.80)), we now have two expressions for the stress power which must hold, namely the definition of stress power in equation (4.73) and the propery for hyperelasticity in equation (4.83) which yields: $$ \begin{equation} \tag{4.90} \omega = T_{ij} D_{ij} = \rho \dot{\psi} = \rho \partd{\psi}{C_ {ij}} \dot{C}_{ij} \end{equation} $$ and by employing the expression for $ \dot{\psi} $ in equation (4.89) we get: $$ \begin{equation} \tag{4.91} \omega = \boldsymbol{T : D} = \left ( 2 \,\rho \boldsymbol{F \partd{\psi}{C} F}^T \right ): \boldsymbol{D} \end{equation} $$ and finally as the hyperelastic potential $ \psi $ per definition is independent of $ \boldsymbol{D} $, we find how the Cauchy stress tensor $ \boldsymbol{T} $ may be derived from the hyperelastic potential $ \psi $ per mass unit for large deformations: $$ \begin{equation} \tag{4.92} \boldsymbol{T} = 2 \,\rho \boldsymbol{F \partd{\psi}{C} F}^T \end{equation} $$ From equations (3.15) and (3.24) we know that the deformation tensors $ \boldsymbol{F} $, $ \boldsymbol{C} $, and $ \boldsymbol{E} $, are functionally related. Thus, from the expression for the Cauchy stress for a hyperelastic material in equation (4.92), we may conclude that $ \boldsymbol{T} = \boldsymbol{T(E)} $, and that an assumption of hyperelasticity for large deformations also implies elasticity. Note that we already came to the same conclusion for small deformations in the section 4.3 Hyperelastic materials and strain energy.

4.3.2 Stress tensors for large deformations

The governing equations are often formulated in $ K_0 $ for large deformations, i.e., $ (X,t) $ and conservation of mass implies $ \rho\,dV =\rho_0 \, dV_0 $ and consequently the volume integrals in Cauchy's equations (2.87) are easily represented in the reference frame by: $$ \begin{equation} \tag{4.93} \int_V \boldsymbol{b} \, \rho \, dV = \int_{V_0} \boldsymbol{b} \, \rho_0 \, dV_0, \quad \int_V \boldsymbol{a} \, \rho \, dV = \int_{V_0} \boldsymbol{a} \, \rho_0 \, dV_0 \end{equation} $$ However, the challenge is how to represent the surface integrals of the $ \int_A \boldsymbol{t} \, dA $. To approach this issue we define a stress vector $ \boldsymbol{t}_0 $ in $ K_0 $ by: $$ \begin{equation} \tag{4.94} \boldsymbol{t}_0 \, dA_0 = \boldsymbol{t} \, dA \quad \Leftrightarrow \quad \int_{A_0} \boldsymbol{t}_0 \, dA_0 = \int_A \boldsymbol{t} \, dA \end{equation} $$ which yields: $$ \begin{equation} \tag{4.95} \boldsymbol{t}_0 = \frac{dA}{dA_0} \, \boldsymbol{t} \end{equation} $$ where $ \boldsymbol{t}_0 $ represents forces in $ K $. Note that $ \boldsymbol{t}_0 $ has same direction as $ \boldsymbol{t} $ but not with the same magnitude. With the definition in equation (4.95) we may now formulate the balance of linear monemntu based on Euler's first axiom in the reference frame $ K_0 $ (or in Lagrangian coordinates): $$ \begin{equation} \tag{4.96} \int_{V_0} \boldsymbol{a} \, \rho_0 \, dV_0 = \int_{A_0} \boldsymbol{t}_0 \, dA_0 + \int_{V_0} \boldsymbol{b} \, \rho_0 \, dV_0 \end{equation} $$ Albeit formulated in the reference configuration, this equation has both volume and surface integrals and is therefore not transferable to a partial differential equation form. Therefore, the first Piola-Kirchhoff stress tensor (PKS) $ \boldsymbol{T}_0 $ is introduced in analogy with the Cauchy stress theorem (see equation (2.65)): $$ \begin{equation} \tag{4.97} \boldsymbol{t}_0 = \boldsymbol{T}_0 \cdot \boldsymbol{n}_0 \end{equation} $$

Equation (4.97) is an invariant linear tensor equation and $ \boldsymbol{n}_0 $ is a unit normal on $ dA_0 $ in the reference configuration $ K_0 $. The first PKS $ \boldsymbol{T}_0 $ is related with the Cauchy stress tensor $ \boldsymbol{T} $ by: $$ \begin{equation} \tag{4.98} \boldsymbol{T}_0 = J \boldsymbol{TF}^{-T}, \qquad \text{where} \quad J = \det \boldsymbol{F} \end{equation} $$ As can be seen from equation (4.98), the first PKS $ \boldsymbol{T}_0 $ is not symmetric, and represents what is commonly called engineering or nominal stress.( CHECK: footnote at end of sentence placed in parenthesis) (i.e., Force in the current configuration divided by the area in the reference configuration.)

The second PKS $ \boldsymbol{S} $ is conventionally defined by: $$ \begin{equation} \tag{4.99} \boldsymbol{S} = \boldsymbol{F}^{-1} \boldsymbol{T}_0 = J \boldsymbol{F}^{-1} \boldsymbol{T F}^{-T} \end{equation} $$ and one may easily show that $ \boldsymbol{S} = \boldsymbol{S}^T $.

By introducing these stress tensors one may derive the Cauchy's equations of motions in the reference configuration $ K_0 $ (see p. 178 [2]).

In the previous the section 4.3.1 Hyperelasticity for large derformations the relation between the Cauchy stress tensor $ \boldsymbol{T} $ and a hyperelastic potential per mass unit $ \psi $ was derived in equation (4.92). An expression for the the second PKS $ \boldsymbol{S} $ from $ \psi $ may be obtained by combining equation (4.99) and (4.92): $$ \begin{equation} \tag{4.100} \boldsymbol{S} = 2 J \rho \partd{\psi}{\boldsymbol{C}} = 2 \rho_0 \partd{\psi}{\boldsymbol{C}}, \qquad\text{where} \quad \rho J = \rho_0 \end{equation} $$ We state without proof that $ \boldsymbol{S} $ also may be express by a hyperelastic potential per unit volume $ \phi = \rho_0 \psi $ by: $$ \begin{equation} \tag{4.101} \boldsymbol{S} = 2 \partd{\phi}{\boldsymbol{C}} \end{equation} $$ which is indeed a simpler expression than the one relating the Cauchy stress tensor $ \boldsymbol{T} $ and the hyperelastic potential in equation (4.101).

4.3.3 Isotropic hyperelastic materials

For isotropic materials, the hyperelastic potential per unit volume $ \phi $ is commonly expressed as a function of the invariants of the Green deformation tensor $ \boldsymbol{C} $, which we denote $ I_1, \ldots, I_3 $, i.e., $ \phi=\phi(I_1,I_2,I_3) $. The motivation for such a formulation is that the mathematical representation will be independent of the particular coorindate system, and thus ensure that the mathematical representation is isostropic. The invariants of a second order tensor are well defined an given by e.g., equations (3.56) to (3.58): $$ \begin{align} I_1 = \tr(\boldsymbol{C}) \nonumber \\ I_2 = \frac{1}{2} ( (\tr \boldsymbol{C})^2 - \| \boldsymbol{C} \|^2 \tag{4.102} \\ I_3 = \det (\boldsymbol{C}) = \det (\boldsymbol{F}^2) = J^2 \nonumber \end{align} $$

For such a formulation the second PKS $ \boldsymbol{S} $ may be derived from the general relation given in equation (4.101) by empolying the chain rule for differentiation: $$ \begin{equation} \tag{4.103} \boldsymbol{S} = 2 \partd{\phi}{\boldsymbol{C}} = 2 \sum_{k=1}^3 \partd{\phi}{I_k} \, \partd{I_k}{\boldsymbol{C}} \end{equation} $$ One may show (see Exercise 2: Invariants): $$ \begin{equation} \tag{4.104} \partd{I_1}{\boldsymbol{C}} = \boldsymbol{1}, \quad \partd{I_2}{\boldsymbol{C}} = I_1 \boldsymbol{1} - \boldsymbol{C}, \quad \partd{I_3}{\boldsymbol{C}}= I_3 \boldsymbol{C}^{-1} \end{equation} $$ Note that the partial derivatives of the invariants are quantities which are computed once and for all, and thus one may establish a more practical representation for how the $ \boldsymbol{S} $ may be computed from $ \phi $, by substitution of equation (4.104) into equation (4.103): $$ \begin{equation} \tag{4.105} \boldsymbol{S} = 2 \left ( \partd{\phi}{I_1} + I_1 \partd{\phi}{I_2} \right ) \, \boldsymbol{1} - 2 \partd{\phi}{I_2} \, \boldsymbol{C} + 2 I_3 \partd{\phi}{I_3} \boldsymbol{C} ^{-1} \end{equation} $$

For incompressible materials we have that $ \det \boldsymbol{F} = J = 1 $ which from equation (4.102) is equivalent to $ I_3 = 1 $. For the incompressible materials it is thefore customary to introduce a modified strain energy function: $$ \begin{equation} \tag{4.106} \bar{\phi} = \phi(\boldsymbol{C}) + p (J-1) \end{equation} $$ where $ p $ denote an undetermined Lagrange multiplier which has to be determined from boundary conditions. As for incompressible fluids, only gradients of pressure determine the solutions of the governing equations for incompressiblesolids, and the level of the pressure (stresses) must be determined from the boundary conditions.

The second PKS may the be found by: $$ \begin{equation} \tag{4.107} \boldsymbol{S} = 2 \, \partd{\bar{\phi}}{\boldsymbol{C}} = 2 \partd{\phi(\boldsymbol{C})}{\boldsymbol{C}} + p J \boldsymbol{C}^{-1} \end{equation} $$ as: $$ \begin{equation} \tag{4.108} \partd{J}{\boldsymbol{C}} = \partd{J}{I_3} \, \partd{I_3}{\boldsymbol{C}} = \partd{\sqrt{I_3}}{I_3} \, I_3 \boldsymbol{C}^{-1} = \frac{J}{2} \, \boldsymbol{C} ^{-1} \end{equation} $$

4.3.4 Example 11: Thin sheet of incompressible material

For a thin sheet of incompressible material (i.e., with $ I_3 = 1 $) we have $ \phi = \phi(I_1,I_2) $. Furthermore, due to the thin sheet we orient our coordinate system in such a way that we may assume zeero stress in direction orthogonal to the sheet plane. From equation (4.107) we get for the $ S_{33} $ component: $$ \begin{equation} \tag{4.109} S_{33} = 2 \, \left ( \partd{\phi}{I_1} + I_1 \partd{\phi}{I_2} \right )- 2 \partd{\phi}{I_2} C_{33} + p \, \boldsymbol{C}_{33}^{-1} = 0 \end{equation} $$ Thus, an expression for the Lagrangian multiplier may be found from equation (4.109) : $$ \begin{equation} \tag{4.110} p = 2 \partd{\phi}{I_2} C_{33}^2 + 2 \, \left ( \partd{\phi}{I_1} + I_1 \partd{\phi}{I_2} \right ) \, C_{33} \end{equation} $$

4.3.5 Example 12: The Mooney-Rivlin material

For isotropic materials one may express the hyperelastic potential per unit mass $ \psi $ as a function of the invariants of the left deformation tensor $ \boldsymbol{B} $ (see equation (3.16)). The hyperelastic potential is formulated accordingly as: $$ \begin{equation} \tag{4.111} \psi = \psi(I_B,II_B,III_B) \end{equation} $$ where $ I_B,II_B $, and $ III_B $ denote the principal invariants of $ \boldsymbol{B} $. By proceeding the same way as for the formulation the left Green deformation tensor $ \boldsymbol{C} $ in equation (4.83), we differentiate the potential with respect to time to get: $$ \begin{equation} \tag{4.112} \dot{\psi} = \partd{\psi}{\boldsymbol{B}} : \dot{\boldsymbol{B}} \end{equation} $$ The expresson for $ \dot{\boldsymbol{B}} $ may be found to be (see Exercise 2: Invariants): $$ \begin{equation} \tag{4.113} \dot{\boldsymbol{B}} = \boldsymbol{LB+BL}^T \end{equation} $$ Substitution of equation (4.113) in equation (4.112): $$ \begin{equation} \tag{4.114} \dot{\psi} = \partd{\psi}{\boldsymbol{B}} : \dot{\boldsymbol{B}} = \partd{\psi}{B_{ij}} \dot{B}_{ij} = \partd{\psi}{B_{ij}} \left ( L_{ik} B_{kj} + B_{ik} L_{jk} \right ) = \left ( 2 \partd{\psi}{\boldsymbol{B}} \boldsymbol{B} \right ) : \boldsymbol{L} \end{equation} $$ where the latter equality comes from the symmetric property of $ \boldsymbol{B} $. From equation (4.90) and (4.114) on may obtain: $$ \begin{equation} \tag{4.115} \omega = \boldsymbol{T:D} = \boldsymbol{T:L} = \rho \dot{\psi} = \left ( 2 \rho \partd{\psi}{\boldsymbol{B}} \boldsymbol{B} \right ) : \boldsymbol{L} \end{equation} $$ and as $ \boldsymbol{L} $ may be chosen arbitrarily, we find that the Cauchy stress tensor can be derived from the strain energy potential as: $$ \begin{equation} \tag{4.116} \boldsymbol{T} = \left ( 2 \rho \partd{\psi}{\boldsymbol{B}} \boldsymbol{B} \right ) \end{equation} $$ In particular, the Mooney-Rivlin material is an example of a material models which has been used to model rubber, which is considered to be isotropic and incompressible, and easily subject to large deformations for typical loading conditions. The Mooney-Rivlin material is defined by the specific elastic energy: $$ \begin{equation} \tag{4.117} \psi = \frac{1}{2} \frac{\mu}{\rho} \, \left ( \frac{1}{2} + \alpha \right ) (I_B - 3) + \frac{1}{2} \frac{\mu}{\rho} \, \left ( \frac{1}{2} - \alpha \right ) (II_B - 3) \end{equation} $$ where $ \alpha $ and $ \mu $ are material parameters (elasticities). The Cauchy stress tensor may for the Mooney-Rivlin material be obatained from equation (4.117) and (4.116): $$ \begin{equation} \tag{4.118} \boldsymbol{T} = \mu \left ( \frac{1}{2} + \alpha \right ) \boldsymbol{B} -\mu \left ( \frac{1}{2} - \alpha \right ) \boldsymbol{B}^{-1} - p \boldsymbol{I} \end{equation} $$

As the Mooney-Rivlin material is incompressible, an isotropic stress $ -p\boldsymbol{I} $, has to be added to the stress tensor. A so-called neo-Hookean material is obtained for $ \alpha=1/2 $ in equation (4.118).