Hyperelasticity

Hyperelasticity or Green's elasticity (from the Greek ὑπέρ hyper "over", ελαστικός elastikos "adaptable" and George Green ) is a material model of elasticity . Elasticity is the property of a body to change its shape under the action of force and to return to its original shape when the force is removed (example: spring). The cause of the elasticity are distortions of the atomic lattice (in metals), the stretching of molecular chains (rubber and plastics) or the change in the mean atomic distance (liquids and gases).

For many materials, the linear elasticity does not exactly describe the observed material behavior. The best-known example with non-linear elastic behavior is rubber , which withstands large deformations and whose reactions can be simulated to a good approximation with hyperelasticity. Biological tissues are also modeled with hyperelasticity. All barotropic, frictionless liquids and gases are, as it were, Cauchy elastic and hyperelastic, which is discussed in Cauchy elasticity . This article deals with solid models. Here, hyperelasticity is the special case of Cauchy elasticity in which the material behavior is conservative .

Ronald Rivlin and Melvin Mooney developed the first solid models of hyperelasticity, the Neo-Hooke Mooney-Rivlin model. Other frequently used material models are the Ogden and Arruda-Boyce models.

description

Macroscopic behavior

Force-displacement diagram in a uniaxial tensile test with non-linear elasticity

The following properties can be observed macroscopically on a hyperelastic body:

• With a given deformation, the reaction forces always have the same value regardless of the previous history.
• If the initial state is unloaded, it is resumed after any deformation when the loads are removed.
• The material behavior is independent of the speed. The speed at which a deformation takes place has no influence on the resistance that the body offers to the deformation.
• In the uniaxial tensile test, loading and unloading always take place along the same path as in the adjacent picture.
• The deformation work involved is completely stored in the body as distortion energy. The material is therefore conservative.

The first four properties characterize the Cauchy elasticity. If the material also has the last property, then the material is hyperelastic.

Time independence

In hyperelasticity, the reaction forces when a body is deformed are determined exclusively by the current deformation. If the initial state is free of forces, this will be resumed after any deformation when the loads are removed. Different deformation paths that ultimately result in the same deformations result in the same reaction forces in the end. The deformation speeds also have no influence on the reactions at the material equation level. In this idealization, the previous history of the material has no influence on the current material behavior. Hyperelasticity is a time-independent material property.

Conservatism

In addition to Cauchy elasticity, the deformation work in hyperelasticity is path-independent, which is expressed in the fact that the deformation work only depends on the start and end point of the deformation path, but not on its course. In the special case of the coincidence of the start and end point, the following results: No work is done or energy is consumed along a closed deformation path. Elaborate work is completely returned by the body until the return to the starting point. The conservatism here also follows from the fact that the deformation performance is exactly the rate of the deformation energy, i.e. the work involved is converted completely (without dissipation) into deformation energy. Deformations are reversible here.

Dissipative processes such as plastic flow or creep are therefore excluded, which is the case with real materials within their elastic limit . Real liquids and gases and some solids (such as iron or glass) are elastic to a good approximation when they move quickly and slightly (e.g. with sound waves). In the case of solids, the elastic limit is adhered to in the case of sufficiently small and slowly occurring deformations, which are present in many applications, especially in the technical field.

Linear hyperelasticity

With sufficiently small deformations, the force-displacement relationship in solids is linear and the elasticity can be described in terms of modulus. Because the force to be applied and the distance covered in the event of a deformation depend largely on the dimensions of the body, the force is related to its effective area and the distance to a suitable dimension of the body. The related force is the tension and the related distance is the elongation . The modules quantify the relationship between the stresses and the strains and are a material property. The modulus of elasticity applies to uniaxial tension , the shear modulus to shear and the compression modulus to all-round tension / pressure . With uniaxial tension, a deformation occurs not only in the direction of tension, but also across it, which quantifies the dimensionless Poisson's ratio. The complete description of the isotropic linear hyperelasticity requires only two of the named quantities, cubic anisotropy three (a modulus of elasticity, a shear modulus and a Poisson's modulus), transversal isotropy already five (two moduli of elasticity, two Poisson's ratio and a shear modulus) and orthotropy nine (three moduli of elasticity, three Poisson's ratio and three shear modulus). However, a maximum of 21 parameters are required to describe a real linear hyperelastic material.

definition

First, the macroscopic behavior of a homogeneous tension rod made of hyperelastic material is used for explanation. The definition of hyperelasticity is made up for by the transition from the macroscopic body to a point in the continuum.

Behavior of a hyperelastic body

A rod (black) is stretched by a force by the amount (red) ${\displaystyle F}$ ${\displaystyle u}$

If a homogeneous rod made of hyperelastic material is stretched axially by an amount as shown in the picture , then a force is required that results from the deformation energy through the derivative ${\displaystyle u}$ ${\displaystyle F}$ ${\displaystyle W}$

${\displaystyle F={\frac {\mathrm {d} W}{\mathrm {d} u}}}$

calculated. In the linear case, the spring constant D is for example

${\displaystyle W={\frac {D}{2}}u^{2}\rightarrow F=Du\,.}$

Three-dimensional continuum

The translation of the behavior of the hyperelastic tension rod into a three-dimensional continuum is done by

• the force F by a stress tensor σ ,
• the displacement u through a strain tensor ε and
• the deformation energy W by the specific deformation energy w

is exchanged. Then the stresses σ are calculated from the derivation from w to ε according to

${\displaystyle {\boldsymbol {\sigma }}=\rho {\frac {\mathrm {d} w}{\mathrm {d} {\boldsymbol {\varepsilon }}}}}$

Here ρ is the density of the material. So that this material model fulfills the principle of material objectivity , the modeling guideline described in Cauchy elasticity must be adhered to, which states that the second Piola-Kirchhoff stress tensor must be used as the stress tensor , which may only depend on the right-hand stretch tensor U. Instead of the right-hand distance tensor, the Green-Lagrangian strain tensor E = ½ ( U · U - 1 ) with the unit tensor 1 or the right-Cauchy-Green tensor C = U · U is used more often: ${\displaystyle {\tilde {\mathbf {T} }}}$

${\displaystyle {\tilde {\mathbf {T} }}=\rho _{0}{\frac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {E} }}=2\rho _{0}{\frac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {C} }}}$

In the Lagrangian representation used here, the density ρ _{0 of} the body is a material parameter that is constant over time.

The volume -related deformation energy is often used instead of the specific deformation energy . Because ρ _{0 is} a constant factor, the formulas that result from the deformation energy related to the mass or the volume can be converted into one another at any time. The representation here follows main. ${\displaystyle {\hat {w}}=\rho _{0}w}$

In Euler's mode of representation, this results in the Cauchy stress tensor:

${\displaystyle {\boldsymbol {\sigma }}:={\frac {1}{\operatorname {det} (\mathbf {F} )}}\mathbf {F} \cdot {\tilde {\mathbf {T} }}\cdot \mathbf {F} ^{\top }=\rho \mathbf {F} \cdot {\frac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {E} }}\cdot \mathbf {F} ^{\top }=2\rho \mathbf {F} \cdot {\frac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {C} }}\cdot \mathbf {F} ^{\top }\,.}$

Here F is the deformation gradient and

${\displaystyle \rho :={\frac {\rho _{0}}{\operatorname {det} (\mathbf {F} )}}}$

the density in the deformed body, which is measured by the determinant det of the deformation gradient.

In the case of isotropic material, according to

${\displaystyle {\boldsymbol {\sigma }}=2\rho {\dfrac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}\cdot \mathbf {b} }$

the left Cauchy-Green tensor b = F · F ^T , which is also a stretch tensor, can also be used, see hyperelasticity # isotropic hyperelasticity in spatial representation below.

Properties of hyperelastic materials

This section elaborates on that in hyperelasticity

• the deformation work depends only on the start and end point of the deformation path, but not on its course,
• no work is performed or energy is consumed along a closed deformation path, i.e. any work expended is completely returned by the body until it returns to the starting point,
• the deformation power is exactly the rate of the deformation energy, that is, the work involved is stored completely (without dissipation) as deformation energy and
• Deformations are reversible.

Deformation work

In the above-mentioned pull rod makes the force along a path from to the work ${\displaystyle F}$ ${\displaystyle u_{0}}$ ${\displaystyle u_{1}}$

${\displaystyle A=\int _{u_{0}}^{u_{1}}F\mathrm {d} u=\int _{u_{0}}^{u_{1}}{\frac {\mathrm {d} W}{\mathrm {d} u}}\mathrm {d} u={[W]}_{u_{0}}^{u_{1}}=W(u_{1})-W(u_{0})\,,}$

d. That is, the work done depends only on the start and end point. In particular, the work done disappears : ${\displaystyle u_{1}=u_{0}}$

${\displaystyle A=W(u_{0})-W(u_{0})=0\,.}$

Similarly, in the continuum, the work of stress is the curve integral over a curve parameterized with time t

${\displaystyle {\begin{aligned}a=&\int _{t_{0}}^{t_{1}}{\tilde {\mathbf {T} }}:{\dot {\mathbf {E} }}\,\mathrm {d} t=\rho _{0}\int _{t_{0}}^{t_{1}}{\frac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {E} }}:{\frac {\mathrm {d} \mathbf {E} }{\mathrm {d} t}}\,\mathrm {d} t=\rho _{0}\int _{t_{0}}^{t_{1}}{\frac {\mathrm {d} w_{0}}{\mathrm {d} t}}\,\mathrm {d} t\\=&\rho _{0}[w_{0}(\mathbf {E} )]_{t_{0}}^{t_{1}}=\rho _{0}[w_{0}(\mathbf {E} (t_{1}))-w_{0}(\mathbf {E} (t_{0}))]\end{aligned}}}$

which proves the independence of the route and conservatism (in special cases ). The mathematical symbol ":" means the Frobenius scalar product and provides the sum of the work increments of the stress components at the components of the deformation gradient. ${\displaystyle \mathbf {E} (t_{1})=\mathbf {E} (t_{0})}$

Deformation performance

The deformation performance in the rod produced by the force is: ${\displaystyle F}$

${\displaystyle L=F{\dot {u}}={\frac {\mathrm {d} W}{\mathrm {d} u}}{\frac {\mathrm {d} u}{\mathrm {d} t}}={\frac {\mathrm {d} W}{\mathrm {d} t}}={\dot {W}}}$

and is the deformation work performed per unit of time. The power output is completely converted into deformation energy without dissipation. The same applies in the continuum

${\displaystyle l_{i}:={\frac {1}{\rho _{0}}}{\tilde {\mathbf {T} }}:{\dot {\mathbf {E} }}={\frac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {E} }}:{\frac {\mathrm {d} \mathbf {E} }{\mathrm {d} t}}={\frac {\mathrm {d} w_{0}}{\mathrm {d} t}}={\dot {w}}_{0}\,,}$

d. In other words , the specific stress power in the case of hyperelasticity is the material time derivative of the specific deformation energy. ${\displaystyle l_{i}}$

The converse also applies: If there is a scalar function so that the specific voltage power is the material time derivative of this function, then the material is hyperelastic. ${\displaystyle w_{0}}$

In Euler's formulation arises

${\displaystyle l_{i}={\frac {1}{\rho _{0}}}{\tilde {\mathbf {T} }}:{\dot {\mathbf {E} }}={\frac {1}{\rho }}{\boldsymbol {\sigma }}:\mathbf {d} }$

with the spatial strain rate tensor

${\displaystyle {\begin{aligned}\mathbf {d} :=&{\frac {1}{2}}(\mathbf {l+l} ^{\top })={\frac {1}{2}}({\dot {\mathbf {F} }}\cdot \mathbf {F} ^{-1}+\mathbf {F} ^{\top -1}\cdot {\dot {\mathbf {F} }}^{\top })\\=&\mathbf {F} ^{\top -1}\cdot {\frac {1}{2}}(\mathbf {F} ^{\top }\cdot {\dot {\mathbf {F} }}+{\dot {\mathbf {F} }}^{\top }\cdot \mathbf {F} )\cdot \mathbf {F} ^{-1}=\mathbf {F} ^{\top -1}\cdot {\dot {\mathbf {E} }}\cdot \mathbf {F} ^{-1}\,,\end{aligned}}}$

which is the symmetrical part of the velocity gradient . ${\displaystyle \mathbf {l} ={\dot {\mathbf {F} }}\cdot \mathbf {F} ^{-1}}$

Sentences about hyperelasticity

The following statements are equivalent:

1. The material is hyper-elastic.
2. The specific voltage power is the material time derivative of the specific deformation energy ${\displaystyle l_{i}}$ ${\displaystyle {\dot {w}}_{0}}$
   ${\displaystyle l_{i}={\dot {w}}_{0}\,.}$
3. The work of the stresses along any path in the expansion space is only dependent on the start and end point of the path but not on its course
   ${\displaystyle \int _{\mathbf {E} _{0}}^{\mathbf {E} _{1}}{\tilde {\mathbf {T} }}:\mathrm {d} \mathbf {E} =\rho _{0}(w_{0}(\mathbf {E} _{1})-w_{0}(\mathbf {E} _{0}))\,.}$
4. The work of the stresses along any closed path in the expansion space disappears
   ${\displaystyle \oint {\tilde {\mathbf {T} }}:\mathrm {d} \mathbf {E} =0\,.}$
5. The specific stress work at any differential expansion increments is equal to the total differential of the specific deformation energy ${\displaystyle \delta \mathbf {E} }$
   ${\displaystyle {\frac {1}{\rho _{0}}}{\tilde {\mathbf {T} }}:\delta \mathbf {E} =\delta w_{0}(\mathbf {E} )\,.}$

In linear hyperelasticity, the stresses as the first derivative of the deformation energy are linear in the strains and the elasticity tensor is constant as the second derivative. Because the sequence of the derivatives is interchangeable with two derivatives, the elasticity tensor is symmetrical and a linear-hyperelastic solid can be described with a maximum of 21 parameters.

Any barotropic elastic fluid is also hyperelastic.

Thermodynamic consistency

→ Main article : Clausius-Duhem inequality

The hyperelasticity is consistent with thermodynamics , as an evaluation of the Clausius-Duhem inequality shows, which represents the second law of thermodynamics in solid mechanics . In the case of an isothermal change of state , the Clausius-Duhem inequality reads in the Lagrangian version

${\displaystyle {\frac {1}{\rho _{0}}}{\tilde {\mathbf {T} }}\cdot {\dot {\mathbf {E} }}-{\dot {\psi }}_{0}\geq 0\,,}$

where represents Helmholtz's free energy . In hyperelasticity, the stress is the derivative of the deformation energy after the elongations and, because the deformation energy is only a function of the elongations, it follows: ${\displaystyle \psi _{0}}$

${\displaystyle {\frac {1}{\rho _{0}}}{\tilde {\mathbf {T} }}:{\dot {\mathbf {E} }}-{\dot {\psi }}_{0}={\frac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {E} }}:{\dot {\mathbf {E} }}-{\dot {\psi }}_{0}={\dot {w}}_{0}-{\dot {\psi }}_{0}={\frac {\mathrm {d} }{\mathrm {d} t}}(w_{0}-\psi _{0})\geq 0\,.}$

Identification of the deformation energy with the Helmholtz free energy allows the hyperelasticity to be in harmony with the thermodynamics.

Incompressibility

Many rubber elastic bodies show pronounced incompressibility and it is therefore worth looking at this case more closely. Incompressibility lets itself through mathematically

${\displaystyle J:=\operatorname {det} (\mathbf {F} )={\sqrt {\operatorname {det} (\mathbf {C} )}}\equiv 1}$

express why the density is then constant over time:

${\displaystyle \rho =\rho _{0}\,.}$

In order to ensure the incompressibility of a hyperelastic material, the specific deformation energy w _{0 is} extended:

${\displaystyle {\bar {w}}_{0}:=w_{0}-{\frac {p}{\rho _{0}}}(J-1)}$

The pressure p is an additional, non-constitutive variable that is introduced as a Lagrangian multiplier to ensure the secondary condition J ≡ 1. The pressure now results exclusively from the laws of nature and the position of the body. The tensions are here

${\displaystyle {\begin{aligned}{\tilde {\mathbf {T} }}=&2\rho _{0}{\dfrac {\mathrm {d} {\bar {w}}_{0}}{\mathrm {d} \mathbf {C} }}=-p\mathbf {C} ^{-1}+2\rho _{0}{\dfrac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {C} }}\\[2ex]\rightarrow {\boldsymbol {\sigma }}=&{\frac {1}{J}}\mathbf {F} \cdot {\tilde {\mathbf {T} }}\cdot \mathbf {F} ^{\top }=-p\mathbf {1} +2\rho _{0}\mathbf {F} \cdot {\dfrac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {C} }}\cdot \mathbf {F} ^{\top }\end{aligned}}\,.}$

Isotropic hyperelasticity

If the material behavior is not directional, then the material is isotropic . After an excursus in Lagrangian representation , isotropy is discussed in the more common Eulerian approach .

Isotropic hyperelasticity in material representation

In isotropic hyperelasticity, the deformation energy in the Lagrangian representation is a function of the main invariants I _{1,2,3 of} the right Cauchy-Green distance tensor :

${\displaystyle {\tilde {\mathbf {T} }}=\rho _{0}{\frac {\mathrm {d} w_{0}(\mathrm {I} _{1},\mathrm {I} _{2},\mathrm {I} _{3})}{\mathrm {d} \mathbf {C} }}}$

These major invariants hang over

${\displaystyle {\begin{array}{rclclcc}\mathrm {I} _{1}(\mathbf {C} )&=&\operatorname {Sp} (\mathbf {C} )&=&\operatorname {Sp} (\mathbf {F^{\top }\cdot F} )=\mathbf {F} :\mathbf {F} &=&\|\mathbf {F} \|^{2}\\\mathrm {I} _{2}(\mathbf {C} )&=&\operatorname {Sp(cof} (\mathbf {C} ))&=&\operatorname {Sp(cof} (\mathbf {F} )^{\top }\cdot \operatorname {cof} (\mathbf {F} ))&=&\|\operatorname {cof} (\mathbf {F} )\|^{2}\\\mathrm {I} _{3}(\mathbf {C} )&=&\operatorname {det} (\mathbf {C} )&=&\operatorname {det} (\mathbf {F^{\top }\cdot F} )&=&\operatorname {det} (\mathbf {F} )^{2}\end{array}}}$

together with the deformation gradient . The operator designates the track . The cofactor of a tensor is its transposed adjunct , which in the case of invertible tensors, as they are here, ${\displaystyle \mathbf {F} }$ ${\displaystyle \mathrm {Sp} }$

${\displaystyle \operatorname {cof} (\mathbf {A} ):=\operatorname {det} (\mathbf {A} )\mathbf {A} ^{\top -1}}$

reads. The Frobenius norm is defined with the Frobenius scalar product ":": ${\displaystyle \|(\cdot )\|}$

${\displaystyle \mathbf {A} :\mathbf {B} :=\operatorname {Sp} (\mathbf {A^{\top }\cdot B} )\,,\quad \|\mathbf {A} \|:={\sqrt {\mathbf {A} :\mathbf {A} }}}$

The main invariants of the Cauchy-Green tensor on the right are measures for the change in line, surface and volume elements .

Isotropic objective hyperelasticity thus implies a stress-deformation relationship of the shape

${\displaystyle {\tilde {\mathbf {T} }}=\rho _{0}{\frac {\mathrm {d} w_{0}(\|\mathbf {F} \|^{2},\|\operatorname {cof} (\mathbf {F} )\|^{2},\operatorname {det} (\mathbf {F} )^{2})}{\mathrm {d} \mathbf {C} }}\,.}$

The reverse is also true: if this stress-deformation relationship exists, then the material is objective, isotropic and hyperelastic. A hyperelastic material is isotropic and objective if the deformation energy can be written as a function of the dimensions for the change in the line, surface and volume elements.

The above strain energy is polyconvex if it is a convex function in each of its arguments ║ F ║², ║cof ( F ) ║² and det ( F ) ² . If this deformation energy is also a coercive function of each of its arguments, then there is always a deformation that minimizes the deformation energy.

Isotropic hyperelasticity in a spatial representation

In the case of isotropic hyperelasticity, the deformation energy is usually assumed to be an isotropic function of the left Cauchy-Green tensor b = F · F ^T. This one has the time derivative ${\displaystyle w(\mathbf {b} )}$

${\displaystyle {\dot {\mathbf {b} }}={\dot {\mathbf {F} }}\cdot {\underline {\mathbf {F} ^{-1}\cdot \mathbf {F} }}\cdot \mathbf {F} ^{\top }+\mathbf {F} \cdot {\underline {\mathbf {F^{\top }\cdot F} ^{\top -1}}}\cdot {\dot {\mathbf {F} }}^{\top }=\mathbf {l\cdot b} +\mathbf {b\cdot l} ^{\top }\,.}$

The underlined terms are unit tensors inserted for the derivation of the last identity. The potential relation to Cauchy's stress tensor then results from the deformation power

${\displaystyle {\begin{aligned}l_{i}=&{\dfrac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}:{\dot {\mathbf {b} }}={\dfrac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}:(\mathbf {l\cdot b} +\mathbf {b\cdot l} ^{\top })=\left({\dfrac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}\cdot \mathbf {b} \right):\mathbf {l} +\left(\mathbf {b} \cdot {\dfrac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}\right):\mathbf {l} ^{\top }\\=&2\left(\mathbf {b} \cdot {\dfrac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}\right):{\frac {1}{2}}(\mathbf {l+l} ^{\top })=:{\dfrac {1}{\rho }}{\boldsymbol {\sigma }}:\mathbf {d} \\\rightarrow {\boldsymbol {\sigma }}=&2\rho \mathbf {b} \cdot {\dfrac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}=2\rho {\dfrac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}\cdot \mathbf {b} \end{aligned}}}$

The fact that b and d w / d b commute was used here , because the derivative here is an isotropic tensor function of the symmetrical left Cauchy-Green tensor b .

In the case of incompressibility, det ( F ) = √det ( b ) = 1 and therefore

${\displaystyle {\boldsymbol {\sigma }}=2\rho _{0}\mathbf {b} \cdot {\frac {\mathrm {d} {\bar {w}}}{\mathrm {d} \mathbf {b} }}=-p\mathbf {1} +2\rho _{0}\mathbf {b} \cdot {\frac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}\,.}$

With isotropy, the deformation energy depends only on the invariants of the symmetrical and positive definite tensor b or the left stretch tensor v = + √ b , which therefore have positive eigenvalues . The deformation energy is usually given with the eigenvalues ​​λ _1,2,3 of v or the main invariants

I ₁ ( b ) = Sp ( b )

I ₂ ( b ) = ½ (Sp ( b ) ² - Sp ( b ) ²)

I ₃ ( b ) = det ( b )

expressed. There are three formulations:

${\displaystyle {\begin{array}{llcl}1.&{\boldsymbol {\sigma }}&=&2\rho \mathbf {b} \cdot {\dfrac {\mathrm {d} w(\mathrm {I} _{1}(\mathbf {b} ),\mathrm {I} _{2}(\mathbf {b} ),\mathrm {I} _{3}(\mathbf {b} ))}{\mathrm {d} \mathbf {b} }}\\[2ex]2.&{\boldsymbol {\sigma }}&=&2\rho \mathbf {b} \cdot {\dfrac {\mathrm {d} w({\bar {\mathrm {I} }}_{1}(\mathbf {b} ),{\bar {\mathrm {I} }}_{2}(\mathbf {b} ),J)}{\mathrm {d} \mathbf {b} }}\quad {\text{und}}\\[2ex]3.&{\boldsymbol {\sigma }}&=&2\rho \mathbf {b} \cdot {\dfrac {\mathrm {d} w(\lambda _{1},\lambda _{2},\lambda _{3})}{\mathrm {d} \mathbf {b} }}\end{array}}\,.}$

The cross-slashed stretch tensor models the volume-conserving or unimodular portion of the deformation, because its determinant is constant: ${\displaystyle {\bar {\mathbf {b} }}:=J^{\frac {-2}{3}}\mathbf {b} }$

${\displaystyle \operatorname {det} ({\bar {\mathbf {b} }})=\operatorname {det} (J^{\frac {-2}{3}}\mathbf {b} )=J^{-2}\operatorname {det} (\mathbf {b} )=J^{-2}\operatorname {det} (\mathbf {F} )^{2}=J^{-2}J^{2}=1\,.}$

Its invariants are also marked with a slash:

${\displaystyle {\bar {\mathrm {I} }}_{1,2}(\mathbf {b} ):=\mathrm {I} _{1,2}({\bar {\mathbf {b} }})\,.}$

The following chapters explain these variants in detail. In the case of incompressibility, the first two formulations are equivalent. Because there is no dependence on the third main invariant or J , a separate section is devoted to incompressible isotropic hyperelasticity. The effort for the division into unimodular and volumetric parts, which characterizes the second formulation, is only worthwhile in the case of compressibility. The third formulation with the eigenvalues ​​can be applied equally to compressibility and incompressibility.

Isotropic compressible hyperelasticity

Use of the main invariants of b

In the case of compressibility, the deformation energy depends on all three main invariants. The following table gives valid derivatives of these invariants and the deformation energy for symmetric tensors.

Derivatives of the invariants

With the derivatives valid for symmetric tensors ${\displaystyle {\frac {\mathrm {dI} _{1}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}=\mathbf {1} }$ ${\displaystyle {\frac {\mathrm {dI} _{2}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}=\mathrm {I} _{1}(\mathbf {b} )\mathbf {1} -\mathbf {b} }$ ${\displaystyle {\frac {\mathrm {dI} _{3}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}=\mathrm {I} _{3}(\mathbf {b} )\mathbf {b} ^{-1}}$ the derivation of the deformation energy is calculated: Using Cayley-Hamilton's theorem : results ${\displaystyle {\frac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}=\left({\frac {\partial w}{\partial \mathrm {I} _{1}}}+\mathrm {I} _{1}{\frac {\partial w}{\partial \mathrm {I} _{2}}}+\mathrm {I} _{2}{\frac {\partial w}{\partial \mathrm {I} _{3}}}\right)\mathbf {1} -\left({\frac {\partial w}{\partial \mathrm {I} _{2}}}+\mathrm {I} _{1}{\frac {\partial w}{\partial \mathrm {I} _{3}}}\right)\mathbf {b} +{\frac {\partial w}{\partial \mathrm {I} _{3}}}\mathbf {b\cdot b} }$ ${\displaystyle {\begin{array}{lcl}\mathbf {b} ^{3}&=&\mathrm {I} _{1}\mathbf {b} ^{2}-\mathrm {I} _{2}\mathbf {b} +\mathrm {I} _{3}\mathbf {1} \\\mathbf {b} ^{2}&=&\mathrm {I} _{1}\mathbf {b} -\mathrm {I} _{2}\mathbf {1} +\mathrm {I} _{3}\mathbf {b} ^{-1}\\\mathrm {I} _{3}\mathbf {b} ^{-1}&=&\mathbf {b} ^{2}-\mathrm {I} _{1}\mathbf {b} +\mathrm {I} _{2}\mathbf {1} \end{array}}}$ ${\displaystyle {\begin{array}{lcl}\mathbf {b} \cdot {\frac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}&=&\left({\frac {\partial w}{\partial \mathrm {I} _{1}}}+\mathrm {I} _{1}{\frac {\partial w}{\partial \mathrm {I} _{2}}}+\mathrm {I} _{2}{\frac {\partial w}{\partial \mathrm {I} _{3}}}\right)\mathbf {b} -\left({\frac {\partial w}{\partial \mathrm {I} _{2}}}+\mathrm {I} _{1}{\frac {\partial w}{\partial \mathrm {I} _{3}}}\right)\mathbf {b\cdot b} +{\frac {\partial w}{\partial \mathrm {I} _{3}}}\left(\mathrm {I} _{1}\mathbf {b} ^{2}-\mathrm {I} _{2}\mathbf {b} +\mathrm {I} _{3}\mathbf {1} \right)\\&=&\mathrm {I} _{3}{\frac {\partial w}{\partial \mathrm {I} _{3}}}\mathbf {1} +\left({\frac {\partial w}{\partial \mathrm {I} _{1}}}+\mathrm {I} _{1}{\frac {\partial w}{\partial \mathrm {I} _{2}}}\right)\mathbf {b} -{\frac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {b\cdot b} \\&=&\left(\mathrm {I} _{2}{\frac {\partial w}{\partial \mathrm {I} _{2}}}+{\frac {\partial w}{\partial \mathrm {I} _{3}}}\mathrm {I} _{3}\right)\mathbf {1} +{\frac {\partial w}{\partial \mathrm {I} _{1}}}\mathbf {b} -\mathrm {I} _{3}{\frac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {b} ^{-1}\end{array}}\,.}$

As a result, the Cauchy stresses add up

${\displaystyle {\begin{array}{lcl}{\boldsymbol {\sigma }}&=&2\rho \mathbf {b} \cdot {\dfrac {\mathrm {d} w(\mathrm {I} _{1},\mathrm {I} _{2},\mathrm {I} _{3})}{\mathrm {d} \mathbf {b} }}=2{\dfrac {\rho _{0}}{J}}\mathbf {b} \cdot {\dfrac {\mathrm {d} w(\mathrm {I} _{1},\mathrm {I} _{2},\mathrm {I} _{3})}{\mathrm {d} \mathbf {b} }}\\[2ex]&=&2\rho \mathrm {I} _{3}{\dfrac {\partial w}{\partial \mathrm {I} _{3}}}\mathbf {1} +2\rho \left({\dfrac {\partial w}{\partial \mathrm {I} _{1}}}+\mathrm {I} _{1}{\dfrac {\partial w}{\partial \mathrm {I} _{2}}}\right)\mathbf {b} -2\rho {\dfrac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {b\cdot b} \\[2ex]&=&2\rho \left(\mathrm {I} _{2}{\dfrac {\partial w}{\partial \mathrm {I} _{2}}}+{\dfrac {\partial w}{\partial \mathrm {I} _{3}}}\mathrm {I} _{3}\right)\mathbf {1} +2\rho {\dfrac {\partial w}{\partial \mathrm {I} _{1}}}\mathbf {b} -2\rho \mathrm {I} _{3}{\dfrac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {b} ^{-1}\end{array}}\,.}$

Division into unimodular and volumetric parts

In the case of compressibility, the invariants of

${\displaystyle {\bar {\mathbf {b} }}=J^{\frac {-2}{3}}\mathbf {b} \rightarrow \operatorname {det} ({\bar {\mathbf {b} }})=1}$

which has the advantage that the volumetric spherical part and the unimodular, shape-changing part can be modeled separately. It then becomes the invariants

${\displaystyle {\begin{array}{lcl}{\bar {\mathrm {I} }}_{1}(\mathbf {b} )&:=&\mathrm {Sp} ({\bar {\mathbf {b} }})=J^{\frac {-2}{3}}\mathrm {Sp} (\mathbf {b} )\\[1ex]{\bar {\mathrm {I} }}_{2}(\mathbf {b} )&:=&{\frac {1}{2}}[\mathrm {Sp} {({\bar {\mathbf {b} }})}^{2}-\mathrm {Sp} ({\bar {\mathbf {b} }}\cdot {\bar {\mathbf {b} }})]=J^{\frac {-4}{3}}\mathrm {I} _{2}(\mathbf {b} )\\[1ex]J&=&{\sqrt {\mathrm {I} _{3}(\mathbf {b} )}}={\sqrt {\operatorname {det} (\mathbf {b} )}}\end{array}}}$

used. The following table gives valid derivatives of these invariants and the deformation energy for symmetric tensors.

Derivatives of the main invariants for compressibility

The derivatives of the invariants are: ${\displaystyle {\frac {\mathrm {d} J}{\mathrm {d} \mathbf {b} }}={\frac {\mathrm {d} {\sqrt {\operatorname {det} (\mathbf {b} )}}}{\mathrm {d} \mathbf {b} }}={\frac {J}{2}}\mathbf {b} ^{-1}}$ ${\displaystyle {\frac {\mathrm {d} {\bar {\mathrm {I} }}_{1}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}={\frac {\mathrm {d} }{\mathrm {d} \mathbf {b} }}\left(J^{\frac {-2}{3}}\mathrm {I} _{1}(\mathbf {b} )\right)=J^{\frac {-2}{3}}\mathbf {1} -{\frac {{\bar {\mathrm {I} }}_{1}}{3}}\mathbf {b} ^{-1}}$ ${\displaystyle {\frac {\mathrm {d} {\bar {\mathrm {I} }}_{2}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}={\frac {\mathrm {d} }{\mathrm {d} \mathbf {b} }}\left(J^{\frac {-4}{3}}\mathrm {I} _{2}(\mathbf {b} )\right)=J^{\frac {-2}{3}}{\bar {\mathrm {I} }}_{1}\mathbf {1} -J^{\frac {-4}{3}}\mathbf {b} -{\frac {2}{3}}{\bar {\mathrm {I} }}_{2}\mathbf {b} ^{-1}}$ It follows from this: because according to Cayley-Hamilton's theorem, is ${\displaystyle {\begin{array}{lcl}\mathbf {b} \cdot {\frac {\mathrm {d} w({\bar {\mathrm {I} }}_{1},{\bar {\mathrm {I} }}_{2},J)}{\mathrm {d} \mathbf {b} }}&=&\mathbf {b} \cdot \left[{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{1}}}\left(J^{\frac {-2}{3}}\mathbf {1} -{\frac {{\bar {\mathrm {I} }}_{1}}{3}}\mathbf {b} ^{-1}\right)+{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}\left(J^{\frac {-2}{3}}{\bar {\mathrm {I} }}_{1}\mathbf {1} -J^{\frac {-4}{3}}\mathbf {b} -{\frac {2}{3}}{\bar {\mathrm {I} }}_{2}\mathbf {b} ^{-1}\right)+{\frac {\partial w}{\partial J}}{\frac {J}{2}}\mathbf {b} ^{-1}\right]\\&=&\left({\frac {J}{2}}{\frac {\partial w}{\partial J}}-{\frac {{\bar {\mathrm {I} }}_{1}}{3}}{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{1}}}-{\frac {2}{3}}{\bar {\mathrm {I} }}_{2}{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}\right)\mathbf {1} +J^{\frac {-2}{3}}\left({\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{1}}}+{\bar {\mathrm {I} }}_{1}{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}\right)\mathbf {b} -J^{\frac {-4}{3}}{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}\mathbf {b\cdot b} \\&=&\left({\frac {J}{2}}{\frac {\partial w}{\partial J}}-{\frac {{\bar {\mathrm {I} }}_{1}}{3}}{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{1}}}+{\frac {1}{3}}{\bar {\mathrm {I} }}_{2}{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}\right)\mathbf {1} +J^{\frac {-2}{3}}{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{1}}}\mathbf {b} -J^{\frac {2}{3}}{\frac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}\mathbf {b} ^{-1}\end{array}}}$ ${\displaystyle \mathbf {b} ^{2}=\mathrm {I} _{1}\mathbf {b} -\mathrm {I} _{2}\mathbf {1} +\mathrm {I} _{3}\mathbf {b} ^{-1}\,.}$

The result is the Cauchy stress

${\displaystyle {\begin{array}{lcl}{\boldsymbol {\sigma }}&=&2\rho \mathbf {b} \cdot {\dfrac {\mathrm {d} w({\bar {\mathrm {I} }}_{1},{\bar {\mathrm {I} }}_{2},J)}{\mathrm {d} \mathbf {b} }}=2{\dfrac {\rho _{0}}{J}}\mathbf {b} \cdot {\dfrac {\mathrm {d} w({\bar {\mathrm {I} }}_{1},{\bar {\mathrm {I} }}_{2},J)}{\mathrm {d} \mathbf {b} }}\\[2ex]&=&\rho \left(J{\dfrac {\partial w}{\partial J}}-{\dfrac {2{\bar {\mathrm {I} }}_{1}}{3}}{\dfrac {\partial w}{\partial {\bar {\mathrm {I} }}_{1}}}-{\dfrac {4{\bar {\mathrm {I} }}_{2}}{3}}{\dfrac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}\right)\mathbf {1} +2\rho \left({\dfrac {\partial w}{\partial {\bar {\mathrm {I} }}_{1}}}+{\bar {\mathrm {I} }}_{1}{\dfrac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}\right){\bar {\mathbf {b} }}-2\rho {\dfrac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}{\bar {\mathbf {b} }}\cdot {\bar {\mathbf {b} }}\\[2ex]&=&\rho \left(J{\dfrac {\partial w}{\partial J}}-{\dfrac {2{\bar {\mathrm {I} }}_{1}}{3}}{\dfrac {\partial w}{\partial {\bar {\mathrm {I} }}_{1}}}+{\dfrac {2{\bar {\mathrm {I} }}_{2}}{3}}{\dfrac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}\right)\mathbf {1} +2\rho {\dfrac {\partial w}{\partial {\bar {\mathrm {I} }}_{1}}}{\bar {\mathbf {b} }}-2\rho {\dfrac {\partial w}{\partial {\bar {\mathrm {I} }}_{2}}}{\bar {\mathbf {b} }}^{-1}\end{array}}\,.}$

Isotropic incompressible hyperelasticity

In the case of incompressibility, there is no dependency on J because J is constantly equal to one. Hence only the main invariants

${\displaystyle {\begin{aligned}\mathrm {I} _{1}(\mathbf {b} )=&\mathrm {Sp} (\mathbf {b} )\\\mathrm {I} _{2}(\mathbf {b} )=&{\frac {1}{2}}[\mathrm {Sp} {(\mathbf {b} )}^{2}-\mathrm {Sp} (\mathbf {b\cdot b} )]\end{aligned}}}$

used. The following table gives valid derivatives of these invariants and the deformation energy for symmetric tensors.

Derivatives of the main invariants in the case of incompressibility

The derivatives of the two main invariants for symmetric tensors are: ${\displaystyle {\frac {\mathrm {dI} _{1}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}={\frac {\mathrm {d} \mathrm {Sp} (\mathbf {b} )}{\mathrm {d} \mathbf {b} }}=\mathbf {1} }$ ${\displaystyle {\frac {\mathrm {dI} _{2}(\mathbf {b} )}{\mathrm {d} \mathbf {b} }}=\mathrm {I} _{1}(\mathbf {b} )\mathbf {1} -\mathbf {b} }$ It follows: With Cayley-Hamilton's theorem in the case of incompressibility results ${\displaystyle \mathbf {b} \cdot {\frac {\mathrm {d} w(\mathrm {I} _{1},\mathrm {I} _{2})}{\mathrm {d} \mathbf {b} }}=\mathbf {b} \cdot \left({\frac {\partial w}{\partial \mathrm {I} _{1}}}\mathbf {1} +{\frac {\partial w}{\partial \mathrm {I} _{2}}}(\mathrm {I} _{1}\mathbf {1} -\mathbf {b} )\right)=\left({\frac {\partial w}{\partial \mathrm {I} _{1}}}+\mathrm {I} _{1}{\frac {\partial w}{\partial \mathrm {I} _{2}}}\right)\mathbf {b} -{\frac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {b\cdot b} }$ ${\displaystyle \mathbf {b\cdot b} =\mathrm {I} _{1}\mathbf {b} -\mathrm {I} _{2}\mathbf {1} +\mathbf {b} ^{-1}}$ ${\displaystyle \mathbf {b} \cdot {\frac {\mathrm {d} w(\mathrm {I} _{1},\mathrm {I} _{2})}{\mathrm {d} \mathbf {b} }}={\frac {\partial w}{\partial \mathrm {I} _{1}}}\mathbf {b} +\mathrm {I} _{2}{\frac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {1} -{\frac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {b} ^{-1}}$

The representation results from this

${\displaystyle {\boldsymbol {\sigma }}=-p\mathbf {1} +2\rho _{0}\mathbf {b} {\frac {\mathrm {d} w(\mathrm {I} _{1},\mathrm {I} _{2})}{\mathrm {d} \mathbf {b} }}=-p\mathbf {1} +2\rho _{0}\left({\frac {\partial w}{\partial \mathrm {I} _{1}}}+\mathrm {I} _{1}{\frac {\partial w}{\partial \mathrm {I} _{2}}}\right)\mathbf {b} -2\rho _{0}{\frac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {b\cdot b} }$

or

${\displaystyle {\boldsymbol {\sigma }}=-p\mathbf {1} +2\rho _{0}{\frac {\partial w}{\partial \mathrm {I} _{1}}}\mathbf {b} -2\rho _{0}{\frac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {b} ^{-1}}$

where the term was added to the indefinite part of the sphere - p 1 . ${\displaystyle 2\rho _{0}\mathrm {I} _{2}{\frac {\partial w}{\partial \mathrm {I} _{2}}}\mathbf {1} }$

Use of the eigenvalues ​​of the left distance tensor

The eigenvalues λ _{1,2,3 of} the left distance tensor v can also be used as invariants, both in the case of compressibility and in the case of incompressibility. The following table gives valid derivatives of the eigenvalues ​​and the deformation energy for symmetric tensors.

Derivations of the eigenvalues ​​of v  ­

The eigenvalues ​​of are the squares of the eigenvalues of but both tensors have the same eigenvectors which are normalized to the absolute value one and are therefore noted with hats. The eigenvectors are pairwise perpendicular to each other or orthogonalizable because and are symmetrical. The derivation of the eigenvalues ​​results in ${\displaystyle \mathbf {b} =\mathbf {v\cdot v} }$ ${\displaystyle \lambda _{i}}$ ${\displaystyle \mathbf {v} }$ ${\displaystyle {\hat {v}}_{1,2,3}}$ ${\displaystyle \mathbf {b} }$ ${\displaystyle \mathbf {v} }$ ${\displaystyle {\frac {\mathrm {d} \lambda _{i}}{\mathrm {d} \mathbf {b} }}={\frac {1}{2\lambda _{i}}}{\hat {v}}_{i}\otimes {\hat {v}}_{i}}$ (no sum, see distance tensor ). The arithmetic symbol " " calculates the dyadic product . The spectral decomposition results in: ${\displaystyle \otimes }$ ${\displaystyle \mathbf {b} =\sum _{i=1}^{3}\lambda _{i}^{2}{\hat {v}}_{i}\otimes {\hat {v}}_{i}}$ ${\displaystyle 2\mathbf {b} \cdot {\frac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}=2\mathbf {b} \cdot \left(\sum _{i=1}^{3}{\frac {\partial w}{\partial \lambda _{i}}}{\frac {\mathrm {d} \lambda _{i}}{\mathrm {d} \mathbf {b} }}\right)=\sum _{i=1}^{3}\lambda _{i}{\frac {\partial w}{\partial \lambda _{i}}}{\hat {v}}_{i}\otimes {\hat {v}}_{i}}$

So when using the eigenvalues

${\displaystyle {\boldsymbol {\sigma }}=2\rho \mathbf {b} \cdot {\frac {\mathrm {d} w}{\mathrm {d} \mathbf {b} }}=\rho \sum _{i=1}^{3}\lambda _{i}{\frac {\partial w}{\partial \lambda _{i}}}{\hat {v}}_{i}\otimes {\hat {v}}_{i}=\rho _{0}\sum _{i=1}^{3}{\frac {\lambda _{i}}{J}}{\frac {\partial w}{\partial \lambda _{i}}}{\hat {v}}_{i}\otimes {\hat {v}}_{i}\,.}$

The arithmetic symbol " " calculates the dyadic product and are the eigenvectors of normalized to one . In the case of incompressibility can be added ${\displaystyle \otimes }$ ${\displaystyle {\hat {v}}_{1,2,3}}$ ${\displaystyle \mathbf {b} }$

${\displaystyle \lambda _{1}\lambda _{2}\lambda _{3}=\operatorname {det} (\mathbf {F} )=1\quad \rightarrow \quad \lambda _{3}={\frac {1}{\lambda _{1}\lambda _{2}}}}$

can be used.

Special deformation energy functions

Some common strain energy functions are presented below.

Linear hyperelasticity

The specific deformation energy that leads to Hooke's law is

${\displaystyle {\begin{array}{rcl}\rho _{0}w_{0}(\mathbf {E} )&=&G\left[\mathbf {E} :\mathbf {E} +{\frac {\nu }{1-2\nu }}\mathrm {Sp} {(\mathbf {E} )}^{2}\right]\\\rightarrow {\tilde {\mathbf {T} }}&=&\rho _{0}{\dfrac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {E} }}=2G\left[\mathbf {E} +{\frac {\nu }{1-2\nu }}\mathrm {Sp} (\mathbf {E} )\mathbf {1} \right]\end{array}}\,.}$

The material parameter is the shear modulus and the Poisson's ratio . The constant and symmetrical elasticity tensor of the fourth level is calculated from the second derivative according to the distortions ${\displaystyle G}$ ${\displaystyle \nu }$ ${\displaystyle \mathbf {E} }$

${\displaystyle {\begin{array}{rcl}\mathbb {C} &:=&{\dfrac {\mathrm {d} {\tilde {\mathbf {T} }}}{\mathrm {d} \mathbf {E} }}=2G\mathbb {I} +\lambda \mathbf {1\otimes 1} =\mathbb {C} ^{\top }\\\rightarrow {\tilde {\mathbf {T} }}&=&\mathbb {C} :\mathbf {E} \quad \rightarrow \quad w_{0}={\frac {1}{2\rho _{0}}}\mathbf {E} :\mathbb {C} :\mathbf {E} \end{array}}}$

with the unit tensor of the fourth order and the Lamé constant ${\displaystyle \mathbb {I} }$

${\displaystyle \lambda ={\frac {2G\nu }{1-2\nu }}\,.}$

This model generalizes Hooke's law to large deformations, but only provides physically plausible answers for moderate strains. The elongation, which corresponds to the compression of a rod to zero length, results, for example, in vanishing Cauchy stresses. However, it approximates any model of hyperelasticity in the case of small strains in the first order.

Mooney Rivlin model

Stresses in the Mooney-Rivlin model under uniaxial tension as a function of the elongation and the material parameter , see #example below. ${\displaystyle a}$ ${\displaystyle \beta =b}$

The Mooney-Rivlin model provides a second-order approximation for incompressible hyperelastic bodies

${\displaystyle \rho _{0}w={\frac {G}{2}}\left[\left({\frac {1}{2}}+\beta \right)(\mathrm {I} _{1}-3)+\left({\frac {1}{2}}-\beta \right)(\mathrm {I} _{2}-3)\right]}$

${\displaystyle \rightarrow {\boldsymbol {\sigma }}=-p\mathbf {1} +G\left[\left(\beta +{\frac {1}{2}}\right)\mathbf {b} +\left(\beta -{\frac {1}{2}}\right)\mathbf {b} ^{-1}\right]}$

. The invariant part of the Stretches sensor , the parameter is the shear modulus and the dimensionless code with ${\displaystyle \mathrm {I} _{1,2}}$ ${\displaystyle \mathbf {b} }$ ${\displaystyle G>0}$ ${\displaystyle \beta }$

${\displaystyle {\frac {-1}{2}}\leq \beta \leq {\frac {1}{2}}}$

represents second order effects. Often the parameters

${\displaystyle C_{10}=G\left({\frac {1}{2}}+\beta \right),C_{01}=G\left({\frac {1}{2}}-\beta \right)\,,}$

used, both of which must not be negative.

Neo-Hooke model

The Neo-Hooke model is the special case

${\displaystyle \beta ={\frac {1}{2}}\quad \Leftrightarrow \quad C_{01}=0}$

of the Mooney-Rivlin material:

${\displaystyle \rho _{0}w={\frac {C_{10}}{2}}(\mathrm {I} _{1}-3)\rightarrow {\boldsymbol {\sigma }}=-p\mathbf {1} +C_{10}\mathbf {b} \,.}$

By adding a volumetric term and using the invariant instead , a plausible generalization of Hooke's law of matter for compressible elastomers arises even with large elongations: ${\displaystyle {\bar {\mathrm {I} }}_{1}(\mathbf {b} )}$ ${\displaystyle \mathrm {I} _{1}(\mathbf {b} )}$

${\displaystyle \rho _{0}w={\frac {C_{10}}{2}}({\bar {\mathrm {I} }}_{1}-3)+{\frac {K}{2}}\ln {(J)}^{2}\rightarrow {\boldsymbol {\sigma }}=K{\frac {\ln(J)}{J}}\mathbf {1} +{\frac {C_{10}}{J}}\left({\bar {\mathbf {b} }}-{\frac {{\bar {\mathrm {I} }}_{1}}{3}}\mathbf {1} \right)\,.}$

The parameter controls the compressibility . However, other formulations have also been proposed for the volumetric fraction. ${\displaystyle K}$

Ogden model

The Ogden model is a model for incompressible hyperelasticity. This model is formulated in the eigenvalues ​​of the left distance tensor:

${\displaystyle \rho _{0}w=\sum _{i=1}^{n}{\frac {\mu _{i}}{\alpha _{i}}}[\operatorname {Sp} (\mathbf {v} ^{\alpha _{i}})-3]=\sum _{i=1}^{n}{\frac {\mu _{i}}{\alpha _{i}}}(\lambda _{1}^{\alpha _{i}}+\lambda _{2}^{\alpha _{i}}+\lambda _{3}^{\alpha _{i}}-3)\,.}$

The numbers μ _i and α _i represent the 2 n material parameters of this model and the power of a tensor is calculated like a function value.

The special case n = 1 and α ₁ = 2 results in the Neo-Hooke model and n = 2, α ₁ = 2 and α ₂ = −2 form the Mooney-Rivlin model.

Approximation with Taylor polynomials

If w (I ₁ , I ₂ ) at the position I ₁ = 3 and I ₂ = 3, which is the case in the undeformed position at F = 1 , is approximated by a Taylor polynomial , the result is:

${\displaystyle \rho _{0}w=\sum _{i+j=1}^{n}C_{ij}(\mathrm {I} _{1}-3)^{i}(\mathrm {I} _{2}-3)^{j}\,.}$

The numbers C _ij are material parameters. The Neo-Hooke and Mooney-Rivlin models are also included as special cases in this model.

Anisotropic hyperelasticity

In the case of a transversely isotropic material , such as e.g. B. unidirectionally reinforced plastic , a sample can be rotated arbitrarily around the fiber direction, but perpendicular to the fiber only by 180 °, without changing the deformation energy at a given strain. These rotations can be grouped into a set . If two turns off are performed in a row, one turn off is obtained again. With the 0 ° rotation as a neutral element and the reverse rotation as an inverse element, one group represents: the symmetry group of the material. ${\displaystyle {\mathcal {G}}}$ ${\displaystyle {\mathcal {G}}}$ ${\displaystyle {\mathcal {G}}}$ ${\displaystyle {\mathcal {G}}}$

In general, the directional dependence of a material is described by the symmetry group of the material. This group includes all rotations that may take place in the material point without the deformation energy changing with a given elongation. Rotations are mathematically orthogonal tensors (with above). Correspondingly, the symmetry group of the material is the set of orthogonal tensors that is defined by: ${\displaystyle \mathbf {Q} }$ ${\displaystyle \mathbf {Q} \cdot \mathbf {Q} ^{\top }=\mathbf {1} }$ ${\displaystyle {\mathcal {G}}}$ ${\displaystyle \mathbf {Q} }$

${\displaystyle \mathbf {Q} \in {\mathcal {G}}\quad \Leftrightarrow \quad w_{0}(\mathbf {E} )=w_{0}(\mathbf {Q\cdot E\cdot Q} ^{\top })\,.}$

A specific deformation energy with this property can be formulated with tensorial structure variables: ${\displaystyle \mathbf {M} _{i}}$

${\displaystyle w_{0}=w_{0}(\mathbf {E} ,\mathbf {M} _{1},\mathbf {M} _{2},\ldots )\,.}$

However, it is generally not easy to find these structural variables, which can be second, fourth or sixth order tensors.

In the case of transverse isotropy, however, it is easy, because one structure variable is sufficient.

Transversely isotropic hyperelasticity

→ Main article : Transverse isotropy

Illustrative explanation of the transverse isotropy.
The material is rotationally symmetrical with respect to the 1 axis, which is perpendicular to the isotropic 2-3 plane.
A round bar made of this material oriented in this way can be rotated around its longitudinal axis without changing its properties.

In the case of transversely isotropic hyperelasticity, the material has a preferred direction, the 1-direction in the image, in which the material has different properties than perpendicular to it. The material behaves isotropically in the 2-3 plane perpendicular to the preferred direction. The preferred direction is defined with a material line element of length one. The associated structure variable is the symmetric tensor ${\displaystyle {\hat {a}}}$

${\displaystyle \mathbf {A} :={\hat {a}}\otimes {\hat {a}}\,.}$

The specific deformation energy then depends on the five invariants

${\displaystyle {\begin{array}{lcl}\mathrm {I} _{1}=\mathrm {Sp} (\mathbf {E} )&\rightarrow &{\dfrac {\mathrm {dI} _{1}}{\mathrm {d} \mathbf {E} }}=\mathbf {1} \\[2ex]\mathrm {J} _{2}=\mathrm {I} _{1}(\mathbf {E\cdot E} )&\rightarrow &{\dfrac {\mathrm {d} \mathrm {J} _{2}}{\mathrm {d} \mathbf {E} }}=\mathbf {E} \\[2ex]\mathrm {I} _{3}=\mathrm {det} (\mathbf {E} )&\rightarrow &{\dfrac {\mathrm {dI} _{3}}{\mathrm {d} \mathbf {E} }}=\mathrm {I} _{3}\mathbf {E} ^{-1}\\[2ex]\mathrm {I} _{4}=\mathbf {A} :\mathbf {E} &\rightarrow &{\dfrac {\mathrm {dI} _{4}}{\mathrm {d} \mathbf {E} }}=\mathbf {A} \\[2ex]\mathrm {I} _{5}=\mathbf {A} :(\mathbf {E\cdot E} )&\rightarrow &{\dfrac {\mathrm {dI} _{5}}{\mathrm {d} \mathbf {E} }}=\mathbf {A\cdot E} +\mathbf {E\cdot A} \end{array}}}$

from. The second Piola-Kirchhoff tensor can then be derived from

${\displaystyle {\begin{aligned}{\dfrac {1}{\rho _{0}}}{\tilde {\mathbf {T} }}=&{\dfrac {\mathrm {d} w_{0}(\mathrm {I} _{1},\mathrm {J} _{2},\mathrm {I} _{3},\mathrm {I} _{4},\mathrm {I} _{5})}{\mathrm {d} \mathbf {E} }}\\=&{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{1}}}\mathbf {1} +{\dfrac {\partial w_{0}}{\partial \mathrm {J} _{2}}}\mathbf {E} +\mathrm {I} _{3}{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{3}}}\mathbf {E} ^{-1}+{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{4}}}\mathbf {A} +{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{5}}}(\mathbf {A\cdot E} +\mathbf {E\cdot A} )\end{aligned}}}$

be calculated.

In the case of the transversely isotropic linear elasticity, the distortions occur in quadratics: ${\displaystyle w_{0}}$

${\displaystyle \rho _{0}w(\mathbf {E} )={\frac {\lambda }{2}}\mathrm {I} _{1}^{2}+2G_{23}\mathrm {J} _{2}+\mu \mathrm {I} _{1}\mathrm {I} _{4}+2(G_{12}-G_{23})\mathrm {I} _{5}+{\frac {\eta }{2}}\mathrm {I} _{4}^{2}}$

The material parameters in

${\displaystyle {\begin{aligned}\lambda =&{\frac {(\nu _{12}\nu _{21}+\nu _{23})E_{2}}{(1-\nu _{23}-2\nu _{12}\nu _{21})(1+\nu _{23})}}\\\mu =&2\nu _{12}(\lambda +G_{23})-\lambda \\\eta =&{\frac {(1-\nu _{23})E_{1}}{1-\nu _{23}-2\nu _{12}\nu _{21}}}-\lambda -2\mu +2G_{23}-4G_{12}\end{aligned}}}$

and G ₁₂ and G ₂₃ are explained on the page Transverse Isotropy . The second Piola-Kirchhoff stresses are after the derivation of the deformation energy

${\displaystyle {\begin{aligned}{\tilde {\mathbf {T} }}=&\rho _{0}{\dfrac {\mathrm {d} w_{0}(\mathrm {I} _{1},\mathrm {J} _{2},\mathrm {I} _{3},\mathrm {I} _{4},\mathrm {I} _{5})}{\mathrm {d} \mathbf {E} }}\\=&(\lambda \mathrm {I} _{1}+\mu \mathrm {I} _{4})\mathbf {1} +2G_{23}\mathbf {E} +(\mu \mathrm {I} _{1}+\eta \mathrm {I} _{4})\mathbf {A} +2(G_{12}-G_{23})(\mathbf {A\cdot E} +\mathbf {E\cdot A} )\end{aligned}}}$

linear functions of the current strains. Another derivation after the distortions calculates the constant and symmetric elasticity tensor of the fourth level ${\displaystyle \mathbf {E} }$

${\displaystyle {\begin{aligned}\mathbb {C} :={\dfrac {\mathrm {d} {\tilde {\mathbf {T} }}}{\mathrm {d} \mathbf {E} }}=&2G_{23}\mathbb {I} +2(G_{12}-G_{23})\mathbb {A} +\lambda \mathbf {1\otimes 1} \\&+\mu (\mathbf {A\otimes 1} +\mathbf {1\otimes A} )+\eta \mathbf {A\otimes A} \\\rightarrow {\tilde {\mathbf {T} }}=&\mathbb {C} :\mathbf {E} \quad \rightarrow \quad w_{0}={\frac {1}{2\rho _{0}}}\mathbf {E} :\mathbb {C} :\mathbf {E} \end{aligned}}}$

The fourth order tensor is defined as ${\displaystyle \mathbb {A} }$

${\displaystyle {\begin{aligned}\mathbb {A} =&\sum _{i,j,k,l=1}^{3}(A_{ik}\delta _{jl}+A_{lj}\delta _{ik})({\hat {e}}_{i}\otimes {\hat {e}}_{j}\otimes {\hat {e}}_{k}\otimes {\hat {e}}_{l})\\\rightarrow \;\mathbb {A} :\mathbf {H} =&\mathbf {A\cdot H} +\mathbf {H\cdot A} \quad {\text{für alle}}\quad \mathbf {H} \end{aligned}}}$

The quantities are the components of the tensor with respect to the standard basis and is the Kronecker delta . ${\displaystyle A_{ij}}$ ${\displaystyle \mathbf {A} }$ ${\displaystyle {\hat {e}}_{1,2,3}}$ ${\displaystyle \delta _{ij}}$

Orthotropic hyperelasticity

→ Main article : Orthotropy

In the case of orthtropic hyperelasticity, the material has no tensile-shear coupling but three preferred directions, the pairwise perpendicular orthotropic axes in which the material has different properties than perpendicular to it. The symmetry group of this material includes every 180 ° rotations around one of these three axes. The structure variables are defined with the dyadic product of two material line elements of length one: ${\displaystyle {\hat {a}},{\hat {b}}}$

${\displaystyle \mathbf {A} :={\hat {a}}\otimes {\hat {a}}\quad {\textsf {und}}\quad \mathbf {B} :={\hat {b}}\otimes {\hat {b}}}$

whereby it is assumed here . In addition to the five invariants known from transverse isotropy, the specific deformation energy also depends on the invariants ${\displaystyle {\hat {a}}\cdot {\hat {b}}=0}$

${\displaystyle {\begin{array}{lclclcl}\mathrm {I} _{6}&=&\mathbf {B} :\mathbf {E} &\rightarrow &{\dfrac {\partial \mathrm {I} _{6}}{\partial \mathbf {E} }}&=&\mathbf {B} \\[2ex]\mathrm {I} _{7}&=&\mathbf {B} :(\mathbf {E\cdot E} )&\rightarrow &{\dfrac {\partial \mathrm {I} _{7}}{\partial \mathbf {E} }}&=&\mathbf {B\cdot E} +\mathbf {E\cdot B} \end{array}}}$

and two further invariants, which, however, make no contribution and do not need to be considered here. The second Piola-Kirchhoff tensor can now be obtained via the derivative ${\displaystyle {\hat {a}}\cdot {\hat {b}}=0}$

${\displaystyle {\begin{aligned}{\dfrac {1}{\rho _{0}}}{\tilde {\mathbf {T} }}=&{\dfrac {\mathrm {d} w_{0}(\mathrm {I} _{1},{\mathrm {J} }_{2},\mathrm {I} _{3},\mathrm {I} _{4},\mathrm {I} _{\mathrm {5,} }\mathrm {I} _{6},\mathrm {I} _{7})}{\mathrm {d} \mathbf {E} }}\\=&{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{1}}}\mathbf {1} +{\dfrac {\partial w_{0}}{\partial {\mathrm {J} }_{2}}}\mathbf {E} +\mathrm {I} _{3}{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{3}}}\mathbf {E} ^{-1}+{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{4}}}\mathbf {A} +{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{5}}}(\mathbf {A\cdot E} +\mathbf {E\cdot A} )\\&+{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{6}}}\mathbf {B} +{\dfrac {\partial w_{0}}{\partial \mathrm {I} _{7}}}(\mathbf {B\cdot E} +\mathbf {E\cdot B} )\end{aligned}}}$

be calculated.

In orthotropic linear elasticity, the distortions occur in quadratics: ${\displaystyle w_{0}}$

${\displaystyle \rho _{0}w_{0}=a\mathrm {J} _{2}+b\mathrm {I} _{5}+c\mathrm {I} _{7}+{\dfrac {d}{2}}\mathrm {I} _{1}^{2}+{\dfrac {e}{2}}\mathrm {I} _{4}^{2}+{\dfrac {f}{2}}\mathrm {I} _{6}^{2}+g\mathrm {I} _{1}\mathrm {I} _{4}+h\mathrm {I} _{4}\mathrm {I} _{6}+p\mathrm {I} _{1}\mathrm {I} _{6}}$

The coefficients and are nine material parameters of the model. The second Piola-Kirchhoff stresses are then calculated as follows: ${\displaystyle a,b,c,d,e,f,g,h}$ ${\displaystyle p}$

${\displaystyle {\begin{aligned}{\tilde {\mathbf {T} }}=&\rho _{0}{\dfrac {\mathrm {d} w_{0}}{\mathrm {d} \mathbf {E} }}\\=&(d\mathrm {I} _{1}+g\mathrm {I} _{4}+p\mathrm {I} _{6})\mathbf {1} +a\mathbf {E} +b(\mathbf {A\cdot E} +\mathbf {E\cdot A} )+c(\mathbf {B\cdot E} +\mathbf {E\cdot B} )\\&+(e\mathrm {I} _{4}+g\mathrm {I} _{1}+h\mathrm {I} _{6})\mathbf {A} +(f\mathrm {I} _{6}+h\mathrm {I} _{4}+p\mathrm {I} _{1})\mathbf {B} \end{aligned}}\,.}$

The constant and symmetrical elasticity tensor of the fourth level is calculated from the second derivative according to the distortions E

${\displaystyle {\begin{aligned}\mathbb {C} :={\dfrac {\mathrm {d} {\tilde {\mathbf {T} }}}{\mathrm {d} \mathbf {E} }}=&a\mathbb {I} +b\mathbb {A} +c\mathbb {B} +d\mathbf {1\otimes 1} +e\mathbf {A\otimes A} +f\mathbf {B\otimes B} \\&+g(\mathbf {A\otimes 1} +\mathbf {1\otimes A} )+h(\mathbf {A\otimes B} +\mathbf {B\otimes A} )+p(\mathbf {B\otimes 1} +\mathbf {1\otimes B} )\\\rightarrow {\tilde {\mathbf {T} }}=&\mathbb {C} :\mathbf {E} \quad \rightarrow \quad w_{0}={\frac {1}{2\rho _{0}}}\mathbf {E} :\mathbb {C} :\mathbf {E} \end{aligned}}}$

Analogous to the tensor , the fourth order tensor is defined as ${\displaystyle \mathbb {B} }$ ${\displaystyle \mathbb {A} }$

${\displaystyle {\begin{aligned}\mathbb {B} =&\sum _{i,j,k,l=1}^{3}(B_{ik}\delta _{jl}+B_{lj}\delta _{ik})({\hat {e}}_{i}\otimes {\hat {e}}_{j}\otimes {\hat {e}}_{k}\otimes {\hat {e}}_{l})\\\rightarrow \;\mathbb {B} :\mathbf {H} =&\mathbf {B\cdot H} +\mathbf {H\cdot B} \quad {\text{für alle}}\quad \mathbf {H} \end{aligned}}}$

The quantities B _ij are the components of the tensor B with respect to the standard basis . ${\displaystyle {\hat {e}}_{1,2,3}}$

If special and is available, in Voigt's notation is : ${\displaystyle {\hat {a}}={\hat {e}}_{1}}$ ${\displaystyle {\hat {b}}={\hat {e}}_{2}}$

${\displaystyle {\begin{bmatrix}{\tilde {\sigma }}_{11}\\{\tilde {\sigma }}_{22}\\{\tilde {\sigma }}_{33}\\{\tilde {\sigma }}_{23}\\{\tilde {\sigma }}_{13}\\{\tilde {\sigma }}_{12}\end{bmatrix}}=\underbrace {\begin{bmatrix}C_{11}&C_{12}&C_{13}&0&0&0\\&C_{22}&C_{23}&0&0&0\\&&C_{33}&0&0&0\\&&&G_{23}&0&0\\&{\text{sym}}&&&G_{13}&0\\&&&&&G_{12}\end{bmatrix}} _{=:C}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{13}\\2\varepsilon _{12}\end{bmatrix}}}$

The variables are the components of and the components of E with respect to the standard basis. With the coefficients ${\displaystyle {\tilde {\sigma }}_{ij}}$ ${\displaystyle {\tilde {\mathbf {T} }}}$ ${\displaystyle \varepsilon _{ij}}$

${\displaystyle {\begin{aligned}C_{11}=&(1-\nu _{23}\nu _{32}){\frac {E_{1}}{D}}=a+2b+d+e+2g\\C_{22}=&(1-\nu _{31}\nu _{13}){\frac {E_{2}}{D}}=a+2c+d+f+2p\\C_{33}=&(1-\nu _{12}\nu _{21}){\frac {E_{3}}{D}}=a+d\\C_{12}=&\nu _{13}\nu _{23}{\frac {E_{3}}{D}}+\nu _{12}{\frac {E_{2}}{D}}=d+g+h+p\\C_{23}=&\nu _{21}\nu _{31}{\frac {E_{1}}{D}}+\nu _{23}{\frac {E_{3}}{D}}=d+p\\C_{13}=&\nu _{32}\nu _{12}{\frac {E_{2}}{D}}+\nu _{31}{\frac {E_{1}}{D}}=d+g\\D=&1-\nu _{12}\nu _{21}-\nu _{23}\nu _{32}-\nu _{13}\nu _{31}-2\nu _{12}\nu _{23}\nu _{31}\end{aligned}}}$

respectively

${\displaystyle {\begin{aligned}a=&2(G_{23}+G_{13}-G_{12})\\b=&2(G_{12}-G_{23})\\c=&2(G_{12}-G_{13})\\d=&(1-\nu _{12}\nu _{21}){\frac {E_{3}}{D}}-2(G_{23}+G_{13}-G_{12})\\=&C_{33}-2(G_{23}+G_{13}-G_{12})\\e=&(1-\nu _{23}\nu _{32}-2\nu _{31}){\frac {E_{1}}{D}}-2\nu _{32}\nu _{12}{\frac {E_{2}}{D}}+(1-\nu _{12}\nu _{21}){\frac {E_{3}}{D}}-4G_{13}\\=&C_{11}+C_{33}-2C_{13}-4G_{13}\\f=&(1-\nu _{12}\nu _{21}-2\nu _{23}){\frac {E_{3}}{D}}-2\nu _{21}\nu _{31}{\frac {E_{1}}{D}}+(1-\nu _{13}\nu _{31}){\frac {E_{2}}{D}}-4G_{23}\\=&C_{22}+C_{33}-2C_{23}-4G_{23}\\g=&\nu _{31}{\frac {E_{1}}{D}}+\nu _{32}\nu _{12}{\frac {E_{2}}{D}}-(1-\nu _{12}\nu _{21}){\frac {E_{3}}{D}}+2(G_{23}+G_{13}-G_{12})\\=&C_{13}-C_{33}+2(G_{23}+G_{13}-G_{12})\\h=&(1-\nu _{12}\nu _{21}-(1-\nu _{13})\nu _{23}){\frac {E_{3}}{D}}+\nu _{12}(1-\nu _{32}){\frac {E_{2}}{D}}-\nu _{31}(1+\nu _{21}){\frac {E_{1}}{D}}-2(G_{23}+G_{13}-G_{12})\\=&C_{33}+C_{12}-C_{13}-C_{23}-2(G_{23}+G_{13}-G_{12})\\p=&\nu _{21}\nu _{31}{\frac {E_{1}}{D}}-(1-\nu _{12}\nu _{21}-\nu _{23}){\frac {E_{3}}{D}}+2(G_{23}+G_{13}-G_{12})\\=&C_{23}-C_{33}+2(G_{23}+G_{13}-G_{12})\end{aligned}}}$

the flexibility matrix of the orthotropic material is obtained by inverting the matrix C:

${\displaystyle C^{-1}={\begin{bmatrix}{\dfrac {1}{E_{1}}}&{\dfrac {-\nu _{12}}{E_{1}}}&{\dfrac {-\nu _{13}}{E_{1}}}&0&0&0\\&{\dfrac {1}{E_{2}}}&{\dfrac {-\nu _{23}}{E_{2}}}&0&0&0\\&&{\dfrac {1}{E_{3}}}&0&0&0\\&&&{\dfrac {1}{G_{23}}}&0&0\\&\mathrm {sym} &&&{\dfrac {1}{G_{13}}}&0\\&&&&&{\dfrac {1}{G_{12}}}\end{bmatrix}}}$

That’s included ${\displaystyle i,j=1,2,3}$

• E _{i is} the modulus of elasticity in direction i
• G _{ij is} the shear modulus in the ij plane
• ν _{ij is} the Poisson's ratio in direction j (effect) when loaded in i- direction (cause)

The dimensions of the modules are force per area while the Poisson's contraction numbers are dimensionless.

example

A block (green) is stretched with the factors a, b and c in the x, y or z direction (white)

A homogeneous block of incompressible hyperelastic material is stretched homogeneously with the factors in the x, y and z directions as in the picture . With the block remains in its original state. Then the deformation gradient, the left Cauchy-Green tensor and the left stretch tensor result ${\displaystyle a,b,c}$ ${\displaystyle a,b,c=1}$

${\displaystyle \mathbf {F} ={\begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}}\rightarrow \mathbf {b} ={\begin{pmatrix}a^{2}&0&0\\0&b^{2}&0\\0&0&c^{2}\end{pmatrix}}\rightarrow \mathbf {v} ={\begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}}}$

as well as the invariants and eigenvectors:

${\displaystyle {\begin{array}{lll}\mathrm {I} _{1}(\mathbf {b} )=a^{2}+b^{2}+c^{2}\,,&\mathrm {I} _{2}(\mathbf {b} )=a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}\,,&\mathrm {I} _{3}(\mathbf {b} )=a^{2}b^{2}c^{2}\\[1ex]{\bar {\mathrm {I} }}_{1}(\mathbf {b} )={\frac {a^{2}+b^{2}+c^{2}}{{\sqrt[{3}]{abc}}^{2}}}\,,&{\bar {\mathrm {I} }}_{2}(\mathbf {b} )={\frac {a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}}{{\sqrt[{3}]{abc}}^{4}}}\,,&J=abc\\\lambda _{1}=a\,,&\lambda _{2}=b\,,&\lambda _{3}=c\\{\hat {v}}_{1}={\begin{pmatrix}1\\0\\0\end{pmatrix}}\,,&{\hat {v}}_{2}={\begin{pmatrix}0\\1\\0\end{pmatrix}}\,,&{\hat {v}}_{3}={\begin{pmatrix}0\\0\\1\end{pmatrix}}\end{array}}\,.}$

Because of the assumed incompressibility , three tests are carried out with the elongations in the table for the data on the material models, see also the images below. ${\displaystyle abc=J=1,{\bar {\mathrm {I} }}_{1}=\mathrm {I} _{1},{\bar {\mathrm {I} }}_{2}=\mathrm {I} _{2}\,.}$

attempt	a	b	c
Uniaxial train	${\displaystyle a}$	${\displaystyle {\frac {1}{\sqrt {a}}}}$	${\displaystyle {\frac {1}{\sqrt {a}}}}$
Planar train	${\displaystyle a}$	${\displaystyle 1}$	${\displaystyle {\frac {1}{a}}}$
Biaxial pull	${\displaystyle a}$	${\displaystyle a}$	${\displaystyle {\frac {1}{a^{2}}}}$

The Mooney-Rivlin model then delivers

${\displaystyle {\begin{array}{rcl}\rho _{0}w&=&{\dfrac {C_{10}}{2}}(\mathrm {I} _{1}-3)+{\dfrac {C_{01}}{2}}(\mathrm {I} _{2}-3)\\[2ex]\rightarrow \rho _{0}{\dfrac {\mathrm {d} w}{\mathrm {dI} _{1}}}&=&{\dfrac {C_{10}}{2}}={\dfrac {G}{2}}\left({\dfrac {1}{2}}+\beta \right)\\[2ex]\rightarrow \rho _{0}{\dfrac {\mathrm {d} w}{\mathrm {dI} _{2}}}&=&{\dfrac {C_{01}}{2}}={\dfrac {G}{2}}\left({\dfrac {1}{2}}-\beta \right)\end{array}}}$

the tensions

${\displaystyle {\begin{array}{rcl}{\boldsymbol {\sigma }}&=&-p\mathbf {1} +2\rho _{0}{\dfrac {\mathrm {d} w}{\mathrm {dI} _{1}}}\mathbf {b} -2\rho _{0}{\dfrac {\mathrm {d} w}{\mathrm {dI} _{2}}}\mathbf {b} ^{-1}\\[3ex]\rightarrow {\begin{pmatrix}\sigma _{xx}&\sigma _{xy}&\sigma _{xz}\\\sigma _{xy}&\sigma _{yy}&\sigma _{yz}\\\sigma _{xz}&\sigma _{yz}&\sigma _{zz}\end{pmatrix}}&=&-p{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}+C_{10}{\begin{pmatrix}a^{2}&0&0\\0&b^{2}&0\\0&0&c^{2}\end{pmatrix}}-C_{01}{\begin{pmatrix}a^{-2}&0&0\\0&b^{-2}&0\\0&0&c^{-2}\end{pmatrix}}.\end{array}}}$

The pressure p is calculated from the boundary conditions present in the tests.

Uniaxial tensile test

A block (green) is stretched uniaxially in the x-direction (white)

Here you have

${\displaystyle b=c={\frac {1}{\sqrt {a}}}}$

and from the boundary condition σ _yy = σ _zz = 0 one determines

${\displaystyle p=C_{10}a^{-1}-C_{01}a}$

and receives so

${\displaystyle \sigma _{xx}=C_{10}(a^{2}-a^{-1})+C_{01}(a-a^{-2})\,.}$

Planar tensile test

A block (green) is stretched planar in the x-direction (white)

Here you have

${\displaystyle b=1\quad {\textsf {und}}\quad c={\frac {1}{a}}}$

and from the boundary condition σ _zz = 0 one determines

${\displaystyle p=C_{10}a^{-2}-C_{01}a^{2}}$

and receives so

${\displaystyle {\begin{aligned}\sigma _{xx}=&(C_{10}+C_{01})(a^{2}-a^{-2})\,,\\\sigma _{yy}=&C_{10}(1-a^{-2})+C_{01}(a^{2}-1)\,.\end{aligned}}}$

Biaxial tensile test

A block (green) is stretched biaxially in the x and y directions (white)

Here you have

${\displaystyle a=b\quad {\textsf {und}}\quad c={\frac {1}{a^{2}}}}$

and from the boundary condition σ _zz = 0 one determines

${\displaystyle p=C_{10}a^{-4}-C_{01}a^{4}}$

and receives so

${\displaystyle \sigma _{xx}=\sigma _{yy}=C_{10}(a^{2}-a^{-4})+C_{01}(a^{4}-a^{-2})\,.}$

See also

• Elasticity theory
• Continuum mechanics
• Potential (physics)
• Hypo-elasticity
• Formula collection tensor algebra
• Tensoranalysis formula collection

References and footnotes

1. ↑ ^{a b} e.g. in Holzapfel (2000).
2. ↑ ^{a b c d e f g h i j k l} The Frechet derivative of a function according to the limited linear operator of the - in all directions - as long as it exists the Gâteaux derivative corresponds, so ${\displaystyle f}$ ${\displaystyle x}$ ${\displaystyle {\mathcal {A}}}$ ${\displaystyle h}$

   ${\displaystyle {\mathcal {A}}(h)=\left.{\frac {\mathrm {d} }{\mathrm {d} s}}f(x+sh)\right|_{s=0}=\lim _{s\rightarrow 0}{\frac {f(x+sh)-f(x)}{s}}\quad {\text{für alle}}\quad h}$

   applies. In it is scalar, vector or tensor valued but and similar. Then will too ${\displaystyle s\in \mathbb {R} \,,f,x\,{\textsf {und}}\,h}$ ${\displaystyle x}$ ${\displaystyle h}$

   ${\displaystyle {\mathcal {A}}={\frac {\mathrm {d} f}{\mathrm {d} x}}}$

   written.
3. ↑ Haupt (2000), p. 326ff
4. ↑ PG Ciarlet (1988), Theorem 4.4-1
5. ↑ John M. Ball : Convexity conditions and existence theorems in non-linear elasticity , Archive for Rational Mechanics and Analysis 63 (1977), pp. 337-403
6. ↑ ^{a b} The function value f ( T ) of a tensor T is calculated with its principal axis transformation, formation of the function value of the diagonal elements and inverse transformation.
7. ↑ Parisch (2003), p. 164
8. ↑ Friedrich Gruttmann: Theory and numerics of thin-walled fiber composite structures . Habilitation thesis at the Faculty of Civil Engineering and Surveying at the University of Hanover, report no. F 96/1, University of Hanover, 1996 ( online ).
9. ↑ Holzapfel (2000), p274

literature

• Horst Parisch: Solid body continuum mechanics. From the basic equations to the solution with finite elements . Teubner, Stuttgart 2003, ISBN 3-519-00434-8 .
• Gerhard A. Holzapfel: Nonlinear Solid Mechanics. A Continuum Approach for Engineering . Wiley, Chichester 2010, ISBN 978-0-471-82319-3 . (EA Chichester 2000)
• Philippe Ciarlet : Mathematical Elasticity, Vol. 1: Three-Dimensional Elasticity . North-Holland, Amsterdam 1988, ISBN 0-444-70259-8 . (Studies in mathematics and its applications; Vol. 20)
• Peter Haupt: Continuum Mechanics and Theory of Materials . Springer, Berlin 2000, ISBN 3-540-43111-X .