Generalization of the ordinary state-based peridynamic model for isotropic linear viscoelasticity

Rolland Delorme; Ilyass Tabiai; Louis Laberge Lebel; Martin Lévesque

doi:10.1007/s11043-017-9342-3

Mechanics of Time-Dependent Materials

November 2017, Volume 21, Issue 4, pp 549–575 | Cite as

Generalization of the ordinary state-based peridynamic model for isotropic linear viscoelasticity

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Authors and affiliations

Rolland DelormeEmail author
Ilyass Tabiai
Louis Laberge Lebel
Martin Lévesque

Article

First Online: 15 February 2017

Received: 09 July 2016

Accepted: 30 January 2017

343 Downloads

Abstract

This paper presents a generalization of the original ordinary state-based peridynamic model for isotropic linear viscoelasticity. The viscoelastic material response is represented using the thermodynamically acceptable Prony series approach. It can feature as many Prony terms as required and accounts for viscoelastic spherical and deviatoric components. The model was derived from an equivalence between peridynamic viscoelastic parameters and those appearing in classical continuum mechanics, by equating the free energy densities expressed in both frameworks. The model was simplified to a uni-dimensional expression and implemented to simulate a creep-recovery test. This implementation was finally validated by comparing peridynamic predictions to those predicted from classical continuum mechanics. An exact correspondence between peridynamics and the classical continuum approach was shown when the peridynamic horizon becomes small, meaning peridynamics tends toward classical continuum mechanics. This work provides a clear and direct means to researchers dealing with viscoelastic phenomena to tackle their problem within the peridynamic framework.

Keywords

Linear viscoelasticity Relaxation Creep Peridynamics Classical continuum mechanics Thermodynamics

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Notes

Acknowledgements

Fruitful discussions with Marc Alexander Schweitzer, Serge Prud’homme and Patrick Diehl are gratefully acknowledged. The help with the mathematics given by Valentine Roos, Mathieu and Francisque Delorme is also gratefully acknowledged.

Appendix A: Peridynamics states

Let $\boldsymbol{x} \in \mathcal{B}$ and $\boldsymbol{x'} \in \mathcal{H}_{\boldsymbol{x}}$ denote two material points in the reference position. Let $\boldsymbol{\xi}$ be the vector defined by $\boldsymbol{\xi} = \boldsymbol{x'} - \boldsymbol{x}$ and called the bond connected to $\boldsymbol{x}$. The family ℋ of $\boldsymbol{x}$ is defined by

$$ \mathcal{H} = \bigl\lbrace \boldsymbol{\xi} \in \bigl( \mathbb{R}^{3} \setminus \boldsymbol{0} \bigr) \mid \bigl(\boldsymbol{ \xi} + \boldsymbol{x} = \boldsymbol{x'} \bigr) \in ( \mathcal{H}_{\boldsymbol{x}} \cap \mathcal{B} ) \bigr\rbrace . $$

(51)

ℋ and $\mathcal{H}_{\boldsymbol{x}}$ are schematized in Fig. 2. Let $\underline{a}$ be a scalar state such as

$$ \forall \boldsymbol{\xi} \in \mathcal{H}, \quad \underline{a} \langle \boldsymbol{\xi} \rangle \in \mathbb{R} . $$

(52)

Let $\underline{\mathbf{A}}$ be a vector state such as

$$ \forall \boldsymbol{\xi} \in \mathcal{H}, \quad \underline{ \mathbf{A}} \langle \boldsymbol{\xi} \rangle \in \mathbb{R}^{3} . $$

(53)

The identity state $\underline{\mathbf{X}}$ is a vector state such as

$$ \forall \boldsymbol{\xi} \in \mathcal{H}, \quad \underline{ \mathbf{X}} \langle \boldsymbol{\xi} \rangle = \boldsymbol{\xi} \in \mathcal{H} \subset \mathbb{R}^{3} . $$

(54)

Let $\vert \underline{\mathbf{A}} \vert$ be a magnitude state such as

$$ \forall \boldsymbol{\xi} \in \mathcal{H}, \quad \vert \underline{\mathbf{A}} \vert \langle \boldsymbol{\xi} \rangle = \sqrt{\underline{ \mathbf{A}} \langle \boldsymbol{\xi} \rangle \cdot \underline{\mathbf{A}} \langle \boldsymbol{\xi} \rangle} \in \mathbb{R} . $$

(55)

The dot products of two scalar states or two vectors states are

$$\begin{aligned} \underline{a} \cdot \underline{b} =& \displaystyle{ \int _{\mathcal{H}}} \underline{a} \langle \boldsymbol{\xi} \rangle \cdot \underline{b} \langle \boldsymbol{\xi} \rangle \mathrm{d}V_{\boldsymbol{\xi}} \, \in \mathbb{R} , \end{aligned}$$

(56)

$$\begin{aligned} \underline{\mathbf{A}} \cdot\underline{\mathbf{B}} =& \displaystyle{ \int _{\mathcal{H}}} \underline{\mathbf{A}} \langle \boldsymbol{\xi} \rangle \cdot \underline{\mathbf{B}} \langle \boldsymbol{\xi} \rangle \mathrm{d}V_{\boldsymbol{\xi}} \, \in \mathbb{R} . \end{aligned}$$

(57)

The norm of $\underline{\mathbf{A}}$ is a scalar defined by

$$ \Vert \underline{\mathbf{A}} \Vert = \sqrt{ (\underline{ \mathbf{A}} \cdot \underline{\mathbf{A}} )} . $$

(58)

A state field is defined as a state valued function of objects (e.g. position or time). The dependencies are in square brackets, e.g., for a vector state field $\underline{\mathbf{A}} [ \boldsymbol{x},t ]$.

The derivative of a function of states is generalized as the Fréchet derivative denoted $\triangledown\varPsi$. For instance, assume that $\varPsi$ is a scalar function of a vector state. Then $\triangledown\varPsi$ is defined as

$$ \varPsi(\underline{\mathbf{A}} + \Delta\underline{\mathbf{A}}) = \varPsi(\underline{\mathbf{A}}) + \triangledown\varPsi(\underline{\mathbf{A}}) \cdot \Delta\underline{\mathbf{A}} + o \bigl(\vert\vert \Delta\underline{\mathbf{A}} \vert \vert\bigr) $$

(59)

where $\Delta\underline{\mathbf{A}}$ is an increment added to $\underline{\mathbf{A}}$.

More details as regards states can be found in Silling et al. (2007) and Silling and Lehoucq (2010).

Appendix B: Constitutive theories in PD

B.1 Fundamental balance laws and material models

Fundamental balance laws in PD and their corresponding expressions in the CCM are recalled in Table 2. Refer to Silling et al. (2007) and Silling and Lehoucq (2010), for more details.

Table 2

Fundamental balance laws for an elastic material

Balance law	Peridynamics	Classical continuum mechanics (Infinitesimal strain theory)
Linear momentum	$\begin{array}{l} \rho(\boldsymbol{x}) \ddot{\boldsymbol{y}}(\boldsymbol{x},t) - \boldsymbol{b}(\boldsymbol{x},t) \\ \quad = \displaystyle{\int_{\mathcal{B}}} {\underline{\mathbf{T}}[\boldsymbol{x},t]} \langle \boldsymbol{x'} - \boldsymbol{x} \rangle - {\underline{\mathbf{T}}[\boldsymbol{x'},t]} \langle \boldsymbol{x} - \boldsymbol{x'} \rangle \, \mathrm{d}V_{\boldsymbol{x'}} \end{array}$	$\rho(\boldsymbol{x}) \ddot{\boldsymbol{y}}(\boldsymbol{x},t) - \boldsymbol{b}(\boldsymbol{x},t) = \mathbf{div} \,\boldsymbol{\sigma}(\boldsymbol{x},t)$
Angular momentum	$\displaystyle{\int_{\mathcal{H}}} {\underline{\mathbf{Y}}[\boldsymbol{x},t]} \langle \boldsymbol{x'} - \boldsymbol{x} \rangle \times {\underline{\mathbf{T}}[\boldsymbol{x},t]} \langle \boldsymbol{x'} - \boldsymbol{x} \rangle \, \mathrm{d}V_{\boldsymbol{x'}} = \boldsymbol{0}$	$\boldsymbol{\sigma} = {\boldsymbol{\sigma}}^{\top}$
First law of thermodynamics	$\dot{u} = \underline{\mathbf{T}} \cdot \underline{\dot{\mathbf{Y}}} + h + r $	$\dot{u} = \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}} + h + r $
Second law of thermodynamics (Clausius–Duhem inequality)	$\left\{ \begin{array} {l} \underline{\mathbf{T}} \cdot \underline{\dot{\mathbf{Y}}} - \dot{\varTheta}\eta - \dot{\psi} \geq 0 \\ \boldsymbol{\psi} = \boldsymbol{\psi}(\underline{\mathbf{Y}},\underline{\dot{\mathbf{Y}}},\varTheta)\\ \boldsymbol{\eta} = \boldsymbol{\eta}(\underline{\mathbf{Y}},\underline{\dot{\mathbf{Y}}},\varTheta) \end{array}\right. $	$\left\{ \begin{array} {l} \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}} - \dot{\varTheta}\eta - \dot{\psi} \geq 0 \\ \boldsymbol{\psi} = \boldsymbol{\psi}(\boldsymbol{\varepsilon},\dot{\boldsymbol{\varepsilon}},\varTheta)\\ \boldsymbol{\eta} = \boldsymbol{\eta}(\boldsymbol{\varepsilon},\dot{\boldsymbol{\varepsilon}},\varTheta) \end{array} \right. $
Thermodynamics restrictions (Elastic materials)	$\begin{array} {l} \biggl(\underline{\mathbf{T}} - \dfrac{\partial \psi}{\partial \underline{\mathbf{Y}}} \biggr) \cdot \underline{\dot{\mathbf{Y}}} - \dfrac{\partial \psi}{\partial \underline{\dot{\mathbf{Y}}}} \cdot \underline{\ddot{\mathbf{Y}}} - \bigl(\eta - \dfrac{\partial \psi}{\partial \varTheta} \bigr) \dot{\varTheta} \geq 0 \\[9pt] \quad \Longrightarrow \dfrac{\partial \psi}{\partial \underline{\dot{\mathbf{Y}}}} = \underline{\mathbf{0}} \Longrightarrow \boldsymbol{\psi} = \boldsymbol{\psi}(\underline{\mathbf{Y}},\varTheta)\\[6pt] \quad \Longrightarrow \eta = \dfrac{\partial \psi}{\partial \varTheta} \bigg\vert _{\underline{\mathbf{Y}}}\\[9pt] \quad \Longrightarrow \underline{\mathbf{T}} = \dfrac{\partial \psi}{\partial \underline{\mathbf{Y}}} \bigg\vert _{\varTheta} \end{array} $	$\begin{array} {l} \biggl(\boldsymbol{\sigma} - \dfrac{\partial \psi}{\partial \boldsymbol{\varepsilon}} \biggr) : \dot{\boldsymbol{\varepsilon}} - \dfrac{\partial \psi}{\partial \dot{\boldsymbol{\varepsilon}}} : \ddot{\boldsymbol{\varepsilon}} - \biggl(\eta - \dfrac{\partial \psi}{\partial \varTheta} \biggr) \dot{\varTheta} \geq 0 \\[9pt] \quad \Longrightarrow \dfrac{\partial \psi}{\partial \dot{\boldsymbol{\varepsilon}}} = \boldsymbol{0} \Longrightarrow \boldsymbol{\psi} = \boldsymbol{\psi}(\boldsymbol{\varepsilon},\varTheta)\\[6pt] \quad \Longrightarrow \eta = \dfrac{\partial \psi}{\partial \varTheta} \bigg\vert _{\boldsymbol{\varepsilon}}\\[9pt] \quad \Longrightarrow \boldsymbol{\sigma} = \dfrac{\partial \psi}{\partial \boldsymbol{\varepsilon}} \bigg\vert _{\varTheta} \end{array} $

u, internal energy density; Θ, uniform temperature; h, rate of heat transfer; η, entropy density; r, rate of heat source; ψ = u − Θη, free energy density

The PD angular momentum given in Table 2 is

$$ \displaystyle{ \int _{\mathcal{H}}} {\underline{\mathbf{Y}}\,[\boldsymbol{x},t]} \bigl\langle \boldsymbol{x'} - \boldsymbol{x} \bigr\rangle \times {\underline{ \mathbf{T}}[\boldsymbol{x},t]} \bigl\langle \boldsymbol{x'} - \boldsymbol{x} \bigr\rangle \, \mathrm{d}V_{\boldsymbol{x'}} = \boldsymbol{0}. $$

(60)

Equation (60) is satisfied if $\underline{\mathbf{T}} = \underline{t} \, \underline{\mathbf{M}}$ (i.e., if $\underline{\mathbf{T}}$ is collinear to $\underline{\mathbf{Y}}$) for any $\boldsymbol{\xi} = \boldsymbol{x'} - \boldsymbol{x}$ in ℋ. The material is called ordinary and $\underline{t}$ is its scalar force state field. An ordinary state-based material in which

$$ \underline{\mathbf{T}}[\boldsymbol{x},t] \bigl\langle \boldsymbol{x'} - \boldsymbol{x} \bigr\rangle = \boldsymbol{t_{v}} \bigl(\boldsymbol{x'},\boldsymbol{x},t\bigr) = \dfrac{\boldsymbol{f_{v}}(\boldsymbol{x'},\boldsymbol{x},t)}{2} $$

(61)

is called a bond-based material. Bond-based theory yields isotropic materials having a Poisson ratio of $1/4$. A material that is not ordinary, but satisfies Eq. (60), is called non-ordinary. Figure 9 schematizes these material models.

Fig. 9
Diagram of *bond-based*, *ordinary* and *non-ordinary* material

B.2 Constitutive model for isotropic elastic materials

B.2.1 Elastic material

An elastic material is defined by its free energy density $\psi$, which only depends on $\underline{\mathbf{Y}}$. Define $\psi = \varOmega(\underline{\mathbf{Y}})$ as the strain energy density. Since $\varOmega$ has a single variable, the force state (see Table 2) becomes

$$ \underline{\mathbf{T}} = \underline{\mathbf{T}}(\underline{ \mathbf{Y}}) = \dfrac{\partial\psi}{\partial\underline{\mathbf{Y}}} = \dfrac{\mathrm{d}\varOmega}{\mathrm{d}\underline{\mathbf{Y}}} = \triangledown\varOmega . $$

(62)

B.2.2 Isotropy

A material is isotropic if and only if, for any $\mathbf{Q}$ in the set of proper orthogonal second-order tensors

$$ \underline{\mathbf{T}}(\underline{\mathbf{Y}}\,\mathbf{Q}) \langle \boldsymbol{\xi} \rangle = \underline{\mathbf{T}}(\underline{\mathbf{Y}}) \langle \mathbf{Q}\boldsymbol{\xi} \rangle . $$

(63)

Physically, Eq. (63) reflects the fact that the force state is invariant by applying rotations before deformations.

B.2.3 Isotropic elastic solid

Let $\vartheta$ be a non-local dilatation

$$ \vartheta = \vartheta(\underline{e}) =\dfrac{3}{m} ( \underline{w} \, \underline{x}) \cdot \underline{e} $$

(64)

where $\underline{w}$ is the scalar influence state used to weight the bonds influence on the force state calculation and $\underline{x} = \vert \underline{\mathbf{X}} \langle \boldsymbol{\xi} \rangle \vert = \vert \boldsymbol{\xi} \vert = \xi$. $m$ is the weighted volume defined as

$$ m = (\underline{w} \, \underline{x}) \cdot \underline{x} = \displaystyle{ \int _{\mathcal{H}}} \underline{w} \langle \boldsymbol{\xi} \rangle \xi^{2} \, \mathrm{d}V_{\boldsymbol{\xi}}. $$

(65)

Finally, $\underline{e}$ is the scalar extension state defined by

$$ \underline{e} = \bigl\vert \underline{\mathbf{Y}} \langle \boldsymbol{ \xi} \rangle \bigr\vert - \underline{x} . $$

(66)

The scalar extension state can be divided into a spherical part $\underline{e}^{s}$ and a deviatoric part $\underline{e}^{d}$ such as

$$\begin{aligned} \underline{e} & = \underline{e}^{s} + \underline{e}^{d} \\ & = \dfrac{1}{3} \vartheta \, \underline{x} + \underline{e}^{d}. \end{aligned}$$

(67)

The spherical part represents the isotropic expansion while the deviatoric part depicts shear deformations.

Suppose an isotropic elastic material defined by its strain energy density $\varOmega$ (Silling et al. 2007)

$$\begin{aligned} \varOmega(\underline{\mathbf{Y}}) & = \varOmega(\, \underline{e}\,) \\ & = \dfrac{k \, \vartheta^{2}(\,\underline{e}\,)}{2} + \dfrac{\alpha^{d}}{2} \bigl(\underline{w}\, \underline{e}^{d}\bigr) \cdot \underline{e}^{d}. \end{aligned}$$

(68)

Consider the following results:

$$\begin{aligned} &\Delta\underline{e} = \Delta\underline{e}^{s} + \Delta \underline{e}^{d} , \end{aligned}$$

(69a)

$$\begin{aligned} &\bigl(\underline{w} \, \underline{e}^{d}\bigr) \bullet \Delta\underline{e}^{s} = 0 , \end{aligned}$$

(69b)

$$\begin{aligned} & \Delta\underline{e} = \dfrac{\underline{\mathbf{Y}}}{\vert \,\underline{\mathbf{Y}} \,\vert} \Delta\underline{ \mathbf{Y}} + o \bigl(\Vert \Delta\underline{\mathbf{Y}} \Vert\bigr) , \end{aligned}$$

(69c)

where $\Delta\underline{e}$, $\Delta\underline{e}^{s}$ and $\Delta\underline{e}^{d}$ are increments resulting from an incremental change to the deformation state $\Delta\underline{\mathbf{Y}}$. Combining Eqs. (59), (62) and (69a)–(69c) leads to the constitutive model of an isotropic elastic material given by the force state

$$ \underline{\mathbf{T}}(\,\underline{\mathbf{Y}}\,) = \biggl( \dfrac{3 k \vartheta}{m} \underline{w} \, \underline{x} + \alpha^{d} \underline{w} \, \underline{e}^{d} \biggr) \underline{\mathbf{M}} . $$

(70)

It is shown that, for small and homogeneous deformations, $\varOmega$ is equivalent to the strain energy density defined for an isotropic linearly elastic solid in CCM. Furthermore, the force state in Eq. (70) corresponds to an ordinary material since $\underline{\mathbf{T}} = \underline{t} \, \underline{\mathbf{M}}$.

Appendix C: Essential proofs

C.1 Conditions to have $\mathbf{H}(W) > \boldsymbol{0}$

Equation (21a) implies that

$$\mathbf{H}(W) > \boldsymbol{0} \quad \Longleftrightarrow \quad \forall \boldsymbol{\mathcal{E}} \neq \boldsymbol{\mathcal{E}}_{\mathit{ref}}, \quad W >0. $$

When Eqs. (26a), (26b) and (28) are combined, the conditions to have $\mathbf{H}(W) > \boldsymbol{0}$ become

C.2 Simultaneous diagonalization of $\mathbf{B}_{\kappa}$, $\mathbf{L^{2}_{\kappa}}$ and $\mathbf{L^{3}_{\kappa}}$

The simultaneous diagonalizability theorem says that a set of diagonalizable matrices are simultaneously diagonalizable if and only if the matrices commute pairwise.

Since $\mathbf{B}_{\kappa}$, $\mathbf{L^{2}_{\kappa}}$ and $\mathbf{L^{3}_{\kappa}}$ are all real symmetric matrices, they are diagonalizable. Using Eqs. (23b), (26b) and (26c), one obtains

Thus, $\mathbf{B}_{\kappa}$, $\mathbf{L^{2}_{\kappa}}$ and $\mathbf{L^{3}_{\kappa}}$ commute pairwise and they are simultaneously diagonalizable.

Appendix D: Tools for calculations in PD

D.1 Free energy density $\varOmega$ spherical part

$$ \bigl( \underline{w} \, \underline{e}^{s} \bigr) \cdot \underline{e}^{s} = \biggl(\underline{w} \, \dfrac{1}{3} \, \vartheta \, \underline{x} \biggr) \cdot \dfrac{1}{3} \, \vartheta \, \underline{x} = \dfrac{1}{9} ( \underline{w} \, \underline{x} \cdot \underline{x} ) \vartheta^{2} = \dfrac{m}{9} \vartheta^{2} = \dfrac{m}{9} \mathrm{tr}^{2}(\boldsymbol{\varepsilon}) . $$

(73)

D.2 Free energy density $\varOmega$ deviatoric part

It is assumed that ℋ is a sphere such as the weighted volume $m$ is defined by

$$\begin{aligned} m =& \bigl(\underline{w} \, \underline{x}\bigr) \cdot \underline{x} = \displaystyle{ \int _{\mathcal{H}}} \underline{w} \langle \boldsymbol{\xi} \rangle \xi^{2} \, \mathrm{d}V_{\boldsymbol{\xi}} = \displaystyle{ \int _{0}^{\delta}}\displaystyle{ \int _{0}^{2\pi}}\displaystyle{ \int _{0}^{\pi}} \, \underline{w} \langle \boldsymbol{\xi} \rangle \, \xi^{2} \, \xi^{2} \sin(\phi)\,\mathrm{d}\phi \,\mathrm{d}\theta \,\mathrm{d}\xi \\ =& 4 \pi \displaystyle{ \int _{0}^{\delta}} \, \underline{w} \langle \boldsymbol{\xi} \rangle \xi^{4} \, \mathrm{d}\xi . \end{aligned}$$

(74)

Define $\underline{e}^{d} = (\frac{\xi_{i}}{\xi} \varepsilon^{d}_{ij} \frac{\xi_{j}}{\xi} ) \xi$, where Einstein’s convention has been used. Combining that result with Eq. (56) leads to

Each term of the sum given in Eq. (75a) is calculated. Note that no summations on $i$, $j$, $k$ and $l$ take place in the following line.

$$\begin{aligned} &\forall (i,j,k,l) \in [\![1;3]\!]^{4} \mid \lbrace i=j \mbox{ or } k=l \rbrace \\ &\quad \Longrightarrow \quad \varepsilon_{ij}^{d} \varepsilon_{kl}^{d} \displaystyle{ \int _{0}^{\delta}}\displaystyle{ \int _{0}^{2\pi}}\displaystyle{ \int _{0}^{\pi}} \underline{w} \langle \boldsymbol{\xi} \rangle \xi_{i} \xi_{j} \xi_{k} \xi_{l} \sin(\phi)\,\mathrm{d}\phi \,\mathrm{d}\theta \,\mathrm{d}\xi = 0. \end{aligned}$$

(76)

$$\begin{aligned} & \forall (i,j,k,l) \in [\![1;3]\!]^{4} \mid \lbrace i=k \mbox{ and } j=l \rbrace \\ &\quad \Longrightarrow \quad \varepsilon_{ij}^{d} \varepsilon_{kl}^{d} \displaystyle{ \int _{0}^{\delta}}\displaystyle{ \int _{0}^{2\pi}}\displaystyle{ \int _{0}^{\pi}} \underline{w} \langle \boldsymbol{\xi} \rangle \xi_{i} \xi_{j} \xi_{k} \xi_{l} \sin(\phi)\,\mathrm{d}\phi \,\mathrm{d}\theta \,\mathrm{d}\xi = \dfrac{m}{15} \varepsilon_{ij}^{d} \varepsilon_{ij}^{d}. \end{aligned}$$

(77)

$$\begin{aligned} & \forall (i,j,k,l) \in [\![1;3]\!]^{4} \mid \lbrace i=l \mbox{ and } j=k \rbrace \\ &\quad \Longrightarrow \quad \varepsilon_{ij}^{d} \varepsilon_{kl}^{d} \displaystyle{ \int _{0}^{\delta}}\displaystyle{ \int _{0}^{2\pi}}\displaystyle{ \int _{0}^{\pi}} \underline{w} \langle \boldsymbol{\xi} \rangle \xi_{i} \xi_{j} \xi_{k} \xi_{l} \sin(\phi)\,\mathrm{d}\phi \,\mathrm{d}\theta \,\mathrm{d}\xi = \dfrac{m}{15} \varepsilon_{ij}^{d} \varepsilon_{ji}^{d}. \end{aligned}$$

(78)

Combining Eqs. (75a)–(78), knowing that $\varepsilon_{ij}^{d} = \varepsilon_{ji}^{d}$ and summing on $i$, $j$, $k$ and $l$ lead to

$$\begin{aligned} \bigl(\underline{w} \, \underline{e}^{d} \bigr) \cdot \underline{e}^{d} & = \dfrac{m}{15} \varepsilon_{ij}^{d} \varepsilon_{kl}^{d} (\delta_{ik}\delta_{jl} + \delta_{il} \delta_{jk} ) \\ & = 2 \dfrac{m}{15} \mathbf{dev}(\boldsymbol{\varepsilon}) : \mathbf{dev}(\boldsymbol{\varepsilon}). \end{aligned}$$

(79)

Appendix E: Analytical solution of a 1D PD elastic self-equilibrated bar

The analytical solution of a 1D PD elastic self-equilibrated bar under a static load, such as that described in Fig. 7, is derived. The development relies on the assumptions that the horizon $\delta = \Delta x$, the weighted volume is constant with $m = 2 {\Delta x}^{2} \Delta V = 2 A {\Delta x}^{3}$ and the elastic modulus is $C = \alpha m$. It is also assumed that the point $x_{n}$ in the middle of the bar is such as $x_{n} = 0$. Thus, it is obvious that $x_{n} = y_{n} = 0$ due to the applied load.

The discretized PD equation of motion (Eqs. (47a), (47b)) applied to $\boldsymbol{x}_{\boldsymbol{0}}$ leads to

$$ \underline{\mathbf{T}}[\boldsymbol{x}_{\boldsymbol{0}}] \langle \boldsymbol{x_{1}} - \boldsymbol{x}_{\boldsymbol{0}} \rangle - \underline{ \mathbf{T}}[\boldsymbol{x_{1}}] \langle \boldsymbol{x}_{\boldsymbol{0}} - \boldsymbol{x_{1}} \rangle \Delta V + \boldsymbol{b}( \boldsymbol{x}_{\boldsymbol{0}}) = \boldsymbol{0} . $$

(80)

As the problem is in 1D, Eq. (80) can be cast in scalar form as follows:

$$\begin{aligned} &\alpha \biggl[ \bigl( \vert y_{1} - y_{0} \vert - \vert x_{1} - x_{0} \vert \bigr) \dfrac{y_{1} - y_{0}}{\vert y_{1} - y_{0} \vert} - \bigl( \vert y_{0} - y_{1} \vert - \vert x_{0} - x_{1} \vert \bigr) \dfrac{y_{0} - y_{1}}{\vert y_{0} - y_{1} \vert} \biggr] \Delta V - b \\ &\quad = 0 , \end{aligned}$$

(81a)

$$\begin{aligned} & -y_{0} + y_{1} = \dfrac{b}{2 \alpha \Delta V} + \Delta x, \end{aligned}$$

(81b)

$$\begin{aligned} & -y_{0} + y_{1} = \dfrac{F \Delta x}{A C} + \Delta x , \end{aligned}$$

(81c)

where one is reminded that $F = b \Delta V$ is the load and $A$ is the bar’s cross sectional area. Applying the same method to every other material point, with the exception of $\boldsymbol{x_{n}}$, results in

$$\begin{aligned} &\forall i \in [\![1;n-1]\!] \cup [\![n+1 ;2n-1]\!], \qquad y_{i-1} - 2 y_{i} + y_{i+1} = 0 , \end{aligned}$$

(82)

$$\begin{aligned} & y_{2n-1} + y_{2n} = - \biggl( \dfrac{F \Delta x}{A C} + \Delta x \biggr) . \end{aligned}$$

(83)

Equations (81a)–(81c) to (83) form a system of linear equations that can be expressed and solved as in Bobaru et al. (2009), i.e. using the following matrix form:

$$\begin{aligned} & \mathbf{A} \boldsymbol{Y} = \boldsymbol{U}, \end{aligned}$$

(84a)

$$\begin{aligned} & \begin{pmatrix} -1 & 1 & 0 & \cdots & 0 & 0 & & \cdots & & 0\\ 1 & -2 & 1 & \ddots & & & & & & \\ 0 & \ddots & \ddots & \ddots & & \vdots & & 0 & & \vdots \\ \vdots & \ddots & \ddots & \ddots & & & & & & \\ 0 & \ddots & \ddots & 1 & -2 & 0 & & \cdots & & 0\\ 0 & & \cdots & & 0 & -2 & 1 & \ddots & \ddots & 0\\ & & & & & & \ddots & \ddots & \ddots & \vdots\\ \vdots & & 0 & & \vdots & & \ddots & \ddots & \ddots & 0 \\ & & & & & & \ddots & 1 & -2 & 1 \\ 0 & & \cdots & & 0 & 0 & \cdots & 0 & 1 & -1 \\ \end{pmatrix} \begin{pmatrix} y_{0}\\ y_{1}\\ y_{2}\\ \vdots\\ y_{n-1}\\ y_{n+1}\\ \vdots\\ y_{2n-2}\\ y_{2n-1}\\ y_{2n}\\ \end{pmatrix} = \begin{pmatrix} u\\ 0\\ 0\\ \vdots\\ 0\\ 0\\ \vdots\\ 0\\ 0\\ -u\\ \end{pmatrix} , \end{aligned}$$

(84b)

where $u = \dfrac{F \Delta x}{A C} + \Delta x$. Finally, $\boldsymbol{Y} = {\mathbf{A}}^{-1} \boldsymbol{U}$ with

$$\begin{aligned} {\mathbf{A}}^{-1}\! =\! \begin{pmatrix} -n & -n+1 & -n+2 & \cdots & -1 & 0 & & \cdots & & 0\\ -n+1 & -n+1 & \vdots & & & & & & & \\ -n+2 & \cdots & -n+2 & & \vdots & \vdots & & 0 & & \vdots \\ \vdots & & & & & & & & & \\ -1 & & \cdots & & -1 & 0 & & \cdots & & 0\\ 0 & & \cdots & & 0 & -1 & & \cdots & & -1\\ & & & & & & & & & \\ \vdots & 0 & & & \vdots & \vdots & & -n+2 & \cdots & -n+2 \\ & & & & & & & \vdots & -n+1 & -n+1 \\ 0 & \cdots & & & 0 & -1 & \cdots & -n+2 & -n+1 & -n \\ \end{pmatrix} \!. \end{aligned}$$

(85)

The strain $\varepsilon_{\mathit{PD}}$ in the middle of the bar is derived as follows:

$$\begin{aligned} \varepsilon_{\mathit{PD}} =& \frac{\vert y_{n+1} - y_{n-1} \vert - \vert x_{n+1} - x_{n-1} \vert}{\vert x_{n+1} - x_{n-1} \vert} , \end{aligned}$$

(86a)

$$\begin{aligned} \varepsilon_{\mathit{PD}} =& \frac{u - \Delta x}{\Delta x} , \end{aligned}$$

(86b)

$$\begin{aligned} \varepsilon_{\mathit{PD}} =& \frac{F}{A C}, \end{aligned}$$

(86c)

$$\begin{aligned} \varepsilon_{\mathit{PD}} =& \frac{\sigma_{0}}{C} . \end{aligned}$$

(86d)

Thus, Eqs. (86a)–(86d) means that the strain derived from a Riemann sum to approximate a 1D PD elastic self-equilibrated bar under a static load and with the horizon shrunk to the local limiting case (i.e. $\delta = \Delta x$) is equal to the strain obtained within CCM. The same analytical approach could have been performed for a 1D PD viscoelastic self-equilibrated bar. $u$ in vector $\boldsymbol{U}$ would have been different as it would include viscoelastic parameters, implying more complex calculations outside of the paper’s objectives. Nevertheless the symmetries of the problem, i.e. the symmetry of the discretization using $2n+1$ nodes and the symmetry of the loading, lead to the same conclusion.

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Authors and Affiliations

Rolland Delorme
- 1
Email authorView author's OrcID profile
Ilyass Tabiai
- 1
Louis Laberge Lebel
- 2
Martin Lévesque
- 1

1.Laboratory for Multiscale Mechanics, Mechanical Engineering Dept.Polytechnique MontrealMontrealCanada
2.Advanced Composite and Fiber Structures Laboratory, Mechanical Engineering Dept.Polytechnique MontrealMontrealCanada

Balance law	Peridynamics	Classical continuum mechanics (Infinitesimal strain theory)
Linear momentum	\(\begin{array}{l} \rho(\boldsymbol{x}) \ddot{\boldsymbol{y}}(\boldsymbol{x},t) - \boldsymbol{b}(\boldsymbol{x},t) \\ \quad = \displaystyle{\int_{\mathcal{B}}} {\underline{\mathbf{T}}[\boldsymbol{x},t]} \langle \boldsymbol{x'} - \boldsymbol{x} \rangle - {\underline{\mathbf{T}}[\boldsymbol{x'},t]} \langle \boldsymbol{x} - \boldsymbol{x'} \rangle \, \mathrm{d}V_{\boldsymbol{x'}} \end{array}\)	\(\rho(\boldsymbol{x}) \ddot{\boldsymbol{y}}(\boldsymbol{x},t) - \boldsymbol{b}(\boldsymbol{x},t) = \mathbf{div} \,\boldsymbol{\sigma}(\boldsymbol{x},t)\)
Angular momentum	\(\displaystyle{\int_{\mathcal{H}}} {\underline{\mathbf{Y}}[\boldsymbol{x},t]} \langle \boldsymbol{x'} - \boldsymbol{x} \rangle \times {\underline{\mathbf{T}}[\boldsymbol{x},t]} \langle \boldsymbol{x'} - \boldsymbol{x} \rangle \, \mathrm{d}V_{\boldsymbol{x'}} = \boldsymbol{0}\)	\(\boldsymbol{\sigma} = {\boldsymbol{\sigma}}^{\top}\)
First law of thermodynamics	\(\dot{u} = \underline{\mathbf{T}} \cdot \underline{\dot{\mathbf{Y}}} + h + r \)	\(\dot{u} = \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}} + h + r \)
Second law of thermodynamics (Clausius–Duhem inequality)	\(\left\{ \begin{array} {l} \underline{\mathbf{T}} \cdot \underline{\dot{\mathbf{Y}}} - \dot{\varTheta}\eta - \dot{\psi} \geq 0 \\ \boldsymbol{\psi} = \boldsymbol{\psi}(\underline{\mathbf{Y}},\underline{\dot{\mathbf{Y}}},\varTheta)\\ \boldsymbol{\eta} = \boldsymbol{\eta}(\underline{\mathbf{Y}},\underline{\dot{\mathbf{Y}}},\varTheta) \end{array}\right. \)	\(\left\{ \begin{array} {l} \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}} - \dot{\varTheta}\eta - \dot{\psi} \geq 0 \\ \boldsymbol{\psi} = \boldsymbol{\psi}(\boldsymbol{\varepsilon},\dot{\boldsymbol{\varepsilon}},\varTheta)\\ \boldsymbol{\eta} = \boldsymbol{\eta}(\boldsymbol{\varepsilon},\dot{\boldsymbol{\varepsilon}},\varTheta) \end{array} \right. \)
Thermodynamics restrictions (Elastic materials)	\(\begin{array} {l} \biggl(\underline{\mathbf{T}} - \dfrac{\partial \psi}{\partial \underline{\mathbf{Y}}} \biggr) \cdot \underline{\dot{\mathbf{Y}}} - \dfrac{\partial \psi}{\partial \underline{\dot{\mathbf{Y}}}} \cdot \underline{\ddot{\mathbf{Y}}} - \bigl(\eta - \dfrac{\partial \psi}{\partial \varTheta} \bigr) \dot{\varTheta} \geq 0 \\[9pt] \quad \Longrightarrow \dfrac{\partial \psi}{\partial \underline{\dot{\mathbf{Y}}}} = \underline{\mathbf{0}} \Longrightarrow \boldsymbol{\psi} = \boldsymbol{\psi}(\underline{\mathbf{Y}},\varTheta)\\[6pt] \quad \Longrightarrow \eta = \dfrac{\partial \psi}{\partial \varTheta} \bigg\vert _{\underline{\mathbf{Y}}}\\[9pt] \quad \Longrightarrow \underline{\mathbf{T}} = \dfrac{\partial \psi}{\partial \underline{\mathbf{Y}}} \bigg\vert _{\varTheta} \end{array} \)	\(\begin{array} {l} \biggl(\boldsymbol{\sigma} - \dfrac{\partial \psi}{\partial \boldsymbol{\varepsilon}} \biggr) : \dot{\boldsymbol{\varepsilon}} - \dfrac{\partial \psi}{\partial \dot{\boldsymbol{\varepsilon}}} : \ddot{\boldsymbol{\varepsilon}} - \biggl(\eta - \dfrac{\partial \psi}{\partial \varTheta} \biggr) \dot{\varTheta} \geq 0 \\[9pt] \quad \Longrightarrow \dfrac{\partial \psi}{\partial \dot{\boldsymbol{\varepsilon}}} = \boldsymbol{0} \Longrightarrow \boldsymbol{\psi} = \boldsymbol{\psi}(\boldsymbol{\varepsilon},\varTheta)\\[6pt] \quad \Longrightarrow \eta = \dfrac{\partial \psi}{\partial \varTheta} \bigg\vert _{\boldsymbol{\varepsilon}}\\[9pt] \quad \Longrightarrow \boldsymbol{\sigma} = \dfrac{\partial \psi}{\partial \boldsymbol{\varepsilon}} \bigg\vert _{\varTheta} \end{array} \)

Generalization of the ordinary state-based peridynamic model for isotropic linear viscoelasticity

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