Nonlinear mechanics of thermoelastic accretion

Abstract

In this paper, we formulate a theory for the coupling of accretion mechanics and thermoelasticity. We present an analytical formulation of the thermoelastic accretion of an infinite cylinder and of a two-dimensional block. We develop numerical schemes for the solution of these two problems, and numerically calculate residual stresses and observe a strong dependence of the final mechanical state on the parameters of the accretion process. This suggests the possibility to predict and control thermal accretion processes of soft materials by manipulating thermal parameters, and therefore, to realize additively manufactured soft objects with the desired characteristics and performances.

Appendices

A A nonlinear thermoelastic constitutive model

In this appendix, we present a thermoelastic model following the works of Chadwick [3], Ogden [37,38,39], and Holzapfel and Simo [24]. See [43] for more details. For a homogeneous isotropic solid, we denote by $\kappa _0$, $\mu _0$ and $\beta _0$, the bulk modulus, the shear modulus, and the volumetric coefficient of thermal expansion at $T_0$, respectively, with $\beta (X,T) = \frac{\partial }{\partial T}{\text {tr}} \varvec{\omega }(X,T)$, while in the isotropic case $\beta (X,T) = 3\frac{\partial }{\partial T}\omega (X,T) = 3 \alpha (X,T)$ in 3D, and $\beta (X,T) = 2\frac{\partial }{\partial T}\omega (X,T) = 2 \alpha (X,T)$ in 2D. We consider the following constitutive model

$$\begin{aligned} \bar{{\mathcal {W}}}(T,\tilde{I},J)= \frac{\mu _0}{2}\frac{T}{T_0} (\tilde{I}-3)+\frac{\kappa _0}{2}\frac{T}{T_0}(J-1)^2 -{\kappa _0}\beta _0(J-1)\left( T-T_0\right) -\rho \int \limits _{T_0}^Tc_E(\eta )\frac{T-\eta }{\eta } {{\mathrm {d}}}\eta , \end{aligned}$$

(51)

where $\tilde{I}=J^{-2/3}I$, $I={\text {tr}}{\varvec{C}}$, ${J} = \sqrt{\det {\varvec{C}}}$, and $c_E$ is the specific heat capacity at constant strain. In the incompressible case, we have the constraint $J=1$ associated with the pressure field p as the Lagrange multiplier and the constitutive model transforms to read

$$\begin{aligned} \bar{{\mathcal {W}}}(T,{I})=\frac{\mu _0}{2}\frac{T}{T_0}({I}-3)-\rho \int \limits _{T_0}^Tc_E(\eta )\frac{T-\eta }{\eta } {{\mathrm {d}}}\eta , \end{aligned}$$

(52)

plus the Lagrange multiplier part $p(J-1)^2$. Note that the shear modulus is linear in temperature, i.e.,

$$\begin{aligned} \mu (T)=\mu _0 \frac{T}{T_0} . \end{aligned}$$

(53)

The simplest thermoelastic model for the expansion coefficients is given by a constant $\alpha$, viz.

$$\begin{aligned} \alpha (T)= \alpha _0,~~~~\omega (T) = \omega _0 \frac{T}{T_0},~~~~\omega _0=\alpha _0 T_0 . \end{aligned}$$

(54)

B A remark on the material metric

Equation (2) is a generalization of the isotropic case that appeared in [40]. In section 2.1 of [43], due to a mis-manipulation of the musical operators for raising and lowering indices, the representation of the material metric using the $({}^{1}_{1})$-tensor $\varvec{\omega }$ was mistakenly presented as

$$\begin{aligned} {\varvec{G}}(X,T)= {\varvec{G}}_0(X) e^{2\varvec{\omega }(X,T)}. \end{aligned}$$

(55)

Indeed, this representation may violate the symmetry requirement for the Riemannian metric ${\varvec{G}}$, while the representation (2) ensures its symmetry. In what follow, we provide a correction of the proof appearing in section 2.1 of [43]. The manifold $({\mathcal {B}},{\varvec{G}}_0)$—which corresponds to the stress-free temperature field $T_0=T_0(X)$—is flat. Hence, there exists a local coordinate chart $\{Y^A\}$ in which

$$\begin{aligned} {\varvec{G}}_0=\delta _{AB}{\mathrm {dY}}^A\otimes {\mathrm {dY}}^B. \end{aligned}$$

(56)

Following Ozakin and Yavari [40], the temperature-dependent material metric can be written as

$$\begin{aligned} {\varvec{G}}(X,T)=\sum _K e^{2\omega _K(X,T)}{\mathrm {dY}}^K\otimes {\mathrm {dY}}^K, \end{aligned}$$

(57)

where $\{\omega _{A}\}_{A=1,2,3}$ describes the thermal expansion properties of the material such that $\omega _K$ is related to the thermal expansion coefficient $\alpha _K$ in the direction $\frac{\partial }{\partial Y^K}$ by

$$\begin{aligned} \alpha _K(X,T)=\frac{\partial \omega _K}{\partial T}(X,T), \end{aligned}$$

(58)

and $\omega _K(X,T_0)=0$. Let the change of basis between $\{Y^A\}_{A=1,2,3}$ and some arbitrary local coordinate chart $\{X^A\}_{A=1,2,3}$ be written as

$$\begin{aligned} {\mathrm {dY}}^K=A^K{}_J{\mathrm {dX}}^J, \end{aligned}$$

(59)

which also reads as

$$\begin{aligned} \frac{\partial }{\partial Y^K}=(A^{-1})^I{}_K\frac{\partial }{\partial X^I}. \end{aligned}$$

(60)

Then it follows that

$$\begin{aligned} {\varvec{G}}=\left( \sum _K e^{2\omega _K} A^K{}_I A^K{}_J\right) {\mathrm {dX}}^I\otimes {\mathrm {dX}}^J. \end{aligned}$$

(61)

Let $\varvec{\omega }$ be the $({}^{1}_{1})$-tensor

$$\begin{aligned} \varvec{\omega }=\sum _K \omega _K \frac{\partial }{\partial Y^K}\otimes {\mathrm {dY}}^K = \sum _K (A^{-1})^I{}_K {\omega _K} A^K{}_J \frac{\partial }{\partial X^I} \otimes {\mathrm {dX}}^J. \end{aligned}$$

(62)

However, in $\{ X^A\}$, one has

$$\begin{aligned} {\varvec{G}}_0= \delta _{AB} A^A{}_I A^B{}_J {\mathrm {dX}}^I\otimes {\mathrm {dX}}^J. \end{aligned}$$

(63)

Hence, the material metric transforms to

$$\begin{aligned} {\varvec{G}}(X,T)= e^{\omega ^L{}_I} (G_0)_{LM}~\!e^{\omega ^M{}_J} {\mathrm {dX}}^I\otimes {\mathrm {dX}}^J, \end{aligned}$$

(64)

where $(G_0)_{LM}$ are the components of ${\varvec{G}}_0$ in $\{ X^A\}$. This is the coordinate representation of (2). Note that even though the general representation for the material metric used in [43] had this symmetry fallacy, the results of the paper and the examples were not affected and remain valid.

C Christoffel symbols for the accreting cylinder

First note that denoting with $(\Gamma _0)^A{}_{BC}$ the Christoffel symbols relative to ${\varvec{G}}_0$ as in (6), one obtains

$$\begin{aligned} \Gamma ^A{}_{BC} = (\Gamma _0)^A{}_{BC} + (G_0)^{AD} [ (G_0)_{DB} \omega _{,C} + (G_0)_{DC} \omega _{,B} - (G_0)_{BC} \omega _{,D} ] . \end{aligned}$$

(65)

Therefore, as ${\varvec{G}}_0^{\sharp }:{\varvec{G}}_0={\text {dim}}{\mathcal {B}}$, one has

$$\begin{aligned} \Gamma ^B{}_{BC} = (\Gamma _0)^B{}_{BC} + ({\text {dim}}{\mathcal {B}})~\omega _{,C} . \end{aligned}$$

(66)

Under the assumption of a uniform material one has $\omega _{,C} = \frac{\partial \omega }{\partial T} \,T_{,C} = \alpha \,T_{,C}$, i.e., ${\mathrm {d}}\omega = \alpha \,{\mathrm {d}}T$, so one writes

$$\begin{aligned} \Gamma ^B{}_{BC} = (\Gamma _0)^B{}_{BC} + ({\text {dim}}{\mathcal {B}}) \alpha \,T_{,C} . \end{aligned}$$

(67)

Now we compute the Christoffel symbols for the material metric ${\varvec{G}}$ of (25). The only nonzero Christoffel symbols are

$$\begin{aligned} \begin{aligned} \Gamma ^R{}_{RR}&= T_{,R}\alpha (T),\\ \Gamma ^R{}_{\Theta \Theta }&= -{\bar{r}}(R) \left[ {\bar{r}}'(R) + {\bar{r}}(R) T_{,R} \alpha (T) \right] ,\\ \Gamma ^\Theta {}_{R\Theta }&= \frac{{\bar{r}}'(R)}{{\bar{r}}(R)} + T_{,R}\alpha (T). \end{aligned} \end{aligned}$$

(68)

Now one can compute the material divergence ${\text {Div}}[K(T) {\varvec{G}}^\sharp {\mathrm {dT}}]$ for the heat equation as

$$\begin{aligned} \begin{aligned} {\text {Div}}[K(T) {\varvec{G}}^\sharp {\mathrm {dT}}]&= \left[ K(T) G^{AB} T_{,B} \right] _{|A}\\&= \left[ K(T) G^{AB}\right] _{|A} T_{,B} + K(T) G^{AB} \left[ T_{,B}\right] _{|A}\\&= \left[ K(T)\right] _{|A} G^{AB} T_{,B} + K(T) G^{AB} \left[ T_{,AB} - \Gamma ^C{}_{AB} T_{,C}\right] \\&= \frac{{\mathrm {dK}}}{{\mathrm {dT}}} T_{,R}{}^2 G^{RR} + K(T) \left[ T_{,RR}G^{RR} -\left( \Gamma ^R{}_{RR}G^{RR}+\Gamma ^R{}_{\Theta \Theta }G^{\Theta \Theta } \right) T_{,R} \right] \\&= \left[ \frac{{\mathrm {dK}}}{{\mathrm {dT}}} T_{,R}{}^2 + \frac{K(T)}{{\bar{r}}(R)}\frac{\partial }{\partial R}\left( {\bar{r}}(R)\frac{\partial T}{\partial R}\right) \right] e^{-2\omega (T)}. \end{aligned} \end{aligned}$$

(69)

D The reduced form of the Clausius–Duhem inequality in thermoelastic accretion

In this appendix, we discuss the restrictions that the second law of thermodynamics imposes on constitutive equations. In particular, we correct a mistake in [43],¹¹ which fortunately did not affect any of the results or conclusions of that work. The localized form of the Clausius-Duhem inequality reads

$$\begin{aligned} \rho \dot{{\mathcal {N}}}\ge \rho \frac{R}{T}-\text {Div}\left( \frac{{\varvec{H}}}{T}\right) +\rho \frac{\partial {\mathcal {N}}}{\partial {\varvec{G}}}\!:\!\dot{{\varvec{G}}}, \end{aligned}$$

(70)

where ${\mathcal {N}}={\mathcal {N}}(X,T,{\varvec{C}}^\flat ,{\varvec{G}})$ is the specific entropy. Expanding Eq. (70) and multiplying by $T>0$ one obtains

$$\begin{aligned} \rho T \left( \frac{\partial {\mathcal {N}}}{\partial T}{\dot{T}} +\frac{\partial {\mathcal {N}}}{\partial {\varvec{C}}^{\flat }}\!:\!\dot{{\varvec{C}}}^{\flat } \right) \ge \rho R-\text {Div}{\varvec{H}} +\frac{1}{T}\left<{\mathrm {d}}T,{\varvec{H}}\right> , \end{aligned}$$

(71)

where in a local coordinate chart $\{X^A\}$, the 1-form ${\mathrm {d}}T$ has the representation ${\mathrm {d}}T=\frac{\partial T}{\partial X^A}dX^A$. The specific free energy function has the form $\Psi =\Psi (X,T,{\varvec{C}}^\flat ,{\varvec{G}})$. The internal energy is defined as the Legendre transform of the free energy with respect to the conjugate variables T and ${\mathcal {N}}$, i.e., ${\mathcal {E}}=T{\mathcal {N}}+\Psi$, and hence, ${\mathcal {E}}={\mathcal {E}}(X,{\mathcal {N}},{\varvec{C}}^\flat ,{\varvec{G}})\,$. Therefore¹²

$$\begin{aligned} \frac{\partial {\mathcal {E}}}{\partial {\varvec{G}}}=T\frac{\partial {\mathcal {N}}}{\partial {\varvec{G}}} +\frac{\partial \Psi }{\partial {\varvec{G}}} ,\quad \frac{\partial {\mathcal {E}}}{\partial {\varvec{C}}^\flat }=T\frac{\partial {\mathcal {N}}}{\partial {\varvec{C}}^\flat } +\frac{\partial \Psi }{\partial {\varvec{C}}^\flat }. \end{aligned}$$

(72)

The localized balance of energy reads [43]

$$\begin{aligned} \rho \dot{{\mathcal {E}}}={\varvec{S}}\!:\!{\varvec{D}}-\text {Div}\,{\varvec{Q}}+\rho R +\rho \frac{\partial {\mathcal {E}}}{\partial {\varvec{G}}}\!:\!\dot{{\varvec{G}}}. \end{aligned}$$

(73)

This can be rewritten in terms of the specific entropy as

$$\begin{aligned} \rho \left( {\mathcal {N}}+\frac{\partial \Psi }{\partial T}\right) {\dot{T}}+\rho T \frac{\partial {\mathcal {N}}}{\partial T} {\dot{T}} +\rho T \frac{\partial {\mathcal {N}}}{\partial {\varvec{C}}^\flat }\!:\!\dot{{\varvec{C}}}^{\flat } +\left( \rho \frac{\partial \Psi }{\partial {\varvec{C}}^\flat }-\frac{1}{2}{\mathbf {S}} \right) \!:\!\dot{{\varvec{C}}}^{\flat } =\rho R-\text {Div}\,{\varvec{H}}. \end{aligned}$$

(74)

Substituting (74) into (71) one obtains

$$\begin{aligned} \rho \left( {\mathcal {N}}+\frac{\partial \Psi }{\partial T}\right) {\dot{T}} +\left( \rho \frac{\partial \Psi }{\partial {\varvec{C}}^\flat } -\frac{1}{2}{\mathbf {S}} \right) \!:\!\dot{{\varvec{C}}}^{\flat } +\frac{1}{T}\left<{\mathrm {d}}T,{\varvec{H}}\right> \le 0. \end{aligned}$$

(75)

This inequality must hold for all deformations $\varphi$ and metrics ${\varvec{G}}\,$. Therefore¹³

$$\begin{aligned} {\mathcal {N}}=-\frac{\partial \Psi }{\partial T},~~~{\mathbf {S}}=2\rho \frac{\partial \Psi }{\partial {\varvec{C}}^\flat } , ~~~ \left<{\mathrm {d}}T,{\varvec{H}}\right> \le 0 . \end{aligned}$$

(76)