A Stochastic Multi-scale Approach for Numerical Modeling of Complex Materials—Application to Uniaxial Cyclic Response of Concrete

Abstract

In complex materials, numerous intertwined phenomena underlie the overall response at macroscale. These phenomena can pertain to different engineering fields (mechanical, chemical, electrical), occur at different scales, can appear as uncertain, and are nonlinear. Interacting with complex materials thus calls for developing nonlinear computational approaches where multi-scale techniques that grasp key phenomena at the relevant scale need to be mingled with stochastic methods accounting for uncertainties. In this chapter, we develop such a computational approach for modeling the mechanical response of a representative volume of concrete in uniaxial cyclic loading. A mesoscale is defined such that it represents an equivalent heterogeneous medium: nonlinear local response is modeled in the framework of Thermodynamics with Internal Variables; spatial variability of the local response is represented by correlated random vector fields generated with the Spectral Representation Method. Macroscale response is recovered through standard homogenization procedure from Micromechanics and shows salient features of the uniaxial cyclic response of concrete that are not explicitly modeled at mesoscale.

Appendices

Appendix 1: On the Ergodicity in Correlation of the Random Fields Simulated with Eq. (41)

From Eq. (41), $j \in [1,\ldots ,m]$:

$$\begin{aligned} g_j(\mathbf {x}; \theta ) = 2 \sqrt{\varDelta \kappa _1 \varDelta \kappa _2}&\sum _{l=1}^m \sum _{\alpha =1}^2 \sum _{n_1=0}^{N_1-1} \sum _{n_2=0}^{N_2-1} \vert H_{jl}(\varvec{\kappa }_{n_1n_2}^{\alpha }) \vert \nonumber \\&\times \cos (\varvec{\kappa }_{n_1n_2}^{\alpha } \cdot \mathbf {x}- \varphi _{jl}(\varvec{\kappa }_{n_1n_2}^{\alpha }) + \varPhi _{l, n_1n_2}^{\alpha }(\theta )) \end{aligned}$$

(105)

On the one hand, because the random phases $\varPhi (\theta )$ are independent and uniformly distributed over $[0 \, , \, 2\pi ]$, the ensemble correlation function of two sample functions $g_j(\varvec{\xi };\theta )$ and $g_k(\varvec{\xi };\theta )$ reads:

$$\begin{aligned} R_{jk}(\varvec{\xi })&:= \mathbb {E}\left[ g_j(\mathbf {x}; \theta ) \ g_k(\mathbf {x}+ \varvec{\xi }; \theta ) \right] \nonumber \\&= \frac{1}{4\pi ^2} \int _0^{2\pi } \int _0^{2\pi } g_j(\mathbf {x}; \theta ) \ g_k(\mathbf {x}+\varvec{\xi }; \theta ) \ d\varPhi ^a \, d\varPhi ^b \end{aligned}$$

(106)

On the other hand, the spatial correlation of the two sample functions $g_j(\varvec{\xi };\theta )$ and $g_k(\varvec{\xi };\theta )$ over a 2D area of size $L^0_1 \times L^0_2$ reads:

$$\begin{aligned} \tilde{R}_{jk}(\varvec{\xi })&:= \langle g_j(\mathbf {x}; \theta ) \ g_k(\mathbf {x}+ \varvec{\xi }; \theta ) \rangle _{L^0_1 \times L^0_2} \nonumber \\&:= \frac{1}{L^0_1 \ L^0_2} \int _0^{L^0_1} \int _0^{L^0_2} g_j(\mathbf {x}; \theta ) \ g_k(\mathbf {x}+\varvec{\xi }; \theta ) \ dx_1 \, dx_2 \end{aligned}$$

(107)

Then, we have from Eq. (105):

$$\begin{aligned}&g_j(\mathbf {x}; \theta ) \ g_k(\mathbf {x}+ \varvec{\xi }; \theta ) = 4 \varDelta \kappa _1 \varDelta \kappa _2 \nonumber \\&\qquad \qquad \qquad \qquad \times \sum _{l^a=1}^m \sum _{\alpha ^a=1}^2 \sum _{n_1^a=0}^{N_1-1} \sum _{n_2^a=0}^{N_2-1} \sum _{l^b=1}^m \sum _{\alpha ^b=1}^2 \sum _{n_1^b=0}^{N_1-1} \sum _{n_2^b=0}^{N_2-1} \vert H_{jl^a}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}) \vert \nonumber \\&\qquad \qquad \qquad \qquad \times \vert H_{kl^b}(\varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}) \vert \ A_{jkl^al^b}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}; \mathbf {x}, \varvec{\xi }; \theta ) \end{aligned}$$

(108)

where we introduced

$$\begin{aligned}&A_{jkl^al^b}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}; \mathbf {x}, \varvec{\xi }; \theta ) = \cos (\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a} \cdot \mathbf {x}- \varphi _{jl^a}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}) + \varPhi _{l^a, n_1^a n_2^a}^{\alpha ^a}(\theta )) \nonumber \\&\qquad \qquad \times \cos (\varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b} \cdot (\mathbf {x}+ \varvec{\xi }) - \varphi _{kl^b}(\varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}) + \varPhi _{l^b, n_1^b n_2^b}^{\alpha ^b}(\theta )) \end{aligned}$$

(109)

Using now the relation $\cos \beta \, \cos \gamma = \frac{1}{2} \left\{ \cos (\beta + \gamma ) + \cos (\beta - \gamma ) \right\}$, it comes:

$$\begin{aligned}&A_{jkl^al^b}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}; \mathbf {x}, \varvec{\xi }; \theta ) = \frac{1}{2} \Big \{ \cos \left( (\varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b} + \varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}) \cdot \mathbf {x}+ \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b} \cdot \varvec{\xi }- \varphi _{jl^b}(\varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}) \right. \nonumber \\&\qquad \left. -\, \varphi _{kl^a}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}) + \varPhi _{l^b, n_1^b n_2^b}^{\alpha ^b}(\theta ) + \varPhi _{l^a, n_1^a n_2^a}^{\alpha ^a}(\theta ) \right) + \cos \left( (\varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b} - \varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}) \cdot \mathbf {x}\right. \nonumber \\&\qquad \left. +\, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b} \cdot \varvec{\xi }- \varphi _{jl^b}(\varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}) + \varphi _{kl^a}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}) + \varPhi _{l^b, n_1^b n_2^b}^{\alpha ^b}(\theta ) - \varPhi _{l^a, n_1^a n_2^a}^{\alpha ^a}(\theta ) \right) \Big \} \end{aligned}$$

(110)

To calculate the ensemble correlations $R_{jk}(\varvec{\xi })$, we have to calculate:

$$\begin{aligned} B_{jkl^al^b}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}; \mathbf {x}, \varvec{\xi }) = \int _0^{2\pi } \int _0^{2\pi } A_{jkl^al^b}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}; \mathbf {x}, \varvec{\xi }; \theta ) \ d\varPhi ^a \, d\varPhi ^b \end{aligned}$$

(111)

Because functions $A_{jkl^al^b}$ are periodic of period $2\pi$, functions $B_{jkl^al^b} = 0$ except in the case where $\varPhi _{l^b, n_1^b n_2^b}^{\alpha ^b}(\theta ) = \varPhi _{l^a, n_1^a n_2^a}^{\alpha ^a}(\theta )$, that is as $n_1^a = n_1^b = n_1$ and $n_2^a = n_2^b = n_2$ and $\alpha ^a = \alpha ^b = \alpha$ and $l^a = l^b = l$. This yields:

$$\begin{aligned} B_{jkl^al^b}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}; \mathbf {x}, \varvec{\xi }) = 2 \pi ^2 \cos \left( \varvec{\kappa }_{n_1 n_2}^{\alpha } \cdot \varvec{\xi }- \varphi _{jl}(\varvec{\kappa }_{n_1 n_2}^{\alpha }) + \varphi _{kl}(\varvec{\kappa }_{n_1 n_2}^{\alpha }) \right) \end{aligned}$$

(112)

and, finally:

$$\begin{aligned} R_{jk}(\varvec{\xi }) = 2 \varDelta \kappa _1 \varDelta \kappa _2 \sum _{l=1}^m \sum _{\alpha =1}^2 \sum _{n_1=0}^{N_1-1} \sum _{n_2=0}^{N_2-1} \cos \left( \varvec{\kappa }_{n_1 n_2}^{\alpha } \cdot \varvec{\xi }- \varphi _{jl}(\varvec{\kappa }_{n_1 n_2}^{\alpha }) + \varphi _{kl}(\varvec{\kappa }_{n_1 n_2}^{\alpha }) \right) \end{aligned}$$

(113)

Then, to calculate the spatial correlations $\tilde{R}_{jk}(\varvec{\xi })$, we have to calculate:

$$\begin{aligned} \tilde{B}_{jkl^al^b}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}; \varvec{\xi }; \theta ) = \int _0^{L^0_1} \int _0^{L^0_2} A_{jkl^al^b}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}; \mathbf {x}, \varvec{\xi }; \theta ) \ dx_1 \, dx_2 \end{aligned}$$

(114)

Because functions $A_{jkl^al^b}$ are periodic of period $L^0_1 \times L^0_2$, and with the condition that $H_{jk}=0$ as any $\kappa _i = 0$, for any $(j,k) \in [1,\ldots ,m]^2$ and for any $i \in [1,\ldots ,d]$, functions $\tilde{B}_{jkl^al^b}$ are equal to zero, except if $\varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b} = \varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}$, that is $n_1^a = n_1^b = n_1$ and $n_2^a = n_2^b = n_2$ and $\alpha ^a = \alpha ^b = \alpha$, in which case:

$$\begin{aligned} \tilde{B}_{jkl^al^b}(\varvec{\kappa }_{n_1^a n_2^a}^{\alpha ^a}, \varvec{\kappa }_{n_1^b n_2^b}^{\alpha ^b}; \varvec{\xi }; \theta )&= \frac{L^0_1 \ L^0_2}{2} \cos \left( \varvec{\kappa }_{n_1 n_2}^{\alpha } \cdot \varvec{\xi }- \varphi _{jl^b}(\varvec{\kappa }_{n_1 n_2}^{\alpha }) \right. \nonumber \\&\qquad \left. +\,\varphi _{kl^a}(\varvec{\kappa }_{n_1 n_2}^{\alpha }) + \varPhi _{l^b, n_1 n_2}^{\alpha }(\theta ) - \varPhi _{l^a, n_1 n_2}^{\alpha }(\theta ) \right) \end{aligned}$$

(115)

With this expression of the functions $\tilde{B}_{jkl^al^b}$, we do not have $\tilde{R}(\varvec{\xi }) = R(\varvec{\xi })$. However, when wave-number shifts are introduced so that wave numbers $\varvec{\kappa }$ become dependent on the index l (as in [7, 26]), the condition $l^a = l^b = l$ has to be added for $\tilde{B}_{jkl^al^b}$ not to be equal to zero. Consequently, $\varPhi _{l^b, n_1 n_2}^{\alpha }(\theta ) = \varPhi _{l^a, n_1 n_2}^{\alpha }(\theta )$ in Eq. (115) and we finally recover $R(\varvec{\xi }) = \tilde{R}(\varvec{\xi })$, meaning that sample fields $g_j(\varvec{\xi }; \theta )$ are ergodic in correlation.

Appendix 2: Material Model at Mesoscale for the Numerical Applications

For the one-dimensional material model at mesoscale used in the numerical applications shown in Sect. 4, we use $h(\sigma )=\vert \sigma \vert$ in the definition of the criterium function (see Eq. 11 in Sect. 2.2):

$$\begin{aligned} \phi _{n+1} = \vert \sigma _{n+1} \vert - \sigma _y \quad \Rightarrow \quad \nu _{n+1} := \frac{\partial \phi _{n+1}}{\partial \sigma _{n+1}} = sign(\sigma _{n+1}) \end{aligned}$$

(116)

Then, Eq. (61) can be written as:

$$\begin{aligned} \sigma _{n+1} = \sigma ^{trial}_{n+1} - D^{-1}_n \gamma _{n+1} sign(\sigma _{n+1}) \end{aligned}$$

(117)

Multiplying both sides of Eq. (117) by $sign(\sigma _{n+1})$, it comes:

$$\begin{aligned} \vert \sigma _{n+1} \vert = \sigma ^{trial}_{n+1} sign(\sigma _{n+1}) - D^{-1}_n \gamma _{n+1} \end{aligned}$$

(118)

Multiplying now both sides of Eq. (118) by $sign(\sigma ^{trial}_{n+1})$, it comes:

$$\begin{aligned} \left( \vert \sigma _{n+1} \vert + D^{-1}_n \gamma _{n+1} \right) sign(\sigma ^{trial}_{n+1}) = \vert \sigma ^{trial}_{n+1} \vert sign(\sigma _{n+1}) \end{aligned}$$

(119)

Setting $\gamma _0 = 0$ and $D_0 > 0$, $\left( \vert \sigma _{n+1} \vert + D^{-1}_n \gamma _{n+1} \right)$ necessarily is non-negative because $\dot{\gamma } \ge 0$ and $\dot{D} \vert \sigma \vert = r \, \dot{\gamma } \ge 0$. Consequently:

$$\begin{aligned} sign(\sigma _{n+1}) = sign(\sigma ^{trial}_{n+1}) \end{aligned}$$

(120)

Then, we have from Eqs. (117) and (116):

$$\begin{aligned}&\vert \sigma _{n+1} \vert = \vert \sigma ^{trial}_{n+1} \vert - D^{-1}_n \gamma _{n+1} \end{aligned}$$

(121)

$$\begin{aligned}&\phi _{n+1} = \phi ^{trial}_{n+1} - D^{-1}_n \gamma _{n+1} \end{aligned}$$

(122)

with $\phi ^{trial}_{n+1} = \vert \sigma ^{trial}_{n+1} \vert - \sigma _y$, from which we can calculate $\gamma _{n+1}$ in case of inelastic evolution:

$$\begin{aligned} \phi _{n+1} = 0 \quad \Rightarrow \quad \gamma _{n+1} = D_n \, \phi ^{trial}_{n+1} \end{aligned}$$

(123)

Appendix 3: Translation from Gaussian to Uniform Distributions

Let $\mathfrak {a}_1$ and $\mathfrak {a}_2$ be two independent normal Gaussian variables: $(\mathfrak {a}_1, \mathfrak {a}_2) \sim \mathscr {N}(0,1)$. Then $\mathfrak {b} = \exp (-(\mathfrak {a}_1^2 + \mathfrak {a}_2^2) \slash 2)$ is a random variable with uniform distribution in $[0, \, 1]$: $\mathfrak {b} \sim \mathscr {U}(0,1)$. Indeed:

$$\begin{aligned} \Pr [\mathfrak {b} \le b] = \frac{1}{2\pi } \int _{ \{ (a_1, a_2) \vert e^{-\frac{1}{2}(a_1^2 + a_2^2)} \le b \} } e^{-\frac{1}{2}(a_1^2 + a_2^2)} da_1 da_2 \end{aligned}$$

(124)

Then:

• if $b > 1$, $\Pr [\mathfrak {b} \le b] = 1$ because $e^{-\frac{1}{2}(a_1^2 + a_2^2)} \le 1$, $\forall (a_1, a_2) \in \mathbb {R}^2$;

• if $b < 0$, $\Pr [\mathfrak {b} \le b] = 0$ because $e^{-\frac{1}{2}(a_1^2 + a_2^2)} > 1$, $\forall (a_1, a_2) \in \mathbb {R}^2$;

• and, if $0\le b \le 1$, we can rewrite relation (124) with polar coordinates as:

$$\begin{aligned} \Pr [\mathfrak {b} \le b] = \frac{1}{2\pi } \int _0^{2\pi } \int _{\sqrt{-2 \ln b}}^{+\infty } \ e^{-\frac{1}{2} r^2} \ r \, dr d\theta = \left[ -e^{-\frac{1}{2} r^2}\right] _{\sqrt{-2 \ln b}}^{+\infty } = b \end{aligned}$$