Modeling plasticity by non-continuous deformation

Abstract

Plasticity and failure theories are still subjects of intense research. Engineering constitutive models on the macroscale which are based on micro characteristics are very much in need. This study is motivated by the observation that continuum assumptions in plasticity in which neighbour material elements are inseparable at all-time are physically impossible, since local detachments, slips and neighbour switching must operate, i.e. non-continuous deformation. Material microstructure is modelled herein by a set of point elements (particles) interacting with their neighbours. Each particle can detach from and/or attach with its neighbours during deformation. Simulations on two- dimensional configurations subjected to uniaxial compression cycle are conducted. Stochastic heterogeneity is controlled by a single “disorder” parameter. It was found that (a) macro response resembles typical elasto–plastic behaviour; (b) plastic energy is proportional to the number of detachments; (c) residual plastic strain is proportional to the number of attachments, and (d) volume is preserved, which is consistent with macro plastic deformation. Rigid body displacements of local groups of elements are also observed. Higher disorder decreases the macro elastic moduli and increases plastic energy. Evolution of anisotropic effects is obtained with no additional parameters.

Appendices

Appendix 1

Analytical approximation for $\langle M \rangle$ using PRS of 3 $\times$ 3 cells

Expression for $\langle M \rangle$ is obtained by calculating the distance distributions between two random points (particles) on a ‘perpendicular-line’ (Fig. 16a), and a ’diagonal-line’ (Fig. 16b). Then, calculating the discrete PDF for $n = 0,1,2, \ldots ,8$ interactions and evaluating the mean.

Fig. 16

Schematic sketch of distance between two random points (black dots) on: a ‘perpendicular-line’ interaction (solid black line). b ‘diagonal-line’ interaction (dashed black line)

Let two random points $(x^{1}_{1}, x^{2}_{1})$ and $(x^{1}_{2}, x^{2}_{2})$, where the $(_{\mathrm{I}})$ is the particle and $(^{\mathrm{i}})$ is the coordinate as shown in Fig. 16a. Also define r as the distance between the two points. Assuming that $x^{1}_{1} and x^{1}_{2}$ are independent and evenly distributed in the interval $(1-\surd \lambda , 1+\surd \lambda )$ and also $x^{2}_{1} and x^{2}_{2}$ in $(0,\surd \lambda )$, the PDF for the event $(x^{1}_{1}-x^{2}_{1})^{2}+(x^{1}_{2}-x^{2}_{2})^{2} \le r^{2}$ is the convolution $f_{\mathrm{p}}$ of $g_{1}$ and $g_{2}$ [14] given by:

$$\begin{aligned} g_1 \left( {q=\left( {\frac{x_1^1 -x_1^2 }{\alpha }} \right) ^{2}} \right)= & {} \left\{ {{\begin{array}{l} {\frac{1}{2\beta ^{2}}\left( {\frac{\beta -1}{\sqrt{q}}+1} \right) \quad ,\left( {1-\beta } \right) ^{2}\le q\le 1} \\ {\frac{1}{2\beta ^{2}}\left( {\frac{\beta +1}{\sqrt{q}}-1} \right) \quad ,1\le q\le \left( {\beta +1} \right) ^{2}} \\ \end{array} }} \right\} , \end{aligned}$$

(25)

$$\begin{aligned} g_2 \left( {q=\left( {\frac{x_2^1 -x_2^2 }{\alpha }} \right) ^{2}} \right)= & {} \left\{ {\frac{1}{\beta ^{2}}\left( {\frac{\beta }{\sqrt{q}}-1} \right) \quad ,0\le q\le \lambda } \right\} , \end{aligned}$$

(26)

where ${ \beta =}\sqrt{\lambda }$. Convolving (25) with (26),

$$\begin{aligned} f_p =\frac{1}{\lambda ^{2}}\left\{ {{\begin{array}{ll} {2\beta \sqrt{r^{2}-\left( {1-\beta } \right) ^{2}}-\left( {r+1-\beta } \right) ^{2}-\beta \left( {1-\beta } \right) \left( {T\left( {\frac{1-\beta }{r}} \right) +\frac{\pi }{2}} \right) ,}&{}\quad {1-\beta \le r\le \sqrt{\beta ^{2}+\left( {1-\beta } \right) ^{2}}} \\ {\left( {1-\beta } \right) \left( {\frac{\beta ^{2}}{\left( {1-\beta } \right) }+2\left( {r-\sqrt{r^{2}-\beta ^{2}}} \right) +\beta \left( {T\left( {\frac{\beta }{r}} \right) -\frac{\pi }{2}} \right) } \right) ,}&{} {\sqrt{\beta ^{2}+\left( {1-\beta } \right) ^{2}}\le r\le 1} \\ {\begin{array}{l} \beta ^{2}+2\left( {r^{2}-2\beta \sqrt{r^{2}-1}-\left( {1-\beta } \right) \sqrt{r^{2}-\beta ^{2}}+1} \right) \\ +\left( {1+\beta } \right) \left( {\frac{\pi \beta -4r}{2}} \right) +\beta \left( {1-\beta } \right) T\left( {\frac{\beta }{r}} \right) +2\beta T\left( {\frac{1}{r}} \right) , \\ \end{array}}&{} {1\le r\le \sqrt{1+\beta ^{2}}} \\ {\left( {1+\beta } \right) \left( {\frac{\pi \beta }{2}-2r-\frac{\beta ^{2}}{\left( {1+\beta } \right) }+2\sqrt{r^{2}-\beta ^{2}}-\beta T\left( {\frac{\beta }{r}} \right) } \right) ,}&{} {\sqrt{1+\beta ^{2}}\le r\le 1+\beta } \\ {\begin{array}{l} 2\beta \sqrt{r^{2}-\left( {1+\beta } \right) ^{2}}+\left( {1+\beta } \right) \left( {2\sqrt{r^{2}-\beta ^{2}}-1-\beta } \right) \\ -\left( {r^{2}+\beta ^{2}} \right) -\beta \left( {1+\beta } \right) \left( {T\left( {\frac{\beta }{r}} \right) +T\left( {\frac{1+\beta }{r}} \right) } \right) , \\ \end{array}}&{} {1+\beta \le r\le \sqrt{\beta ^{2}+\left( {1+\beta } \right) ^{2}}} \\ \end{array} }} \right\} , \end{aligned}$$

(27)

where $T(x)=\hbox {sin}^{-1}(1-2x^{2})$ for $0 \le x \le 1$. Notice that since q in (25) and (26) is the square of the distance, a change of variables is performed in (27) to obtain r.

The probability that a particle will interact with its neighbour located within a circle of radius R is,

$$\begin{aligned} P_1 =\mathrm{Prob}\left( r\le \frac{R}{\alpha }\right) =\int \limits _0^{R/\alpha } {f_\mathrm{p} \left( {\lambda ,r} \right) r\mathrm{d}r} \end{aligned}$$

(28)

which we assume is the same for all ‘perpendicular-line’ interactions appears in Fig. 3.

Next, consider two random points on diagonal-line: $(x^{1}_{1}, x^{2}_{1})$ and $(x^{1}_{2}, x^{2}_{2})$ as shown in Fig. 16b. Assuming that $x^{1}_{1} and x^{1}_{2}$ are independent and evenly distributed in the interval $(1-\surd \lambda , 1+\surd \lambda )$ and also $x^{2}_{1} and x^{2}_{2}$ in the same interval, the PDF for the event $(x^{1}_{1}-x^{2}_{1})^{2}+(x^{1}_{2}-x^{2}_{2})^{2} \le r^{2}$ is the convolution $f_{d}$ of $g_{1}$ with itself:

$$\begin{aligned} f_\mathrm{d} =\frac{1}{\lambda ^{2}}\left\{ {{\begin{array}{ll} {\frac{1}{2}\left( {2\left( {1-\beta } \right) -\sqrt{r^{2}-\left( {1-\beta } \right) ^{2}}} \right) ^{2}-\left( {1-\beta } \right) ^{2}\left( {\frac{1}{2}-T\left( {\frac{1-\beta }{r}} \right) } \right) ,}&{} \quad {\sqrt{2}\left( {1-\beta } \right) \le r\le \sqrt{1+\left( {1-\beta } \right) ^{2}}} \\ {\begin{array}{l} \left( {1-\beta } \right) \left( {4\sqrt{r^{2}-1}-2T\left( {\frac{1}{r}} \right) } \right) -\left( {\frac{4+3r^{2}}{2}} \right) \\ +2\left( {1+\beta } \right) \sqrt{r^{2}-\left( {1-\beta } \right) ^{2}}-\left( {1-\beta ^{2}} \right) \left( {1+T\left( {\frac{1-\beta }{r}} \right) } \right) , \\ \end{array}}&{} {\sqrt{1+\left( {1-\beta } \right) ^{2}}\le r\le \sqrt{2}} \\ {\begin{array}{l} \frac{4+r^{2}}{2}-\left( {1-\beta } \right) ^{2}-\left( {1-\beta ^{2}} \right) T\left( {\frac{1-\beta }{r}} \right) \\ +2\left( {1+\beta } \right) \left( {\sqrt{r^{2}-\left( {1-\beta } \right) ^{2}}-2\sqrt{r^{2}-1}+2T\left( {\frac{1}{r}} \right) } \right) , \\ \end{array}}&{} {\sqrt{2}\le r\le \sqrt{2\left( {1+\beta ^{2}} \right) }} \\ {\begin{array}{l} \left( {1+\beta } \right) \left( {\beta +2-4\sqrt{r^{2}-1}+2T\left( {\frac{1}{r}} \right) } \right) \\ +\left( {1-\beta ^{2}} \right) T\left( {\frac{1+\beta }{r}} \right) +\left( {1-\beta } \right) \left( {\frac{1}{2}-2\sqrt{r^{2}-\left( {1+\beta } \right) ^{2}}} \right) , \\ \end{array}}&{} {\sqrt{2\left( {1+\beta ^{2}} \right) }\le r\le \sqrt{1+\left( {1+\beta } \right) ^{2}}} \\ {2\left( {1+\beta } \right) \sqrt{r^{2}-\left( {1+\beta } \right) ^{2}}-\frac{r^{2}}{2}-\left( {1+\beta } \right) ^{2}\left( {1+T\left( {\frac{1+\beta }{r}} \right) } \right) ,}&{} {\sqrt{1+\left( {1+\beta } \right) ^{2}}\le r\le \sqrt{2}\left( {1+\beta } \right) } \\ \end{array} }} \right\} . \end{aligned}$$

(29)

The probability that a particle will interact with its neighbour located within a circle of radius R is,

$$\begin{aligned} P_2 =\mathrm{Prob}\left( r\le \frac{R}{\alpha }\right) =\int \limits _0^{R/\alpha } {f_\mathrm{d} \left( {\lambda ,r} \right) r\mathrm{d}r} \end{aligned}$$

(30)

again, we assume $P_{2}$ the same for all ‘diagonal-line’ interactions appears in Fig. 3.

Given the probabilities (28) and (30), we can precede to derive an expression for the discrete PDF for $n=0,1,2,{\ldots },8$ interactions. This task is analogical to: playing two separate games (independently) where in each game the player is trying to fill 4 empty cells with 4 tries, each with different probability to fill: $P_{1}$ and $P_{2}$, respectively. What is the probability of having $n = 0, 1, 2,$ etc. cells filled? Let the number of interactions from each type (‘perpendicular’ or ‘diagonal’) be $M_{1}, M_{2}$ respectively. The $M_{i}{} { \sim Binomial}(k = 4,P_{\mathrm{i}})$ and thus,

$$\begin{aligned} \begin{array}{l} \Pr \left\{ {M_1 +M_2 =n} \right\} \\ =\sum \limits _{m=\max \left\{ {0,n-4} \right\} }^{\min \left\{ {4,n} \right\} } {\Pr \left\{ {M_1 =n-m} \right\} \Pr \left\{ {M_2 =m} \right\} } \\ =\sum \limits _{m=\max \left\{ {0,n-4} \right\} }^{\min \left\{ {4,n} \right\} } \left( {{\begin{array}{l} 4 \\ {n-m} \\ \end{array} }} \right) P_1 ^{n-m}\left( {1-P_1 } \right) ^{4-n+m}\left( {{\begin{array}{l} 4 \\ m \\ \end{array} }} \right) \\ \quad \times P_2 ^{m}\left( {1-P_2 } \right) ^{4-m} \\ \end{array} \end{aligned}$$

(31)

for $n=0,1,2,{\ldots },8$. The mean of the discrete PDF in (31) is equal to,

$$\begin{aligned} \left\langle M \right\rangle =\sum _{n=0}^8 {n\Pr \left\{ {M_1 +M_2 =n} \right\} =4\left( {P_1 +P_2 } \right) }. \end{aligned}$$

(32)

Substitute Eqs. (28) and (30) into (32) and divide by 8, which is the number of interactions at $\lambda =0$, we get (4).

Appendix 2

Analytic approximation for $\langle E_{22} \rangle$ using a PRS of 2 $\times$ 2 cells

Define PRS of 2 $\times$ 2 cells where each pair of particles interacts, as shown in Fig. 17.

Fig. 17

PRS of 2 $\times$ 2 cells filled with 4 particles (white circles), where each pair of particles interacts in accordance to ‘perpendicular-line’ (solid black line) and ‘diagonal-line’ (dashed black line) interaction, respectively

The structure is subjected to the following symmetry boundary conditions:

$$\begin{aligned} u_1^1 =u_1^2 =u_2^2 =0. \end{aligned}$$

(33)

For small deformations, the displacement field of the structure u is given by the equilibrium equations:

$$\begin{aligned} \mathbf{u=K}^{\mathbf{-1}}{} \mathbf{F}=\mathbf{SF}, \end{aligned}$$

(34)

where F is the external load vector

$$\begin{aligned} \mathbf{F}=\frac{f}{2}\left[ {{\begin{array}{lllll} 0&{} 0&{} {-1}&{} 0&{} {-1} \\ \end{array} }} \right] ^\mathrm{T} \end{aligned}$$

(35)

and K is the global stiffness matrix,

$$\begin{aligned} \mathbf{K}=\frac{k_0 }{2}\left[ {{\begin{array}{lllll} {\cos ^{2}\theta _1 +2\cos ^{2}\theta _4 +\cos ^{2}\theta _5 }&{} {-2\cos ^{2}\theta _4 }&{} {-\sin 2\theta _4 }&{} {-\cos ^{2}\theta _5 }&{} {-\frac{1}{2}\sin 2\theta _5 } \\ .&{} {\cos ^{2}\theta _2 +2\cos ^{2}\theta _4 +\cos ^{2}\theta _6 }&{} {\frac{1}{2}\sin 2\theta _2 +\sin 2\theta _4 +\frac{1}{2}\sin 2\theta _6 }&{} {-\cos ^{2}\theta _6 }&{} {-\frac{1}{2}\sin 2\theta _6 } \\ .&{} .&{} {\sin ^{2}\theta _2 +2\sin ^{2}\theta _4 +\sin ^{2}\theta _6 }&{} {-\frac{1}{2}\sin 2\theta _6 }&{} {-\sin ^{2}\theta _6 } \\ .&{} .&{} .&{} {2\cos ^{2}\theta _3 +\cos ^{2}\theta _5 +\cos ^{2}\theta _6 }&{} {\sin 2\theta _3 +\frac{1}{2}\sin 2\theta _5 +\frac{1}{2}\sin 2\theta _6 } \\ {\mathrm{sym}.}&{} .&{} .&{} .&{} {2\sin ^{2}\theta _3 +\sin ^{2}\theta _5 +\sin ^{2}\theta _6 } \\ \end{array} }} \right] \nonumber \\ \end{aligned}$$

(36)

${ \theta }_{m}$ is the angle of interaction m relative to ${{\varvec{e}}}_{1}$. From periodic geometry considerations, the stiffness of interaction k,

$$\begin{aligned} k=\left\{ {{\begin{array}{ll} {k_0 \quad \quad \mathrm{for}\quad m=3,4} \\ {{k_0 }/2 \quad \mathrm{for}\quad m=1,2,5,6} \\ \end{array} }} \right\} . \end{aligned}$$

(37)

We classify $\theta$ as ${ \theta }_\mathrm{p}$ and ${ \theta }_\mathrm{d}$ for ‘perpendicular-line’ and ‘diagonal-line’ of interaction, respectively. For the PRS shown in Fig. 17,

$$\begin{aligned} \begin{array}{l} \theta _1 =\theta _6 =\theta _\mathrm{p}; \quad \theta _2 =\theta _5 =\theta _\mathrm{p} +\pi /2 \\ \theta _3 =\theta _\mathrm{d}; \quad \quad \quad \quad \theta _4 =\theta _\mathrm{d} +\pi /2 \\ \end{array} \end{aligned}$$

(38)

${ \theta }_\mathrm{p}$ and ${ \theta }_\mathrm{d}$ have distinctive PDF which depends on $\lambda$ and can be found analytically, similar to the method used in Appendix 1. Substituting (38) into (36) yields a stochastic stiffness matrix:

$$\begin{aligned} \mathbf{K}=\frac{k_0 }{2}\left[ {{\begin{array}{lllll} {1+2\sin ^{2}\theta _\mathrm{d} }&{} {-2\sin ^{2}\theta _\mathrm{d} }&{} {\sin 2\theta _\mathrm{d }}&{} {-\sin ^{2}\theta _\mathrm{p} }&{} {\frac{1}{2}\sin 2\theta _\mathrm{p} } \\ .&{} {1+2\sin ^{2}\theta _\mathrm{d }}&{} {-\sin 2\theta _\mathrm{d }}&{} {-\cos ^{2}\theta _\mathrm{p} }&{} {-\frac{1}{2}\sin 2\theta _\mathrm{p} } \\ .&{} .&{} {1+2\cos ^{2}\theta _\mathrm{d} }&{} {-\frac{1}{2}\sin 2\theta _\mathrm{p }}&{} {-\sin ^{2}\theta _\mathrm{p }} \\ .&{} .&{} .&{} {1+2\cos ^{2}\theta _\mathrm{d }}&{} {\sin 2\theta _\mathrm{d} } \\ \mathrm{sym.}&{} .&{} .&{} .&{} {1+2\sin ^{2}\theta _\mathrm{d} } \\ \end{array} }} \right] .\nonumber \\ \end{aligned}$$

(39)

Substituting Eqs. (39) and (35) into (34), we derive an expression for the stochastic displacement field u. More specifically, the structure’s average displacement in e $_{2}$,

$$\begin{aligned} \bar{{u}}=\frac{u_3^2 +u_4^2 }{2}=\frac{f}{4}\left( {S_{33} +2S_{35} +S_{55} } \right) , \end{aligned}$$

(40)

where $S_{ij}$ are entries from the compliance matrix S.

The initial stochastic stiffness $E_{22}$ is the total force f divided by (40):

$$\begin{aligned} E_{22} =\frac{4}{\left( {S_{33} +2S_{35} +S_{55} } \right) }. \end{aligned}$$

(41)

Dividing (41) by ${E}_{0}=3k_{0}/2$ and calculating the inverse of (39) we derive Eq. (22).