IM_Model

IM (Ianniruberto, G.and Marrucci, G.) Model

The Classic Single Mode IM Differential Model, or DCR-CS Model

ref: Ianniruberto, G.; Marrucci, G. A simple constitutive equation for entangled polymers with chain stretch. Journal of Rheology 2001, 45, 1305-1318.
Define an effective time $\tau_{eff}$ to replace the disengagement time $\tau_d$ in DE model:
The classical DE model of $\mathbf S(t)$ is replaced to be:

$$\begin{align} &\overset{\bigtriangledown }{\mathbf S^2} +2\mathbf S^2(\boldsymbol{\kappa}:\mathbf S)+\frac{2}{\tau}\mathbf S\cdot\left(\mathbf S-\frac{1}{3}\mathbf I\right) = \mathbf 0\\ &\tau = \frac{1}{2\left(\frac{1}{\tau_d}+\boldsymbol{\kappa}:\mathbf S\right)} + \tau_R\\ &\frac{d\lambda}{dt} = \lambda\boldsymbol{\kappa}:\mathbf S - \frac{\lambda F(\lambda)-1}{\tau_R} \\ &F(\lambda) = \left(\frac{\lambda_{max}^2-\frac{\lambda^2}{3}}{\lambda_{max}^2-\lambda^2}\right)\left(\frac{\lambda_{max}^2-1}{\lambda_{max}^2-\frac{1}{3}}\right) \\ &\boldsymbol{\sigma} = 3G_N^0 F(\lambda)\lambda^2\boldsymbol{S} \end{align}$$

Here:
$$\overset{\bigtriangledown }{\mathbf S^2} = \mathbf S\cdot \dot{\mathbf S} + \dot{\mathbf S}\cdot \mathbf S -\boldsymbol{\kappa}\cdot\mathbf S^2-\mathbf S^2\cdot\boldsymbol{\kappa}^T$$

The Classical Single Mode IM Intergral Model

ref: Ianniruberto, G.; Marrucci, G. A simple constitutive equation for entangled polymers with chain stretch. Journal of Rheology 2001, 45, 1305-1318.
Define an effective time $\tau$ to replace the disengagement time $\tau_d$ in DE model:
The classical DE model of $\mathbf S(t)$ is replaced to be:

$$\begin{align} \mathbf S(t) &= \int_{-\infty}^t \left[\frac{1}{\tau(t')}\right] \exp \left[-\int_{t'}^t \frac{dt''}{\tau(t'')}\right] \mathbf Q\left[\mathbf E(t,t')\right] dt' \\ \tau &= \frac{1}{2\left(\frac{1}{\tau_d}+\boldsymbol{\kappa}:\mathbf S\right)} +\tau_R \\ \frac{d\lambda}{dt} &= \lambda\boldsymbol{\kappa}:\mathbf S - \frac{\lambda F(\lambda)-1}{\tau_R} \\ F(\lambda)&=\left(\frac{\lambda_{max}^2-\frac{\lambda^2}{3}}{\lambda_{max}^2-\lambda^2}\right)\left(\frac{\lambda_{max}^2-1}{\lambda_{max}^2-\frac{1}{3}}\right) \\ \boldsymbol{\sigma} &= 3G_N^0F(\lambda)\lambda^2\boldsymbol{S} \end{align}$$

The Classic Multi Mode IM Integral Model

ref: Costanzo, S.; Huang, Q.; Ianniruberto, G.; Marrucci, G.; Hassager, O.; Vlassopoulos, D. Shear and Extensional Rheology of Polystyrene Melts and Solutions with the Same Number of Entanglements. Macromolecules 2016, 49, 3925-3935.

$$\begin{align} \mathbf S_i(t) &= \int_{-\infty}^t \left[\frac{1}{\tau_i(t')}\right] \exp \left[-\int_{t'}^t \frac{dt''}{\tau_i(t'')}\right] \mathbf Q\left[\mathbf E(t,t')\right] dt' \\ \tau_i(t) &= \frac{1}{2\left(\frac{1}{\tau_{i,eq}}+\boldsymbol{\kappa}:\mathbf S_i\right)} + \tau_R \\ \frac{d\lambda}{dt} &= \lambda\boldsymbol{\kappa}:\overline{\mathbf S} - \frac{\lambda F(\lambda)-1}{\tau_R} \\ F(\lambda)&=\left(\frac{\lambda_{max}^2-\frac{\lambda^2}{3}}{\lambda_{max}^2-\lambda^2}\right)\left(\frac{\lambda_{max}^2-1}{\lambda_{max}^2-\frac{1}{3}}\right) \\ \overline{\mathbf S} &= \int_{-\infty}^t \left[\frac{1}{\tau_d(t')}\right] \exp \left[-\int_{t'}^t \frac{dt''}{\tau_d(t'')}\right] \mathbf Q\left[\mathbf E(t,t')\right] dt'\\ \tau_d(t) &= \frac{\sum_iG_i\tau_i^2(t)}{\sum_i G_i\tau_i(t)}\\ \boldsymbol{\sigma} &= C_Q F(\lambda)\lambda^2\sum{G_i\boldsymbol{S}_i} \end{align}$$

Here, $C_Q = 6$ if $\mathbf Q = \frac{\mathbf B^{1/2}}{\mathrm{Tr}\mathbf B^{1/2}}$

IM Tumbling Multi Mode Model

ref: Costanzo, S.; Huang, Q.; Ianniruberto, G.; Marrucci, G.; Hassager, O.; Vlassopoulos, D. Shear and Extensional Rheology of Polystyrene Melts and Solutions with the Same Number of Entanglements. Macromolecules 2016, 49, 3925-3935.

$$\begin{align} \mathbf S_i(t) &= \int_{-\infty}^t \left[\frac{1}{\tau_i(t')}\right] \exp \left[-\int_{t'}^t \frac{dt''}{\tau_i(t'')}\right] \mathbf Q\left[\mathbf E(t,t')\right] dt' \\ \tau_i(t) &= \frac{1}{2\left(\frac{1}{\tau_{i,eq}}+\boldsymbol{\kappa}:\mathbf S_i\right)} + \tau_R\\ \frac{d\lambda}{dt} &= \lambda\boldsymbol{\kappa}:\overline{\mathbf S} - \frac{\lambda F(\lambda)-1}{\tau_R} \\ F(\lambda)&=\left(\frac{\lambda_{max}^2-\frac{\lambda^2}{3}}{\lambda_{max}^2-\lambda^2}\right)\left(\frac{\lambda_{max}^2-1}{\lambda_{max}^2-\frac{1}{3}}\right) \\ \overline{\mathbf S} &= \int_{-\infty}^t \left[\frac{1}{\tau_d(t')}\right] \exp \left[-\int_{t'}^t \frac{dt''}{\tau_d(t'')}\right] \mathbf Q\left[\mathbf E(t,t')\right] dt'\\ \tau_d(t) &= \frac{\sum_iG_i\tau_i^2(t)}{\sum_i G_i\tau_i(t)}\\ \boldsymbol{\sigma} &= C_Q F(\lambda)\lambda^2\sum{G_i\boldsymbol{S}_i} \end{align}$$

Here, $C_Q = 6$ if $\mathbf Q = \frac{\mathbf B^{1/2}}{\mathrm{Tr}\mathbf B^{1/2}}$
Tumbling term of $\lambda$ (shear case only):

$$\begin{align} \phi (t) &= \cos(2\pi \omega t)\exp (-\beta t)\\ \frac{d\lambda}{dt} &= \boldsymbol{\kappa}:\overline{\mathbf S}\phi(t)\lambda - \frac{F(\lambda)\lambda -1}{\tau_R}\\ \omega &= \frac{Wi_R^{-0.2}}{8\pi}\dot\gamma \\ \beta &= \frac{Wi_R^{-0.2}}{8}\dot\gamma \end{align}$$

For step shear:

$$\begin{align} \mathbf S(t) =& \int_{0}^t \left[\frac{1}{\tau(t')}\right] \exp \left[-\int_{t'}^t \frac{dt''}{\tau(t'')}\right] \mathbf Q\left[\mathbf E(t,t')\right] dt' \\ &+\int_{-\infty}^0 \left[\frac{1}{\tau(0)}\right] \exp \left[-\int_{t'}^t \frac{dt''}{\tau(t'')}\right] \mathbf Q\left[\mathbf E(t,0)\right] dt' \end{align}$$

The second parts becomes:

$$\begin{align} \mathbf S(t) =& ...\\ &+\int_{-\infty}^0 \frac{1}{\tau(0)} \exp \left[-\int_{t'}^{0} \frac{dt''}{\tau(0)}-\int_0^t\frac{dt''}{\tau(t'')}\right] \mathbf Q\left[\mathbf E(t,0)\right] dt' \\ =& ...\\ &+\int_{-\infty}^0 \frac{1}{\tau(0)} \exp \left[\frac{t'}{\tau(0)}-\int_0^t\frac{dt''}{\tau(t'')}\right] \mathbf Q\left[\mathbf E(t,0)\right] dt' \\ =& ...\\ &+ \exp \left[\frac{t'}{\tau(0)}-\int_0^t\frac{dt''}{\tau(t'')}\right] \mathbf Q\left[\mathbf E(t,0)\right] \bigg |_{t'=-\infty}^{t'=0}\\ =& ...\\ &+ \exp \left[-\int_0^t\frac{dt''}{\tau(t'')}\right] \mathbf Q\left[\mathbf E(t,0)\right] \end{align}$$

Approximation of nonlinear strain measure $\mathbf Q\left[\mathbf E(t,t')\right]$

1. A highly accurate approximation for this strain measure is:
   ref: chapter 11.3 of Dealy, J. M. Larson, R. G. Read, D. J. "Structure and rheology of molten polymers, from structure to flow behavior and back again." 2018, Carl Hanser Verlag GmbH Co KG.

$$\mathbf Q \approx \left(\frac{5}{J-1}\right)\mathbf B - \left[\frac{5}{(J-1)(I_2+13/4)^{1/2}}\right]\mathbf C$$

here:

$$J\equiv I_1+2(I_2+13/4)^{1/2}$$

$$I_1\equiv \mathrm{Tr}(\mathbf B)$$

$$I_2\equiv \mathrm{Tr}(\mathbf C)$$

$\mathbf B$ is the Finger tensor, and $\mathbf C$ is the Cauchy tensor.
2. Marrucci et al. proposed another approximation of this strain measure, which gives a much improved prediction of the normal stress ratio in shear, namely $-N_2/N_1=1/4$ in the limit of small strains, as compared to Doi-Edwards value of 1/7. ref: Milner, S. T. Improved model of nonaffine strain measure. Journal of Rheology 2001, 45, 1023-1028.

$$\mathbf Q(\mathbf E)=\frac{\mathbf C^{-1/2}}{\mathrm{Tr}(\mathbf{C^{-1/2}})}$$

here: $\mathbf C^{-1}=\mathbf E\cdot\mathbf E^T$ is the finger tensor. $\mathbf E$ is the inverse of displacement gradient tensor.

Finger Tensor $\mathbf B$ and Cauchy Tensor $\mathbf C$

For simple shear:

$$\mathbf F (t_0,t_1)=\begin{pmatrix} 1 & \gamma (t_1)-\gamma (t_0) & 0\\ 0 & 1 & 0\\ 0 & 0 &1 \end{pmatrix}$$

$$\mathbf E (t_0,t_1)=\begin{pmatrix} 1 & \gamma (t_0)-\gamma (t_1) & 0\\ 0 & 1 & 0\\ 0 & 0 &1 \end{pmatrix}$$

The Cauchy Tensor, also known as right Cauchy–Green deformation tensor, $\mathbf C$:

$$\begin{align} \mathbf C &= \mathbf F^{T} \cdot \mathbf F \\ &=\begin{pmatrix} 1 & \gamma & 0\\ \gamma & 1+\gamma^2 & 0\\ 0 & 0 &1 \end{pmatrix} \end{align}$$

The Finger Tensor, also known as left Cauchy–Green deformation tensor, $\mathbf B$:

$$\begin{align} \mathbf B &= \mathbf F\cdot\mathbf F^{T} \\ &=\begin{pmatrix} 1+\gamma^2 & \gamma & 0\\ \gamma & 1 & 0\\ 0 & 0 &1 \end{pmatrix} \end{align}$$

Sometimes, Finger strain tensor is also defined by inverse of Cauchy tensor as (ref: Milner, S. T. Journal of Rheology 2001, 45, 1023-1028.):

$$\mathbf B = \mathbf C^{-1} = \mathbf E\cdot \mathbf E^T = \begin{pmatrix} 1+\gamma^2 & -\gamma & 0\\ -\gamma & 1 & 0\\ 0 & 0 &1 \end{pmatrix}$$

Here, the diagonal is the same, while the $B_{0,1}, B_{0,2}, B_{1,2}$ parts are negative.