Constitutive Equation: General Principles

Abstract

In isothermal flow where the conservation of energy is not relevant, there are four scalar balance equations (one conservation of mass and three conservation of linear momentum), and there are 10 scalar variables (3 velocity components, one pressure, and 6 independent stress components – thanks to the conservation of angular momentum, the stress tensor is symmetric). Clearly we do not have a mathematically well-posed problem until 6 extra equations are specified.

Problems

Problem 4.1

In a simple shear deformation of a linear elastic material (4.7), the displacement field takes the form

$$\begin{aligned} v_{1}=\gamma y,\;\;\;v_{2}=0,\;\;\;v_{3}=0, \end{aligned}$$

(4.89)

where $\gamma$ is the amount of shear. Find the elastic stress, in particular, the shear stress. This justifies calling $\mu$ the shear modulus.

Problem 4.2

In a uni-axial extension of a linear elastic material (4.7), the displacement field is given by

$$\begin{aligned} v_{1}=\varepsilon x,\;\;\;v_{2}=-\nu \varepsilon y,\;\;\;v_{3}=-\nu \varepsilon z, \end{aligned}$$

(4.90)

where $\varepsilon$ is the elongational strain, and $\nu$ is the amount of lateral contraction due to the axial elongation, called Poisson’s ratio. If the lateral stresses are zero, show that

$$\begin{aligned} \nu =\frac{\lambda }{{2\left( {\lambda +\mu }\right) }},\;\;\;\mu =\frac{E}{{ 2\left( {1+\nu }\right) }}, \end{aligned}$$

(4.91)

where E is the Young’s modulus, i.e., $T_{xx}=E\varepsilon .$

Problem 4.3

Show that a material element $d\mathbf {X}=dX\mathbf {P},$ where $\mathbf {P}$ is a unit vector, is stretched according to

$$ dx^{2}=dX^{2}\mathbf {C}:\mathbf {PP}, $$

where $\mathbf {C}$ is the right Cauchy-Green tensor. Thus when $\mathbf {P}$ is randomly distributed in space, the average amount of stretch is

$$ \left\langle {dx^{2}}\right\rangle =\frac{1}{3}dX^{2}\mathrm {tr}\mathbf {C}, $$

and therefore $\frac{1}{3}\mathrm {tr}\mathbf {C}$ can be used as a definition of the Weissenberg number.

Problem 4.4

Let $\mathbf {f}$ be a vector-valued, isotropic polynomial of a symmetric tensor $\mathbf {S}$ and a vector $\mathbf {v}.$ Use the integrity basis in (4.36) to prove that

$$ \mathbf {f}(\mathbf {S},\mathbf {v})=[f_{0}\mathbf {1}+f_{1}\mathbf {S}+f_{2} \mathbf {S}^{2}]\mathbf {v}, $$

where the scalar valued coefficients are polynomials in the six invariants involving only $\mathbf {S}$ and $\mathbf {v}$ in the list (4.36).

Problem 4.5

Consider a simple shear deformation of a rubber-like material (4.55), where

$$\begin{aligned} x=X+\gamma Y,\;\;\;y=Y,\;\;\;z=Z. \end{aligned}$$

(4.92)

Show that the Finger strain tensor $\mathbf {B}$ and its inverse are given by

$$\begin{aligned} \left[ \mathbf {B}\right] =\left[ {\begin{array}{*{20}c} {1 + \gamma ^2 }&{} \gamma &{} 0 \\ \gamma &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 \\ \end{array}}\right] ,\;\;\;\left[ {\mathbf {B}^{-1}}\right] =\left[ {\begin{array}{*{20}c}1 &{} { - \gamma } &{} 0 \\ { - \gamma } &{} {1 + \gamma ^2 }&{} 0 \\ 0 &{} 0 &{} 1 \\ \end{array}}\right] . \end{aligned}$$

(4.93)

Consequently, show that the stress tensor is given by

$$\begin{aligned} \left[ \mathbf {T}\right] =-P\left[ \mathbf {I}\right] +\left[ { \begin{array}{*{20}c} {\beta _1 \left( {1 + \gamma ^2 } \right) - \beta _2 }&{} {\left( {\beta _1 + \beta _2 } \right) \gamma } &{} 0 \\ {\left( {\beta _1 + \beta _2 } \right) \gamma } &{} {\beta _1 - \beta _2 \left( {1 + \gamma ^2 }\right) } &{} 0 \\ 0 &{} 0 &{} {\beta _1 - \beta _2 } \\ \end{array}}\right] , \end{aligned}$$

(4.94)

where P is the hydrostatic pressure. Thus, show that the shear stress and the normal stress differences are

$$\begin{aligned} S=\left( {\beta _{1}+\beta _{2}}\right) \gamma ,\;\;\;N_{1}=\left( {\beta _{1}+\beta _{2}}\right) \gamma ^{2},\;\;\;N_{2}=-\beta _{2}\gamma ^{2}. \end{aligned}$$

(4.95)

Deduce that the linear shear modulus of elasticity is

$$\begin{aligned} G=\lim _{\gamma \rightarrow 0}\left( {\beta _{1}+\beta _{2}}\right) . \end{aligned}$$

(4.96)

The ratio

$$\begin{aligned} \frac{{N_{1}}}{S}=\gamma \end{aligned}$$

(4.97)

is independent of the material properties. Such a relation is called universal.

Problem 4.6

In a uniaxial elongational deformation of a rubber-like material (4.55), where (in cylindrical coordinates)

$$\begin{aligned} R=\lambda ^{1/2}r,\;\;\;\varTheta =\theta ,\;\;\;Z=\lambda ^{-1}z, \end{aligned}$$

(4.98)

show that the inverse deformation gradient is

$$\begin{aligned} \left[ {\mathbf {F}^{-1}}\right] =\left[ {\begin{array}{*{20}c}{\frac{{ \partial R}}{{\partial r}}} &{} {\frac{1}{r}\frac{{\partial R}}{{\partial \theta }}} &{} {\frac{{\partial R}}{{\partial z}}} \\ 0 &{} {\frac{R}{r}} &{} 0 \\ {\frac{{\partial Z}}{{\partial r}}} &{} {\frac{1}{r}\frac{{\partial Z}}{{\partial \theta }}} &{} {\frac{{\partial Z}}{{\partial z}}}\\ \end{array}}\right] =\left[ {\begin{array}{*{20}c} {\lambda ^{1/2}}&{} 0 &{} 0 \\ 0 &{} {\lambda ^{1/2} } &{} 0 \\ 0 &{} 0 &{} {\lambda ^{ - 1} }\\ \end{array}} \right] . \end{aligned}$$

(4.99)

Consequently, the Finger strain tensor $\mathbf {B}$ and its inverse $\mathbf { B}^{-1}$ are

$$\begin{aligned} \left[ {\mathbf {B}^{-1}}\right] =\left[ {\mathbf {F}^{-T}\mathbf {F}^{-1}} \right] =\left[ {\begin{array}{*{20}c} \lambda &{} 0 &{} 0 \\ 0 &{} \lambda &{} 0 \\ 0 &{} 0 &{} {\lambda ^{ - 2} } \\ \end{array}}\right] ,\;\;\;\left[ \mathbf {B} \right] =\left[ {\begin{array}{*{20}c} {\lambda ^{ - 1}}&{} 0 &{} 0 \\ 0 &{} {\lambda ^{ - 1} } &{} 0 \\ 0 &{} 0 &{} {\lambda ^2 } \\ \end{array}}\right] . \end{aligned}$$

(4.100)

Thus the total stress tensor for a rubber-like material (4.55) is

$$\begin{aligned} \left[ \mathbf {T}\right] =-P\left[ \mathbf {I}\right] +\left[ { \begin{array}{*{20}c} {\beta _1 \lambda ^{ - 1} - \beta _2 \lambda }&{} 0 &{} 0 \\ 0 &{} {\beta _1 \lambda ^{ - 1} - \beta _2 \lambda }&{} 0 \\ 0 &{} 0 &{} {\beta _1 \lambda ^2 - \beta _2 \lambda ^{ - 2} } \\ \end{array}}\right] . \end{aligned}$$

(4.101)

Under the condition that the lateral tractions are zero, i.e., $T_{rr}=0$, the pressure can be found, and thus show that the tensile stress is

$$\begin{aligned} T_{zz}&=-P+\beta _{1}\lambda ^{2}-\beta _{2}\lambda ^{-2}=\beta _{1}\lambda ^{2}-\beta _{2}\lambda ^{-2}-\beta _{1}\lambda ^{-1}+\beta _{2}\lambda \nonumber \\&=\left( {\lambda ^{2}-\lambda ^{-1}}\right) \left( {\beta _{1}+\beta _{2}\lambda ^{-1}}\right) . \end{aligned}$$

(4.102)

This tensile stress is the force per unit area in the deformed configuration. As $r=\lambda ^{1/2}R,$ the corresponding force per unit area in the undeformed configuration is

$$\begin{aligned} T_{ZZ}=T_{zz}\lambda ^{-1}=\left( {\lambda -\lambda ^{-2}}\right) \left( { \beta _{1}+\beta _{2}\lambda ^{-1}}\right) . \end{aligned}$$

(4.103)

Problem 4.7

Show that

$$\begin{aligned} \mathbf {C}\left( \tau \right) = \mathbf {F}\left( t \right) ^{T} \mathbf {C} _{t} \left( \tau \right) \mathbf {F}\left( t \right) . \end{aligned}$$

(4.104)

Problem 4.8

Consider a simple shear flow

$$\begin{aligned} u=\dot{\gamma }y,\;\;\;v=0,\;\;\;w=0. \end{aligned}$$

(4.105)

Show that the stress tensor in the second-order model is given by

$$\begin{aligned} \left[ \mathbf {S}\right] =\eta _{0}\left[ {\begin{array}{*{20}c}0 &{} {\dot{\gamma }} &{} 0 \\ {\dot{\gamma }} &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array}}\right] +\left( {\nu _{1}+\nu _{2}}\right) \left[ {\begin{array}{*{20}c}{\dot{\gamma }^2 } &{} 0 &{} 0 \\ 0 &{} {\dot{\gamma }^2 }&{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array}}\right] - \frac{{\nu _{1}}}{2}\left[ {\begin{array}{*{20}c} 0 &{} 0 &{} 0 \\ 0 &{} {2\dot{\gamma }^2 }&{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array}}\right] . \end{aligned}$$

(4.106)

Thus, the three viscometric functions are

$$\begin{aligned} S=\eta _{0}\dot{\gamma },\;\;\;N_{1}=S_{11}-S_{22}=\nu _{1}\dot{\gamma } ^{2},\;\;\;N_{2}=S_{22}-S_{33}=\nu _{2}\dot{\gamma }^{2}. \end{aligned}$$

(4.107)

Problem 4.9

Consider a second-order fluid in an elongational flow

$$\begin{aligned} u = \dot{\varepsilon }x,\;\;\;v = - \frac{{\dot{\varepsilon }}}{2}y,\;\;\;w = - \frac{{\dot{\varepsilon }}}{2}z. \end{aligned}$$

(4.108)

Show that the stress is given by

$$\begin{aligned} \left[ \mathbf {S} \right] = \eta _{0} \left[ {\begin{array}{*{20}c}{2\dot{\varepsilon }} &{} 0 &{} 0 \\ 0 &{} { - \dot{\varepsilon }} &{} 0 \\ 0 &{} 0 &{} { - \dot{\varepsilon }} \\ \end{array} } \right] + \left( {\nu _{1} + \nu _{2} } \right) \left[ {\begin{array}{*{20}c} {4\dot{\varepsilon }^2 } &{} 0 &{} 0 \\ 0 &{} {\dot{\varepsilon }^2 } &{} 0 \\ 0 &{} 0 &{} {\dot{\varepsilon }^2 } \\ \end{array} } \right] - \frac{{\nu _{1} }}{2}\left[ {\begin{array}{*{20}c} {4\dot{\varepsilon }^2 }&{} 0 &{} 0 \\ 0 &{} {\dot{\varepsilon }^2 } &{} 0 \\ 0 &{} 0 &{} {\dot{\varepsilon }^2 }\\ \end{array} } \right] . \end{aligned}$$

(4.109)

Consequently the elongational viscosity is given by

$$\begin{aligned} \eta _{E} = \frac{{S_{xx} - S_{yy} }}{{\dot{\varepsilon }}} = 3\eta _{0} + 3\left( {\frac{{\nu _{1} }}{2} + \nu _{2} } \right) \dot{\varepsilon }. \end{aligned}$$

(4.110)

Problem 4.10

Show that, for potential flows,

$$\begin{aligned} \nabla \cdot \mathbf {A}_{1}^{2}={\frac{{1}}{{2}}}\nabla (tr\ \mathbf {A} _{1}^{2}),\quad \ \nabla \cdot \mathbf {A}_{2}={\frac{{3}}{{4}}}\nabla (\text {tr} \ \mathbf {A}_{1}^{2}). \end{aligned}$$

(4.111)

Problem 4.11

For steady two-dimensional incompressible flows, a stream function $\psi =\psi (x, y)$ can be defined such that the velocity components u and v can be expressed as

$$\begin{aligned} u={\frac{{\partial \psi }}{{\partial y}}},\ v=-{\frac{{\partial \psi }}{{ \partial x}}}. \end{aligned}$$

(4.112)

Obtain the stresses in a second-order fluid in terms of $\psi$. Substitute the stresses into the equations of motion and eliminating the pressure term through the use of the equality of the mixed partial derivatives, i.e., $p_{,xy}=p_{, yx},$ to obtain

$$\begin{aligned} \eta \triangle ^{2}\psi -{\frac{{\nu }_{1}}{{2}}}\mathbf {v}\cdot \nabla (\triangle ^{2}\psi )=0, \end{aligned}$$

(4.113)

where $\triangle ^{2}$ is the two-dimensional biharmonic operator. Deduce that a Newtonian velocity field is also a velocity field for the second-order fluid. The result is due to Tanner [83].

Problem 4.12

In a simple shear flow,

$$ u= \dot{\gamma }\left( t\right) y,\;\;\;v=0,\;\;\;w=0, $$

show that the path lines $\varvec{\xi }\left( \tau \right) =\left( {\xi ,\psi ,\zeta }\right)$ are given by

$$\begin{aligned} \xi \left( \tau \right) =x+y\gamma \left( {t,\tau }\right) ,\;\;\;\psi \left( \tau \right) =y,\;\;\;\zeta \left( \tau \right) =z, \end{aligned}$$

(4.114)

where

$$\begin{aligned} \gamma \left( {t,\tau }\right) =\int _{t}^{\tau }{\dot{\gamma }\left( s\right) ds. } \end{aligned}$$

(4.115)

Find the relative strain tensor, the stress tensor for a finite viscoelastic integral fluid, and show that the shear stress and the normal stress differences are given by

$$\begin{aligned} S_{12}\left( t\right)&=\int _{0}^{\infty }{\mu \left( s\right) \gamma \left( { t, t-s}\right) ds},\;\;\; \end{aligned}$$

(4.116)

$$\begin{aligned} N_{1}\left( t\right)&=\int _{0}^{\infty }{\mu \left( s\right) \left[ { \gamma \left( {t, t-s}\right) }\right] ^{2}ds}=-N_{2}. \end{aligned}$$

(4.117)

Investigate the case where the shear rate is constant and sinusoidal in time for the memory function in (4.88).