Volume preserving and non-preserving strains
1.3 Strain and strain rate
We now refine and quantify our description of deformation in soft materials. Both strain and stress are 2nd-rank tensor quantities as each relates two vectors. In the case of strain, these two vectors can be chosen to be an embedded spatial vector and its displacement by the deformation. In the deformation of a continuous body, any embedded vector X(r) is transformed to a new vector X’(r) (Figure 2).
Figure 2. Illustrating the deformation of a continwm body.
We define the strain tensor by this transformation, as follows:
-
X‘
=- E
*x .
(2)For small enough deformations we may write
E
=+ e
so that the field of embedded displacement vectors U obeys U EX’ - X
= gTX. -Fora uniform deformation we may write this BS g = V u or e;j = Viuj. Volume-preserving deformations have det(5) - = 1, which becomes, for small displacements, the condition that n ( g )-
= 0.82 Tom McLeish
If the strain is time dependent, there is a velocity field v(r) that generates the local
= Vv. Embedded vectors now change deformation rate through its spatial gradient
with time, X' = X'(t), so that v(X') =
4
- 3 X' af;dxr(t +
6 t ) =(1
-+
-k 6 t ).
X'(t) =(1 +
$ 6 t ) *g t )
'x .
(3)But we also have Xr(t
+
6t) =E(t +
6 t ).
X by definition ofE.
Comparing this with Equation 3 and defining the timederivative of E ( t ) - by the usuallimit, givesaE _ - - - $ * E
dt
- - (4)Note that the tensor $ is by definition aVu/dt = Vv where v(r) is the velocity field, defined above. TherefGre we can identify
6
- withK. -
The differential Equation 4 is just a tensorial version of the familiar first-order linear equation
af /at
=K f ,
and, in the case where - is constant (steady flow), has the solution-
E(t) = exp(K:t) 3 ( 5 )
where we use the initial condition that E(0) - =
1,
- and where the exponential of a tensor is defined by its series expansion1 1
1 + M + -
M2+
-M3+ .. .. .
exp(M) - - - 2!= 3!"
1.3.1 Examples
There are two very important examples of volume-preserving deformations in soft matter, shear and extension. Shear occurs in sliding, or lubricating flows; extension in stretching flows such as the forming of fibres and films.
Shear
A shear flow with velocity along x and gradient along y, of shear rate
.4
= avz/ay, has a deformation rate tensor (in Cartesians)==+
E ( t ) =exp(Kt) =o
1o .
.-(
- ; i) - - (::I)
The shear flow does not generate exponential separation of embedded points (as Equa- tion 5 might suggest) because the displacements of embedded vectors are always orthog- onal to the vectors themselves.
Uniaxial extension
In contrast, a uniaxial extensional deformation (as occurs when pulling out a thread, during fibre-spinning for example), with extension along the x-axis and extension rate t = dv,/dx, gives exponential separation of points and has the representation
Rheology of linear and branched polymers 83
Planar extension
There is a second important extensional flow, termed planar extension (as occurs in some film-forming processes). This has a neutral direction in which there is no deformation (in common with shear flow), yet exponential separation of embedded points (in common with extensional flow):
K = [ t -i 0 0 0 )
=+
g(tl-( ,it 0!).
- -
0 0 0
In such a deformation (with
t
positive) fluid is pulled inward along f y and stretched outward along fz, with z neutral.1.4 Stress
In deformed matter, forces are transmitted across any surface embedded in the material.
The stress, like strain, is tensorial because both the locally transmitted force per unit area, and the local surface element (characterised by its normal) are vectors. We therefore define the stress tensor
a
so that the force dF acting across a small area element d A of unit normal n is givenTy (Figure 3)Alternatively, oiJ gives the i-th Cartesian component of the force per unit area across the j-th face of a small cube embedded locally in the material.
Like many physical rank-2 tensors, q is symmetric. Indeed, the torque (in the z- direction) on a small cube of side 1 is
13pzg -
oYz). These two stress components must cancel, because the moment of inertia of such a cube scales as M12 N 15, which would otherwise lead to a divergent angular acceleration as E 4 0.d F = a . n d A . - (6)
Figure 3. Illustrating the definition of the stress tensor.
1.4.1 Examples Hydrostatic pressure
For a fluid at static equilibrium, one has =
-PI,
so that p = -(1/3)Tr(q). The pressure field is not normally interesting in soft matt&, as it acts as a Lagrange multiplier for the conservation of volume.84 Tom McLeish
Newtonian fluid
The simplest constitutive equation for a fluid is that suggested by the tensorial symmetry of the strain rate and stress tensors, and the requirement of symmetry in E. These, combined with an assumption of linear response, and of incompressible flow, give in general = 17
(K + ET).
This characterises the fluid by a single number only, 7, the viscosityTSo in the case of simple shearNote that the pressure term just adds to the stress arising from the shear. Often we speak of the ‘deviatoric stress’
a’
=a-
(1/3) Tr(g)l, which captures the stress arising from the material structure as it responds to a volumepreserving deformation. We will often drop the prime in what follows.Rubbery solid
A similar argument can be applied to an isotropic solid, allowing for the fact that the stress is now linearly proportional to the strain itself, rather than the strain rate. At small strains, this gives
a
= G(e+ g T ) for the deviatoric stress; G is the elastic modulus.An extension of this model tolarge strains, useful for rubbery solids, is to write
a -
= G(E- ET).
Expanding for smalle
=E -
recovers the previous result to linear order, plus a cGnGibution to the isotropic pressure.A Maxwell model
The ‘rubbery solid’ constitutive equation just found may be generalised to a continuously- deformed material with a single viscoelastic relaxation time T , by writing the following:
- -
-Thus, if
K
vanishes (for example after a step-strain measurement: see below), the devia- toric stress decays to zero like exp[-t/r]. In steady shear (with ;i = av,/dy) the stress tensor becomes:G ( l
+
2 ( ~ ; 1 ) ~ ) Gr;1 0O G
-
-
U =(
Gr+0
Exercise: Check this last result, and think about the physics of the proposed consti- tutive equation and the consequences of the predicted ‘first normal stress difference’:
OZI - Oyy
#
0.1.5 Rheometry
Rheometers are designed to impose on a material either shear fiow (easy) or extensional flow (more difficult). A rotational device that generates a spatially uniform shear flow is the ‘cone-and-plate’ rheometer, Figure 4.
Rheology of linear and branched polymers 85
Figure 4.
between cone (white) and plate (grey).
Schematic of a cone-and-plate shear rheometer. The sample (black) lies
The cone is rotated at instantaneous angular velocity w. The material in the gap at distance r from the axis has a velocity in the tangential direction that is zero at the bottom plate and wr at the top plate. (We assume non-slip boundary conditions.) The local separation of the plates obeys h ( r ) = a r where a is the angle between plate and cone, which must be small. If so, the local shear rate is
+
= ave/az = w r / a r = w / a : a uniform shear field. Maintaining such uniformity is especially important in non-linear deformation, where the material response may differ for different strains and strain rates.The shear force is measured from the torque.on the rotor, and normal stress differences can also, in principle, be monitored (e.g. from the upthrust on the cone).
Figure 5. Schematic of a moving-belt extensional rheometer (sample in black).
Extensional rheometers (Figure 5 ) have been much harder to develop to the point where reproducible data is obtainable. This is due to the necessity of free surfaces over most of the sample in an extensional flow. However, extensional rheometry gives an important measure of the non-linear flow of many materials, that is independent of the shear response. For example, branched entangled polymers (see Section 3 below) may be strain-hardening in extension (the effective ‘viscosity’, which is the ratio of stress to strain rate, increases with strain), but strain-softening in simple shear. An illustration is in Figure 6. Here the extensional stress difference oxx - ouu, divided by the extensional strain rate i, is plotted against time, for two experiments at constant strain rate (started suddenly at time zero). On the same graph are the corresponding transient shear ex- periments, showing This way of representing data ensures that the curves within each set superimpose at early times, when the deformations are purely linear. The upper curves show strong extensional ‘hardening’ at the higher of two extension rates, but no hardening at the lower rate. The lower curves (for shear) all show softening. In this
86 Tom McLeish
lo'
t - - 1 ,
:
Figure 6. Time dependent shear (lower curues) and extensional (upper curues) stresses normalised b y deformation rates ouer a range of rates. Lines are from a non-linear gen- eralisation of a model of branched polymers discussed in Section 3.
case the polymer melt is composed of monodisperse molecules of identical ('H-shaped') branched structure.