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All I wanna know bout Viscous Stress
(but I never dare to ask !)
You will find the basic facts about viscous phenomenon - no details -. If you wanna know more just email me or feel free to ask in the Discussion Forum. I purposely erased all the bibliographical references and detailed equations to keep the text simple and easy to read.
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Done and updated by Sébastien Dartevelle, The WebMaster, Saturday November 22 2003.
After all, what is viscosity?
Hmmmm, there are many different ways to answer to this question. When I was a High-School student, my science teacher said that viscosity is a property that owns any fluid and makes it somehow resistant to the flowing process. When I was Bachelor student, I got another picture of the viscosity. I was told that the viscosity is the tendencies towards uniformity of mass-velocity (momentum) throughout the flow. Such tendencies are explained by kinetic theory which attributes this phenomenon to the motion of molecules or grains from point to point.
I have been long a wee bit confused about those two views or explanations of the phenomenon of viscosity because I could not reconcile what was for me two different stories. But believe me, it is the same story but seen from a different point of view. The "resistance to flow" point of view is a global macroscopic force balance view, while the "mass-velocity uniformity" is a molecular-granular microscopic view of the viscosity phenomenon.
Herewith, I will only consider the viscosity concept for granular materials, and since granular theory has evolved from gas-kinetic theory, the following paragraphs also apply to the gas viscosity concept.
A moving fluid exhibit evidences of internal friction (friction is a good word, it gives us somehow the feeling of resistance), called viscosity. To see this, lets consider a moving fluid where the velocity is minimal at the bottom and tend to be higher towards the top of the fluid:
[Figure 1a: the Y axis is perpendicular to your screen, hence the XY plane is parallel to your table and the ZX plane is nothing but your screen]
[Figure 1b: Variation of the velocity in the X-direction along the Z axis]
[Be aware that those two figures do not exactly say the same thing, see their respective axis. However, those two figures show that the X-velocities (velocities parallel to the X-direction) are not the same along the vertical Z-direction. Those unequal velocities in one specific direction is a "velocity gradient", namely the gradient of X-velocity along the Z-direction]
Now, lets imagine we can subdivided the moving fluid into many very thin layers parallel to the XY plane (your table in the Fig. 1a). If the height of any individual layers is small enough, we can see the motion of the individual layer as that of a rigid body. This means that each layer can be characterized with a single velocity along the Z-direction. Obviously, in this particular example, each layer moves with a different velocity. Now, there are many, many physical processes that can cause those velocity differences (or velocity gradient) between fluid layers but, one thing is sure, because of this velocity gradient, viscosity will act, and act as long as those velocity differences exist. Be aware!! Viscosity is not responsible of this velocity gradient, it rather tends to counteract it.
Since some layers move a little bit faster, and others a little bit slower, we might expect that each layer will experience a forward drag from the layer above (and will exert a backward drag on this upper layer) and, at the same time each, layer experience a backward drag from the layer below (and will exert a forward drag on the lower layer). This is represented on the following figure which focuses on the j-layer. In Green, we have what the j-layer does or exerts to its neighboring layers and, in Black what the j-layer experiences or suffers from its neighbors:
[Figure 2: What the j-layer does and what it suffers]
[in Black what it suffers from its neighbors and in Green what it exerts to its neighbors]
Now, what we have in the Figure 2 is a quite complicate situation, so, lets adopt the following convention:
- If the frictional forces acting on a plane of a given layer are parallel to it, they are referred as shearing viscous force (see Fig. 2),
- If the frictional force acting on a plane are normal to it, they are referred as normal viscous force (not shown on Fig. 2),
- Now, since those forces act on a surface, it is much more useful to work with a new quantity, call stress, which is a Force per unit of Surface Area (N/m2). Hence, we have shearing stress and normal stress. If you try to imagine Fig. 2 in 3D instead, you will see that we can deal with many different planes (XY, XZ, YZ) and directions (X, Y, Z). We need a further convention:
- To properly define a stress, we need to specify its magnitude (which is the magnitude of the force), its direction (which is the direction of the force), and the surface on which the stress acts (which will be given by the normal vector of that surface). For instance, Tzx is the stress in the X-direction acting on the plane XY (whose normal vector is parallel to the Z-direction). Hence, in Tzx, the first subscript (z) gives the direction of the normal vector of the XY surface, and the second subscript (x) gives the direction of the force. Tzz, is the stress in the Z-direction acting on the XY plane (whose normal is in the Z-direction). Tzz is an example of a normal stress. All this can be easily understood from Figure 3:
[Figure3: Principal directions of stress acting on the XY plane. The first subscript (Z) indicates the normal direction of the XY plane, the second subscript indicates the main direction of the stress acting on the plane. Note that ous sign convention is postive in compression.]
Whenever the two subscripts are the same (Txx, Tyy, Tzz), we deal with normal stress (the force is perpendicular to the plane, like, for instance, the pressure). Whenever the two subscripts are different (Txy, Txz, Tzy, Tzx, so forth), we deal with shearing stress (the force is tangential to the plane). Be aware that the order of the two subscripts is fundamentally important, e.g., xy is the not the same as yx !!!
Now, lets go back to Fig. 2. It illustrates the forces acting along the top and bottom planes of a given layer (j-layer). Recall that the flow velocity increases upward. Hence, the j-layer exerts a backward shearing stress, , on the faster moving fluid layer above (it tends to decelerate the layer j+1). But, according to Newtons Third Law, the fluid above (j+1) also exerts a forward stress () equal in magnitude to . Similarly, the j-layer exerts a forward stress () on the layer below, while the layer below (j-1) exerts an equal in magnitude but opposed backward stress on the j-layer (). What we have is nasty, since the j-layer is sheared by the forces in Black on Fig. 2. At the top, the force tends to make the j-layer moving faster, while at its bottom, the force tends to decelerate it. The shearing forces generated by the velocity difference try to reduce those velocity differences between adjoining layers of fluid. It has been found empirically that the decelerating viscous stress is proportional to the velocity gradient (assuming we have a Newtonian fluid):
Eq.1 is the Newtons Law of viscosity. In this equation, the proportionality factor () of this empirical law is called the coefficient of viscosity of the fluid. The negative sign in Eq.1 (that is not often seen in the literature) expresses the fact that the viscous forces generated by the velocity gradient is trying to reduce the physical effect which generated it. In other words, a force must be applied to the lower wall to keep it at rest (or viscosity effects will take over and make it move faster) and a force must be applied to the upper wall to maintain its uniform motion (or viscosity effect will take over and make it move slower). Hence, we feel this notion of resistance to the natural flowing. If we dont do anything, viscosity effects will take over until the velocity gradient dwindles.
What is clear is that the fluid viscosity tends to reduce the velocity difference between adjacent fluid layers. In terms of vector, the viscous shearing are opposite to the velocity direction (as they are positive in the opposite direction of the velocity vectors, see Fig.2).
Be aware that viscosity effects do not necessarily make the fluid stop after some time, but rather, it only reduces the velocity gradient within the fluid. Actually, for understanding the full effect of viscosity, we must also understand what happen at the boundaries (e.g., on the ground). For properly modeling viscous flows, we must set up with great care the boundary conditions (i.e., free-slip wall, no-slip wall, or partial-slip wall). This is beyond the scope of this web page, I will present boundary models later and elsewhere, when I will have time for this
In a nutshell, viscous stress try to stop relative motion between different small part of the fluid I can even say whenever there is a strain rate (i.e., deformation that changes with time and is caused by the velocity gradient) within the fluid, a viscous stress act to reduce that rate-of-strain. By the way, Eq. 1 says that in a Newtonian fluid, the viscous stress is simply a linear function of the rate-of-strain (the higher the rate-of-strain, the higher the viscous stress). If the fluid is non-Newtonian (like granular flow) the relationship between strain and viscous stress is not simply linear, but more complicate
Now, I am nor sure how you feel here, but I can tell you that I still feel not quite at ease Why? you may ask. Well, the problem is that we know what does viscosity but we still do not know why it is like that, we did not have explain the deep reason of the viscosity phenomenon, how it works. But this can only be done with a molecular-granular kinetic model, which lead us to the next paragraph.
Kinetic theory attributes the tendencies towards uniformity of velocity within the fluid to the motion of gas molecules or grains from one layer to another. This tends to equalize conditions at the two ends by transporting to a further end amount of momentum (mass-velocity) that is characteristic of the starting-point.
The motions of grains or gas molecules in the Y and Z-directions is totally random, while in the X-direction the average velocity of those grains or molecules is equal to the bulk motion of the global fluid (ux). Gas-molecules or grains randomly cross the interface between two adjoining layers in both directions with equal frequency:
[Figure 4: In blue, grains or gas molecules moving from the upper level to the lower one with more momentum, and in red molecules or grains from the lower level moving to the upper one with less momentum]
Of course, in addition to this vertical random motion, each molecule or grain has a velocity in the X-direction, in other words it has some X-momentum. Therefore, each molecule or grain that crosses from above carries with it momentum (and energy too) to the lower layer and so does the grains or molecules moving from below to the upper layer. But the grains or molecules from above carry out a somewhat larger average X-momentum to the layer below, while the molecules or grains from below carry out a somewhat smaller average X-momentum to the layer above. As a result of this microscopic kinetic process, macroscopic momentum is slowly transferred from the upper layer to the lower one. Viscosity tends to equalize momentums throughout the fluid, hence, equalize velocities throughout the fluid. We see why the upper layer exerts a forward drag to the layer below.
According to Newton Second Law, the rate of change of the X-momentum of the j-layer is equal to the force exerted on it. The net transfer of X-momentum per unit of time and unit of surface area is equal to the shearing viscous stress. That is exactly what says Eq. 1 in a qualitative way. Somehow, the net variation of X-momentum of the j-layer must be equal to the sum of all the forces acting on it (e.g., in Green, -Tzx from above and Tzx from below in Fig.2).
So, now we know that viscous stresses try to reduce the relative motion between near parts of the fluids. In other words, whenever there is a strain rate (i.e., a velocity gradient within the fluid), a stress acts to reduce the nasty strain And the higher the viscosity, the higher will be the stress to reduce the strain rate. Remember Eq. 1 we wrote earlier? It is a very simple one as it considers only two-dimensions, but it can be generalized over the entire 3D space. This equation simply states that the stress () is proportional to the velocity gradient (), the constant of proportionality is the viscosity. Ok, but we have considered two directions (X and Z). In 3D, the Newtonian Law is written as:
where now we must consider the shear and normal stresses acting in all the possible directions of all the surfaces of a given infinitesimal reference volume. So, even if we dont know anything about tensor, we may feel that is the stress very much indeed, and D must be somehow related to the velocity gradient () as it was in Eq. 1. Also, since is a scalar, it is clear that D is also a tensor, which is named the rate-of-deformation, or usually, rate-of-strain tensor. Its name says all! However, D is not directly equal to the gradient of the velocity but rather to a somehow more complicate function. Why it is so? Well, lets have a deeper look into those tensors ().
Lets imagine a very, very small (infinitesimal) volume of fluid which shows no spin. As I explained earlier, there are two different sort of stresses or forces acting upon a plane, normal and tangential (shear). The stress tensor can be written as a 3x3 matrix in which the diagonal elements are the normal stresses and off-diagonal element are the tangential stresses:
where ex, ey, ez are unit vectors in the X, Y and Z-directions respectively. In the 3x3 matrix, the first line indicates all the forces in the 3-directions of space (X, Y, Z, second indices) acting on a surface whose normal is parallel to the X-direction (first indices), the second lines is all the 3-directional forces acting on a surface whose normal is parallel to the Y-direction, and the third lines is for all the forces acting on a surface whose normal is parallel to the Z-direction. The first column indicates the different values of forces systematically parallel to the X-direction (but acting on different planes), the second column is all the values of the forces parallel to the Y-direction (but on three different planes), and the third column is all the forces parallel to the Z-direction. The diagonal elements are the normal stress (they have the same two indices), and may be associated to a "pressure", while the off-diagonal elements represents the shear forces. Now since this volume is spinless, we must have:
The off-diagonal elements which have the same pair of indices are equal, you may transpose columns and rows in this matrix, it wont change anything. This tensor is said to be symmetric. The physical reasons is because if the shear stress having the same indices were not equal, the fluid volume would spin.
Now, lets see the velocity gradient tensor as it can be also written as a tensor (but a little bit different than the stress tensor):
In the 3x3 matrix, the first line indicates the variation along the X-direction of the three velocities (ux, uy, and uz), the second line is the variation along the Y-direction of the same three velocities, and the third line indicates their variation along the Z-direction. Therefore, a column considers the variation of a given velocity (ux, uy or uz) in the three directions of space.
This velocity gradient tensor is not exactly similar to the stress tensor in that the velocity gradient tensor is not symmetric, if you transpose rows and columns you will have something totally different. The physical interpretation of this is that the velocity gradient measures not only the deformation of the fluid body (by compression and/or shearing), but also the relative orientation of the fluid body (rotation). When Ill have some more time, I will demonstrate this. Therefore, we cannot write a direct mathematical relationship between stress tensor () and velocity gradient () since the stress tensor doesnt care at all of the rotation of the body, which has nothing to do with pure deformation. However, the velocity gradient tensor like any similar tensor can be broken into two parts, one symmetric (D) and the other one, antisymmetric (S). Hence, we can simply write the velocity gradient tensor as the sum of those two parts:
The symmetric tensor is the rate-of-strain (or rate-of-deformation) tensor (thats what we are looking for) and is equal to (for now, Im not gonna write the unit tensor matrices, which are therefore implied):
where the "T" describes the operation of Transpose (i.e., switching rows and columns). It can be seen that D is indeed a symmetrical tensor as (look inside the 3x3 matrix). You may be confused by the minus sign in Eq.7. Well, I should demonstrate this but for now it simply indicates that the rate-of-deformation is positive whenever deformation is compressive. The antisymmetric tensor is the vorticity or spin tensor and is equal to:
I will not comment this spin tensor any further as it has nothing to do with the scope of this page (viscous phenomenon). Again, note the minus sign in Eq.8, which indicates that the spin tensor is positive clockwise in the inward direction.
So, I guess you can see where Im heading for, the stress tensor is not a direct function of the velocity gradient but rather a function of the symmetric part of the velocity gradient tensor, i.e., D. Indeed, D is not only symmetric as but represents all the rate-of-deformation suffered by the fluid body. Youpie! We have all we need to write the Newtons law of viscosity in terms of tensors (in 3D), the stress is proportional to the rate-of strain:
Now, notice the minus sign. Recall Eq. 1, where we have already seen this minus sign. In Eq. 1, it expresses two main ideas. The first one is that momentum flux "moves" from region of high momentum to region of lower momentum in order to equalize momentum throughout the whole flow. The second idea is that the stress is positive in the opposite direction of increasing velocity to precisely counteract this velocity increase. Now, from Eq. 9, we can add a new idea to the meaning of this minus sign. It can be demonstrated (Ill do that whenever Ill have time but you may have a good clue if you read till the bottom of this page) that whenever we are in compression, D is positive and so is the compressive stress tensor. This means that if I compress the fluid body, the compressive stress will be positive, and the rate-of-compression will be positive as well.
The idea of this minus sign in that particular place in Eq. 1 and Eq. 9 is extremely powerful, smart and physically sane. However, it is sometimes sad too see many people confusing the exact meaning of this minus sign and/or too many times forgetting to add this sign. For example, it can be sometimes seen in some (bad) books that the stress tensor is defined positive in compression but the rate-of-strain is defined negative in compression or that the stress tensor is positive in compression and in the next chapter positive in extension, all this is horribly confusing, complicate and pure non-sense. So, we should take care of the meaning of the directions of the stress tensor AND at the same time the rate-of-strain tensor. Of course, all this is a matter of convention, but this one is really the best. Note that Eq. 9 is not yet quite accurate (see next paragraph).
The very last thing you may ask is how from this complicate tensorial equation (Eq.9) we can come back to the simple Newtons equation seen in Eq. 1. Well thats not too difficult to do. Eq. 1 says that the stress in the plane XY is proportional to the X-velocity gradient in the Z-direction. So, it means we are only interested in the velocity in the X-direction and its variation along the Z-direction. So, lets assume in this case that the X-velocity is only a function of the Z-direction, and that all the other velocities are equal to zero:
ux = f(z); uy = 0 and uz = 0
It follows from Eq. 7 that:
Finally, we have formally the original Eq.1:
where the partial derivative is unnecessary since ux only varies in the Z-direction, and therefore could be replaced by a total derivative as in Eq. 1.
Ok. We have accomplished tremendous things, we know what is viscosity and why. We also know the relationship between strain and stress (I should say rate-of-strain and stress). Basically, I explained that from a microscopic point of view, viscosity is due to the random motion of gas molecules (or grains in a granular flow) between neighboring fluid layers. This random motion also carries some momentum, and therefore is responsible for homogenizing the momentum throughout the whole flow. This momentum flux explains the stress within the fluid. The problem is that this nice picture is not quite accurate yet, as we have two different viscosities, and both dont do exactly the same thing to the fluid. Those two different viscosities are the bulk viscosity (or dilatational viscosity) and shear viscosity (or dynamic viscosity). In consequence, Eq. 9 is not yet very accurate.
The shear viscosity, or sometimes called the dynamic viscosity, measures the random chaotic motion of molecules or grains, that is what we have been reviewing here on this page so far. Usually, in many, many fluid dynamic books, it is the only viscosity we have to worry about. Indeed, for most of the fluid, the bulk viscosity is equal or approximated to zero (which is called Stokes hypothesis) and/or the fluid is incompressible. Actually, in Eq. 9, it is exactly what I have assumed (because I wanted simple first). However, in a gas made of huge polyatomic molecules, the bulk viscosity will not be equal to zero. Indeed, such a molecules may vibrate and also rotate. Since these molecules may have a dimension large enough relative to their mean free path, their vibration and/or rotation may somehow affect their translation along their mean free path. This vibrational and rotational effects are somehow measured by the bulk viscosity. Those supplementary motions required energy (as the translational motion measured by the shear viscosity) which will be taken at the expenses of the translational mode. In granular matter, the bulk viscosity is also not equal to zero, particularly for concentrated granular matter. Of course, in the granular matter case, the bulk viscosity cannot be explained by the vibration/rotation of the grains. Actually, the granular bulk viscosity is simply proportional to the granular shear viscosity and is simply related to the random motion of the grains.
Why do we call those viscosities bulk (or dilatational) and shear (or dynamical)? Good question.
Well, the reason is that the shear viscosity is only associated with the shear stress and shear deformation, it has nothing to do with compressive stress or compressive deformation, while the bulk viscosity is only associated with the change of volume of the fluid parcel. The bulk viscosity is only associated with the normal viscous stress.
Soooooo, whenever we have a shear-rate-of-strain, we have a proportional shear stress where the proportionality constant is the shear viscosity (). Whenever we have a pure volumetric-rate-of-strain, we have a proportional normal stress, where the proportionality constant is the bulk viscosity (). In gas kinetic theory, the shear viscosity is often called the dynamic viscosity as it is simply associated to the random chaotic motion of the gas molecules. In granular kinetic theory, we use the name "shear viscosity" rather than dynamic viscosity as the granular bulk viscosity is also associated to the chaotic random motion of the grains. Hence, for granular matter, the name "dynamic viscosity" would be pretty loose and unclear as it could refer to the shear viscosity as well as the bulk viscosity.
So what now? Well it means that we can specifically associate normal deformation with normal stress and shear deformation with shear stress. Lets see this
The rate-of-strain tensor, D, represents all the possible deformation a fluid body can suffer; namely, volumetric (compressive/extensive) deformations and shear deformations. In terms of tensors, it can be shown that D can be broken into a spherical part () and deviatoric part (). Where only represents pure volumetric-rate-of-deformation (compression/extension, no shear) and is simply proportional to the sum of the diagonal element of D (i.e., trace of D, which is ID), while only represents pure shear-rate-of-deformation (no volume deformation):
where is the divergence of the velocity field, and is equal and opposite to the sum of the diagonal element of D (= -Dxx-Dyy-Dzz). If the flow is divergent and represents an extensional state, while if it represents a compressive state. However, because of our sign convention, at the very end, is positive in compression and negative in extension. You may see from Eq.11c that the deviator of D is equal to D minus one-third the trace of D or minus the spherical part of D. Hence the deviator of D can only represent the shear-rate-of-deformation rate (as we have subtracted all the volumetric-rate-of-deformation). In other words, the deviator of a tensor is said to be traceless; i.e., the sum of all the diagonal element of is zero. In terms of the stress tensor we have:
It should be noted that the shear viscosity is more often symbolized by the simple Greek letter (mu) instead of . Eq.12b is the utmost complete mathematical relationship between stress and rate-of-strain for a Newtonian flow.
Also in the case the divergence of the velocity is equal to zero (), then the flow is said to incompressible, and the bulk viscosity is irrelevant. In that very particular case, you may see that there is no difference between the deviatoric part of the rate-of-strain tensor and the rate-of-strain tensor itself (). Hence, there is no difference between the deviatoric part of the viscous stress tensor and the viscous stress tensor itself ().
Very often, in many fluid dynamic treatise the equations I have been reviewing are not written that way as they rather use another viscosity concept, called the second coefficient of viscosity. What is this second coefficient of viscosity? Lets see this and rearrange Eq. 12b in gathering all the terms containing the divergence of the velocity.
and is the second coefficient of viscosity and represents a combination of all the viscous effects associated with the volumetric-rate-of-strain (the divergence of the velocity). But the problem now is that this equation is less clear because D is indeed the total rate-of-strain and also contains the volumetric-rate-of-strain in addition to the shear-rate-of-strain. Thats why I prefer Eq. 12 as it clearly splits shear effects from the volumetric ones, which is not the case in Eq.13. The second coefficient of viscosity concept is rather a mathematical trick than a truly physical one (but sometimes it may be useful to do so, e.g., for easily computing the work done by viscous dissipation).
If you wonder what is the first coefficient of viscosity well that is simply another name for the shear viscosity. Also, in most treatise book of fluid dynamic, all this doesnt matter as it is assumed (most of the time) that the bulk viscosity is approximately zero. Hence the second coefficient of viscosity is proportional to the shear viscosity In this case, the shear viscosity is the only one you have to worry about. However, as I said earlier, most of the time you cannot do that for granular matters.
Maybe an important thing?. The SI unit of viscosity (bulk, shear, second) is Pa s (Pascal second), that is also kg/m s.
From the simple Eq. 1 or from the more comprehensive relations in Eq. 12a, it looks like that there is a linear (proportional) relationship between viscous stress and rate-of-strain. In other words, if the components of the rate-of-strain-tensor are increased by some factor, then the components of the viscous stress tensor will increase accordingly and most importantly proportionally. That is true if we have a linear rheology, i.e., if the constant of proportionality is indeed held constant,. This typical for Newtonian and Bingham rheologies.
If you read the frictional course in this website (which starts here), you will soon find out that within frictional granular flows (i.e., plastic rheology) viscous stress is first-order rate-of-strain independent. In other words, if you increase the rate-of-strain, you will not increase the viscous frictional stress. This means, for a given normal applied stress and a given volumetric grain concentration, it is impossible to have shear stress within a frictional granular flow higher than its yield shear stress (i.e., the flow is at yield and flow plastically). Within plasticity, the linear relationship between viscous stress and rate-of-strain cannot hold anymore. Once at yield, you may have a very high rate-of-strain or very low, it wont matter, the stress remains unchanged. Therefore, from Eq. 12a, the shear viscosity () and the bulk viscosity () must be somehow non-linear functions of the rate-of-strain, so that when the rate-of-strain increases the viscosity functions decrease to maintain the viscous stress constant (more details on this page, Constitutive Equations for frictional Granular Flows). For granular frictional flows, plasticity leads to a non-linear rheology.
There is no problem of having a viscosity not constant anymore and of being a non-linear function of the rate-of-strain itself. Such rheology is non-linear and completely non-Newtonian indeed. But some "linear" people are deeply troubled by that and have accused me of "abusing" the concept of viscosity and/or its word within the rigid-plastic theory. That is because the only thing they know (and want to hear) is a proportional relationship between viscous stress and rate-of-strain (i.e., Newtonian). Within Computational Fluid Dynamic, there is no problem of having non-linear rheologies and, as a matter of fact, such rheologies are quite common in this Universe. I agree that within elasto-plasticity the word "modulus" is rather used (bulk modulus for bulk viscosity and shear modulus for shear viscosity) but it wont change the basic fact that viscous stress is developed to reduce the relative motions between near parts of the fluid, even though, at yield, plastic viscous stress will remain constant no matter what happen to the rate-of-strain.
The conclusion is that there is much more than simple linear rheologies (e.g., Newtonian and Bingham) and nobody should be afraid of dealing with more complex (but richer and more interesting) non-linear rheologies as we do throughout this website.
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