It makes a difference because the stress is a symmetric tensor that has at least orthorhombic symmetry. Thus if you try to apply a shear stress to a surface of the rigid plate, automatically the symmetric shear stress is generated, because the stress tensor is symmetric. Thus in reality, although one may think one is only applying a shear stress to one surface, one is actually applying symmetric shear stresses to a conjugate set of surfaces, and that means one is applying an orthorhombic tensor as a boundary condition. In applying an orthorhombic stress to an isotropic material, the effect can have a symmetry no lower than that of the combined causes, according to the symmetry principle. Thus the effect must have at least orthorhombic symmetry, which is the symmetry of the combined causes (orthorhombic stress and isotropic material properties). Thus applying an (orthorhombic) stress boundary condition to an isotropic material cannot produce a monoclinic effect (monoclinic velocity gradient).
The only way to get a monoclinic effect is to apply monoclinic boundary conditions, which one can do by applying monoclinic velocity boundary conditions, or in effect a monoclinic velocity gradient tensor. The effect then has a higher symmetry (orthorhombic stress) than the combined causes, which have a monoclinic symmetry (monoclinic velocity gradient with isotropic material properties). This is permitted by the symmetry principle, which states that the effect symmetry must include those symmetry elements that are common to the combined causes, but can have other symmetry elements as well, i.e. the effect symmetry can be higher than that of the combined causes.
So I would argue that to produce a simple shear between two parallel plates, although you may think you are only applying a shear stress to the rigid plate, in fact you must be applying a velocity boundary condition in order for the resulting deformation to be a simple shear, because applying only an (orthorhombic) stress to an isotropic material cannot give you a monoclinic effect.
The boundary conditions must affect the entire deformation. Solving the equations of motion basically requires an integration, which leaves undetermined constants (or functions for partial differential equations) in the solution. The presence of these undetermined constants or functions in the solution means that there is a whole class of different solutions that can satisfy the differential equations, depending on the value of the constants or the form of the functions. It is the application of boundary conditions that allows one to pick out the unique values of the constants or forms of the functions, and these then identify the unique solution to the integrated equations from among this whole class of possible solutions. The compatibility conditions eliminate those solutions that would result in an overlapping or a separation of the material.
That is how it seems to me, at least.
Cheers,
rob
On Mar 27, 2011, at 12:57 PM, Dazhi Jiang wrote:
> Rob,
>
> I don't have Twiss & Moores at hand now. But I don't understand why it would
> make a difference in principle if the boundary condition is a stress one.
> For the Newtonian fluid between parallel rigid plates example, one can as
> well view it as a constant traction boundary condition problem: the shear
> stress at the fluid-plate interface is kept constant. Nothing would change.
>
> It is all the requirements (mechanic, kinematic, rheology, and boundary
> conditions) that the velocity field must satisfy that make it generally
> asymmetric. At a point far away from the boundaries, kinematic compatibility
> requirement is perhaps more important than the boundary constraints.
>
> Cheers,
> Dazhi
>
> -----Original Message-----
> From: Tectonics & structural geology discussion list
> [mailto:[log in to unmask]] On Behalf Of Robert J. Twiss
> Sent: Sunday, March 27, 2011 1:32 PM
> To: [log in to unmask]
> Subject: Re: a plea and a new question?
>
> For a discussion that amplifies Dazhi's points, see Twiss & Moores,
> Structural Geology, 2nd Ed., Section 18.1 (p.544-546). We can consider the
> boundary conditions as defining the 'cause' of the mechanical process
> because these are the conditions that are externally imposed on the
> deforming body. For a mechanically isotropic body, the symmetry principle
> [see Twiss & Moores, Section 17.8-ii, p.537] shows that if the stress is the
> cause of a deformation (stress boundary conditions), the resulting
> deformation can never have the low monoclinic symmetry of a simple shear.
> Only if velocity boundary conditions are applied can a simple shear result
> [Twiss & Moores, Section 17.8-iv (p.539)].
>
> I might add that it is important to understand the distinction between a
> real material and the continuum model of that material. The continuum
> formulation of deformation of a material is simply a mathematical
> idealization, a model, of the physical system, and should not be confused
> with the actual physical system itself. For the mathematical idealization,
> we can imagine taking a limit as we shrink a cube or a tetrahedron to an
> infinitesimal point. For a real material, however, such a process becomes
> meaningless as the size of the volume decreases because of the inherent
> discontinuities and heterogeneities in a real material. Thus we must always
> keep in mind what the correspondence is between an infinitesimal point in a
> mathematically idealized continuum and what that point represents in the
> real material.
>
> In particular, the value of a field quantity such as force at a point in a
> continuum is a mathematical idealization that is meant to represent an
> average of all real forces in the real material over a local volume around
> that point. This is the basic approach of statistical mechanics. When, in
> the mathematical idealization, we allow a volume to shrink to zero so that
> all moment arms vanish and torques become zero, we are using a
> mathematically convenient technique to express the physical situation that
> within a small local volume around a point in space, whatever torques there
> may be will average out to zero.
>
> Thus it is best to keep in mind the statistical mechanical basis for the
> relationship between a continuum model and what it is designed to represent
> in the real world. If the continuum model does not represent the real world
> adequately, we are free to use a different model. For example, a micropolar
> continuum model could provide a better representation of the behavior of a
> granular material than the classical continuum model.
>
> rob
>
>
> On Mar 27, 2011, at 6:19 AM, Dazhi Jiang wrote:
>
>> I'd like to add a few more lines to what I sent around yesterday (below)
>> after reading R. J. Twiss's email.
>>
>> In applying continuum mechanics, we assume that the continuum assumption
> is
>> valid for the problem. One may refer to many textbooks for this
> assumption.
>> Where this assumption is not valid, other formulations are necessary. But
>> John Waldron's question still must and can be answered in the context of
>> classic continuum mechanics.
>>
>> Imagine a simple case where a Newtonian fluid is constrained between two
>> parallel rigid plates moving parallel to each other. The velocity field in
>> the fluid is a perfect progressive simple shear and is everywhere
>> monoclinic. But the stress tensor is everywhere orthorhombic. Where does
>> this unparallelism arise? I think the answer is that the velocity field is
>> not just driven by stress (the "deformation driven by stress" thinking).
> It
>> must satisfy the compatibility requirement and the boundary conditions as
>> well.
>>
>> As we know, a complete set of equations for a continuum mechanics problem
>> includes: mechanic laws which ensure stress equilibrium and require that
> the
>> stress tensor be symmetric, constitutive equations (relating stress and
>> strain and strain rate etc.), kinematics (strain and compatibility etc.),
>> and the boundary conditions.
>>
>> Cheers,
>> Dazhi
>>
>> -----Original Message-----
>> From: Tectonics & structural geology discussion list
>> [mailto:[log in to unmask]] On Behalf Of Dazhi Jiang
>> Sent: Saturday, March 26, 2011 8:32 PM
>> To: [log in to unmask]
>> Subject: Re: a plea and a new question?
>>
>> John,
>>
>> Here is how I look at the stress and strain problem you have.
>>
>> First, to say that deformation is driven by stress is incorrect, or at
>> least, incomplete. Let's limit ourselves to infinitesimal elastic
>> deformation first. One can say the strain is driven by the stress (through
>> the Hooke's law). Or equivalently, the other way around (left side equal
> to
>> right side of the Hooke's law). But deformation must be defined by the
>> complete displacement field, of which strain is only the symmetrical part.
>> The antisymmetric part of the displacement field is the rotation. Now to
>> answer your question, what determines the displacement field? It is the
>> combination of mechanical laws (balance of linear momentum, angular
>> momentum), stress-strain relation, compatibility, and the boundary
>> conditions. How does the vanishingly small cube 'know' about the boundary
>> conditions of the system in which it sits? It is through compatibility
>> requirement. The possible displacement field for a continuous body
>> deformation must make all parts compatible.
>>
>> The about explanation applies to the deformation of any continuous body.
> For
>> a viscous body, just replace the displacement field by the velocity field.
>>
>> When one moves from infinitesimal deformation to accumulative deformation,
>> one simply deals with the time integration of the displacement/velocity
>> field.
>>
>> Hope this helps.
>> Dazhi
>>
>> ______________________________________________
>> Dr. Dazhi Jiang, Associate Professor
>> Department of Earth Sciences
>> The University of Western Ontario
>> London, Ontario
>> Canada N6A 5B7
>> Tel: (519) 661-3192
>> Fax: (519) 661-3198
>> www.uwo.ca/earth/people/faculty/jiang.html
>> ___________________________________________________
>>
>> -----Original Message-----
>> From: Tectonics & structural geology discussion list
>> [mailto:[log in to unmask]] On Behalf Of John Waldron
>> Sent: Saturday, March 26, 2011 5:51 PM
>> To: [log in to unmask]
>> Subject: a plea and a new question?
>>
>> I have a plea and a question.
>>
>> I have too found the discussion on this list very informative over time,
> and
>> I would like it to remain so. However, the last 18 posts have been about
>> the behaviour of people, not rocks. Public statements of opinion, or
>> announcements of intent to leave the list, however well-intentioned, may
>> contribute to the problem; more list members will be tempted to leave
>> because they don't want to read this stuff. So, I would make a plea based
>> on my experience on the Canadian list mentioned by Jürgen. If you intend
> to
>> leave the list, I would urge you to leave quietly, or to make your
> opinions
>> on individuals (on whichever side of the argument) known in private emails
>> or to the list owner (he may not thank me for this), rather than to the
>> whole list. I intend to stay on, and hope there will be enough expertise
>> left in the list to make it as informative in the future as it has been in
>> the past.
>>
>> In that spirit (and lest I contribute to the same problem) I would like to
>> ask a question, that has been raised in my head by some of Dr. Koenemann's
>> comments. Like many members, I work in general field-based structural
>> geology, and am not an expert in continuum mechanics. However, I do teach
>> the basics of stress and strain in my undergraduate and graduate classes,
>> typically to students with even less background in physics and mathematics
>> than mine. Like most of us who teach this stuff, I take my students
> through
>> the hypothetical vanishingly small cubic element of a solid under stress,
>> and represent the three components of stress (or more properly traction)
> on
>> each surface so as to fill out the 9 components of the stress tensor.
>>
>> Then comes the part that always leaves me with nagging doubts. There is
> an
>> argument in all the texts that the shear stresses sigma-x-y and sigma-y-x
>> are identical, based on the case that there is no net moment about the z
>> axis in this vanishingly small cube. When applied to all the off-diagonal
>> elements, this leads to a symmetrical stress tensor with 6 independent
>> terms, in contrast to the asymmetric deformation gradient tensor with 9
>> terms. I am uncomfortable with this contrast, which seems
>> counter-intuitive. If deformation is driven by stress, and the stress
>> tensor only controls the six terms that describe distortion (or distortion
>> rate) then how is the rotational part of deformation controlled? I
> realize
>> that rotation can be constrained by setting appropriate boundary
> conditions,
>> but my discomfort is that that vanishingly small cube doesn't 'know' about
>> the boundary conditions of the system in which it sits, so what controls
> its
>> rotation if not the state of stress? So I always end my lecture with the
>> feeling that the argument is sleight of hand - I have used phrases like
>> 'arguments beyond the scope of this course lead to...', without feeling
> that
>> I actually have a proper grasp of those arguments.
>>
>> This may be something that can be very simply answered, and that I simply
>> missed out on in my own education. However, Dr. Koenemann's discourses
>> raised the idea that we should be able to explain stress-strain
>> relationships in terms of forces that act along bonds between atoms, not
>> infinite imaginary surfaces within continua, so I am tempted to wonder
>> whether there are elements of his argument that might lead to a resolution
>> of my question, perhaps by including a rotational element into the
>> description of stress. If anyone has any suggestions or explanations that
>> help to make this make sense, and help me make sense of this to my
> students,
>> it would be most welcome.
>>
>> John Waldron
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