Rob,

I don't think you give any reason why the velocity BC cannot be regarded
equivalently as a shear traction BC other than that the latter perspective
does not come to terms with the symmetry principle. I do feel that this is
forced on, merely by the symmetry principle.

One can calculate the stress at the interface between the plates and the
fluid, and fictitiously remove the plates but apply the stress condition
everywhere on the boundary. Nothing would change in the mechanics of the
system. But as you claimed this is not allowed by the symmetry principle.
Isn't this odd?

If the symmetry principle is to be used only locally, then how is it applied
to the converging shear zone case? What is this triclinic cause, at every
locality, for the triclinic effect (velocity field)? 

In solving a continuum mechanics problem, we have momentum equation,
constitutive equations, kinematics, energy equation, and boundary
conditions, but never is there a symmetry equation. In fact I wonder if it
is a universally applicable.

Cheers,
Dazhi
  

-----Original Message-----
From: Tectonics & structural geology discussion list
[mailto:[log in to unmask]] On Behalf Of Robert J. Twiss
Sent: Tuesday, March 29, 2011 12:43 PM
To: [log in to unmask]
Subject: Re: a plea and a new question?

Dazhi,

I agree with you that the material must deform according to the whole set of
requirements.  But I think you misunderstand me when you say "we must regard
a shear traction boundary condition as a velocity condition, merely to make
the symmetry principle apply."  I am not trying to force the interpretation
of the boundary conditions to fit the symmetry principle.  I am trying to
use the symmetry principle to understand what the boundary conditions really
must be. 

The fact that a rigid plate is used to apply the boundary conditions for
simple shear necessarily implies that the displacements (or velocities) at
the boundary are constrained.  The material is not free to deform at the
boundary in any manner other than what is imposed by the motion of the
plate.  If it really were stress boundary conditions that were being
applied, there should be no restriction on the displacements (or velocities)
of the deforming material.  That is why I think that one is really imposing
displacement boundary conditions, not stress boundary conditions.

You raise a good point in your thought experiment for convergent shear.  The
conditions you propose, however, require the deformation to be
inhomogeneous, if the fluid has to stick to the rigid plates but still
accommodate the convergence with a constant-volume flow.  Applying the
symmetry principle to the global deformation is only valid if the boundary
conditions are uniform and if the resulting global deformation is
homogeneous, because that is the only situation in which the local and the
global conditions are the same.

Perhaps we have imposed this discussion on the list members long enough now
to stop postings and continue as a private discussion if necessary.  Perhaps
we will just agree to disagree?  But I have enjoyed the exchange!

Cheers,

rob

On Mar 28, 2011, at 5:00 AM, Dazhi Jiang wrote:

> Rob,
> I would like to emphasize that the material deforms according to the 
> whole set of requirements (constitutive equations, compatibility, 
> mechanical laws, and boundary conditions). You suggest that we must 
> regard a shear traction boundary condition as a velocity condition, 
> merely to make the symmetry principle apply. If that is what the principle
is about, I find it annoying.
> 
> Now let's imagine another situation with the fluid between two rigid
plates.
> Suppose the plates are vertical and extend infinitely along strike. 
> Add a boundary normal convergence component to the boundary velocity 
> in addition to strike-slip shear and suppose the fluid can move 
> vertically. This is the model Robin & Gruden (1994, JSG) considered 
> for transpression. How does the symmetry principle apply here? In this 
> case, the velocity boundary condition is monoclinic. But the velocity 
> field is triclinic everywhere, except in the middle of the zone.
> 
> Again, to me at least, I can use a stress BC to achieve the same 
> velocity
> field: a normal stress plus a shear stress, in other words, an oblique 
> traction BC.
> 
> I have to admit that although I have known symmetry principle since my 
> PhD time, I have never applied it in my work.
> 
> Cheers,
> Dazhi
> 
> 
> -----Original Message-----
> From: Tectonics & structural geology discussion list 
> [mailto:[log in to unmask]] On Behalf Of Robert J. Twiss
> Sent: Monday, March 28, 2011 12:58 AM
> To: [log in to unmask]
> Subject: Re: a plea and a new question?
> 
> Dazhi,
> 
> But when you apply a stress boundary condition, you do not prescribe 
> what the deformation will be; the material will deform according to 
> the constitutive equations.  And the material only knows that an 
> orthorhombic stress is applied to the boundary.  What, then, requires 
> the material to react by deforming in a simple shear as opposed to a 
> pure shear, for example?  What would be the difference between the 
> situation you describe and a situation in which an orthorhombic stress 
> that is applied to the same material produces a pure shear?
> 
> My response would be that when an orthorhombic stress is applies to an 
> isotropic material, the deformation (the effect) can have no lower 
> symmetry than orthorhombic, and thus cannot be a simple shear.  To get 
> a simple shear, you have to apply monoclinic boundary conditions, and 
> you cannot do that by applying a stress.
> 
> I do not understand where you see the circularity in this.  If you 
> accept the symmetry principle as valid, it seems to me this result is 
> necessary and unambiguous.
> 
> Cheers,
> 
> rob
> 
> On Mar 27, 2011, at 6:42 PM, Dazhi Jiang wrote:
> 
>> Frankly, that seems a bit like a circular argument. Also, according 
>> to
> your argument, we can altogether get rid of the so-called stress or 
> traction boundary condition. For if shear traction boundary condition 
> must be in fact regarded as a velocity BC, why not the pure shear case 
> as well? But a prescribed velocity BC and a prescribed boundary 
> traction are different boundary constraints.
>> Thanks for sharing your thoughts.
>> Dazhi
>> Sent from my BlackBerry device
>> 
>> -----Original Message-----
>> From: "Robert J. Twiss" <[log in to unmask]>
>> Sender: Tectonics & structural geology discussion list 
>> <[log in to unmask]>
>> Date: Sun, 27 Mar 2011 15:11:18
>> To: <[log in to unmask]>
>> Reply-to: Tectonics & structural geology discussion list 
>> <[log in to unmask]>
>> Subject: Re: a plea and a new question?
>> 
>> 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