Rob Twiss schrieb:
>
> I. Correction and clarification of my previous posting
> In my last posting, in haste I incorrectly substituted e(i,j), the
> infinitesimal strain tensor, for d(i,j), the instantaneous strain rate
> tensor (the symmetric part of the velocity gradient tensor), and 'work' for
> 'rate of work' or 'stress power'. This error, however, does not affect my
> conclusions, as amplified in item III below.
OK, but this is not the point. It is my claim that the entire theoretical
concept is out of sync with general theoretical physics. In this context, why
are terms as e.g. "strain rate tensor" used? Rate with regard to what? It must
be time. Thus an equation of motion is implied. Why is an equation of motion
used instead of an equation of state? Alternatively, how are they transformed
into one another?
We will return to this question below.
> II. Koenemann's error regarding the definition of divergence
> If instead, we temporarily understand the letter 'd' to
> indicate partial differentiation, then the divergences of the stress and
> strain tensors would be, respectively,
>
> dt(i,j)/dx(j) and de(i,j)/dx(j) (both summed
> on j)
>
> Koenemann wrote:
> '... the trace of a tensor is its divergence.'
>
> This is just not correct! Koenemann does not seem to realize that the
> divergence by definition involves partial differentiation.
Yes I do, see below.
> Again using 'd'
> to denote partial differentiation, the divergence of an arbitrary vector
> function f(i) and of an arbitrary second rank tensor function g(i,j), in
> index notation and expanded notation, are:
>
> df(i)/dx(i) = df(1)/dx(1) + df(2)/dx(2) + df(3)/dx(3)
>
> dg(i,j)/dx(j) = dg(i,1)/dx(1) + dg(i,2)/dx(2) + dg(i,3)/dx(3)
>
> Thus the divergence of a vector-valued function (such as f(i) ) is a
> scalar, and the divergence of a tensor-valued function (such as g(i,j) ) is
> a vector, as indicated by the fact that dg(i,j)/dx(j) has three components
> for i=1:3. The trace of g(i,j), however, is a scalar g(i,i) (summed over
> i=1:3).
> g(i,i) = g(1,1) + g(2,2) + g(3,3)
> The infinitesimal strain e(i,j) is the symmetric part of the
> displacement gradient tensor. If we take u(i) to be the displacement
> vector, and we use 'd' to indicate partial differentiation, then
> e(i,j) = 0.5 [ du(i)/dx(j) + du(j)/dx(i) ]
> Thus the trace (the first scalar invariant) of the strain tensor is the
> divergence of the displacement vector, NOT the divergence of the strain
> itself, because
> e(i,i) = e(1,1) + e(2,2) + e(3,3)
> e(i,i) = du(1)/dx(1) + du(2)/dx(2) + du(3)/dx(3)
> which is identical in form to the divergence of a general vector-valued
> function f(i) given above.
But dt_ij/dx_j = 0 (equilibrium condition, see Landau & Lifschitz Vol.6, p.7).
And e_ii = du_1/dx_1 + du_2/dx_2 + du_3/dx_3 = 0 is still a straightforward
statement to the effect that all displacements cancel. If so, it does not matter
which work function is being used; the total work is zero.
> III. Koenemann's error regarding the work
> Let us accept that we can obtain the work associated with an
> elastic deformation by integrating t(i,j) d(i,j) over the time during which
> the deformation accumulates, and that the work will involve the product
> t(i,j) e(i,j).
>
> Koenemann wrote:
> 'Both t(i,j) and e(i,j) are "symmetric", i.e. orthogonal, i.e. the off-diago=
> nal
> terms can be taken to be zero, hence if e(i,i) = 0 and/or t(i,i) = 0,=
> there is
> no way out: t(i,j) e(i,j) = 0.'
>
> Mathematically this is just plain wrong, as we can see immediately from
> simple algebra.
> Let us assume that the stress and strain components are expressed
> in principal coordinates, and that the principal axes of stress and those
> of infinitesimal strain are parallel. Then the off-diagonal terms are
> indeed zero. If we further assume that the traces (the first scalar
> invariant) of the stress and of the infinitesimal strain respectively are
> both zero, then we have:
> (1) e(i,i) = e(1,1) + e(2,2) + e(3,3) = 0
> (2) t(i,i) = t(1,1) + t(2,2) + t(3,3) = 0
> Equation (1) expresses the condition for constant-volume deformation for
> infinitesimal strain (but not, by the way, for finite strain, which also
> includes non-linear terms), and equation (2) implies that the mean normal
> stress is zero.
There is a little more behind it. Since t is a force (per area) in the sense of
Newton f = ma, (2) is also a mass conservation condition. And if the mean normal
stress is zero it means that the sum of the forces is zero, that is, no net work
is done. We do not realy need to consider e_ij in considering work; it suffices
entirely to consider the forces; if they sum to zero, there cannot be a net
nonzero work.
> In principal components, we have for the product t(i,j)
> e(i,j) summed over i=1:3 and j=1:3
> (3) t(i,j) e(i,j) = t(1,1) e(1,1) + t(2,2) e(2,2) + t(3,3) e(3,3)
> All other terms resulting from this summation involve off-diagonal
> components (i not equal to j) which are zero in principal coordinates.
> Solving (1) for e(1,1), solving (2) for t(1,1) and writing the product
> t(1,1) e(1,1) gives:
> (4) e(1,1) = -e(2,2) - e(3,3)
> (5) t(1,1) = -t(2,2) - t(3,3)
> (6) t(1,1) e(1,1) = t(2,2) e(2,2) + t(3,3) e(3,3) + t(2,2)
> e(3,3) + t(3,3) e(2,2)
> Substituting (6) for the first term on the right of (3), we can rewrite (3) =
> as,
> (7) t(i,j) e(i,j) = 2 t(2,2) e(2,2) + 2 t(3,3) e(3,3) + t(2,2)
> e(3,3) + t(3,3) e(2,2)
> (8) t(i,j) e(i,j) = t(2,2) [2e(2,2) + e(3,3)] + t(3,3) [e(2,2)
> + 2e(3,3)]
> This is not zero! Thus it is clear that equations (1) and (2) do NOT lead
> to the requirement that the product t(i,j) e(i,j) must be zero, it is
> incorrect to argue that it does, and all results relying on that error are
> wrong.
For simplicity, let's assume that all terms containing subscript 2 are zero
(plane strain etc.). It follows that e_11 = - e_33, ditto for t_ij. Assume that
contraction takes place along x_3, extension along x_1. That is, both e_33 and
t_33 are negative. Necessarily their product is positive, that is,
t_11 e_11 + t_33 e_33 = 2 t_11 e_11 = q > 0 always. What does that mean? It
means that q is independent of the sign of the components of t_ij and/or e_ij.
It holds for all cases where like components e_ii and t_ii (no sum on i) have
the same sign, be it for compression both along x_1 and x_3 as well as for
expansion both along x_1 and x_3, as well as for compression along one, but
expansion along the other - and for all these cases q is the same, even though
in two cases volume work is involved, in the last case it is not.
Let's compare with reality.
- First, if volume work is done either way, the required energy for
any unit change of length is so much larger than for an anisotropic
volume-constant deformation that the argument must be spurious.
- Second, at least in my uneducated opinion it matters whether volume
work is positive or negative, so a theory that is insensitive to signs is
not to be trusted.
- Third, consider real materials subjected to anisotropic loading.
Twiss's argument implies that q > 0 for all materials. Let's perform
a slow volume-neutral deformation on a gas and a solid. In the case of
the solid measurable work is done; in the case of the gas this is not so.
However, the theory is applied to gases. On the other hand, if we let
the solid go it will instantly return to its initial configuration whereas
the gas will not.
The difference is: in the case of the solid we have done work upon the bonds in
the solid, and an elastic potential builds up. In the case of the gas we have
done work only by pushing around single atoms which behave like discrete bodies
in free space, and an elastic potential cannot build up. Since indeed all inward
paths of mass along x_3 is equalled by mass moving outward along x_1, the total
work must cancel. This is what indeed we observe, but this is not what Twiss's
theory predicts.
The term t_ij e_ij is physically meaningless.
[snip]
> Perhaps also the error arises from a misconception about the
> significance of diagonalizing the stress and strain tensors. Expressing
> the stress components in principal coordinates does NOT change the state of
> stress in the material. [snip]
That's not the point. The point is that shear stresses are not represented in
the tensor, nor in a consideration of work as long as the trace of the stress
tensor is to have any physical meaning. Twiss's theory is unable to consider
work done by shear forces or shear stress. I do not even understand how shear
stress does work. In the case of a coaxial deformation there must be a sinistral
stress vector for any dextral stress vector; if one of them has the tendency to
cause a sinistral rotational acceleration, the other should do the opposite, but
the effect must cancel; and since the equilibrium condition requires that no
external rotation takes place no net rotational work is done. Or do I again
misunderstand something?
And sooner or later I want it to be discussed whether or not the stress tensor
exists. I do not like to speculate about voodoo physics.
[snip]
> IV. Comments on the thermodynamics of continuous media
[snip]
> ... for example viscous materials ...
Do me the favor and exclude irreversible processes from this discussion, simply
to keep a minimum focus for those who listen in. There is a lot to be said here,
but it would lead far too far. Stress is an elastic, time-independent,
reversible phenomenon, it relates directly to elasticity.
>> Koenemann wrote with respect to Eringen's discussion of
>> thermodynamics of continuous media:
>> 'Chapter 4 starts with a rather idiosyncratic statement of the First Law:
>> dK/dt + dE/dt = w + q
>> (Eringen used Newton's dot notation) where K is the kinetic energy and E is
>> the internal energy.
>> 'This is not what I learned. The First Law is Delta E = w + q, or
>> dE = dw + dq.
>> Nowhere does the First Law say anything about kinetic energy.'
>
> In fact Eringen refers to the first equation as the 'conservation
> of global energy' (surely kinetic energy must be included in any general
> energy conservation law),
Short outline of theoretical physics
Consider a discrete body in free space. It has a velocity and a direction, a
kinetic energy E_kin and a potential energy E_pot. Bernoulli's law of energy
conservation requires that E_kin + E_pot = const. If we have n discrete bodies
we have a kinetic system, and the const of all bodies is the total energy H of
the system. All bodies together have an average velocity which is a measure of
the total energy of the system. Work is done during collision of two bodies
which thus accelerate one another (+ or -), that is, work is work done _in_ the
system, and work is done against inertia.
As long as it stays that way the system is conservative (H = invariant). For
this case an equation of motion (Newton's 2nd law) is required in order to
understand force; Newton's third law is the proper equilibrium condition, and
work is path-indpendent in Euclidean space (ie. real space). This is the state
of the art of 1842, the year in which the Hamiltonian was discovered (by
Hamilton, of course - not Alexander, but William).
Note that the precise nature of the mass is irrelevant, what matters is only its
weight in [kg] which I call the inertial mass.
Up to here the theory is called Newtonian mechanics. Enter electromagnetic
potentials (EMP). All atoms repel one another without touching one another, and
this potential is not a function of the weight, but it exists per atom. It is
thus proportional to mass, but it may change with chemical species, hence the
inertial properties of mass are irrelevant, we must count the mass in atoms. For
simplicity we count it in mol.
The EMP is not included in H, hence we get a new sum: H + EMP = U which is the
internal energy of the kinetic system; it now is also a thermodynamic system.
For illustration, either H or EMP may be disregarded, depending on the problem.
If we consider rocks in outer space their EMP is irrelevant; if we consider a
gas at room density both EMP and H matter; if we consider the inside of a solid
H is irrelevant because the atoms are bonded anyway, bonds are electromagnetic
forces, and they are internally balanced in the unloaded state.
If we now change the energy U of the system we may do so either by changing the
average velocity of the n bodies by adding kinetic energy from outside; this is
called a heat influx. Or we change the volume of the system, this we call work;
hence dU = dq + dw. Note that this is not possible without interaction of system
and surrounding, work is therefore work done _upon_ the system, it is work done
against the thermodynamic potential U of the system. The proper equilibrium
condition is P_int + P_surr = 0. Since U is a variable the process is said to be
non-conservative (it does not mean that energy is lost, it just means that it is
changed). If the process is reversible it is path-independent in PV-space, but
not in real space.
The global energy conservation law is therefore the First Law. Kinetic energy is
included way down at the bottom,
U + dU = ((E_kin + E_pot) + EMP) + dw + dq
In the case of solids we must also consider the fact that solids have an
internal pressure; a mol of gas has a V = 22,4 l but a mol of solid has eg. V =
6 cm^3, hence the solid has an internal pressure of 500 bar which is internally
balanced. If the external pressure changes from 1 to 2 bar, the internal
pressure changes from 500 to 501 bar. Furthermore, because of the bonds the
single atoms cannot move freely, but for all practical purposes we can consider
them fixed in space such that their E_kin = 0. And since we are essentially
dealing with electromagnetic potentials and work done upon a system we cannot
apply an equation of motion, but we _must_ use an equation of state.
Finally, since solids have internal bonds they can sustain an anisotropic loaded
state indefinitely. That is, it is possible to do work upon a system of solid
and cause a change of state without changing its volume, just by changing its
shape. This is an elastic deformation.
Questions?
It is characteristic of present-day continuum mechanics that the profound
difference between Newtonian mechanics and thermodynamics, between conservative
and non-conservative, work done in a system and work done upon a system, is
blurred; Eringen's "global law" is a very good example. This mixup is probably
the worst mistake one can possibly make in classical physics this side of
Einstein. It is typical of continuum mechanics that it often explicitly refers
to the inertial mass which is useful only and only in the context of Newtonian
mechanics; whereas the thermodynamic mass is dimensionless and counted in atoms,
and the density is n/V.
I grant that Eringen at least concedes in principle that elastic loading is a
thermodynamic process; in the major articles and books of Truesdell there is no
mention of the First Law whatsoever.
[snip]
> V. Futility of debating results that rest on erroneous analysis
> Koenemann's 'theory' apparently leads to the dimensionally
> impossible conclusion that force per unit area scales differently from
> energy per unit volume. It must therefore contain an error.
Prove me wrong.
Falk Koenemann
_____________________________________________________________________
| Dr. Falk H. Koenemann Aachen, Germany |
| |
| Email: [log in to unmask] Phone: *49-241-75885 |
| |
| URL: http://home.t-online.de/home/peregrine/hp-fkoe.htm |
| stress elasticity deformation of solids plasticity strain |
|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~|
| The rain, it raineth on the Just |
| And on the Unjust fella. |
| But chiefly on the Just because |
| The Unjust stole the Just's umbrella. |
|_____________________________________________________________________|
|
