1
$\begingroup$

From AN INTRODUCTION TO CONTINUUM MECHANICS by J.N.REDDY

A rigid-body motion is one in which all material particles of the continuum $\mathcal{B}$ undergo the same linear and angular displacements. However, a deformable body is one in which the material particles can move relative to each other. Then the deformation of a continuum can be determined only by considering the change of distance between any two arbitrary but infinitesimally close points of the continuum.

Why can the deformation of a continuum be determined only by considering the change of distance between any two arbitrary but infinitesimally close points of the continuum?

I think one can determine the deformation of a continuum by (for example) directly measuring change of distance between any two arbitrary points of the continuum.

Although, I know that we will use infinitesimally close points for defining the deformation gradient tensor, IMHO, the sentence "Then the deformation of a continuum can be determined only by considering the change of distance between any two arbitrary but infinitesimally close points of the continuum" is not correct.

$\endgroup$
  • 1
    $\begingroup$ I don't know continuum mechanics at all but it is a common occurrence that "pathologically behaving" stuff behave nicely "infinitesimally". For example, transformations that form a smooth group might not commute, but "infinitesimal" transformations do. Parallel transport provided by a connection is path-dependent, but "infinitesimal" parallel transport isn't. If you consider how two distant points get displaced, it gives no idea about what deformation happens, since many deformations can give the same displacement. "Infinitesimally" separated points are in a sense.... $\endgroup$ – Bence Racskó Feb 23 '17 at 18:26
  • $\begingroup$ ... adjacent, so this ambiguity is removed. $\endgroup$ – Bence Racskó Feb 23 '17 at 18:27
  • 2
    $\begingroup$ I think "only" here actually means that "you don't have to consider points separated by a large distance". In other words, I guess he means that "infinitesimal are enough"! $\endgroup$ – Peaceful Feb 23 '17 at 18:36
  • 1
    $\begingroup$ The relative position of two points a finite distance apart may remain the same, even if the material "in between the points" deforms. And don't forget pathological cases like a body whose deformation $u(x)$ close to $x = 0$ follow a pattern like $u(x) = x \sin (1/x)$. $\endgroup$ – alephzero Feb 23 '17 at 21:16
2
$\begingroup$

It is a calculus requirement.

To solve such problems, you have to consider strain, and stress, which are related to the derivative of the displacement.

To be infinitesimally close means close enough that the change in distance, divided by the original distance, equals the derivative of the displacement field.

That is a really great text, by the way.

$\endgroup$
0
$\begingroup$

Okay I'm going to get a lot of hate for this answer, by both physicists and mathematicians, but for the sake of keeping this simple and easy to understand intuitively, I'm going to use a very crude example:

Think of this: Rigid object = a block of wood Flexible object ("deformed") = a block of goo

You push the wood and what happens? Every part of it moves together.

You push the goo and what happens? Parts of the goo move fast and change direction, whilst parts of it stay fairly still.

Now there's a risk that if we picked a random 2 points on the block of goo (let's say the very top and the very bottom), by complete coincidence, they might just happen to move at the exact same speed and direction. Does that make the goo rigid? No, it means you got lucky.

To get rid of this "luck" element, we have to pick 2 points that are infinitesimally close to each other. Because we'd expect all the points in the immediate surroundings to be moving at a different speed and direction, if the object is truly flexible all around.

I hope this helps you to understand this theory better.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.