Time-dependent axial deformation

[Ryan Gardner]

TL;DR Summary

In the attached PDF I have documented an attempt to derive a time-dependent equation for axial deformation from Newton's 2nd law and the fact that forces propagate at the speed of sound in solid materials.

This is highly speculative, and I very much doubt that it is actually correct. If anybody knows of a correct equation for time-dependent axial deformation, or at least how to go about deriving a more correct equation, I would greatly appreciate any feedback.

 

[JBA]

"the fact that forces propagate at the speed of sound in solid materials"

A contact shock wave will propagate through a rod at the speed of sound of the material; but, I am not sure that is true for an applied force which is applied at both ends of the object and will simultaneously and uniformly stress the rod through out its entire length (ignoring material property or physical variations). Either way "t = L / speed of sound"; so, while I admire the effort you put into your project, I cannot see a real application or contribution from your derived equation.

 

[Ryan Gardner]

Thank you for your reply. I'm not exactly sure what you're getting at here; the whole point of this is that the applied force is not applied simultaneously at both ends of the object. At the instant the force is applied Newton's 3rd law is violated and the object is not in equilibrium, presumably until the resulting stress wave has traveled through the object. Consequently, the fixed support at the other end and the initially applied force will not act simultaneously until the object is in equilibrium, at which point deformation will cease.

If reaction forces responded to applied forces instantaneously it would be impossible for any deformation to ever occur in any situation, though brittle failure could occur if the forces were sufficiently great. Instantaneous force propagation implies instantaneous energy transport, which is equivalent to a violation of energy conservation as well as special relativity.

 

[JBA]

Other than a force from an impact force being resisted by the inertia of a body you can not create a force on an object without a simultaneous force from the opposite end.
From here I am going to leave to the other forum members to address this with you.
With only one other viewer than my self there appears to be any interest though.

 

[Chestermiller]

What @JBA is getting at is that the deformation of the rod is not homogeneous. One part of the rod is deforming, while the far end does not even know that anything is happening yet. To describe this behavior, you can't just use ordinary differential equations, because the deformation is varying both with time and spatial position. In this case, the behavior is described by a partial differential equation known as the "wave equation."

The derivation of the wave equation goes something like this. Let x be the location of a material cross section of the rod at time zero. At time t, this same material cross section is at location x+u(x,t), where u is the "displacement" of the cross section at time t. Let's consider the section of the rod between material cross sections x and $x+\Delta x$. The mass of this section of the rod is $\rho A\Delta x$, where $\rho$ is the density and A is the cross sectional area. The tensile forces acting on the two ends of this cross section at time t are $\sigma(x+\Delta x,t)A$ and $\sigma(x,t)A$, where $\sigma$ is the tensile stress. The acceleration of the mass is $\partial^2 u/\partial t^2$. So a force balance on the mass is:
$$\rho A \Delta x\frac{\partial^2 u}{\partial t^2}=\sigma(x+\Delta x,t)A-\sigma(x,t)A$$If we divide this equation by $A\Delta x$ and take the limit as $\Delta x$ approaches zero, we obtain: $$\rho\frac{\partial^2 u}{\partial t^2}=\frac{\partial \sigma}{\partial x}=E\frac{\partial \epsilon}{\partial x}$$where E is Young's modulus and $\epsilon$ is the local strain in the rod. The local strain in the rod is related kinematically to the local displacement by the equation $$\epsilon=\frac{\partial u}{\partial x}$$So from these equation, it follows that:
$$\rho\frac{\partial^2 u}{\partial t^2}=E\frac{\partial ^2 u}{\partial ^2 x}$$or, equivalently, $$\frac{\partial^2 u}{\partial t^2}=c^2\frac{\partial ^2 u}{\partial ^2 x}$$where c is the speed of sound in the rod material:$$c=\sqrt{\frac{E}{\rho}}$$This partial differential equation is called the "wave" equation. We can also readily show that not only does the displacement u satisfy the wave equation, but so also does the strain $\epsilon$ and the stress $\sigma$.

 

[Ryan Gardner]

I've already addressed this, but it seems that I will need to do so again.

There is no such thing as a force that propagates with infinite speed. That is an absolute fact. << Snide comment removed by a Mentor >> The gravitational force and electrical force propagate at the speed of light, and it has recently been experimentally confirmed that gravitational waves travel at the speed of light. Mechanical forces are transmitted through solid bodies at the speed of sound. Forces are necessary to transmit energy through a continuous medium. All energy is information, and vice versa. It is not possible for energy, whether it be light, sound, or anything else to be transmitted through any medium faster than the speed of light, and certainly not with infinite speed. This also means that it is not possible for information to be transmitted with infinite speed.

When you hit an object with a hammer, there is an equal and opposite force exerted on the hammer by the object - after an extremely short amount of time has passed. This appears to take place instantly, but it does not. If an object is sitting on a table and you hit that object with a hammer, it will be impossible to deform that object if there are forces acting upon every molecule in that object such that the net force on each molecule at every instant in time is zero. Do you care to explain to me how a solid body can deform when the net force acting upon every molecule in that body is zero, so that none of the molecules could have possibly accelerated, or gained a non-zero velocity so that they could be removed from their original positions?

If a solid body that is rigidly supported is suddenly acted upon by a constant force, the rigid support cannot respond to that force until the force begins to act upon the support, or upon the solid body at the location of the support. It takes some time for information (in the form of a stress wave) about the externally applied forces to be transmitted to a location where a reaction force can be generated. If forces could be transmitted with infinite speed in solid bodies none of the molecules in any solid body could ever be accelerated, so that no deformation could ever occur.

This is extremely basic physics. You cannot deform an object without displacing the molecules in that object, which requires that they be accelerated; which requires that for a brief period of time (at least) they experience a non-zero net force. Now, if all molecules were accelerated by the same amount then the object would merely change position in space without experiencing any deformation, so that deformation requires that some molecules experience greater displacement than others.

Not surprisingly, the molecules closer to the location where the initial external force is applied experience a greater amount of displacement, which is due to the nature of their acceleration. The molecules nearest to the rigid support do not accelerate at all, because they are simultaneously acted upon (at all points in time) by the initial external force and the reaction force, which are of equal magnitude.

 

[berkeman]

Thread closed briefly for Moderation...

 

[berkeman]

After a bit of editing to remove some snide comments by the OP, the thread is re-opened.

This is extremely basic physics.

Yep, it is. And we are trying to be patient and discuss what you have written. We do not allow personal theory development here at the PF, but what you are trying to understand and quantify in your summary paper is basic stuff.

The post by @Chestermiller is probably the best thing for you to read and understand in detail. The propagation of the impulse from one end of the object is described by the wave equation, and the last part of this introductory sentence in your paper is wrong:

If a constant external force is exerted on the end of a long slender member
in the longitudinal direction, the portion of the member's total mass that will
be influenced by the external force will vary with time until information about
the force has traveled to the opposite end of the member. If this opposite end
of the member is fixed, then the deformation of the member will cease when
information about the external force reaches the fixed support and the support
exerts an opposite force of equal magnitude upon the entire member, which
will behave as a single entity at that time.

Can you guess why?

 

[Ryan Gardner]

Thank you, Chestermiller, for providing a coherent, intelligible and thorough response to my original post.

 

[Dale]

At the instant the force is applied Newton's 3rd law is violated and the object is not in equilibrium

This is not correct. In Newtonian mechanics Newton’s 3rd law is never violated, including in non equilibrium dynamics with non rigid bodies. Were it ever violated then momentum would not be conserved.

If reaction forces responded to applied forces instantaneously it would be impossible for any deformation to ever occur in any situation

This is not correct. I am not sure why you believe it, so it is a little hard to correct. Again, if 3rd law forces were not instantaneous then momentum would not be conserved. At the point of contact between two objects there is always an equal and opposite force on each, including during deformation and including the time while any stress wave is still propagating across the object.

Instantaneous force propagation implies instantaneous energy transport, which is equivalent to a violation of energy conservation as well as special relativity.

Third law forces don’t have to propagate. They are at the point of interaction or contact, so the distance is 0. Relativity is not violated by a force that propagates 0 distance in 0 time.

 

[Ryan Gardner]

This is not correct. In Newtonian mechanics Newton’s 3rd law is never violated, including in non equilibrium dynamics with non rigid bodies. Were it ever violated then momentum would not be conserved.

This is not correct. I am not sure why you believe it, so it is a little hard to correct. Again, if 3rd law forces were not instantaneous then momentum would not be conserved. At the point of contact between two objects there is always an equal and opposite force on each, including during deformation and including the time while any stress wave is still propagating across the object.

Third law forces don’t have to propagate. They are at the point of interaction or contact, so the distance is 0. Relativity is not violated by a force that propagates 0 distance in 0 time.

You stated that in Newtonian mechanics Newton's 3rd law is never violated. This, of course, would seem to be true by definition. However, which version of Newtonian mechanics are you talking about? Newtonian mechanics as originally envisioned by Newton, or the mechanics that evolved through the 1700's and 1800's? Newton himself assumed that gravity (and presumably all other forces) propagated with infinite speed, which is false. By the early 1800's physicists were no longer comfortable with this assumption, and this is at least partly why the idea of a time-dependent force field through which forces can propagate was developed.

Let's say an external force is applied to the end of a cantilever beam, such that the force is only directly applied to the outermost layer of atoms at the end of the beam (assuming a perfectly orderly crystal lattice structure). Obviously, it takes zero amount of time for information about that force to be communicated to the end of the beam at which the force is applied. However, it takes some (very small) amount of time for information about the applied force to travel from the first layer of atoms to the second layer of atoms; and until this transfer of information takes place the second layer of atoms cannot respond to the applied force so that the positions of the atoms in the first layer will not change relative to the positions of the atoms in the second layer.

It is true that in this particular scenario Newton's 3rd law is not violated if you consider the reaction force to be the force exerted by the end of the beam on whatever is responsible for the externally applied force at the location where this force is applied, and it is correct to say that this is the case. However, the individual who initially responded to my original post did not seem to consider this to be the case. If you begin with the assumption that the cantilever beam is perfectly rigid, so that forces propagate through it with infinite speed, then the "reaction force" is effectively exerted by the fixed support at the end of the beam opposite that at which the external force is applied. If I were to rewrite my second post in this thread I would not say anything about the violation of Newton's 3rd law, and in my original post I said nothing about Newton's 3rd law. The individual who first responded to my initial post seems to have been convinced that it is absolutely impossible for a fixed object to only experience a force at it's unfixed end even for an infinitesimal amount of time, so that he was essentially thinking of the object in question as being perfectly rigid and forever in a state of equilibrium. In my second post I attempted to demonstrate that this is false by showing that if this actually were the case, then if the object were not in a state of equilibrium this would essentially constitute a violation of Newton's 3rd law (because it would not be important to differentiate between the reaction force at the point of force application and the reaction force at the fixed end of the object, because they would always be equal at all points in time), when in reality these reaction forces are not always equal.

In general, when I have spoken about a reaction force, I have been speaking of the reaction force at the far end of the beam after the externally applied force has propagated through the length of the beam. I was not talking about Newton's 3rd law when I made the following statement:

"If reaction forces responded to applied forces instantaneously it would be impossible for any deformation to ever occur in any situation."

In other words, if the beam were perfectly rigid, the fixed support would immediately respond to the applied force, so that the beam would always remain in a state of equilibrium, so that no deformation could occur. This statement has nothing to do with Newton's 3rd law.

"Instantaneous force propagation implies instantaneous energy transport, which is equivalent to a violation of energy conservation as well as special relativity."

If a particle disappears and immediately reappears at another location, so that it travels with infinite speed, since the particle does not have a continuous trajectory there is no way to prove that the particle that reappeared is the same particle that initially disappeared, so that instantaneous transport of matter (and energy or information in general) is essentially equivalent to violation of energy conservation from a local perspective (although if the entire universe is the system under consideration then the total energy in the universe is conserved at all points in time) because it appears that matter is destroyed at one location and created out of nothing at another location. This also has nothing to do specifically with Newton's 3rd law.

Almost everything in this thread has resulted from the apparent fact that the individual who initially responded to my original post did not seem to understand what I was was saying, and my clumsy attempt to provide additional clarification. Very little of this has anything to do with my original post, which makes no mention of Newton's 3rd law.

 

[Dale]

In general, when I have spoken about a reaction force, I have been speaking of the reaction force at the far end of the beam

Usually the term “reaction force” means the 3rd law pair as in the formulation of Newton’s third law “for every action there is an equal and opposite reaction”. I would not use this term at all, but especially not to describe the force at the opposite support point. I can see now that several of your posts were intended very differently than they were understood. Just avoid the word “reaction” entirely.

You stated that in Newtonian mechanics Newton's 3rd law is never violated. This, of course, would seem to be true by definition. However, which version of Newtonian mechanics are you talking about?

Whenever I talk about any theory I am always talking about today’s modern understanding of that theory. This forum is about science as it is understood and practiced by professional scientists today. Historical versions of a theory are only of historical interest, and that is not the focus of this forum. However, it doesn’t really matter here since Newton’s 3rd law is a basic part of every historical and current version of Newtonian mechanics that I am aware of.

If I were to rewrite my second post in this thread I would not say anything about the violation of Newton's 3rd law

Excellent! That is good to hear.

 

[Ryan Gardner]

I don't especially like the term reaction force, but it seems to be very commonly used in the mechanical engineering community to mean exactly what I meant it to mean, which is why I used it. I also dislike the term Newtonian mechanics because Newton had very little to do with much of the development of classical mechanics. Classical fluid mechanics, for example, didn't really have a solid foundation until the early 20th century.