If a uniform beam was in space and only was applied a force on one side (Stress)

[gladius999]

If a uniform beam was in space (no gravity or resistance of any kind) and only was applied a force F on only one side, does the inside of the beam feel a stress and deform? If so, what does the stress equal? Can you explain why/why not the beam feels a stress or not?

I know that that if applied a force on both sides, the stress would just be F/cross sectional area of beam. But in this case the force is only on one side.

I believe that if I imagine the beam to consist of 2 parts joined up (part A and part B), if say a pulling force was applied to the part A, part A would pull on part B. But then part B would also exert a force equal in magnitude but opposite in direction on part A (Newton's third law). Would'nt part A feel a tensile stress then as it would have a force pulling it in one direction and another pulling it towards B in the opposite direction? Can someone correct me on this?

Thank you very much

 

[viscousflow]

Well if you want a true answer to this question you have to define your constraints. If the element was simply floating in space and a force applied on one side then there would be no stress.

If it is simply supported on one end and a force applied to another the resulting stress far away from the point of application is F/A. (In technical terms this element is now a rod).

Even if you left the rod floating and applied precisely the same force on both ends the stress would remain F/A. At any cut along the rod, forces must remain in equilibrium for this to remain true.

Imagine the rod as one continuous element and at any point equilibrium must hold. So all sections should either be in a tensile or compressive state.

 

[AlephZero]

If the element was simply floating in space and a force applied on one side then there would be no stress.

That is wrong. The element would accelerate. From Newton's laws of motion there is a force on each particle causing it to accelerate. That force comes from the gradient (rate of change) of stress in the body. In other words, it you consider a small part of the body, the different stresses on each side of it create different forces acting on it, and the difference between the forces = mass x acceleration.

Even if you left the rod floating and applied precisely the same force on both ends the stress would remain F/A.

No. If the force is applied at one end of the rod in the direction along its length, the stress is F/A at the end where the force is applied, and decreases linearly to 0 at the other end.

Imagine the rod as one continuous element and at any point equilibrium must hold. So all sections should either be in a tensile or compressive state.

The rod is not in equilibrium. It is accelerating.

 

[gladius999]

Well if you want a true answer to this question you have to define your constraints. If the element was simply floating in space and a force applied on one side then there would be no stress.

If it is simply supported on one end and a force applied to another the resulting stress far away from the point of application is F/A. (In technical terms this element is now a rod).

Even if you left the rod floating and applied precisely the same force on both ends the stress would remain F/A. At any cut along the rod, forces must remain in equilibrium for this to remain true.

Imagine the rod as one continuous element and at any point equilibrium must hold. So all sections should either be in a tensile or compressive state.

That is wrong. The element would accelerate. From Newton's laws of motion there is a force on each particle causing it to accelerate. That force comes from the gradient (rate of change) of stress in the body. In other words, it you consider a small part of the body, the different stresses on each side of it create different forces acting on it, and the difference between the forces = mass x acceleration.No. If the force is applied at one end of the rod in the direction along its length, the stress is F/A at the end where the force is applied, and decreases linearly to 0 at the other end.The rod is not in equilibrium. It is accelerating.

Thanks for both your replies. It appears to me that there seems to be some contradiction between you two.

So what I'm getting here is, that if a uniform bar is in free space(with no gravity, resiistance) and no end fixed but only one end being pulled by a force F, the bar does feel a stress at the end being pulled at a pressure of F/A and decreases along the bar linearly to zero?

I think it makes sense to have a stress in the bar even if only pulled by one side because if we view the bar as consisting of many small bars added together, when a force is applied to one end, all of the bars must feel a pulling force, but each bar also has to pull the bar behind it, therefore each bar should be feeling a tensile stress?
Can someone point out the flaws in my reasoning?

 

[viscousflow]

That is wrong. The element would accelerate. From Newton's laws of motion there is a force on each particle causing it to accelerate. That force comes from the gradient (rate of change) of stress in the body. In other words, it you consider a small part of the body, the different stresses on each side of it create different forces acting on it, and the difference between the forces = mass x acceleration.

Ok true, within the region of application yes there is a stress, however, very small.

No. If the force is applied at one end of the rod in the direction along its length, the stress is F/A at the end where the force is applied, and decreases linearly to 0 at the other end.

You are correct however, I mentioned two forces equal and opposite. Equal and opposite forces applied at both ends make the rod a two-force member. Regardless if its on Earth or in space.
In the region of application the stress is not F/A, sufficiently far away it is ( Re: Saint Venant's region...).

The rod is not in equilibrium. It is accelerating.

Maybe I should've been more specific, forces inside the rod must maintain equilibrium up to fracture, i.e. regardless of the reference frame forces inside add up to zero as long as it is not buckling, warping or under any other effect which it is continually deforming...

 

[gladius999]

Ok true, within the region of application yes there is a stress, however, very small.


You are correct however, I mentioned two forces equal and opposite. Equal and opposite forces applied at both ends make the rod a two-force member. Regardless if its on Earth or in space.
In the region of application the stress is not F/A, sufficiently far away it is ( Re: Saint Venant's region...).



Maybe I should've been more specific, forces inside the rod must maintain equilibrium up to fracture, i.e. regardless of the reference frame forces inside add up to zero as long as it is not buckling, warping or under any other effect which it is continually deforming...

Good sir, would you say my reasoning(shown below) is correct?:


I think it makes sense to have a stress in the bar even if only pulled by one side because if we view the bar as consisting of many small bars added together, when a force is applied to one end, all of the bars must feel a pulling force, but each bar also has to pull the bar behind it, therefore each bar should be feeling a tensile stress?

 

[Dale]

Thanks for both your replies. It appears to me that there seems to be some contradiction between you two.

yes, there is a contradiction. Viscousflow is wrong, AlephZero is right.

 

[DaveC426913]

Any time any force is applied to any object, the object will undergo stress and will deform. The motion will only propagate through the object at the object's speed of sound, which is very finite.

The only circumstance in which this would not occur is in an non-Einsteinian universe, where ideally-rigid objects can exist and forces can be transmitted at infinite speed.