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Imagine you have a simple, uniform beam floating out in space and you apply a force to the end of it. Somehow, you are able to constantly apply this force normal to the surface such that the beam is experiencing a constant net torque (about it's center of mass, I believe).

enter image description here

Question: Assuming the beam is real (not idealized as being rigid), will the beam bend under such a constant torque? If so, how can we quantify this bending?

(And yes, I realize the beam will translate under this unbalanced force, but I'm really more interested in any potential bending).


Context on why this even matters:

I'm trying to model a system where a beam is on a pivot. An applied force at the end of the beam rotates it clockwise (like pictured above), while a torque applied counterclockwise at the pivot will oppose that motion. The only thing is... the torque from the applied force is larger than the torque at the pivot, meaning there will be some net torque in the clockwise direction.

In order to determine how much the beam will deflect in this situation, I was considering modeling this as a superposition of two different beams: one which is in mechanical equilibrium where the pivot torque and part of the applied torque are equal and opposite (in which case it's not too difficult to find the stresses developed in the beam) and one which is simply a beam pivoted at one end and which has a single force applied at the other (this force would be whatever portion of the force was not counteracted by the torque applied at the pivot).

I figured the bending of the beam could be attributed to the mechanical equilibrium portion only, while the angular acceleration of the beam could be attributed to the net torque portion only (in which case the calculations are simple). But, based on the question I asked above, I now feel like the simple net torque on a beam model might also bend it somewhat, in which case I would want to know how to quantify that.


Edit 1

  • Added some updated thoughts below
  • Removed some of the context section above, as it didn't seem to add much

Taking inspiration from Claudio's answer:

Consider a Cantilevered Beam with a force on the end of it. Suppose, for the sake of argument, that the beam has a large mass at its end, which means that this force down on the end of it can be treated as the gravitational force pulling down on this beam.

This beam has shear and moment reactions at the fixed support. According to the equivalence principle, this can equally be treated as a free floating beam with the same reactions, but with no gravitational force.

enter image description here

This is close to what I want, but in the picture I provided in my original question there was no moment. What I'm really looking for is something more like this:

enter image description here

This is a beam with a pin on one end that it pivots about. The pin provides a reaction force to oppose the applied force due to gravity (we know these are equal and opposite because the beam's center of mass does not translate due to these forces - only due to the net torque), but the pin does not provide a moment reaction, which means there will be a net torque and the beam will spin about the pin.

Apply the equivalence principle, and you have a free floating beam with a single shear force on one end - exactly the situation that I proposed.

So, what does this tell us about how it will bend?

My intuition tells me that each of these situations would bend like this:

enter image description here

But that can't be true since the equivalence principle tells us that these two beams are the exact same thing. Therefore, however the first beam bends, so too will the second beam, which leads to something like:

enter image description here

The fact that the free floating beam would bend like that seems very strange to me... but according to the equivalence principle it is indeed what happens. I can even take a crack at quantifying it.

If you draw the shear/bending moment diagrams for the pinned beam with the upward shear on the left and the downward shear due to gravity on the right, you'll find that it is always in positive shear and has a constantly increasing (and always positive) internal bending moment. The contour of the beam will take the form:

$$y' = \frac{mgx^2}{2EI}$$

which is a concave-up parabola.

At first, this doesn't seem to match our "normal" case since that's concave down as you move away from the pivot and towards the applied force (see the previous image). But in the case of the free-floating beam (which is really a more true representation of the former), our pivot is the beam's center of mass, in which case moving away from the pivot (the CM) and towards the applied force, it will be concave up (shown in the previous image).

Thus, the equations seem to gel with the intuition.

I suppose my question with this edit is does this seem like a reasonable application of the theory in Claudio's answer?

Edit 2: Some assumptions I made.

  1. The mass of the beam is negligible compared to some point mass at the end. So if we write all of the forces due to gravity, we can leave out the beam's weight without changing the results much.
  2. Re: gravity will cease being a pure shear on the beam as the beam rotates; in my original question I stipulated that the force would always be normal to the beam surface, so it would create a constant torque. The utility of the equivalence ratio (in my view) is that it allows me to take a snapshot of the internal shear/bending moment when the force is first applied (when the beam is experiencing one pure shear at its end). I can then convert the situation back to a non-gravitational force out in space that is always normal to the beam and assume that the shear/bending moment snapshot I took a moment ago will hold even while the beam rotates (because in the end, both situations are fundamentally just a force shearing a beam).
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The beam will undergo damped vibrations around a deformed configuration. The equations of motion of a deformable body are given in Section 2.3 here, for example, and are quite a bit more complex than those for either rigid-body kinematics or elastic deformation in an inertial frame.

As a clarifying analogy, consider the simpler case of an axial load suddenly applied to the end of a real spring. We obtain a decaying transient response (governed by the spring mass distribution, spring constant and damping coefficient) superposed over a steady-state response (governed by the spring mass distribution and spring constant alone). The spring mass distribution depends on the geometry and material density, the spring constant depends on the geometry and material stiffness, and the damping coefficient depends on the internal friction, also known as material hysteresis. Since a load through the center of mass of a free-floating object induces linear acceleration, by the equivalence principle, the behavior is identical to that of a spring with mass that's suddenly placed vertically on a surface in a gravity field: transient settling (involving oscillation if the spring is underdamped) dies down to a compressed state.

The motion I'm referring to is something like this (source):

This analogy seems useful in helping us predict the behavior of the beam in the actual problem without spending hours working out the equations of motion and deformation: the entire beam will undergo a damped transient response to an equilibrium position of curvature and shear, superposed atop accelerating translation and accelerating rotation from the off-balance lateral load.

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  • $\begingroup$ That makes sense. I agree, solving this problem by analogy would certainly be preferable, but do you know of anyway to quantify this deformation (at least, the steady state deformation). $\endgroup$
    – Chrøme
    Commented May 19, 2021 at 1:35
  • $\begingroup$ Yes; solve the equations. It's not a trivial problem. Your edit above equates gravity with a point load at the end of a cantilever beam, but gravity acts on the entire beam. It also changes direction constantly and at a varying rate because the offset load induces angular acceleration (in addition to linear acceleration). Furthermore, the accumulating angular speed tends to straighten the beam even as the offset load tends to curve it. $\endgroup$ Commented May 19, 2021 at 17:50
  • $\begingroup$ Okay - you're right... doesn't look like a trivial problem, but I appreciate at least being aware of the equations of motion. It remains to be seen whether I'll actually go through with it. All of your comments on my edit were valid- I've addressed the relevant assumptions I made with another edit to my question. Regarding the beam straightening as it builds up angular speed... this is something I hadn't considered before. Luckily, the system I'm modeling wouldn't allow the beam to reach speeds or angular displacements where those centripetal effects would become significant. $\endgroup$
    – Chrøme
    Commented May 19, 2021 at 18:58
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We can use the equivalence principle from standard relativity to realize that, from the point of view of the stress and strain, this is equivalent to a cantilever beam when the force is initially applied.

That is due to the fact that a cantilever beam can be thought as being accelerated upward by the ground, through the contact point, instead of being attracted by the force of gravity.

The force "F" of the picture can be taken as "mg" for some "g".

After the beam starts to rotate, centripetal forces will produce additional tensile stress, mainly in the direction of its length.

Edit
My idea of comparing to a cantilever beam was different, but now I realize that it doesn't work. Suppose gravity is off, and the beam is carefully placed so that it is touches only one end. There is no force on the beam and it is at rest.

Now turn gravity on. It starts to fall (except the fixed end) and rotate. This situation is similar to a beam in the outer space, and a force F applied to one end.

The problem is that initially the curvature would be upward, while for the cantilever beam it is downward. The transient geometry is different from the final one.

It seems that you have to solve the problem by elasticity theory, and it is not possible to reduce to a static situation.

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  • $\begingroup$ I added a (slightly long... sorry!) edit to my original question to try and get a better grip on your answer. Would you mind taking a look at it and seeing if there are any holes in my reasoning? $\endgroup$
    – Chrøme
    Commented May 19, 2021 at 17:16

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