Shear Deformation - Point Moment 7

[Celt83]

This is probably something that is very easy and I am way overthinking it.

I've done the unit force method which shows that the shear deformation from a point moment is 0 at all locations on a simple span. I've done the same thing to show that the shear component of the slope at any location on the beam is constant = k M / A G L. Here is where I get hung up if the slope is constant and non-zero why is the deflection 0 from a math stand point, the mechanics make sense to me but I'm lost in the math on this one.

I have a great mechanics of materials book by Timoshenko and Young which for me so far has the best break down of shear deformations I've been able to find and in there they present that the slope due to shear is simply dy/dx = tau,max / G = k V_x / A G, which aligns with my unit method solution of k M / A G L.

integrating that once yields: y,shear = k M x / G A L + C1

Initial conditions are x=0, y=0 and x=L, y=0 the first condition yields C1 = 0, however the second condition yields C1 = k M / A G ... so I get that something is missing here which would should result in y,shear = 0 but I am at a loss.

would greatly appreciate any insight on this one.

Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[cal91]

I disagree with this

Lets take the debate to an example.

I propose that the deformed case 1,2 & 3 cubes below are identical in every way except for a rigid body rotation. Case 1 doesn't have "horizontal" shear strain, and case 2 doesn't have "vertical" shear strain. All three have the exact same shear strain. In fact strain is a scalar quantity, and has no direction (even though it can be positive or minus).

It is not until the boundary conditions determine the rigid body rotation that we see what the final orientation is.

 

[cal91]

Likewise, (as far as shear strain, stress, deformation goes) case 1 and case 2 are identical except for a rigid body rotation.

 

[rb1957]

I don't see case 2 deflecting that way. Maybe if the load was applied near one support and reacted at the other.

The difference I see between case 1 and case 2 is that for deflections (and I mean the beam bending deflection) to be the same, then case 2 deflection are determined with respect to the tangent at pt1. In case 1 the deflection is monotonic increasing from 0 at pt1. In case 2 I see the beam sagging down at pt1, reaching zero then increasing to pt2.

another day in paradise, or is paradise one day closer ?

 

[IDS]

I'm a bit confused by this. Looking only at the shear component of curvature, do we not normally have discontinuities of curvature? Option one -- the correct option repeated below -- also has two different slopes in each of the beam halves and in the central section. Regardless, to get hung up on continuity is to not entertain the crux of the mental experiment.

OK, my terminology was confusing. I was referring to the "slope" as the angle to the vertical of the plane defining the section. Since under shear deformation these planes remain parallel (by definition of "shear deformation"), then if the plane is fixed against rotation at one end, all the others must remain vertical.

I disagree that "getting hung up on continuity is to not entertain the crux of the mental experiment". It seems to me that considerations of continuity + the definition of shear strain entirely define how the beam will move, without bringing energy into it at all.

I will look at the stability discussion later.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BAretired]

But cal91 is talking about deformation due to shear whereas rb1957 is talking about bending deformation.

BA

 

[KootK]

I disagree that "getting hung up on continuity is to not entertain the crux of the mental experiment".

If you're stuck on the continuity thing in a hard way, then the mental experiment is pointless. Let's not discuss it any further..

 

[IDS]

The difference I see between case 1 and case 2 is that for deflections (and I mean the beam bending deflection) to be the same, then case 2 deflection are determined with respect to the tangent at pt1. In case 1 the deflection is monotonic increasing from 0 at pt1. In case 2 I see the beam sagging down at pt1, reaching zero then increasing to pt2.

The sketches just show shear deflections, with opposite faces of any segment remaining parallel.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

I propose that the deformed case 1,2 & 3 cubes below are identical in every way except for a rigid body rotation.

This is the way in which I believe transverse & longitudinal shear to be different. The key is recognizing the relationship between the strains and the undeformed member shape.

 

[IDS]

KootK - I think the stability question is interesting, but from the graph you posted I'm not really sure what we are looking at.

Would you like to start a new thread?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

I think the stability question is interesting, but from the graph you posted I'm not really sure what we are looking at

We're looking at a generic stability problem characterized by multiple instability peaks and stability troughs. One simple example:

- Two columns side by side, same height and end fixities.

- Right column much stiffer than left.

- Left and right columns joined at mid-height by a non-rigid bar such that right column braces left for a while.

- Load left column until it goes Euler critical and starts to buckle, straining against the bracing bar.

- Keep adding load until eventually both columns "buckle" first mode even though there is no load on the right column.

Would you like to start a new thread?

I'm grateful for the offer but will pass. I've only got enough energy available to maintain presence on one deep dive thread this week and, for now, this is it.

 

[IDS]

If you're stuck on the continuity thing in a hard way, then the mental experiment is pointless. Let's not discuss it any further..

Well yes, other than providing insight into how the non-intuitive behaviour under shear deflections follows directly from the definition of shear strain and strain compatibility, all entirely pointless.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

Well yes, other than providing insight into how the non-intuitive behaviour under shear deflections follows directly from the definition of shear strain and strain compatibility, all entirely pointless.

I didn't say that that continuity was pointless. Not once. What I said was that my mental experiment, which I was hoping you'd join me in discussing, goes nowhere and is pointless if you refuse to consider any scenario that violates continuity. If you refuse the starting premises of the argument flat out, then you've declined to engage and there's simply nothing to discuss.

Yes, I understand that continuity requirements rule out certain deformation modes. The question that I've been trying to drill down to is WHY does nature avoid continuity violations? Obviously, when continuity is violated, most things don't just explode or cease to exist. Rather, damage occurs. Things can in fact be non-continuous. My hypothesis is that, given the choice, materials organize themselves in a way that favors continuity because such damage would produce a condition in which expended energy is not minimized. I'm not suggesting that continuity doesn't matter. Rather, I'm proposing energetic concerns as the reason for why continuity matters.

Check out this shameful, non-continuous beam and all of the energy that it must have took to pulverize that heap of concrete below it.

 

[IDS]

I didn't say that that continuity was pointless. Not once. What I said was that my mental experiment, which I was hoping you'd join me in discussing, goes nowhere and is pointless if you refuse to consider any scenario that violates continuity. If you refuse the starting premises of the argument flat out, then you've declined to engage and there's simply nothing to discuss.

I know you didn't say that continuity was pointless.

I didn't say (or even imply) that you had said that continuity was pointless.

What you did say was that consideration of continuity was pointless, and I pointed out that it was not pointless.

As for your thought experiment, and I'm not sure which particular one of the many you are referring to, where did I say that it goes nowhere and is pointless?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

I didn't say (or even imply) that you had said that continuity was pointless. What you did say was that consideration of continuity was pointless, and I pointed out that it was not pointless.

Ah, I see. Night and day difference. Rather than trying to figure out which of my comments gave you that impression, I'll just clarify how I do feel about considering continuity as straight forwardly as I can:

1) In general continuity is fundamental to our craft and should be considered.

2) In the context of a thought experiment considering what might happen if continuity were not enforced, it is pointless to enforce continuity.

As for your thought experiment, and I'm not sure which particular one of the many you are referring to...

The one shown below. If you've not even been clear on which thought experiment I've been discussing, then I can understand why you've been confused about my comments on continuity. Those comments absolutely needed to be taken in the context of the scenario that they pertained to.

...where did I say that it goes nowhere and is pointless?

You didn't. I did. Then, as now, I'm ready to let this one die off as it's certainly not provoked any of the interesting discussion that I'd hoped it might. And that's okay. Some ideas naturally dead end in any long running technical discussion.

If you're stuck on the continuity thing in a hard way, then the mental experiment is pointless. Let's not discuss it any further..

 

[cal91]

KootK & IDS, just a quick input about why nature tries to minimize energy.

It's really just because that is what we observe from nature. We really only try to align our theoretical, mathematical models with the reality we observe. Answering why nature tries to minimize energy would be similar to asking why gravity exists. You can't really answer why, it is just how our universe works.

I copied responses from "annav" on the physics stack exchange, as she gets across what I'm trying to say.

https://physics.stackexchange.com/q...oes-a-system-try-to-minimize-potential-energy

The only answer to "why" questions about principles in physics is "because the theoretical models dependent on it have been found to describe all known data and can predict new ones".

Why questions in physics when they hit postulates and laws, is like asking why for an axiom in mathematics. Principles are part of the definition of the theory, and physics theories are established when validated continuously by the data.

One cannot explain this principle except by attributing it to observations that forced us to use it axiomatically.

Its only explanation is that it is necessary for the theoretical mathematical model to correctly predict the data.

It is an assumption in the theoretical modelling which is justified by the validation given to the model by the observations.

If you change the axioms of Euclidean geometry, you get, for example, spherical geometry. The axiom is tied up with the geometry intrinsically.

It explains it in the sense that without it one would have a different theory

I know this "just because" answer won't satisfy some people. But that's all I got.

KootK when I have time later today I'll respond to you about vertical vs horizontal shear strain

 

[KootK]

I'm good with that being as far as the drill down can go. After all, I don't really even know what energy is other than it's work based definition.

Part of what I've been trying to get at with the energy/continuity business is that I really just see energy minimization as a further drill down on the continuity concept.

Axioms

1) Nature minimizes energy.

2) Continuity violations cause damage which consumes energy.

Premises

3) 1 + 2 --> Nature will enforce continuity when possible.

Conclusion

4) 3 ---> Loaded materials will organize themselves such that continuity is maintained.

 

[cal91]

I think that makes sense... continuity is only broken when the strength of materials is surpassed.

 

[BAretired]

If two equal and opposite vertical point loads are placed on a simple span, separated by a distance b, the shearing force and shear deflection diagram are shown below. The formula dy/dx = kV/AG is used to determine slope in the three zones 0<x<a, a<x<a+b and a+b<x<L.

dy_a/dx = kPb/LAG; y_a = kPab/LAG

dy_b/dx = kP(b-L)/LAG; y_a+b = kP[ab+(b-L)b]/LAG = kP(ab-(a+c)b)/LAG = -kPbc/LAG

dy_c/dx = kPb/LAG; y_a+b+c = y_L = -kPbc/LAG + kPbc/LAG = 0

The two point loads produce a moment but not a point moment. If dimension b approaches zero, the slope, dy/dx approaches zero in Range a and c and shear deflection approaches zero throughout the length of beam.

Edit: k factor was omitted in figure for deflection at a+b.

BA

 

[KootK]

That's interesting BA. I'll suggest a refinement as shown below whereby the point loads are not kept constant but, rather, increased as they come together to keep the applied moment constant. Some features of this arrangement:

1) The strain pattern would flip from a vertical shear strain to a horizontal shear strain once the loads coalesce. Like a singularity.

2) Given that this implies that the point loads, and the mid region shear strains, would become infinite as "A" shrinks to zero, the approach doesn't represent a real condition that would satisfy other material constraints.

3) There can't actually be such a thing as a truly concentrated moment that is the product of a vertical force couple. Any true concentrated moment requires a horizontal force couple and consideration of member depth.

 

[BAretired]

There can't actually be such a thing as a truly concentrated moment that is the product of a vertical force couple. Any true concentrated moment requires a horizontal force couple and consideration of member depth.

This may be true when the concentrated moment is in the span, but getting back to the OP, suppose you add a cantilever of length 'A' to the end of the beam with gravity load P at the end. The moment is P*A at the support. From the perspective of the unloaded span, how does this moment differ from a concentrated moment applied at the end? The reactions are P(L+A)/L and P*A/L and the shear is P*A/L across the unloaded span and there is no horizontal force couple.

Or are you saying that the bending stresses at the support create a horizontal couple?

I have to admit that I am having trouble seeing why the slope of the unloaded span (according to theory) has a value other than zero when we know it should be zero...very frustrating.

BA