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:

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[rb1957]

guys, I get that a shear applied mid-span creates shear deflections. But how does shear at a support create deflection due to shear ?

We've agreed that the simple beam has no shear deflection. But the unloaded overhang does ?? I just don't get it.

Say you had a simple beam, and applied point loads above the supports. How does the beam deflect ? (obviously it doesn't)
Could the answer be any different if the beam over hung the supports ? (no)

Say you had a simple beam with a point load applied mid-span. Ok, there are shear deflections.
Now, add an over hang … this changes the shear deflections ??

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

 

[KootK]

I think this lines up with your sketches Koot.

I agree, it's completely analogous to my previous sketches. Which is, frankly, why I don't feel that it pushes the conversation forward beyond mere superposition and physical reasoning. In my opinion, it doesn't move thing any closer to the "why" that I was seeking with my energy theory.

I submit the sketch below for consideration. It's a constant shear beam that isn't straight. I don't, for a second, believe that this would actually happen but, without invoking something beyond the entry level arguments, I don't see how it can be proven that it couldn't happen.

 

[Celt83]

This may not add much to the conversation but its helping me a bit. Sketch below I just looked at free motion of a small section vs restrained motion.

Open Source Structural Applications:

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

 

[BAretired]

The right half in KootK's beam cannot "decide to shear longitudinally" if the left support is fixed. Otherwise it would create a discontinuity where the left and right halves meet.

Shear deformation in a prismatic element is a small change in the right angle at each corner of the element. If one vertical face is held vertical as is the case in a cantilever, the angle change must be measured from a vertical plane, so shear deflection must be vertical. If the horizontal face remains horizontal, as it does in a simple beam with moment at one end, the angle change must be measured from a horizontal plane, so deflection must be horizontal.

In the case of a simply supported beam with end moment and an unloaded cantilever at the opposite end, the slope of the cantilever is determined by the slope of the beam face to which it is attached. For shear deformation, the portion of beam between supports could be regarded as fixed against rotation.

BA

 

[KootK]

Otherwise it would create a discontinuity where the left and right halves meet.

I'm not so sure that is a discontinuity from the perspective of shear alone. Shear strain remains constant across slope change, as it should.

The right half in KootK's beam cannot "decide to shear longitudinally" if the left support is fixed.

That's kind of my point. Nothing in nature just "decides". Hence why I proposed my energy theorem as a way to have the "choice" based on a principle of some sort rather than on randomness, gut feel, or anthropomorphism.

If one vertical face is held vertical as is the case in a cantilever, the angle change must be measured from a vertical plane, so shear deflection must be vertical. If the horizontal face remains horizontal, as it does in a simple beam with moment at one end, the angle change must be measured from a horizontal plane, so deflection must be horizontal.

I agree, and I've been saying/showing as much from the beginning. But I submit that the challenge remaining is to prove it. I've taken my shot at that.

 

[BAretired]

KootK,

I am not sure what exactly you are saying we need to prove. Seems to me it comes from the definition of shear stress and shear strain.

Your last diagram combining left and right half is correct for a beam with a concentrated gravity load at midspan. This requires a reversal of shear. Assuming P is the applied load, each reaction is P/2. The shear is positive to the left and negative to the right by the usual definition. In the case of a moment applied at one end, there is no shear reversal so your diagram has a discontinuity at the midpoint.

BA

 

[KootK]

I am not sure what exactly you are saying we need to prove.

Firstly, nobody need to prove anything unless they're interested in digging a little deeper with me. However, what I'm interested in proving / sorting is how it is that shear manifests itself transversely in some situations and longitudinally in others. And I think that's reasonable given that was the impetus for this thread in the first place.

You and IDS seem to feel that you you've got this sorted out without needing to resort to additional investigation like my energy theorem. I understand and respect that you feel that way but I do not. I don't share your opinions that basic physical reasoning and the definition of shear strain are sufficient.

Your last diagram combining left and right half is correct for a beam with a concentrated gravity load at midspan.

My last diagram is 100% in reference to a situation with a concentrated end moment, constant shear, and no concentrated load at mid span.

In the case of a moment applied at one end, there is no shear reversal so your diagram has a discontinuity at the midpoint.

I get that this looks like a discontinuity if you consider the usual case of flexure dominating the deflected shape. However, I think that one needs to set aside those conventional expectations to really dig into the shear dominant situations. After all, does the case shown below, which I believe we agree upon, not also present as though it has a deflected shape discontinuity?

Seems to me it comes from the definition of shear stress and shear strain.

I disagree. See the second sketch below. I believe that both halves of my fictional beam satisfy the basic definition of shear stress and shear strain. Yet additional information would need to be introduced in order to preclude such a situation, as we know must ultimately be the case. The conventional mathematics only seem to tell us what the angular strain will be in our beam element. The mathematics do not seem to tell us whether those strains will be longitudinal or transverse. And, again, that was pretty much the impetus the the thread to begin with. If the basic definitions of shear stress and strain were getting the job done here, we wouldn't be having this conversation to begin with.

 

[IDS]

Kootk - I had a response in mind, but I would just be repeating what baretired said, so I gave him a little pink star instead.

I would add two things though:

Shear strain remains constant across slope change, as it should.

No, it reverses in sign. If there was no change in shear strain the bottom face of the elements meeting at the section would have the same slope.

Regarding what constitutes a "proof", certainly any hypothesised mechanism must be consistent with energy conservation, but personally I don't see showing a mechanism as being consistent as explaining how that mechanism happens. You have to look at the geometry for that.

De gustibus non est disputandum





Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

No, it reverses in sign.

No it doesn't. And that fact that you think that it does indicates that you're not properly grasping my hypothetical example. And that's probably my fault for not explaining it sufficiently. The sketch below is my attempt to rectify that.

If there was no change in shear strain the bottom face of the elements meeting at the section would have the same slope.

They do have the same slope if one recognizes that the slope of the sheared element is not always the same as the slope of the member itself. Since the shear stress is constant across the beam, the shear strain is constant across the beam. And I submit that my hypothetical beam does show constant strain across the beam. Shear strain and slope are not the same thing and should not be confused. Shear strain is just the angular change in the shape of the element which, as I illustrated above, is the same for both halve of my funky beam.

certainly any hypothesised mechanism must be consistent with energy conservation, but personally I don't see showing a mechanism as being consistent as explaining how that mechanism happens.

I disagree. I'm no professional philosopher but it seems to me that, all over the world of Newtonian physics, we take energy conservation to be inviolate. This is how we prove that a basket ball dropped at the top of a half pipe will eventually come to rest at the bottom. Minimum potential energy. Good enough for Newton and 8th grade physics, good enough for me.

You have to look at the geometry for that.

I disagree again. Your "look at the geometry" has two logical holes in it:

1) It involves a rotation that doesn't actually take place physically and;

2) It rests on your unproven assertion that constant shear strain makes for a straight member from a shear deformation perspective.

As such, I believe that my energy explanation has significantly greater validity than your geometric explanation. The business of imaginary rotations and physical reasoning is, as evidenced here, leading us to have to go back to the drawing board with each new case considered to make our theory match observed behavior. In formulating a more fundamental principle, such as the energy hypothesis, we have a tool that should apply universally. I would think that alone would give it more weight.

De gustibus non est disputandum

I don't consider this a matter taste. Some theories survive scrutiny better than others. It's science / evolution. The winners float to the top.

 

[BAretired]

In the case we are addressing in the OP, the beam is supported at each end, hence fixed against rotation. The angular change of any element of that beam is measured against a horizontal line.

BA

 

[KootK]

In trying to elaborate upon the concept below, I discovered my own mistake in it. As such, withdraw the kinky-beam shown below as a valid counter example to the notion that that a constant shear beam is straight edged top and bottom.

 

[KootK]

In the case we are addressing in the OP, the beam is supported at each end, hence fixed against rotation. The angular change of any element of that beam is measured against a horizontal line.

No argument there. As indicated in my last post, which I must have be prepping while you were posting yours, I see that there was an error in my kinky-beam counter example.

 

[IDS]

In your last example (5May 0:06) you have ended up with an inclined beam, fixed in position at the right hand end, rather than the left. You need to rotate it to horizontal, converting the "vertical" shear deformation into a "horizontal" deformation.

You have said that the statement that a straight beam under uniform shear deformation must remain straight has not been proved, but I think it is self evident:
- Under shear deformation the opposite faces of any segment remain parallel (by definition of "shear deformation").
- An equal shear force will cause the same change in angle of the corners.
- Therefore a straight beam will remain straight.
- If the bottom face of a straight beam is fixed in position at both ends, then it must remain horizontal if it is to remain straight.

Regarding the philosophical question, I am reminded of a question raised by my high school physics teacher (about 53 years ago):
When a beam of light passes an interface between two materials where its speed changes, the path it takes is such as to minimise the travel time between any pair of start and end points. The question is, when it reaches the interface, how does the light beam know where the end point is, and what its speed through that material will be, so that it can change its direction by precisely the right amount to minimise the travel time?

My answer would now be that the light doesn't "know" anything. The mechanics of wave motion happen to have that end result, which is also the only result consistent with conservation of energy.

Likewise showing that a particular mode of deformation is consistent with conservation of energy does not "explain why" that mode of deformation happens, but if a hypothesised mechanism is shown to be consistent with energy conservation, then that does show that that mechanism may b possible (it also has to satisfy strain compatibility and force equilibrium, of course).

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[KootK]

In your last example (5May 0:06) you have ended up with an inclined beam, fixed in position at the right hand end, rather than the left. You need to rotate it to horizontal, converting the "vertical" shear deformation into a "horizontal" deformation.

Agreed. That really goes back to several of my sketches early on in this thread.

You have said that the statement that a straight beam under uniform shear deformation must remain straight has not been proved, but I think it is self evident:
- Under shear deformation the opposite faces of any segment remain parallel (by definition of "shear deformation").
- An equal shear force will cause the same change in angle of the corners.
- Therefore a straight beam will remain straight.
- If the bottom face of a straight beam is fixed in position at both ends, then it must remain horizontal if it is to remain straight.

Having no counter point to offer, sure, I accept this the proof that you intended it to be. That said, for me, different kinds of proofs have different levels of utility. My summary of your proof would go something like this: stuff strains per definitions and this is the only way that it fits the boundary conditions. That's true but, for me, it doesn't aid my physical understanding of the situation. It's a bit like that variational calculus proof that proves that the straightest distance between two points is a straight line. It's true and it's elegant, but it's not making me feel as thought I understand my world any better as I'm walking about.

For me, an energy explanation is much more satisfying because I feel that I have a grasp on a principle that nature adheres to and that I understand how the molecules of a sheared element will tend to organize themselves to be in compliance with that principle. While I consider this to be an improvement over bare geometrical reasoning, it's still not wholly satisfying for me as since I don't fully understand why nature prefers to minimize potential energy. I believe that it's linked to entropy etc but entropy itself is not a concept that I wholly "feel". As such, am forced to accept it as an axiom.

My ideal form of proof would be to be able to draw differential sheared elements, similar to what celt has done, and to be able to load and constrain them such that I could point to one or the other and be able to say "this one will shear transversely" or "this one will shear longitudinally". I feel that should be possible but don't yet know how to do it myself.

but if a hypothesised mechanism is shown to be consistent with energy conservation, then that does show that that mechanism may b possible...

I think that statement needs to be expressed more strongly. In many applications, the mechanism that minimizes energy isn't just a possible mechanism, it's the correct mechanism. In fact, as far as I know (I'm no general physicist), there are many applications in which energy minimization is our only tool for discerning how the world organizes itself.

 

[KootK]

Consider this case:

1) I think we all agree that transverse and longitudinal shear strains are both possible under the right circumstances.

2) So here, how will the sheared elements "decide" to organize themselves?

3) Yes, I understand that option two has a discontinuity. But "discontinuity" is just a word. What is it physically about a discontinuity that makes a discontinuity indicative of an inadmissible solution?

4) I argue that, in the physical world, the discontinuity would represent a bunch of localized, energy wasting damage. And since nature prefers to to minimize unnecessary energy expenditures, the sheared molecules of the beam organize themselves in a way that is consistent with that preference (option one).

Personally, I don't know how to justify the choice of option one over option two without invoking energy arguments. If other methods exist, I'd love to hear about them.

 

[rb1957]

I thought shear deflection was due to the shear in the web, something not account for in Euler beam deflection.

But then you realise that generally the beam web has different shear stress on both sides of the shear load.

You guys seem to be saying that it's due to the curvature of the beam … that sections are being rotated ? But why isn't this captured in Euler beam deflection ?

Maybe because the deflected beam sections are no longer normal to the beam longitudinal axis ?

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

 

[IDS]

rb1957 - The Euler curvature is 0, since plane sections remain both plane and parallel. The rotation is due to the change in the corner angles, which changes at any point load.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IDS]

For me, an energy explanation is much more satisfying because I feel that I have a grasp on a principle that nature adheres to and that I understand how the molecules of a sheared element will tend to organize themselves to be in compliance with that principle.

I was going to say that my quote on not debating matters of taste was stretching things a bit, but maybe not. I really think that our preferences really do amount to a matter of taste. To me "energy" is a rather abstract concept, whereas considering geometry, strain and stress really does give "a grasp on a principle that nature adheres to and that I understand how the molecules of a sheared element will tend to organize themselves."

I think that statement needs to be expressed more strongly. In many applications, the mechanism that minimizes energy isn't just a possible mechanism, it's the correct mechanism. In fact, as far as I know (I'm no general physicist), there are many applications in which energy minimization is our only tool for discerning how the world organizes itself.

But energy by itself is not enough. We also need to look at strain compatibility. Can you give some examples of applications where energy minimisation is the only tool for discerning how the world organizes itself?

For your example of a point load at mid-span, the model with two different slopes in each half can't work because it results in the central section having two different slopes, but each section can only have one slope, since the beam is continuous.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

"The Euler curvature is 0" … no, the Euler derivation understand the beam deflects in some "curved" manner".

but I get the distinction between a section normal to the undeflected beam axis and one normal to the deflected axis.

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

 

[IDS]

rb1957 - If we are looking just at shear deflection, which we are, then the Euler curvature is zero, because Euler bending theory does not include shear deflections.

In particular, if one end of beam is fixed against rotation of the vertical axis, then all vertical sections remain vertical, because under shear deflections parallel planes remain parallel.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/