Beam bending refresher 2

[rcfraz37]

So its been a while since I did beam bending back in school and now I'm having some trouble.

The regular equations, I believe, assume fixed ends, at least the ones I can find. Trouble is I am looking for the length by which a beam gets shorter as it bends, and my end points rotate.

See pic
A is rigidly affixed translationally but allows the beam to rotate freely. B prevents Translation in the Y direction but no other Direction.

I'm trying to make my design driven because we keep changing materials so I'm going with F= force (assuming a uniform load across the length) L= length I=moment of inertia and E= modulus of elasticity.

I tried finding the deformation with (5*F*L^4)/(384*E*I)=d. and then the slope with w*L^3/(24*E*I). and using those to back out the geometry of the arc but the math isn't right.

Am I heading in the right direction?

Any help would be a life saver

 

[rb1957]

I just noticed your expressions are wrong ...
slope looks right for the end of a simply supported beam under a UDL, w ... = w*L^3/(24*E*I)
but deflection uses "F" = "wL". deflection should be (5*w*L^4)/(384*E*I) or (5*F*L^3)/(384*E*I)

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[GregLocock]

SAE have a leaf spring design manual that includes nomograms for shackle angle, ie exial displacement of the end of the beam. However that seems an expensive solution to an obvious problem - the first assumption I'd make under SS at one end, SS+roller at the other, is that the neutral axis doesn't change in length, so if you can approximate the deflected shape to a known curve than Robert's yer aunty.

In fact it will stretch slightly as axial loads become part of the solution, but I think you can check whether axial strain is likely to be significant by considering it as a taut cable, once you know the deflection at the middle.

Cheers

Greg Locock


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[Denial]

There are at least two possibly relevant second-order effects.

The first, the one I think the OP is chasing, is how much the distance between the beam's end points reduces because the loaded beam "bridges" between its end points by a curved path rather than a straight-line path.[ ] Ways to calculate this effect have been described by rb1957 in the 09Oct24@20:37 and 10Oct24@05:02 posts.

The second, raised by GregLocock in his 11Oct24@02:08 post, is that once the beam's lateral displacement becomes large enough the beam experiences axial tensile forces.[ ] These forces will cause the beam to elongate.[ ] Greg suggests getting a "feel" for this effect by considering the beam as a catenary cable, but I don't think this is correct.[ ] For a simply supported beam under the action of a UDL, simple static equilibrium tells us that the magnitude of this secondary (¿tertiary?) tension at any point is given by the product of the vertical end reaction and the sine of the beam's slope at that point.[ ] The elongation follows directly.

We could go on, Alice like, chasing our rabbit down the hole of third and fourth order effects.[ ] Poisson effects changing the shape of the cross-section, anyone?


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[sup]Engineering mathematician / analyst.[ ] See my profile for more details.[/sup]

 

[rb1957]

how will axial loads develop in the beam, if one end is a roller. It is not like a taut cable, with two external axial reactions.

Yes, the assumption is that the length of the beam doesn't change, so the bow deflection reduces the horizontal distance between the ends. Even with large deflections, how does the beam generate external axial reactions.

It could be I've misunderstood this "catch" ... so that maybe end B is initially a roller, but then becomes axially constrained.

I'd love to know what the OP is trying to accomplish with this design.

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Denial]

How will axial loads develop in the beam, if one end is a roller?

Equilibrium![ ] Consider that roller end.[ ] It has just one external force acting on it:[ ] the reaction, which is vertical.[ ] It has two internal forces acting on it:[ ] the beam's end axial force (parallel to the beam's INCLINED centreline), and the beam's end shear force (normal to the beam's inclined centreline).[ ] These three forces are in equilibrium.[ ] Consider equilibrium in the direction of the beam's inclined centreline.[ ] The reaction force has a component in that direction, and this component can can only be balanced by an axial force in the beam.

[sub][ ]—————————————————————————————————[/sub]
[sup]Engineering mathematician / analyst.[ ] See my profile for more details.[/sup]

 

[rb1957]

oh, so not a roller joint ? the whole point of the roller is to release the axial freedom. The reaction at B is vertical.

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[GregLocock]

There's no axial force at the ends, there's some sort of maximum near the 25% and 75% points. Think of membrane type structures. Sort of

Cheers

Greg Locock


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[rb1957]

ok, make a FBD of the beam, including end B. how can there be an axial reaction at B ? what is balancing it ? there is no shear force into the beam (from some web or other). if bending creates a net axial force on the section, IDK how, maybe plastic bending ??, this would be reacted at end A.

Unless of course end B changes from a roller to a properly pinned reaction, maybe as some "catch" engages.

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[GregLocock]

The beam is curved when under load. At the end the vertical force at the support is resisted by an axial force (beam coordinates) and the shear force. These resolve to give a pure vertical force at the support, usual triangle of forces stuff. Ignore my 25% 75% guess it was wrong. My guess is the effect of this on the deflected distance between the supports is only affected by this to a tiny extent but without dimensions it is impossible to say.

The bigger effect I am sure is the tendency for the neutral axis to remain at constant length, since this is clearly visible on leaf springs.

Cheers

Greg Locock


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[mfgenggear]

This way out of my league. That said why not calculated as rb said.
Calculate the beam configuration.
Do a manual test under load, measure the actual deflection. Use sensors
To measure the movement or moment.
Is that practical.

 

[mfgenggear]

Measure the elasticity, prior to plastic deformation. And failure.
Do a 3D model first.

 

[rb1957]

oh ok ... you're saying that resolving the vertical load into deflected beam axes will give an axial component. ok, and change all the moment calculation too ...

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Stress_Eng]

It looks like there is concern that the change in length will be small and insufficient. The beam could be modified to magnify the change in length. First thing that comes to mind is to utilise not just the change in length but also the rotation at the end by introducing an arm at 90 degrees to the beam axis. Utilise the deflection at the end of the arm.

 

[rb1957]

if the OP wants a large translation, I'd create a mechanism (4 bar linkage ?) that would do it.

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[mfgenggear]

Yes it appears what the OP objective may not be practical. The answer may well be a mechanism or machine.

 

[IFRs]

Great discussion!
My 2 cents:
[ol 1]
[li]the most productive path forward of all the above is to test it with actual loading[/li]
[li]the best first order approximation is the difference between a straight line and curved length using uniform load defletion to etermine a circular or elliptical shape.[/li]
[li]when it snaps through (which it will) that beam end probably needs a taper cut so when the beam goes back to straight it does not jam up. There are many real life situations where this effect happens - when you push something until it snaps through, some designed that way, others foruitous or unfortunate.[/li]
[li]second and higher order effects may complicate reality, particulrly temperature, the stiffness of the hinge and the other side of the snap, friction and wear on the snapping parts.[/li]
[li]I worry about wear on both sides of the snap, I guess it matters how many times this is supposed to happen. I have seen spring loaded rollers or detents used to accomplish this task, perhaps that is possible here?[/li]
[/ol]

 

[mfgenggear]

Yes measure the actual results from the load

 

[Compositepro]

No one seems to be considering the neutral axis of said beam. As the beam flexes the tension face lengthens and the compression face contracts. The final straight cord length of the tension face must become less than the distance between supports at the trigger point. A thin beam will trigger sooner than a thick one.

 

[Denial]

To sort this out in my own mind I followed SWComposite's 09Oct24@21:03 suggestion and resorted to the Strand7™ software.[ ] I set up a simple horizontal beam as follows.
» No units (all data to use internally consistent units).
» Overall length 20, comprising 20 end-to-end beamlets each of length 1.[ ] (Strand7's demo version has 20 beam elements as an upper limit).
» Overall beam runs from (0,[ ]0,[ ]0) to (20,[ ]0,[ ]0)
» Beam cross section is square, with side length of 1 giving A=1.
» Beam material has E=25000, Poisson's ration of zero, density of zero.
» Beam has pin support at LH end.
» Beam has roller support at RH end, acting in the Y (up/down) direction.
» Loading is a single point load of (0,[ ]-1,[ ]0) on each of the 19 internal nodes.[ ] This gives a substantial "linear elastic" deflection, thereby making any second order effects even more obvious.

Under a conventional linear static analysis the results were as follows.
» Midpoint deflection (X,[ ]Y,[ ]Rot°) of (0,[ ]-0.998,[ ]0°).
» LH end deflection (0,[ ]0,[ ]-9.14°), RH end (0,[ ]0,[ ]+9.14°).
» LH end reactions (0,[ ]9.5,[ ]–), RH end reactions (–,[ ]9.5,[ ]–).
» Beam internal actions (axial, shear, bending) at LH end (0,[ ]9.5,[ ]0), mirrored at RH end.
» Beam internal actions at beam midpoint (0,[ ]0,[ ]95.0).

Under a non-linear large-displacement analysis the results were:
» Midpoint deflection of (-1.05,[ ]-3.99,[ ]0°).
» LH end deflection (0,[ ]0,[ ]-38°), RH end (-2.09,[ ]0,[ ]38°).
» LH end reactions (0,[ ]9.5,[ ]–), RH end reactions (–,[ ]9.5,[ ]–).
» Beam internal actions (axial, shear, bending) at LH end (5.8,[ ]7.5,[ ]0), mirrored at RH end.
» Beam internal actions at beam midpoint (0,[ ]0,[ ]85.0).
The beam's axial tension is greatest at the ends, declining to zero at the midpoint.

These results are consistent with the behaviour I hypothesised in my early post.


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[GregLocock]

Good stuff - now increase A but leave I the same. This will give you the sensitivity to the axial force induced elongation.

Cheers

Greg Locock


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