Non linear analysis of a simply supported beam under UDL 3

[Samuee]

If we consider a simply supported beam under uniform distributed load,

when it deform, the vertical load will transform to vertical and horizontal support reaction (As the member is already tilted at the support side).

As such, support should experience axial pulling force by the beam.

I wanna know the above illustration is correct or not.

 

[BAretired]

If the beam is supported by a pin support at one end and a horizontal roller support at the other, there are no horizontal reactions under gravity loading.

If the beam is supported by a pin support at one end and a sloping roller support at the other, there will be a horizontal reaction at the pin under gravity loading.

BA

 

[dik]

and if pinned-pinned, there will be a slight horizontal component.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[JStephen]

Roark's Formulas For Stress and Strain has a section, "Beams Restrained Against Horizontal Displacement at the Ends". In the 5th Edition, this is Table 12, Page. 171, specifically Case 3.
In a lot of cases, the ends may be relatively restrained, but still not rigid enough to develop the forces calculated on this assumption.
In some of the plate-bending problems, considering this effect will actually give much higher load capacity for a given stress, so it can conservatively be neglected, especially if the fixity of the ends is questionable.

 

[goutam_freelance]

It appears the pin jointed support will have some horizontal load. W is equivalent of uniformly distributed load. Here theta is not small.

Engineers, think what we have done to the environment !

https://www.linkedin.com/in/goutam-das-59743b30/

 

[JoshPlumSE]

If we consider a simply supported beam under uniform distributed load, when it deform, the vertical load will transform to vertical and horizontal support reaction (As the member is already tilted at the support side).

As such, support should experience axial pulling force by the beam.

Yes and no.

Situation 1 = small deflection theory
When the deflection in the beam is small compared to the cross section of the beam, then any horizontal load that develops is inconsequential. So low that it cannot really be measured. Theoretically, it might exist, but you won't be able to measure it.

Situation 2 = large deflection theory
Let's make the same setup, but instead of using a real beam (like a wide flange or such), we're going to use a rod or cable. Something that has very little bending resistance, but has a pretty large tensile strength. The cable's moment of inertia is very, very low. Therefore, any transverse loading applied to it must be resisted by the axial stiffness of the cable in its deflected shape. This behavior is a "geometrically" non-linear since the transverse stiffness changes greatly as deflection increases.

Now, I present these as two very separate theories. Because that's how I think of them in terms of what kind of an analysis I'm going to perform. The vast majority of the time, the structures I deal with are okay with situation #1.... Maybe with some P-Delta corrections as the only form of non-linearity. However, there are times when a non-linear analysis is necessary.

Keep in mind that there are different types of non-linear analysis too. So, you have to really know your program and your structure to be sure the type of non-linear analysis the program is performing is appropriate for your structure.

 

[BAretired]

It appears the pin jointed support will have some horizontal load. W is equivalent of uniformly distributed load. Here theta is not small.

It may appear that way, but it is not true. The support is a pin, so the reaction is vertical. If the the roller rests on a surface with slope other than zero, there will be an equal and opposite horizontal reaction at each end. Equilibrium requires that the sum of horizontal forces is zero.



BA

 

[BAretired]

and if pinned-pinned, there will be a slight horizontal component.

If the supports are pin/pin, it is not a simple span beam. The horizontal component may be large, depending on the stiffness of the member. If it is a cable, load will be carried as a catenary, not as a beam.

If it starts with zero deflection and the load is substantial, the cable must stretch in order to gain enough deflection to sustain the load.

BA

 

[BAretired]

@JoshPlumSE,

Can you elaborate on Situation 2? I am not aware of that theory.

BA

 

[IDS]

Can you elaborate on Situation 2? I am not aware of that theory.

I suspect you are

If you have a "beam" with zero bending resistance, and high tensile strength, that's normally known as a cable, and catenary theory applies.

This question is a good example of some of the things neglected by simple beam theory.

For the system to work as shown the beam must have a bending resistance, so must have a finite depth.

If it is supported by a frictionless roller at one end, the reaction at both ends must be vertical, so there will be no nett axial force on the section.

If it is supported at the base by pins at both ends:
- For small deflections there will be additional compression on the base, so a nett compression force on the beam.
- For large deflections the reduction in the horizontal length of the beam will be greater than the increase in length at the base, so there will be a nett tensile force on the beam.
- At some stage there will be an unstable transition between the two.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BAretired]

Okay Doug, but if it is supported by pins at each end, it is not really a simply supported beam as suggested by the thread title.

BA

 

[goutam_freelance]

On further thoughts it appears that the configuration requires pin jointed supports at both ends, which is departure from simply supported configuration. With only roller support an one end, the roller may move away due to horizontal load along supporting surface.

However practically speaking there will always be some horizontal restraints at both the supports, frictional or otherwise.
So under large deflections with pin jointed supports there will be axial tension (or compression) along the beam with UDL. This is especially clear with the following:

 

[rb1957]

"(As the member is already tilted at the support side)" ... what do you mean by this ? The reaction is not normal to the beam.

If the beam is pinned-pinned (as opposed to pinned-roller) than there is "some" lateral reactions, due to the strain in the length of the beam (the deflected beam is longer than the original beam, therefore lateral loads are applied to the supports. You could analyze this beam by taking advantage of the axis of symmetry ... a cantilever with a pinned support.

A classical simply supported beam has a pinned-roller supports so that under vertical loads (distributed or point) the reactions are vertical, as required by equilibrium (as shown by a free body diagram.

If you have a pinned-pinned support then you have a redundancy (the axial load in the beam due to the strain along the beam).

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

 

[JoshPlumSE]

BAretired -

When I say large deflection theory, what I mean is that the geometry of the structure changes significantly enough to affect the stiffness of the structure. I presented the two theories as completely separate, but there are gradations between the two. I'll list the way I think of them:

1) Pure small deflection theory: The load vector caused deflections based on the initial stiffness of the structure only.
2) Small Deflection theory w/ a P-Delta correction: You solve the structure based on pure small deflection theory. Then, based on the member forces and displacement of the structure, you add a simple correction (either to the load vector or to the stiffness matrix) to account for the P-Delta effect.
3) An iterative geometrically non-linear analysis: This would be similar to doing a stiffness matrix correction per P-Delta, but is done iteratively. You apply a certain percentage of the load calculate the deflection, see how it changes the stiffness matrix, apply the next bit of load to the new stiffness matrix and see how much more deflection occurs.... continue until the entire load has been applied or there is divergence. Usually used for structured underdoing very large deformations.... Think a cable stayed tower that is very slender. Fabric structures, cable structures.

 

[goutam_freelance]

If you have a pinned-pinned support then you have a redundancy

This indeterminacy can be taken care of by displacement compatibility condition, like slope=0 at mid point etc.

Engineers, think what we have done to the environment !

https://www.linkedin.com/in/goutam-das-59743b30/

 

[BAretired]

Thanks JoshPlumSE. At the moment, it is a bit beyond my grasp, but something to keep in mind for unusual cases in the future.

BA

 

[wcfrobert]

Very detailed answer from JoshPlumSE. Here is my take:

[ul]
[li]The horizontal reaction exists, but is negligible for small deflections which is about 99% of the structures we deal with. Furthermore, simply-supported beam usually refer to pin-roller supports. It's hard to actually fully "clamp" a beam such that the ends do not translate at all[/li]
[li]For large-deflection problems like tensile structures, these horizontal reactions are what keeps the structure up. A tiny wire has negligible moment of inertia, yet it can resist a lot of load (e.g. horizontal lifelines) because it can deflect and form an axial load path. [/li]
[li]Most software do not pick up the horizontal component. Large-deflection analyses are difficult to do because you have to adjust the load increment dynamically to "form-find" the shape. Stiffness only starts to increase when the axial load path can start form (think about a sagging cable being deflected, the stiffness is almost zero initially which is tough for most programs to handle)[/li]
[/ul]

[ul]
[li]Here is a screenshot of a test model in RISA. The top left beam is pin-pin, the beam below that is pin-roller. Both does not have a horizontal reaction because of small deflection assumption[/li]
[li]On the right, you can see a 2-D plate model with the same load and cross-section and pin-roller support. No horizontal reaction[/li]
[li]On the bottom is a 2-D plate model with pin-pin support. Look at the stress contour, notice how some arching action formed because of pin-pin rather than the conventional flexure theory. But realize that a pin-support assumes infinite rigidity to prevent horizontal translation, this is often not the case[/li]
[li]I didn't model it here, but you can also move the pin support to the neutral axis of the plates, in which case no horizontal reactions will form either[/li]

[/ul]

 

[BAretired]

wcfrobert,

Your diagram on the bottom left brings to mind a recent discussion, which was not resolved at the time. You show a horizontal force of 33k acting at the bottom of the plate. In other words, the plate has an axial compression of 33k and a moment of 33*t/2 at each end where t is the thickness of the plate (or depth of beam).

I agree that the horizontal reactions are zero when moved to align with the neutral axis. If they were located at the top of the plate, the axial load would be tensile. End moments would exist as before.

These results were obtained using RISA, so an iterative, non-linear analysis does not seem to be involved unless the initial geometry is modified to include deflections of previous runs, then terminated when the results are deemed sufficiently close.



BA

 

[rb1957]

a pinned-pinned beam will have an axial reaction due to the axial strain in the beam, whether the constraint point is on the NA or not.

This requires a large deflection FE solution (not the typical small deflections) and could be (probably is) non-linear.

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

 

[BAretired]

a pinned-pinned beam will have an axial reaction due to the axial strain in the beam, whether the constraint point is on the NA or not.

Correct, but at the NA, the net strain would be very small. There has to be a point slightly above the NA where the net strain is zero.

BA