# A beam with many supportpoints

• posted

If I put a beam on four support points that are placed at unequal distances, and try to calculate the forces that keeps it in balance (by looking at vertical forces and moments), then I seem to get equations that results in forces that can't be uniquely determined. It seems to give a solution that is dependant on two parameters.

Is this really the case? How do you do if you really need to calculate the forces?

• posted

Yes it's really the case - a mass can be supported under gravity loads at only three points if the support is to be completely determinate.

Brian Whatcott Altus, OK

• posted

A 3-D solid requires three points not in the same plane for support, and their forces are what is called 'statically determinate", i.e. can be calculated by the theory of statics only.

But Gunnar is talking about a beam, which is considered a 2-D solid where all forces are in the same plane. In this case, only two support points are required.

For beams with more supports, say 3 or 4, the support forces are called statically indeterminate to the first and 2nd degree respectively. They can be still determined by considering the elasticity of the beam.

• posted

How can three points not be in the same plane?

Don Kansas City

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Yes. You have more unknowns than equations just using statics.

Consider stiffnesses, symmetry (if present) or other assumptions.

Beam deflection by integration, energy methods and numerical methods are some ways to find the 4 reactions by considering stiffnesses.

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I need to add an important practical point. The support reactions of statically indeterminate beams with more supports than necessary are sensitive to settlement or inaccurate alignments. The reactions determined by stiffness and statics calculations assume perfect alignments. For stiff beams, the reactions are very sensitive to settlements, so be careful. They are not so sensitive for flexible beams, e.g. long pipelines with many supports.

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I've heard that one way is to "replace" the support points with very stiff dampers and do a numerical calculation. Is that ok to do?

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Lion seems to have serious reservations about the computability of a 3 legged stool. [whose feet are all in the same plane]

Novel!

:-)

Brian Whatcott

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No. Dampers are for dynamics. You could use stiff springs, but what's the point? Get yourself a small finite element program, apply the appropriate constraints, and calculate the reactions. Many people who read this newsgroup could solve your problem in five minutes.

• posted

Hi Gunnar:

As has been mentioned, a two-dimensional beam with more than two supports is "statically indeterminate". That literally means that there is no practical way to calculate the loads in more than two supports using only external measurements, no matter how fancy your computer program might be.

Here's a simple way to think about it. Think of a beam supported at both ends. You can use simple statics to determine the reaction at both supports with just about any loading conditions on top of the beam. OK, now let's add a third support, somewhere between the other two, that just touches the beam, but does not support any weight. Now you have a beam with three supports, but it is impossible to tell which ones support what portion of the load. In this case we happen to know that the center support isn't supporting anything, but there is no way to tell that by exterior measurement.

What if the left support were the one that was just touching the beam and the two right supports were carrying the full load? It would look the same. What if the right two supports shared their end of the load

50/50? 60/40? 90/10? There's no way to tell without actually measuring the load on at least one of the supports.

Don Kansas City

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Ah, now I see the issue.

One is using the term "support" to imply a line or compact area pad support. In kinematics, one starts from point supports - not very practical for beams - still three points of this kind will support a beam, but two of THESE supports will allow the beam to topple.

:-)

Brian W.

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Right. You were right also, but in three dimensions. The same situation arises in your case if you add a fourth leg to a table.

Don Kansas City

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Points?

I looked over your earlier posts again, to see where we diverged. I was using the term "point" to mean a sensibly zero dimensional support. And you began using the term "point" to mean a line or area contact.

Defining terms is always good.....

:-)

Brian Whatcott

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"Don A. Gilmore" wrote in news:Rqrse.15832\$ snipped-for-privacy@tornado.rdc-kc.rr.com:

Tara, Boom!

Actually they must not be colinear.

Cheers

Greg Locock

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Ha. It was wrongly described and thanks for the correction.

• posted

OP-

A beam with more than two simple supports is statically indeterminate.

You have to use other / additional methods, flexibility, stiffnes, energy.......

What you're describing is a "continuous beam" situation.

Look for a shareware continuous beam program.

cheers Bob

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For a planar problem toppling isn't possible. That rotation is removed for it to stay in a plane.

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Take a look at Gunnar's original post, and tell me where "planar problem" is specified. He seems to have been talking about a real beam, originally.

Brian Whatcott Altus OK

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I was speaking in general terms above. What you stated is true for a

3D problem. That's why I worded my comment as I did.
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I used to have a good beam program- it was a demo, but I used it for what I needed and it was totally adequate. 'beam demo' and its bro, column demo. The beam one, you can determine your # of supports, loads, whether point loads or distributed, your E, your I, etc. and get out your shear and moment diagrams (which are nicely done in the program) and numerical answers. IIRC, the beams had to be symmetrical. I haven't had it on my 'puter for a while. I just googled them, and there are versions of both still available. good luck, karinne

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