Simply Supported Beam Deflection

[333OnlyHalfEvil]

Hello everyone:

Simple question on beam deflections for simply supported beams only. Is the max deflection located at the point of max bending moment? If so, is there an easy to use formula for it? Or do I have to do the integral of the bending moment every time? Trying to figure out if I can skip the integral part for simply supported brand with somewhat complicated shear and moment diagrams.

Thanks in advance!

 

[jayrod12]

You've pretty much got it, would need to do the integral.

Potentially superposition could work for you to simplify the calculations, however that's generally only useful if you've still got fairly regular loading scenarios.

How mission critical are the results of your analysis? If you don't need extreme precision, I often will make some simplifying modifications to the loading on the conservative side so I can just use general formulae for deflection.

 

[Celt83]

Maximum deflections occur at the roots of the slope/rotation curve.

for deflection you need to integrate twice starting from the curvature diagram.

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[BAretired]

If you don't like to perform integration, the Conjugate Beam method is easiest for a simple beam.
To put it simply, the Conjugate Beam is loaded with the M/EI diagram.

Slope and deflection of the real beam is equal to shear and moment respectively of the Conjugate Beam.

BA

 

[rb1957]

a short cut is to analyze the beam as though it is loaded mid-span, then you can use the canned formulas for dmax.

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

 

[BAretired]

How does that work, rb1957? You put all the load at midspan? That would be very conservative.

BA

 

[JStephen]

To expand on rb1957's comment-
Do you actually need to know the deflection? Or just show that it doesn't exceed a maximum limit?
In general, if you have several different loadings, each of those loadings will have a maximum deflection that it alone would cause. You can't just add up those individual maximum deflectons as the maximums occur at different locations. But, the maximum deflection should be less than the sum of the individual maximums, if that's enough information. And I say this, assuming all loads are in the same direction, if you have some up and some down, that might take some additional thought.

 

[jayrod12]

what rb is suggesting is why I asked how critical the results were. I would agree with BA that putting all the load at midspan would be overly conservative though. I tend to try to smooth out the loading to a uniform load that results in similar max moments and shears as the actual loading. i.e. take your actual moments and shears and back-calculate for what UDL would give each of those, and then check deflection using that information. In that scenario, the calculated uniform load from the actual moment holds more weight to me than the shear.

 

[dik]

For symmetric loading, I would suggest the two were concurrent.

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

-Dik

 

[BAretired]

One approximation would be to consider all loads to be carried on two carts, each of length L/2. Calculate the load P at the middle support, then calculate deflection as PL[sup]3[/sup]/48EI.

Edit: I haven't tried it, but it should be reasonably close.
Edit #2: Tried it. It's not close enough and is not always conservative.

Maximum deflection does not necessarily occur at the point of maximum moment. See handbook for unsymmetrical point load on a beam.

BA

 

[rb1957]

yes, conservative. clearly there are other options but any symmetric loading would have max deflections on the CL.

Mind you, how hard is it to do the integration ? do it once for a single point load at x = a, then you can superimpose loads. "Easy".

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

 

[phamENG]

I usually do what jayrod is suggesting. Find the maximum moment, if it's in the middle third of the span I find an equivalent uniform load that gives me the same moment and calculate that deflection to make sure I'm not close to the limit. If it's sensitive, I do the 'hard' work.

This only works for bending dominated deflection - steel beams, shallow/long wood and concrete, etc. If shear deflection is dominant or even more than negligible, you may need to be more careful.

 

[rb1957]

man! that sounds like a lot more "work", but I guess once you have s/sheets set up it's "easy". But hard to review.

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

 

[IDS]

I'm puzzled why people want to find some approximation.

Why not just feed the actual loads into a spreadsheet or your favoured beam analysis program, with a reasonably large number of subdivisions, and read off the maximum deflection?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[phamENG]

IDS - I find it helpful during initial design to get in the ball park. I can rough out loading on a beam in a few lines of hand writing and arrive at an approximate beam size faster than I can open a beam analysis calculator. Sometimes that's enough (deflections comes out to L/900 or something). Or I run a full analysis after I've worked everything else out and get a better defined number.

 

[BAretired]

Simple question on beam deflections for simply supported beams only. Is the max deflection located at the point of max bending moment? If so, is there an easy to use formula for it? Or do I have to do the integral of the bending moment every time? Trying to figure out if I can skip the integral part for simply supported brand with somewhat complicated shear and moment diagrams.

Maximum deflection is not necessarily at the point of maximum moment, but it is probably close. If the maximum moment is known, it is possible to set an upper boundary on deflection by assuming the maximum moment occurs throughout the span, which means that deflection follows a circular arc. So the easy formula would be:

Δ[sub]max[/sub] = M[sub]max[/sub]L[sup]2[/sup]/8EI.

It's a little conservative for uniform load, more conservative for a point load at midspan, but if an upper limit is wanted, it's conservative for any load distribution. If more accuracy is required, the Moment Area or integration method must be used.

BA

 

[JStephen]

Consider a pinned-pinned beam with applied moment at one end. Deflection at the point of maximum moment is then zero.

 

[rb1957]

or consider a beam with a load at "a" (not on the mid-span) ... max deflection is not at max moment.

And Roark already has this case.

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

 

[rb1957]

are you there "333" ?

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

 

[BAretired]

Consider a pinned-pinned beam with applied moment at one end. Deflection at the point of maximum moment is then zero.

Good point!

BA