Double point load beam deflection 3

[MrMaturity]

Hi everyone,

Sorry if this has been answered before, search didn't turn up any results.

Anyway, taken some time during this Christmas break to study up on some structural mechanics and make sure I haven't forgotten everything. I've been having a look at the deflection of a simply supported beam with two equal, unevenly spaced point loads.

I can't seem to find a formula for that exact condition so I have used the formula for a single point load and then applied superposition to find the resultant deflection.

The formula I am using is P*L^3*Alpha/48*E*I where Alpha= 3*b/L - 4b^3/L^3 when a>b

The problem that I am facing is that I've had a look at a colleagues work as an example and they have not used superposition, they have simply found the Alpha for each of the point loads then applied it with in the one formula, as below.

(Alpha1*P1 + Alpha2*P2)*L^3/48*E*I

I am trying to work out if I've gone insane and am forgetting something basic or if my colleague has made a mistake.


Thanks in advance

 

[KootK]

I'm not familiar with the formula but algebraicallt, the two approaches appear equivalent.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[SlideRuleEra]

Should have read your question more carefully. Now I see you are looking for deflection. Can't help, post deleted. Sorry.

 

[rb1957]

I think adding the alphas is conservative, since max deflection occurrs at different places for the different loads.

If being conservative is ok, then it's ok.

If being "right" is needed (or to appreciate the degree of conservatism), then plot out the displaced shape for both loads. Your equation is for max deflection, calculating the expression for deflection as a function of x isn't hard (for a SS beam with a single point load, double differentiation of the momnet, and once you've done it you have the expression for future work as well).

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

 

[rb1957]

another short-cut I've seen is to replace the two (or more) point loads as a UDL ... maybe investigate to see how conservative (or not) this assumption is.

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

 

[BAretired]

A fairly simple method of finding deflection at any point is to draw the bending moment diagram for the actual beam, then use the Conjugate Beam Method to find rotation and deflection at any point on the span. The Conjugate Beam method for a simple beam is a useful way of applying area-moment principles.

The conjugate beam has the same span as the actual beam and is loaded with the M/EI diagram of the actual beam. The conjugate beam shear at any point is the actual beam slope at that point. The conjugate beam moment at any point is the actual beam deflection at that point.

BA

 

[Archie264]

Do you have the AISC Steel Book? If so it's shown as condition 9 in Table 3-23.

 

[MrMaturity]

Thanks for the replies everyone. It seems that I had had a major brain malfunction and was forgetting my high school maths, apologies for the silly question.

@KootK - You are absolutely correct, both options are equivalent.

@BAretired - Thanks for the hint about the Conjugate Beam Method. Works a treat.

 

[KootK]

No sweat Mr.M. We've all been there.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[rb1957]

"both options are equivalent."

well yes, the two approaches in the OP are equivalent, but also incorrect. It seems to me that you're superimposing maximum deflection from the two loads which is IMHO incorrect (as max deflection occurs at different places on the beam for the two loads). I suspect that the conjugate beam analysis solves this (and should give a different answer to the original approach), as would calculating the deflected shape for the two loads and superimposing these.

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

 

[BAretired]

I agree with rb1957; superposition cannot be used in this case.

Archie264 referred to the AISC Steel Book condition 9, Table 3-23 which I do not have; however I do have the CISC Handbook of Steel Construction and note that Example 11 of "Beam Diagrams and Formulas" is for 2 unequal concentrated loads unsymmetrically placed. The CISC reference provides shears and moments but not deflections for this case.

What formula does the AISC Book provide for deflection?

BA

 

[SlideRuleEra]

BA - Like Archie264, the diagrams in the AISC manual are what I referred to in my edited post. AISC does not give a deflection formula for two loads (either equal- Diagram 10, or unequal- Diagram 11) that are unsymmetrical placed.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[Archie264]

Whoops, my apologies; AISC Steel Book, Table 3-23, condition 9 refers to two equal concentrated loads symmetrically placed. I had thought that was what was under discussion. By rereading the original question more carefully this time I see that what is under discussion is two equal concentrated loads unevenly placed. Table 3-23 has a deflection formula for the former but not the latter.

 

[SlideRuleEra]

It's not an exact solution, but this problem can be bounded pretty closely by considering two cases where deflection can be calculated. This won't give the location of maximum deflection but does give an upper an lower bound on its magnitude.

Case 1: Two equal loads, symmetrically placed at the distance closest to beam ends. Actual deflection is greater than the calculated value.

Case 2: Two equal loads, symmetrically placed at the distance closest to the beam center. Actual deflection is less than the calculated value.

For extreme cases, say distance "x" is near the beam center and distance "y" is near the beam end, engineering judgement can help give a tighter band than the calcs alone.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[rb1957]

geez,

1) won't conjugate beam do this properly ?

2) you can superimpose the deflected shapes of the beam under the different loads, and it ain't that hard to calc.

3) if you can live with the conservatism of superimposing the maximum deflection from each load then ok.

4) if the loads are reasonably uniform, how inaccurate is a UDL ?

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

 

[jayrod12]

1) it will but when running numbers by hand, depending on the level of precision required, the above noted approximations make more sense.
2) Again, you can but not easily by hand
3) Most of the time, this is acceptable, I rarely try to push the limits on deflection anyway.
4) We haven't been provided with enough info to know, that would be drastically dependent on span and load locations.

 

[BAretired]

If the magnitude and location of loads P1 and P2 are known, deflection can be easily calculated at any point in the span using Conjugate Beam Theory. The location of maximum deflection can be found by differentiation or by trial and error.

If the magnitude and location of loads P1 and P2 are not known, the location and magnitude of maximum deflection cannot be simply expressed. This is probably why the steel handbooks do not include an expression for deflection for this case.

BA

 

[IDS]

I have set up a spreadsheet to do the "conjugate beam" analysis for any two loads anywhere on a simply supported beam. It is also set up to use the Excel solver to adjust the location of the central point to find the point of zero slope, and hence maximum deflection. You can download from:

Conjugatebeam.xlsb

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[Archie264]

Nice work.

 

[BAretired]

Indeed!

BA