Beam Formulas for Multiple Point Loads.
Beam Formulas for Multiple Point Loads.
(OP)
Can anyone out there please help me?
For ages now I've been looking for simple formulas for calculating Mmax for simple beams with more than 2 point loads, for example multiple beams on girders. Text books etc only ever seem to list formulas for 2 loads.
What if I have 5, 6, 7 etc?
I did find one set of notes on-line that gave me the formula for 4 equally spaced point loads (equal loads): 3*Pl*L/5, but would like to know for more than 4 loads.
Appreciate any assistance.
Paul.
For ages now I've been looking for simple formulas for calculating Mmax for simple beams with more than 2 point loads, for example multiple beams on girders. Text books etc only ever seem to list formulas for 2 loads.
What if I have 5, 6, 7 etc?
I did find one set of notes on-line that gave me the formula for 4 equally spaced point loads (equal loads): 3*Pl*L/5, but would like to know for more than 4 loads.
Appreciate any assistance.
Paul.






RE: Beam Formulas for Multiple Point Loads.
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RE: Beam Formulas for Multiple Point Loads.
But to me - that ain't 5 equal point loads equally spaced on a simply supported beam..?
RE: Beam Formulas for Multiple Point Loads.
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RE: Beam Formulas for Multiple Point Loads.
I take it there are no "standard" formulas then...
RE: Beam Formulas for Multiple Point Loads.
For n equal concentrated loads of P spaced at s = L/(n+1),
w = P/s
Mmax = wL2/8 when n is odd
or wL2/8 - w.s2/8 when n is even.
If the loads are not equal, draw the shear force and bending moment diagram.
BA
RE: Beam Formulas for Multiple Point Loads.
The problem with standard formulae, is that you have to create then anyway!
To be accurate, that formula would require that the first load is S/2 from the support at each end, where s = L/n wouldn't it!
RE: Beam Formulas for Multiple Point Loads.
I tend to agree that searching for the standard formulae for such simple calculations is probably more time consuming than just preparing the shear force and bending moment diagrams.
Your second comment is not correct. Loads spaced at S result in a bending moment diagram which just touches the uniform load curve at every point load. When n is odd, no correction is needed. When n is even, there is a small deduction at midspan.
BA
RE: Beam Formulas for Multiple Point Loads.
n = 5 (odd)
w = 100/2 = 50 plf
Mmax = 50(12)2/8 = 900'#.
BA
RE: Beam Formulas for Multiple Point Loads.
I was looking at it only from the conversion to a UDL load point of view, did not look at the actual results.
I would have always created a UDL equally about each point load, so 1' either side of the point loads. So your UDL loading would start and stop 1' from the supports at each end.
In your modified loading you are actually applying the loading over the full frame, so the total load applied is actually 600, not the 500 calculated from the sum of the point loads. That is why I thought it would give the wrong results.
But is does give the correct moment at midspan the way you have done it. Unfortunately maximum shear is wrong by 20% - 300# instead of 250# but that was not the question! My solution gave the correct maximum shears but the Mmax = 875, so needed w.s^2/8 added for the odd case to get Mmax = 900, whereas yours needs to subtract it in the even number of loads case.
Actually the old British Steel Designers Manual gives the solution as Mmax = (n^2)Pl/8(n-1) (odd) and (n-1)PL/8 (even) where n is the number of forces and L = (n+1)s, while peak shear is np/2.
But that is the danger of trying to take short-cuts rather than doing the full analysis. That short cut only gives the maximum moment and the moment at each point load, no shears etc. You need a different formula for them.
The complexities are all too complicated for me, Better to just do the analysis as agreed above
Truckdesigner
There is a rule for the location of the maximum moment. It is the point of 0 shear which is the centroid of the applied loads.
RE: Beam Formulas for Multiple Point Loads.
Out of the office now, but will mull over it next week...
RE: Beam Formulas for Multiple Point Loads.
I use excel so I can make these points quite small and hence very accurate.
If there is a varying load lets say a trinaglular shape or parapolic, then I typically use numerical integration (Simpson's rule) also plotting to the discrete points mentioned.
"God grant me the serenity, to accept the things I cannot change, the courage to change the things I can, and the wisdom to know the difference" -Reinhold Niebuhr
RE: Beam Formulas for Multiple Point Loads.
RE: Beam Formulas for Multiple Point Loads.
RE: Beam Formulas for Multiple Point Loads.
When I design a beam to carry joists, I consider the load to be uniform over the span, whether or not the number of joists is odd or even. The small error introduced when n is even is not significant, particularly when n is large.
When I design a beam to carry joists supporting a trapezoidal area of floor or roof, I consider the load to be a distributed load varying from w1 to w2.
BA
RE: Beam Formulas for Multiple Point Loads.
RE: Beam Formulas for Multiple Point Loads.
No, nothing generic.
RE: Beam Formulas for Multiple Point Loads.
BA
RE: Beam Formulas for Multiple Point Loads.
RE: Beam Formulas for Multiple Point Loads.
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.
RE: Beam Formulas for Multiple Point Loads.
For a "hand" calculation, where we have access to an intelligent computing device inside our heads, yes. On the other hand if you want to program a dumb external computing device to do it, it actually gets quite complex if you have non-uniform distributed loads over part of the length of the beam. Splitting the beam into short sections, either loaded over the full length or just at the ends, then solving the resulting matrix equations, is actually a reasonably simple way to do it, with the added advantage that it becomes trivial to extend the system to deal with continuous beams, and relatively simple to extend to 2D frame analysis.
My continuous beam spreadsheet Here goes the opposite way. It treats the beam between end supports as a simply supported beam, then finds the internal reaction loads required to get zero (or specified) deflection at the supports, but the complexity of the code for both approaches is probably about the same, and the computation time is negligible either way.
Buggar - there is a Visual Basic forum here, also VBA and spreadsheet forums which cover VBA, which is just Basic built into MS Office.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Beam Formulas for Multiple Point Loads.
RE: Beam Formulas for Multiple Point Loads.
A point load over a support has no effect on the moment or shear diagram! Leaving it out would have no effect on the calculations. The "simplification" to equivalent UDL's is inherently dangerous (but not a lot) as it requires "fiddles in some areas to get the correct answer, so you need to know if your answer is correct before you start.
RE matrix, why would you do it for a simply supported beam unless you had to do it a lot of times and your loads were not symmetric or you already had a computer program to do it (that's a good idea if you do a lot of them!)?
RE: Beam Formulas for Multiple Point Loads.
With multiple variable loads at multiple variable locations, it seems to me that a generic formula involving summations and boolean functions could be written, but I believe it would be more cumbersome than simply calculating the maximum moment as the area under the shear diagram while the value of shear remains positive.
rapt,
A point load over each support has no effect on the moment or shear diagram but it does affect the end reactions which are consistent with that of a uniform load. I see nothing "inherently dangerous" about the simplification, particularly if the number of point loads is large.
I agree that a computer program using matrix inversion will save time when there are a lot of beams to design, but it doesn't answer the question posed by the OP. Moreover, those who do not understand statics should not be using such programs. That would be dangerous.
BA
RE: Beam Formulas for Multiple Point Loads.
There is lots of ‘dangerous’ out there these days, and it seems to pass as real engineering. It just staggers my imagination these days that almost nobody wants to really understand the fundamentals of what they are doing. Just give them a formula that they can plug numbers into or the exact code paragraph which tells them what to do, without any understanding of the intent of that section of the code, and they are on to their next problem without having learned a damn thing about real engineering. Just plug and grind without any understanding of the basics. And, better yet, if you can offer a computer program which does that math and takes the code into account, then they are really happy, no need to even lift a pencil. I think this is really a sad state of affairs.
RE: Beam Formulas for Multiple Point Loads.
1) Construct a shear diagram to ascertain point of max moment.
2) Apply formula below to each point load for moment contribution at max moment location.
Of course, now you've arrived at a place where the effort involved would start to rival that required to just do it from fundamentals or run it through a coded algorithm.
If one requires max deflection, and a midspan estimate isn't sufficient, then you're definitely in software territory.
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.
RE: Beam Formulas for Multiple Point Loads.
Completely agree. Knowledge and understanding of statics is essential before using anything automated.
I continually emphasis with RAPT software that designers should not be using it until they can regularly complete the same designs by hand calculation and have a full understanding of the design code they are using and the detailing requirements for structures. No computer program that I know of can completely analyse, design and detail concrete members, especially more complex concrete member variations. The designer must be able to do it and understand it all so that they can apply the results of the automated calculations in their final design decisions and detailing of the structure.
Unfortunately the opposite seems to be becoming the norm, where if you do not know how to do it, software will do it for you.
RE: Beam Formulas for Multiple Point Loads.
A confused student is a good student.
Nathaniel P. Wilkerson, PE
www.medeek.com
RE: Beam Formulas for Multiple Point Loads.
BA
RE: Beam Formulas for Multiple Point Loads.
I suppose you could, but evaluating the step functions directly (which is what Macaulay's method does) seems much simpler.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Beam Formulas for Multiple Point Loads.
You are probably right, Doug. I was just thinking of the remark made by KootK:
If the deflection can be described as a continuous function by means of a Fourier Series, perhaps the maximum deflection can be determined without resorting to software. But in any case, it is easy enough to find the maximum deflection in the standard way by simply testing values each side of the peak.
BA
RE: Beam Formulas for Multiple Point Loads.
Yes, since we are talking about a simply supported beam, then the reactions are easily calculated and the beam diagrams can be procedurally generated by hand. The math gymnastics I mentioned, aren't appropriate until you have an indeterminate beam. Of course, now that I've developed the tool, I have no problem using it for the simple problems too.
RE: Beam Formulas for Multiple Point Loads.
have we considered, as a simplification of the problem, replacing many point loads with a uniform load ?
another day in paradise, or is paradise one day closer ?
RE: Beam Formulas for Multiple Point Loads.
In the hand drawn example I provided a total of 500 lb. is distributed over a 12 ft. long beam, the maximum bound is 1500 ft-lb (one 500 lb. point load at the center, PL/4).
The minimum bound is 750 ft-lb for an infinite number of equal point loads, equally spaced that total 500 lb. (actually a UDL).
Any number of equally spaced, equal point loads that total 500 lb. will fall between 750 ft-lb and 1500 ft-lb. The more points, the closer the approximation of a UDL, therefore the closer to 750 ft-lb. Semi-graphical integral calculus.
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RE: Beam Formulas for Multiple Point Loads.
RE: Beam Formulas for Multiple Point Loads.
Sure. This is precisely what I do for the general case. It's somewhat approximate but, then, what isn't?
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.
RE: Beam Formulas for Multiple Point Loads.
if you have multiple loads on a beam, then you can add together the effects of each individual load.
this is (really) a simple problem.
another day in paradise, or is paradise one day closer ?
RE: Beam Formulas for Multiple Point Loads.
I don't think that superposition is approximate. I think that, if you're stepping through values at discrete points to find the maximums, it's quite likely that you'll end up with something close to, but not exactly, the maximum. Such is the case with all things FEM-ish.
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.
RE: Beam Formulas for Multiple Point Loads.
RE: Beam Formulas for Multiple Point Loads.