Beam Formulas for Multiple Point Loads. 1

[truckdesigner]

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.

 

[SlideRuleEra]

A hand-drawn Shear and Moment Diagram will do that:

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[truckdesigner]

Thanks,

But to me - that ain't 5 equal point loads equally spaced on a simply supported beam..?

 

[SlideRuleEra]

Here you go, Mmax = 900 Ft-Lb:

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[truckdesigner]

Excellent. Thank you.

I take it there are no "standard" formulas then...

 

[BAretired]

Standard formulas? Sure there are!

For n equal concentrated loads of P spaced at s = L/(n+1),
w = P/s
Mmax = wL[sup]2[/sup]/8 when n is odd
or wL[sup]2[/sup]/8 - w.s[sup]2[/sup]/8 when n is even.

If the loads are not equal, draw the shear force and bending moment diagram.

BA

 

[rapt]

BAretired,

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!

 

[BAretired]

rapt,

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

 

[BAretired]

In the example which SlideRuleEra provided,
n = 5 (odd)
w = 100/2 = 50 plf

Mmax = 50(12)[sup]2[/sup]/8 = 900'#.


BA

 

[rapt]

BAretired,

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.

 

[truckdesigner]

Thanks folks.

Out of the office now, but will mull over it next week...

 

[pete600]

When I have multiple point loads or distributed loads on beam I typicall plot the shear and moment at discrete points (let's say 0.1 point) and superimpose the results for each point load.

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

 

[UcfSE]

AISC has a download "Design Aid for Deflection of Simple Beams Under Concentrated Loads" that covers deflection of simple beams with any point load in any positions; they do not have to be equal in magnitude or symmetric. I've spot checked several conditions against software and gotten good agreement. The paper only gives deflection equations though, no moment or shear.

 

[BUGGAR]

I have a Basic program for that. Does anyone use Basic? Can I post it here?

 

[BAretired]

Ordinarily, a load of P/2 exists over each support. This was not shown in SlideRuleEra's sketch, presumably because they did not affect the maximum moment but when they are considered, the reactions are 300# at each support which agrees with the uniform load assumption. The shear value will not agree at every point along the span because a stepped function is not a smooth curve.

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

 

[Archie264]

This is a statics problem, which is to say, something taught in a fundamental introductory engineering course. If we can't quickly do this by hand should we really be designing beams where life safety is involved?

 

[Teguci]

Sorry. If you have multiple variable loads at multiple variable locations, there is no generic formula to cover this situation. Even at 2 point loads you have to solve a system of equations involving these 4 variables. I have a spreadsheet that can take these multiple variables, runs through a unit force procedure to create a matrix, finds the inverse of the matrix to solve for reactions loads, reaction moments, initial deflection and initial rotation, then plugs everything back in to the beam and graphs the shear, moment, rotation and deflection along the length of the beam. I also have it calculate the solutions for the cubic and quartic functions in order to solve for local maximums and minimums (but I don't have that on my output page yet).

No, nothing generic.

 

[BAretired]

Creating a matrix and inverting it seems to be a rather sophisticated technique for solving a relatively simple problem.

BA

 

[allgoodnamestaken]

This is basic university stuff. Drawing a load/shear/moment diagram should take you a couple minutes each time, maybe.

 

[KootK]

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.