Distributed Load vs. Point Load 2

[PatBethea]

I had a SE size a beam for me that will be used to support three steel trusses (columns will sit under the two end trusses). In his calculations, he assumed that the load from the trusses was distributed over the length of the beam. The trusses are approximately 14' apart. I'm trying to understand this. Were the trusses 16" or 24" OC, a distributed load is easy for me to see. When sizing a beam or a header, is there a truss spacing at which you would start to consider the loading a point load as opposed to a distributed one?

 

[MiketheEngineer]

14' apart is a definite point load.

Was the distrbuted load for something else??

 

[COEngineeer]

14' apart... yikes I dont think the moment diagram nor the shear will even be close if you put it right next to distributed load diagram.

 

[MillR]

I agree that 3 loads at 14' ought to be checked as point loads.

To the other part of your question - at what point can you look at a load as distributed? - it is a question of the length of the beam and the number and spacing of the loads. For example, a 140' beam with 10 point loads at 14' on center might be assumed a distributed load. In any case it's a judgment you develop by checking it both ways a few times.

 

[COEngineeer]

yeah, I agree with Millr, 140' with 10 point load at 14' o.c. will probably have moment diagram pretty darn close. The maximum moment is slightly lower on the distributed one.

 

[PatBethea]

Thanks, all.

I think I will take his calculated loads and create shear and bending moment diagrams in EXCEL. I will then convert the distributed loads to point loads and do the same thing. Then I'll take printouts of the diagrams back to the SE, explain my concern, and ask him to re-evaluate his design. Sound reasonable?

 

[COEngineeer]

Excel? sounds like a lot of work. You dont have a structural analysis program? Good luck!

 

[trainguy]

As far as I can see it, it's just 1 point load in the center, according to your original post. Maybe you'll have a small distrib. load due to the deck or whatever else may be there.

In either case, you don't need a computer, or Excel to do this.

Just use superposition.

Point load - M= PL/4 V=P/2

Distrib. load - M = w L^2/8 V = w L / 2

Hope this helps.

tg

 

[dozer]

I like computers as much as the next person but take a step back here, it's an easy hand calc. For a simply supported beam, 56 ft long with three equally spaced loads P, the moment at the center is 28P.

If we assume the total load of 3P is uniformly distributed along the beam, the moment at the center is 21P or 75% of the true moment.

You tell me if he's being unconservative. OK, OK, I'll tell you. Yes, he is.

 

[csd72]

If the spacing of the trusses is half the span, then the moment ends up being the same.

The reason is that only half of the uniformly distributed load goes to the central point load (the other 2 quarters go to the trusses over the columns.

Therefore

P=wl/2
M = Pl/4 = wl^2/8 (same as a uniform load)

Compare the deflection:

Pl^3/(48EI) = wl^4/(96EI)

Compare this to the udl deflection of

5wl^4/(384EI) = wL^4/(76.8EI)

Point load deflection is actually less!

Shear is also half of the udl case.

The structural engineers calculations are conservative and safe.

 

[dik]

If the span is broken into an even number of equal spaces, the solution should be 'exact'. If broken into an odd number of spaces, then the solution is conservative. for anything other than 3 equal spaces, the degree of conservatism is small.

Dik

 

[JStephen]

Why not just call him up and ask why it's done that way?

 

[COEngineeer]

I am not following you guys. Equally spaced point loads are always more conservative than distributed load. The only time point load is less conservative is if you put it close to the column.

 

[csd72]

COEngineer,

What you say is correct if your point loads add up to the same amount of load as the udl.

But in cases such as trusses, part of the udl will be supported by trusses outside the span. (i.e half of each purlin/batten load goes to each truss).

Follow my maths above, it is correct!

Shear will be higher though, if trusses were spaced close to one end.

 

[civilperson]

Two differing interpretations of the problem are used in the previous posts. One analysis assumes a 56 foot span with 3 points loads at L/4, L/2 and 3L/4 and the other assumes a 28 foot span with points loads at the ends and at L/2. csd72 assumes something different all together. A restatement of the problem will have all engineers getting the same answer.

 

[UcfSE]

You consider point loads as distributed when it doesn't make a significant difference. For instance, the difference between the calculated moment using a single point load and using the point load as distributed over the beam is 100%. that's significant. When you have 7 point loads, the difference is about 14%, much more reasonable but perhaps not enough for some.

 

[csd72]

Civilperson and UcfSE,

You both need to sketch the area up(including the trusses on each side of these trusses (i.e. 5 trusses total).

Follow the loads from roof level all the way down to the beam.

With a 28 foot span and trusses at 14 foot, each truss only picks up 14 foot worth of load. Your point load is then equal to wl/2 as noted above assuming w is taken as the average roof load per foot.

All my numbers are correct for this case.

 

[UcfSE]

csd72, I was not addressing you and I do not need to review your "maths". My post was directed to the OP who asked a general question.

 

[PEinc]

If, as described in the OP, the 3 trusses are 14' c.c. and there are 2 support columns, one each under the 1st and 3rd trusses, there is only a point load from the 2nd truss at the center of a 28' span.

M = PL/4 which will be greater than the moment from a distributed load. P = the load from the center truss. L = 2 x 14' = 28' (not 56').

As trainguy said, you don't need a computer. You can use M = PL/4 and V = P/2. You can also easily draw the shear and moment diagrams to get the design values.

 

[PEinc]

Example (with made up truss loads):

Load per truss = 10 kips
3 trusses at 14' c.c.
Support columns under 1st & 3rd trusses. Span = 28'.

M = 10 kips x 28 / 4 = 70 kip-ft
V = 5 kips
R1 = R2 = 15 kips

If the SE designed as a distributed load, he MAY have used the following:

Total load = 3 ea. x 10 kips = 30 kips
w = 30 kips / 28' = 1.071 klf
M = 1/8 x 1.071 klf x (28 x 28) = 105 kip-ft (> 70 kip-ft)
V = 0.5 x 28' x 1.071 (or = 1/2 x (3 x 10 kips) = 15 kips
V = 15 kips (> 5 kips)
R1 = R2 = 15 kips

Big differences between the point load analysis and the distributed load analysis (assuming the SE distributed the loads like I assumed).

As JStephen said, call and ask the SE.