## Distributed Load vs. Point Load

## Distributed Load vs. Point Load

(OP)

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?

## RE: Distributed Load vs. Point Load

Was the distrbuted load for something else??

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

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.

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

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?

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

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

## RE: Distributed Load vs. Point Load

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.

## RE: Distributed Load vs. Point Load

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.

## RE: Distributed Load vs. Point Load

Dik

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

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.

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

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.

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

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.

## RE: Distributed Load vs. Point Load

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.

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

Dik

## RE: Distributed Load vs. Point Load

The reason I did not just go back to the SE for clarification was that I wanted to get it straight in my head first.

COEngineer was correct - getting the shear and bending moment diagrams out of Excel was alot more work than I anticipated.

Yes, Michfan, these trusses are holding up a roof.

PEinc - you got about the same result in 7 or 8 lines that it took me all afternoon to arrive at. Using point loading, my maximum moment was 142 kip-ft. The SE got a maximum moment of about 200 kip-ft using distributed loading. This result was counter-intuitive to my way of thinking. However, I am now convinced that his beam will work.

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

I still don't quite understand why a distributed load was used for the three truss loads. True, it gave a higher bending moment and maximum shear which results in a larger beam being selected. So it was more conservative. But was it too conservative? I'm not going to ask that it be changed, but it seems to me like a point-loaded beam gives a more realistic analysis. I guess it was just a short cut to save time on the part of the SE.

## RE: Distributed Load vs. Point Load

You are looking at it from a mechanical perspective where you may be doing a design for 100000 cars(say) and even 1% would make a lot of difference.

In Structural engineering we are always designing 1 off structures and the additional time it takes to cut things down to the bare bones usually outweighs the savings. The material cost is only a proportion of the total build cost.

knowing the shortcuts is a big part of our profession.

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load

This will not work.

Turn over an envelope:

Use easy numbers:

Draw a 10'x10' bay with columns at each corner.

Bisect the bay with one intermediate beam.

Use 50psf roof load

Analyze the south beam.

Using the point load:

Wtrib for intermediate beam = 5'

RXN from intermediate beam = wl/2 = 50x5x10/2=1250#

PL/4=1250*10/4= 3125#'

Dist. Load:

Wtrib of south beam=5'

w=50x5=250plf (NOT the 1250 from above divided by 10)

wl^2/8=250(10^2)/8=3125"'

Same moment. Size beam, check shear and deflection. done.

Take it further if you wish... try non-square bays, try multiple intermediate beams, draw the moment diagrams, and here's what you'll find:

Even number of spaces between intermediate beams means the udl moment will equal the point load moment. Odd spaces mean a slight disparity, but the UDL moment is slightly more conservative, but it is much easier and quicker. I always use UDL unless there are 3 spaces.

dik's above post said the same thing, only more succintly.

## RE: Distributed Load vs. Point Load

## RE: Distributed Load vs. Point Load