distributed load vs. point load

huangyhg

distributed load vs. point load
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?
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14' apart is a definite point load.
was the distrbuted load for something else??
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.
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.
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.   
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?
excel? sounds like a lot of work.  you dont have a structural analysis program?  good luck!
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
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.
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.
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