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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Distributed Load vs. Point Load 2
- Thread starter PatBethea
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- #25
Thanks for all the replies.
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.
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.
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- #27
The truss reactions were roughly 16 kips. There was a small (relative to the truss load) distributed load that made up the difference (also, one of the columns is not centered directly under a truss - it is offset to the inside by 18"). I didn't mention it before because I didn't think it mattered in the overall scheme. I was trying to understand what was done.
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.
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.
PatBethea,
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.
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.
I interpret the problem as a beam with a truss at midspan of the beam and a truss at each end where the beam is supported. I would design the support beam using a discreet point load at midspan equal to the middle truss reaction, not a uniformly distributed load. The moment would be equal to PL/4, where L = the beam span.
What the SE did was perfectly acceptable. Provided that you follow the correct tributary areas for each element, the PL/4 moment from the truss rxn is equal to the wl^2/8 moment that the supporting beam sees. From the posts above, the most common mistake is dividing the point load by the length of the beam to get the equivalent distributed 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.
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.
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