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Old 22nd October 2013, 07:24 PM   #1
king295
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Default Correct inputting of loads

Sorry if this has been dealt with previously.

I have a floor joist system which supports roof loading as a point load.

What is the correct method of inputting the loading if the floor joists are at 400mm centres for instance. I am getting conflicting advice.

Do I times or divide the roof laoding by 0.4 (centres of the joists) to create the point load.

Help/clarifcation would be appreciated.
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Old 22nd October 2013, 10:10 PM   #2
Tony Bryer
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Default Re: Correct inputting of loads

I'm not quite sure of the question. If you've got individual joists bearing on a trimmer etc at 400-600c/s, such loads are treated as a UDL or part UDL (note that the loading in kN/m run so the general case load will be load/m2 x supported joist span/2. If this isn't what you've got can you explain in a little more detail please.
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Old 23rd October 2013, 06:51 AM   #3
motorphotos
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Default Re: Correct inputting of loads

If you have a floor joist which say takes a line load of studwork with say 2m of roof

loading would be

udl floor 2.1 x 0.4
Point load stud 0.5 x 0.4 x 2.3 say for height of stud
point load roof 1.66 x 0.4 x 2

the point load need to be positioned from one end

probably not too clear
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Old 23rd October 2013, 10:34 AM   #4
Andy Hudson
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Default Re: Correct inputting of loads

If I am understanding the question correctly you have a floor joist system spaced at 400c/c which is carrying roof loads which occur in a transverse direction to the span of the floor joists creating a series of point loads across the line of the joists. ( presumably there must also be a stud wall or other support member that transmits the load from the roof onto the floor joist which must also be accounted for)

But simply if the roof load is say 2.15kN/m run then when supported transversely by the floor joists at 400c/c then each individual point load on each joist will be = 2.15 x 0.4 = 0.86kN point loads.

If you think about this you have 2.5 joists in every metre of floor width therefore 0.86 x 2.5 = 2.15kN/m

Hope I have understood your question correctly.
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