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shelf with partially distributed load... how does it approximate?

  • Thread starter Thread starter mir
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mir

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Hi.
I have a doubt about how to make welding calculations in a shelf with a load distributed only on a part of the shelf itself.

the shelf is reinforced but I would first like to make the calculations on the welding, that is to see if the welding would be enough to hold the load.

as you see in the pdf attached step from a load on area to a linear load... but I don't know if I do well.

Thanks for any help
 

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I don't know if I understand the question.
a shelf has a load, you want the binding reactions of the wall? Do you know that binding reactions are independent of load distribution?
 
the moment in the disaster depends on the distribution (and position) of the load
Actually, no, you just need to know where the result of your distributed load is, and multiply it by the arm compared to the ink point. things are as you say if you want to know the trend of the cut and the stinging moment along the shelf.
 
I would say there is something wrong with your calculations; to me is the following:

Linear load: 43.12 n/mm (44*0.98)
Total load (cut vert.): 34712 n (half the central team takes it and half the other two).
mf max: 2360389 n*mm (always half at the center and the rest at the ends).

I have neglected the mt transferred (also because there is no data to be able to calculate it, and I am not sure to do it correctly) from the ends of the shelf to the teams, but already so you are not far from the real.
n.b. I considered the shelf as two beams stuck at the ends.
Bye.
 
Actually, no, you just need to know where the result of your distributed load is, and multiply it by the arm compared to the ink point. things are as you say if you want to know the trend of the cut and the stinging moment along the shelf.
ok but at the same load applied if I distribute differently the load the binding reactions do not change but the bending moment si ...
 
I said a cabbage, as I found the reactions the shelf should be considered as two beams resting at the ends.
for a more precise calculation should be considered the rigidities of shelf and supports in order to evaluate the moments of reaction to the disaster, but the thing is quite complicated (at least for me).
If you consider the shelf only supported, and divided into two (other approximation), a simplification is introduced but in favor of safety, because the reaction on the central support is greater than that which is actually.
Bye.
 
Yes but now I want to dimension only the welding.... as if reinforcements weren't there. (it is basically to understand if the reinforcements serve!)
 
hmmm... do you think the flender moment doesn't change? Tell me why if it's what you say I have to get there:
What I mean is,

1) as you say, the trend of the bending moment varies along the beam irrespective of how the load is distributed

2) as I say, and as he said at the beginning Roman lightning, the moment at the time of the disaster depends only on the result of the distributed load and from where it is applied.

3) Sol. 1 and 2 must give the same result of course (i.e. the moment at the time of the disaster will have to be equal both to proceed according to 1) and second 2), but sol.2 is simpler.
 
what we say hunter and I is that:
- the moment (internal characteristics) depends on the distribution of the load, and here it does not rain
- the couple (external characteristics) given by the welding is a binding reaction, and as a bond it is not part of the shelf. if it is external to the shelf, to rigor it is called couple and not moment, although the dimensions are the same
- the purpose of this couple is not to drop the shelf, so, how the tensions are distributed within the shelf, it could not give him less. the important thing is to balance the result, including the weight of the shelf.
[..]binding reactions are independent of load distribution[...]
 

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