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Floor Loading of a structure .

  1. Feb 15, 2015 #1
    Been trying to bend my head round something that is no doubt so simple it is unreal...!!

    I am designing a structure that has a total weight of 1.6 tonnes (1600kg), the structure floor area is 8m x 3.5m and sits on 8 pads of 500mm x 500mm.

    To calculate the floor loading do I:

    Divide the weight by the floor area i.e (1600 x 9.81) / (8 x 3.5) = 560.6 N/m2

    or

    Divide the weight by the total floor area of the pads i.e. (1600 x 9.81) / ((0.5 x 0.5) x 8) = 7848 N/m2

    ??

    It is driving me nuts, any help is a massive step forward!
     
  2. jcsd
  3. Feb 15, 2015 #2

    mfb

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    Staff: Mentor

    Neither.
    The force on the individual pads will depend on the weight distribution of the structure, the positions of the pads and some more details of the setup. In the ideal case, all 8 pads get the same force and apply it uniformly to the ground, then the second calculation gives the pressure at the pads.
    The first calculation will always give the average over the whole room, but pressure won't be uniform over the whole surface.
     
  4. Feb 15, 2015 #3
    Many thanks for the reply, I thought as much that it may be neither of those solutions.

    I attach some diagrams of the structure to give a better understanding of the question. In the question above I used roughly generic figures, the images show the final dimensions.

    It has to be located in an area with 3.35kN/m2 maximum floor load....

    The stress model shows that for the load the structure has to withstand I cannot reduce the material sizes to lighten the structure and it is coming in at the 1.6 tonnes mark as it is!

    Hope the images give a better indication.
     

    Attached Files:

  5. Feb 15, 2015 #4

    mfb

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    So ~7m^2 for most of the load. At least the total load looks fine, but don't rely on that. And single pads could still lead to issues.
     
  6. Feb 16, 2015 #5
    Further too....

    Used the simulation software to tell me the reaction forces, then using the area of each pad on the reaction forces to give me the floor loading force.

    2 images attached, one with the original 0.5m square pads and one with 1m square pads under the 'heavy' end....whilst the smaller pads on the end of the structure are over the imposed floor loading limit of 3.35kN/m^2, I can make the pads larger at that end to compensate....my only issue now is from a 'form' point of view....doesn't look very pretty with dirty great steel pads on the floor!

    Ho hum, the scourge of the designer...form/function etc.!
     

    Attached Files:

  7. Feb 17, 2015 #6
    Thanks, to all for letting me be parts of this form.
     
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