How Do I Calculate Floor Loading for My Structure?

  • Thread starter Saint.V8
  • Start date
  • Tags Tags
    Structure
5 replies · 5K views · last reply
Saint.V8
Messages
5
Reaction score
3
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!
 
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.
 
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.
 
Attachments
  • 3D render of a blue steel frame structure with a rectangular top platform, cross-bracing on the vertical posts, and diagonal ground braces on a base frame.
    Platform_001.jpg
    38.9 KB · Views: 736
  • Two-view engineering drawing of a steel frame: left, a front elevation with X-bracing between two uprights, 3585 across and a 500 offset; right, a side elevation with a 2000 top width, 4870 height, 1630 and two 2442.5 bays, a sloped diagonal brace, and an overall length of 7015.
    Platform_002.jpg
    15.5 KB · Views: 669
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.
 
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.!
 
Attachments
  • SolidWorks simulation screenshot of a steel frame structure on base pads, with reaction force labels at each pad showing kN/m² values and a note that each pad is 0.5m x 0.5m.
    Forece_Res_001.jpg
    55.1 KB · Views: 836
  • SolidWorks Simulation screenshot of a steel frame structure with reaction force arrows at its base pads, showing force values in N and kN/m² pressure labels, and a platform weight of 1578.9kg (15.489kN).
    Forece_Res_002.jpg
    57.4 KB · Views: 1,018
Thanks, to all for letting me be parts of this form.