Calculation of Reinforced Framework for Floor?

[fredl]

Hello!
I live in a house with wooden floors and a wooden staircase.

Given a distributed load of 2500 kg/m2, what should a wooden framework for the floor look like? Is it even feasible? There is a crawl space underneath with about 30-40 cm of air between the ground and the subfloor. According to the drawing, it is currently 2"x9" with a span of 365 cm between the sills (60 cm apart between the joists). It is a single-story brick house with an unfinished attic. There is parquet flooring on top of 1" planks.

Can any structural engineer provide input? The fireplace is in the same room but in the opposite corner. The foundation of the fireplace is reinforced with steel beams and extra piers.

 

[huggan]

Normally, one designs for 200 kg/sqm. Why 2500 kg/sqm?

 

[fredl]

Normally, designs are made for 200 kg/sqm. Why 2500 kg/sqm?

It's not my idea; it's a case officer at an authority who thinks it's a good idea. I would like to have some support for my argument that it's not feasible.

 

[fredl]

Can clarify by noting that the surface to be loaded is 2x 640 x 553 mm, so it is not the entire framework that will have such a large load.

 

[huggan]

Counting point load as you now describe is something completely different from distributed load. Can you explain what this is about?

 

[fredl]

Calculating point load as you describe now is something completely different from distributed load. Can you explain what this is about?

Does it matter? Two solid objects causing this load (according to above). Is it possible to (with reasonable means) reinforce according to given regulations so that it can handle the load?

 

[huggan]

The background is not important, mostly because I'm curious about what is so thin but still weighs 2500 kg/sqm, or is 2 mm a typo?

However, how to calculate the load spread out or point looks completely different.

 

[fredl]

2mm? That describes 2x the base area.

 

[fredl]

Found the answer myself in the Boverket documents. This can be calculated as congestion load with 4.0 kN/m2 during dimensioning. Other than tearing up the entire floor and doubling up the joists along with an additional support, it does not meet the requirements.