Question for a structural engineer?

Question

Ok so a commonly used design LL is 40 psf. So a 12' x 15' bedroom would have a total capacity of 7200 lbs. I understand this part but is that load a uniform load throughout the bedroom? Or can it also be a 7200 lb point load anywhere in the bedroom. thanks

Answer 1

That would be a uniform load over the whole floor, divided up among the floor joists. 40 psf is pretty heavy for a bedroom.  More likely it would be 30. Building codes usually specify a concentrated load to be applied at any point, separately from (not in addition to) the uniform load.  For a bedroom this would be around 300 lbs.  If you are building something, you need to get the local building code and read it carefully.

Answer 2

12' x 15' = 180ft².

7,200lb / 180ft² = 40psf, meaning that each ft² will take a load of 40 pounds.

(= 7,200lb / (180ft² x 144in²/ft²) = 7,200lb / 29,920in².

= 0.28psi).

Answer 3

You would design the floor for 40 psf. The floor would consist of typically wood member that would be evenly spaced somewhere between 12" o.c. to 24 " o.c. For 24" o.c. each member would be designed for 40 psf x 24" to get 80 pounds per lineal foot (plf). For a uniformly loaded beam (15' long) at 80 plf an equivalent point load would be about 1200 lbs. at the midpoint of the member. So to design the floor for a 7200 lbs point load you be over designing the flooring members.

Its best to design the floor as a uniform load.