Help with beam calculations

[Magnus_Nordmark]

Hello everyone!
I'm considering building a basement under our new house, but I'm a bit uncertain about the dimensions of the two beams that are green in the diagram below. Can anyone help calculate what bending resistance/moment of inertia the beams need to have? It's possible with either laminated wood or steel. The floor will have a living room with normal load.

Grateful for help!

 

[Magnus_Nordmark]

Help help help! Can someone give me a little push? What I basically need is the weight of a wood floor structure per m2 and the distributed load one would calculate in a residential building. Does anyone know?

Wood floor structure of:
22mm chipboard
220 joists
195 mineral wool
sparse 28x70 c/c30
2x13mm gypsum.

 

[anaitis]

Help help help! Can someone give me a little assistance? What I basically need is the weight of a timber floor per m2 and the distributed load typically considered for a residential house. Does anyone know?

Timber floor of:
22mm chipboard
220 battens
195 mineral wool
sparse 28x70 c/c30
2x13mm plasterboard.

One usually considers 2 kN/m2 as the live load and maybe 0.5 kN/m2 for the dead load.
You might get a lower value for the dead load if you calculate in detail, I usually don’t bother.

 

[Krawk]

Useful load in a typical residence is 0.5kN in fixed load and 1.5kN (psi=0.33) free.
The self-weight of the floor can be easily calculated.
Consider 6-7kN/m3 for wood, 8-9kN/m3 for plaster, and 0.3kN/m3 for insulation.

Don’t forget to account for the continuity of the floor joists over the beams.

Edit: ahh, anaitis beat me before I hit send.
Edit2: Switched the loads. Correct now.

 

[petersk]

0.5kN/m2 bound and 1.5kN/m2 is it.

In failure it becomes 1.3*0.5+1.3*1.5=2.6kN/m2
In service 1.0*0.5+0.33*1.5=1kN/m2

excl self-weight

 

[Krawk]

Ahh, yes, of course. That's what happens when you move things around to make it "prettier"
I have edited my post to avoid confusing anyone.

However, you can't treat bound and free loads as separate. They are two load components of one variable load.

So.
Ultimate [LK1]: 1.3*(0.5+1.5)=2.6kN/m2
Service [LK8]: 1.0*(0.5+1.5)=2.0kN/m2

A big parenthesis here:
(Since the area for free load is greater than 45m2, you can reduce the occupancy load by 0.7. However, I usually don't do this when calculating roughly/simplified.

Reduced loads:
Ultimate [LK1]: 0.7*2.6=1.82kN/m2
Service [LK8]: 0.7*2.0=1.4kN/m2

To use these loads, you should account for the worst case regarding continuity. That is, in this case, you have bound load in all 3 sections, free load in 2 out of 3 sections.

Line loads on the beam then become:
Ultimate [LK1]: 1.3*4*0.7*(1.1*0.5+1.2*1.5)=8.55kN/m
Service [LK8]: 1.0*4*0.7(1.1*0.5+1.2*1.5)=6.58kN/m
Self-weight of the floor [LK1, LK8]: 1.0*1.1*0.5=0.55kN/m )

Okay, enough talking on a Saturday morning!

Edit: sausage fingers

 

[Magnus_Nordmark]

The line loads on the beam are then:
Failure [LK1]: 1.3*4*0.7*(1.1*0.5+1.2*1.5)=8.55kN/m
Serviceability [LK8]: 1.0*4*0.7(1.1*0.5+1.2*1.5)=6.58kN/m
Self-weight of floor [LK1, LK8]: 1.0*1.1*0.5=0.55kN/m )

Thanks for those numbers! As a newcomer to structural calculations, which of these line loads should I use as the design load? Is it 8.55kN/m, 8.55+0.55kN/m or 8.55+6.58+0.55kN/m that should be used in the deflection calculation?

Is there any recommendation on how large the maximum deflection should be in such a case? I have seen standards for floors at L/400 as the maximum deflection. Is it the same when calculating such a beam?

Thanks again!