Hi, I wonder if there is any table for what, for example, a 145*45 rule can handle, for constructions cc 60. I am planning to make a floor and wonder what dimension I should use.
 
This is a whole science. The choice of rule depends on the type of wood, number of knots, residential/office, how much deflection one can accept, etc...

Fortunately, there is an overview dimensioning available at.
http://www.traguiden.se/
Log in and check under planning.
You can also read and learn a lot about wood in general.
Good Luck.
/CC
 
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nissjohan
  • Laddar…
oh oh that seems tricky?
 
For me too, but you can stick your neck out ;)
If you have two fixed points with cc600 between them, and place a regel/råplan 145x45 on its edge over that, I believe you can load 1 ton on the regel without it breaking.
Maybe someone wants to increase a bit :)
 
Mention the span you need, and there is probably someone here who will quickly answer which timber dimension you require!

/Ingenjören
 
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Sandsjöberg
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The span I do not know. just wanted to check in general. but let's say it's 5m 4m 3m 2m?
 
It's not the fracture toughness that is used for design, but rather how much deflection is allowed. You don't want a floor to flex too much, and before a joist breaks under load, you'd have a halfpipe in the room.
 
If we say that we have a uniformly distributed load over the beam, the formula for deflection looks like this:

    5 * Q * L³
f =     ___________

   384 * E * I


f = deflection (m)

Q = The distributed load ( N/m )

L = Length of the beam ( mm )

E = Modulus of elasticity for the material ( N/mm² ) = 7.0 * 10³ for K12 structural timber

I = Moment of inertia (mm^4) = ( Width * Height ³ ) / 12

Something like that, clear as crystal, right? :D

(Formula and values are taken from my teflon-coated head, so don't bet the house on their accuracy ;))
 
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UppGiven
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The formula for deflection of distributed load shall be:
                 4
     5 x q x L
f = ------------
     384 x E x I

E = 5.2 N/mm[sup]2[/sup] for K12, load type A, climate class 0
The deflection should not be greater than L/300 for total load (L/500 for variable load only)

In addition, it is common to ensure that the deflection of a point load of 100kg (P = 1kN) is less than 1.5mm to avoid the floor feeling wobbly.

           3
    P x L
f = ------------
   48 x E x I

These formulas work for larger spans where deflection is generally decisive.
However, they do not address the strength which must be checked separately (Moment, shear capacity, supports, fixings etc. depending on construction).
The values of E also vary with the timber's strength class, climate class, and load type. E=5.2 N/mm[sup]2[/sup] applies for load type A, K12 and climate class 0 or 1, which means standard construction timber class K12, indoor environment, and residential load (fixed load portion).

/Ingenjören
 
Oh oh, I just wanted to know what dimensions are commonly used in different constructions? Let's say I buy the timber at byggmax, I need a load-bearing floor, and the length should be 5m, 4m, 3m, 2m, 1m?

What should I get then?
 
I usually use 2 x 6" in framing, has worked just fine so far anyway ::)

With small spans, you might be able to reduce the size, but it's much easier to use the same dimension throughout.
 
The deflection is calculated separately for each load type and the different deflections are added together. For a floor structure in a residential building, there are typically 3 load types P=self-weight, A=useful fixed load, B=useful free load. The modulus of elasticity varies with the load type.

In deflection calculations, one can account for interaction with the flooring particleboard if it is screw- or nail-glued to the floor joists. In the calculation of deflection, this interaction is usually not considered. Lateral load distribution can also be considered in deflection calculations.

Example
Floor joist 45*220, K24, center distance 60cm, residential house indoors, normal self-weight, supported on two ends. 22mm flooring particleboard. No lateral torsional buckling. No holes in the beam. Shear deformations are not considered.

Screwed flooring particleboard. No lateral load distribution. Max span 3.1m. Deflection decisive for the span.

Screwed flooring particleboard. Lateral load distribution (particleboard). Max span 3.3m. Deflection decisive for the span.

Screw-glued flooring particleboard. No lateral load distribution. Max span 3.8m. Deflection decisive for the span.

Screw-glued flooring particleboard. Lateral load distribution (particleboard). Max span 3.9m. Deflection decisive for the span (permanent damage, max deflection L/300 or max 20mm).
 
it seems tricky this. I am thinking of expanding about 3.5m and it should have an upper floor too, I was thinking of using 145*45 or 170*45, is that enough or? there won't be any load-bearing walls on 3.5m?
 
Hi! I'll continue here.
I have a project where I plan to attach a 45x195 regel that is 5 meters long at both ends. How much load can I then place on the middle of the regel?
/Anton
 
thorstis said:
Hi! I'll continue here.
I have a project where I intend to attach a 45x195 beam that is 5 meters at both ends. How much can I then load on the middle of the beam?
/Anton
More information is needed to be able to answer the question.
Load until what happens? The beam snaps, bends to a certain limit, suffers permanent damage? What is the beam made of? Classification?
 
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