Load-bearing capacity 220mm K24

[ronolo]

I would need to know how much a 220mm K24 that is 215cm long can be loaded. Is there a table to look at somewhere? I can't find anything.

 

[huggan]

The bending moment at the midpoint of the load P is M=PL/2, where L is the length.

The bending stress is S=M/W, where the section modulus W=bh^2/6. b and h are the width and height of the beam.

Thus, S=3PL/bh^2. P in Newtons and all dimensions in millimeters will give you the bending stress in Megapascal (MPa). S should be less than the beam's bending strength.

S = 24
L = becomes 2150 mm
b = 45 mm
h = 220 mm

24=(3*P*2150)/(45*220^2)
24=(3*P*2150)/2178000
24*2178000=3*P*2150
52272000/2150=3*P
24313/3=P
8104=P

Thus, a beam can handle a load of 810kg at the middle with a span of 2150 mm. If you want to know how much it can handle per m^2, simply multiply 810 by the number of beams and then divide by the square meter area of the roof. Using cc 600 mm, the roof can support a load of 690 kg per square meter.

 

[menhir]

Was just in because I had the same question and found this thread at the top of the page. Thanks for the formula and calculation! It can be used without major issues. However, I have three additional questions:
1. Is there any smart free program similar to those available for Glulam and Kerto for calculations of "ordinary" studs?
2. Is there a standard for how much a floor should support in kg/sqm in houses versus vacation homes?
3. What quality class is assumed for unspecified timber at the building market?

 

[huggan]

A roof must withstand a snow load of at least 250kg.

S for pine and spruce is 87 and 75, respectively.

 

[menhir]

Thanks huggan! Do you know what load a floor structure should also be able to handle?

 

[Gabbe1]

Thanks huggan! Do you know what load a floor structure should also handle?

I'm not entirely sure but I seem to remember that it is often calculated with 200kg/m2

 

[huggan]

Depends on the snow zone, but I always calculate with 250kg to be sure that it holds. But in this case, he certainly has no problem with being too low.

 

[menhir]

The snow zone does not affect what a floor joist should withstand, does it? Or? I'm a bit lost, but I interpret your answers as:
- a roof should be designed to handle 250kg snow load/sqm
- a floor joist should be designed to handle 200kg/sqm
Is that correctly understood by me?

 

[hakaner]

The numerical example in post #2 provides the failure load, right?
- For floor structures, isn't the allowable load usually limited by the deflection?
=> Maximum allowable deflection determines the allowable load... (often)

Use the program "dimensionering" at:
http://www.byggbeskrivningar.se/

 

[huggan]

No, it doesn't. I made a small mistake there. I'm talking ceiling, he's talking floor. That's correct.

 

[hempularen]

Interesting calculation. Now I'm on very thin ice, but in my opinion, 810kg sounds like a low failure load. Is that really correct? I would have guessed that you can load such a short span with considerably more before it breaks.

 

[menhir]

I wasn't aware of the program Dimensionering on byggbeskrivningar.se before. That was a great tip! It's just a pity that you can't calculate with unspecified timber, but understandable since there isn't a definitive value for such timber that can be guaranteed.

 

[Ingenjören]

The bending moment at the midpoint of the load P is M=PL/2, where L is the length.

The bending stress is S=M/W, where the section modulus W=bh^2/6. b and h are the width and height of the beam.

Hence, S=3PL/bh^2. P in Newtons and all dimensions in millimeters, you get the bending stress in Megapascals (MPa). S should be less than the bending strength of the beam.

S = 24
L = becomes 2150 mm
b = 45 mm
h = 220 mm

24=(3*P*2150)/(45*220^2)
24=(3*P*2150)/2178000
24*2178000=3*P*2150
52272000/2150=3*P
24313/3=P
8104=P

Thus, a beam can handle a load of 810kg in the middle with a span of 2150 mm. If you want to know how much it can handle per m^2, you just multiply 810 by the number of beams and then divide by the square meterage of the roof. If you have cc 600 mm, the roof can handle a load of 690 kg per sqm.

I just want to comment that the moment of a point load in the middle of a simply supported beam is NOT PL/2 but PL/4. The permissible design bending stress of K24 (C24) indoors in safety class 3 is 12.4MPa. This gives a permissible point load of 8.4kN = 840kg as the main load, which corresponds to a usual load of about 650kg.

Looking at what it can be loaded with before it breaks (no safety factors on either the load or material), it ends up at 1600kg.

However, do not attempt to load an unbraced beam with this load as it will buck much earlier...

As someone mentioned, it's usually relatively uninteresting how much you can load a floor beam since sway and deflection normally dimension simply supported floor beams.

/Engineer

 

[ronolo]

Thank you so much for the help. The rule in question is one of the floor joists in a floor where I want to place a stove. If I interpret the calculations correctly, it means that I can easily place a stove weighing 1000 kg if I can distribute the weight on 2 of the joists? Cross bracing and floor chipboard should probably distribute the weight a bit more.

Does my interpretation of your calculations seem correct?

 

[Forbo]

Thank you very much for your help. The beam in question is one of the floor joists in a floor where I want to place a stove. If I interpret the calculations correctly, it means that I can easily place a stove weighing 1000 kg if I can distribute the weight on 2 of the joists? Cross-bracing and floor chipboard should distribute the weight a little further.

Is my interpretation of your calculations correct?

Someone else will have to answer that... but remember that the floor chipboards can't handle just anything... if you install a stove that is small/narrow enough to fit between the joists, etc... the chipboards can handle "only" 2-300 kg...