steel beam

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My SE is saying that we need a column to support a beam - as the length (7.5meters) is too long without having one). the beam will hold up celing joists for 2 rooms and be 'load bearing' under a purlin line - to stop sag..

The load should be evenly distributed along most of the steel..

The builder is saying he has fitted up to 8 meters without a column - and we would just need a bigger steel.

I have asked the SE - for no columns - he has 'repositioned a coloumn' .. it was in the middle of a bathroom - but is now near a wall.. we could create a built in cupboard to 'hide' the column.. its about 60cms out from the wall.. I'm not sure if this would look nice.. or be a bit odd..

Any ideas if a longer steel - would work - is it worth asking the SE again (I dont hold much hope) or finding another SE ..

is there a rule of thumb for the lenght of a beam?


TIA
 
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yes masses of room above -its in the loft cavity.. would a 400mm beam cost more than a smaller beam and a column?
 
And it it will do that without sticking out of the roof? What is it holding up then?

You will need to price up the beam and compare it. Don't forget to factor in the crane hire at about £800 or so, and the extra builder's charges. But your builder would know about that presumably
 
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Why is the engineer reluctant to do away with the column? If you don't know you need him to clarify it.
 
He's a bit old school and I think he's erring on the side of caution..

He just said the beam would be too big and too heavy.. and then found somewhere the column would fit..
 
cheapest thing at this stage would be to ask another SE. I went through 4 before I found a guy who would even accept speccing a steel for a 4m span on a wall 400mm thick. I now have a 203 and a 305 bolted together side by side. Incidentally, the opening was supported by just 3 pairs of acrows and 3 beams 127x76, on their sides, when the hole was being cut, so you could say that erring on the side of caution is de rigeur for anyone with PI insurance
 
I must be a bit thick, you have a beam 7.5m long which is "unsafe", but with a column .6m from the end, so beam is 6.9 long its safe. So that is less then 10% change in length. So you need a beam that is 10% stiffer, which is about 3% taller/thicker. . . I would point this out to your SE. Is there some other reason for the column?
Frank
 
Deflection might be the issue here. Even a small reduction in span can allow a considerable reduction of Ixx., resulting in a smaller or lighter section.
If the span is 10% less, the Ixx can be reduce by a factor of at least 1.3 and maybe more with the reduced load, which gives a considerable saving.
 
δ = FL^3 ∕ 48EI This gives a deflection of 6.9/7.5^3 ~ .778. But as E ~h^4, so if you increase the height of the beam by 1/.778^4 this would get you back to the original deflection. Which is 6%. Sorry about the font :( Blow me down its sorted its self??? (http://www.clag.org.uk/beam.html)
Frank
 
δ = FL^3 ∕ 48EI This gives a deflection of 6.9/7.5^3 ~ .778. But as E ~h^4, so if you increase the height of the beam by 1/.778^4 this would get you back to the original deflection. Which is 6%. Sorry about the font :( Blow me down its sorted its self??? (http://www.clag.org.uk/beam.html)
Frank
Your deflection equation seems to assume a central point load, but we don't know the loading condition?
We don't even know F?
And how is E related to h?
Sorry, but am I missing something?
 
δ = FL^3 ∕ 48EI This gives a deflection of 6.9/7.5^3 ~ .778. But as E ~h^4, so if you increase the height of the beam by 1/.778^4 this would get you back to the original deflection. Which is 6%. Sorry about the font :( Blow me down its sorted its self??? (http://www.clag.org.uk/beam.html)
Frank
Your deflection equation seems to assume a central point load, but we don't know the loading condition?
We don't even know F?
And how is E related to h?
Sorry, but am I missing something?
Tony, you are as confused as I am...

The flexural stiffness of a beam is EI (in MN/mm²). As far as I can work out a beam of 7.5m is 1.28 times less stiff than a beam of 6.9m (7.5³ / 6.9³). Or 1/1.28 = 0.778.

Frank, 0.778 is correct but only comes into the calculation in the case of a UDL if exactly the same load is applied. Obviously less load will be applied to the beam if a UDL simply because the beam is shorter (same load per metre, less load in total).
If a point load the 0.778 factor only comes into play if the point load is proportionally the same distance along the beam. Obviously it won't be as the beam is shorter (same load, closer to end of beam).

If we assume the load is a UDL then the factor is 0.716 so 28.4% less deflection, so for every 10mm of deflection of the 7.5m beam, the 6.9m beam will deflect 7.16mm. Probably similar for the point load.

*EDIT* I do see what you are saying though, that if the beam is increased in height slightly the deflection will reduce. However I think the formula for this is (7.5³/6.9³)^0.3333 or simply 7.5/6.9.
So (and this only works for solid sections) if the section is 7.5/6.9 = 1.08 times deeper, the deeper 7.5m beam would deflect the same amount as the shallower 6.9m beam.

Does anyone follow this?! :confused:
 
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As I see it I is the moment of inertia or in English, a fiddle factor that gives a "stiffness" of a beam depending on its cross sectional shape, H,I, rectangular etc. So to find I, you put in the height of the beam, the cross section of all the bits etc. One constant is the height of the cross section which is always h^4. This means that for a beam if you increase its height by 2, you increase its stiffness by a factor of 16. or put it another way, you decrease its deflection by 16. Try a long plank, flat or on edge. When its on edge it always fails by twisting, this is why the beams we use are H section the horizontal bits are to stop the vertical bit twisting. However I digress, I was just thinking that a very marginal increase in the beam dimensions would accomadate the increase in length from 6.9 to 7.5m.
Frank
 

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