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Can't build house so making shepherd's hut / site hut


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Last chance for me to change my mind on the timbers I'm looking to use for the base. The trailer has two rails that go the full length of the trailer and are 1270mm apart. The width of the hut is 2200mm or 2400mm, not decided yet. If I presume 2400mm, that means the hut overhangs the sides of the rails by 565mm on each side. Do you think 4x2 timber will be strong enough for that cantilever? They are on ~400mm centres and the panels and weight of the hut itself is likely to be equivalent to less than 50kg bearing down on the end of the 4x2. Adding that someone could stand against the wall and that could add as much as another 100kg. Does 150kg on the end of a 4x2 cantilever 600mm long sound OK?

 

There is lots of safety margin in that as the person's weight wouldn't be solely supported by the one beam and the walls etc will be much less than 50kg (the panels way <20kg per 400mm).

 

 

 

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image.png.80e75036d746afff1e24091810662ed5.png

 

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How well are the front wheels fixed to the body?  I remember some old farm trailers (a rulley was the Yorkshire name for them when I used to drag them around the Wolds) being a bit unstable when cornering; I suspect you won't be moving it far, but you might want to consider the extra leverage effect on stability as well as deflection.

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15 minutes ago, Roundtuit said:

How well are the front wheels fixed to the body?  I remember some old farm trailers (a rulley was the Yorkshire name for them when I used to drag them around the Wolds) being a bit unstable when cornering; I suspect you won't be moving it far, but you might want to consider the extra leverage effect on stability as well as deflection.

Good thought, it will be pretty top heavy

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HEALTH WARNING: AMATEUR AT WORK!!

 

Right so I am going to have a go at playing Structural Engineer. I've found a copy of BS5268 - Structural use of timber. Table 8 of this has "Grade stresses and moduli of elasticity for various strength classes" and lists:

  • C16 - bending parallel to grain 5.3N/mm2 and minimum Y=5800 N/mm2
  • C24 - bending parallel to grain 7.5N/mm2 and minimum Y=7200 N/mm2

 

My cantilever length is L ~ 600mm

My maximum moment with 150kg at ~600mm is M = 150kg*9.81N/kg*600mm = 9E+5 Nmm

A beam's second moment of inertia is I = b*h^3 / 12   (where b with its cross section width and h its cross section height)

The resulting maximum deflection is d_max = M*L^2 / 2*Y*I

The bending stress is sigma = M*y / I where y is the distance to the neutral axis so h/2, so sigma = M*h / 2*I

For C24, Y = 7200 N/mm2

 

4x2

h = 100mm, b = 47mm

Second moment of inertia: I = b*h^3 / 12 = 47*100^3 / 12 = 4E+6 mm4

The maximum deflection: d_max = M*L^2 / 2*Y*I = 9E+5 * 600^2 / (2 * 7200 * 4E+6) = 5.6mm.

Bending stress: sigma= 9E+5 * 100 / 2 * 4E+6 = 11.25N/mm2 > 7.5N/mm2   SNAP!

 

5x2

h = 125mm, b = 47mm

Second moment of inertia: I = 7.6E+6 mm4

The maximum deflection: d_max = 3.0mm

Bending stress: sigma = 7.4N/mm2 < 7.5N/mm2   OK

 

However, that is based on permanent loading, for short term loading you can multiply the permissible stress by a loading duration factor, K_3 = 1.5 (from Table 17 of BS5268). That raises the limit to 7.5 * 1.5 = 11.25 N/mm2 which makes the 4x2 OK (just). There are other factors (like the depth factor) which I could make fractionally greater than 1. My scenario is very pessimistic and in reality the 100kg person wouldn't be out at 600mm and their weight wouldn't be carried solely by one beam.

 

I need to think carefully about how I secure the beams to the chassis so as not to weaken them. I'm thinking like this, but possibly only 2 screws holding the blocks on so that the screws are closer to the neutral axis.

 

image.png.10f3b716cd9e8aa77f86916f727ed44b.png

 

Do these calculations look correct?

 

HEALTH WARNING: AMATEUR AT WORK!!

 

Cantilever Compare

http://www.learneasy.info/MDME/MEMmods/MEM09155A-CAE/020-Compare/Compare_to_formulas.html

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One thing you have not mentioned yet is the ring beam around the perimiter.  This joins all the joist ends together.  So it has the effect of spreading the load on one beam to adjacent beams.  So someone jumping on the floor at the edge, the load won't be placed on one beam, but shared by several.  Likewise a high static load on one will be shared by adjacent beams.

 

Our static caravan which is much the same construction, though with a shorter cantilever is made of timbers that look more like 3" by 2"  I think it has something like 2" steel angle around the edge as a "ring beam"

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ProDave, you're right I'm sure it would make a big difference and lower the maximum stress considerably. I did mention "their weight wouldn't be carried solely by one beam", but that perhaps understates it. I'm happy that C24 4x2 would be sufficiently strong, but wanted to crunch some worst case numbers to get some confidence. I may double up  a couple of the beams at the doorway though.

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7 minutes ago, SteamyTea said:

Could you add another 4 by 2 below the chassis and join them up with ply.  In effect make a large box section.  You would have to get creative around the chassis.

It would also be somewhere to stuff full of insulation.

 

I think I understand what you are suggesting, effectively making a timber I-beam. From the calculations I am content that it will be stiff enough without that.

 

There will be a fair amount of cold bridging, but the plan is to fill the gaps between the 4x2 noggins with 100mm Celotex then with 18mm CDX ply on top and 9mm CDX ply beneath. If I make it too warm then a log burner would turn it in to an oven. Probably already the case actually as not many shepherd huts will have a 100mm PIR insulated envelope. That's better insulated than we are planning for the house.

Edited by MortarThePoint
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41 minutes ago, MortarThePoint said:

effectively making a timber I-beam

Well a box.

42 minutes ago, MortarThePoint said:

If I make it too warm then a log burner would turn it in to an oven

Pop some solar panels in the roof, then some elements in the floor.

Any combustion technology is frowned on.

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Actually looking again at the static 'van, I see there are steel cross members periodically.  So imagine say every 5th wooden joist being replaced with a same size C section steel beam.  That plus the ring beam gives additional support to the walls at the edges.

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53 minutes ago, Jeremy Harris said:

Would an unlocked copy of BeamCalc help?  It's an Excel spreadsheet that does simple beam calculations quickly and easily, for both point and distributed loads. 

 

That was the first program that every architect I knew wrote on their BBC Micro ?.

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3 minutes ago, Ferdinand said:

 

That was the first program that every architect I knew wrote on their BBC Micro ?.

 

 

Still as useful now as it's ever been, and the Excel version is fairly easy to use, and pretty comprehensive, in that it will handle about a dozen different beam configurations, as well as a handful of different cross sections.

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1 hour ago, Jeremy Harris said:

Would an unlocked copy of BeamCalc help?  It's an Excel spreadsheet that does simple beam calculations quickly and easily, for both point and distributed loads. 


is that the one where the unlock key is hidden behind the logo on the page ..??? ?

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1 hour ago, PeterW said:


is that the one where the unlock key is hidden behind the logo on the page ..??? ?

 

 

I can't remember, TBH, as I unlocked it years ago.  I have a feeling it may have been something as simple as that, as it didn't take long to get it to work, IIRC.  The neat thing about BeamCalc is that will handle pretty much any type of simply supported beam, either supported at two points or cantilevered, with point loads or distributed loads.  It works with rectangular, circular, elliptical or I beams, either solid or hollow, so covers pretty much any fairly simple case.  It also contains material properties for lots of stuff, metals, different types of timber, glass and carbon fibre reinforced composites, plastics etc.

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I've realised the figures for the deflection were wrong as I should have used:

    d_max = W*L^3 / 3*Y*I            []NB: I used Y for Young's modulus but table uses E]

as set out on the second line of the BEAM BENDING table. I didn't read the table properly as I assumed M=W*L applied to the deflection as well, but it doesn't ( M = W*(L - x) ).

 

That changes the deflections figures to:

  • 4x2 : d_max = W*L^3 / 3*Y*I = (150*9.81) * 600^3 / (3 * 7200 * 4E+6) = 3.7mm
  • 5x2 d_max = W*L^3 / 3*Y*I = (150*9.81) * 600^3 / (3 * 7200 * 7.6E+6) = 2.0mm
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  • 1 year later...

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