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I've tried using various calculators but that has just confused me even more so I'll describe my scenario and hopefully I can get some answers. In the picture above, each item is described below

A: Timber Loft Joist, Width 3.5m, Height 225mm, Depth 45mm. Either end of each joist sits on top of a block wall 100mm thick and 2.5m in height. From this link https://www.engineeringtoolbox.com/floor-joist-capacity-d_1832.html, it says that these joists can support 3kN/m2.

B: OSB board, 18mm thick

C: Mild Steel plate, Width 400mm, Length 400mm, Depth ?mm

D: Load, 120x120mm, Weight 150kg

With my very limited engineering knowledge, I'm assuming that I need C to spread the 150kg load from D so that it doesn't punch a hole through B. If this is the correct approach, what would be an adequate thickness for C to accomplish this. Any thoughts or suggestions would be greatly appreciated as it'll help me with knowing the viability of my project.

chillydk147
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    Your sketch needs more detail as to what is the attachment to the plate. is it going to shake in an earthquake? Otherwise, any plate thicker than 3mm would do. also, you need to put wood blocks under the load between the joists. if you need the calculations, use $m=150*200$ and $\sigma= 150*200/((400*4^2)/6)$. a safety factor of 2.8 is ok. – kamran Feb 09 '23 at 21:09
  • Thanks Karman, there wouldn't be any attachment to the plate and we don't have earthquakes here in Ireland just perpetual rain. Ok so I'm assuming a wooden block will help to ease the load on the 2 joists. I don't understand some of the values in the formula you provided, like where have the 200 and 4 squared come from? – chillydk147 Feb 10 '23 at 09:46

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This can be handled like a beam resting on two pin supports, with good approximation. A plate can support a bit more than a beam. I suppose that the OSB alone, along the major axis, would be strong enough to support this load but just to be on the safe side.

let's pick a 4mm thick plate with:

  • M = moment kg.mm
  • S = beam section modulus =$bh^2/6$
  • h = plate thickness, 4mm

$$ M=Pl/4=150kg∗400mm/4=15000kgmm$$ $$\sigma= M/S=15000/ \frac{bh^2}{6}=15000/ (\frac{400*4^2}{6})=15000/1066=14.06kg/mm^2<25kg/mm^2 \text{ allow stress of mild steel}$$

So the safety factor is $ 1.75 >1.6 \ $and is okay.

In my comment, I had recommended SF of 2.8 but with your adding more info 1.6 is okay.

The blocking under the plate between joists is for joists' lateral support and is required by most codes.

kamran
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  • Thanks Karman, these calculations put my mind at ease so I will now go for a 4mm mild steel plate option. To give a little more context as to what the load represents. The load represents 1 of 8 legs of a 12x6ft snooker table with a maximum of 2 legs spread across any given joist – chillydk147 Feb 11 '23 at 10:17
  • @chillydk147, it is prudent to check the overall span of the joists and their strength too. – kamran Feb 11 '23 at 14:44
  • Totally agree, the house is 9.5m wide with a central solid block corridor 2m wide. There are 3 joists, 2 are 3.5m which connect with a smaller joist that sits on top of the corridor walls. My plan is to place the table legs as close to the corridor walls and as close to the joists underneath, I know my original picture shows the leg in between 2 joists but I'd preferably have as many legs close to being above a joist – chillydk147 Feb 11 '23 at 21:03