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I am thinking of constructing a wooden bed/desk/roof/shed/ship/bridge/rocket

  • How can I calculate the maximum load a rectangular horizontal wooden beam of dimensions L x H x W can safely support if the beam is adequately supported at both ends?
    • assuming worst case - load concentrated at center
    • for commonly available types of wood (e.g. Spruce)
  • 2
    Damnit, now I want to build a bed/desk/roof/shed/ship/bridge/rocket. – Comintern Oct 14 '15 at 4:07
7

There are lots of span calculators available online, which help you determine what size lumber to use in home or deck construction. For example

http://www.awc.org/codes-standards/calculators-software/reversecalc

and

http://www.awc.org/codes-standards/calculators-software/spancalc

You could try to figure out what the live loads and dead loads for the bed are and go from there.

Shortcut - I might try to get by with 2x4's spaced 2' or less apart if they run side to side across a single mattress, but I'd want 2x6's if they run long ways, or for a full or larger.

  • Great resources! – AndyT Oct 14 '15 at 9:08
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    These are excellent resources (+1) for uniformly loaded beams, especially for floors and roofs. I am waiting to see if anyone also finds something for loads concentrated at a point. – RedGrittyBrick Oct 15 '15 at 13:38
3

Here are two documents I've found helpful, giving specs for southern yellow pine, which is the wood typically used in treated lumber for its added strength compared to SPF pines.

  • These documents are for uniformly loaded lumber in a repetitive installation. Op is seeking “maximum load”. – Lee Sam Nov 15 '17 at 7:44
2

Your question referred to a simple central load. So the formula here seems useful:

https://en.wikipedia.org/wiki/Flexural_strength

Looking up the bending strength of Spruce here:

http://workshopcompanion.com/KnowHow/Design/Nature_of_Wood/3_Wood_Strength/3_Wood_Strength.htm

We get 10,200 psi.

Assuming a 6ft length of 4x4 that actually measures 3.5 inches square, and plugging these numbers into the formula, we get:

10,200 psi * (2 * 3.5in * 3.5in^2) / (3 * 72in) = 4049.306 pounds

This appears to be the point at which your beam will deform.

  • 4
    @ LRU I think you’ve calculated the load at which point the beam will “yield” or “fail” in bending. You need to factor a “safety factor” into your calculations. I use working stress, not ultimate strength. Usually shear governs for short spans, and bending governs on longer spans. I get about 1,000 lbs. (not 4,000 lbs.) before horizontal shear failure. – Lee Sam Nov 15 '17 at 6:47
  • any references in metric system? – arthur Aug 24 '18 at 8:55

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