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

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




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
  • 3
    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

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

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


Looking up the bending strength of Spruce here:


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.

  • 5
    @ 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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