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I'm new to woodworking and I'm in the planning phase of making a bench like the one pictured below. I'm planning on storing very heavy boxes on this and would also like to use it as a stool to reach high shelves.

Is there a standard practice in determining how much weight a bench can hold? Is there any quick estimation practices that usually give accurate results?

The bench will be made all out of Douglas-fir 2x4 studs, except for the four legs which will be Douglas-fir 4x4 posts. The only fasteners I plan on using are steel 2.5 in. common nails. How can I find the weight this bench can realistically hold?

enter image description here

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    Build the bench, then keep adding weight until it breaks. Then rebuild the bench. Commented Mar 24, 2019 at 13:41
  • Better build 2 identical benches at once - you will save on time and operations.
    – Kromster
    Commented Oct 28, 2019 at 12:09

5 Answers 5

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The 4x4 posts used for the legs support virtually "infinite" weight longitudinally. Now, of course, given a naive design, each leg would only transmit force onto one 4x2 beam (which are then maybe nailed together with some kind of ledge or so), which isn't optimal. You most definitely want to connect the beams with a dovetail batten or a similar construct close to the legs, if stability matters. Four 10mm carriage bolts that go through each leg and the dovetail batten are a good solution.
Glueing and nailing a thick batten onto the beams might also "work", but alltogether that's not such a great plan since it rigidly fixes the cross-piece which will put strain on the longitudinal bonding when the batten expands (and that's kind of inevitable).

Something like this (that's my garden table, it confirmedly supports 8 adult persons dancing on top of it without shaking):

dovetail batten on table

Also, you need to make sure that the bench is somewhat resilient against sheering. The cross-beam that is shown in the picture is already a good approach, but nails as the only fastener will not do. No way.
I would join every two distinct pieces with glue and two 10mm hardwood dovels as the very minimum. Preferrably though, mortise and tenon.

As for bearing capacity on the bearing/sitting surface, a single 4x2 douglasia beam that is 180cm long and supported at both ends easily sustains my body weight, so the easy answer on that account would be "no worries, this will do". However, you can do orders of magnitude better or worse here too, with only "small" modifications.

As pointed out by rob, you will ideally want to make a laminate bearing surface by glueing the 4x2 beams together. A small ledge in the middle as shown in the picture will work as the absolute minimum, but will not be nearly as good.
The reasoning behind this is that the weight is better distributed over all beams, not just on the ones the weight is on.

However, note that using the 4x2 beams in the way depicted in the image is not optimal for bearing capacity, and the horizontal laminate is sub-optimal from the point of view of stability.
The bearing capacity of a rectangular beam is proportional to 1/12 · h3b which means that a single 4x2 beam (upright) equals four 2x4 beams (flat).

4x2 = 4 * 2x4 enter image description here

You can therefore greatly increase (double or triple) overall stability and bearing capacity by simply laminating another one or two upright beams under the bearing surface. Or, you can laminate a thin ledge to one or several of the beams longitudinally, preferrably set back a bit so one can't see it.
For this, it is absolutely important that the beams are glued together carefully on the entire length.

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  • Very nice answer. In the last sentence, do you mean the upright support beams need to be glued to the flat ones above, or are you talking about the flat ones as a group? Commented Mar 21, 2015 at 15:20
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    @ChristopherCreutzig: Glueing the flat ones together helps, but not as much as glueing a supportive beam below. The most stable way is the "T" shape as depicted in the last image on the right side. Glueing 4 flat pieces together will give 4x the strength, but glueing a single support below will add the same (so both together would be 8x, and using 2 support beams would give 12x). Support beams must be glued on the entire length to be most useful since that works as (h1+h2)^3 rather than (h1^3)+(h2^3).
    – Damon
    Commented Mar 21, 2015 at 15:51
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    please provide evidence to support the claim that it "supports 8 adult persons dancing on top of it without shaking."
    – aaron
    Commented May 15, 2017 at 20:01
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When determining the load bearing capacity of any structure, you'll need to take into account several different factors:

  • Acceptable deflection
  • Maximum allowable bending stress
  • Column compressive and buckling strengths

For calculating deflection, @rob's suggestion of using the Sagulator is a good start. You don't need to have an in-depth grasp of all the engineering concepts going on behind the scenes to get a 90% answer. The Sagulator is simply inputting your furniture's dimensions into a beam deflection equation and spitting out the answer. These equations rely on the assumption the material remains within its elastic limit. You can test this in the Sagulator by inputting a 5,000 lb loading for a 1/2" thick pine shelf. The deflection shown will be greater than the length of the shelf, since the equation doesn't factor in the ultimate strength of the material.

Maximum bending stress allows you to calculate how much loading your furniture can take without breaking. The maximum load a horizontal beam can sustain depends on the span, how the load is applied, how the ends are supported, the material, and the cross-sectional geometry. Luckily, there are several online resources like Engineer's Edge and the Engineering Toolbox that can help with generalizations, so you don't have to derive each condition.

For a uniform loading on a beam with fixed-fixed end conditions (that means the joints are connected at each end by glue, or at least by more than one fastener), the maximum bending stress will be located at the ends, at a magnitude of

sigma = M*y/I

where

M is the maximum moment,
I is the cross section moment of area
y is the distance from the neutral axis. 

For a rectangular cross section, , and I = 1/12*b*h^3, where h is the beam thickness, and b is the beam width. The y is simply h/2.

You can calculate your maximum M with this calculator.

The maximum allowable stress (or Modulus of Rupture) for many wood species can be found by searching Google for wood material properties.

Don't forget your safety factor on this part!

Column compressive and buckling strengths won't really play into your design considerations unless you're planning on using long, spindly legs. With 4x4 legs, you won't have this problem. With thinner legs, you can calculate the critical centric loading (aligned with the axis of the leg, applied in the center of the leg) with the following formula:

F=(pi^2*EI)/(KL)^2

where

 E is the material modulus of elasticity
 I is the area moment of inertia of the leg cross section,
 K is the column effective length factor,
 L is the unsupported length

For unsupported legs attached only at a table apron or the top of your bench, K will be 2, since it's essentially a fixed-free end condition. For a legs connected with a brace at the bottom, K will be closer to 0.5, and column compressive strength is more of a factor.

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Just to find the load bearing capacity of the horizontal bench top, I'd treat it like a shelf and run the numbers through some sort of load-bearing calculator like the sagulator. The load-bearing capacity will also depend on whether the weight is mostly toward the center or if it is evenly distributed. Ideally you would laminate (thoroughly glue) several 2x4s together to make a strong solid panel for the top of the bench. In that case you can treat it as one solid board in your load calculator. If you don't laminate the top pieces together, I'd run calculations for a single 2x4 of the appropriate length, as well as multiplying that answer by 2 to 4, depending on how many of the 2x4s any item on the bench is likely to span.

Once you calculate the load-bearing capacity, the other factor to consider is racking--in this case, most likely side-to-side movement or wobbling. You mention that your only mechanical fasteners will be nails, but you don't mention whether you intend to use glue or joinery. The cleats that attach the top to the legs will help somewhat against racking, but gluing them in addition to nailing them can increase the racking resistance dramatically. Glued mortise and tenon joints at various locations on the bench would make it even stronger.

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  • When using sagulator be sure to first set the "Shelf attachment" button to "floating". Also realize that the failing values for loading are based on aesthetics (shelves and such lose visual appeal when the sag too much), rather than catastrophic failure (breaking).
    – Ast Pace
    Commented Feb 2, 2016 at 21:26
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You can always take the Calvin & Hobbes approach!

enter image description here

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When using the Sagulator, be sure to explicitly add the weight of the wood itself. This is not documented, and Sagulator feedback is disabled. However, one can confirm this by comparing a 1 pound total load to a 2 pound total load, on a 120' span made of 1" x 12" walnut. (This is surely beyond the elastic limit of walnut; we're simply debugging the tool here.) The computed sag doubles, when in fact the wood itself weights 400 pounds. Adding a pound or two should have an inconsequential effect; the effective sag will be due to the weight of the wood itself.

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    Welcome to WSE. This is a very helpful observation, but it does not actually answer the question. It could be better placed as a comment to another answer addressing the use of the sagulator.
    – Ashlar
    Commented Mar 26, 2019 at 0:48

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