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How much load can a (wood) board support, if it is supported only at the ends? has already been asked, but I found the answers unsatisfactory because they focused mostly on the asker's particular setup and barely mentioned anything about the properties of the wood. One answer references PSI which seems completely irrelevant to me. Another answer references the Sagulator, but the Sagulator does not help you determine the breaking weight.

In this answer, @Ecnerwal tells how much weight the asker's particular boards can hold, but doesn't mention how he calculated that or includes references.

This is an important generalize-able question

without a satisfactory answer on this site and without a "layman's" answer on the internet. It has been asked twice on here already and seems like a basic tenant of building anything that's more than decorational.

How do I calculate the force (weight) a board can reasonably withstand across its grain?

The answer should include:

  1. An algebra-level equation
  2. A lookup table for the relevant properties of different species of wood
  3. I should only have to come up with the pounds (lbs) or kilograms(kg) of my force. I do not know how to measure the kPa that my lawn mower exerts.
  4. Any relevant notes

In @Doresdoom's answer, he includes a link to Mechanical Properties of Wood and says the relevant variable is the Modulus of Elasticity (E) found in Table 4-3a. This PDF could be used to satisfy bullet point 2 of my question or not; I don't know if it is helful, hence why I'm asking this question.

I am not asking about the compression PSI a board can withstand, the force with the grain, the pulling tension, etc. I know this question can get complicated fast, but we're dealing with what's called a simple beam. This should be as simple as possible for all the homeowner/backyard engineers out there.

Example: There is a 10' wide moat. I lay a nominal 2" x 12" x 12' (so 1.5" x 11.5" x 12') across. The board is Shortleaf Pine. It is not fixed to either side, just laying there. My soldiers line up in order of weight and cross one at a time. They tiptoe and move as smoothly as possible so they can be considered a static load (don't go off-topic here). Once in the very middle, the soldier represents a worst case scenario. How much will the soldier weigh that will break the board?

Wood obviously has variations, but so does rope and they somehow come up with strength measurements for that. The key here is to generalize. Obviously, I could get a terrible board so I wouldn't really base someone's life off of these calculations.

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Zach Mierzejewski
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  • After learning more about this type of problem, I think it is necessary to clarify that soldiers feet will exert a uniform force over 12" x 12" area. And also, PSI *does* seem to be related. – Zach Mierzejewski Jun 25 '15 at 18:33
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    Breaking force =/= the load that can be supported. Strictly speaking, it's the load that can't be supported. Safe loads involve safety factors, which are important with a variable natural material like wood. And tension matters greatly in what loads can be supported (it's beam mechanics - everything below the centerline is in tension, and that's usually where an overloaded wood beam fails first.) In short, you are asking for a short-answer to a question that requires a textbook. – Ecnerwal Jun 25 '15 at 21:38
  • @Ecnerwal Fair point about the safety factors. You're supposed to do the same thing with rope. I've been poring over a few Statics books and beam PDFs for several hours so I realize it takes a lot, but more than half that stuff is not relevant to a home owner. A home owner is never going to create an angled, cantilever beam shelf. The only two applicable versions of a beam to home owners is supported on two sides and either floating or fixed with the load centered (as a worst case scenario). I'm only asking for one chapter of the giant textbook! – Zach Mierzejewski Jun 25 '15 at 22:15
  • [kPa](https://en.wikipedia.org/wiki/Pascal_%28unit%29) is pressure, which is what force applied against a static object creates. So I'm at a loss as to what you mean by "I should only have to come up with the pounds (lbs) or kilograms(kg) of my force". You link to the stress tables, so I'm guessing you'll get a better answer as to how to calculate kPa on [Mathematics SE](http://math.stackexchange.com/) or [Physics SE](http://physics.stackexchange.com/). – Comintern Jun 26 '15 at 02:07
  • @Comintern Yeah, I think this question is dead. I was trying to get an equation that your average Joe could put into a calculator. People can easily look up or measure the weight of objects. I wanted to include whatever conversions were necessary for them, such as weight to PSI (kPa). For the area part of PSI, my example was the soldiers feet are spread over a 144 in^2 section. – Zach Mierzejewski Jun 26 '15 at 17:40
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    Here, backyard engineers beam calc cheatsheet: http://faculty.arch.tamu.edu/media/cms_page_media/4198/NS8-2beamdiagrams.pdf The failure modes are usually either shear or moment. A third failure mode for code and comfort is deflection. You need to solve all three to figure out which failure mode "governs". – Damon Dec 17 '15 at 07:37
  • A lot of weight. Hope that helps you. Every piece of wood if different, figure that in. Knots on the wood, big lose of strength. You don't have to a rocket scientist to figure that out. Stand it on ends and you can hold a car. –  Feb 24 '16 at 02:40
  • [Amercan Wood Council](http://www.awc.org/) is a source for wood structure strength. – Fiasco Labs Feb 24 '16 at 06:03
  • I guess no answers because so much variables. What is your goal? To allow soldiers able to cross safely? If so, then breaking will happen long after bending. You specified "supported only at the ends", then qualified by saying "not fixed, just laying". In this case, the board will bend. If the board is suspended by an inch on either side, the board could bend so much that it - and the soldier - will fall in to the bottomless chasm below, yet will not break. If you stipulated that the board is anchored on both sides, now bending is irrelevant. All other variables must be factored. – Andrew Jay Aug 16 '17 at 16:36

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You need to calculate the bending moment and once you know the max bending moment you can fairly easily calculate if you have exceeded the bending design value of the beam. In most situations with wood beams bending strength will be the way the beam will fail so I am not worrying about shear for now. Shear is usually only an issue in short(in length) and tall beams. Here are some useful links. A beam bending moment/deflection/shear calculator

https://clearcalcs.com/freetools/beam-analysis/us

Wood reference design values

https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.cecalc.com/WoodDesign/Graphics/Wood_Reference_Design_Values.pdf&ved=2ahUKEwjmsK7Xt-X3AhWvgGoFHQsaAoUQFnoECBMQAQ&usg=AOvVaw0PyH-PX8EIqTykErrwB21n

And one more link that is the most useful in explaining how to go about your calculations

https://www.cedengineering.com › ...PDF Advanced Wood Beam Design - CED Engineering

Also 2nd moment of area(moment of inertia) formula for rectangular beam where beam will be loaded vertically is height cubed times width divided by twelve. Your units are all in inches and the unit for moment of inertia is in^4

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