# How much load can a (wood) board support, if it is supported only at the ends?

How much load can a board support, if it is supported only at the ends?

At the moment my specific problem is hanging a tire swing between two trees, and depending on which trees I choose they could be 10, 12, or 14 feet apart (center to center). If I securely attach a 2x4 to the trees and hang the swing from the middle, would it be at risk of breaking from 2 children playing on it (so I'd guess around 300lbs max with them bouncing and jumping onto it)

But my bigger question is the general case: how do you calculate (or where do you look up) what kind of loads can be supported by the wood in question? (other examples would be a 1x10 shelf with supports 24" apart, or 1/2" plywood on a 2' x 4' frame). I'm not an engineer (well, software, but that doesn't count here), but I can do simple math (linear algebra, trig, calculus) and have a basic understanding of physics.

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Thanks for all the great answers and crash course in materials engineering! –  evil otto Aug 8 '11 at 4:39
About as much as a woodchuck could if a woodchuck could chuck wood (come on, you know you were thinking it!!) ;) –  diyaddict Sep 22 '14 at 6:39

I'd go with @Aarthi's load bearing table resource for a general idea of what's reasonable.

Beam Deflection Formulas

Beam Deflection and Stress Calculator

Area Moments of Inertia

Using the Parallel Axis Theorem

Wood Material Properties (Modulus of Elasticity (E) found in Table 4-3a)

For the dynamic loading, you'll want to do something similar to the fun I had on this question.

...and you may want to consult a good Mechanics of Materials book. (cheaper paperback international edition on Ebay)

As @Ian points out, the problem is not a simple one and is best solved by simply using what's worked for other people in the past. Go take a look at the swings at your local park and use the same size of beam, provided the span is comparable.

Also, if you're really worried, you could always make the rope into a 'Y' to eliminate bending stress on the beam, leaving it solely in shear. This way, the beam is bearing the compression load from the lateral tension on the 'Y', which will keep the trees from bowing toward each other.

Diagram:

|      |______________________|      |
|      |  |                 | |      |
| tree |  |                 | | tree |
|      |__|_________________|_|      |
|      |   \               /  |      |
|      |    \             /   |      |
|      |     \           /    |      |
|      |      \         /     |      |
|      |       \       /      |      |
|      |        \     /       |      |
|      |         \   /        |      |
|      |          \ /         |      |
|      |           Y          |      |
|      |           |          |      |
|      |           |          |      |
|      |           |          |      |
...more rope and trees...
|      |           |          |      |
|      |           |          |      |
|      |         -----        |      |
|      |       /  ___  \      |      |
|      |      |  /   \  |     |      |
|      |      |  \___/  |     |      |
|      |       \       /      |      |
|      |         -----        |      |

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I like the Y solution. It puts all the vertical load on the trees –  Chris Cudmore Aug 5 '11 at 20:52
That Y mounting is an awesome idea. It will also make the actual hanging easier, since it turns out to be rather difficult to reach the middle of a 10-foot span from a straight ladder against a tree. –  evil otto Aug 5 '11 at 20:53
This is a fundamental of anchor building in rock climbing. To be safe we require the angle in the Y to be < 90 degrees, otherwise the attachment points are being pulled inward too much, instead of mostly downward (a load direction that rock climbing bolts aren't set for). The same physics applies here. –  Dave Aug 11 '12 at 15:10
Dave, that's a very good point if you're going to anchor straight into the tree with a lag bolt. However, by using a board between trees, the board bears the compressive force, and the bolts holding the board to the tree are still in shear. Minimizing the angle of the "Y" will still reduce the compressive force on the board, but it's not as critical this way. –  Doresoom Aug 15 '12 at 18:13
ms paint and photoshop are a beautiful thing... try it sometime to draw a picture. –  diyaddict Sep 22 '14 at 7:14

"...the short answer is, don't use the 2 x 4 for your swing..."

Otis' question was posted long ago, but I thought this may help other people researching similar structural questions. I'm not an engineer, but I've worked with wood for 30+ years.

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In the real world I would definitely go with the example of the commercially designed swing set. Its design deals with static and dynamic factors, it shows a tried situation. All it needs is how to use a tape measure.

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This might be more appropriate as a comment since it doesn't really answer the question directly.. –  Steven Feb 6 '13 at 18:11

Lots of interesting answers for how to figure out the "correct" answer, but hopefully this will help a bit.

We purchased a commercial play set similar to this.

To cover a 12' span, they use three 2x6" beams laminated together - glue, nails and finally carriage bolts.

This is to support 2 swings and a set of rings.

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There is an online calculator called the Sagulator that estimates the deflection in a shelf given its dimensions, the type of wood, and the load.

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Keep in mind that your load is not static, but dynamic, and the stresses will multiply during the movement of the swing. Also, the movement of the swing will apply stresses against the short dimension of the beam, which it was never intended to support. Looking at the swing sets available at building centers, I've never seen a main support beam smaller than 4 x 6.

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I was thinking a bit of the dynamic/load-multiplying stresses, but the other comment that #300 is too low is probably correct. I could obviously use bigger pieces, but those are correspondingly heavier and more expensive. I'm not saying I'm trying to do things on the cheap, but I want to know what my trade-offs are. –  evil otto Aug 5 '11 at 19:39

This is a fairly complex problem to answer starting from scratch as it has multiple components so I will just summarise the calculations that will need to be done.

In terms of the stresses in the plank, typically as a minimum you will need to calculate the following forces:

• Bending moments
• Shear forces
• Bearing stresses
• Deflections

These will need to be calculated for different load cases including different locations for the weights on the beam as different positions for the load will give different worst case results. The method of calculation for the stresses will vary depending on the structural details you adopt at the supports but in the case you describe will probably be based on what is known as a simply supported beam.

Having calculated the forces in the beam, you will need to calculate some geometric properties of the beam in order to calculate the stresses. Typical geometric properties will be Second Moment of Area (for bending moments), Shear Area (for shear force), and Bearing Area (for bearing stresses). Again the calculation of these properties will depend on the detailing you choose, as will the use of these properties to calculate the stresses.

The final calculations you will have to do will be the stresses that the wood can withstand. Again, this is somewhat complex because wood, being an organic material, has different strengths in different loading conditions with factors such as grain direction, load type, load duration, wood type, etc. all affecting the calculation. You will also need to include an appropriate factor of safety in the calculations.

Having said all of that, this is over the top for most domestic applications and for the most part basing the size on what has worked in similar circumstances before is normally adequate.

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You can't really use simply supported for this model though. It's more realistically fixed-fixed, which is indeterminate for simple beam equations. –  Doresoom Aug 5 '11 at 19:41
This gives me some great food for thought and things to look for in other references for the actual calculations. Going off what has worked in similar circumstances is a good rule of thumb, but it requires you to have similar circumstances to compare to, which I don't. –  evil otto Aug 5 '11 at 19:42
@Ian Turner +1 Very! good answer. –  Mike Perry Aug 5 '11 at 20:53
The nice thing about using simply supported beam for a simple plank is that the bending forces and deflections will be maximum values. If you use a fixed-fixed model there will be some worse bending moments at the ends but they won't be any greater than the mid-span moment for a simply supported case, but that is fine if your beam is the same thickness all along. –  Ian Turner Aug 5 '11 at 20:59
@evil otto, the simply supported beam equations and the fixed fixed beam equations are actually really simple and you won't need to use differential equations. raeng.org.uk/education/diploma/maths/pdf/exemplars_engineering/… might be a good start. –  Ian Turner Aug 5 '11 at 21:02

I'm not a hundred percent sure this answers your question, but I will say this much: 300lbs is actually much, MUCH too low of an estimate, if all of my trolling on this site is anything to go by. Also note that it's not weight but force (ie Newtons) that you need to be looking at.

Second, this document should answer your load-bearing questions. It's kind of technical, though, from what I can tell.

Finally, here is a similar question from DIYChatroom.com.

Wood bears a toleration of roughly 625 pounds per square inch (PSI) of a compression load. Concrete can bear 3,000 PSI of a compression load. Steel can bear 30,000 PSI of a compression load.

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That document looks like a really good reference, thanks! Also, weight (e.g., pounds) IS force. We just think of it as mass, being earth-centric and all. –  evil otto Aug 5 '11 at 19:41
A "pound" is either a unit of mass or a unit of force, depending on who you ask. If you want to consider it a unit of mass, the acceleration portion of the force equation (f=m*a) is pretty obvious... it's gravity... problem solved. 300 lbs is indeed too low, but there was a qualifier: "bouncing and jumping." If we assume that "bouncing and jumping" increases the forces two-fold, then that seems pretty reasonable. I'd also include a safety factor of 2 or so in there, meaning that the swing should be able to hold 1,200 lbs static load. –  Michael Aug 6 '11 at 10:38
Also, material strength is tricky business. How much compression (push together) strength does rope have? None. But it has plenty of tensile (pull apart) strength. Concrete is the opposite -- it has great compression strength, but relatively poor tensile strength, which is why it needs reinforcement from rebar, high-tensile strength fibers, etc. Between figuring out material strength, lever arm lengths, accounting for the fact that the swing won't always be straight down, etc., the load calculations on the swing are quite a bit more difficult than one might initially imagine. –  Michael Aug 6 '11 at 10:52
@Michael - Not sure if you're saying the same thing, but the #300 was may idea of a safety margin - the static weight would be around #100, so I doubled that for the bouncing and swinging and added 50% margin. –  evil otto Aug 8 '11 at 4:38
@warren - they can build their own swing then!! –  evil otto Aug 10 '11 at 4:21