I'd go with @Aarthi's load bearing table resource for a general idea of what's reasonable.
If you're looking for equations though, you can start with these:
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...
| | | | |
| | | | |
| | ----- | |
| | / ___ \ | |
| | | / \ | | |
| | | \___/ | | |
| | \ / | |
| | ----- | |
The maximum load you can place on a relatively slender column such as a table leg is usually dictated by the buckling strength, rather than the compressive strength. The critical load can be calculated using Euler's formula applied to a fixed-free column:
Pcr= (pi^2*EI)/(f*Le^2)
where E is the modulus of elasticity
I is the area moment of inertia (w^4/12 in this case)
Le is the effective length (2L in a fixed-free case)
f is the safety factor
Naively assuming that the entire load is uniformly distributed, the loadings on each leg are centric (perfectly rigid table top), and the loading on each leg is 1/6th of the total (six-legged tables are statically indeterminate), then each leg is subject to a loading of 654-818 Newtons.
Plugging in a length of 1 m, a safety factor of 10 (lots of assumptions here) and one of the following values of modulus of elasticity:
- 9 GPa (pine)
- 11 GPa (oak)
we get a minimum leg width dimension of:
- 3.9 cm (pine)
- 3.7 cm (oak)
I'd still stick with your initial plans for 6 cm square legs though. There are a lot of assumptions made for those calculations, including no lateral loading, no deflection of the table top, no knots or imperfections in the wood, and overlooking the fact that wood is anisotropic.
If you're working with a different type of wood, here's a good chart for modulus of elasticity (PDF) (don't forget to convert from MPSI to GPa!)
Best Answer
Short answer: it will sag a lot.
Using The Sagulator I get a deflection of about 0.9" at the center, which will clearly be noticeable. (I'm assuming your "5/4 maple" is actually going to end up 1" thick when the finishing is done.)
Also I'm not sure what you are going to be using the shelving for but if this is going to be for books you should probably assume 30 lbs per foot, which would give you a deflection of 1.3".
(You can use that calculator to estimate sag for a variety of shelving configurations, but make sure you pick accurate values for all the inputs. Also note the footnote that says that long-term sag will be 50% higher than the calculated value.)