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...
| | | | |
| | | | |
| | ----- | |
| | / ___ \ | |
| | | / \ | | |
| | | \___/ | | |
| | \ / | |
| | ----- | |
Best Answer
2x6 DF #2 Supported o the ends can handle about 11 lb. per Ln. Ft. total load. This would include snow. would recommend at least 2x8 if you are okay with sagging from wood creep after many years. 4x8, (2) 2x8, or 2x10 would be better.