Machinery's Handbook is confusing because it's just a reference, not a how-to -- it assumes the reader has studied structural engineering and just needs to look up the formulas and constants. But calculating deflection of a point load on a simple beam is extremely easy. There are only three things you need from the book: 1) the modulus of elasticity (known by the symbol "E") of the material, 2) the moment of intertia of the beam shape (symbol "I"), and 3) the formula to calculate the deflection for your configuration.
All of these are in the chapter "Strength of Materials". Look at the section with tables for "Steel Wide-Flange Sections". Wide-flange is the standard construction shape that most people colloquially call "I beams". I-beams are actually a different and mostly obsolete shape but still available, as someone else pointed out. Wide-flange beams are known by a designation like "W6x25", "W14x136", etc. The first number is the height and the second number is the weight per foot. There are typically around a dozen different weights for each height. The weight, and structural properties, vary because the width and the thickness of the flanges varies for each size within a given height. The Wide-Flange table in the Handbook lists the full dimensions of each size, along with the moment of inertia for each shape.
Moment of Inertia (I) is too complicated to explain and not necessary to understand beyond knowing that it is a number that reflects how rigid a given cross section is. For example, a solid round is more rigid than a tube of the same diameter, and has a different I (the specific value which of course varies, depending on diameter and wall thickness) For your purposes, you just need to find out the I value for the beams you might choose to use. These are in the Handbook tables.
Modulus of elasticity (E) is a very simple concept. It is merely a measure of how much a material stretches under load. Every linear elastic material has a constant E, and you just need to know what it is and plug it into the formulas. (A linear elastic material is one that will strech a given amount per given load. In other words if you applied n pounds of tension to a wire and it stretched x inches, and you apply 2n pounds and it stretches 2x, it is linear.) Steel is linear elastic, and its E value is 2.9E7. It varies slightly depending on alloy, but not enough to be relevant here.
Then there are tables titled "Stresses and Deflections in Beams", which has a series of drawings showing beam and load configurations. For example, a simple beam with a single load in the middle; a simple beam with a single load at some point that is not the middle; a beam with a uniforming distributed load; etc. The pictures are largely self-explanatory. Foe each configuration, the table lists formulas to calculate stress and deflection at any given point along the length of the beam, as well as handy shortcut formulas for common things, like maximum deflection at the center of a beam.
So, with I and E (which describe the material and the shape of the beam), plus your parameters of length and load, you have all the ingredients to do the calcs. Note that for a simple point load at the center of a beam, the maximum deflection is simply the formula:
W * length^3 / 48 * E * I
So if I pick a W6x12, its moment of inertia (I value) is 21.7. Plugging your parameters in, gives us this:
3000 * 108^3 / 48 * 2.9E7 * 21.7 = 0.125"
3,000 pounds on the center of a 108" long W6x12 will cause it to sag 1/8" in the middle.
So that's how you interpret what you saw in Machinery's Handbook.
Here's the formula applied to the example Forrest cited from some handbook:
I of S5x10 = 12.3 (I-beams are designated "Snxn")
3800 * 120^3 / 48 * 2.9E7 * 12.3 = 0.384"
His book said it would sag .4", so we're basically dead-on here.
In closing, I would note that your question was what size beam is "needed to support", but the need must be defined. Probably you mean "need not to break and fall on your head." But note that no material starts out rigid and then just breaks. First, metals flex under load and return to prior shape when the load is removed; this is the elastic range. Second, when the load is high enough, the material flexes under load but does not return to prior shape, ie it bends. Third, when the load is really high, the material may break.
3,000 pounds is not a lot of weight relative to Wide-Flange beams, and even small ones will not bend, let alone break under a static load (3,000 pounds dropping 50' and hitting a W4x13 would be a different story of course.) So in many applications, deflection is the design constraint, not failure. For example, the Uniform Building Code specifies minimum building floor joist sizes not to a safety standard, but to a comfort standard -- people don't like to feel the floor under them flex, even if it is perfectly safe.
So in summary, something from an S5x10 to a S7x15.3 (like John has) would be in order. A middle ground is a S6x12 which is twice as strong as a S5x10, and about 60% as strong as the 7 incher. None of these pose a failure risk in a relatively static load, home shop environment -- they will just vary in the amount of flex.