In case the size of the beam is known, its safe span can, of course, be found by reversing the above procedure, or if the depth of beam and span is settled, we can find the necessary thickness and distance between centres; in this way the Table, of course, covers every problem.

Table XIII is calculated for wooden girders of all sizes. Any thickness not given in the table can be obtained by taking the line for a girder of same depth, but one inch thick and multiplying by the thickness. For very short spans, look out for danger of horizontal shearing (see formula 13); where this danger exists, pass vertical bolts through ends of girder, or bolt thin iron plates, or straps, or even boards with vertical grain, to each side of girder, at ends.

The use of this table is very simple. The vertical columns to the left give the safe uniform loads on girders (sufficiently stiff not to crack plastering) for different woods: these apply to the dotted parts of curves. The columns on the right-hand side give the same, but apply to the parts of curves drawn in full lines.

If we have a 6" X 16" Georgia pine beam of 20 feet span and want to know what it will carry, we select the curve marked at its upper end G X 16 = 96; we follow this curve till it intersects the vertical line 20' 0"; as this is in the part of curve drawn full, we pass horizontally to the right and find under the column marked " Georgia Pine," 7980, which is the safe, uniform load in pounds. Supposing, however, we had simply settled the span, say 8 feet, and load, say 7000 pounds, and wished to select the most economical girder, being, we will say, limited to the use of white pine: the span not being great we will expect to strike the dotted part of curve, and therefore select the fourth (white pine) column to the left. We pass down to the nearest figure to 7000 and then pass horizontally to the right till we meet the vertical 8 feet line; this we find is, as we expected, at the dotted part, and therefore our selection of the left column was right. We follow the curve to its upper end and find it requires a girder 4" X 12" = 48 square inches. Now, can we use a cheaper girder? of course, all the lines under and to the right of our curve are stronger, so that if either has a smaller sectional area, we will use it. The next curve we find is a 6" X 10" = 60"; then comes a 4" X 14" = 56"; then an 8" X 10" = 80"; then a 6" X 12"= 72" and so on; as none has a smaller area we will stick to our 4" X 12" girder, provided it is braced or supported sideways. If not, to avoid twisting or lateral flexure, we must select the next cheapest section, where the thickness is at least equal to half the depth;1 the cheapest section beyond our curve that corresponds to this, we find is the 6" X 10" girder, which we should use if not braced sideways. In the smaller sections of girders where the difference between the loads given from line to line is proportionally great, a safe load should be assumed between the two, according to the proximity to either line at which the curve cuts the vertical. The point where the curve cuts the bottom horizontal line of each part is the length of span for which the safe load opposite the line is calculated.

### Table XV. Iron I-Beam Girders, Braced Sideways

For Steel Beams, add one-quarter to safe uniform load on Iron Beams: but length of span in feet must not exceed twice the depth of beam in inches, or deflection will be too great for plastering.

Where a different load than 90 pounds per square root, must be provided for, we can either increase the thickness of beams as found in Table XII, or decrease their distance from centres, cither in proportion to the additional amount of load. Or, if we wish to be more economical, we can calculate the safe uniform load on each floor beam, and consider it as a separate girder, supported sideways, using of course, Table XIII.