This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.
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.
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.
The Tables XIV and XV are very similar to the foregoing, but calculated for wrought-iron I-beams. Table XIV gives the size of beams and distance from centres required to carry different loads per square foot of floor, 150 pounds per square foot of floor (including the weight of construction), however, being the usual load allowed for in churches, office-buildings, public halls, etc., where the space between beams is filled with arched brickwork, or straight hollow-brick arches, and then covered over with concrete. A careful estimate, however, should be made of the exact weight of construction per square foot, including the ironwork, and to this should be added 70 pounds per square foot, which is the greatest load likely ever to be produced if packed solidly with people. Furniture rarely weighs as much, though heavy safes should be provided for separately. The load on roofs should be 30 pounds additional to the weight of construction, to provide for the weight of snow or wind. Look out for tanks, etc., on roofs. Plastered ceilings hanging from roofs add about 10 pounds per square foot, and slate about the same. Where a different load than given in the Table must be provided for, the distance between centres of beams can be reduced, proportionally from the next greater load; or the weight on each beam can be figured and the beam treated as a girder, supported sideways, in that case using Table XV. Both tables are calculated for the beams not to deflect sufficiently to crack plastering. The use of Table XIV is very simple. Supposing we have a span of 24 feet and a load of 150 pounds per square foot. We pass down the vertical line 24' 0" and strike first the 12"-96 pounds beam, which (for 150 pounds) is opposite (and three-quarter way between) 3' 0" and 3' 4" therefore 3' 3" from centres. The next beam is the 12"-120 pounds beam 4' 0" from centres; then the 12"-125 pounds beam 4' 1" from centres; then the 15"- 125 pounds beam 5' 0" from centres and so on. It is simply a question, therefore, which "distance from centres" is most desirable and as a rule in fireproof buildings it is desirable to keep these as near alike as possible, so as not to have too many different spans of beam arches and centres. If economy is the only question, we divide the weight of beam by its distance from centres, and the curve giving the smallest result is, of course, the cheapest. Supposing, however, that we desire all distances from centres alike, say 5 feet. In that case we pass down the 150-pound column to and then along the horizontal line 5' 0" till we strike the vertical (span) line, in this case 24' 0", and then take the cheapest beam to the right of the point of intersection. Thus, in our case the nearest beam would be 15"- 125 pounds; next comes 12 1/4" -170 pounds; then 15"-150 pounds, etc. As the nearest beam is the lightest in this case, we should select it. The weight of a beam is always given per yard of length. The reason for this is that a square inch of wrought-iron, one yard long, weighs exactly 10 pounds. Therefore if we know the weight per yard in pounds we divide it by ten to obtain the exact area of cross-section in square inches; or if we know the area, we multiply by ten and obtain the exact weight per yard.