For load at any point of beam.

w1= b.d2.L/72.M.N . (k/f) (74)

u = b.d2/36.L. (k/f) (75)

For load concentrated at end of cantilever.

w = b.d2/72.L. (k/f) (76)

For load at any point of cantilever.

w1=b.d2/72.Y.(k/f) (77)

Where w = safe uniform load, in pounds.

Wherew = safe centre load on beam, in pounds; or safe load at end of cantilever, in pounds.

Where w1= safe concentrated load, in pounds, at any point.

Where Y= length, in feet, from wall to concentrated load (in cantilever) .

Where M and N = the respective lengths, in feet, from concentrated load on beam to each support.

Where L= the length, in feet, of span of beam, or length of cantilever.

Where b = the breadth of beam, in inches.

Where d = the depth of beam, in inches.

Where(k/f) =the safe modulus of rupture, per square inch, of the material of beam or cantilever (see Table IV).

The above formulae are for rectangular wooden beams supported against lateral flexure (or yielding sideways). Where beams or girders are not supported sideways the thickness should be equal to at least half of the depth.

The above formulae make no allowance for deflection, and except in cases, such as factories, etc., where strength only need be considered and not the danger of cracking plastering, or getting floors too uneven for machinery, are really of but little value. They are so easily understood that the simplest example will answer:

Example. Take a 3" X 10" hemlock timber and 9 feet long {clear span), loaded in different ways, what will it safely carry? taking no account of deflection.

No allowance for deflection.

The safe modulus of rupture ( k/f ) for hemlock from Table IV is

= 750 pounds.

If both ends are supported and the load is uniformly distributed the beam will safely carry, (Formula 72): u= 3.102/9.9. 750=2778 pounds.

If both ends are supported and the load concentrated at the centre, the beam will safely carry, (Formula 73): w = 3.102/18.9 .750 = 1389 pounds.

If both ends are supported and the load is concentrated at a point

I, distant four feet from one support (and five feet from the other) the beam will safely carry, (Formula 74): w1 = 3.102.9/72.4.5. 750 = 1406 pounds.

If one end of the timber is built in and the other end free and the load uniformly distributed, the cantilever will safely carry, (Formula

75): u= 3.102/36.9 750 = 694 pounds.

If one end is built in and the other end free, and the load concentrated at the free end, the cantilever will safely carry, (Formula 76): w = 3.102/72.9. 750 = 347 pounds.

If one end is built in and the other end free, and the load concentrated at a point I, which is 5 feet from the built-in end, the cantilever will safely carry, (Formula 77): w1=3.102/72.5 750 = 625 pounds.

Where, however, the span of the beam, in feet, greatly exceeds the depth in inches (see Table VIII), and regard must be had to deflection, the formulae (28) and (29), also (37) to (42) should always be used, inserting for i its value from Table I, section No. 2, or: - i = b.d3/12

Where b - the thickness of timber, in inches.

Where d = the depth of timber, in inches.

Where i - the moment of inertia of the cross-section, in inches.

Table IX, however, gives a much easier method of calculating wooden beams, allowing for both rupture and deflection, and Formula) (72) to (77) have only been given here, as they are often erroneously given in text-books, as the only calculations necessary for beams.

To still further simplify to the architect the labor of calculating wooden beams or girders, the writer has constructed Tables XII and XIII.

Table XII is calculated for floor beams of dwellings, offices' churches, etc., at 90 pounds per square foot, including weight of construction. The beams are supposed to be cross-bridged.

Table XIII is for isolated girders, or lintels, uniformly loaded, and supported sideways. When not supported sideways decrease the load, or else use timber at least half as thick as it is deep. In no ease will beams or girders (with the loads given) deflect sufficiently to crack plastering. For convenience Table XII has been divided into two parts, the first part giving beams of from 5' 0" to 15' 0" span, the second part of from 15' 0" to 29' 0" span.

The use of the table is very simple and enables us to select the most economical beam in each case. For instance, we have say a span of 21' 6". We use the second part of Table XII. The vertical dotted line between 21' 0" and 22' 0" is, of course, our line for 21' 6". We pass our finger down this line till we strike the curve. To the left opposite the point at which we struck the curve, we read:

21.6 spruce, W. P. 56 - 4-14-14 or: at 21' 6' span we can use spruce or white-pine floor beams, of 56 inches sectional area each, viz.: 4" thick, 14" deep and 14" from centres. Of course we can use any other beam below this point, as they are all stronger and stiffer, but we must not use any other beam above this point. Now then, is a 4" X 14" beam of spruce or white pine, and 14" from centres the most economical beam. We pass to the columns at the right of the curve and there read in the first column 48". This means that while the sectional area of the beam is 56 square inches, it is equal to only 48 square inches per square foot of floor, as the beams are more than one foot from centres. In this column the areas are all reduced to the "area per square foot of floor," so that we can see at a glance if there is any cheaper beam below our point. We find below it, in fact, many cheaper beams, the smallest area (per square foot of floor) being, of course, the most economical. The smallest area we find is 36, 0 or 36 square inches of section per square foot of floor (this we find three times, in the sixteenth, twenty-ninth and thirty-first lines from the bottom). Passing to the left we find they represent, respectively, a Georgia pine beam, 3" thick, 16" deep and 16" from centres; or a Georgia pine beam 3" thick, 14" deep and 14" from centres; or a white oak beam 3" thick, 16" deep and 16" from centres. If therefore, we do not consider depth, or distance from centres, it would simply be a question, which is cheaper, 48 inches (or four feet "board-measure") of white pine or spruce, or 36 inches (or three feet "board-measure") of either white oak or Georgia pine. The four other columns on the right-hand side, are for the same purpose, only the figures for each kind of wood are in a column by themselves; so that, if we are limited to any kind of wood we can examine the figures for that wood by themselves. Take our last case and suppose we are limited to the use of hemlock; now from the point where our vertical line (21' 6") first struck the curve, we pass to the right-hand side of Table, to the second column, which is headed "Hemlock." From this point we seek the smallest figure below this level, but in the same column; we find, that the first figure we strike, viz: 41, 2 is the smallest, so we use this; passing along its level to the left we find it represents a hemlock beam of 48 square inches cross-section, or 3" thick, 16" deep and 14" from centres.