divided by 1028, the product of the rate of deflection by the value of E, as found in the preceding example, equals 114, the value of U. Now by Rule XXVII., 114, the value of U, divided by 0-589, equals 19-42, the square root of which is 4407, and the square root of this square root is 2-099, the diameter required-very nearly 2 inches.

Example.- When the beam is round and laid inclining, the weight concentrated at the middle. A round beam of white pine, 20 feet long between bearings, and laid inclining so that the horizontal distance between bearings is 18 feet, is required to support 1250 pounds at the middle, with a deflection not to exceed 0.05 inch per foot; what must be its diameter? By Rule XXVI. for load at middle, modified by Rule XXIX. for inclined beams, 1250, the weight, multiplied by 20, the length, and by 18, the horizontal distance, equals 450,000. The rate of deflection, 0.05, multiplied by 1750, the value of E, Table II., for white pine, equals 87.5. The above 450,000 divided by 87.5, equals 5142.86, the value of M. Now by Rule XXXII. for round beams, 5142-86, the value of M, divided by 0.589, equals 8731-5, the square root of which is 93.44, and the square root of this square root is 9.667, the diameter required equal to 93 inches.

Example.-Same as in preceding example, but the weight equally distributed. By Rule XXVII., five-eighths of the weight is to be used instead of the whole weight, therefore 8731-5, the result in the last example just previous to taking the square root, multiplied by, equals 5457-2, the square root of which is 73-87, and the square root of this square root is 8-59, the diameter required-nearly 8 inches.

Example.-Same as in the next but one preceding example, but the weight per foot superficial, and supported by two or more beams. A series of round hemlock poles or beams, 10 feet long clear bearing, laid inclining so as that the horizontal distance between the supports equals 7 feet, and laid 2 feet

and 6 inches apart from centres, are required to support 20 pounds per superficial foot without regard to the amount of deflection, provided that the elasticity of the material be not injured; what must be their diameter? By Rule XXVIII. for weight per foot superficial, modified by Rule XXIX. for inclined beams, 2-5, the distance from centres, multiplied by 100, the square of the length, and by 7, the horizontal distance between bearings, and by five-eighths of the weight, 125, equals 21,875. The greatest value of n, Table II., for hemlock, 0.08794, multiplied by 1240, the value of E, Table II., for hemlock, equals 109-0456. The above 21,875, divided by 109-0456, equals 2006, the value of U. Now by Rule XXXII., the above 2006, divided by 0·589, equals 340-6, the square root of which is 18:46, and the square root of this square root is 4.296, the diameter required-equal to 4% inches nearly.

329. The greater the depth of a beam in proportion to the thickness, the greater the strength. But when the difference between the depth and the breadth is great, the beam must be stayed, (as at Fig. 228,) to prevent its falling over and breaking sideways. Their shrinking is another objection to deep beams; but where these evils can be remedied, the advantage of increasing the depth is considerable. The following rule is, to find the strongest form for a beam out of a given quantity of timber.

Rule.-Multiply the length in feet by the decimal, 0·6, and divide the given area in inches by the product; and the square of the quotient will give the depth in inches.

Example.-What is the strongest form for a beam whose given area of section is 48 inches, and length of bearing 20 feet? The length in feet, 20, multiplied by the decimal, 0·6, gives 12; the given area in inches, 48, divided by 12, gives a quotient of 4, the square of which is 16-this is the depth in inches; and the breadth must be 3 inches. A beam 16 inches

Inclining

by 3 would bear twice as much as a square beam of the same area of section; which shows how important it is to make beams deep and thin. In many old buildings, and even in new ones, in country places, the very reverse of this has been practised; the principal beans being oftener laid on the broad side than on the narrower one.

The foregoing rules, stated algebraically, are placed in the following table.

In the above table, b equals the breadth, and d the depth of cross-section of beam; n equals the rate of deflection per foot of the length; b, d and n, all in inches. Also, l equals the length c the distance between two parallel beams measured from the

centres of their breadth, and h equals the horizontal distance between the supports of an inclined beam; 7, c and h, all in feet. Again, w equals the weight on the beam, ƒ equals the weight upon each superficial foot of a floor or roof, supported by two or more beams laid parallel and at equal distances apart; and f in pounds. And r is any decimal, chosen at pleasure, in proportion to unity, as b is to d-from which proportion b = dr. E is a constant the value of which is found in Table II.

330.—To ascertain the scantling of the stiffest beam that can be cut from a cylinder. Let d a c b, (Fig. 223,) be the section, and e the centre, of a given cylinder. Draw the diameter, ab; upon a and b, with the radius of the section, describe the arcs, de and e c; join d and a, a and c, e and b, and b and d; then the rectangle, d a c b, will be a section of the beam required.

331.-Resistance to Rupture.-The resistance to deflection having been treated of in the preceding articles, it now remains to speak of the other branch of resistance to cross strains, namely, the resistance to rupture. When a beam is laid horizontally and supported at each end, its strength to resist a cross strain, caused by a weight or vertical pressure at the middle of its length, is directly as the breadth and square of the depth and inversely as the length. If the beam is square, or the depth equal to the breadth, then the strength is directly

as the cube of a side of the beam and inversely as the length, and if the beam is round the strength is directly as the cube of the diameter and inversely as the length.

When the weight is concentrated at any point in the length, the strength of the beam is directly as the length, breadth, and square of the depth, and inversely as the product of the two parts into which the length is divided by the point at which the weight is located.

When the beam is laid not horizontal but inclining, the strength is the same as in each case above stated, and also in proportion, inversely as the cosine of the angle of inclination with the horizon, or, which is the same thing, directly as the length and inversely as the horizontal distance between the points of support.

When the weight is equally diffused over the length of a beam, it will sustain just twice the weight that could be sustained at the middle of its length.

A beam secured at one end only, will sustain at the other end just one-quarter of the weight that could be sustained at its middle were the beam supported at each end.

These relations between the strain and the strength exist in all materials. For any particular kind of material,

S, representing a constant quantity for all materials of like strength. The superior strength of one kind of material over another is ascertained by experiment; the value of S being ascertained by a substitution of the dimensions of the piece tried for the symbols in the above formula. Having thus obtained the value of S, the formula, by proper inversion, becomes useful in ascertaining the dimensions of a beam that will require a certain weight to break it; or to ascertain the weight that will be required to break a certain beam. It will be observed in the preceding formula, that if each of the dimensions of the