that may arise. Any deflection within the limits of the elasticity of the material, may be given to beams used for some purposes, while others require to be restricted to that amount. of deflection that shall not be perceptible to a casual observer. Let n represent, in the decimal of an inch, the rate of deflection per foot of the length of the beam; then the product of n, multiplied by the number of feet contained in the length of the beam, will equal the total deflection, n l. Now, if n l be substituted for p in the formulas, (26,) (27) and (28,) they will be rendered more available for general use. For example, let this substitution be made in (26,) and there results = (29.) where is in feet, and b, d and n in inches; and for (27)— where the notation is as before, with also 8 and D in inches. In these formulas, w represents the weight in' pounds concen trated at the middle of the length of the beam. If the weight, instead thereof, is equally distributed over the length of the beam, then, since of it concentrated at the middle will deflect a beam to the same depth that the whole does when equally distributed, (Art. 281,) therefore— where w equals the whole of the equally distributed load, Again, if the load is borne by more beams than one, laid parallel to each other-as, for example, a series or tier of floor beams-and the load is equally distributed over the supported surface or floor; then, if f represents the number of pounds of the load contained on each square foot of the floor, or the pounds' weight per foot superficial, and c represents the distance in feet between each two beams, or rather the distance from their centres, and 7 the length of the beam in feet, in the clear, between the supports at the ends; then cl will equal the area of surface supported by one of the beams, and fel will represent the load borne by it, equally distributed over its length. Now, if this representation of the load be substituted for w in (32,) (33) and (34) there results— Practical Rules and Examples. 323. To ascertain the weight, placed upon the middle of a beam, that will cause a given deflection. Rule XX.-Multiply the area of the cross-section of the beam by the square of the depth and by the rate of the deflection, all in inches; multiply the product by the value of E, Table II., and divide this product by the square of the length in feet, and the quotient will be the weight in pounds required. Example.-What weight can be supported upon the middle of a Georgia pine girder, ten feet long, eight inches broad, and ten inches deep, the deflection limited to three-tenths of an inch, or at the rate of 0·03 of an inch per foot of the length ? Here the area equals 8 x 1080; the square of the depth equals 10 x 10 = 100: 80 x 100 = 8,000; this by 0.03, the rate of deflection, the product is 240; and this by 2970, the value of E for Georgia pine, Table II., equals 712,800. This product, divided by 100, the square of the length, the quotient, 7,128, is the weight required in pounds. Rule XXI.-Where the beam is square the weight may be found by the preceding rule or by this:-Multiply the square of the area of the cross-section by the rate of deflection, both in inches, and the product by the value of E, Table II., and divide this product by the square of the length in feet, and the quotient will be the weight required in pounds. Example.-What weight placed on the middle of a spruce beam will deflect it seven-tenths of an inch, the beam being 20 feet long, 6 inches broad, and 6 inches deep? Here the area is 6 x 6 = 36, and its square is 36 × 36 = 1296; the rate of deflection is equal to the total deflection divided by the 0.7 = 0.035; therefore, 1296 x 0.035 45.36, and = length, = 20 this by 1550, the value of E for spruce, Table II., equals 70,308. This, divided by 400, the square of the length, equals 175.77, the required weight in pounds. Rule XXII.-When the beam is round find the weight by this rule:-Multiply the square of the diameter of the crosssection by the square of the diameter, and the product by the rate of deflection, all in inches, and this product by 0.589 times the value of E, Table II. This last product, divided by the square of the length in feet, will give the required weight in pounds. Example.-What weight on the middle of a round white pine beam will cause a deflection of 0·028 of an inch per foot, the beam being 10 inches diameter and 20 feet long? The square of the diameter equals 10 x 10 = 100; 100 × 100 = 10,000; this by the rate, 0-028, 280, and this by 0-589 x 1750, the value of E, Table II., for white pine, equals 288,610. This last product, divided by 400, the square of the length, equals 721-5, the required weight in pounds. = 324.-To ascertain the weight that will produce a given de flection, when the weight is equally distributed over the length of the beam. Rule XXIII.--The rules for this are the same as the three preceding rules, with this modification, viz., instead of the square of the length, divide by five-eighths of the square of the length. 325. In a series or tier of beams, to ascertain the weight per foot, equally distributed over the supported surface, that will cause a given deflection in the beam. Rule XXIV.-The rules for this are the same as Rules XX., XXI., and XXII., with this modification, viz., instead of the square of the length, divide by the product of the distance apart in feet between each two beams, (measured from the centres of their breadths,) multiplied by five-eighths of the cube of the length, and the quotient will be the required weight in pounds that may be placed upon each superficial foot of the floor or other surface supported by the beams. In this and all the other rules, the weight of the material composing the beams, floor, and other parts of the constructions is understood to be a part of the load. Therefore from the ascertained weight deduct the weight of the framing, floor, plastering, or other parts of the construction, and the remainder will be the neat load required. Example. In a tier of white pine beams, 4 x 12 inches, 20 fect long, placed 16 inches or 13 feet from centres, what weight per foot superficial may be equally distributed over the floor covering said beams-the rate of deflection to be not more than 0.025 of an inch per foot of the length of the beams. Proceeding by Rule XX. as above modified, the area of the cross-section, 4 x 12, equals 48; this by 144, the square of the depth, equals 6912, and this by 0.025, the rate of deflection, equals 172.8. Then this product, multiplied by 1750, the value of E, Table II., for white pine, equals 302,400. The distance between the centres of the beams is 13 feet, the cube of the length is 8,000, and by of 8,000 equals 6,6663. The above 302,400, divided by 6,6663, the quotient, 45·36, equals the required weight in pounds per foot superficial. The weight of beams, floor plank, cross-furring, and plastering occurring under every square foot of the surface of the floor, is now to be ascertained. Of the timber in every 16 inches by 12 inches, there occurs 4 x 12 inches, one foot long; this equals one-third of a cubic foot. Now, by proportion, if 16 inches in width contains of a cubic foot, what will 12 inches in width contain ? × 12 = 12 3 = 1 of a cubic foot. The floor plank (Georgia pine) is 12 x 12 inches, and 1 inches = T furring strips, 1 x 2 inches, placed 12 inches from centres, there will occur one of a foot long in every superficial foot. Now, since in a cubic foot there is 144 rods, one inch square and one foot long, therefore, this furring strip, 1 × 2 × 12 inches, equals of a cubic foot. The weight of the timber and furring strips, being of white pine, may be estimated together! + q 2 = }} + q2 = 4 of a cubic foot. White pine varies from 23 to 30 pounds. If it be taken at 30 pounds, the beam and furring together will weigh 30 × 2 pounds, equals 7.92 pounds. Georgia pine may be taken at 50 pounds per cubic foot;* the weight of the floor plank, then, is 50 × = 5.21 pounds. A superficial foot of lath and plastering will weigh about 10 lbs. Thus, the white pine, 7.92, Georgia pine, 5-21, and the plastering, 10, together equal 23·13 pounds; this from 45:36, as before ascertained, leaves 22-23, say 224 pounds, the neat weight per foot superficial that may be equally distributed over the floor as its load. To get the weight of wood or any other material, multiply its specific gravity by 62 5. For Spe cic Gravities see Tables I., II., and III. and the Appendix for Weight of Materials. |