When several beams are laid in a tier, placed at equal distances apart, as in a tier of floor-beams, it is desirable to know what should be their size in order to sustain a load equally distributed over the floor.

If the distance apart at which they are placed, measured from the centres of the beams, be multiplied by the length of the beams between bearings, the product will equal the area of the floor sustained by one beam; and if this area be multiplied by the weight upon a superficial foot of the floor, the product will equal the total load uniformly distributed over the length of the beam; or, if c be put to represent the distance apart between the centres of the beams in feet, and l represent the length in feet of the beam between bearings, and f equal the pounds per superficial foot on the floor, then the product of these, or c f l, will represent the uniformly distributed load on a beam; but this load was before represented by U (Art. 128); therefore, we have cfl= U, and they may be substituted for it in (26.) and (27.). Thus we have -

bd2= cflal, / 2B

or -

bd2= acfl2. (28.)

2 B.

Therefore, to ascertain the size of floor-beams to sustain safely a given load per superficial foot, we have -

Rule XXIII. - Multiply the given weight per superficial foot by the factor of safety, by the distance between the centres of the beams in feet, and by the square of the length in feet; divide the product by twice the value of B for the material of the beams, in Table III., and the quotient will be equal to the breadth into the square of the depth. Now, to obtain the breadth, divide said quotient by the square of the depth, and this quotient will be the required breadth. But if, instead of the breadth, the depth be required, divide the aforesaid quotient by the breadth; then the square root of this quotient will be the required depth.

Example. - What should be the size of white-pine floor-beams 20 feet long, placed 16 inches from centres, to sustain safely 90 pounds per superficial foot, including the weight of the materials of construction - the beams, flooring, plastering, etc.? The value of B for white pine is 500; the factor of safety may be put at 5. Then, by the rule, we have 90 x 5 x 16/12 x 202 = 240000. This divided by (2 x 500 =) 1000 gives 240. Now, for the breadth, if the depth be fixed at 9 inches, then 240 divided by (92 = ) 81 gives a quotient of 2.963. The beams therefore should be, say, 3x9. But if the breadth be fixed, say, at 2.5 inches, then 240 divided by 2.5 gives a quotient of 96, the square root of which is 9.8 nearly. The beams in this case would require therefore to be, say, 2 1/2 x 10 inches.

N. B. - It is well to observe that the question decided by Rule XXII. is simply that of strength only. Floor-beams computed by it will be quite safe against rupture, but they will in most cases deflect much more than would be consistent with their good appearance. Floor-beams should be computed by the rules which include the effect of deflection. (See Art. 152.)

131. - Levers: Load at One End. - The beams so far considered as being exposed to transverse strains have been supposed to be supported at each end. When a piece is held firmly at one end only, and loaded at the other, it is termed a lever; and the load which a piece so held and loaded will sustain is equal to one fourth that which the same piece would sustain if it were supported at each end and loaded at the middle. Or, the strain in a beam supported at each end caused by a given weight located at the middle is equal to that in a lever of the same breadth and depth, when the length of the latter is equal to one half that of the beam, and the load at its end is equal to one half of that at the middle of the beam. Or, when P represents the load at the end of the lever, and n its length, then W= 2 P, and l - 2n. Substituting these values of Wand l in equation (20.), we have -

2 P = Bbd2/2 an from which -

P= Bbd2/4 an (29.)

Hence, to ascertain the weight which may be safely sustained at the end of a lever, we have -

Rule XXIV. - Multiply the breadth of the lever by the square of its depth, and by the value of B for the material of the lever, in Table III.; divide the product by the product of four times the length in feet into the factor of safety, and the quotient will be the required weight in pounds.

Example. - What weight can be safely sustained at the end of a maple lever of which the breadth is 2 inches, the depth is 4 inches, and the length is 6 feet? The value of B for maple, in Table III., is 1100; therefore, by the rule, 2 x42 x 1100 = 35200. And, taking the factor of safety at 5, 4x5x6= 120, and 35200 divided by 120 gives a quotient of

293.33. or 293 1/3 pounds.

N. B. - When a lever is loaded with a weight uniformly distributed over its length, it will sustain just twice the load which can be sustained at the end.

##### Levers To Sustain Given Weights

132 - Levers: Breadth or Depth. - By a proper transposition of the factors in (29.), we obtain -

bd2 = 4 Pan/B (30.)

Hence, to ascertain the breadth or depth of a lever to sustain safely a given weight, we have -

Rule XXV. - Multiply four times the given weight by the length of the lever, and by the factor of safety; divide the product by the value of B for the material of the lever, in Table III., and the quotient will be equal to the breadth multiplied by the square of the depth. Now, if the breadth be required, divide said quotient by the square of the depth, and this quotient will be the required breadth; but if, instead of the breadth, the depth be required, divide the said quotient by the breadth; then the square root of this quotient will be the required depth.

Example. - What should be the size of a cherry lever 5 feet long to sustain safely 250 pounds at its end? Proceeding by the rule, taking the factor of safety at 5, we have 4x250x5x5 = 25000. The value of B for cherry, in Table III., is 650; and 25000 divided by 650 gives a quotient of 38.46. Now, if the depth be fixed at 4, then 38.46 divided by (4x4 =) 16 gives a quotient of 2.4, the required breadth. But if the breadth be fixed at 2, then 38.46 divided by 2 gives a quotient of 19.23, the square root of which is 4.38, the required depth. Therefore, the lever maybe 2.4x4, or 2 x 4 3/8 inches.