When the load is spread out uniformly over the length of a beam, the beam will require just twice the weight to break it that would be required if the weight were concentrated at the centre.
Therefore, we have W = U/2, where U represents the distributed load. Substituting this value of W in equation
(20.), we have -
U/2 = Bbd2/al
U= 2 Bbd2/al. (26.)
Therefore, to ascertain the weight which may be safely sustained, when uniformly distributed over the length of a beam, we have -
Rule XXI. - Multiply twice the breadth by the square of the depth, and by the value of B for the material of the beam, in Table III., and divide the product by the product of the length in feet by the factor of safety, and the quotient will be the required weight in pounds.
Example. - What weight uniformly distributed may be safely sustained upon a hemlock beam 4x9 inches, and 20 feet long? The value of B for hemlock, in Table III., is 450; therefore, by the rule, 2x4x92 x450 = 291600. Taking the factor of safety at 4, we have 4x20 = 80, the product by which the former product is to be divided. This division produces a quotient of 3645, the required weight.
129. - Breadth or Depth: Load Uniformly Distributed. -
By a proper transposition of factors in (26.), we obtain -
bd2 = Ual, (27.)
an expression giving the value of the breadth into the square of the depth. From this, therefore, to ascertain the breadth or the depth of a beam to sustain safely a given weight uniformly distributed over the length of a beam, we have -
Rule XXII. - Multiply the given weight by the factor of safety, and by the length; divide the product by the product of twice the value of B for the material of the beam, in Table III., and the quotient will be equal to the breadth into the square of the depth. Now, to find the breadth, divide the said quotient by the square of the depth; but if, instead of the breadth, the depth be required, then divide said quotient by the breadth, and the square root of this quotient will be the required depth.
Example. - What should be the size of a white-pine beam 20 feet long to sustain safely 10,000 pounds uniformly distributed over its length? The value of B for white pine, in Table III., is 500. Let the factor of safety be taken at 4. Then, by the rule, 10000 x 4 x 20 = 800000; this divided by (2 x 500 =)
1000 gives a quotient of 800. Now, if the depth be fixed at 12, then the said quotient, 800, divided by (12 x 12=) 144 gives 59, the required breadth of beam; and the beam may be, say, 5 3/4 x 12. Again, if the breadth is fixed, say, at 6, and the depth is required, then the said quotient, 8co, divided by 6 gives 133 1/3, the square root of which, 11.55, is the required depth. The beam in this case should therefore be, say, 6 x 11 3/4 inches.