This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

When the load is at the middle of a beam it exerts the greatest possible strain; at any other point the strain would be less. The strain decreases gradually as it approaches one of the bearings, and when arrived at the bearing its effect upon the beam as a cross-strain is zero. The effect of a weight upon a beam is in proportion to its distance from one of the bearings, multiplied by the portion of the load borne by that bearing.

The load upon a beam is divided upon the two bearings, as shown at Art. 88. The weight which is required to rupture a beam is in proportion to the breadth and square of the depth, b d2, as before shown, and also in proportion to the length divided by 4 times the rectangle of the two parts into which the load divides the length, or1/4 mn(see Fig. 35).

This, when the load is at the middle, may be put as

l =I/l, a result coinciding with the relation before

4x1/2lx1/2l

given in Art. 122, viz.: "The resistance is inversely in proportion to the length." The total resistance, therefore, putting the two statements together, is in proportion to b d22l/4 m n

There are, therefore, these two ratios, viz., W: b d2l/4 m n and B: 1, from which we have the proportion -

B: I:: W: b d2l/4m n,

from which we have -

W= Bbd2l/4mn . (23.)

This is the relation at the point of rupture, and when - is used instead of B, the expression gives the safe weight. Therefore -

W= Bbd2l/4amn (24.)

is an expression for the safe weight. Now, to ascertain the weight which may be safely borne by a beam at any point in its length, we have -

Rule XIX. - Multiply the breadth by the square of the depth, by the length in feet, and by the value of B for the material of the beam, in Table III.; divide the product by the product of four times the factor of safety into the rectangle of the two parts into which the centre of gravity of the weight divides the beam, and the quotient will be the required weight in pounds.

Example. - What weight may be safely sustained at 3 feet from one end of a Georgia-pine beam which is 4 x 10 inches, and 20 feet long? The value of B for Georgia pine, in Table III., is 850; therefore, by the rule, 4 x 102 x 20 x 850 = 6800000. Taking the factor of safety at 4, we have 4x4x3x17=816. Using this as a divisor with which to divide the former product, we have as a quotient 8333 pounds, the required weight.

127. - Breadth or Depth: Load at any Point. - By a proper transposition of the factors of (24.) we obtain -

bd2 = 4 Wamn/B1, (25.)

an expression showing the product of the breadth into the square of the depth; hence, to ascertain the breadth or depth of a beam to sustain safely a given weight located at any point on the beam, we have -

Rule XX. - Multiply four times the given weight by the factor of safety, and by the rectangle of the two parts into which the load divides the length; divide the product by the product of the length into the value of B for the material of the beam, found in Table III., and the quotient will be equal to the product of the breadth into the square of the depth. Now, to obtain the breadth, divide this product by the square of the depth, and the quotient will be the required breadth. But if, instead of the breadth, the depth be desired, divide the said product by the breadth; then the square root of the quotient will be the required depth.

Example. - What should be the breadth (the depth being 8) of a white-pine beam 12 feet long to safely sustain 3500 pounds at 3 feet from one end? Also, what should be its depth when the breadth is 3 inches? By the rulemaking the factor of safety at 4, 4 x 3500 x 4 x 3 x 9 = 1512000. The value of B for white pine, in Table III., is 500; therefore, 500 x 12 = 6000; with this as divisor, dividing 1512000, the quotient is 252. Now, to obtain the breadth when the depth is 8, 252 divided by (8 x 8 - ) 64 gives a quotient of 3.9375, the required breadth; or the beam may be, say, 4 x 8. Again, when the breadth is 3 inches, we have for the quotient of 252 divided by 3 = 84, and the square root of 84 is 9.165, or 9 1/6 inches. For this case, therefore, the beam should be, say, 3 x 9 1/4 inches.

Continue to: