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

To ascertain the sectional area of a beam or rod that will sustain a given weight safely, when applied as a tensile strain, we have -

Rule XII I. - Multiply the given weight in pounds by the factor of safety; divide the product by the value of T, Table II., and the quotient will be the area required in inches; or -

A= Wa/T. (16.)

This is the area of uncut fibres. If the piece is to be cut for mortices, or for any other purpose, then for this an adequate addition is to be made to the result found by the rule.

Example - A. rafter produces a thrust horizontally of 80,000 pounds; the tie-beam is to be of oak: what must be the area of the cross-section of the tie-beam in order to sustain the rafter safely? The given weight, 80000, multiplied by 10, as a factor of safety, gives 800000; this divided by 19500, the value of T, Table II., the quotient, 41, is the area of uncut fibres. This should have usually one half of its amount added to it as an allowance for cutting; therefore, 41+21 = 62. The tie-beam may be 6 x 10 1/2 inches.

Another Example. - A tie-rod of American refined wrought iron is required to sustain safely 36,000 pounds: what should be its area of cross-section? Taking 7 as the factor of safety, 7x36000= 252000; and this divided by 60000, the value of T, Table II., gives a quotient of 4. 2 inches, the required area of the rod.

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