To ascertain the area of surface that is required to sustain a given weight safely, when the weight tends to split off the part pressed against, by causing, in case of fracture, one surface to slide on the other, we have -
Rule IV. - Divide the given weight in pounds by the value of H, Table I.; multiply the quotient by the factor of safety, and the product will be the required area in inches; or -
A = Wa/H. (4.)
Example. - The load on a rafter causes a horizontal thrust at its foot of 40,000 pounds, tending to split off the end of the tie-beam: what must be the length of the tie-beam beyond the line where the foot of the rafter is framed into it, the tie-beam being of Georgia pine, and 9 inches thick? The weight, or horizontal thrust, 40000, divided by 840, the value of H, Table I., set opposite Georgia pine, gives a quotient of 47.619, and this multiplied by 3, as a factor of safety, gives a product of 142.857. This, the area of surface in inches, divided by 9, the breadth of the surface strained (equal to the thickness of the tie-beam), the quotient, 15.87, is the length in inches from the end of the tie-beam to the rafter joint, say 16 inches.
104. - Tcuons and Splices - A knowledge of this kind of resistance of materials is useful, also, in ascertaining the length of framed tenons, so as to prevent the pin, or key, with which they are fastened from tearing out; and, also, in cases where tie-beams, or other timber under a tensile strain, are spliced, this rule gives the length of the joggle at each end of the splice.