To ascertain the weight that can be sustained safely by a post the height of which is at least ten times its least side if rectangular, or ten times its diameter if round, the direction of the pressure coinciding with the axis, we have -

Rule VII. - Divide the height of the post in inches by the diameter, or least side, in inches; multiply the quotient by itself, or take its square; multiply the square by the value of e, in Table III., set opposite the kind of material of which the post is made; to the product add the half of itself; to the sum add unity (or one); multiply this sum by the factor of safety, and reserve the product for use, as below. Now multiply the area of cross-section of the post in inches by the value of C, in Table I., set opposite the material of the post,and divide the product by the above reserved product; the quotient will be the required weight in pounds; or -

W= AC (7.)

a(1+3/2 er2)

Example: A Round Post. - What weight may be safely placed upon a post of Georgia pine 10 inches diameter and 10 feet high, the pressure coinciding with the axis of the post? The height of the post, (10 x 12 =) 120 inches, divided by 10, its diameter, gives a quotient of 12; this multiplied by itself gives 144, its square; and this by .00109, the value of e for Georgia pine, in Table III., gives . 15696; to which adding its half, the sum is 0.23544; to which adding unity, the sum is 1.23544; and this multiplied by 7, as a factor of safety, the product is 8.648, the reserved divisor. Now the area of the post is (see Table of Areas of Circles, in the Appendix, opposite its diameter, 10) 78.54; this multiplied by 9500, the value of C for Georgia pine, in Table I., gives a product of 746130; which divided by 8.648, the above reserved divisor, gives a quotient of 86278, the required weight in pounds.

Another Example: A Rectangular Post. - What weight may be safely placed upon a white-pine post 10x 12 inches, and 15 feet high, the pressure coinciding with the axis of the post? Proceeding according to the rule, we find the height of the post to be 180 inches, which divided by 10, the least side of the post, gives 18; this multiplied by itself gives 324, its square; which multiplied by .0014, the value of e for white pine, in Table III., gives .4536; to which adding its half, the sum is .6804; to which adding unity, the sum is 1 .6804; and this multiplied by 8, as a factor of safety, the product is 13.4432, the reserved divisor. Now the area of the post, (10x 12 =) 120 inches, multiplied by 6650, the value of C for white pine, in Table I., gives a product of 798,000, and this divided by 13.4432, the above reserved divisor, the quotient, 59360, is the required weight in pounds.

111. - Diameter of the Post: when Round. - To ascertain the size of a round post to sustain safely a given weight, when the height of the post is at least ten times the diameter; the direction of the pressure coinciding with the axis of the post; we have -

Rule VIII. - Multiply the given weight by the factor of safety, and divide the product by 1.5708 times the value of C for the material of the post, found in Table I.; reserve the quotient, calling its value G. Now multiply 432 times the value of e for the material of the post, found in Tablc III., by the square of the height in feet, and by the above quotient G; to the product add the square of G; extract the square root of the sum, and to it add the value of G; then the square root of this sum will be the required diameter;

or-

G= Wa/1.5708 C. (8)

(9.)

Example. - What should be the diameter of a locust post 10 feet high to sustain safely 40,000 pounds, the pressure coinciding with the axis? Proceeding by the rule, the given weight multiplied by 6, taken as a factor of safety, equals 240000. Dividing this by 1.5708 times 11700, the value of C for locust, in Table I., the quotient, 13.06, is the value of G, the square of which is 170.53. Now, the value of e for locust, in Table III., is .0015. This multiplied by 432, by 100, the square of the height, and by the above value of G, gives a product of 846.2; which added to 170.53, the above square of G, gives the sum of 1016.73. To 31.89, the square root of this, add the above value of G; then 6.7, the square root of this sum, is the required diameter of the post. The post therefore requires to be 6.7, say 6 3/8 inches diameter.

112. - Side of the Post: when Square. - To ascertain the side of a square post to sustain safely a given weight, when the height of the post is at least ten times the side; the pressure coinciding with the axis; we have -

Rule IX. - Multiply the given weight by the factor of safety, and divide the product by twice the value of C for the material of the post, found in Table I.; reserve the quotient, calling its value G.. Now multiply 432 times the value of e for the material of the post, found in Table III., by the square of the height in feet, and by the above quotient G; to the product add the square of G; extract the square root of the sum, and to it add the value of G; then the square root of this sum will be the required side of the post; or -

G=Wa/2C. (10.)

(11.)

Example. - What should be the side of a Georgia-pine square post 15 feet high to sustain safely 50,000 pounds, the pressure coinciding with the axis of the post? Proceeding by the rule, 50,000 pounds multiplied by 6, as a factor of safety, gives 300000; this divided by 2 x 9500 (the value of C) = 19000, the quotient, 15.789, is the value of G. The value of e for Georgia pine is .00109; the square of the height is 225; then, 432 times .00109 by 225 and by 15.789 (the above value of G) gives a product of 1672.86; the square of G equals 249.31; this added to 1672.86 gives a sum of 1922.17, the square root of which is 43.843; which added to 15.789, the value of G, gives 59.632, the square root of which is 7.722, the required side of the post. The post, therefore, requires to be, say, 7 3/4 inches square.