tais resistance is useful in limiting the load on a post according to the kind of material contained, not in the post, but in the timber upon which the post presses.

293.—In Table II. are the results of experiments made tc test the resistance of materials to flexure: first, the flexure produced by compression, the force acting on the ends of the fibres longitudinally; secondly, the flexure arising from the effects of a cross strain, the force acting on the side of the fibres transversely, the beams being laid on chairs or rests. Of white oak, No. 1, there were eight specimens, of 2 by 4 inches, and 34 feet long, seasoned more than a year after they were prepared for experiment. Of the other kinds of wood there were from three to five specimens of each, of 14 by 24 inches, and from 1 to 2 feet long. Of the cast iron there were six specimens, of 1 inch square and 1 foot long; and of the wrought iron there were five specimens of American, three of by 2 inches, and two of 13 inches square, and three specimens of common English, by 2 inches; the eight specimens being each 19 inches long, clear bearing. In each case the result is the average of the stiffness of the several specimens. The numbers contained in the second column are the weights producing the first degree of flexure in a post or strut, where the post or strut is one foot long and one inch square; so, likewise, the numbers in the fifth column, and which are represented in the rules by E, are the weights required to deflect a beam one inch, where the beam is one foot long, clear bearing, and one inch square. (See remarks upon this, Art. (321.) The numbers in the third column are equal to one-half of those in the second. The numbers contained in the fourth column, and represented by n in the rules, show the greatest rate of deflection that the material may be subjected to without injury. This rate multiplied by the length in feet, equals the total deflection within the limits of elasticity.

294.-To find the weight that can be safely sustained by a post, when the height of the post is less than ten times the diameter if round, or ten times the thickness if rectangular, and the direction of the pressure coinciding with the axis.

Rule I.-Multiply the area of the cross-section of the post, in inches, by the value of C in Table I., the product will be the required weight in pounds.

A C=w.

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(1.)

Example.-A Georgia pine post is 6 feet high, and in crosssection, 8 x 12 inches, what weight will it safely sustain? The area 8 x 12 96 inches; this multiplied by 1700, the value of C, in the table, set opposite Georgia pine, the result, 163,200, is the weight in pounds required. It will be observed that the weight would be the same for a Georgia pine post of any height less than 10 times 8 inches = 80 inches 6 feet 8

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inches, provided its breadth and thickness remain the same, 12 and 8 inches.

295.-To find the area of the cross-section of a post to sus tain a given weight safely, the height of the post being less than ten times the diameter if round, or ten times the least side if rectangular; the pressure coinciding with the axis.

Rule II.-Divide the given weight in pounds by the value of C, in Table I., and the product will be the required area in inches

Example.-A weight of 38,400 pounds is to be sustained by a white pine post 4 feet high, what must be its area of section in order to sustain the weight safely? Here, 38,400 divided by 1200, the value of C, in Table I., set opposite white pine, gives a quotient of 32; this, therefore, is the required area, and such a post may be 5 × 6-4 inches. To find the least side, so that it shall not be less than one-tenth of the height, divide the height, reduced to inches, by 10, and make the least side to exceed this quotient. The area, divided by the least side. so determined, will give the wide side. If, however, by this process, the first side found should prove to be the greatest, then the size of the post is to be found by Rule VII., VIII., or IX.

296.-If the post is to be round, by reference to the Table of Circles in the Appendix, the diameter will be found in the column of diameters, set opposite to the area of the post found in the column of areas, or opposite to the next nearest area. For example, suppose the required area, as just found by the example under Rule II., is 32; by reference to the column of areas, 33-183 is the nearest to 32, and the diameter set opposite is 65. The post may, therefore, be 63 inches diameter.

Second Case.

297. To ascertain the weight that can be sustained safely

by a post whose height 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.

Rule III.- When the post is round the weight may be found by this rule: Multiply the square of the diameter in inches by the square of the diameter in inches, and multiply the product by 0-589 times the value of B, in Table II., divide this product by the square of the height in feet, and the quotient will be the required weight in pounds.

Example. What weight will a Georgia pine post sustain safely, whose diameter is 10 inches and height 10 feet? The square of the diameter is 100; 100 x 100 = 10,000. And × 10,000 by 0-589 times 4830, the value of B, Table II., set opposite Georgia pine, = 28,448,700, and this divided by 100, the square of the height, equals 284,487, the weight required, in pounds.

Rule IV-If the post be rectangular the weight is found. by this rule: Multiply the area of the cross-section of the post by the square of the thickness, both in inches, and by the value of B, Table II. Divide the product by the square of the height in feet, and the quotient will be the required weight in pounds.

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Example.-What weight will a white pine post sustain safely, whose height is 12 feet, and sides 8 and 12 inches respectively? The area 8 x 1296 inches; the square of the thickness, 8, 64. The area by the square of the thickness, 96 × 64, 6144; and this by 1175, the value of B, for white pine, equals 7,219,200. This, divided by 144, the square of the height = 50,133, the required weight in pounds.

Rule V.-If the post be square, the weight is found by this rule: Multiply the value of B, Table II., by the square of the area of the post in inches, and divide the product by the square of the height in feet, and the quotient will be the required weight in pounds.

Example.-What weight will a white oak post sustain safely, whose height is 9 feet, and sides each 6 inches? The value of B, set opposite white oak, is 3475; this, by (36 × 36 =) 1296, the square of the area, equals 4,503,600. This product, divided by 81, the square of the height, gives for quotient, 55,600, the required weight in pounds.

298. To ascertain the size of a post to sustain safely a given. weight when the height of the post is at least ten times the least side or diameter.

Rule VI.- When the post is to be round or cylindrical, the size may be obtained by this rule: Divide the weight in pounds by 0-589 times the value of B, Table II., and extract the square root of the product; multiply the square root by the height in feet, and the square root of this product will be the diameter of the post in inches.

Example.-What must be the diameter of a locust post, 10 feet high, to sustain safely 40,000 pounds? Here 0.589 times 5,460, the value of B for locust, Table II., equals 3215-9. The weight, 40,000, divided by 3215-9, equals 12-438. The square root of this, 3.5268, multiplied by 10, the height, equals 35.268, and the square root of this is 5.9386 or 515 inches, the required diameter of the post.

Rule VII.-If the post is to be rectangular, the size may be obtained by this rule: Multiply the square of the height in