Table II. - Weight in Pounds per Lineal Foot for Steel Pipe Weight I cubic inch Steel = .2833 pound 501.8 503.8 505.7 507.6 488.3 490.I 492.0 493.8 509.5 511.4 513.4 515.3 Table II. - Weight in Pounds per Lineal Foot for Steel Pipe Weight I cubic inch Steel = .2833 pound TABLE OF THE PROPERTIES OF TUBES AND ROUND BARS Plan of Table. This table was planned with a view of stating the properties of tubes and pipe in the best form for application to practice. The scheme is based upon the fact Fig. 132 In order to be able to apply this table to the solution of problems in tubular mechanics, it will only be necessary, in addition to having the above fundamental relation clearly in mind, to remember that the table states the proper ties of a series of solid round bars, each one foot long, whose diameters advance by .or inch to 16 inches, and thereafter by % inch. Calculation of Table. The table was calculated on an eight-slot Burkhardt machine, making use of the following data: C = TD = 3.1415927 D = circumference of a cross-section in inches. A = D2 = 0.78539816 D2 = area of cross-section in square inches. S π 4 C 12 = D= 0.26179939 D= cylindrical surface in square feet per foot length. V = 12 A = 3 πD2 = 9.4247780 D2 = volume in cubic inches per foot length. = 2.6700396 D2 weight of a round steel bar W = 0.2833 V = 3.3996 A 16 = = 0.0625 D2 = radius of gyration of cross-section, squared. y=D distance of farthest fiber from the axis of a round bar in = inches. The last value stated in each of the above formulæ is the one actually used in making the calculations. The machine calculations, except for the moment of inertia, I, were all carried out to the respective degrees of accuracy indicated by the constants of the above formulæ. Each result was then contracted to a lesser number of significant figures for the reason explained below. The moment of inertia, I, was obtained by multiplying the area of cross-section, A, by the corresponding radius of gyration squared, R2, both being taken to six significant figures. Precision of Tabular Statement. While entering the calculated values in this table, care was taken to have the precision of statement just sufficient to meet the demands of practice. The number of significant figures given in the different columns corresponding to any tabular diameter is based upon the assumption that diameters are measured to the nearest one-thousandth of an inch, thus involving a possible error of 0.0005 inch. This error in the diameter of a round bar will give rise to corresponding errors in its volume, weight, moment of inertia, and other properties. An investigation has shown these resulting errors to be as follows: For C, 0.00157 inch; for A, 0.000785 D; for S, 0.000131 square inch; for V, 0.00942 D; for W, 0.00267 D; for R2, 0.0000625 D; for I, 0.000098 D3; and for y, 0.00025 inch. Checking of Tabular Values. Each individual entry of this table has been calculated twice, and wherever a difference was found a third independent calculation was made to decide which of the two values in question was in error. The second calculation was made after the table had been traced, and all errors found were corrected directly on the tracings. A set of blue-prints was then made, and this was finally checked by the well-known method of differences. APPLICATION OF TABLE TO ROUND BARS 6.00 For the properties of round bars use the different tabular values direct. Thus for a round steel bar 6.35 inches in diameter, turn to the table, page 436, headed inches, and opposite 6.35, in column D, take the required properties from the table as follows: For circumference of cross-section, 19.949 inches; for area of crosssection, 31.669 square inches; for cylindrical surface, 1.6624 square feet per foot length; for volume, 380.03 cubic inches per foot length; for weight of steel bar, 107.66 pounds per foot length; for moment of inertia of cross-section, 79.81, from which the polar moment of inertia, being equal to twice the moment of inertia, is 79.81 X 2, or 159.62; for distance from axis of the bar to the most remote fiber, 3.175 inches; and for the square of the radius of gyration of cross-section, 2.5202. The table is applicable to diameters when stated in inches and hundredths to 16 inches and thereafter when stated in inches and eighths. When diameters are stated to thousandths of an inch, interpolate in per lineal foot, of a round steel bar 6.356 inches diameter, add to the tabular weight corresponding to 6.35, six-tenths of the difference of weights corresponding to diameters of 6.36 and 6.35; thus, difference of these weights is 108.00- 107.66 0.34; and six-tenths of this difference is 0.34 X 0.6 = 0.204; which added to the weight corresponding to 6.35 diameter gives 107.66+ 0.204 = 107.86 pounds per lineal foot as the weight of a bar 6.356 inches in diameter. Similarly all the other properties may be obtained; thus, moment of inertia, I, 79.81 +0.6 (80.32 79.81) 79.81 +0.31 = 80.12. = When diameters are stated to sixteenths, thirty-seconds, or sixty-fourths, above 16 inches interpolate similarly. Thus the weight of a round bar 18532 inches in diameter, since this diameter lies between 18% and 1814, will be (weight for 18%) + weight for 18%) or 877.15 +4 of 5/32-18 (889.29 (weight for 18 877.15) = 877.15 + To Find Diameter of Bar Corresponding to a Given Property. This is accomplished by taking the diameter opposite the tabular property nearest to that stated. For example, to find what diameter of round bar will correspond to a moment of inertia of 46, look down column I of the table until 45.91 is reached, which is the nearest tabular value, and then read opposite, in column D, 5.53 inches as the diameter required. Similarly a round bar of 15 square inches cross-sectional area will have a diameter of 4.37 inches, as read opposite 14.999 in column A. APPLICATION OF TABLE TO TUBES AND PIPE Let it be required to find the properties of a tube having outside and inside diameters of 7.62 and 7.02 inches respectively. It will be observed that according to the plan of this table (see page 419) the different properties of a tube may be grouped as follows: (1) The circumference, surface, fluid capacity, and distance of the farthest fiber from the axis are to be used direct as taken from the table. For the above example these will be as follows: From the table, column C, the outside circumference, opposite 7.62, is 23.939 inches; and the inside circumference, opposite 7.02, is 22.054 inches; from column S, similarly the outside and inside surfaces are found to be respectively 1.9949 and 1.8378 square feet per foot length of tube; from column V, the fluid capacity will be found opposite 7.02, the inside diameter, and is 464.46 cubic inches per foot length; while from column y, the distance of the farthest fiber from the axis of the tube will be found opposite the outside diameter, 7.62, and is 3.810 inches. Fig. 135 (2) The area of cross-section, volume of wall, weight, and moment of |