The first step is to ascertain the number of cubic inches in the casting. To illustrate the method, consider the column shown in the figure to be divided into several parts, as given below. With irregular shapes, as at e, follow the principle of "give and take," reducing the shape to an equivalent one, easy to calculate. The operations are then as follows:

 Cu. In. Cylinder a, excluding cap. Net height = 12 ft. =(l 1/2 in.+1 in.) = 141.5 in.; net area = 10" circle - 8 1/2" circle = 21.8 sq. in. Contents = 21.8 sq. in. X 141.5 in. = 3,084.7 Base b = 19 in. X19 in. X 1 1/2 in. = 541.5 cu. in.; deduct area of 10" circle X 1 1/2"- 117.8cu. in.; also, for corners (5" square-5" circle)X 1 1/2" = 8.1 cu. in. Net contents = 415.6
 Triangular ribs c = 1/2 (4 1/2 in.x 7 in.)x1 1/8 in.x 4 = Brackets d. Horizontal part d is, very nearly, 5 in. X 5 in. X 1 1/2 in. = 37 1/2 cu. in. Vertical part c: To compensate for swelled portion next cylinder (see e") sider vertical portion as triangular, with sides 5 in. X 7 in.; its contents are J (5 in. X 7 in.) x 1 1/8 in. = 19.7 cu. in. Each bracket = 57.2 cu. in: 4 brackets =57.2 X 4 = 70.9 228.8 Lugs f = 11 in. X 5 in. X 1 in. = 55 cu. in.; deduct 1/2 (4" square - 4" circle) X 1" = 1.7 cu. in.: also, two 3/4" holes = .9 cu. in. Net contents of each lug, 52.4 cu. in.; 4 lugs = 52.4 X 4 = 209.6 Cap g = (16" X 16" - 4" circle) X 1 1/2" = 865.2 Carried forward, 4,374.8
 Brought forward, 4,374.8 Upper part of cap h = (16" square - 12" square) x 1" thick - deduction for bevel; 256 cu. in. - 144 cu. in. = 112 cu. in. Portion to be deducted consists of 4 wedges, each 16 in. X 1 in. at the back, 13 in. at the edge, which is 1 1/2 in. from back; by rule for wedges, deduction is [J X (16 in. + 16 in. + 13 in.) X 1 in. x 1 1/2 in.] X 4 = 45 cu. in. Net contents of part h = 112 - 45 = 67.0 Total contents of column = 4,441.8

One cu. ft. of cast iron weighs 450 lb., or 1 cu. in. weighs .26 lb.-called 1/4 lb., roughly. Hence the weight is 4,442 X .26 = 1,155 lb. This calculation is much more detailed than is usual; generally, no account is taken of rounded corners, small fillets, holes, etc.

When a casting or a rolled shape of steel, etc. is of uniform cross-section throughout - as a rail, I beam, or channel - the weight may be very expeditiously determined after calculating the sectional area. A cubic foot of wrought iron weighs 480 lb.; hence, a piece 1 yd. long and 1 in. square contains 36 cu. in., and weighs 36/1728 of 480 lb.= 10 lb.; or, if 1 ft. long and 1 in. square, weighs 10/3= 3 1/3 lb. Therefore, if the sectional area be known or calculated, the weight per foot may be found by taking 10/3 the area, or by multiplying the latter by 3 1/3. This result will be in pounds, and, if multiplied by the total length in feet of the member, will give the whole weight.

If the member is made of steel, which weighs 490 lb. per cu. ft., the weight per foot may be determined by multiplying the sectional area by 3.4. For cast iron the multiplier is 3 1/8.

Example

What is the weight per foot of a 20" steel I beam, having a sectional area of 26.4 sq. in.? 26.4 X 3.4 = 89.76 lb., or, practically, 90 lb. By reference to Table XIX, page90, it will be seen that this is the weight given.

If the weight per foot is given, and it is desired to find the sectional area, the former, divided by 10/3, or multiplied by .3, will give the required area. For steel, the multiplier is .294, and for cast iron .32.