Quality calculation of common materials

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The formulas below are provided for reference by those involved. When purchasing or assessing the weight of materials, if weighing is done, the actual weight should be used; otherwise, the weight can be estimated using the formulas below. (The formulas have had their calculation steps simplified to ensure accuracy.) Make sure to convert all units of measurement to millimeters, so that the resulting value is in kilograms. 1 meter = 1m = 1000 millimeters = 1000mm). 1. Solid rod material: Assuming the diameter of the rod is 150 mm and its length is 3000 mm (i.e., 3 meters), its weight is calculated as follows: Weight = Diameter × Diameter × Length × Coefficient 6.165 ÷ 1,000,000 = 150*150*3000*6.165/100,0000 = 416.1375 kilograms. This formula can also be used to calculate the weight of materials with circular holes or other cylindrical solid steel shapes. 2. Seamless pipes: Assuming the outer diameter of the pipe is 168 mm, the wall thickness is 8 mm, and the length is 6 meters (i.e., 6000 mm), its weight is calculated as follows: Weight = (Outer diameter – Wall thickness) × Wall thickness × Pipe length × 4 × Coefficient 6.165 ÷ 1,000,000 = (168–8) × 8 × 4 × 6000 × 6.165/1,000,000 = 189.388 kilograms. Alternatively, the inner diameter can be calculated first using the outer diameter and wall thickness; in this case, the inner diameter is 168–8×2 = 152 mm. The weight can then be calculated as: Weight = (Outer diameter × Outer diameter – Inner diameter × Inner diameter) × Pipe length × Coefficient 6.165 ÷ 1,000,000 = (168×168–152×152) × 6000 × 6.165/1,000,000 = 189.388 kilograms. This formula can also be used to calculate the weight of ring-shaped objects, such as flanges and cylindrical objects. Assuming the outer diameter of the flange is 300 mm, the inner diameter is 200 mm, and the thickness is 25 mm, its weight is as follows: Weight = (Outer diameter² – Inner diameter²) × Thickness × Coefficient 6.165 ÷ 1,000,000 = (300×300 – 200×200) × 25 × 6.165/1,000,000 = 7.7 kilograms. If the weight of the holes in the flange is deducted, assuming the diameter of each hole is 22 mm and there are 20 such holes, then the weight to be deducted is: Weight to be deducted for 20 holes = Hole diameter² × Thickness × Coefficient 6.165 ÷ 1,000,000 × Number of holes = 22×22×25×6.165/1,000,000 × 20 = 1.49 kilograms. Therefore, the net weight of the flange is approximately 7.7 – 1.49 = 6.2 kilograms.

3. Square steel plate: Assuming the width of the steel plate is 2 meters (i.e., 2000 mm), the length is 6 meters (i.e., 6000 mm), and the thickness is 20 mm, its weight is: Weight = Length × Width × Thickness × Coefficient 7.85 ÷ 1,000,000 = 2000×6000×20×7.85/1,000,000 = 1884 kilograms. Note: The coefficient of 6.165/100,0000 for the companies mentioned above is derived from: π/4 × density of steel, which is 7.85/100,0000; thus, 3.1415926/4 * 7.85/100,0000 = 6.165/100,0000. The coefficient of 7.85/100,0000 comes from the fact that the theoretical density of steel is 7.85/100,0000 kilograms per cubic millimeter. When the material is not steel (cast steel is treated as steel), for example, in the case of cast iron, the coefficient 6.165 is changed to 5.65 and the coefficient 7.85 is changed to 7.2. For materials such as copper (including copper alloys), the coefficient 6.165 is changed to 6.99 and the coefficient 7.85 is changed to 8.9. The coefficient of 1 million (100 0000) used in the division within the formula remains unchanged in both of the above cases. This post was last edited by Sugihai on 2008-1-18 10:54.]