Practical Hand Book for Millwrights Describing the Practical Planning and Arrangement of Mill Buildings: Strength of Materials ... and All Subjects Directly Connected with the Installation of Mill Machinery; Etc., Etc |
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Practical Hand Book for Millwrights Describing the Practical Planning and ... Calvin Franklin Swingle No preview available - 2017 |
Practical Hand Book for Millwrights Describing the Practical Planning and ... Calvin Franklin Swingle No preview available - 2018 |
Practical Hand Book for Millwrights Describing the Practical Planning and ... Calvin Franklin Swingle No preview available - 2014 |
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angle antimony arms babbitt Babbitt metal beam bearing bed-plate bending boiler bolts bottom brick cast iron cement cent Cloth compression concrete crank crank pin cubic cylinder damp depth diam diameter distance drive pipe engine equal factor of safety feet per minute flange floor foot foundation friction gear girder hangers head header heat heavy holes horizontal horsepower hydraulic ram I-BEAMS joints journal leather belts length located lubrication machine material metal method millwright motor multiply pitch placed plate pounds pressure proportions pulley pulse valve purlins rafters Rape oil revolutions per minute roof rope SAFE LOADS sand screw shaft shown in Fig side span speed square inch steam steel strain strands strength stress struts supported surface TABLE teeth tensile stress tension thickness timber tion truss tube turbine velocity vertical wall wheel width wrought iron
Popular passages
Page 51 - ... in the proportion of one part of cement, two parts of sand and four parts of stone or gravel, or in such proportion as to produce a maximum density.
Page 411 - The quantities of water discharged during the same time by different apertures under different heights of water in the reservoir, are to one another in the compound ratio of the areas of the apertures, and of the square roots of the heights in the reservoirs.
Page 221 - ... of importance and should be considered for each case. The ropes commonly used for the transmission of power in factories or mills vary from 3 to 5 inches in circumference. No matter what the diameter of the pulley may be, ropes of 1| inches diameter should not be exceeded for main drives, and l| inches diameter for secondary drives.
Page 413 - To find the lateral pressure of water upon the side of a tank, multiply in inches, the area of the submerged side, by the pressure due to one-half the depth. Example. — Suppose a tank to be 12 feet long and 12 feet deep. Find the pressure on the side of the tank. 144 X 144 = 20,736 square inches area of side.
Page 396 - Pipes.—It is frequently desired to know what number of pipes of a given size are equal in carrying capacity to one pipe of a larger size. At the same velocity of flow the volume delivered by two pipes of different sizes is proportional to the squares of their diameters; thus, one 4-inch pipe will deliver the same volume as four 2-inch pipes. With the same head, however, the velocity is less in the smaller pipe and the volume delivered varies about as the square root of the fifth power (ie, as the...
Page 78 - To find the dimensions of purlins: Rule.— Multiply the cube of the length of the purlin in feet by the distance the purlins are apart in feet, and the fourth root of the product will give the depth in inches, and the depth multiplied by 0.6 will give the thickness.
Page 411 - To find the velocity in feet per minute necessary to discharge a given volume of water in a given time, multiply the number of cubic feet of water by 144 and divide the product by the area of the pipe in inches.
Page 194 - Weishtof material (Ibs. per cub. in.). (inches). ANALYSIS. If a flexible band be wrapped completely about a pulley, and a heavy stress be put upon each end of the band, the rim of the pulley will tend to collapse just like a boiler tube with steam pressure on the outside of it. A compressive stress is induced which is very nearly evenly distributed over the cross-section of the rim, except at points where the arms are connected thereto. At these points the arms, acting like rigid posts, take this...
Page 413 - To find the head which will produce a given velocity of water through a pipe of a given diameter and length: Multiply the square of the velocity, expressed in feet per second, by the length of pipe multiplied by the quotient obtained by dividing 13.9 by the diameter of the pipe in inches, and divide the result obtained by 2,500. The final amount will give the head in feet. Example.— The horizontal length of pipe is 1,200 feet, and the diameter is 4 inches.