This section is from the book "The English And American Mechanic", by B. Frank Van Cleve. Also available from Amazon: The English And American Mechanic.
The strains on the principal parts can be ascertained with great ease in the following manner - the strength being proportioned accordingly.
To find the Strain on the Post. Weight suspended, tons X Projection, feet / Height of post above ground, feet =
Strain on top of post, tons.
The post can then be calculated as a beam, twice as long as this height from ground, with twice the weight on the middle. [See Beams.]
Usually ¼ of cylinder diameter when the stroke is ½ that of piston. 1/3 " ¼
To find the proper size, under any circumstances, capable of supplying twice the quantity ordinarily used for injection.
Cub. ft. water per hour used in oylinder in form of steam _
Stroke of pump, ft. X strokes per minute of pump in square feet.
Solid. Metal round brass equal to ½ diameter of bearing.
General thickness web, etc., equal to ¼ diameter of bearing. With Feathers. Width at lightest equal to diameter of bearing. Thickness equal to 1/6 " "
The friction of metal on metal, without unguents, May be taken at 1/6 of the weight up to 40 lbs. per square inch. " 1/5 " " 100 "
Brass on cast iron ¼ " " 800 "
Wrought on cast iron 1/3 " " 500 "
With tallow at 1/10 of the weight. " olive oil at 1/13 "
800 lbs. per inch forces out the oil. Friction of journals under ordinary circumstances 1/10 of weight. " well oiled, sometimes only 1/60 "
(Revolutions per min)² X dia. in ft. X weight / 5870 = centrifugal force in terms of weight.
Area, sq. ins. X Length, ft. X 3 1/7 = Weight, lbs. cast iron.
Multiplier for Cast iron................. 3.156 or 31.
Wrought iron........... 3.312 or 3 1/3.
" Lead...................... 4.854.
" Brass...................... 3.644.
" Copper................... 3.87.
Or, Area, sq. ins. X 10 = lbs. per yard for wrought iron.
Area, sq. ins. X Length, ft. / 31.9 = Weight, cwts. cast iron. For wrought iron, divide by 33.6.
Wheels which act upon each other in the same plane.
Divide circumference at the pitch-line by the number of teeth.
A wheel 40 ins. in diameter requires 75 teeth ; what is its pitch ?
3.1416X40 / 75 = 1.6755 ins.
 
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