A.—­Some of those beams were of wood and some of cast iron.  The wood beams were so proportioned that the thickness was 1/58th of the circumference, and the depth 1/375.  The side of the beam, supposing it square, was found by multiplying the diameter of the cylinder by the length of the stroke, and extracting the cube root of the quotient, which will be the depth or thickness of the beam.  This rule allows a beam 16 feet long to bend 1/8th of an inch, and a beam 32 feet long to bend 1/4 of an inch.  For cast iron beams the square of the diameter of the cylinder, multiplied by the length between the centres, is equal to the square of the depth, multiplied by the thickness.

352. Q.—­What law does the strength of beams and shafts follow?

A.—­In the case of beams subjected to a breaking force, the strength with any given cohesion of the material will be proportional to the breadth, multiplied by the square of the depth; and in the case of revolving shafts exposed to a twisting strain, the strength with any given cohesive power of the material will be as the cube of the diameter.

353. Q.—­How is the strength of a cast iron shaft to resist torsion determined?

A.—­Experiments upon the force requisite to twist off cast iron necks show that if the cube of the diameter of neck in inches be multiplied by 880, the product will be the force of torsion which will twist them off when acting at 6 inches radius; on this fact the following rule is founded:  To find the diameter of a cast iron fly wheel shaft:—­multiply the square of the diameter of the cylinder in inches, by the length of the crank in inches, and extract the cube root of the product, which multiply by 0.3025, and the result will be the proper diameter of the shaft in inches at the smallest part, when of cast iron.

354. Q.—­What was Mr. Watt’s rule for the necks of his crank shafts?

A.—­Taking the pressure on the piston at 12 lbs. pressure on the square inch, and supposing this force to be applied at one foot radius, divide the total pressure of the piston reduced to 1 foot of radius by 31.4, and extract the cube root of the quotient, which is the diameter of the shaft:  or extract the cube root of 13.7 times the number of cubic feet of steam required to make one revolution, which is also the diameter of the shaft.

355. Q.—­Can you give any rule for the strength of the teeth of wheels?

A.—­To find the proper dimensions for the teeth of a cast iron wheel:—­ multiply the diameter of the pitch circle in feet by the number of revolutions to be made per minute, and reserve the product for a divisor; multiply the number of actual horses power to be transmitted by 240, and divide the product by the above divisor, which will give the strength.  If the pitch be given to find the breadth, divide the above strength by the square of the pitch in inches; or if the breadth be