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iron is taken in place of cast iron in the last example, what must be its diameter ?

1509 x 9 v

= 9.89 inches diameter nearly. 14

3

ON TORSION OR TWISTING.

The strength of bodies to resist torsion, or wrenching asunder, is directly as the cubes of their diameters; or, if square, as the cube of one side; and inversely as the force applied multiplied into the length of the lever.

Hence the rule.-1. Multiply the strength of an inch bar by experiment (as in the following table) by the cube of the diameter, or of one side in inches; and divide by the radius of the wheel, or length of the lever also in inches; and the quotient will be the ultimate strength of the shaft or bar, in libs. avoirdupois nearly

2.-Multiply the force applied in pounds by the length of the lever in inches, and divide the product by one-third of the ultimate strength of an inch bar, (as in the table,) and the cube root of the quotient will be the diameter, or side of a square in inches; that is capable of resisting that force permanently. The following table contains the result of experiments on inch bars, of various metals, in libs. avoirdupois.

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EXAMPLE 1-What weight, applieil on the end of a 5 feet lever, will wrench asunder a 3 inch round bar of cast iron ? 11943 x 33

= 5374 libs. avoirdupois nearly.

60 EXAMPLE 2.-Required the side of a square bar of wrought iron, capable of resisting the twist of 600 lbs. on the end of a lever 8 feet long. 600 x 96

= 2 inches nearly.

5120 In the case of revolving shafts for machinery, &c. the strength is directly as the cubes of their diameters and revolutions, and inversely as the resistance they have to overcome; hence,

From practice, we find that a 40-horse power steam engine, making 25 revolutions per minute, requires a shaft (if made of wrought iron) to be 8 inches dia. meter; now the cube of 8, multiplied by 25, and divided by 40, = 320; which serves as a constant multiplier for all others in the same proportion.

EXAMPLE 3.-What must be the diameter of a wrought iron shaft for an engine of 65-horse power, making 23 revolutions per minute?

65 x 320 3

= 9.67 inches diameter nearly.

23 Mr. Robertson Buchanan, in his essay on shafts, gives 400 as a constant multiplier for cast iron shafts that are intended for first movers in machinery;

200 for second movers, and

100 for shafts connecting smaller machinery, &c. Example 1.-- T'he velocity of a 30-horse power steam engine is intended to be 19 revolutions per minute. Required the diameter of bearings for the fly wheel shaft.

400 x 30 3 ✓

= 8.579 inches diameter nearly. 19

EXAMPLE 2.-Required the diameter of the bearings of shafts, as second movers from a 30-horse engine; their velocity being 36 revolutions per minute.

200 x 30 3 v

= 5.5 inches dianieter.

36 NOTE.—When shafting is intended to be of wrought iron, use 160 for second movers; and 80 for shafts connecting smaller machinery.

OF THE MECHANICAL POWERS.

When power is applied to overcome weight, or force to overcome resistance, the machines employed are called mechanic powers, and the application of such, the science of mechanics.

The power and weight are said to balance each other, or to be in equilibrio, when the effort of the one to produce motion in one direction is equal to the effort of the other to produce it in an opposite direction; or when the weight opposes that degree of resistance which is precisely required to destroy the action of the power.

The momentum or quantity of force of any moving body is the result of the quantity of matter multiplied by the velocity by which it is moved ; and when the product arising from the multiplication of the particular quantities of matter in any two bodies by their respective velocities are equal, their momentum will be so too.

And it holds universally true, that when two bodies are suspended upon any machine, so as to act contrary to each other, if the machine be put in motion, and the perpendicular ascent of one body, multiplied into its weight, be equal to the perpendicular descent of the other, multiplied into its weight, those bodies, however unequal they be in weight, will balance each other in all situations; for, as the whole ascent of the one is performed in the same time as the whole descent of the other, their respective velocities must be as the spaces they move through ; and the excess of weight in the one is compensated by the excess of velocity in the other. Upon this principle it is easy to compute the power of any machine, either simple or compound; for it is only finding how much swifter

I

98

the power moves than the weight; and just so much is the power increased by the help of the machine.

The simple machines, usually called mechanic powers, are six in number, namely, the Lever, the Wheel and Axle, the Pulley, the Inclined Plane, the Wedge, and the Screw.

There are three kinds of levers, caused by the different situations of the weights, props, and powers.

1.-When the weight is at one end, the power at the other, and the prop somewhere between.

2.-When the prop is at one end, the power at the other, and the weight between. And,

3.-When the prop is at one end, the weight at the other, and the

power

between. Thus, 1.

3.

2.

P

Р
96

B
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80

80

80

W

W

W

Р

from the prop;

In the first and second kind, the advantage gained is as the distance of the power from the prop, to the distance of the weight from the prop.

In the third kind, that there may be a balance between the power and the weight, the intensity of the power must exceed the intensity of the weight, just as much as the distance of the weight from the prop

exceeds the distance of the power
that is, P X AF = W x BF; or the power

and weight are reciprocally as the distances at which they act.

Or, in other words, Multiply the weight given by the distance from the prop, and divide by the distance from the power; the quotient will be the power or weight required.

EXAMPLES 1, 2, and 3.

Required the power necessary to counterpoise a weight of 80 lbs. on each of the three levers, whose

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