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EXAMPLE.-Let the moving force be equal to 7 horses, and the number of turns per minute 11, (see the case cited in Art.

201.) then by the rule

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the cube root of 146'09 is 5.267 nearly; or practically 5.3 inches should be the diameter of the shaft were it of cast iron. And 2.06 x 5.3 10.918 inches, or nearly 11 inches for the diameter of a fir shaft. It seems that a shaft of 9 inches square, of fir, was found equal to the strain; and one 11 inches diameter is at least stronger. I have made these calculations directly from the theory of equal cohesion, but it is so well a known fact that the lateral cohesion of fir is vastly inferior to the direct cohesion, that in the rule for fir shafts to resist torsion, an increase of diameter should be allowed by considering the number of horses' power about more than it is intended to be; at least, till experiment shall have given the precise effect of lateral còhesion in decreasing the force of shafts to resist torsion.

[H. 201.] If a shaft have to sustain both lateral stress and torsion, then the sum of the straining forces must be taken; and hence by Art. [A. 195,] and [A. 201,] we have + = ds. But in this equation it.

240 N W12

N

2 d

is difficult to calculate the value of the diameter, as it is what algebraists call an equation of the fourth degree. This difficulty may however be easily avoided by considering 2 d to be 2 only, for then the error will always be in excess, except when dis less than unity; and it is much better to be in excess than defect, Consequently we have as a practical rule (240+ W)+

=

N

H

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d, the diameter of the shaft in inches, when of cast iron. Where is the length in feet between the bearings, H the number of horses which are equal the power of the first mover, N the number of revolutions to be made by the shaft in a minute, and W the lateral stress in cwts.

EXAMPLE. Suppose that a cylindrical shaft of cast iron is to make 34 revolutions per minute, the power of the first mover

being equal to 3 horses, the length of the shaft 8 feet, and the lateral stress 3 cwts., when reduced to the middle point, then

3×82

(240 X 3 + 3x2) = (21∙18 + 96)*

34

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2

4.893 inches. (Ed.)

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202. With regard to the making of patterns of cast-iron shafts, I beg leave to refer the reader to what I have said in Essay I. page 219 to 229, relative to the making of patterns for cast-iron wheels, which is, in a great measure, applicable to those of shafts. Nor do I recollect any thing on this subject to add here; only I would remind the millwright to make the allowance for contraction of metal of one-eighth of an inch to the foot in the pattern.

203. The following table contains the dimensions of shafts subject to torsion, and to considerable lateral pressure, as they were executed by a respectable millwright. It will serve to shew the sizes of the parts as found in practice sufficiently strong,

and may be found useful to compare with those which would be produced, calculating on the principles laid down in this Essay.

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Column 6th, therefore, shews the diameters which these journals ought to have, were 400 used as the multiplier.-(See Art. 171.)

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APPENDIX.

COHESIVE STRENGTH OF DIFFERENT METALS.

204. "We shall take for the measure of cohesion the number of pounds avoirdupois which are just sufficient to tear asunder a rod or bundle of one inch square. From this it will be easy to compute the strength corresponding to any other dimension.

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