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i, j, d,h, g (fig. 71), and also to torsion due to the force e ƒ acting at B with radius of crank. Make bn = radius of crank, and from n draw a line perpendicular to b n cutting b d in p; then n, p = 12 tons × the polar distance in feet = 12 x

9.5
12

= 9.5 foot

tons, which equals the required torsional moment, and c d = 15

tons the polar distance in feet = 15 x

9.5
12

= 11.7 foot

tons, which equals the maximum bending moment to which the crank pin is subjected.

Combine these two into an equivalent twisting moment.

E. T. M. M. + √M2 + T2 = 11·7 + √11·72 + 9.52 = 26.8 foot-tons, and the diameter of crank pin

3 E. T. M. (in lbs.)

140 (safe moment for steel).

In this example the diameter of crank pin

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In practice, however, the crank pin is never made smaller than the diameter of the crank neck, but generally larger (a good proportion is one-eighth of itself larger), and its length is determined by the maximum working pressure to which it is advisable to subject it so as to eliminate any fear of heating during long This should not exceed 600 lbs. per square pressure inch of crank pin bearing surface, and is best taken as 500 lbs.

runs.

The maximum tangential pressure on crank pin being 34,000 lbs., and dividing this by 500 gives 68 square inches as the required area of bearing surface.

The diameter of crank neck being 73 inches, and making the diameter of pin one-eighth larger we have 84 inches for its diameter; and as the net bearing surface of a journal equals one-third of its circumference × its length, or its diameter x the length, 68 = 8.25 inches. 8.25

.. The length of the crank pin will be

The crank arm E G is subject to bending, due to the force ef acting at E, for which the bending moment diagram is found

by making the angle x at E' = the angle y at b, then G' Չ = the maximum bending strain; and to torsion due to the same force acting with the leverage E G and for which amount =gh. The strains in this arm are small, and are consequently neglected in proportioning the size of the crank arm, the strength being the same as the crank arm D F.

The crank arm D F is subject to bending due to the maximum tangential pressure acting at E, which equals the force cl = D'r 24 tons x polar distance in feet

=

24 ×

9.5

12

=

=

19 foot-tons = the maximum bending moment;

and also to torsion, due to the force a ƒ acting with the leverage A D, which equals i j = 9 tons × polar distance in feet

9.5

= 9 x = 7·1 foot-tons, the required torsional moment.

12

Combine these two into an equivalent twisting moment.
E. T. M. = M. + √ M2 + T2 = 19 + √192 + 7·12 = 39.3 foot-

tons.

To find the size of the crank arm assume it to be a cantilever fixed at one end and loaded at the other, the equivalent equivalent twisting moment bending moment taken = 2

In

this case

39.3
2

= 19.65 foot-tons

= 44016 foot-lbs.

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A 1-inch square steel bar 1 foot long, fixed at one end and loaded at the other, will break with a maximum bending moment of 1,200 foot-lbs., or with a factor of safety of 10 foot-lbs. safe bending moment.

= 120

The safe bending moment of any bar will equal b × d2 × 120 ft.-lbs. where b equals breadth and d equals depth,

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By assuming one dimension for crank arm, the other may readily be found. Thus, assuming the breadth of crank web, which is often made 8 of the diameter of crank shaft, then 7.5 x 8 6 inches equals breadth of crank arm; the other

44016

dimension will be

=

44016
= 10.1 inches.
62 + 120 4320

In the above example the flywheel and driving pulley are on one side of the engine, with the outer end of crank shaft carried by a third bearing, and, strictly speaking, the bending moment diagram should also be drawn for weight of flywheel, &c. ; but in most cases this moment will be less than the combined equivalent twisting moment as found for shaft A D, so that the diameter of the crank shaft as found from the latter will be more than ample to cover the extra weight of flywheel and driving pulley,

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also in engines of this class the crank shaft is enlarged between the engine main bearing and outer bearing to compensate for loss of section owing to the cutting of key bed. This enlargement may be made 4 inch for every inch in diameter of crank shaft. (See Flywheels, page 72.)

G

With fig. 72 is given a table of sizes of crank shafts, of the slab form, from actual practice of some of the best makers, and with fig. 73 a table of sizes for the bent' forms.

FIG. 73

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When dealing with the Robey gas engines mention was made of the method by which the reciprocating parts were balanced, by means of counter weights strapped to the webs. The method of building up and fixing them is clearly shown on figs. 75 and 76.

FIG. 75

FIGS. 75, 76.-ROBEY BALANCE CRANK

FIG. 76

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