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of one flywheel only; and in engines giving over 40 E.H.P. the crank shaft should be enlarged where the flywheel is fitted, which enlargement may be continued sufficiently far to take the driving pulley. There should be a journal and outer bearing at the end of the crank shaft with the flywheel and pulley between it and the engine crank neck. This makes a much better arrangement than taking the whole power of the engine off at a distance from the bearing by means of an overhung crank shaft. In large engines, where the crankshaftisenlarged, say, a quarter of an inch for every inch diameter of engine crank neck bearing, and where the systems shown in fig. 69 are used, the size of the key may be as follows: Breadth of key = diameter of enlarged shaft + 4, depth of key. = half the breadth, and the depth of key-way in shaft may behalf the depth of key. But where there is no enlargement of shaft then the keys should be larger, especially in their depth.
Fig. 70 is a diagram giving the sizes of the keys as generally used. The following formulae for flywheels of single-acting ‘ Otto cycle engines is based on the practical experience gained by . Mr. F. W. Lanchester with over 500 engines, so far as the constants are concerned. It is assumed for purposes of convenience that flywheels of different sizes are geometrically proportional in their various parts. Let W = weight of wheel in cwts. = outside diameter of rim in feet. = volume in cubic feet swept by piston. mean effective pressure of working stroke. = constant. = revolutions per minute. Then W = }, x , x v × P. | ". D2 R2 W V P
That is, W =
Value of ‘C’ from Practice.—For engines for ordinary purposes, C = 125; for engines for electric lighting purposes, C = 250; Messrs. Crossley, for electric lighting engines, make C = 400. -
In practice the constants 125 to 250 are sufficient when the engine is fitted with good governor gear. Unfortunately for the efficiency of many engines flywheels are used which are too heavy when the governor is in action to respond quickly to the governor control, which is a more important factor than storing energy in the flywheel when steady running is required. But as this result is difficult to obtain without very ‘close ’ governing, many makers hide their inability by laying great stress on the heavy flywheels fitted to their engines.
CHAPTE R VI
IN the various engines dealt with from page to page attention has been drawn to the types of crank shafts used on them, Crossley's and Robey's using upon all sizes the form known as the “slab’ crank, and the other makers having arbitrarily and with no uniformity as to size of engine, adopted both this and the ‘bent' forms; and it would appear that in many cases the makers seem at a loss to know how to determine the sizes, if the divergencies from any possible formulae in their own manufactures can be taken as evidence. It is necessary in proportioning the crank shaft to take into account the combined strains to which it is subjected, one of which is due to the twisting action caused by the pressure acting on the crank pin, and the other to bending owing to the distance of centre of crank pin from centre of bearings. In some formulae for determining the size of crank shafts it is given that the latter force may be neglected and the diameter found for the twisting action only. By this rule the diameter of the shaft will be = &/i. & for steel, where s = load on piston (area of cylinder × maximum pressure of explosion), l = length of stroke in feet; but this rule is only approximate, as the bending moment is a very considerable item, and is well shown by the following graphic method. Suppose an engine has a cylinder of 19 inches bore x 30 inches stroke, with a maximum tangential force of 120 lbs. per square inch (as found in figs. 61 and 62), which may be taken as a constant in all engines using the same compression, a maximum pressure of 34,000 lbs. is exerted on the crank pin. By plotting out the centre line of the crank shaft A B (see fig. 71), together with centre line of crank pin C F G and centre lines of crank shaft bearings A and B, projecting these down to the horizontal line a b c, continue point c downwards, and set off c d = maximum force on crank pin, viz. 15 tons. And on a line projected downwards through a, representing the centre
join a d, and from e draw a line e o parallel to b d, cutting line a d at the point o. From o draw of perpendicular to a e, when ef will equal the load on the bearing B, and f a, the load on the bearing A, o f being the polar distance for the bending moment diagram a, b, d. In this engine; the power being all taken from the side A D of the crank shaft, the side B F has nothing to drive but the cross shaft of the engine, therefore the shaft B E is subject only to a bending strain, as shown by the right-hand side of bending moment diagram g, h, b. - ... The shaft A D is subject to a bending strain, as shown by the left-hand side of the bending moment diagram a, i, j, and to torsion due to the force on the crank pin–Viz. 15 tons x radius of crank. Make d k parallel to c a and = polar distance o f, and draw a line through c k, which line continue through k ; now draw line l m = radius of crank and parallel to d k. Then c l = 24 tons x the polar distance of in
feet = 24 × o = 19 foot-tons, the required torsion moment, r . K and i j = 9 tons x the polar distance in feet = 9 x o
= 7-1 foot-tons, the maximum bending moment to which the shaft A D is subjected. Combining these into an equivalent torsion moment by the following rule :
o E. T. M = equivalent twisting moment
... E. T. M. = 7-1 + V 7.1% + 192 = 27:37 foot-tons the diameter of the crank shaft.
3/E. T. M. in lbs. ^\/ 140 (safe moment for steel).
6.1308 - v/ 140 = Z 437 = 73 inches. The above moment for steel gives a factor of safety of 10, and is based upon a bar 1 inch in diameter breaking with 1,400 lbs. at the end of a lever 1 foot long. The crank pin F, C, G is subject to bending, as shown in the diagram
In this example the diameter of crank shaft =