CHAPTER VI

CRANK SHAFTS

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 formulæ 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 formulæ 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 for steel, where s = load on piston

the shaft will be = 3/8x1

120

length

(area of cylinder × maximum pressure of explosion), 1 = 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 × 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

line of one main bearing, set off a e = c d, and from b, which represents the centre line of the other main bearing, draw b d,

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 e ƒ will equal the load on the bearing B, and ƒ a, the load on the bearing A, o ƒ 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 × radius of crank. Make d k parallel to c a and = polar distance of, and draw a line through c k, which line continue through k; now draw line 7 m = radius of crank and parallel to d k. Then c 24 tons x the polar distance of in

9.5
12

feet = 24 × = 19 foot-tons, the required torsion moment,

9 tons the polar distance in feet = 9 ×

9.5 12

= 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:

E. T. M = equivalent twisting moment

=

In this example the diameter of crank shaft =

61308 140

3/437 = 7 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

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 î, p

9.5
12

= 12 tons × the polar distance in feet = 12 × = 9.5 foot

tons, which equals the required torsional moment, and c d = 15 9.5 tons the polar distance in feet = 15 × = 11.7 foot12

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 =268 foot-tons, and the diameter of crank pin

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 runs. This pressure should not exceed 600 lbs. per square inch of crank pin bearing surface, and is best taken as 500 lbs.

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 71⁄2 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

.. The length of the crank pin will be = 8.25 inches.

The crank arm E G is subject to bending, due to the force e ƒ 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' q = 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 × polar distance in feet 19 foot-tons = the maximum bending moment;

= 24 ×

9.5
12

=

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

= 9 x

9.5
12

=

7.1 foot-tons, the required torsional moment. Combine these two into an equivalent twisting moment. E. T. M. = M. + √ M2 + T2 = 19+ 192 + 7·12 393 foottons.

=

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

=

= 19.65 foot-tons = 44016 foot-lbs.

In

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 = 120 foot-lbs. safe bending moment.

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

= b x d2.

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 86 inches equals breadth of crank arm; the other