Part 8 (1/2)

in which

_r_ = _L_ _l_; _L_ = lead of spiral; _l_ = lead of hob thread.

For example, if a hob has a pitch circ.u.mference of 3.25, a single thread of 0.75 inch lead, and 6 spiral flutes, what compensating gears would be required?

The lead _L_ of the spiral flutes is first determined by dividing the square of the circ.u.mference _C_ of the hob at the pitch line by the lead _l_ of the hob thread. Thus lead _L_ = _C^2_/_l_, or, in this case, _L_ = 3.25^2/0.75 = 14 inches, approximately. Then _r_ = 14 0.75 = 18-2/3.

Inserting these values in the formula for ratio R,

18-2/3 + 1 19-2/3 19-2/3 3 59 _R_ = ---------- = ------ = ---------- = -- 18-2/3 18-2/3 18-2/3 3 56

Hence, the compensating gears will have 56 and 59 teeth, respectively, the latter being the driver. As the gears for 6 flutes listed on the regular index plate are, stud-gear 60 teeth, cam-shaft gear 40 teeth, the entire train of gears would be as follows: Gear on stud, 60; _driven_ intermediate gear, 56; _driving_ intermediate gear, 59; cam-shaft gear, 40. It will be understood that the position of the driving gears or the driven gears can be transposed without affecting the ratio.

=Cla.s.ses of Fits Used in Machine Construction.=--In a.s.sembling machine parts it is necessary to have some members fit together tightly, whereas other parts such as shafts, etc., must be free to move or revolve with relation to each other. The accuracy required for a fitting varies for different cla.s.ses of work. A shaft that revolves in its bearing must be slightly smaller than the bearing so that there will be room for a film of lubricant. A crank-pin that must be forced into the crank-disk is made a little larger in diameter than the hole, to secure a tight fit.

When a very accurate fitting between two cylindrical parts that must be a.s.sembled without pressure is required, the diameter of the inner member is made as close to the diameter of the outer member as is possible. In ordinary machine construction, five cla.s.ses of fits are used, _viz_; running fit, push fit, driving fit, forced fit and shrinkage fit. The running fit, as the name implies, is employed when parts must rotate; the push fit is not sufficiently free to rotate; the other cla.s.ses referred to are used for a.s.sembling parts that must be held in fixed positions.

=Forced Fits.=--This is the term used when a pin, shaft or other cylindrical part is forced into a hole of slightly smaller diameter, by the use of a hydraulic press or other means. As a rule, forced fits are restricted to parts of small and medium size, while shrinkage fits have no such limitations and are especially applicable when a maximum ”grip”

is desired, or when (as in the construction of ordnance) accurate results as to the intensity of stresses produced in the parts united are required. The proper allowance for a forced fit depends upon the ma.s.s of metal surrounding the hole, the size of the work, the kind and quality of the material of which the parts are composed and the smoothness and accuracy of the pin and bore. When a pin or other part is pressed into a hole a second time, the allowance for a given tonnage should be diminished somewhat because the surface of the bore is smoother and the metal more compact. The pressure required in a.s.sembling a forced fit will also vary for cast hubs of the same size, if they are not uniform in hardness. Then there is the personal factor which is much in evidence in work of this kind; hence, data and formulas for forced fit allowances must be general in their application.

=Allowance for Forced Fits.=--The allowance per inch of diameter usually ranges from 0.001 inch to 0.0025 inch, 0.0015 being a fair average.

Ordinarily, the allowance per inch decreases as the diameter increases; thus the total allowance for a diameter of 2 inches might be 0.004 inch, whereas for a diameter of 8 inches the total allowance might not be over 0.009 or 0.010 inch. In some shops the allowance is made practically the same for all diameters, the increased surface area of the larger sizes giving sufficient increase in pressure. The parts to be a.s.sembled by forced fits are usually made cylindrical, although sometimes they are slightly tapered. The advantages of the taper form are that the possibility of abrasion of the fitted surfaces is reduced; that less pressure is required in a.s.sembling; and that the parts are more readily separated when renewal is required. On the other hand, the taper fit is less reliable, because if it loosens, the entire fit is free with but little axial movement. Some lubricant, such as white lead and lard oil mixed to the consistency of paint, should be applied to the pin and bore before a.s.sembling, to reduce the tendency of abrasion.

Allowances for Different Cla.s.ses of Fits

(Newall Engineering Co.)

+-----+--------------------------------------------------------------+ | | Tolerances in Standard Holes[1] | |Cla.s.s+------------+---------+---------+---------+---------+---------+ | | Nominal | Up to | 9/16”-1”| 1-1/16”-| 2-1/16”-| 3-1/16”-| | | Diameters | 1/2” | | 2” | 3” | 4” | +-----+------------+---------+---------+---------+---------+---------+ | | High Limit | +0.0002 | +0.0005 | +0.0007 | +0.0010 | +0.0010 | | A | Low Limit | -0.0002 | -0.0002 | -0.0002 | -0.0005 | -0.0005 | | | Tolerance | 0.0004 | 0.0007 | 0.0009 | 0.0015 | 0.0015 | +-----+------------+---------+---------+---------+---------+---------+ | | High Limit | +0.0005 | +0.0007 | +0.0010 | +0.0012 | +0.0015 | | B | Low Limit | -0.0005 | -0.0005 | -0.0005 | -0.0007 | -0.0007 | | | Tolerance | 0.0010 | 0.0012 | 0.0015 | 0.0019 | 0.0022 | +-----+------------+---------+---------+---------+---------+---------+ | Allowances for Forced Fits | +-----+------------+---------+---------+---------+---------+---------+ | | High Limit | +0.0010 | +0.0020 | +0.0040 | +0.0060 | +0.0080 | | F | Low Limit | +0.0005 | +0.0015 | +0.0030 | +0.0045 | +0.0060 | | | Tolerance | 0.0005 | 0.0005 | 0.0010 | 0.0015 | 0.0020 | +-----+------------+---------+---------+---------+---------+---------+ | Allowances for Driving Fits | +-----+------------+---------+---------+---------+---------+---------+ | | High Limit | +0.0005 | +0.0010 | +0.0015 | +0.0025 | +0.0030 | | D | Low Limit | +0.0002 | +0.0007 | +0.0010 | +0.0015 | +0.0020 | | | Tolerance | 0.0003 | 0.0003 | 0.0005 | 0.0010 | 0.0010 | +-----+------------+---------+---------+---------+---------+---------+ | Allowances for Push Fits | +-----+------------+---------+---------+---------+---------+---------+ | | High Limit | -0.0002 | -0.0002 | -0.0002 | -0.0005 | -0.0005 | | P | Low Limit | -0.0007 | -0.0007 | -0.0007 | -0.0010 | -0.0010 | | | Tolerance | 0.0005 | 0.0005 | 0.0005 | 0.0005 | 0.0005 | +-----+------------+---------+---------+---------+---------+---------+ | Allowances for Running Fits[2] | +-----+------------+---------+---------+---------+---------+---------+ | | High Limit | -0.0010 | -0.0012 | -0.0017 | -0.0020 | -0.0025 | | X | Low Limit | -0.0020 | -0.0027 | -0.0035 | -0.0042 | -0.0050 | | | Tolerance | 0.0010 | 0.0015 | 0.0018 | 0.0022 | 0.0025 | | | High Limit | -0.0007 | -0.0010 | -0.0012 | -0.0015 | -0.0020 | | Y | Low Limit | -0.0012 | -0.0020 | -0.0025 | -0.0030 | -0.0035 | | | Tolerance | 0.0005 | 0.0010 | 0.0013 | 0.0015 | 0.0015 | | | High Limit | -0.0005 | -0.0007 | -0.0007 | -0.0010 | -0.0010 | | Z | Low Limit | -0.0007 | -0.0012 | -0.0015 | -0.0020 | -0.0022 | | | Tolerance | 0.0002 | 0.0005 | 0.0008 | 0.0010 | 0.0012 | +-----+------------+---------+---------+---------+---------+---------+

[1] Tolerance is provided for holes, which ordinary standard reamers can produce, in two grades, Cla.s.ses A and B, the selection of which is a question for the user's decision and dependent upon the quality of the work required; some prefer to use Cla.s.s A as working limits and Cla.s.s B as inspection limits.

[2] Running fits, which are the most commonly required, are divided into three grades: Cla.s.s X for engine and other work where easy fits are wanted; Cla.s.s Y for high speeds and good average machine work; Cla.s.s Z for fine tool work.

=Pressure for Forced Fits.=--The pressure required for a.s.sembling cylindrical parts depends not only upon the allowance for the fit, but also upon the area of the fitted surfaces, the pressure increasing in proportion to the distance that the inner member is forced in. The approximate ultimate pressure in pounds can be determined by the use of the following formula in conjunction with the accompanying table of ”Pressure Factors.”

=Pressure Factors=

+-----+-----++-----+-----++-----+-----++------+------++------+------+ |Diam-|Pres-||Diam-|Pres-||Diam-|Pres-||Diam- |Pres- ||Diam- |Pres- | |eter,|sure ||eter,|sure ||eter,|sure ||eter, |sure ||eter, |sure | |In- |Fac- ||In- |Fac- ||In- |Fac- ||In- |Fac- ||In- |Fac- | |ches | tor ||ches | tor ||ches | tor ||ches |tor ||ches |tor | +-----+-----++-----+-----++-----+-----++------+------++------+------+ |1 | 500 ||3-1/2| 132 ||6 | 75 || 9 | 48.7 ||14 | 30.5 | |1-1/4| 395 ||3-3/4| 123 ||6-1/4| 72 || 9-1/2| 46.0 ||14-1/2| 29.4 | |1-1/2| 325 ||4 | 115 ||6-1/2| 69 ||10 | 43.5 ||15 | 28.3 | |1-3/4| 276 ||4-1/4| 108 ||6-3/4| 66 ||10-1/2| 41.3 ||15-1/2| 27.4 | |2 | 240 ||4-1/2| 101 ||7 | 64 ||11 | 39.3 ||16 | 26.5 | |2-1/4| 212 ||4-3/4| 96 ||7-1/4| 61 ||11-1/2| 37.5 ||16-1/2| 25.6 | |2-1/2| 189 ||5 | 91 ||7-1/2| 59 ||12 | 35.9 ||17 | 24.8 | |2-3/4| 171 ||5-1/4| 86 ||7-3/4| 57 ||12-1/2| 34.4 ||17-1/2| 24.1 | |3 | 156 ||5-1/2| 82 ||8 | 55 ||13 | 33.0 ||18 | 23.4 | |3-1/4| 143 ||5-3/4| 78 ||8-1/2| 52 ||13-1/2| 31.7 ||.... | .... | +-----+-----++-----+-----++-----+-----++------+------++------+------+

a.s.suming that _A_ = area of fitted surface; _a_ = total allowance in inches; _P_ = ultimate pressure required, in tons; _F_ = pressure factor based upon a.s.sumption that the diameter of the hub is twice the diameter of the bore, that the shaft is of machine steel, and the hub of cast iron, then,

_A_ _a_ _F_ _P_ = --------------- 2

_Example:_--What will be the approximate pressure required for forcing a 4-inch machine steel shaft having an allowance of 0.0085 inch into a cast-iron hub 6 inches long?

_A_ = 4 3.1416 6 = 75.39 square inches;

_F_, for a diameter of 4 inches, = 115 (see table of ”Pressure Factors”). Then,

_P_ = (75.39 0.0085 115)/2 = 37 tons, approximately.