GearAssignment/Mechanical

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Chapter Outline

13–1 Types of Gears 666

13–2 Nomenclature 667

13–3 Conjugate Action 669

13–4 Involute Properties 670

13–5 Fundamentals 670

13–6 Contact Ratio 676

13–7 Interference 677

13–8 The Forming of Gear Teeth 679

13–9 Straight Bevel Gears 682

13–10 Parallel Helical Gears 683

13–11 Worm Gears 687

13–12 Tooth Systems 688

13–13 Gear Trains 690

13–14 Force Analysis—Spur Gearing 697

13–15 Force Analysis—Bevel Gearing 701

13–16 Force Analysis—Helical Gearing 704

13–17 Force Analysis—Worm Gearing 706

Gears—General13

665

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666 Mechanical Engineering Design

This chapter addresses gear geometry, the kinematic relations, and the forces transmit- ted by the four principal types of gears: spur, helical, bevel, and worm gears. The forces transmitted between meshing gears supply torsional moments to shafts for motion and power transmission and create forces and moments that affect the shaft and its bearings. The next two chapters will address stress, strength, safety, and reli- ability of the four types of gears.

13–1 Types of Gears Spur gears, illustrated in Fig. 13–1, have teeth parallel to the axis of rotation and are used to transmit motion from one shaft to another, parallel, shaft. Of all types, the spur gear is the simplest and, for this reason, will be used to develop the primary kinematic relationships of the tooth form. Helical gears, shown in Fig. 13–2, have teeth inclined to the axis of rotation. Helical gears can be used for the same applications as spur gears and, when so used, are not as noisy, because of the more gradual engagement of the teeth during meshing. The inclined tooth also develops thrust loads and bending couples, which are not present with spur gearing. Sometimes helical gears are used to transmit motion between nonparallel shafts. Bevel gears, shown in Fig. 13–3, have teeth formed on conical surfaces and are used mostly for transmitting motion between intersecting shafts. The figure actually illustrates straight-tooth bevel gears. Spiral bevel gears are cut so the tooth is no longer straight, but forms a circular arc. Hypoid gears are quite similar to spiral bevel gears except that the shafts are offset and nonintersecting.

Figure 13–1 Spur gears are used to transmit rotary motion between parallel shafts.

Figure 13–2 Helical gears are used to transmit motion between parallel or nonparallel shafts.

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Gears—General 667

Worms and worm gears, shown in Fig. 13–4, represent the fourth basic gear type. As shown, the worm resembles a screw. The direction of rotation of the worm gear, also called the worm wheel, depends upon the direction of rotation of the worm and upon whether the worm teeth are cut right-hand or left-hand. Worm gearsets are also made so that the teeth of one or both wrap partly around the other. Such sets are called single- enveloping and double-enveloping worm gearsets. Worm gearsets are mostly used when the speed ratios of the two shafts are quite high, say, 3 or more.

13–2 Nomenclature The terminology of spur-gear teeth is illustrated in Fig. 13–5. The pitch circle is a theoretical circle upon which all calculations are usually based; its diameter is the pitch diameter. The pitch circles of a pair of mating gears are tangent to each other. A pinion is the smaller of two mating gears. The larger is often called the gear. The circular pitch p is the distance, measured on the pitch circle, from a point on one tooth to a corresponding point on an adjacent tooth. Thus the circular pitch is equal to the sum of the tooth thickness and the width of space.

Figure 13–3 Bevel gears are used to transmit rotary motion between intersecting shafts.

Figure 13–4 Worm gearsets are used to transmit rotary motion between nonparallel and nonintersecting shafts.

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668 Mechanical Engineering Design

The module m is the ratio of the pitch diameter to the number of teeth. The custom- ary unit of length used is the millimeter. The module is the index of tooth size in SI. The diametral pitch P is the ratio of the number of teeth on the gear to the pitch diameter. Thus, it is the reciprocal of the module. Since diametral pitch is used only with U.S. units, it is expressed as teeth per inch. The addendum a is the radial distance between the top land and the pitch circle. The dedendum b is the radial distance from the bottom land to the pitch circle. The whole depth ht is the sum of the addendum and the dedendum. The clearance circle is a circle that is tangent to the addendum circle of the mat- ing gear. The clearance c is the amount by which the dedendum in a given gear exceeds the addendum of its mating gear. The backlash is the amount by which the width of a tooth space exceeds the thickness of the engaging tooth measured on the pitch circles. You should prove for yourself the validity of the following useful relations:

P 5 N d

(13–1)

m 5 d N

(13–2)

p 5 pd N

5 pm (13–3)

pP 5 p (13–4)

where P 5 diametral pitch, teeth per inch

N 5 number of teeth

d 5 pitch diameter, in or mm

m 5 module, mm

p 5 circular pitch, in or mm

Figure 13–5 Nomenclature of spur-gear teeth.

Addendum

Dedendum

Clearance

Bott om

la nd

Fillet radius

Dedendum circle

Clearance circle

Tooth thickness

Fac e w

idt h

Width of space

Face

Top land

Addendum circle

Pitch circle

Flank Circular pitch

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Gears—General 669

13–3 Conjugate Action The following discussion assumes the teeth to be perfectly formed, perfectly smooth, and absolutely rigid. Such an assumption is, of course, unrealistic, because the appli- cation of forces will cause deflections. Mating gear teeth acting against each other to produce rotary motion are similar to cams. When the tooth profiles, or cams, are designed so as to produce a constant angular-velocity ratio during meshing, these are said to have conjugate action. In theory, at least, it is possible arbitrarily to select any profile for one tooth and then to find a profile for the meshing tooth that will give conjugate action. One of these solutions is the involute profile, which, with few exceptions, is in universal use for gear teeth and is the only one with which we will be concerned. When one curved surface pushes against another (Fig. 13–6), the point of contact occurs where the two surfaces are tangent to each other (point c), and the forces at any instant are directed along the common normal ab to the two curves. The line ab, representing the direction of action of the forces, is called the line of action. The line of action will intersect the line of centers O-O at some point P. The angular-velocity ratio between the two arms is inversely proportional to their radii to the point P. Circles drawn through point P from each center are called pitch circles, and the radius of each circle is called the pitch radius. Point P is called the pitch point. Figure 13–6 is useful in making another observation. A pair of gears is really a pair of cams that act through a small arc and, before running off the involute contour, are replaced by another identical pair of cams. The cams can run in either direction and are configured to transmit a constant angular-velocity ratio. If involute curves are used, the gears tolerate changes in center-to-center distance with no variation in con- stant angular-velocity ratio. Furthermore, the rack profiles are straight-flanked, making primary tooling simpler. To transmit motion at a constant angular-velocity ratio, the pitch point must remain fixed; that is, all the lines of action for every instantaneous point of contact must pass through the same point P. In the case of the involute profile, it will be shown that all points of contact occur on the same straight line ab, that all normals to the tooth profiles at the point of contact coincide with the line ab, and, thus, that these profiles transmit uniform rotary motion.

A O

Figure 13–6 Cam A and follower B in contact. When the contacting surfaces are involute profiles, the ensuing conjugate action produces a constant angular-velocity ratio.

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670 Mechanical Engineering Design

13–4 Involute Properties An involute curve may be generated as shown in Fig. 13–7a. A partial flange B is attached to the cylinder A, around which is wrapped a cord def, which is held tight. Point b on the cord represents the tracing point, and as the cord is wrapped and unwrapped about the cylinder, point b will trace out the involute curve ac. The radius of the curvature of the involute varies continuously, being zero at point a and a maximum at point c. At point b the radius is equal to the distance be, since point b is instantaneously rotating about point e. Thus the generating line de is normal to the invo- lute at all points of intersection and, at the same time, is always tangent to the cylinder A. The circle on which the involute is generated is called the base circle. Let us now examine the involute profile to see how it satisfies the requirement for the transmission of uniform motion. In Fig. 13–7b, two gear blanks with fixed centers at O1 and O2 are shown having base circles whose respective radii are O1a and O2b. We now imagine that a cord is wound clockwise around the base circle of gear 1, pulled tight between points a and b, and wound counterclockwise around the base circle of gear 2. If, now, the base circles are rotated in different directions so as to keep the cord tight, a point g on the cord will trace out the involutes cd on gear 1 and ef on gear 2. The involutes are thus generated simultaneously by the tracing point. The tracing point, therefore, represents the point of contact, while the portion of the cord ab is the generating line. The point of contact moves along the generating line; the generating line does not change position, because it is always tangent to the base circles; and since the generating line is always normal to the involutes at the point of contact, the requirement for uniform motion is satisfied.

13–5 Fundamentals Among other things, it is necessary that you actually be able to draw the teeth on a pair of meshing gears. You should understand, however, that you are not doing this for manufacturing or shop purposes. Rather, we make drawings of gear teeth to obtain an understanding of the problems involved in the meshing of the mating teeth.

Figure 13–7 (a) Generation of an involute; (b) involute action.

Base circle

Pitch circle

c a

P e d

Pitch circle

Gear 1

Gear 2 Base circle

(a) (b)

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Gears—General 671

First, it is necessary to learn how to construct an involute curve. As shown in Fig. 13–8, divide the base circle into a number of equal parts, and construct radial lines OA0, OA1, OA2, etc. Beginning at A1, construct perpendiculars A1B1, A2B2, A3B3, etc. Then along A1B1 lay off the distance A1A0, along A2B2 lay off twice the distance A1A0, etc., producing points through which the involute curve can be constructed. To investigate the fundamentals of tooth action, let us proceed step by step through the process of constructing the teeth on a pair of gears. When two gears are in mesh, their pitch circles roll on one another without slip- ping. Designate the pitch radii as r1 and r2 and the angular velocities as v1 and v2, respectively. Then the pitch-line velocity is

V 5 0 r1v1 0 5 0 r2v2 0 Thus the relation between the radii on the angular velocities is

` v1

v2 ` 5 r2

r1 (13–5)

Suppose now we wish to design a speed reducer such that the input speed is 1800 rev/min and the output speed is 1200 rev/min. This is a ratio of 3:2; the gear pitch diameters would be in the same ratio, for example, a 4-in pinion driving a 6-in gear. The various dimensions found in gearing are always based on the pitch circles. Suppose we specify that an 18-tooth pinion is to mesh with a 30-tooth gear and that the diametral pitch of the gearset is to be 2 teeth per inch. Then, from Eq. (13–1), the pitch diameters of the pinion and gear are, respectively,

d1 5 N1

P 5

18 2

5 9 in d2 5 N2

P 5

30 2

5 15 in

The first step in drawing teeth on a pair of mating gears is shown in Fig. 13–9. The center distance is the sum of the pitch radii, in this case 12 in. So locate the pinion and gear centers O1 and O2, 12 in apart. Then construct the pitch circles of radii r1 and r2. These are tangent at P, the pitch point. Next draw line ab, the common tangent, through the pitch point. We now designate gear 1 as the driver, and since it is rotating counterclockwise, we draw a line cd through point P at an angle f to the common tangent ab. The line cd has three names, all of which are in general use. It is called the pressure line, the generating line, and the line of action. It represents the direction in which the resultant force acts between the gears. The angle f is called the pressure angle, and it usually has values of 20 or 25°, though 141

2 ° was once used.

Figure 13–8 Construction of an involute curve.

Base circle InvoluteA4

A0 B1

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672 Mechanical Engineering Design

Next, on each gear draw a circle tangent to the pressure line. These circles are the base circles. Since they are tangent to the pressure line, the pressure angle determines their size. As shown in Fig. 13–10, the radius of the base circle is

rb 5 r cos f (13–6)

where r is the pitch radius. Now generate an involute on each base circle as previously described and as shown in Fig. 13–9. This involute is to be used for one side of a gear tooth. It is not necessary to draw another curve in the reverse direction for the other side of the tooth, because we are going to use a template which can be turned over to obtain the other side. The addendum and dedendum distances for standard interchangeable teeth are, as we shall learn later, 1yP and 1.25yP, respectively. Therefore, for the pair of gears we are constructing,

a 5 1 P

5 1 2

5 0.500 in b 5 1.25

P 5

1.25 2

5 0.625 in

Using these distances, draw the addendum and dedendum circles on the pinion and on the gear as shown in Fig. 13–9. Next, using heavy drawing paper, or preferably, a sheet of 0.015- to 0.020-in clear plastic, cut a template for each involute, being careful to locate the gear centers prop- erly with respect to each involute. Figure 13–11 is a reproduction of the template used to create some of the illustrations for this book. Note that only one side of the tooth profile is formed on the template. To get the other side, turn the template over. For some problems you might wish to construct a template for the entire tooth.

Figure 13–9 Circles of a gear layout.

Base circle

Dedendum circle

Pitch circle Base circle

Involute

Addendum circles

Pitch circle

Dedendum circle

Involute

Figure 13–10 Base circle radius can be related to the pressure angle f and the pitch circle radius by rb 5 r cos f.

Pitch circle

Pressure line

Base circle

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Gears—General 673

To draw a tooth, we must know the tooth thickness. From Eq. (13–4), the circular pitch is

p 5 p

P 5 p

2 5 1.57 in

Therefore, the tooth thickness is

t 5 p

2 5

1.57 2

5 0.785 in

measured on the pitch circle. Using this distance for the tooth thickness as well as the tooth space, draw as many teeth as desired, using the template, after the points have been marked on the pitch circle. In Fig. 13–12 only one tooth has been drawn on each gear. You may run into trouble in drawing these teeth if one of the base circles happens to be larger than the dedendum circle. The reason for this is that the involute begins at the base circle and is undefined below this circle. So, in drawing gear teeth, we usually draw a radial line for the profile below the base circle. The actual shape, however, will depend upon the kind of machine tool used to form the teeth in manufacture, that is, how the profile is generated. The portion of the tooth between the clearance circle and the dedendum circle includes the fillet. In this instance the clearance is

c 5 b 2 a 5 0.625 2 0.500 5 0.125 in

The construction is finished when these fillets have been drawn.

Figure 13–11 A template for drawing gear teeth.

21 O2

Figure 13–12 Tooth action.

Angle of approach

Angle of recess

Pressure line

Dedendum circle Base circle Pitch circle Addendum circle

Angle of recess

Pinion (driver)

Addendum circle

Pitch circle

Base circle

Dedendum circle Gear

(driven)

Angle of approach

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674 Mechanical Engineering Design

Referring again to Fig. 13–12, the pinion with center at O1 is the driver and turns counterclockwise. The pressure, or generating, line is the same as the cord used in Fig. 13–7a to generate the involute, and contact occurs along this line. The initial contact will take place when the flank of the driver comes into contact with the tip of the driven tooth. This occurs at point a in Fig. 13–12, where the addendum circle of the driven gear crosses the pressure line. If we now construct tooth profiles through point a and draw radial lines from the intersections of these profiles with the pitch circles to the gear centers, we obtain the angle of approach for each gear. As the teeth go into mesh, the point of contact will slide up the side of the driving tooth so that the tip of the driver will be in contact just before contact ends. The final point of contact will therefore be where the addendum circle of the driver crosses the pressure line. This is point b in Fig. 13–12. By drawing another set of tooth profiles through b, we obtain the angle of recess for each gear in a manner similar to that of finding the angles of approach. The sum of the angle of approach and the angle of recess for either gear is called the angle of action. The line ab is called the line of action. We may imagine a rack as a spur gear having an infinitely large pitch diameter. Therefore, the rack has an infinite number of teeth and a base circle which is an infinite distance from the pitch point. The sides of involute teeth on a rack are straight lines making an angle to the line of centers equal to the pressure angle. Figure 13–13 shows an involute rack in mesh with a pinion. Corresponding sides on involute teeth are parallel curves; the base pitch is the constant and fundamental distance between them along a common normal as shown in Fig. 13–13. The base pitch is related to the circular pitch by the equation

pb 5 pc cos f (13–7)

where pb is the base pitch. Figure 13–14 shows a pinion in mesh with an internal, or ring, gear. Note that both of the gears now have their centers of rotation on the same side of the pitch point. Thus the positions of the addendum and dedendum circles with respect to the pitch circle are reversed; the addendum circle of the internal gear lies inside the pitch circle. Note, too, from Fig. 13–14, that the base circle of the internal gear lies inside the pitch circle near the addendum circle. Another interesting observation concerns the fact that the operating diameters of the pitch circles of a pair of meshing gears need not be the same as the respective design pitch diameters of the gears, though this is the way they have been constructed in Fig. 13–12. If we increase the center distance, we create two new operating pitch circles having larger diameters because they must be tangent to each other at the pitch

Figure 13–13 Involute-toothed pinion and rack.

Circular pitch

Base pitch

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Gears—General 675

point. Thus the pitch circles of gears really do not come into existence until a pair of gears are brought into mesh. Changing the center distance has no effect on the base circles, because these were used to generate the tooth profiles. Thus the base circle is basic to a gear. Increasing the center distance increases the pressure angle and decreases the length of the line of action, but the teeth are still conjugate, the requirement for uniform motion trans- mission is still satisfied, and the angular-velocity ratio has not changed.

Figure 13–14 Internal gear and pinion.

Pitch circle

Base circle

!2 Base circle

Pitch circle

Pressure line

Dedendum circle

Addendum circle

EXAMPLE 13–1 A gearset consists of a 16-tooth pinion driving a 40-tooth gear. The diametral pitch is 2, and the addendum and dedendum are 1yP and 1.25yP, respectively. The gears are cut using a pressure angle of 20°. (a) Compute the circular pitch, the center distance, and the radii of the base circles. (b) In mounting these gears, the center distance was incorrectly made 1

4 in larger. Compute the new values of the pressure angle and the pitch-circle diameters.

Solution

Answer (a) p 5 p

P 5 p

2 5 1.571 in

The pitch diameters of the pinion and gear are, respectively,

dP 5 NP

P 5

16 2

5 8 in dG 5 NG

P 5

40 2

5 20 in

Therefore the center distance is

Answer dP 1 dG

2 5

8 1 20 2

5 14 in

Since the teeth were cut on the 20° pressure angle, the base-circle radii are found to be, using rb 5 r cos f,

Answer rb(pinion) 5 8 2

cos 20° 5 3.759 in

Answer rb(gear) 5 20 2

cos 20° 5 9.397 in

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676 Mechanical Engineering Design

(b) Designating d9P and d9G as the new pitch-circle diameters, the 1 4-in increase in the

center distance requires that

d¿P 1 d¿G

2 5 14.250 (1)

Also, the velocity ratio does not change, and hence

d¿P d¿G

5 16 40

(2)

Solving Eqs. (1) and (2) simultaneously yields

Answer d¿P 5 8.143 in d¿G 5 20.357 in

Since rb 5 r cos f, using either the pinion or gear, the new pressure angle is

Answer f¿ 5 cos21 rb(pinion)

d¿Py2 5 cos21

3.759 8.143y2

5 22.59°

13–6 Contact Ratio The zone of action of meshing gear teeth is shown in Fig. 13–15. We recall that tooth contact begins and ends at the intersections of the two addendum circles with the pressure line. In Fig. 13–15 initial contact occurs at a and final contact at b. Tooth profiles drawn through these points intersect the pitch circle at A and B, respectively. As shown, the distance AP is called the arc of approach qa, and the distance PB, the arc of recess qr. The sum of these is the arc of action qt. Now, consider a situation in which the arc of action is exactly equal to the cir- cular pitch, that is, qt 5 p. This means that one tooth and its space will occupy the entire arc AB. In other words, when a tooth is just beginning contact at a, the previ- ous tooth is simultaneously ending its contact at b. Therefore, during the tooth action from a to b, there will be exactly one pair of teeth in contact. Next, consider a situation in which the arc of action is greater than the circular pitch, but not very much greater, say, qt < 1.2p. This means that when one pair of teeth is just entering contact at a, another pair, already in contact, will not yet have reached b. Thus,

Figure 13–15 Definition of contact ratio.

Lab

Motion

Addendum circle

Pressure line

Pitch circle Addendum circle

Arc of approach qa

Arc of recess qr

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Gears—General 677

for a short period of time, there will be two teeth in contact, one in the vicinity of A and another near B. As the meshing proceeds, the pair near B must cease contact, leaving only a single pair of contacting teeth, until the procedure repeats itself. Because of the nature of this tooth action, either one or two pairs of teeth in contact, it is convenient to define the term contact ratio mc as

mc 5 qt

p (13–8)

a number that indicates the average number of pairs of teeth in contact. Note that this ratio is also equal to the length of the path of contact divided by the base pitch. Gears should not generally be designed having contact ratios less than about 1.20, because inaccuracies in mounting might reduce the contact ratio even more, increasing the possibility of impact between the teeth as well as an increase in the noise level. An easier way to obtain the contact ratio is to measure the line of action ab instead of the arc distance AB. Since ab in Fig. 13–15 is tangent to the base circle when extended, the base pitch pb must be used to calculate mc instead of the circular pitch as in Eq. (13–8). If the length of the line of action is Lab, the contact ratio is

mc 5 Lab

p cos f (13–9)

in which Eq. (13–7) was used for the base pitch.

13–7 Interference The contact of portions of tooth profiles that are not conjugate is called interference. Consider Fig. 13–16. Illustrated are two 16-tooth gears that have been cut to the now obsolete 141

2 ° pressure angle. The driver, gear 2, turns clockwise. The initial and final

points of contact are designated A and B, respectively, and are located on the pressure line. Now notice that the points of tangency of the pressure line with the base circles C and D are located inside of points A and B. Interference is present. The interference is explained as follows. Contact begins when the tip of the driven tooth contacts the flank of the driving tooth. In this case the flank of the driving tooth first makes contact with the driven tooth at point A, and this occurs before the involute portion of the driving tooth comes within range. In other words, contact is occurring below the base circle of gear 2 on the noninvolute portion of the flank. The actual effect is that the involute tip or face of the driven gear tends to dig out the noninvo- lute flank of the driver. In this example the same effect occurs again as the teeth leave contact. Contact should end at point D or before. Since it does not end until point B, the effect is for the tip of the driving tooth to dig out, or interfere with, the flank of the driven tooth. When gear teeth are produced by a generation process, interference is automati- cally eliminated because the cutting tool removes the interfering portion of the flank. This effect is called undercutting; if undercutting is at all pronounced, the undercut tooth is considerably weakened. Thus the effect of eliminating interference by a gen- eration process is merely to substitute another problem for the original one. The smallest number of teeth on a spur pinion and gear,1 one-to-one gear ratio, which can exist without interference is NP. This number of teeth for spur gears is

1Robert Lipp, “Avoiding Tooth Interference in Gears,” Machine Design, Vol. 54, No. 1, 1982, pp. 122–124.

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678 Mechanical Engineering Design

given by

NP 5 2k

3 sin2 f (1 1 21 1 3 sin2 f) (13–10)

where k 5 1 for full-depth teeth, 0.8 for stub teeth and f 5 pressure angle. For a 20° pressure angle, with k 5 1,

NP 5 2(1)

3 sin2 20° (1 1 21 1 3 sin2 20°) 5 12.3 5 13 teeth

Thus 13 teeth on pinion and gear are interference-free. Realize that 12.3 teeth is pos- sible in meshing arcs, but for fully rotating gears, 13 teeth represents the least number. For a 141

2 ° pressure angle, NP 5 23 teeth, so one can appreciate why few 141

2 °-tooth

systems are used, as the higher pressure angles can produce a smaller pinion with accompanying smaller center-to-center distances. If the mating gear has more teeth than the pinion, that is, mG 5 NGyNP 5 m is more than one, then the smallest number of teeth on the pinion without interference is given by

NP 5 2k

(1 1 2m) sin2 f (m 1 2m2 1 (1 1 2m) sin2 f) (13–11)

Figure 13–16 Interference in the action of gear teeth.

Driving gear 2

Driven gear 3

Base circle

Interference is on flank of driver during approach

This portion of profile is not an involute

Addendum circlesPressure line

D B

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Gears—General 679

For example, if m 5 4, f 5 20°,

NP 5 2(1)

[1 1 2(4)] sin2 20° [4 1 242 1 [1 1 2(4)] sin2 20°] 5 15.4 5 16 teeth

Thus a 16-tooth pinion will mesh with a 64-tooth gear without interference. The largest gear with a specified pinion that is interference-free is

NG 5 N2

P sin2 f 2 4k2

4k 2 2NP sin2 f (13–12)

For example, for a 13-tooth pinion with a pressure angle f of 20°,

NG 5 132 sin2 20° 2 4(1)2

4(1) 2 2(13) sin2 20° 5 16.45 5 16 teeth

For a 13-tooth spur pinion, the maximum number of gear teeth possible without interference is 16. The smallest spur pinion that will operate with a rack without interference is

NP 5 2(k)

sin2 f (13–13)

For a 20° pressure angle full-depth tooth the smallest number of pinion teeth to mesh with a rack is

NP 5 2(1)

sin2 20° 5 17.1 5 18 teeth

Since gear-shaping tools amount to contact with a rack, and the gear-hobbing process is similar, the minimum number of teeth to prevent interference to prevent undercutting by the hobbing process is equal to the value of NP when NG is infinite. The importance of the problem of teeth that have been weakened by undercutting cannot be overemphasized. Of course, interference can be eliminated by using more teeth on the pinion. However, if the pinion is to transmit a given amount of power, more teeth can be used only by increasing the pitch diameter. Interference can also be reduced by using a larger pressure angle. This results in a smaller base circle, so that more of the tooth profile becomes involute. The demand for smaller pinions with fewer teeth thus favors the use of a 25° pressure angle even though the frictional forces and bearing loads are increased and the contact ratio decreased.

13–8 The Forming of Gear Teeth There are a large number of ways of forming the teeth of gears, such as sand casting, shell molding, investment casting, permanent-mold casting, die casting, and centrifu- gal casting. Teeth can also be formed by using the powder-metallurgy process; or, by using extrusion, a single bar of aluminum may be formed and then sliced into gears. Gears that carry large loads in comparison with their size are usually made of steel and are cut with either form cutters or generating cutters. In form cutting, the tooth space takes the exact form of the cutter. In generating, a tool having a shape different from the tooth profile is moved relative to the gear blank so as to obtain the proper tooth shape. One of the newest and most promising of the methods of forming teeth is called cold forming, or cold rolling, in which dies are rolled against steel blanks to form the teeth. The mechanical properties of the metal are greatly improved by the rolling process, and a high-quality generated profile is obtained at the same time.

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680 Mechanical Engineering Design

Gear teeth may be machined by milling, shaping, or hobbing. They may be fin- ished by shaving, burnishing, grinding, or lapping. Gears made of thermoplastics such as nylon, polycarbonate, acetal are quite pop- ular and are easily manufactured by injection molding. These gears are of low to moderate precision, low in cost for high production quantities, and capable of light loads, and can run without lubrication.

Milling Gear teeth may be cut with a form milling cutter shaped to conform to the tooth space. With this method it is theoretically necessary to use a different cutter for each gear, because a gear having 25 teeth, for example, will have a different-shaped tooth space from one having, say, 24 teeth. Actually, the change in space is not too great, and it has been found that eight cutters may be used to cut with reasonable accuracy any gear in the range of 12 teeth to a rack. A separate set of cutters is, of course, required for each pitch.

Shaping Teeth may be generated with either a pinion cutter or a rack cutter. The pinion cutter (Fig. 13–17) reciprocates along the vertical axis and is slowly fed into the gear blank to the required depth. When the pitch circles are tangent, both the cutter and the blank rotate slightly after each cutting stroke. Since each tooth of the cutter is a cutting tool, the teeth are all cut after the blank has completed one rotation. The sides of an invo- lute rack tooth are straight. For this reason, a rack-generating tool provides an accurate method of cutting gear teeth. This is also a shaping operation and is illustrated by the drawing of Fig. 13–18. In operation, the cutter reciprocates and is first fed into the gear blank until the pitch circles are tangent. Then, after each cutting stroke, the gear blank

Figure 13–17 Generating a spur gear with a pinion cutter. (Courtesy of Boston Gear Works, Inc.)

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Gears—General 681

and cutter roll slightly on their pitch circles. When the blank and cutter have rolled a distance equal to the circular pitch, the cutter is returned to the starting point, and the process is continued until all the teeth have been cut.

Hobbing The hobbing process is illustrated in Fig. 13–19. The hob is simply a cutting tool that is shaped like a worm. The teeth have straight sides, as in a rack, but the hob axis must be turned through the lead angle in order to cut spur-gear teeth. For this reason, the teeth generated by a hob have a slightly different shape from those generated by a rack cutter. Both the hob and the blank must be rotated at the proper angular-velocity ratio. The hob is then fed slowly across the face of the blank until all the teeth have been cut.

Figure 13–18 Shaping teeth with a rack. (This is a drawing-board figure that J. E. Shigley executed over 35 years ago in response to a question from a student at the University of Michigan.)

Gear blank rotates in this direction

Rack cutter reciprocates in a direction perpendicular to this page

Figure 13–19 Hobbing a worm gear. (Courtesy of Boston Gear Works, Inc.)

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682 Mechanical Engineering Design

Finishing Gears that run at high speeds and transmit large forces may be subjected to additional dynamic forces if there are errors in tooth profiles. Errors may be diminished somewhat by finishing the tooth profiles. The teeth may be finished, after cutting, by either shav- ing or burnishing. Several shaving machines are available that cut off a minute amount of metal, bringing the accuracy of the tooth profile within the limits of 250 min. Burnishing, like shaving, is used with gears that have been cut but not heat-treated. In burnishing, hardened gears with slightly oversize teeth are run in mesh with the gear until the surfaces become smooth. Grinding and lapping are used for hardened gear teeth after heat treatment. The grinding operation employs the generating principle and produces very accurate teeth. In lapping, the teeth of the gear and lap slide axially so that the whole surface of the teeth is abraded equally.

13–9 Straight Bevel Gears When gears are used to transmit motion between intersecting shafts, some form of bevel gear is required. A bevel gearset is shown in Fig. 13–20. Although bevel gears are usu- ally made for a shaft angle of 90°, they may be produced for almost any angle. The teeth may be cast, milled, or generated. Only the generated teeth may be classed as accurate. The terminology of bevel gears is illustrated in Fig. 13–20. The pitch of bevel gears is measured at the large end of the tooth, and both the circular pitch and the pitch diameter are calculated in the same manner as for spur gears. It should be noted that the clearance is uniform. The pitch angles are defined by the pitch cones meeting at the apex, as shown in the figure. They are related to the tooth numbers as follows:

tan g 5 NP

NG tan G 5

NP (13–14)

Figure 13–20 Terminology of bevel gears.

Back-cone radius, rb

Cone dista nce A o

Face

Pitch angle

Uniform clearance

Pitch diameter DG

Back cone

Pitch angle

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Gears—General 683

where the subscripts P and G refer to the pinion and gear, respectively, and where g and G are, respectively, the pitch angles of the pinion and gear. Figure 13–20 shows that the shape of the teeth, when projected on the back cone, is the same as in a spur gear having a radius equal to the back-cone distance rb. This is called Tredgold’s approximation. The number of teeth in this imaginary gear is

N¿ 5 2prb

p (13–15)

where N9 is the virtual number of teeth and p is the circular pitch measured at the large end of the teeth. Standard straight-tooth bevel gears are cut by using a 20° pres- sure angle, unequal addenda and dedenda, and full-depth teeth. This increases the contact ratio, avoids undercut, and increases the strength of the pinion.

13–10 Parallel Helical Gears Helical gears, used to transmit motion between parallel shafts, are shown in Fig. 13–2. The helix angle is the same on each gear, but one gear must have a right-hand helix and the other a left-hand helix. The shape of the tooth is an involute helicoid and is illustrated in Fig. 13–21. If a piece of paper cut in the shape of a parallelogram is wrapped around a cylinder, the angular edge of the paper becomes a helix. If we unwind this paper, each point on the angular edge generates an involute curve. This surface obtained when every point on the edge generates an involute is called an involute helicoid. The initial contact of spur-gear teeth is a line extending all the way across the face of the tooth. The initial contact of helical-gear teeth is a point that extends into a line as the teeth come into more engagement. In spur gears the line of contact is parallel to the axis of rotation; in helical gears the line is diagonal across the face of the tooth. It is this gradual engagement of the teeth and the smooth transfer of load from one tooth to another that gives helical gears the ability to transmit heavy loads at high speeds. Because of the nature of contact between helical gears, the contact ratio is of only minor importance, and it is the contact area, which is proportional to the face width of the gear, that becomes significant. Helical gears subject the shaft bearings to both radial and thrust loads. When the thrust loads become high or are objectionable for other reasons, it may be desirable to use double helical gears. A double helical gear (herringbone) is equivalent to two helical gears of opposite hand, mounted side by side on the same shaft. They develop opposite thrust reactions and thus cancel out the thrust load. When two or more single helical gears are mounted on the same shaft, the hand of the gears should be selected so as to produce the minimum thrust load.

Figure 13–21 An involute helicoid.

Involute

Base cylinder

Edge of paper

Base helix angle

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684 Mechanical Engineering Design

Figure 13–22 represents a portion of the top view of a helical rack. Lines ab and cd are the centerlines of two adjacent helical teeth taken on the same pitch plane. The angle c is the helix angle. The distance ac is the transverse circular pitch pt in the plane of rotation (usually called the circular pitch). The distance ae is the normal circular pitch pn and is related to the transverse circular pitch as follows:

pn 5 pt cos c (13–16)

The distance ad is called the axial pitch px and is related by the expression

px 5 pt

tan c (13–17)

Since pn Pn 5 p, the normal diametral pitch is

Pn 5 Pt

cos c (13–18)

The pressure angle fn in the normal direction is different from the pressure angle ft in the direction of rotation, because of the angularity of the teeth. These angles are related by the equation

cos c 5 tan fn

tan ft (13–19)

Figure 13–23 illustrates a cylinder cut by an oblique plane ab at an angle c to a right section. The oblique plane cuts out an arc having a radius of curvature of R. For the condition that c 5 0, the radius of curvature is R 5 Dy2. If we imagine the angle c to be slowly increased from zero to 90°, we see that R begins at a value of Dy2 and increases until, when c 5 90°, R 5 q. The radius R is the apparent pitch radius of

Figure 13–22 Nomenclature of helical gears.

Section B–B

b d

a c

e px

A BA

Section A–A

(a)

(b)

(c)

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Gears—General 685

a helical-gear tooth when viewed in the direction of the tooth elements. A gear of the same pitch and with the radius R will have a greater number of teeth, because of the increased radius. In helical-gear terminology this is called the virtual number of teeth. It can be shown by analytical geometry that the virtual number of teeth is related to the actual number by the equation

N¿ 5 N

cos3 c (13–20)

where N9 is the virtual number of teeth and N is the actual number of teeth. It is necessary to know the virtual number of teeth in design for strength and also, some- times, in cutting helical teeth. This apparently larger radius of curvature means that few teeth may be used on helical gears, because there will be less undercutting.

Figure 13–23 A cylinder cut by an oblique plane.

! b

D +

(a)

(b)

EXAMPLE 13–2 A stock helical gear has a normal pressure angle of 20°, a helix angle of 25°, and a transverse diametral pitch of 6 teeth/in, and has 18 teeth. Find: (a) The pitch diameter (b) The transverse, the normal, and the axial pitches (c) The normal diametral pitch (d ) The transverse pressure angle

Solution

Answer (a) d 5 N Pt

5 18 6

5 3 in

Answer (b) pt 5 p

Pt 5 p

6 5 0.5236 in

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686 Mechanical Engineering Design

Answer pn 5 pt cos c 5 0.5236 cos 25° 5 0.4745 in

Answer px 5 pt

tan c 5

0.5236 tan 45°

5 1.123 in

Answer (c) Pn 5 Pt

cos c 5

6 cos 25°

5 6.620 teeth/in

Answer (d ) ft 5 tan21 atan fn

cos c b 5 tan21 a tan 20°

cos 25° b 5 21.88°

Just like teeth on spur gears, helical-gear teeth can interfere. Equation (13–19) can be solved for the pressure angle ft in the tangential (rotation) direction to give

ft 5 tan21 atan fn

cos c b

The smallest tooth number NP of a helical-spur pinion that will run without interfer- ence2 with a gear with the same number of teeth is

NP 5 2k cos c

3 sin2 ft (1 1 21 1 3 sin2 ft ) (13–21)

For example, if the normal pressure angle fn is 20°, the helix angle c is 30°, then ft is

ft 5 tan21 a tan 20° cos 30°

b 5 22.80°

NP 5 2(1) cos 30°

3 sin2 22.80° (1 1 21 1 3 sin2 22.80°) 5 8.48 5 9 teeth

For a given gear ratio mG 5 NG yNP 5 m, the smallest pinion tooth count is

NP 5 2k cos c

(1 1 2m) sin2 ft [m 1 2m2 1 (1 1 2m) sin2 ft ] (13–22)

The largest gear with a specified pinion is given by

NG 5 N2

P sin2 ft 2 4k2 cos2 c

4k cos c 2 2NP sin2 ft (13–23)

For example, for a nine-tooth pinion with a pressure angle fn of 20°, a helix angle c of 30°, and recalling that the tangential pressure angle ft is 22.80°,

NG 5 92 sin2 22.80° 2 4(1)2 cos2 30°

4(1) cos 30° 2 2(9) sin2 22.80° 5 12.02 5 12

The smallest pinion that can be run with a rack is

NP 5 2k cos c

sin2 ft (13–24)

2Op. cit., Robert Lipp, Machine Design, pp. 122–124.

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Gears—General 687

For a normal pressure angle fn of 20° and a helix angle c of 30°, and ft 5 22.80°,

NP 5 2(1) cos 30°

sin2 22.80° 5 11.5 5 12 teeth

For helical-gear teeth the number of teeth in mesh across the width of the gear will be greater than unity and a term called face-contact ratio is used to describe it. This increase of contact ratio, and the gradual sliding engagement of each tooth, results in quieter gears.

13–11 Worm Gears The nomenclature of a worm gearset is shown in Fig. 13–24. The worm and worm gear of a set have the same hand of helix as for crossed helical gears, but the helix angles are usually quite different. The helix angle on the worm is generally quite large, and that on the gear very small. Because of this, it is usual to specify the lead angle l on the worm and helix angle cG on the gear; the two angles are equal for a 90° shaft angle. The worm lead angle is the complement of the worm helix angle, as shown in Fig. 13–24. In specifying the pitch of worm gearsets, it is customary to state the axial pitch px of the worm and the transverse circular pitch pt, often simply called the circular pitch, of the mating gear. These are equal if the shaft angle is 90°. The pitch diam- eter of the gear is the diameter measured on a plane containing the worm axis, as shown in Fig. 13–24; it is the same as for spur gears and is

dG 5 NG pt

p (13–25)

Figure 13–24 Nomenclature of a single- enveloping worm gearset.

Axial pitch px Lead angle !

Pi tc

h di

am et

er d

Lead L

Pitch diameter dw

Worm

Root diameter

"W, helix angle

Pitch cylinder Helix

Worm gear

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688 Mechanical Engineering Design

Since it is not related to the number of teeth, the worm may have any pitch diam- eter; this diameter should, however, be the same as the pitch diameter of the hob used to cut the worm-gear teeth. Generally, the pitch diameter of the worm should be selected so as to fall into the range

C0.875

3.0 # dW #

C0.875

1.7 (13–26)

where C is the center distance. These proportions appear to result in optimum horse- power capacity of the gearset. The lead L and the lead angle l of the worm have the following relations:

L 5 px NW (13–27)

tan l 5 L pdW

(13–28)

13–12 Tooth Systems3

A tooth system is a standard that specifies the relationships involving addendum, dedendum, working depth, tooth thickness, and pressure angle. The standards were originally planned to attain interchangeability of gears of all tooth numbers, but of the same pressure angle and pitch. Table 13–1 contains the standards most used for spur gears. A 141

2 ° pressure angle

was once used for these but is now obsolete; the resulting gears had to be compara- tively larger to avoid interference problems. Table 13–2 is particularly useful in selecting the pitch or module of a gear. Cutters are generally available for the sizes shown in this table. Table 13–3 lists the standard tooth proportions for straight bevel gears. These sizes apply to the large end of the teeth. The nomenclature is defined in Fig. 13–20. Standard tooth proportions for helical gears are listed in Table 13–4. Tooth pro- portions are based on the normal pressure angle; these angles are standardized the

3Standardized by the American Gear Manufacturers Association (AGMA). Write AGMA for a complete list of standards, because changes are made from time to time. The address is: 1001 N. Fairfax Street, Suite 500, Alexandria, VA 22314-1587; or, www.agma.org.

Tooth System Pressure Angle !, deg Addendum a Dedendum b

Full depth 20 1yP or m 1.25yP or 1.25m

1.35yP or 1.35m

221 2 1yP or m 1.25yP or 1.25m

1.35yP or 1.35m

25 1yP or m 1.25yP or 1.25m

1.35yP or 1.35m

Stub 20 0.8yP or 0.8m 1yP or m

Table 13–1

Standard and Commonly Used Tooth Systems for Spur Gears

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689

Diametral Pitch P (teeth/in)

Coarse 2, 21 4, 21

2, 3, 4, 6, 8, 10, 12, 16

Fine 20, 24, 32, 40, 48, 64, 80, 96, 120, 150, 200

Module m (mm/tooth)

Preferred 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 40, 50

Next Choice 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14, 18, 22, 28, 36, 45

Table 13–2

Tooth Sizes in General Uses

Table 13–3

Tooth Proportions for 20° Straight Bevel-Gear Teeth

Item Formula

Working depth hk 5 2.0yP

Clearance c 5 (0.188yP) 1 0.002 in

Addendum of gear aG 5 0.54

P 1

0.460

P(m90)2

Gear ratio mG 5 NGyNP

Equivalent 90° ratio m90 5 mG when G 5 90°

m90 5 BmG cos g

cos G when G fi 90°

Face width F 5 0.3A0 or F 5 10 P

, whichever is smaller

Minimum number of teeth Pinion 16 15 14 13

Gear 16 17 20 30

Table 13–4

Standard Tooth Proportions for Helical Gears

Quantity* Formula Quantity* Formula

Addendum 1.00 Pn

External gears:

Dedendum 1.25 Pn

Standard center distance D 1 d

Pinion pitch diameter NP

Pn cos c Gear outside diameter D 1 2a

Gear pitch diameter NG

Pn cos c Pinion outside diameter d 1 2a

Normal arc tooth thickness† p

Pn 2

2 Gear root diameter D 2 2b

Pinion base diameter d cos ft Pinion root diameter d 2 2b

Internal gears:

Gear base diameter D cos ft Center distance D 2 d

Base helix angle tan21 (tan c cos ft) Inside diameter D 2 2a

Root diameter D 1 2b

*All dimensions are in inches, and angles are in degrees. †Bn is the normal backlash.

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690 Mechanical Engineering Design

same as for spur gears. Though there will be exceptions, the face width of helical gears should be at least 2 times the axial pitch to obtain good helical-gear action. Tooth forms for worm gearing have not been highly standardized, perhaps because there has been less need for it. The pressure angles used depend upon the lead angles and must be large enough to avoid undercutting of the worm-gear tooth on the side at which contact ends. A satisfactory tooth depth, which remains in about the right proportion to the lead angle, may be obtained by making the depth a proportion of the axial circular pitch. Table 13–5 summarizes what may be regarded as good prac- tice for pressure angle and tooth depth. The face width FG of the worm gear should be made equal to the length of a tangent to the worm pitch circle between its points of intersection with the addendum circle, as shown in Fig. 13–25.

13–13 Gear Trains Consider a pinion 2 driving a gear 3. The speed of the driven gear is

n3 5 ` N2

N3 n2 ` 5 ` d2

d3 n2 ` (13–29)

where n 5 revolutions or rev/min

N 5 number of teeth

d 5 pitch diameter

Equation (13–29) applies to any gearset no matter whether the gears are spur, helical, bevel, or worm. The absolute-value signs are used to permit complete freedom in choosing positive and negative directions. In the case of spur and parallel helical gears, the directions in the viewing plane ordinarily correspond to the right-hand rule—positive for counterclockwise rotation and negative for clockwise rotation. Rotational directions are somewhat more difficult to deduce for worm and crossed helical gearsets. Figure 13–26 will be of help in these situations. The gear train shown in Fig. 13–27 is made up of five gears. Considering gear 2 to be the primary driving gear, the speed of gear 6 is

n6 5 2 N2

N3 N3

N4 N5

N6 n2 (a)

Hence we notice that gear 3 is an idler, that its tooth numbers cancel in Eq. (a), and hence that it affects only the direction of rotation of gear 6. We notice, furthermore,

Table 13–5

Recommended Pressure Angles and Tooth Depths for Worm Gearing

Lead Angle !, Pressure Angle Addendum Dedendum deg Fn, deg a bG

0–15 141 2 0.3683px 0.3683px

15–30 20 0.3683px 0.3683px

30–35 25 0.2865px 0.3314px

35–40 25 0.2546px 0.2947px

40–45 30 0.2228px 0.2578px

Figure 13–25 A graphical depiction of the face width of the worm of a worm gearset.

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Gears—General 691

that gears 2, 3, and 5 are drivers, while 3, 4, and 6 are driven members. We define the train value e as

e 5 product of driving tooth numbers product of driven tooth numbers

(13–30)

Note that pitch diameters can be used in Eq. (13–30) as well. When Eq. (13–30) is used for spur gears, e is positive if the last gear rotates in the same sense as the first, and negative if the last rotates in the opposite sense. Now we can write

nL 5 enF (13–31)

where nL is the speed of the last gear in the train and nF is the speed of the first. As a rough guideline, a train value of up to 10 to 1 can be obtained with one pair of gears. Greater ratios can be obtained in less space and with fewer dynamic problems by compounding additional pairs of gears. A two-stage compound gear train, such as shown in Fig. 13–28, can obtain a train value of up to 100 to 1. The design of gear trains to accomplish a specific train value is straightfor- ward. Since numbers of teeth on gears must be integers, it is better to determine

Figure 13–26 Thrust, rotation, and hand relations for crossed helical gears. Note that each pair of drawings refers to a single gearset. These relations also apply to worm gearsets. (Reproduced by permission, Boston Gear Division, Colfax Corp.)

Driver

(a) (b)

Thrust bearing

Right hand

Thrust bearing

Driver

Driver Driver

Figure 13–27 A gear train.

+ + + +

N4 N5 N6

2 3 4 5

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692 Mechanical Engineering Design

them first, and then obtain pitch diameters second. Determine the number of stages necessary to obtain the overall ratio, then divide the overall ratio into portions to be accomplished in each stage. To minimize package size, keep the portions as evenly divided between the stages as possible. In cases where the overall train value need only be approximated, each stage can be identical. For example, in a two-stage compound gear train, assign the square root of the overall train value to each stage. If an exact train value is needed, attempt to factor the overall train value into integer components for each stage. Then assign the small- est gear(s) to the minimum number of teeth allowed for the specific ratio of each stage, in order to avoid interference (see Sec. 13–7). Finally, applying the ratio for each stage, determine the necessary number of teeth for the mating gears. Round to the nearest integer and check that the resulting overall ratio is within acceptable tolerance.

Figure 13–28 A two-stage compound gear train.

EXAMPLE 13–3 A gearbox is needed to provide a 30:1 (61 percent) increase in speed, while minimiz- ing the overall gearbox size. Specify appropriate tooth numbers.

Solution Since the ratio is greater than 10:1, but less than 100:1, a two-stage compound gear train, such as in Figure 13–28, is needed. The portion to be accomplished in each stage is 130 5 5.4772. For this ratio, assuming a typical 20° pressure angle, the minimum number of teeth to avoid interference is 16, according to Eq. (13–11). The number of teeth necessary for the mating gears is

Answer 16130 5 87.64 < 88

From Eq. (13–30), the overall train value is

e 5 (88y16)(88y16) 5 30.25

This is within the 1 percent tolerance. If a closer tolerance is desired, then increase the pinion size to the next integer and try again.

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Gears—General 693

It is sometimes desirable for the input shaft and the output shaft of a two-stage compound gear train to be in-line, as shown in Fig. 13–29. This configuration is called a compound reverted gear train. This requires the distances between the shafts to be the same for both stages of the train, which adds to the complexity of the design task. The distance constraint is

d2y2 1 d3y2 5 d4y2 1 d5y2

Figure 13–29 A compound reverted gear train. 5

2 2

EXAMPLE 13–4 A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the overall gearbox size. Specify appropriate teeth numbers.

Solution The previous example demonstrated the difficulty with finding integer numbers of teeth to provide an exact ratio. In order to obtain integers, factor the overall ratio into two integer stages.

e 5 30 5 (6)(5)

N2yN3 5 6 and N4yN5 5 5

With two equations and four unknown numbers of teeth, two free choices are available. Choose N3 and N5 to be as small as possible without interference. Assuming a 20° pressure angle, Eq. (13–11) gives the minimum as 16. Then

N2 5 6N3 5 6(16) 5 96

N4 5 5N5 5 5(16) 5 80

The overall train value is then exact.

e 5 (96y16)(80y16) 5 (6)(5) 5 30

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694 Mechanical Engineering Design

The diametral pitch relates the diameters and the numbers of teeth, P 5 Nyd. Replacing all the diameters gives

N2y(2P) 1 N3y(2P) 5 N4y(2P) 1 N5y(2P)

Assuming a constant diametral pitch in both stages, we have the geometry condition stated in terms of numbers of teeth:

N2 1 N3 5 N4 1 N5

This condition must be exactly satisfied, in addition to the previous ratio equations, to provide for the in-line condition on the input and output shafts.

EXAMPLE 13–5 A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the overall gearbox size. The input and output shafts should be in-line. Specify appro- priate teeth numbers.

Solution The governing equations are

N2yN3 5 6

N4yN5 5 5

N2 1 N3 5 N4 1 N5

With three equations and four unknown numbers of teeth, only one free choice is available. Of the two smaller gears, N3 and N5, the free choice should be used to minimize N3 since a greater gear ratio is to be achieved in this stage. To avoid inter- ference, the minimum for N3 is 16. Applying the governing equations yields

N2 5 6N3 5 6(16) 5 96

N2 1 N3 5 96 1 16 5 112 5 N4 1 N5

Substituting N4 5 5N5 gives

112 5 5N5 1 N5 5 6N5

N5 5 112y6 5 18.67

If the train value need only be approximated, then this can be rounded to the nearest integer. But for an exact solution, it is necessary to choose the initial free choice for N3 such that solution of the rest of the teeth numbers results exactly in integers. This can be done by trial and error, letting N3 5 17, then 18, etc., until it works. Or, the problem can be normalized to quickly determine the minimum free choice. Beginning again, let the free choice be N3 5 1. Applying the governing equations gives

N2 5 6N3 5 6(1) 5 6

N2 1 N3 5 6 1 1 5 7 5 N4 1 N5

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Substituting N4 5 5N5, we find

7 5 5N5 1 N5 5 6N5

N5 5 7y6

This fraction could be eliminated if it were multiplied by a multiple of 6. The free choice for the smallest gear N3 should be selected as a multiple of 6 that is greater than the minimum allowed to avoid interference. This would indicate that N3 5 18. Repeating the application of the governing equations for the final time yields

N2 5 6N3 5 6(18) 5 108

N2 1 N3 5 108 1 18 5 126 5 N4 1 N5

126 5 5N5 1 N5 5 6N5

N5 5 126y6 5 21

N4 5 5N5 5 5(21) 5 105

Thus,

Answer N2 5 108

N3 5 18

N4 5 105

N5 5 21

Checking, we calculate e 5 (108y18)(105y21) 5 (6)(5) 5 30. And checking the geometry constraint for the in-line requirement, we calculate

N2 1 N3 5 N4 1 N5

108 1 18 5 105 1 21

126 5 126

Unusual effects can be obtained in a gear train by permitting some of the gear axes to rotate about others. Such trains are called planetary, or epicyclic, gear trains. Planetary trains always include a sun gear, a planet carrier or arm, and one or more planet gears, as shown in Fig. 13–30. Planetary gear trains are unusual mechanisms because they have two degrees of freedom; that is, for constrained motion, a planetary train must have two inputs. For example, in Fig. 13–30 these two inputs could be the motion of any two of the elements of the train. We might, say, specify that the sun gear rotates at 100 rev/min clockwise and that the ring gear rotates at 50 rev/min counterclockwise; these are the inputs. The output would be the motion of the arm. In most planetary trains one of the elements is attached to the frame and has no motion. Figure 13–31 shows a planetary train composed of a sun gear 2, an arm or

Gears—General 695

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696 Mechanical Engineering Design

carrier 3, and planet gears 4 and 5. The angular velocity of gear 2 relative to the arm in rev/min is

n23 5 n2 2 n3 (b)

Also, the velocity of gear 5 relative to the arm is

n53 5 n5 2 n3 (c)

Dividing Eq. (c) by Eq. (b) gives

n53

n23 5

n5 2 n3

n2 2 n3 (d )

Equation (d) expresses the ratio of gear 5 to that of gear 2, and both velocities are taken relative to the arm. Now this ratio is the same and is proportional to the tooth numbers, whether the arm is rotating or not. It is the train value. Therefore, we may write

e 5 n5 2 n3

n2 2 n3 (e)

This equation can be used to solve for the output motion of any planetary train. It is more conveniently written in the form

e 5 nL 2 nA

nF 2 nA (13–32)

Figure 13–30 A planetary gear train.

Sun gear Arm

2 4

30T

80T

20T Planet gear

Ring gear

Figure 13–31 A gear train on the arm of a planetary gear train.

Arm

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Gears—General 697

where nF 5 rev/min of first gear in planetary train

nL 5 rev/min of last gear in planetary train

nA 5 rev/min of arm

EXAMPLE 13–6 In Fig. 13–30 the sun gear is the input, and it is driven clockwise at 100 rev/min. The ring gear is held stationary by being fastened to the frame. Find the rev/min and direction of rotation of the arm and gear 4.

Solution Let nF 5 n2 5 2100 rev/min, and nL 5 n5 5 0. For e, unlock gear 5 and fix the arm. Then, planet gear 4 and ring gear 5 rotate in the same direction, opposite of sun gear 2. Thus, e is negative and

e 5 2aN2

N4 b

aN4

N5 b 5 2a20

30 b a30

80 b 5 20.25

Substituting this value in Eq. (13–32) gives

20.25 5 0 2 nA

(2100) 2 nA

Answer nA 5 220 rev/min 5 20 rev/min clockwise

To obtain the speed of gear 4, we follow the procedure outlined by Eqs. (b), (c), and (d). Thus

n43 5 n4 2 n3 n23 5 n2 2 n3

and so

n43

n23 5

n4 2 n3

n2 2 n3 (1)

But

n43

n23 5 2

20 30

5 2 2 3

(2)

Substituting the known values in Eq. (1) gives

2 2 3

5 n4 2 (220)

(2100) 2 (220)

Solving gives

Answer n4 5 133 1 3 rev/min 5 33 13 rev/min counter-clockwise

13–14 Force Analysis—Spur Gearing Before beginning the force analysis of gear trains, let us agree on the notation to be used. Beginning with the numeral 1 for the frame of the machine, we shall designate the input gear as gear 2, and then number the gears successively 3, 4, etc., until we arrive

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698 Mechanical Engineering Design

at the last gear in the train. Next, there may be several shafts involved, and usually one or two gears are mounted on each shaft as well as other elements. We shall des- ignate the shafts, using lowercase letters of the alphabet, a, b, c, etc. With this notation we can now speak of the force exerted by gear 2 against gear 3 as F23. The force of gear 2 against shaft a is F2a. We can also write Fa2 to mean the force of shaft a against gear 2. Unfortunately, it is also necessary to use super- scripts to indicate directions. The coordinate directions will usually be indicated by the x, y, and z coordinates, and the radial and tangential directions by superscripts r and t. With this notation, Ft

43 is the tangential component of the force of gear 4 acting against gear 3. Figure 13–32a shows a pinion mounted on shaft a rotating clockwise at n2 rev/min and driving a gear on shaft b at n3 rev/min. The reactions between the mating teeth occur along the pressure line. In Fig. 13–32b the pinion has been separated from the gear and the shaft, and their effects have been replaced by forces. Fa2 and Ta2 are the force and torque, respectively, exerted by shaft a against pinion 2. F32 is the force exerted by gear 3 against the pinion. Using a similar approach, we obtain the free-body diagram of the gear shown in Fig. 13–32c. In Fig. 13–33, the free-body diagram of the pinion has been redrawn and the forces have been resolved into tangential and radial components. We now define

Wt 5 Ft 32 (a)

as the transmitted load. This tangential load is really the useful component, because the radial component Fr

32 serves no useful purpose. It does not transmit power. The applied torque and the transmitted load are seen to be related by the equation

T 5 d 2

Wt (b)

where we have used T 5 Ta2 and d 5 d2 to obtain a general relation. The power H transmitted through a rotating gear can be obtained from the stan- dard relationship of the product of torque T and angular velocity v.

H 5 Tv 5 (Wt dy2)v (13–33)

Figure 13–32 Free-body diagrams of the forces and moments acting upon two gears of a simple gear train.

Gear

Pinion

Fb3

F23

Tb3

Ta2

Fa2

F32

3 b

(a)

(c)

(b)

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Gears—General 699

While any units can be used in this equation, the units of the resulting power will obviously be dependent on the units of the other parameters. It will often be desirable to work with the power in either horsepower or kilowatts, and appropriate conversion factors should be used. Since meshed gears are reasonably efficient, with losses of less than 2 percent, the power is generally treated as constant through the mesh. Consequently, with a pair of meshed gears, Eq. (13–33) will give the same power regardless of which gear is used for d and v. Gear data is often tabulated using pitch-line velocity, which is the linear velocity of a point on the gear at the radius of the pitch circle; thus V 5 (dy2)v. Converting this to customary units gives V 5 pdny12 (13–34)

where V 5 pitch-line velocity, ft/min

d 5 gear diameter, in

n 5 gear speed, rev/min

Many gear design problems will specify the power and speed, so it is convenient to solve Eq. (13–33) for Wt. With the pitch-line velocity and appropriate conversion factors incorporated, Eq. (13–33) can be rearranged and expressed in U.S. customary units as

Wt 5 33 000

H V

(13–35)

where Wt 5 transmitted load, lbf

H 5 power, hp

V 5 pitch-line velocity, ft/min

The corresponding equation in SI units is

Wt 5 60 000H pdn

(13–36)

where Wt 5 transmitted load, kN

H 5 power, kW

d 5 gear diameter, mm

n 5 speed, rev/min

Figure 13–33 Resolution of gear forces.

Fa2

F t a2

F r 32

F t 32

F r a2

F32

Ta2

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700 Mechanical Engineering Design

EXAMPLE 13–7 Pinion 2 in Fig. 13–34a runs at 1750 rev/min and transmits 2.5 kW to idler gear 3. The teeth are cut on the 20° full-depth system and have a module of m 5 2.5 mm. Draw a free-body diagram of gear 3 and show all the forces that act upon it.

Solution The pitch diameters of gears 2 and 3 are

d2 5 N2m 5 20(2.5) 5 50 mm

d3 5 N3m 5 50(2.5) 5 125 mm

From Eq. (13–36) we find the transmitted load to be

Wt 5 60 000H pd2n

5 60 000(2.5) p(50)(1750)

5 0.546 kN

Thus, the tangential force of gear 2 on gear 3 is F t 23 5 0.546 kN, as shown in Fig. 13–34b.

Therefore

Fr 23 5 Ft

23 tan 20° 5 (0.546) tan 20° 5 0.199 kN

and so

F23 5 Ft

cos 20° 5

0.546 cos 20°

5 0.581 kN

Since gear 3 is an idler, it transmits no power (torque) to its shaft, and so the tangential reaction of gear 4 on gear 3 is also equal to Wt. Therefore

F t 43 5 0.546 kN Fr

43 5 0.199 kN F43 5 0.581 kN

and the directions are shown in Fig. 13–34b. The shaft reactions in the x and y directions are

Fx b3 5 2(Ft

23 1 Fr 43) 5 2(20.546 1 0.199) 5 0.347 kN

Fy b3 5 2(Fr

23 1 Ft 43) 5 2(0.199 2 0.546) 5 0.347 kN

Figure 13–34 A gear train containing an idler gear. (a) The gear train. (b) Free-body of the idler gear.

c b

50T

20T30T

x b

F t 23

F23

F r 23

Fb3

F y b3

F x b3

F r 43

F t 43F43

20°

(a) (b)

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13–15 Force Analysis—Bevel Gearing In determining shaft and bearing loads for bevel-gear applications, the usual practice is to use the tangential or transmitted load that would occur if all the forces were con- centrated at the midpoint of the tooth. While the actual resultant occurs somewhere between the midpoint and the large end of the tooth, there is only a small error in making this assumption. For the transmitted load, this gives

Wt 5 T rav

(13–37)

where T is the torque and rav is the pitch radius at the midpoint of the tooth for the gear under consideration. The forces acting at the center of the tooth are shown in Fig. 13–35. The resultant force W has three components: a tangential force Wt, a radial force Wr, and an axial force Wa. From the trigonometry of the figure,

Wr 5 Wt tan f cos g

Wa 5 Wt tan f sin g (13–38)

The resultant shaft reaction is

Fb3 5 2(0.347)2 1 (0.347)2 5 0.491 kN

These are shown on the figure.

Gears—General 701

Figure 13–35 Bevel-gear tooth forces.

" rav

WrWa

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The three forces Wt, Wr, and Wa are at right angles to each other and can be used to determine the bearing loads by using the methods of statics.

702 Mechanical Engineering Design

EXAMPLE 13–8 The bevel pinion in Fig. 13–36a rotates at 600 rev/min in the direction shown and transmits 5 hp to the gear. The mounting distances, the location of all bearings, and the average pitch radii of the pinion and gear are shown in the figure. For simplicity, the teeth have been replaced by pitch cones. Bearings A and C should take the thrust loads. Find the bearing forces on the gearshaft.

2 1 2

6 3

3.88

1 2

1 5 16

3 5

1.293

A B

8 15-tooth pinion P = 5 teeth /in

45-tooth gear

(a)

(b)

T F y

F z C Wt

F x C

5 8

1 2

2 1.293

3.88

y F z D

F x D

Figure 13–36 (a) Bevel gearset of Ex. 13–8. (b) Free-body diagram of shaft CD. Dimensions in inches.

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Solution The pitch angles are

g 5 tan21 a3 9 b 5 18.4° G 5 tan21 a9

3 b 5 71.6°

The pitch-line velocity corresponding to the average pitch radius is

V 5 2prPn

12 5

2p(1.293)(600) 12

5 406 ft /min

Therefore the transmitted load is

Wt 5 33 000H

V 5

(33 000)(5) 406

5 406 lbf

and from Eq. (13–38), with G replacing g, we have

Wr 5 Wt tan f cos G 5 406 tan 20° cos 71.6° 5 46.6 lbf

Wa 5 Wt tan f sin G 5 406 tan 20° sin 71.6° 5 140 lbf

where Wt acts in the positive z direction, Wr in the 2x direction, and Wa in the 2y direction, as illustrated in the isometric sketch of Fig. 13–36b. In preparing to take a sum of the moments about bearing D, define the position vector from D to G as

RG 5 3.88i 2 (2.5 1 1.293)j 5 3.88i 2 3.793j

We shall also require a vector from D to C:

RC 5 2(2.5 1 3.625)j 5 26.125j

Then, summing moments about D gives

RG 3 W 1 RC 3 FC 1 T 5 0 (1)

When we place the details in Eq. (1), we get

(3.88i 2 3.793j) 3 (246.6i 2 140j 1 406k)

1 (26.125j) 3 (Fx C i 1 Fy

C j 1 F z Ck) 1 T j 5 0

(2)

After the two cross products are taken, the equation becomes

(21540i 2 1575j 2 720k) 1 (26.125F z C i 1 6.125F x

Ck) 1 T j 5 0

from which

T 5 1575j lbf ? in Fx C 5 118 lbf Fz

C 5 2251 lbf (3)

Now sum the forces to zero. Thus

FD 1 FC 1 W 5 0 (4)

When the details are inserted, Eq. (4) becomes

(Fx D i 1 Fz

D k) 1 (118i 1 Fy C j 2 251k) 1 (246.6 i 2 140j 1 406k) 5 0 (5)

Gears—General 703

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13–16 Force Analysis—Helical Gearing Figure 13–37 is a three-dimensional view of the forces acting against a helical-gear tooth. The point of application of the forces is in the pitch plane and in the center of the gear face. From the geometry of the figure, the three components of the total (normal) tooth force W are

Wr 5 W sin fn

Wt 5 W cos fn cos c (13–39)

Wa 5 W cos fn sin c

where W 5 total force

Wr 5 radial component

Wt 5 tangential component, also called the transmitted load

Wa 5 axial component, also called the thrust load

First we see that Fy C 5 140 lbf, and so

Answer FC 5 118i 1 140j 2 251k lbf

Then, from Eq. (5),

Answer FD 5 271.4i 2 155k lbf

These are all shown in Fig. 13–36b in the proper directions. The analysis for the pinion shaft is quite similar.

704 Mechanical Engineering Design

Figure 13–37 Tooth forces acting on a right-hand helical gear.

!t "

Tooth element

Pitch cylinder

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Gears—General 705

Usually Wt is given and the other forces are desired. In this case, it is not difficult to discover that

Wr 5 Wt tan ft

Wa 5 Wt tan c

W 5 Wt

cos fn cos c

(13–40)

EXAMPLE 13–9 In Fig. 13–38 an electric motor transmits 1-hp at 1800 rev/min in the clockwise direc- tion, as viewed from the positive x axis. Keyed to the motor shaft is an 18-tooth heli- cal pinion having a normal pressure angle of 20°, a helix angle of 30°, and a normal diametral pitch of 12 teeth/in. The hand of the helix is shown in the figure. Make a three-dimensional sketch of the motor shaft and pinion, and show the forces acting on the pinion and the bearing reactions at A and B. The thrust should be taken out at A.

Solution From Eq. (13–19) we find

ft 5 tan21

tan fn

cos c 5 tan21

tan 20° cos 30°

5 22.8°

Also, Pt 5 Pn cos c 5 12 cos 30° 510.39 teeth/in. Therefore the pitch diameter of the pinion is dp 5 18y10.39 5 1.732 in. The pitch-line velocity is

V 5 pdn 12

5 p(1.732)(1800)

12 5 816 ft /min

The transmitted load is

Wt 5 33 000H

V 5

(33 000)(1) 816

5 40.4 lbf

From Eq. (13–40) we find

Wr 5 Wt tan ft 5 (40.4) tan 22.8° 5 17.0 lbf

Wa 5 Wt tan c 5 (40.4) tan 30° 5 23.3 lbf

W 5 Wt

cos fn cos c 5

40.4 cos 20° cos 30°

5 49.6 lbf

10 in 3 in

36T

18T

Figure 13–38 The motor and gear train of Ex. 13–9.

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13–17 Force Analysis—Worm Gearing If friction is neglected, then the only force exerted by the gear will be the force W, shown in Fig. 13–40, having the three orthogonal components Wx, Wy, and Wz. From the geometry of the figure, we see that

These three forces, Wr 5 17.0 lbf in the 2y direction, Wa 5 23.3 lbf in the 2x direc- tion, and Wt 5 40.4 lbf in the 1z direction, are shown acting at point C in Fig. 13–39. We assume bearing reactions at A and B as shown. Then Fx

A 5 Wa 5 23.3 lbf. Taking moments about the z axis,

2(17.0)(13) 1 (23.3)(0.866) 1 10Fy B 5 0

or Fy B 5 20.1 lbf. Summing forces in the y direction then gives Fy

A 5 3.1 lbf. Taking moments about the y axis, next

10Fz B 2 (40.4)(13) 5 0

or Fz B 5 52.5 lbf. Summing forces in the z direction and solving gives Fz

A 5 12.1 lbf. Also, the torque is T 5 Wtdpy2 5 (40.4)(1.732y2) 5 35 lbf ? in. For comparison, solve the problem again using vectors. The force at C is

W 5 223.3i 2 17.0j 1 40.4k lbf

Position vectors to B and C from origin A are

RB 5 10i RC 5 13i 1 0.866j

Taking moments about A, we have

RB 3 FB 1 T 1 RC 3 W 5 0

Using the directions assumed in Fig. 13–39 and substituting values gives

10i 3 (Fy B j 2 Fz

Bk) 2 T i 1 (13i 1 0.866j) 3 (223.3i 2 17.0j 1 40.4k) 5 0

When the cross products are evaluated we get

(10Fy Bk 1 10Fz

B j) 2 T i 1 (35i 2 525j 2 201k) 5 0

obtaining T 5 35 lbf ? in, Fy B 5 20.1 lbf, and Fz

B 5 52.5 lbf. Next,

FA 5 2FB 2 W, and so FA 5 23.3i 2 3.1j 1 12.1k lbf.

706 Mechanical Engineering Design

Figure 13–39 Free-body diagram of motor shaft of Ex. 13–9. Forces in lbf.

F z A

F y A

F z B

F y B

B x

Wr Wt

F x A

10 in 3 in dp!2 ! 1.732!2 ! 0.866 in

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Gears—General 707

W x 5 W cos fn sin l

W y 5 W sin fn (13–41)

W z 5 W cos fn cos l

We now use the subscripts W and G to indicate forces acting against the worm and gear, respectively. We note that W y is the separating, or radial, force for both the worm and the gear. The tangential force on the worm is W x and is W z on the gear, assum- ing a 90° shaft angle. The axial force on the worm is W z, and on the gear, Wx. Since the gear forces are opposite to the worm forces, we can summarize these relations by writing

WWt 5 2WGa 5 Wx

WWr 5 2WGr 5 W y (13–42)

WWa 5 2WGt 5 Wz

It is helpful in using Eq. (13–41) and also Eq. (13–42) to observe that the gear axis is parallel to the x direction and the worm axis is parallel to the z direction and that we are employing a right-handed coordinate system. In our study of spur-gear teeth we have learned that the motion of one tooth relative to the mating tooth is primarily a rolling motion; in fact, when contact occurs at the pitch point, the motion is pure rolling. In contrast, the relative motion between worm and worm-gear teeth is pure sliding, and so we must expect that friction plays an important role in the performance of worm gearing. By introducing a coefficient of friction f, we can develop another set of relations similar to those of Eq. (13–41). In Fig. 13–40 we see that the force W acting normal to the worm-tooth profile pro- duces a frictional force Wf 5 f W, having a component f W cos l in the negative x direc- tion and another component f W sin l in the positive z direction. Equation (13–41) therefore becomes

W x 5 W(cos fn sin l 1 f cos l)

W y 5 W sin fn (13–43)

W z 5 W(cos fn cos l 2 f sin l)

Figure 13–40 Drawing of the pitch cylinder of a worm, showing the forces exerted upon it by the worm gear.

f W sin !

f W cos ! !

Wf = f W

x Wz

Pitch helix

Pitch cylinder

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708 Mechanical Engineering Design

Equation (13–42), of course, still applies. Inserting 2WGt from Eq. (13–42) for Wz in Eq. (13–43) and multiplying both sides by f, we find the frictional force Wf to be

Wf 5 f W 5 f WGt

f sin l 2 cos fn cos l (13–44)

A useful relation between the two tangential forces, WWt and WGt, can be obtained by equating the first and third parts of Eqs. (13–42) and (13–43) and eliminating W. The result is

WWt 5 WGt

cos fn sin l 1 f cos l f sin l 2 cos fn cos l

(13–45)

Efficiency h can be defined by using the equation

h 5 WWt (without friction)

WWt (with friction) (a)

Substitute Eq. (13–45) with f 5 0 in the numerator of Eq. (a) and the same equation in the denominator. After some rearranging, you will find the efficiency to be

h 5 cos fn 2 f tan l cos fn 1 f cot l

(13–46)

Selecting a typical value of the coefficient of friction, say f 5 0.05, and the pressure angles shown in Table 13–5, we can use Eq. (13–46) to get some useful design infor- mation. Solving this equation for lead angles from 1 to 30° gives the interesting results shown in Table 13–6. Many experiments have shown that the coefficient of friction is dependent on the relative or sliding velocity. In Fig. 13–41, VG is the pitch-line velocity of the gear and VW the pitch-line velocity of the worm. Vectorially, VW 5 VG 1 VS; consequently, the sliding velocity is

VS 5 VW

cos l (13–47)

Lead Angle L, Efficiency H, deg %

1.0 25.2

2.5 45.7

5.0 62.6

7.5 71.3

10.0 76.6

15.0 82.7

20.0 85.6

30.0 88.7

Table 13–6

Efficiency of Worm Gearsets for f 5 0.05

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Gears—General 709

Published values of the coefficient of friction vary as much as 20 percent, undoubtedly because of the differences in surface finish, materials, and lubrication. The values on the chart of Fig. 13–42 are representative and indicate the general trend.

Gear

Worm above

Gear axis

Worm axis

Figure 13–41 Velocity components in worm gearing.

0 400 800 1200 1600 2000 0

0.02

0.04

0.06

0.08

0.10

Sliding velocity VS , ft /min

C oe

ffi ci

en t o

f fr

ic ti

on ,

Figure 13–42 Representative values of the coefficient of friction for worm gearing. These values are based on good lubrication. Use curve B for high-quality materials, such as a case-hardened steel worm mating with a phosphor- bronze gear. Use curve A when more friction is expected, as with a cast-iron worm mating with a cast-iron worm gear.

EXAMPLE 13–10 A 2-tooth right-hand worm transmits 1 hp at 1200 rev/min to a 30-tooth worm gear. The gear has a transverse diametral pitch of 6 teeth/in and a face width of 1 in. The worm has a pitch diameter of 2 in and a face width of 21

2 in. The normal pressure angle is 141

2 °. The materials and quality of the gearing to be used are such that curve

B of Fig. 13–42 should be used to obtain the coefficient of friction. (a) Find the axial pitch, the center distance, the lead, and the lead angle. (b) Figure 13–43 is a drawing of the worm gear oriented with respect to the coordinate system described earlier in this section; the gear is supported by bearings A and B. Find the forces exerted by the bearings against the worm-gear shaft, and the output torque.

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710 Mechanical Engineering Design

Solution (a) The axial pitch is the same as the transverse circular pitch of the gear, which is

Answer px 5 pt 5 p

P 5 p

6 5 0.5236 in

The pitch diameter of the gear is dG 5 NGyP 5 30y6 5 5 in. Therefore, the center distance is

Answer C 5 dW 1 dG

2 5

2 1 5 2

5 3.5 in

From Eq. (13–27), the lead is

L 5 px NW 5 (0.5236)(2) 5 1.0472 in

Answer Also using Eq. (13–28), we find

Answer l 5 tan21

L pdW

5 tan21

1.0472 p(2)

5 9.46°

(b) Using the right-hand rule for the rotation of the worm, you will see that your thumb points in the positive z direction. Now use the bolt-and-nut analogy (the worm is right-handed, as is the screw thread of a bolt), and turn the bolt clockwise with the right hand while preventing nut rotation with the left. The nut will move axially along the bolt toward your right hand. Therefore the surface of the gear (Fig. 13–43) in contact with the worm will move in the negative z direction. Thus, viewing the gear in the negative x direction, the gear rotates clockwise about the x axis The pitch-line velocity of the worm is

VW 5 pdWnW

12 5 p(2)(1200)

12 5 628 ft/min

Figure 13–43 The pitch cylinders of the worm gear train of Ex. 13–10.

1 in2 1

2 in2 1

1200 rev/min

Gear pitch cylinder

Worm pitch cylinder

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Gears—General 711

The speed of the gear is nG 5 ( 2 30) (1200) 5 80 rev/min. Therefore the pitch-line

velocity of the gear is

VG 5 pdGnG

12 5 p(5)(80)

12 5 105 ft/min

Then, from Eq. (13–47), the sliding velocity VS is found to be

VS 5 VW

cos l 5

628 cos 9.46°

5 637 ft /min

Getting to the forces now, we begin with the horsepower formula

WWt 5 33 000H

VW 5

(33 000)(1) 628

5 52.5 lbf

This force acts in the negative x direction, the same as in Fig. 13–40. Using Fig. 13–42, we find f 5 0.03. Then, the first equation of Eq. (13–43) gives

W 5 W x

cos fn sin l 1 f cos l

5 52.5

cos 14.5° sin 9.46° 1 0.03 cos 9.46° 5 278 lbf

Also, from Eq. (13–43),

W y 5 W sin fn 5 278 sin 14.5° 5 69.6 lbf

W z 5 W(cos fn cos l 2 f sin l)

5 278(cos 14.5° cos 9.46° 2 0.03 sin 9.46°) 5 264 lbf

We now identify the components acting on the gear as

WG a 5 2Wx 5 52.5 lbf

WGr 5 2W y 5 269.6 lbf

WGt 5 2Wz 5 2264 lbf

A free-body diagram showing the forces and torsion acting on the gearshaft is shown in Fig. 13–44. We shall make B a thrust bearing in order to place the gearshaft in compression. Thus, summing forces in the x direction gives

Answer Fx B 5 252.5 lbf

Taking moments about the z axis, we have

Answer 2(52.5)(2.5) 2 (69.6)(1.5) 1 4Fy B 5 0 Fy

B 5 58.9 lbf

Taking moments about the y axis,

Answer (264)(1.5) 2 4Fz B 5 0 Fz

B 5 99 lbf

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Summing forces in the y direction,

Answer 269.6 1 58.9 1 Fy A 5 0 Fy

A 5 10.7 lbf

Similarly, summing forces in the z direction,

Answer 2264 1 99 1 Fz A 5 0 Fz

A 5 165 lbf

We still have one more equation to write. Summing moments about x,

Answer 2(264)(2.5) 1 T 5 0 T 5 660 lbf ? in

It is because of the frictional loss that this output torque is less than the product of the gear ratio and the input torque.

712 Mechanical Engineering Design

Figure 13–44 Free-body diagram for Ex. 13–10. Forces are given in lbf.

1 2

2 in

1 2

1 in

1 2

2 in

69. 6

52.5

69.6

G y

T x

F z A

F y A

F y B

F z B

F x B

PROBLEMS Problems marked with an asterisk (*) are linked to problems in other chapters, as summarized in Table 1–2 of Sec. 1–17, p. 34.

13–1 A 17-tooth spur pinion has a diametral pitch of 8 teeth/in, runs at 1120 rev/min, and drives a gear at a speed of 544 rev/min. Find the number of teeth on the gear and the theoretical center- to-center distance.

13–2 A 15-tooth spur pinion has a module of 3 mm and runs at a speed of 1600 rev/min. The driven gear has 60 teeth. Find the speed of the driven gear, the circular pitch, and the theoretical center-to-center distance.

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Gears—General 713

13–3 A spur gearset has a module of 6 mm and a velocity ratio of 4. The pinion has 16 teeth. Find the number of teeth on the driven gear, the pitch diameters, and the theoretical center-to-center distance.

13–4 A 21-tooth spur pinion mates with a 28-tooth gear. The diametral pitch is 3 teeth/in and the pressure angle is 20°. Make a drawing of the gears showing one tooth on each gear. Find and tabulate the following results: the addendum, dedendum, clearance, circular pitch, tooth thick- ness, and base-circle diameters; the lengths of the arc of approach, recess, and action; and the base pitch and contact ratio.

13–5 A 20° straight-tooth bevel pinion having 14 teeth and a diametral pitch of 6 teeth/in drives a 32-tooth gear. The two shafts are at right angles and in the same plane. Find: (a) The cone distance (b) The pitch angles (c) The pitch diameters (d) The face width

13–6 A parallel helical gearset uses a 20-tooth pinion driving a 36-tooth gear. The pinion has a right-hand helix angle of 30°, a normal pressure angle of 25°, and a normal diametral pitch of 4 teeth/in. Find: (a) The normal, transverse, and axial circular pitches (b) The normal base circular pitch (c) The transverse diametral pitch and the transverse pressure angle (d) The addendum, dedendum, and pitch diameter of each gear

13–7 A parallel helical gearset consists of a 19-tooth pinion driving a 57-tooth gear. The pinion has a left-hand helix angle of 30°, a normal pressure angle of 20°, and a normal module of 2.5 mm. Find: (a) The normal, transverse, and axial circular pitches (b) The transverse diametral pitch and the transverse pressure angle (c) The addendum, dedendum, and pitch diameter of each gear

13–8 To avoid the problem of interference in a pair of spur gears using a 20° pressure angle, specify the minimum number of teeth allowed on the pinion for each of the following gear ratios. (a) 2 to 1 (b) 3 to 1 (c) 4 to 1 (d) 5 to 1

13–9 Repeat Prob. 13–8 with a 25° pressure angle.

13–10 For a spur gearset with f 5 20°, while avoiding interference, find: (a) The smallest pinion tooth count that will run with itself (b) The smallest pinion tooth count at a ratio mG 5 2.5, and the largest gear tooth count pos-

sible with this pinion (c) The smallest pinion that will run with a rack

13–11 Repeat problem 13–10 for a helical gearset with fn 5 20° and c 5 30°.

13–12 The decision has been made to use fn 5 20°, Pt 5 6 teeth/in, and c 5 30° for a 2:1 reduction. Choose the smallest acceptable full-depth pinion and gear tooth count to avoid interference.

13–13 Repeat Problem 13–12 with c 5 45°.

13–14 By employing a pressure angle larger than standard, it is possible to use fewer pinion teeth, and hence obtain smaller gears without undercutting during machining. If the gears are full-depth spur gears, what is the smallest possible pressure angle f that can be obtained without under- cutting for a 9-tooth pinion to mesh with a rack?

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714 Mechanical Engineering Design

13–15 A parallel-shaft gearset consists of an 18-tooth helical pinion driving a 32-tooth gear. The pinion has a left-hand helix angle of 25°, a normal pressure angle of 20°, and a normal module of 3 mm. Find: (a) The normal, transverse, and axial circular pitches (b) The transverse module and the transverse pressure angle (c) The pitch diameters of the two gears

13–16 The double-reduction helical gearset shown in the figure is driven through shaft a at a speed of 700 rev/min. Gears 2 and 3 have a normal diametral pitch of 12 teeth/in, a 30° helix angle, and a normal pressure angle of 20°. The second pair of gears in the train, gears 4 and 5, have a normal diametral pitch of 8 teeth/in, a 25° helix angle, and a normal pressure angle of 20°. The tooth numbers are: N2 5 12, N3 5 48, N4 5 16, N5 5 36. Find: (a) The directions of the thrust force exerted by each gear upon its shaft (b) The speed and direction of shaft c (c) The center distance between shafts

Problem 13–16 Dimensions in inches.

x z

A B

4 3

1 1 4

3 4

2 1

1 1 42 1

b a

3 4

13–17 Shaft a in the figure rotates at 600 rev/min in the direction shown. Find the speed and direction of rotation of shaft d.

Problem 13–17

3 5 6 7

40T

20T, ! = 30° RH

17T, ! = 30° RH 20T 60T

c d

8T, ! = 60° RH

13–18 The mechanism train shown consists of an assortment of gears and pulleys to drive gear 9. Pulley 2 rotates at 1200 rev/min in the direction shown. Determine the speed and direction of rotation of gear 9.

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Gears—General 715

13–19 The figure shows a gear train consisting of a pair of helical gears and a pair of miter gears. The helical gears have a 171

2 ° normal pressure angle and a helix angle as shown. Find:

(a) The speed of shaft c (b) The distance between shafts a and b (c) The pitch diameter of the miter gears

6-in dia.

10-in dia. 18T

38T

20T

36T

48T

Worm 3T • R.H.

Problem 13–18

13–20 A compound reverted gear train is to be designed as a speed increaser to provide a total increase of speed of exactly 45 to 1. With a 20° pressure angle, specify appropriate numbers of teeth to minimize the gearbox size while avoiding the interference problem in the teeth. Assume all gears will have the same diametral pitch.

13–21 Repeat Prob. 13–20 with a 25° pressure angle.

13–22 Repeat Prob. 13–20 for a gear ratio of exactly 30 to 1.

13–23 Repeat Prob. 13–20 for a gear ratio of approximately 45 to 1.

13–24 A gearbox is to be designed with a compound reverted gear train that transmits 25 horsepower with an input speed of 2500 rev/min. The output should deliver the power at a rotational speed in the range of 280 to 300 rev/min. Spur gears with 20° pressure angle are to be used. Determine

a A

45° 8 normal DP, 12T, 23° !

540 rev/min

4P, 32T 2 5 8

1 1 2

21 3 8

5 1 4

3 3 4

1 1 4

32T

40T

5 8

Problem 13–19 Dimensions in inches.

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716 Mechanical Engineering Design

suitable numbers of teeth for each gear, to minimize the gearbox size while providing an out- put speed within the specified range. Be sure to avoid an interference problem in the teeth.

13–25 The tooth numbers for the automotive differential shown in the figure are N2 5 16, N3 5 48, N4 5 14, N5 5 N6 5 20. The drive shaft turns at 900 rev/min. (a) What are the wheel speeds if the car is traveling in a straight line on a good road surface? (b) Suppose the right wheel is jacked up and the left wheel resting on a good road surface.

What is the speed of the right wheel? (c) Suppose, with a rear-wheel drive vehicle, the auto is parked with the right wheel resting

on a wet icy surface. Does the answer to part (b) give you any hint as to what would hap- pen if you started the car and attempted to drive on?

13–26 The figure illustrates an all-wheel drive concept using three differentials, one for the front axle, another for the rear, and the third connected to the drive shaft. (a) Explain why this concept may allow greater acceleration. (b) Suppose either the center or the rear differential, or both, can be locked for certain road

conditions. Would either or both of these actions provide greater traction? Why?

Problem 13–25

To rear wheel

Planet gears

2 3

Drive shaft

Ring gear

Front differential

Center differential

Rear differential

Driveshaft

Problem 13–26 The Audi “Quattro concept,”

showing the three differentials that provide permanent all-wheel drive.

(Reprinted by permission of Audi of America, Inc.)

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Gears—General 717

13–27 In the reverted planetary train illustrated, find the speed and direction of rotation of the arm if gear 2 is unable to rotate and gear 6 is driven at 12 rev/min in the clockwise direction as viewed from the bottom of the figure.

13–28 In the gear train of Prob. 13–27, let gear 6 be driven at 85 rev/min counterclockwise (as viewed from the bottom of the figure) while gear 2 is held stationary. What is the speed and direction of rotation of the arm?

13–29 Tooth numbers for the gear train shown in the figure are N2 5 12, N3 5 16, and N4 5 12. How many teeth must internal gear 5 have? Suppose gear 5 is fixed. What is the speed of the arm if shaft a rotates at 320 rev/min counterclockwise as viewed from the left side of the figure?

20T 30T

16T

Problem 13–27

13–30 The tooth numbers for the gear train illustrated are N2 5 20, N3 5 16, N4 5 30, N6 5 36, and N7 5 46. Gear 7 is fixed. If shaft a is turned through 10 revolutions, how many turns will shaft b make?

a b

2Problem 13–29

a b

2 7

Problem 13–30

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718 Mechanical Engineering Design

13–31 Shaft a in the figure has a power input of 75 kW at a speed of 1000 rev/min in the counterclock- wise direction. The gears have a module of 5 mm and a 20° pressure angle. Gear 3 is an idler. (a) Find the force F3b that gear 3 exerts against shaft b. (b) Find the torque T4c that gear 4 exerts on shaft c.

13–32 The 24T 6-pitch 20° pinion 2 shown in the figure rotates clockwise at 1000 rev/min and is driven at a power of 25 hp. Gears 4, 5, and 6 have 24, 36, and 144 teeth, respectively. What torque can arm 3 deliver to its output shaft? Draw free-body diagrams of the arm and of each gear and show all forces that act upon them.

13–33 The gears shown in the figure have a module of 12 mm and a 20° pressure angle. The pinion rotates at 1800 rev/min clockwise and transmits 150 kW through the idler pair to gear 5 on shaft c. What forces do gears 3 and 4 transmit to the idler shaft?

a x

51T

34T

17T

Problem 13–31

2 4

3 5

Fixed

+ + +Problem 13–32

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Gears—General 719

13–34 The figure shows a pair of shaft-mounted spur gears having a diametral pitch of 5 teeth/in with an 18-tooth 20° pinion driving a 45-tooth gear. The power input is 32-hp at 1800 rev/min. Find the direction and magnitude of the forces acting on bearings A, B, C, and D.

a x

32T

18T

48T

Problem 13–33

3 in 3 in

TinAB

Problem 13–34

5 5 8 5 5

5 5 8

1 7 8

5 8

15 1 4

11 1 2

Key 4× 5 8

× 1 4

3 4

Problem 13–35 NEMA No. 364 frame; dimensions

in inches. The z axis is directed out of the paper.

13–35 The figure shows the electric-motor frame dimensions for a 30-hp 900 rev/min motor. The frame is bolted to its support using four 3

4-in bolts spaced 111 4 in apart in the view shown and

14 in apart when viewed from the end of the motor. A 4 diametral pitch 20° spur pinion hav- ing 20 teeth and a face width of 2 in is keyed to and flush with the end of the motor shaft. This pinion drives another gear whose axis is in the same xz plane and directly behind the motor shaft. Determine the maximum shear and tensile forces on the mounting bolts based on 200 percent overload torque. Does the direction of rotation matter?

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720 Mechanical Engineering Design

13–36 Continue Prob. 13–24 by finding the following information, assuming a diametral pitch of 6 teeth/in. (a) Determine pitch diameters for each of the gears. (b) Determine the pitch line velocities (in ft/min) for each set of gears. (c) Determine the magnitudes of the tangential, radial, and total forces transmitted between

each set of gears. (d) Determine the input torque. (e) Determine the output torque, neglecting frictional losses.

13–37 A speed-reducer gearbox containing a compound reverted gear train transmits 35 horsepower with an input speed of 1200 rev/min. Spur gears with 20° pressure angle are used, with 16 teeth on each of the small gears and 48 teeth on each of the larger gears. A diametral pitch of 10 teeth/in is proposed. (a) Determine the speeds of the intermediate and output shafts. (b) Determine the pitch line velocities (in ft/min) for each set of gears. (c) Determine the magnitudes of the tangential, radial, and total forces transmitted between

each set of gears. (d) Determine the input torque. (e) Determine the output torque, neglecting frictional losses.

13–38* For the countershaft in Prob. 3–72, p. 152, assume the gear ratio from gear B to its mating gear is 2 to 1. (a) Determine the minimum number of teeth that can be used on gear B without an interference

problem in the teeth. (b) Using the number of teeth from part (a), what diametral pitch is required to also achieve

the given 8-in pitch diameter? (c) Suppose the 20° pressure angle gears are exchanged for gears with 25° pressure angle, while

maintaining the same pitch diameters and diametral pitch. Determine the new forces FA and FB if the same power is to be transmitted.

13–39* For the countershaft in Prob. 3–73, p. 152, assume the gear ratio from gear B to its mating gear is 5 to 1. (a) Determine the minimum number of teeth that can be used on gear B without an interference

problem in the teeth. (b) Using the number of teeth from part (a), what module is required to also achieve the given

300-mm pitch diameter? (c) Suppose the 20° pressure angle for gear A is exchanged for a gear with 25° pressure angle,

while maintaining the same pitch diameters and module. Determine the new forces FA and FB if the same power is to be transmitted.

13–40* For the gear and sprocket assembly analyzed in Prob. 3–77, p. 153, information for the gear sizes and the forces transmitted through the gears was provided in the problem statement. In this problem, we will perform the preceding design steps necessary to acquire the information for the analysis. A motor providing 2 kW is to operate at 191 rev/min. A gear unit is needed to reduce the motor speed by half to drive a chain sprocket. (a) Specify appropriate numbers of teeth on gears F and C to minimize the size while avoiding

the interference problem in the teeth. (b) Assuming an initial guess of 125-mm pitch diameter for gear F, what is the module that

should be used for the stress analysis of the gear teeth?

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Gears—General 721

(c) Calculate the input torque applied to shaft EFG. (d) Calculate the magnitudes of the radial, tangential, and total forces transmitted between

gears F and C.

13–41* For the gear and sprocket assembly analyzed in Prob. 3–79, p. 153, information for the gear sizes and the forces transmitted through the gears was provided in the problem statement. In this problem, we will perform the preceding design steps necessary to acquire the information for the analysis. A motor providing 1 hp is to operate at 70 rev/min. A gear unit is needed to double the motor speed to drive a chain sprocket. (a) Specify appropriate numbers of teeth on gears F and C to minimize the size while avoiding

the interference problem in the teeth. (b) Assuming an initial guess of 10-in pitch diameter for gear F, what is the diametral pitch

that should be used for the stress analysis of the gear teeth? (c) Calculate the input torque applied to shaft EFG. (d) Calculate the magnitudes of the radial, tangential, and total forces transmitted between

gears F and C.

13–42* For the bevel gearset in Probs. 3–74 and 3–76, pp. 152 and 153 respectively, shaft AB is rotating at 600 rev/min and transmits 10 hp. The gears have a 20° pressure angle. (a) Determine the bevel angle g for the gear on shaft AB. (b) Determine the pitch-line velocity. (c) Determine the tangential, radial, and axial forces acting on the pinion. Were the forces

given in Prob. 3–74 correct?

13–43 The figure shows a 16T 20° straight bevel pinion driving a 32T gear, and the location of the bearing centerlines. Pinion shaft a receives 2.5 hp at 240 rev/min. Determine the bearing reac- tions at A and B if A is to take both radial and thrust loads.

21 2 3 1

1 1 2

2 1 2

Problem 13–43 Dimensions in inches.

13–44 The figure shows a 10 diametral pitch 18-tooth 20° straight bevel pinion driving a 30-tooth gear. The transmitted load is 25 lbf. Find the bearing reactions at C and D on the output shaft if D is to take both radial and thrust loads.

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722 Mechanical Engineering Design

13–45 The gears shown in the figure have a normal diametral pitch of 5 teeth/in, a normal pressure angle of 20°, and a 30° helix angle. The transmitted load is 800 lbf. The pinion rotates coun- terclockwise about the y axis, as viewed from the positive y axis. Find the force exerted by each gear on its shaft.

a x

1 2

5 8

9 16

Problem 13–44 Dimensions in inches.

2 3 y

a b 18T, LH 32T, RH

Problem 13−45

13–46 The gears shown in the figure have a normal diametral pitch of 5 teeth/in, a normal pressure angle of 20°, and a 30° helix angle. The transmitted load is 800 lbf. Gear 2 rotates clockwise about the y axis, as viewed from the positive y axis. Gear 3 is an idler. Find the forces exerted by gears 2 and 3 on their shafts.

Problem 13−46

2 3 4

24T 18T16T a b c

13–47 A gear train is composed of four helical gears with the three shaft axes in a single plane, as shown in the figure. The gears have a normal pressure angle of 20° and a 30° helix angle. Gear 2 is the driver, and is rotating counterclockwise as viewed from the top. Shaft b is an idler and the transmitted load from gear 2 to gear 3 is 500 lbf. The gears on shaft b both have a normal diametral pitch of 7 teeth/in and have 54 and 14 teeth, respectively. Find the forces exerted by gears 3 and 4 on shaft b.

2 3 4

RH LH

LH RH

a b c

Problem 13–47

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Gears—General 723

13–48 In the figure for Prob. 13–34, pinion 2 is to be a right-hand helical gear having a helix angle of 30°, a normal pressure angle of 20°, 16 teeth, and a normal diametral pitch of 6 teeth/in. A motor delivers 25-hp to shaft a at a speed of 1720 rev/min clockwise about the x axis. Gear 3 has 42 teeth. Find the reaction exerted by bearings C and D on shaft b. One of these bearings is to take both radial and thrust loads. This bearing should be selected so as to place the shaft in compression.

13–49 Gear 2, in the figure, has 16 teeth, a 20° transverse pressure angle, a 15° helix angle, and a module of 4 mm. Gear 2 drives the idler on shaft b, which has 36 teeth. The driven gear on shaft c has 28 teeth. If the driver rotates at 1600 rev/min and transmits 6 kW, find the radial and thrust load on each shaft.

13–50 The figure shows a double-reduction helical gearset. Pinion 2 is the driver, and it receives a torque of 1200 lbf ? in from its shaft in the direction shown. Pinion 2 has a normal diametral pitch of 8 teeth/in, 14 teeth, and a normal pressure angle of 20° and is cut right-handed with a helix angle of 30°. The mating gear 3 on shaft b has 36 teeth. Gear 4, which is the driver for the second pair of gears in the train, has a normal diametral pitch of 5 teeth/in, 15 teeth, and a normal pressure angle of 20° and is cut left-handed with a helix angle of 15°. Mating gear 5 has 45 teeth. Find the magnitude and direction of the force exerted by the bearings C and D on shaft b if bearing C can take only a radial load while bearing D is mounted to take both radial and thrust loads.

Problem 13–49

b c

4 3

90°

5 3

3 2 2

x T2

3 1 4

1 1 2

3 1 4

A B

Problem 13–50 Dimensions in inches.

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724 Mechanical Engineering Design

13–51 A right-hand single-tooth hardened-steel (hardness not specified) worm has a catalog rating of 2000 W at 600 rev/min when meshed with a 48-tooth cast-iron gear. The axial pitch of the worm is 25 mm, the normal pressure angle is 141

2 °, the pitch diameter of the worm is 100 mm,

and the face widths of the worm and gear are, respectively, 100 mm and 50 mm. Bearings are centered at locations A and B on the worm shaft. Determine which should be the thrust bearing (so that the axial load in the shaft is in compression), and find the magnitudes and directions of the forces exerted by both bearings.

13–52 The hub diameter and projection for the gear of Prob. 13–51 are 100 and 37.5 mm, respectively. The face width of the gear is 50 mm. Locate bearings C and D on opposite sides, spacing C 10 mm from the gear on the hidden face (see figure) and D 10 mm from the hub face. Choose one as the thrust bearing, so that the axial load in the shaft is in compression. Find the output torque and the magnitudes and directions of the forces exerted by the bearings on the gearshaft.

13–53 A 2-tooth left-hand worm transmits 3 4 hp at 600 rev/min to a 36-tooth gear having a transverse

diametral pitch of 8 teeth/in. The worm has a normal pressure angle of 20°, a pitch diameter of 11

2 in, and a face width of 11 2 in. Use a coefficient of friction of 0.05 and find the force exerted

by the gear on the worm and the torque input. For the same geometry as shown for Prob. 13–51, the worm velocity is clockwise as viewed from the positive z axis.

13–54 Write a computer program that will analyze a spur gear or helical-mesh gear, accepting fn, c, Pt, NP, and NG; compute mG, dP, dG, pt, pn, px, and ft; and give advice as to the smallest tooth count that will allow a pinion to run with itself without interference, run with its gear, and run with a rack. Also have it give the largest tooth count possible with the intended pinion.

y B

50 100

Worm pitch cylinder

Gear pitch cylinder

Problem 13–51 Dimensions in millimeters.

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Chapter Outline

14–1 The Lewis Bending Equation 726

14–2 Surface Durability 735

14–3 AGMA Stress Equations 737

14–4 AGMA Strength Equations 739

14–5 Geometry Factors I and J (ZI and YJ) 743

14–6 The Elastic Coefficient Cp (ZE) 748

14–7 Dynamic Factor Kv 748

14–8 Overload Factor Ko 750

14–9 Surface Condition Factor Cf (ZR) 750

14–10 Size Factor Ks 751

14–11 Load-Distribution Factor Km (KH) 751

14–12 Hardness-Ratio Factor CH (ZW) 753

14–13 Stress-Cycle Factors YN and ZN 754

14–14 Reliability Factor KR (YZ) 755

14–15 Temperature Factor KT (Yu) 756

14–16 Rim-Thickness Factor KB 756

14–17 Safety Factors SF and SH 757

14–18 Analysis 757

14–19 Design of a Gear Mesh 767

Spur and Helical Gears14

725

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726 Mechanical Engineering Design

This chapter is devoted primarily to analysis and design of spur and helical gears to resist bending failure of the teeth as well as pitting failure of tooth surfaces. Failure by bending will occur when the significant tooth stress equals or exceeds either the yield strength or the bending endurance strength. A surface failure occurs when the significant contact stress equals or exceeds the surface endurance strength. The first two sections present a little of the history of the analyses from which current methodology developed. The American Gear Manufacturers Association1 (AGMA) has for many years been the responsible authority for the dissemination of knowledge pertaining to the design and analysis of gearing. The methods this organization presents are in general use in the United States when strength and wear are primary considerations. In view of this fact it is important that the AGMA approach to the subject be presented here. The general AGMA approach requires a great many charts and graphs—too many for a single chapter in this book. We have omitted many of these here by choosing a single pressure angle and by using only full-depth teeth. This simplification reduces the complexity but does not prevent the development of a basic understanding of the approach. Furthermore, the simplification makes possible a better development of the fundamentals and hence should constitute an ideal introduction to the use of the general AGMA method.2 Sections 14–1 and 14–2 are elementary and serve as an examination of the foundations of the AGMA method. Table 14–1 is largely AGMA nomenclature.

14–1 The Lewis Bending Equation Wilfred Lewis introduced an equation for estimating the bending stress in gear teeth in which the tooth form entered into the formulation. The equation, announced in 1892, still remains the basis for most gear design today. To derive the basic Lewis equation, refer to Fig. 14–1a, which shows a rectangular cantilever beam of cross-sectional dimensions F and t, having a length l and a load Wt, uniformly distributed across the face width F. The section modulus Iyc is Ft 2y6, and therefore the bending stress is

s 5 M

Iyc 5

6W tl

Ft2 (a)

Gear designers denote the components of gear-tooth forces as Wt, Wr, Wa or W t, Wr, W a interchangeably. The latter notation leaves room for post-subscripts essential to free-body diagrams. For instance, for gears 2 and 3 in mesh, Wt

23 is the transmitted

11001 N. Fairfax Street, Suite 500, Alexandria, VA 22314-1587. 2The standards ANSI/AGMA 2001-D04 (revised AGMA 2001-C95) and ANSI/AGMA 2101-D04 (metric edition of ANSI/AGMA 2001-D04), Fundamental Rating Factors and Calculation Methods for Involute Spur and Helical Gear Teeth, are used in this chapter. The use of American National Standards is completely voluntary; their existence does not in any respect preclude people, whether they have approved the standards or not, from manufacturing, marketing, purchasing, or using products, processes, or procedures not conforming to the standards. The American National Standards Institute does not develop standards and will in no circumstances give an interpretation of any American National Standard. Requests for interpretation of these standards should be addressed to the American Gear Manufacturers Association. [Tables or other self-supporting sections may be quoted or extracted in their entirety. Credit line should read: “Extracted from ANSI/AGMA Standard 2001-D04 or 2101-D04 Fundamental Rating Factors and Calculation Methods for Involute Spur and Helical Gear Teeth” with the permission of the publisher, American Gear Manufacturers Association, 1001 N. Fairfax Street, Suite 500, Alexandria, Virginia 22314-1587.] The foregoing is adapted in part from the ANSI foreword to these standards.

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Spur and Helical Gears 727

Symbol* Name Where Found

Ce Mesh alignment correction factor Eq. (14–35)

Cf (ZR) Surface condition factor Eq. (14–16)

CH (ZW) Hardness-ratio factor Eq. (14–18)

Cma Mesh alignment factor Eq. (14–34)

Cmc Load correction factor Eq. (14–31)

Cmf Face load-distribution factor Eq. (14–30)

Cp (ZE) Elastic coefficient Eq. (14–13)

Cpf Pinion proportion factor Eq. (14–32)

Cpm Pinion proportion modifier Eq. (14–33)

d Pitch diameter Ex. (14–1)

dP Pitch diameter, pinion Eq. (14–22)

dG Pitch diameter, gear Eq. (14–22)

F (b) Net face width of narrowest member Eq. (14–15)

fP Pinion surface finish Fig. 14–13

H Power Fig. 14–17

HB Brinell hardness Ex. 14–3

HBG Brinell hardness of gear Sec. 14–12

HBP Brinell hardness of pinion Sec. 14–12

hp Horsepower Ex. 14–1

ht Gear-tooth whole depth Sec. 14–16

I (ZI) Geometry factor of pitting resistance Eq. (14–16)

J (YJ) Geometry factor for bending strength Eq. (14–15)

KB Rim-thickness factor Eq. (14–40)

Kf Fatigue stress-concentration factor Eq. (14–9)

Km (KH) Load-distribution factor Eq. (14–30)

Ko Overload factor Eq. (14–15)

KR (YZ) Reliability factor Eq. (14–17)

Ks Size factor Sec. 14–10

KT (Yu) Temperature factor Eq. (14–17)

Kv Dynamic factor Eq. (14–27)

m Module Eq. (14–15)

mB Backup ratio Eq. (14–39)

mF Face-contact ratio Eq. (14–19)

mG Gear ratio (never less than 1) Eq. (14–22)

mN Load-sharing ratio Eq. (14–21)

mt Transverse module Eq. (14–15)

N Number of stress cycles Fig. 14–14

NG Number of teeth on gear Eq. (14–22)

NP Number of teeth on pinion Eq. (14–22)

n Speed, in rev/min Eq. (13–34)

Table 14–1

Symbols, Their Names, and Locations

(Continued)

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728 Mechanical Engineering Design

Symbol* Name Where Found

nP Pinion speed, in rev/min Ex. 14–4

P Diametral pitch Eq. (14–2)

Pd Transverse diametral pitch Eq. (14–15)

pN Normal base pitch Eq. (14–24)

pn Normal circular pitch Eq. (14–24)

px Axial pitch Eq. (14–19)

Qv Quality number Eq. (14–29)

R Reliability Eq. (14–38)

Ra Root-mean-squared roughness Fig. 14–13

rf Tooth fillet radius Fig. 14–1

rG Pitch-circle radius, gear In standard

rP Pitch-circle radius, pinion In standard

rbP Pinion base-circle radius Eq. (14–25)

rbG Gear base-circle radius Eq. (14–25)

SC Buckingham surface endurance strength Ex. 14–3

Sc AGMA surface endurance strength Eq. (14–18)

St AGMA bending strength Eq. (14–17)

S Bearing span Fig. 14–10

S1 Pinion offset from center span Fig. 14–10

SF Safety factor—bending Eq. (14–41)

SH Safety factor—pitting Eq. (14–42)

W t or Wt Transmitted load Fig. 14–1

YN Stress-cycle factor for bending strength Fig. 14–14

ZN Stress-cycle factor for pitting resistance Fig. 14–15

b Exponent Eq. (14–44)

s Bending stress, AGMA Eq. (14–15)

sC Contact stress from Hertzian relationships Eq. (14–14)

sc Contact stress from AGMA relationships Eq. (14–16)

sall Allowable bending stress, AGMA Eq. (14–17)

sc,all Allowable contact stress, AGMA Eq. (14–18)

f Pressure angle Eq. (14–12)

fn Normal pressure angle Eq. (14–24)

ft Transverse pressure angle Eq. (14–23)

c Helix angle Ex. 14–5

*Where applicable, the alternate symbol for the metric standard is shown in parenthesis.

Table 14–1

Symbols, Their Names, and Locations (Continued)

force of body 2 on body 3, and W t 32 is the transmitted force of body 3 on body 2.

When working with double- or triple-reduction speed reducers, this notation is compact and essential to clear thinking. Since gear-force components rarely take exponents, this causes no complication. Pythagorean combinations, if necessary, can be treated with parentheses or avoided by expressing the relations trigonometrically.

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Spur and Helical Gears 729

Referring now to Fig. 14–1b, we assume that the maximum stress in a gear tooth occurs at point a. By similar triangles, you can write

ty2 x

5 l

ty2 or x 5

4l or l 5

4x (b)

By rearranging Eq. (a),

s 5 6W tl

Ft2 5 W t

t2y6l 5

W t

t2y4l 1 4 6

(c)

If we now substitute the value of l from Eq. (b) in Eq. (c) and multiply the numera- tor and denominator by the circular pitch p, we find

s 5 W tp

F(2 3) xp

(d )

Letting y 5 2xy(3p), we have

s 5 W t

F p y (14–1)

This completes the development of the original Lewis equation. The factor y is called the Lewis form factor, and it may be obtained by a graphical layout of the gear tooth or by digital computation. In using this equation, most engineers prefer to employ the diametral pitch in determining the stresses. This is done by substituting p 5 pyP and y 5 pY in Eq. (14–1). This gives

s 5 W tP F Y

(14–2)

where

Y 5 2x P

3 (14–3)

The use of this equation for Y means that only the bending of the tooth is considered and that the compression due to the radial component of the force is neglected. Values of Y obtained from this equation are tabulated in Table 14–2.

Figure 14–1

W t

W r

(a) (b)

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730 Mechanical Engineering Design

The use of Eq. (14–3) also implies that the teeth do not share the load and that the greatest force is exerted at the tip of the tooth. But we have already learned that the contact ratio should be somewhat greater than unity, say about 1.5, to achieve a quality gearset. If, in fact, the gears are cut with sufficient accuracy, the tip-load condition is not the worst, because another pair of teeth will be in contact when this condition occurs. Examination of run-in teeth will show that the heaviest loads occur near the middle of the tooth. Therefore the maximum stress probably occurs while a single pair of teeth is carrying the full load, at a point where another pair of teeth is just on the verge of coming into contact.

Dynamic Effects When a pair of gears is driven at moderate or high speed and noise is generated, it is certain that dynamic effects are present. One of the earliest efforts to account for an increase in the load due to velocity employed a number of gears of the same size, material, and strength. Several of these gears were tested to destruction by meshing and loading them at zero velocity. The remaining gears were tested to destruction at various pitch-line velocities. For example, if a pair of gears failed at 500 lbf tangen- tial load at zero velocity and at 250 lbf at velocity V1, then a velocity factor, designated Kv, of 2 was specified for the gears at velocity V1. Then another, identical, pair of gears running at a pitch-line velocity V1 could be assumed to have a load equal to twice the tangential or transmitted load. Note that the definition of dynamic factor Kv has been altered. AGMA standards ANSI/AGMA 2001-D04 and 2101-D04 contain this caution:

Dynamic factor Kv has been redefi ned as the reciprocal of that used in previous AGMA standards. It is now greater than 1.0. In earlier AGMA standards it was less than 1.0.

Care must be taken in referring to work done prior to this change in the standards.

Table 14–2

Values of the Lewis Form Factor Y (These Values Are for a Normal Pressure Angle of 20°, Full-Depth Teeth, and a Diametral Pitch of Unity in the Plane of Rotation)

Number of Number of Teeth Y Teeth Y

12 0.245 28 0.353

13 0.261 30 0.359

14 0.277 34 0.371

15 0.290 38 0.384

16 0.296 43 0.397

17 0.303 50 0.409

18 0.309 60 0.422

19 0.314 75 0.435

20 0.322 100 0.447

21 0.328 150 0.460

22 0.331 300 0.472

24 0.337 400 0.480

26 0.346 Rack 0.485

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Spur and Helical Gears 731

In the nineteenth century, Carl G. Barth first expressed the velocity factor, and in terms of the current AGMA standards, they are represented as

Kv 5 600 1 V

600 (cast iron, cast profile) (14–4a)

Kv 5 1200 1 V

1200 (cut or milled profile) (14–4b)

where V is the pitch-line velocity in feet per minute. It is also quite probable, because of the date that the tests were made, that the tests were conducted on teeth having a cycloidal profile instead of an involute profile. Cycloidal teeth were in general use in the nineteenth century because they were easier to cast than involute teeth. Equation (14–4a) is called the Barth equation. The Barth equation is often modified into Eq. (14–4b), for cut or milled teeth. Later, AGMA added

Kv 5 50 1 1V

50 (hobbed or shaped profile) (14–5a)

Kv 5 B78 1 1V 78

(shaved or ground profile) (14–5b)

In SI units, Eqs. (14–4a) through (14–5b) become

Kv 5 3.05 1 V

3.05 (cast iron, cast profile) (14–6a)

Kv 5 6.1 1 V

6.1 (cut or milled profile) (14–6b)

Kv 5 3.56 1 1V

3.56 (hobbed or shaped profile) (14–6c)

Kv 5 B5.56 1 1V 5.56

(shaved or ground profile) (14–6d)

where V is in meters per second (m/s). Introducing the velocity factor into Eq. (14–2) gives

s 5 KvW tP

FY (14–7)

The metric version of this equation is

s 5 KvW t

FmY (14–8)

where the face width F and the module m are both in millimeters (mm). Expressing the tangential component of load Wt in newtons (N) then results in stress units of megapascals (MPa). As a general rule, spur gears should have a face width F from 3 to 5 times the circular pitch p. Equations (14–7) and (14–8) are important because they form the basis for the AGMA approach to the bending strength of gear teeth. They are in general use for

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732 Mechanical Engineering Design

estimating the capacity of gear drives when life and reliability are not important con- siderations. The equations can be useful in obtaining a preliminary estimate of gear sizes needed for various applications.

EXAMPLE 14–1 A stock spur gear is available having a diametral pitch of 8 teeth/in, a 11 2-in face, 16 teeth,

and a pressure angle of 20° with full-depth teeth. The material is AISI 1020 steel in as-rolled condition. Use a design factor of nd 5 3 to rate the horsepower output of the gear corresponding to a speed of 1200 rev/m and moderate applications.

Solution The term moderate applications seems to imply that the gear can be rated by using the yield strength as a criterion of failure. From Table A–20, we find Sut 5 55 kpsi and Sy 5 30 kpsi. A design factor of 3 means that the allowable bending stress is 30y3 5 10 kpsi. The pitch diameter is d 5 NyP 5 16y8 5 2 in, so the pitch-line velocity is

V 5 pdn 12

5 p(2)1200

12 5 628 ft/min

The velocity factor from Eq. (14–4b) is found to be

Kv 5 1200 1 V

1200 5

1200 1 628 1200

5 1.52

Table 14–2 gives the form factor as Y 5 0.296 for 16 teeth. We now arrange and substitute in Eq. (14–7) as follows:

W t 5 FYsall

Kv P 5

1.5(0.296)10 000 1.52(8)

5 365 lbf

The horsepower that can be transmitted is

Answer hp 5 WtV

33 000 5

365(628) 33 000

5 6.95 hp

It is important to emphasize that this is a rough estimate, and that this approach must not be used for important applications. The example is intended to help you understand some of the fundamentals that will be involved in the AGMA approach.

EXAMPLE 14–2 Estimate the horsepower rating of the gear in the previous example based on obtain- ing an infinite life in bending.

Solution The rotating-beam endurance limit is estimated from Eq. (6–8), p. 290,

S¿e 5 0.5Sut 5 0.5(55) 5 27.5 kpsi

To obtain the surface finish Marin factor ka we refer to Table 6–3, p. 298, for machined surface, finding a 5 2.70 and b 5 20.265. Then Eq. (6–19), p. 295, gives the surface finish Marin factor ka as

ka 5 aSb ut 5 2.70(55)20.265 5 0.934

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The next step is to estimate the size factor kb. From Table 13–1, p. 688, the sum of the addendum and dedendum is

l 5 1 P

1 1.25

P 5

1 8

1 1.25

8 5 0.281 in

The tooth thickness t in Fig. 14–1b is given in Sec. 14–1 [Eq. (b)] as t 5 (4lx)1y2 when x 5 3Yy(2P) from Eq. (14–3). Therefore, since from Ex. 14–1 Y 5 0.296 and P 5 8,

x 5 3Y 2P

5 3(0.296)

2(8) 5 0.0555 in

then

t 5 (4lx)1y2 5 [4(0.281)0.0555]1y2 5 0.250 in

We have recognized the tooth as a cantilever beam of rectangular cross section, so the equivalent rotating-beam diameter must be obtained from Eq. (6–25), p. 297:

de 5 0.808(hb)1y2 5 0.808(Ft)1y2 5 0.808[1.5(0.250)]1y2 5 0.495 in

Then, Eq. (6–20), p. 296, gives kb as

kb 5 a de

0.30 b20.107

5 a0.495 0.30

b20.107

5 0.948

The load factor kc from Eq. (6–26), p. 298, is unity. With no information given con- cerning temperature and reliability we will set kd 5 ke 5 1. In general, a gear tooth is subjected only to one-way bending. Exceptions include idler gears and gears used in reversing mechanisms. We will account for one-way bending by establishing a miscellaneous-effects Marin factor kf. For one-way bending the steady and alternating stress components are sa 5 sm 5 sy2 where s is the largest repeatedly applied bending stress as given in Eq. (14–7). If a material exhibited a Goodman failure locus,

S¿e 1

Sut 5 1

Since Sa and Sm are equal for one-way bending, we substitute Sa for Sm and solve the preceding equation for Sa, giving

Sa 5 S¿eSut

S¿e 1 Sut

Now replace Sa with sy2, and in the denominator replace S9e with 0.5Sut to obtain

s 5 2S¿eSut

0.5Sut 1 Sut 5

2S¿e 0.5 1 1

5 1.33S¿e

Now kf 5 syS9e 5 1.33S9eyS9e 5 1.33. However, a Gerber fatigue locus gives mean values of

S¿e 1 aSm

Sut b2

5 1

Spur and Helical Gears 733

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Setting Sa 5 Sm and solving the quadratic in Sa gives

Sa 5 S2

2S¿e a21 1 B1 1

4S¿2 e

S2 ut b

Setting Sa 5 sy2, Sut 5 S9ey0.5 gives

s 5 S¿e

0.52 [21 1 21 1 4(0.5)2] 5 1.66S¿e

and kf 5 syS9e 5 1.66. Since a Gerber locus runs in and among fatigue data and Goodman does not, we will use kf 5 1.66. The Marin equation for the fully corrected endurance strength is

Se 5 kakbkckdkekf S¿e 5 0.934(0.948)(1)(1)(1)1.66(27.5) 5 40.4 kpsi

For stress, we will first determine the fatigue stress-concentration factor Kf. For a 20° full-depth tooth the radius of the root fillet is denoted rf, where

rf 5 0.300

P 5

0.300 8

5 0.0375 in

From Fig. A–15–6

r d

5 rf

t 5

0.0375 0.250

5 0.15

Since Dyd 5 q, we approximate with Dyd 5 3, giving Kt 5 1.68. From Fig. 6–20, p. 303, q 5 0.62. From Eq. (6–32), p. 303,

Kf 5 1 1 (0.62)(1.68 2 1) 5 1.42

For a design factor of nd 5 3, as used in Ex. 14–1, applied to the load or strength, the maximum bending stress is

smax 5 Kf sall 5 Se

sall 5 Se

Kf nd 5

40.4 1.42(3)

5 9.5 kpsi

The transmitted load W t is

W t 5 FYsall

Kv P 5

1.5(0.296)9 500 1.52(8)

5 347 lbf

and the power is, with V 5 628 ft/min from Ex. 14–1,

hp 5 W tV

33 000 5

347(628) 33 000

5 6.6 hp

Again, it should be emphasized that these results should be accepted only as pre- liminary estimates to alert you to the nature of bending in gear teeth.

734 Mechanical Engineering Design

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Spur and Helical Gears 735

In Ex. 14–2 our resources (Fig. A–15–6) did not directly address stress concentra- tion in gear teeth. A photoelastic investigation by Dolan and Broghamer reported in 1942 constitutes a primary source of information on stress concentration.3 Mitchiner and Mabie4 interpret the results in term of fatigue stress-concentration factor Kf as

Kf 5 H 1 a t r bL at

l bM

(14–9)

where H 5 0.34 2 0.458 366 2f

L 5 0.316 2 0.458 366 2f

M 5 0.290 1 0.458 366 2f

r 5 (b 2 rf)

(dy2) 1 b 2 rf

In these equations l and t are from the layout in Fig. 14–1, f is the pressure angle, rf is the fillet radius, b is the dedendum, and d is the pitch diameter. It is left as an exercise for the reader to compare Kf from Eq. (14–9) with the results of using the approximation of Fig. A–15–6 in Ex. 14–2.

14–2 Surface Durability In this section we are interested in the failure of the surfaces of gear teeth, which is generally called wear. Pitting, as explained in Sec. 6–16, is a surface fatigue failure due to many repetitions of high contact stresses. Other surface failures are scoring, which is a lubrication failure, and abrasion, which is wear due to the presence of foreign material. To obtain an expression for the surface-contact stress, we shall employ the Hertz theory. In Eq. (3–74), p. 138, it was shown that the contact stress between two cylinders may be computed from the equation

pmax 5 2F pbl

(a)

where pmax 5 largest surface pressure

F 5 force pressing the two cylinders together

l 5 length of cylinders

and half-width b is obtained from Eq. (3–73), p. 138, given by

b 5 c 2F pl

(1 2 n2

1)yE1 1 (1 2 n2 2)yE2

1yd1 1 1yd2 d 1y2

(14–10)

where n1, n2, E1, and E2 are the elastic constants and d1 and d2 are the diameters, respectively, of the two contacting cylinders. To adapt these relations to the notation used in gearing, we replace F by W tycos f, d by 2r, and l by the face width F. With these changes, we can substitute the value

3T. J. Dolan and E. I. Broghamer, A Photoelastic Study of the Stresses in Gear Tooth Fillets, Bulletin 335, Univ. Ill. Exp. Sta., March 1942, See also W. D. Pilkey and D. F. Pilkey, Peterson’s Stress-Concentration Factors, 3rd ed., John Wiley & Sons, Hoboken, NJ, 2008, pp. 407–409, 434–437. 4R. G. Mitchiner and H. H. Mabie, “Determination of the Lewis Form Factor and the AGMA Geometry Factor J of External Spur Gear Teeth,” J. Mech. Des., Vol. 104, No. 1, Jan. 1982, pp. 148–158.

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736 Mechanical Engineering Design

of b as given by Eq. (14–10) in Eq. (a). Replacing pmax by sC, the surface compres- sive stress (Hertzian stress) is found from the equation

s2 C 5

pF cos f

1yr1 1 1yr2

(1 2 n2 1)yE1 1 (1 2 n2

2)yE2 (14–11)

where r1 and r2 are the instantaneous values of the radii of curvature on the pinion- and gear-tooth profiles, respectively, at the point of contact. By accounting for load sharing in the value of W t used, Eq. (14–11) can be solved for the Hertzian stress for any or all points from the beginning to the end of tooth contact. Of course, pure roll- ing exists only at the pitch point. Elsewhere the motion is a mixture of rolling and sliding. Equation (14–11) does not account for any sliding action in the evaluation of stress. We note that AGMA uses m for Poisson’s ratio instead of n as is used here. We have already noted that the first evidence of wear occurs near the pitch line. The radii of curvature of the tooth profiles at the pitch point are

r1 5 dP sin f

2 r2 5

dG sin f 2

(14–12)

where f is the pressure angle and dP and dG are the pitch diameters of the pinion and gear, respectively. Note, in Eq. (14–11), that the denominator of the second group of terms contains four elastic constants, two for the pinion and two for the gear. As a simple means of combining and tabulating the results for various combinations of pinion and gear materials, AGMA defines an elastic coefficient Cp by the equation

Cp 5 ≥ 1

p a1 2 n2 P

EP 1

1 2 n2 G

EG b ¥

1y2

(14–13)

With this simplification, and the addition of a velocity factor Kv, Eq. (14–11) can be written as

sC 5 2Cp c KvW t

F cos f a 1

r1 1

1 r2 b d 1y2

(14–14)

where the sign is negative because sC is a compressive stress.

EXAMPLE 14–3 The pinion of Examples 14–1 and 14–2 is to be mated with a 50-tooth gear manu- factured of ASTM No. 50 cast iron. Using the tangential load of 382 lbf, estimate the factor of safety of the drive based on the possibility of a surface fatigue failure. The surface endurance strength of cast iron can be estimated from Sc 5 0.32 HB kpsi for 108 cycles.

Solution From Table A–5 we find the elastic constants to be EP 5 30 Mpsi, nP 5 0.292, EG 5 14.5 Mpsi, nG 5 0.211. We substitute these in Eq. (14–13) to get the elastic coefficient as

Cp 5 ep c 1 2 (0.292)2

30(106) 1

1 2 (0.211)2

14.5(106) d f2(1y2)

5 1817 (psi)1y2

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In addition to the dynamic factor Kv already introduced, there are transmitted load excursions, nonuniform distribution of the transmitted load over the tooth contact, and the influence of rim thickness on bending stress. Tabulated strength values can be means, ASTM minimums, or of unknown heritage. In surface fatigue there are no endurance limits. Endurance strengths have to be qualified as to corresponding cycle count, and the slope of the S-N curve needs to be known. In bending fatigue there is a definite change in slope of the S-N curve near 106 cycles, but some evidence indi- cates that an endurance limit does not exist. Gearing experience leads to cycle counts of 1011 or more. Evidence of diminishing endurance strengths in bending have been included in AGMA methodology.

14–3 AGMA Stress Equations Two fundamental stress equations are used in the AGMA methodology, one for bend- ing stress and another for pitting resistance (contact stress). In AGMA terminology, these are called stress numbers, as contrasted with actual applied stresses, and are

From Example 14–1, the pinion pitch diameter is dP 5 2 in. The value for the gear is dG 5 50y8 5 6.25 in. Then Eq. (14–12) is used to obtain the radii of curvature at the pitch points. Thus

r1 5 2 sin 20°

2 5 0.342 in r2 5

6.25 sin 20° 2

5 1.069 in

The face width is given as F 5 1.5 in. Use Kv 5 1.52 from Example 14–1. Substituting all these values in Eq. (14–14) with f 5 20° gives the contact stress as

sC 5 21817 c 1.52(380) 1.5 cos 20°

a 1 0.342

1 1

1.069 b d 1y2

5 272 400 psi

Table A–24 gives HB 5 262 for ASTM No. 50 cast iron. Therefore SC 5 0.32(262) 5 83.8 kpsi. Contact stress is not linear with respect to the transmitted load [see Eq. (14–14)]. If the factor of safety is defined as the loss-of-function load divided by the imposed load, then the ratio of loads is the ratio of stresses squared. In other words,

n 5 loss-of-function load

imposed load 5

S2 C

s2 C

5 a83.8 72.4 b2

5 1.34

One is free to define factor of safety as SCysC. Awkwardness comes when one com- pares the factor of safety in bending fatigue with the factor of safety in surface fatigue for a particular gear. Suppose the factor of safety of this gear in bending fatigue is 1.20 and the factor of safety in surface fatigue is 1.34 as above. The threat, since 1.34 is greater than 1.20, is in bending fatigue since both numbers are based on load ratios. If the factor of safety in surface fatigue is based on SCysC 5 11.34 5 1.16, then 1.20 is greater than 1.16, but the threat is not from surface fatigue. The surface fatigue factor of safety can be defined either way. One way has the burden of requiring a squared number before numbers that instinctively seem comparable can be compared.

Spur and Helical Gears 737

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738 Mechanical Engineering Design

designated by a lowercase letter s instead of the Greek lower case s we have used in this book (and shall continue to use). The fundamental equations are

s 5 µW tKoKvKs

F KmKB

J (U.S. customary units)

WtKoKvKs

1 bmt

KHKB

YJ (SI units)

(14–15)

where for U.S. customary units (SI units),

W t is the tangential transmitted load, lbf (N) Ko is the overload factor Kv is the dynamic factor Ks is the size factor Pd is the transverse diametral pitch F (b) is the face width of the narrower member, in (mm) Km (KH) is the load-distribution factor KB is the rim-thickness factor J (YJ) is the geometry factor for bending strength (which includes root fillet stress-concentration factor Kf) (mt) is the transverse metric module

Before you try to digest the meaning of all these terms in Eq. (14–15), view them as advice concerning items the designer should consider whether he or she follows the voluntary standard or not. These items include issues such as

• Transmitted load magnitude

• Overload

• Dynamic augmentation of transmitted load

• Size

• Geometry: pitch and face width

• Distribution of load across the teeth

• Rim support of the tooth

• Lewis form factor and root fillet stress concentration

The fundamental equation for pitting resistance (contact stress) is

sc 5 µ CpBWtKoKvKs

dPF Cf

I (U.S. customary units)

ZEBW tKoKvKs

dw1b ZR

ZI (SI units)

(14–16)

where W t, Ko, Kv, Ks, Km, F, and b are the same terms as defined for Eq. (14–15). For U.S. customary units (SI units), the additional terms are

Cp (ZE) is an elastic coefficient, 2lbf/in2 (2N/mm2) Cf (ZR) is the surface condition factor dP (dw1) is the pitch diameter of the pinion, in (mm) I (ZI) is the geometry factor for pitting resistance

The evaluation of all these factors is explained in the sections that follow. The devel- opment of Eq. (14–16) is clarified in the second part of Sec. 14–5.

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Spur and Helical Gears 739

14–4 AGMA Strength Equations Instead of using the term strength, AGMA uses data termed allowable stress numbers and designates these by the symbols sat and sac. It will be less confusing here if we continue the practice in this book of using the uppercase letter S to designate strength and the lowercase Greek letters s and t for stress. To make it perfectly clear we shall use the term gear strength as a replacement for the phrase allowable stress numbers as used by AGMA. Following this convention, values for gear bending strength, designated here as St, are to be found in Figs. 14–2, 14–3, and 14–4, and in Tables 14–3 and 14–4. Since gear strengths are not identified with other strengths such as Sut, Se, or Sy as used elsewhere in this book, their use should be restricted to gear problems. In this approach the strengths are modified by various factors that produce limit- ing values of the bending stress and the contact stress.

Figure 14–2 Allowable bending stress number for through-hardened steels, St. The SI equations are: St 5 0.533HB 1 88.3 MPa, grade 1, and St 5 0.703HB 1

113 MPa, grade 2. (Source: ANSI/AGMA 2001-D04 and 2101-D04.)

Metallurgical and quality control procedure required

150 200 250 300 350 400 450 10

Brinell hardness, HB

A llo

w ab

le b

en di

ng s

tr es

s nu

m be

r, S t

k ps

Grade 1 St = 77.3 HB + 12 800 psi

Grade 2 St = 102 HB + 16 400 psi

Figure 14–3 Allowable bending stress number for nitrided through- hardened steel gears (i.e., AISI 4140, 4340), St. The SI equations are: St 5 0.568HB 1 83.8 MPa, grade 1, and St 5 0.749HB 1 110 MPa, grade 2. (Source: ANSI/AGMA 2001-D04 and 2101-D04.)

Metallurgical and quality control procedures required

250 275 300 325 350 20

Grade 1 St = 82.3HB + 12 150 psi

Grade 2 St = 108.6HB + 15 890 psi

A llo

w ab

le b

en di

ng s

tr es

s nu

m be

r, S t

k ps

Core hardness, HB

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740 Mechanical Engineering Design

250 275 300 325 350 30

Core hardness, HB A

llo w

ab le

b en

di ng

s tr

es s

nu m

be rs

, S t k

ps i

Metallurgical and quality control procedures required

Grade 1 − Nitralloy St = 86.2HB + 12 730 psi

Grade 1 − 2.5% Chrome St = 105.2HB + 9280 psi

Grade 2 − Nitralloy St = 113.8HB + 16 650 psi

Grade 2 − 2.5% Chrome St = 105.2HB + 22 280 psi

Grade 3 − 2.5% Chrome St = 105.2HB + 29 280 psi

Figure 14–4 Allowable bending stress numbers for nitriding steel gears, St. The SI equations are: St 5 0.594HB 1 87.76 MPa Nitralloy grade 1 St 5 0.784HB 1 114.81 MPa Nitralloy grade 2 St 5 0.7255HB 1 63.89 MPa 2.5% chrome, grade 1 St 5 0.7255HB 1 153.63 MPa 2.5% chrome, grade 2 St 5 0.7255HB 1 201.91 MPa 2.5% chrome, grade 3 (Source: ANSI/AGMA 2001-D04, 2101-D04.)

Minimum Allowable Bending Stress Number St, 2

Material Heat Surface psi Designation Treatment Hardness1 Grade 1 Grade 2 Grade 3

Steel3 Through-hardened See Fig. 14–2 See Fig. 14–2 See Fig. 14–2 — Flame4 or induction See Table 8* 45 000 55 000 — hardened4 with type A pattern5

Flame4 or induction See Table 8* 22 000 22 000 — hardened4 with type B pattern5

Carburized and See Table 9* 55 000 65 000 or 75 000 hardened 70 0006

Nitrided4,7 (through- 83.5 HR15N See Fig. 14–3 See Fig. 14–3 — hardened steels)

Nitralloy 135M, Nitrided4,7 87.5 HR15N See Fig. 14–4 See Fig. 14–4 See Fig. 14–4 Nitralloy N, and 2.5% chrome (no aluminum)

Notes: See ANSI/AGMA 2001-D04 for references cited in notes 1–7. 1Hardness to be equivalent to that at the root diameter in the center of the tooth space and face width. 2See tables 7 through 10 for major metallurgical factors for each stress grade of steel gears. 3The steel selected must be compatible with the heat treatment process selected and hardness required. 4The allowable stress numbers indicated may be used with the case depths prescribed in 16.1. 5See figure 12 for type A and type B hardness patterns. 6If bainite and microcracks are limited to grade 3 levels, 70 000 psi may be used. 7The overload capacity of nitrided gears is low. Since the shape of the effective S-N curve is flat, the sensitivity to shock should be investigated before proceeding with the design. [7]

*Tables 8 and 9 of ANSI/AGMA 2001-D04 are comprehensive tabulations of the major metallurgical factors affecting St and Sc of flame-hardened and induction-hardened (Table 8) and carburized and hardened (Table 9) steel gears.

Table 14–3

Repeatedly Applied Bending Strength St at 107 Cycles and 0.99 Reliability for Steel Gears Source: ANSI/AGMA 2001-D04.

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Spur and Helical Gears 741

The equation for the allowable bending stress is

sall 5 µ St

KTKR (U.S. customary units)

YuYZ (SI units)

(14–17)

where for U.S. customary units (SI units),

St is the allowable bending stress, lbf/in2 (N/mm2) YN is the stress-cycle factor for bending stress KT (Yu) are the temperature factors KR (YZ) are the reliability factors SF is the AGMA factor of safety, a stress ratio

Table 14–4

Repeatedly Applied Bending Strength St for Iron and Bronze Gears at 107 Cycles and 0.99 Reliability Source: ANSI/AGMA 2001-D04.

Allowable Bending Material Heat Typical Minimum Stress Number, St,3 Material Designation1 Treatment Surface Hardness2 psi

ASTM A48 gray Class 20 As cast — 5000 cast iron Class 30 As cast 174 HB 8500

Class 40 As cast 201 HB 13 000

ASTM A536 ductile Grade 60–40–18 Annealed 140 HB 22 000–33 000 (nodular) Iron Grade 80–55–06 Quenched and 179 HB 22 000–33 000 tempered

Grade 100–70–03 Quenched and 229 HB 27 000–40 000 tempered

Grade 120–90–02 Quenched and 269 HB 31 000–44 000 tempered

Bronze Sand cast Minimum tensile strength 5700 40 000 psi

ASTM B–148 Heat treated Minimum tensile strength 23 600 Alloy 954 90 000 psi

Notes: 1See ANSI/AGMA 2004-B89, Gear Materials and Heat Treatment Manual. 2Measured hardness to be equivalent to that which would be measured at the root diameter in the center of the tooth space and face width. 3The lower values should be used for general design purposes. The upper values may be used when: High quality material is used. Section size and design allow maximum response to heat treatment. Proper quality control is effected by adequate inspection. Operating experience justifies their use.

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742 Mechanical Engineering Design

The equation for the allowable contact stress sc,all is

sc,all 5 µ Sc

SH ZNCH

KTKR (U.S. customary units)

SH ZNZW

YuYZ (SI units)

(14–18)

where the upper equation is in U.S. customary units and the lower equation is in SI units, Also,

Sc is the allowable contact stress, lbf/in2 (N/mm2) ZN is the stress-cycle factor CH (ZW) are the hardness ratio factors for pitting resistance KT (Yu) are the temperature factors KR (YZ) are the reliability factors SH is the AGMA factor of safety, a stress ratio

The values for the allowable contact stress, designated here as Sc, are to be found in Fig. 14–5 and Tables 14–5, 14–6, and 14–7. AGMA allowable stress numbers (strengths) for bending and contact stress are for

• Unidirectional loading

• 10 million stress cycles

• 99 percent reliability

150 200 250 300 350 400 450 75

100

125

150

175

A llo

w ab

le c

on ta

ct s

tr es

s nu

m be

r, S c

Brinell hardness, HB

Grade 1 Sc = 322 HB + 29 100psi

Grade 2 Sc = 349 HB + 34 300psi

Metallurgical and quality control procedures required

10 00

lb /i

n2Figure 14–5 Contact-fatigue strength Sc at 107 cycles and 0.99 reliability for through-hardened steel gears. The SI equations are: Sc 5 2.22HB 1 200 MPa, grade 1, and Sc 5 2.41HB 1 237 MPa, grade 2. (Source: ANSI/AGMA 2001-D04 and 2101-D04.)

Hardness, Temperature Nitriding, Rockwell C Scale Steel Before Nitriding, °F °F Case Core

Nitralloy 135* 1150 975 62–65 30–35

Nitralloy 135M 1150 975 62–65 32–36

Nitralloy N 1000 975 62–65 40–44

AISI 4340 1100 975 48–53 27–35

AISI 4140 1100 975 49–54 27–35

31 Cr Mo V 9 1100 975 58–62 27–33

*Nitralloy is a trademark of the Nitralloy Corp., New York.

Table 14–5

Nominal Temperature Used in Nitriding and Hardnesses Obtained Source: Darle W. Dudley, Handbook of Practical Gear Design, rev. ed., McGraw-Hill, New York, 1984.

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Spur and Helical Gears 743

The factors in this section, too, will be evaluated in subsequent sections. When two-way (reversed) loading occurs, as with idler gears, AGMA recom- mends using 70 percent of St values. This is equivalent to 1y0.70 5 1.43 as a value of ke in Ex. 14–2. The recommendation falls between the value of ke 5 1.33 for a Goodman failure locus and ke 5 1.66 for a Gerber failure locus.

14–5 Geometry Factors I and J (ZI and YJ) We have seen how the factor Y is used in the Lewis equation to introduce the effect of tooth form into the stress equation. The AGMA factors5 I and J are intended to accomplish the same purpose in a more involved manner. The determination of I and J depends upon the face-contact ratio mF. This is defined as

mF 5 F px

(14–19)

where px is the axial pitch and F is the face width. For spur gears, mF 5 0.

5A useful reference is AGMA 908-B89, Geometry Factors for Determining Pitting Resistance and Bending Strength of Spur, Helical and Herringbone Gear Teeth.

Table 14–6

Repeatedly Applied Contact Strength Sc at 107 Cycles and 0.99 Reliability for Steel Gears Source: ANSI/AGMA 2001-D04.

Minimum Material Heat Surface Allowable Contact Stress Number,2 Sc, psi Designation Treatment Hardness1 Grade 1 Grade 2 Grade 3

Steel3 Through hardened4 See Fig. 14–5 See Fig. 14–5 See Fig. 14–5 —

Flame5 or induction 50 HRC 170 000 190 000 —

hardened5 54 HRC 175 000 195 000 —

Carburized and See Table 9* 180 000 225 000 275 000 hardened5

Nitrided5 (through 83.5 HR15N 150 000 163 000 175 000

hardened steels) 84.5 HR15N 155 000 168 000 180 000

2.5% chrome Nitrided5 87.5 HR15N 155 000 172 000 189 000 (no aluminum)

Nitralloy 135M Nitrided5 90.0 HR15N 170 000 183 000 195 000

Nitralloy N Nitrided5 90.0 HR15N 172 000 188 000 205 000

2.5% chrome Nitrided5 90.0 HR15N 176 000 196 000 216 000 (no aluminum)

Notes: See ANSI/AGMA 2001-D04 for references cited in notes 1–5. 1Hardness to be equivalent to that at the start of active profile in the center of the face width. 2See Tables 7 through 10 for major metallurgical factors for each stress grade of steel gears. 3The steel selected must be compatible with the heat treatment process selected and hardness required. 4These materials must be annealed or normalized as a minimum. 5The allowable stress numbers indicated may be used with the case depths prescribed in 16.1.

*Table 9 of ANSI/AGMA 2001-D04 is a comprehensive tabulation of the major metallurgical factors affecting St and Sc of carburized and hardened steel gears.

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744 Mechanical Engineering Design

Low-contact-ratio (LCR) helical gears having a small helix angle or a thin face width, or both, have face-contact ratios less than unity (mF # 1), and will not be considered here. Such gears have a noise level not too different from that for spur gears. Consequently we shall consider here only spur gears with mF 5 0 and conven- tional helical gears with mF . 1.

Bending-Strength Geometry Factor J (YJ) The AGMA factor J employs a modified value of the Lewis form factor, also denoted by Y; a fatigue stress-concentration factor Kf; and a tooth load-sharing ratio mN. The resulting equation for J for spur and helical gears is

J 5 Y

Kf mN (14–20)

It is important to note that the form factor Y in Eq. (14–20) is not the Lewis factor at all. The value of Y here is obtained from calculations within AGMA 908-B89, and is often based on the highest point of single-tooth contact.

Allowable Contact Material Heat Typical Minimum Stress Number,3 Sc, Material Designation1 Treatment Surface Hardness2 psi

ASTM A48 gray Class 20 As cast — 50 000–60 000 cast iron Class 30 As cast 174 HB 65 000–75 000 Class 40 As cast 201 HB 75 000–85 000

ASTM A536 ductile Grade 60–40–18 Annealed 140 HB 77 000–92 000 (nodular) iron Grade 80–55–06 Quenched and 179 HB 77 000–92 000 tempered

Grade 100–70–03 Quenched and 229 HB 92 000–112 000 tempered

Grade 120–90–02 Quenched and 269 HB 103 000–126 000 tempered

Bronze — Sand cast Minimum tensile 30 000 strength 40 000 psi

ASTM B-148 Heat treated Minimum tensile 65 000 Alloy 954 strength 90 000 psi

Notes: 1See ANSI/AGMA 2004-B89, Gear Materials and Heat Treatment Manual. 2Hardness to be equivalent to that at the start of active profile in the center of the face width. 3The lower values should be used for general design purposes. The upper values may be used when: High-quality material is used. Section size and design allow maximum response to heat treatment. Proper quality control is effected by adequate inspection. Operating experience justifies their use.

Table 14–7

Repeatedly Applied Contact Strength Sc 107 Cycles and 0.99 Reliability for Iron and Bronze Gears Source: ANSI/AGMA 2001-D04.

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Spur and Helical Gears 745

The factor Kf in Eq. (14–20) is called a stress-correction factor by AGMA. It is based on a formula deduced from a photoelastic investigation of stress concentration in gear teeth over 50 years ago. The load-sharing ratio mN is equal to the face width divided by the minimum total length of the lines of contact. This factor depends on the transverse contact ratio mp, the face-contact ratio mF, the effects of any profile modifications, and the tooth deflection. For spur gears, mN 5 1.0. For helical gears having a face-contact ratio mF . 2.0, a conservative approximation is given by the equation

mN 5 pN

0.95Z (14–21)

where pN is the normal base pitch and Z is the length of the line of action in the transverse plane (distance Lab in Fig. 13–15, p. 676). Use Fig. 14–6 to obtain the geometry factor J for spur gears having a 20° pres- sure angle and full-depth teeth. Use Figs. 14–7 and 14–8 for helical gears having a 20° normal pressure angle and face-contact ratios of mF 5 2 or greater. For other gears, consult the AGMA standard.

Figure 14–6 Spur-gear geometry factors J. Source: The graph is from AGMA 218.01, which is consistent with tabular data from the current AGMA 908-B89. The graph is convenient for design purposes.

12 15 17 20 24 30 35 40 45 50 60 80 125 275 ∞

0.20

0.25

0.30

0.35

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

Generating rack 1 pitch

0.35 rT

A dd

en du

m 1.

00 0

2. 40

0 W

ho le

d ep

20°

Pinion addendum 1.000

Gear addendum 1.000

0.40

0.45

0.50

0.55

0.60

1000 170 85 50 35 25 17

Number of teeth in mating gear

L oa

d ap

pl ie

d at

h ig

he st

p oi

nt of

s in

gl e-

to ot

h co

nt ac

Load applied at tip of tooth

Number of teeth for which geometry factor is desired

G eo

m et

ry fa

ct or

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746 Mechanical Engineering Design

Surface-Strength Geometry Factor I (ZI) The factor I is also called the pitting-resistance geometry factor by AGMA. We will develop an expression for I by noting that the sum of the reciprocals of Eq. (14–14), from Eq. (14–12), can be expressed as

1 r1

1 1 r2

5 2

sin ft a 1

dP 1

1 dG b (a)

where we have replaced f by ft, the transverse pressure angle, so that the relation will apply to helical gears too. Now define speed ratio mG as

mG 5 NG

NP 5

dP (14–22)

Figure 14–7 Helical-gear geometry factors J9. Source: The graph is from AGMA 218.01, which is consistent with tabular data from the current AGMA 908-B89. The graph is convenient for design purposes.

Value for Z is for an element of indicated numbers of teeth and a 75-tooth mate

Normal tooth thickness of pinion and gear tooth each reduced 0.024 in to provide 0.048 in total backlash for one normal diametral pitch

Factors are for teeth cut with a full fillet hob

0° 5° 10° 15° 20° 25° 30° 35° 0.30

0.40

0.50

0.60

0.70

500 150 60

Helix angle !

(b)

(a)

G eo

m et

ry fa

ct or

J '

mN =

pN 0.95Z

To ot

h he

ig ht

Generating rack

2. 35

5 P n

A dd

. 1.

0 P n

20°

0.4276 Pnd

rT =

N um

be r o

f t ee

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Spur and Helical Gears 747

Equation (a) can now be written

1 r1

1 1 r2

5 2

dP sin ft mG 1 1

mG (b)

Now substitute Eq. (b) for the sum of the reciprocals in Eq. (14–14). The result is found to be

sc 5 0sC 0 5 Cp ≥ Kv W t

dP F

1 cos ft sin ft

mG 1 1

¥ 1y2

(c)

The geometry factor I for external spur and helical gears is the denominator of the second term in the brackets in Eq. (c). By adding the load-sharing ratio mN, we obtain a factor valid for both spur and helical gears. The equation is then written as

I 5 µ cos ft sin ft

2mN

mG 1 1 external gears

cos ft sin ft

2mN

mG 2 1 internal gears

(14–23)

where mN 5 1 for spur gears. In solving Eq. (14–21) for mN, note that

pN 5 pn cos fn (14–24)

where pn is the normal circular pitch. The quantity Z, for use in Eq. (14–21), can be obtained from the equation

Z 5 [(rP 1 a)2 2 r 2 b P]1y2 1 [(rG 1 a)2 2 r 2

b G]1y2 2 (rP 1 rG) sin ft (14–25)

where rP and rG are the pitch radii and rbP and rbG the base-circle radii of the pinion and gear, respectively.6 Recall from Eq. (13–6), the radius of the base circle is

rb 5 r cos ft (14–26)

Figure 14–8 J9-factor multipliers for use with Fig. 14–7 to find J. Source: The graph is from AGMA 218.01, which is consistent with tabular data from the current AGMA 908-B89. The graph is convenient for design purposes.

The modifying factor can be applied to the J factor when other than 75 teeth are used in the mating element

0° 5° 10° 15° 20° 25° 30° 35° 0.85

0.90

0.95

1.00

1.05

500 150 75 50

Helix angle !

M od

if yi

ng fa

ct or

N um

be r o

f t ee

th in

m at

in g

el em

en t

6For a development, see Joseph E. Shigley and John J. Uicker Jr., Theory of Machines and Mechanisms, McGraw-Hill, New York, 1980, p. 262.

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748 Mechanical Engineering Design

Certain precautions must be taken in using Eq. (14–25). The tooth profiles are not conjugate below the base circle, and consequently, if either one or the other of the first two terms in brackets is larger than the third term, then it should be replaced by the third term. In addition, the effective outside radius is sometimes less than r 1 a, owing to removal of burrs or rounding of the tips of the teeth. When this is the case, always use the effective outside radius instead of r 1 a.

14–6 The Elastic Coefficient Cp (ZE) Values of Cp may be computed directly from Eq. (14–13) or obtained from Table 14–8.

14–7 Dynamic Factor KY As noted earlier, dynamic factors are used to account for inaccuracies in the manu- facture and meshing of gear teeth in action. Transmission error is defined as the departure from uniform angular velocity of the gear pair. Some of the effects that produce transmission error are:

• Inaccuracies produced in the generation of the tooth profile; these include errors in tooth spacing, profile lead, and runout

• Vibration of the tooth during meshing due to the tooth stiffness

• Magnitude of the pitch-line velocity

• Dynamic unbalance of the rotating members

• Wear and permanent deformation of contacting portions of the teeth

• Gearshaft misalignment and the linear and angular deflection of the shaft

• Tooth friction

In an attempt to account for these effects, AGMA has defined a set of quality numbers, Qv.

7 These numbers define the tolerances for gears of various sizes manu- factured to a specified accuracy. Quality numbers 3 to 7 will include most commercial- quality gears. Quality numbers 8 to 12 are of precision quality. The following equations for the dynamic factor are based on these Qv numbers:

Kv 5 µ aA 1 1V A

V in ft/min

aA 1 1200V A

V in m/s (14–27)

where

A 5 50 1 56(1 2 B)

B 5 0.25(12 2 Qv)2y3 (14–28)

7AGMA 2000-A88. ANSI/AGMA 2001-D04, adopted in 2004, replaced the quality number Qv with the transmission accuracy level number Av and incorporated ANSI/AGMA 2015-1-A01. Av ranges from 6 to 12, with lower numbers representing greater accuracy. The Qv approach was maintained as an alternate approach, and resulting Kv values are comparable.

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7 4 9

Gear Material and Modulus of Elasticity EG, lbf/in2 (MPa)* Malleable Nodular Cast Aluminum Tin Pinion Modulus of Steel Iron Iron Iron Bronze Bronze Pinion Elasticity Ep 30 3 106 25 3 106 24 3 106 22 3 106 17.5 3 106 16 3 106

Material psi (MPa)* (2 3 105) (1.7 3 105) (1.7 3 105) (1.5 3 105) (1.2 3 105) (1.1 3 105)

Steel 30 3 106 2300 2180 2160 2100 1950 1900 (2 3 105) (191) (181) (179) (174) (162) (158)

Malleable iron 25 3 106 2180 2090 2070 2020 1900 1850 (1.7 3 105) (181) (174) (172) (168) (158) (154)

Nodular iron 24 3 106 2160 2070 2050 2000 1880 1830 (1.7 3 105) (179) (172) (170) (166) (156) (152)

Cast iron 22 3 106 2100 2020 2000 1960 1850 1800 (1.5 3 105) (174) (168) (166) (163) (154) (149)

Aluminum bronze 17.5 3 106 1950 1900 1880 1850 1750 1700 (1.2 3 105) (162) (158) (156) (154) (145) (141)

Tin bronze 16 3 106 1900 1850 1830 1800 1700 1650 (1.1 3 105) (158) (154) (152) (149) (141) (137)

Poisson’s ratio 5 0.30.

*When more exact values for modulus of elasticity are obtained from roller contact tests, they may be used.

Table 14–8

Elastic Coefficient Cp (ZE), 1psi (1MPa) Source: AGMA 218.01

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750 Mechanical Engineering Design

Figure 14–9 graphically represents Eq. (14–27). The maximum recommended pitch-line velocity for a given quality number is represented by the end point of each Qv curve, and is given by

(Vt)max 5 • [A 1 (Qv 2 3)]2 ft/min

[A 1 (Qv 2 3)]2

200 m/s

(14–29)

14–8 Overload Factor Ko

The overload factor Ko is intended to make allowance for all externally applied loads in excess of the nominal tangential load W t in a particular application (see Figs. 14–17 and 14–18 for tables). Examples include variations in torque from the mean value due to firing of cylinders in an internal combustion engine or reaction to torque variations in a piston pump drive. There are other similar factors such as application factor or service factor. These factors are established after considerable field experi- ence in a particular application.8

14–9 Surface Condition Factor Cf (ZR) The surface condition factor Cf or ZR is used only in the pitting resistance equation, Eq. (14–16). It depends on

• Surface finish as affected by, but not limited to, cutting, shaving, lapping, grinding, shotpeening

• Residual stress

• Plastic effects (work hardening)

Standard surface conditions for gear teeth have not yet been established. When a det- rimental surface finish effect is known to exist, AGMA specifies a value of Cf greater than unity.

Figure 14–9 Dynamic factor Kv. The equations to these curves are given by Eq. (14–27) and the end points by Eq. (14–29). (ANSI/AGMA 2001-D04, Annex A)

Qv = 5 Qv = 6

Qv = 7

Qv = 8

Qv = 9

Qv = 10

Qv = 11

0 2000 4000 6000 8000 10 000 1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

Pitch-line velocity, Vt , ft /min

D yn

am ic

fa ct

or , K

v “Very Accurate Gearing”

8An extensive list of service factors appears in Howard B. Schwerdlin, “Couplings,” Chap. 16 in Joseph E. Shigley, Charles R. Mischke, and Thomas H. Brown, Jr. (eds.), Standard Handbook of Machine Design, 3rd ed., McGraw-Hill, New York, 2004.

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Spur and Helical Gears 751

14–10 Size Factor Ks

The size factor reflects nonuniformity of material properties due to size. It depends upon

• Tooth size

• Diameter of part

• Ratio of tooth size to diameter of part

• Face width

• Area of stress pattern

• Ratio of case depth to tooth size

• Hardenability and heat treatment

Standard size factors for gear teeth have not yet been established for cases where there is a detrimental size effect. In such cases AGMA recommends a size factor greater than unity. If there is no detrimental size effect, use unity. AGMA has identified and provided a symbol for size factor. Also, AGMA sug- gests Ks 5 1, which makes Ks a placeholder in Eqs. (14–15) and (14–16) until more information is gathered. Following the standard in this manner is a failure to apply all of your knowledge. From Table 13–1, p. 688, l 5 a 1 b 5 2.25yP. The tooth thick- ness t in Fig. 14–6 is given in Sec. 14–1, Eq. (b), as t 5 14lx where x 5 3Yy(2P) from Eq. (14–3). From Eq. (6–25), p. 297, the equivalent diameter de of a rectangular section in bending is de 5 0.8081Ft. From Eq. (6–20), p. 296, kb 5 (dey0.3)20.107. Noting that Ks is the reciprocal of kb, we find the result of all the algebraic substitution is

Ks 5 1 kb

5 1.192 aF1Y P b0.0535

(a)

Ks can be viewed as Lewis’s geometry incorporated into the Marin size factor in fatigue. You may set Ks 5 1, or you may elect to use the preceding Eq. (a). This is a point to discuss with your instructor. We will use Eq. (a) to remind you that you have a choice. If Ks in Eq. (a) is less than 1, use Ks 5 1.

14–11 Load-Distribution Factor Km (KH) The load-distribution factor modified the stress equations to reflect nonuniform dis- tribution of load across the line of contact. The ideal is to locate the gear “midspan” between two bearings at the zero slope place when the load is applied. However, this is not always possible. The following procedure is applicable to

• Net face width to pinion pitch diameter ratio FydP # 2

• Gear elements mounted between the bearings

• Face widths up to 40 in

• Contact, when loaded, across the full width of the narrowest member

The load-distribution factor under these conditions is currently given by the face load distribution factor, Cmf, where

Km 5 Cmf 5 1 1 Cmc(Cp f Cpm 1 Cma Ce) (14–30)

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752 Mechanical Engineering Design

where

Cmc 5 e1 for uncrowned teeth 0.8 for crowned teeth

(14–31)

Cpf 5 f F 10dP

2 0.025 F # 1 in

F 10dP

2 0.0375 1 0.0125F 1 , F # 17 in

F 10dP

2 0.1109 1 0.0207F 2 0.000 228F2 17 , F # 40 in

(14–32)

Note that for values of Fy(10dP) , 0.05, Fy(10dP) 5 0.05 is used.

Cpm 5 e1 for straddle-mounted pinion with S1yS , 0.175 1.1 for straddle-mounted pinion with S1yS $ 0.175

(14–33)

Cma 5 A 1 BF 1 CF2 (see Table 14–9 for values of A, B, and C) (14–34)

Ce 5 •0.8 for gearing adjusted at assembly, or compatibility is improved by lapping, or both

1 for all other conditions (14–35)

See Fig. 14–10 for definitions of S and S1 for use with Eq. (14–33), and see Fig. 14–11 for graph of Cma.

Condition A B C

Open gearing 0.247 0.0167 20.765(1024)

Commercial, enclosed units 0.127 0.0158 20.930(1024)

Precision, enclosed units 0.0675 0.0128 20.926(1024)

Extraprecision enclosed gear units 0.00360 0.0102 20.822(1024)

*See ANSI/AGMA 2101-D04, pp. 20–22, for SI formulation.

Table 14–9

Empirical Constants A, B, and C for Eq. (14–34), Face Width F in Inches* Source: ANSI/AGMA 2001-D04.

S1 S 2

Centerline of bearing

Centerline of gear face

Figure 14–10 Definition of distances S and S1 used in evaluating Cpm, Eq. (14–33). (ANSI/AGMA 2001-D04.)

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Spur and Helical Gears 753

14–12 Hardness-Ratio Factor CH (ZW) The pinion generally has a smaller number of teeth than the gear and consequently is subjected to more cycles of contact stress. If both the pinion and the gear are through- hardened, then a uniform surface strength can be obtained by making the pinion harder than the gear. A similar effect can be obtained when a surface-hardened pinion is mated with a through-hardened gear. The hardness-ratio factor CH is used only for the gear. Its purpose is to adjust the surface strengths for this effect. For the pinion, CH 5 1. For the gear, CH is obtained from the equation

CH 5 1.0 1 A¿ (mG 2 1.0) (14–36)

where

A¿ 5 8.98(1023) aHBP

HBG b 2 8.29(1023) 1.2 #

HBP

HBG # 1.7

The terms HBP and HBG are the Brinell hardness (10-mm ball at 3000-kg load) of the pinion and gear, respectively. The term mG is the speed ratio and is given by Eq. (14–22). See Fig. 14–12 for a graph of Eq. (14–36). For

HBP

HBG , 1.2, A¿ 5 0

HBP

HBG . 1.7, A¿ 5 0.006 98

When surface-hardened pinions with hardnesses of 48 Rockwell C scale (Rockwell C48) or harder are run with through-hardened gears (180–400 Brinell), a work hard- ening occurs. The CH factor is a function of pinion surface finish fP and the mating gear hardness. Figure 14–13 displays the relationships:

CH 5 1 1 B¿ (450 2 HBG) (14–37)

0 5 10 15 20 25 30 35 0.0

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

M es

h al

ig nm

en t f

ac to

r, C

m a

Face width, F (in)

Curve 1

Curve 2

Curve 3

Curve 4

Open gearing

Commercial enclosed gear units

Precision enclosed gear units

Extra precision enclosed gear units

For determination of Cma , see Eq. (14–34)

Figure 14–11 Mesh alignment factor Cma. Curve-fit equations in Table 14–9. (ANSI/AGMA 2001-D04.)

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754 Mechanical Engineering Design

where B9 5 0.000 75 exp[20.0112 fP] and fP is the surface finish of the pinion expressed as root-mean-square roughness Ra in m in.

14–13 Stress-Cycle Factors YN and ZN

The AGMA strengths as given in Figs. 14–2 through 14–4, in Tables 14–3 and 14–4 for bending fatigue, and in Fig. 14–5 and Tables 14–5 and 14–6 for contact-stress fatigue are based on 107 load cycles applied. The purpose of the stress-cycle factors YN and ZN is to modify the gear strength for lives other than 107 cycles. Values for these factors are given in Figs. 14–14 and 14–15. Note that for 107 cycles YN 5 ZN 5 1 on each graph. Note also that the equations for YN and ZN change on either side of 107 cycles. For life goals slightly higher than 107 cycles, the mating gear may be experiencing fewer than 107 cycles and the equations for (YN)P and (YN)G can be different. The same comment applies to (ZN)P and (ZN)G.

Figure 14–12 Hardness-ratio factor CH (through-hardened steel). (ANSI/AGMA 2001-D04.)

When

Use CH = 1

HBP HBG

< 1.2,

0 2 4 6 8 10 12 14 16 18 20

1.2

1.3

1.4

1.5

1.6

1.7

1.00

1.02

1.04

1.06

1.08

1.10

1.12

1.14

Single reduction gear ratio mG

H ar

dn es

s- ra

tio fa

ct or

, C H

H B

P H

B G

C al

cu la

te d

ha rd

ne ss

-r at

io ,

Figure 14–13 Hardness-ratio factor CH (surface-hardened steel pinion). (ANSI/AGMA 2001-D04.)

180 200 250 300 350 400 1.00

1.02

1.04

1.06

1.08

1.10

1.12

1.14

1.16

Brinell hardness of the gear, HBG

H ar

dn es

s- ra

tio fa

ct or

, C H

Surface Finish of Pinion, fP, microinches, Ra

fP = 16

fP = 32

fP = 64

When fP > 64 use CH = 1.0

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Spur and Helical Gears 755

14–14 Reliability Factor KR (YZ) The reliability factor accounts for the effect of the statistical distributions of material fatigue failures. Load variation is not addressed here. The gear strengths St and Sc are based on a reliability of 99 percent. Table 14–10 is based on data developed by the U.S. Navy for bending and contact-stress fatigue failures. The functional relationship between KR and reliability is highly nonlinear. When interpolation is required, linear interpolation is too crude. A log transformation to each quantity produces a linear string. A least-squares regression fit is

KR 5 e0.658 2 0.0759 ln(1 2 R) 0.5 , R , 0.99 0.50 2 0.109 ln(1 2 R) 0.99 # R # 0.9999

(14–38)

For cardinal values of R, take KR from the table. Otherwise use the logarithmic inter- polation afforded by Eqs. (14–38).

Figure 14–14 Repeatedly applied bending strength stress-cycle factor YN. (ANSI/AGMA 2001-D04.)

NOTE: The choice of YN in the shaded area is influenced by:

Pitchline velocity Gear material cleanliness Residual stress Material ductility and fracture toughness

YN = 9.4518 N −0.148

YN = 6.1514 N −0.1192

YN = 4.9404 N −0.1045

YN = 3.517 N −0.0817

YN = 1.3558 N −0.0178

YN = 1.6831 N −0.0323

YN = 2.3194 N −0.0538

102 103 104 105 106 107 108 109 1010 0.5

0.6

0.7 0.8 0.9 1.0

2.0

3.0

4.0

5.0

0.5

0.6 0.7 0.8 0.9 1.0

160 HB

Nitrided 250 HB

Case carb.

400 HB

Number of load cycles, N

St re

ss -c

yc le

fa ct

or , Y

N Figure 14–15 Pitting resistance stress-cycle factor ZN. (ANSI/AGMA 2001-D04.)

102 103 104 105 106 107 108 109 1010 0.5

0.6 0.7 0.8 0.9 1.0 1.1

2.0

3.0

4.0

5.0 NOTE: The choice of ZN in the shaded zone is influenced by:

Lubrication regime Failure criteria Smoothness of operation required Pitchline velocity Gear material cleanliness Material ductility and fracture toughness Residual stress

ZN = 2.466 N −0.056

Nitrided ZN = 1.249 N −0.0138

ZN = 1.4488 N −0.023

Number of load cycles, N

St re

ss -c

yc le

fa ct

or , Z

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756 Mechanical Engineering Design

14–15 Temperature Factor KT (YU) For oil or gear-blank temperatures up to 250°F (120°C), use KT 5 Yu 5 1.0. For higher temperatures, the factor should be greater than unity. Heat exchangers may be used to ensure that operating temperatures are considerably below this value, as is desirable for the lubricant.

14–16 Rim-Thickness Factor KB

When the rim thickness is not sufficient to provide full support for the tooth root, the location of bending fatigue failure may be through the gear rim rather than at the tooth fillet. In such cases, the use of a stress-modifying factor KB is recommended. This factor, the rim-thickness factor KB, adjusts the estimated bending stress for the thin-rimmed gear. It is a function of the backup ratio mB,

mB 5 tR

ht (14–39)

where tR 5 rim thickness below the tooth, and ht 5 the tooth height. The geometry is depicted in Fig. 14–16. The rim-thickness factor KB is given by

KB 5 µ 1.6 ln

2.242 mB mB , 1.2

1 mB $ 1.2 (14–40)

Reliability KR (YZ)

0.9999 1.50

0.999 1.25

0.99 1.00

0.90 0.85

0.50 0.70

Table 14–10

Reliability Factors KR (YZ) Source: ANSI/AGMA 2001-D04.

Figure 14–16 Rim-thickness factor KB. (ANSI/AGMA 2001-D04.)

mB = tR ht

For mB < 1.2 KB = 1.6 ln 2.242

mB( (

For mB ≥ 1.2 KB = 1.0

0.5 0.6 0.8 1.0 1.2 2 3 4 5 6 7 8 9 10 0

1.0

1.2

1.4

1.6

1.8

2.0

2.2

2.4

Backup ratio, mB

R im

-t hi

ck ne

ss fa

ct or

, K B

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Spur and Helical Gears 757

Figure 14–16 also gives the value of KB graphically. The rim-thickness factor KB is applied in addition to the 0.70 reverse-loading factor when applicable.

14–17 Safety Factors SF and SH

The ANSI/AGMA standards 2001-D04 and 2101-D04 contain a safety factor SF guarding against bending fatigue failure and safety factor SH guarding against pitting failure. The definition of SF, from Eq. (14–17), for U.S. customary units, is

SF 5 StYNy(KTKR)

s 5

fully corrected bending strength bending stress

(14–41)

where s is estimated from Eq. (14–15), for U.S. customary units. It is a strength-over- stress definition in a case where the stress is linear with the transmitted load. The definition of SH, from Eq. (14–18), is

SH 5 ScZNCHy(KTKR)

sc 5

fully corrected contact strength contact stress

(14–42)

when sc is estimated from Eq. (14–16). This, too, is a strength-over-stress definition but in a case where the stress is not linear with the transmitted load Wt. While the definition of SH does not interfere with its intended function, a caution is required when comparing SF with SH in an analysis in order to ascertain the nature and severity of the threat to loss of function. To render SH linear with the transmitted load, Wt it could have been defined as

SH 5 afully corrected contact strength contact stress imposed

(14–43)

with the exponent 2 for linear or helical contact, or an exponent of 3 for crowned teeth (spherical contact). With the definition, Eq. (14–42), compare SF with S2

H (or S3

H for crowned teeth) when trying to identify the threat to loss of function with confidence. The role of the overload factor Ko is to include predictable excursions of load beyond Wt based on experience. A safety factor is intended to account for unquantifi- able elements in addition to Ko. When designing a gear mesh, the quantity SF becomes a design factor (SF)d within the meanings used in this book. The quantity SF evaluated as part of a design assessment is a factor of safety. This applies equally well to the quantity SH.

14–18 Analysis Description of the procedure based on the AGMA standard is highly detailed. The best review is a “road map” for bending fatigue and contact-stress fatigue. Figure 14–17 identifies the bending stress equation, the endurance strength in bending equation, and the factor of safety SF. Figure 14–18 displays the contact-stress equation, the contact fatigue endurance strength equation, and the factor of safety SH. The equations in these figures are in terms of U.S. customary units. Similar roadmaps can readily be gener- ated in terms of SI units. The following example of a gear mesh analysis is intended to make all the details presented concerning the AGMA method more familiar.

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758 Mechanical Engineering Design

SPUR GEAR BENDING Based on ANSI!AGMA 2001-D04 (U.S. customary units)

dP = NP Pd

V = πdn 12

W t = 33 000 Η V

Gear bending stress equation Eq. (14–15)

Gear bending endurance strength equation Eq. (14–17)

Bending factor of safety Eq. (14–41)

SF =

Pd F

KmKB J

1 [or Eq. (a), Sec. 14–10]; p. 751

Eq. (14–30); p. 751

Eq. (14–40); p. 756

Eq. (14–27); p. 748

Table below

St SF

YN KT KR

St YN /(KT KR) "

= W tKoKvKs"

all ="

0.99(St)107 Tables 14–3, 14–4; pp. 740, 741

Fig. 14–14; p. 755

Table 14–10, Eq. (14–38); pp. 756, 755

1 if T < 250°F

Remember to compare SF with S2 H when deciding whether bending

or wear is the threat to function. For crowned gears compare SF with S 3 H .

Fig. 14–6; p. 745

Table of Overload Factors, Ko

Driven Machine

Power source

Uniform Light shock Medium shock

Uniform

1.00 1.25 1.50

Moderate shock

1.25 1.50 1.75

Heavy shock

1.75 2.00 2.25

Figure 14–17 Roadmap of gear bending equations based on AGMA standards. (ANSI/AGMA 2001-D04.)

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Spur and Helical Gears 759

SPUR GEAR WEAR Based on ANSI!AGMA 2001-D04 (U.S. customary units)

dP = NP Pd

V = πdn 12

W t = 33 000 Η V

Gear contact stress equation Eq. (14–16)

Gear contact endurance strength Eq. (14–18)

Wear factor of safety Eq. (14–42)

!c = Cp W tKoKvKs

!c,all =

SH =

dP F Cf

I( )

Eq. (14–13), Table 14–8; pp. 736, 749

1 [or Eq. (a), Sec. 14–10]; p. 751 Eq. (14–30); p. 751

1/2

Eq. (14–27); p. 748

Eq. (14–23); p. 747

Table below

Sc ZN CH SH KT KR

Sc ZN CH /(KT KR) !c

Fig. 14–15; p. 755

Gear only

Section 14–12, gear only; pp. 753, 754

Table 14–10, Eq. (14–38); pp. 756, 755 1 if T < 250°F

Remember to compare SF with S2 H when deciding whether bending

or wear is the threat to function. For crowned gears compare SF with S 3 H .

Table of Overload Factors, Ko

Driven Machine

Power source

Uniform Light shock Medium shock

Uniform

1.00 1.25 1.50

Moderate shock

1.25 1.50 1.75

Heavy shock

1.75 2.00 2.25

0.99(Sc )107 Tables 14–6, 14–7; pp. 743, 744

Figure 14–18 Roadmap of gear wear equations based on AGMA standards. (ANSI/AGMA 2001-D04.)

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760 Mechanical Engineering Design

EXAMPLE 14–4 A 17-tooth 20° pressure angle spur pinion rotates at 1800 rev/min and transmits 4 hp to a 52-tooth disk gear. The diametral pitch is 10 teeth/in, the face width 1.5 in, and the quality standard is No. 6. The gears are straddle-mounted with bearings immediately adjacent. The pinion is a grade 1 steel with a hardness of 240 Brinell tooth surface and through-hardened core. The gear is steel, through-hardened also, grade 1 material, with a Brinell hardness of 200, tooth surface and core. Poisson’s ratio is 0.30, JP 5 0.30, JG 5 0.40, and Young’s modulus is 30(106) psi. The loading is smooth because of motor and load. Assume a pinion life of 108 cycles and a reliability of 0.90, and use YN 5 1.3558N20.0178, ZN 5 1.4488N20.023. The tooth profile is uncrowned. This is a commercial enclosed gear unit. (a) Find the factor of safety of the gears in bending. (b) Find the factor of safety of the gears in wear. (c) By examining the factors of safety, identify the threat to each gear and to the mesh.

Solution There will be many terms to obtain so use Figs. 14–17 and 14–18 as guides to what is needed.

dP 5 NPyPd 5 17y10 5 1.7 in dG 5 52y10 5 5.2 in

V 5 pdPnP

12 5 p(1.7)1800

12 5 801.1 ft/min

Wt 5 33 000 H

V 5

33 000(4) 801.1

5 164.8 lbf

Assuming uniform loading, Ko 5 1. To evaluate Kv, from Eq. (14–28) with a quality number Qv 5 6,

B 5 0.25(12 2 6)2y3 5 0.8255

A 5 50 1 56(1 2 0.8255) 5 59.77

Then from Eq. (14–27) the dynamic factor is

Kv 5 a59.77 1 1801.1 59.77

b0.8255

5 1.377

To determine the size factor, Ks, the Lewis form factor is needed. From Table 14–2, with NP 5 17 teeth, YP 5 0.303. Interpolation for the gear with NG 5 52 teeth yields YG 5 0.412. Thus from Eq. (a) of Sec. 14–10, with F 5 1.5 in,

(Ks)P 5 1.192 a1.510.303 10

b0.0535

5 1.043

(Ks)G 5 1.192 a1.510.412 10

b0.0535

5 1.052

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Spur and Helical Gears 761

The load distribution factor Km is determined from Eq. (14–30), where five terms are needed. They are, where F 5 1.5 in when needed:

Uncrowned, Eq. (14–30): Cmc 5 1, Eq. (14–32): Cpf 5 1.5y[10(1.7)] 2 0.0375 1 0.0125(1.5) 5 0.0695 Bearings immediately adjacent, Eq. (14–33): Cpm 5 1 Commercial enclosed gear units (Fig. 14–11): Cma 5 0.15 Eq. (14–35): Ce 5 1

Thus,

Km 5 1 1 Cmc(Cpf Cpm 1 CmaCe) 5 1 1 (1)[0.0695(1) 1 0.15(1)] 5 1.22

Assuming constant thickness gears, the rim-thickness factor KB 5 1. The speed ratio is mG 5 NGyNP 5 52y17 5 3.059. The load cycle factors given in the problem state- ment, with N(pinion) 5 108 cycles and N(gear) 5 108ymG 5 108y3.059 cycles, are

(YN)P 5 1.3558(108)20.0178 5 0.977

(YN)G 5 1.3558(108y3.059)20.0178 5 0.996

From Table 14.10, with a reliability of 0.9, KR 5 0.85. From Fig. 14–18, the tem- perature and surface condition factors are KT 5 1 and Cf 5 1. From Eq. (14–23), with mN 5 1 for spur gears,

I 5 cos 20° sin 20°

3.059 3.059 1 1

5 0.121

From Table 14–8, Cp 5 23001psi. Next, we need the terms for the gear endurance strength equations. From Table 14–3, for grade 1 steel with HBP 5 240 and HBG 5 200, we use Fig. 14–2, which gives

(St)P 5 77.3(240) 1 12 800 5 31 350 psi

(St)G 5 77.3(200) 1 12 800 5 28 260 psi

Similarly, from Table 14–6, we use Fig. 14–5, which gives

(Sc)P 5 322(240) 1 29 100 5 106 400 psi

(Sc)G 5 322(200) 1 29 100 5 93 500 psi

From Fig. 14–15,

(ZN)P 5 1.4488(108)20.023 5 0.948

(ZN)G 5 1.4488(108y3.059)20.023 5 0.973

For the hardness ratio factor CH, the hardness ratio is HBPyHBG 5 240y200 5 1.2. Then, from Sec. 14–12,

A¿ 5 8.98(1023) (HBPyHBG) 2 8.29(1023)

5 8.98(1023) (1.2) 2 8.29(1023) 5 0.002 49

Thus, from Eq. (14–36),

CH 5 1 1 0.002 49(3.059 2 1) 5 1.005

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762 Mechanical Engineering Design

(a) Pinion tooth bending. Substituting the appropriate terms for the pinion into Eq. (14–15) gives

(s)P 5 aWtKo Kv

Ks Pd

F Km

J b

P 5 164.8(1)1.377(1.043)

10 1.5

1.22(1)

0.30

5 6417 psi

Substituting the appropriate terms for the pinion into Eq. (14–41) gives

Answer (SF)P 5 aSt YNy(KT

KR) s

b P

5 31 350(0.977)y[1(0.85)]

6417 5 5.62

Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15) gives

(s)G 5 164.8(1)1.377(1.052) 10 1.5

1.22(1)

0.40 5 4854 psi

Substituting the appropriate terms for the gear into Eq. (14–41) gives

Answer (SF)G 5 28 260(0.996)y[1(0.85)]

4854 5 6.82

(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14–16) gives

(sc)P 5 CP aW t Ko Kv

dP F Cf

I b1y2

5 2300 c164.8(1)1.377(1.043)

1.22 1.7(1.5)

0.121 d 1y2

5 70 360 psi

Substituting the appropriate terms for the pinion into Eq. (14–42) gives

Answer (SH)P 5 c Sc ZNy(KT

KR) sc

d P

5 106 400(0.948)y[1(0.85)]

70 360 5 1.69

Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is Ks. Thus,

(sc)G 5 c (Ks)G

(Ks)P d 1y2

(sc)P 5 a1.052 1.043

b1y2

70 360 5 70 660 psi

Substituting the appropriate terms for the gear into Eq. (14–42) with CH 5 1.005 gives

Answer (SH)G 5 93 500(0.973)1.005y[1(0.85)]

70 660 5 1.52

threat in the pinion is from wear. For the gear, we compare (SF)G with (SH)2 G, or 6.96

with 1.522 5 2.31, so the threat in the gear is also from wear.

There are perspectives to be gained from Ex. 14–4. First, the pinion is overly strong in bending compared to wear. The performance in wear can be improved by surface-hardening techniques, such as flame or induction hardening, nitriding, or car- burizing and case hardening, as well as shot peening. This in turn permits the gearset

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Spur and Helical Gears 763

to be made smaller. Second, in bending, the gear is stronger than the pinion, indicating that both the gear core hardness and tooth size could be reduced; that is, we may increase P and reduce the diameters of the gears, or perhaps allow a cheaper material. Third, in wear, surface strength equations have the ratio (ZN)yKR. The values of (ZN)P and (ZN)G are affected by gear ratio mG. The designer can control strength by specify- ing surface hardness. This point will be elaborated later. Having followed a spur-gear analysis in detail in Ex. 14–4, it is timely to analyze a helical gearset under similar circumstances to observe similarities and differences.

EXAMPLE 14–5 A 17-tooth 20° normal pitch-angle helical pinion with a right-hand helix angle of 30° rotates at 1800 rev/min when transmitting 4 hp to a 52-tooth helical gear. The normal diametral pitch is 10 teeth/in, the face width is 1.5 in, and the set has a quality num- ber of 6. The gears are straddle-mounted with bearings immediately adjacent. The pinion and gear are made from a through-hardened steel with surface and core hard- nesses of 240 Brinell on the pinion and surface and core hardnesses of 200 Brinell on the gear. The transmission is smooth, connecting an electric motor and a centrifu- gal pump. Assume a pinion life of 108 cycles and a reliability of 0.9 and use the upper curves in Figs. 14–14 and 14–15. (a) Find the factors of safety of the gears in bending. (b) Find the factors of safety of the gears in wear. (c) By examining the factors of safety identify the threat to each gear and to the mesh.

Solution All of the parameters in this example are the same as in Ex. 14–4 with the exception that we are using helical gears. Thus, several terms will be the same as Ex. 14–4. The reader should verify that the following terms remain unchanged: Ko 5 1, YP 5 0.303, YG 5 0.412, mG 5 3.059, (Ks)P 5 1.043, (Ks)G 5 1.052, (YN)P 5 0.977, (YN)G 5 0.996, KR 5 0.85, KT 5 1, Cf 5 1, Cp 5 23001psi, (St)P 5 31 350 psi, (St)G 5 28 260 psi, (Sc)P 5 106 380 psi, (Sc)G 5 93 500 psi, (ZN)P 5 0.948, (ZN)G 5 0.973, and CH 5 1.005. For helical gears, the transverse diametral pitch, given by Eq. (13–18), p. 684, is

Pt 5 Pn cos c 5 10 cos 30° 5 8.660 teeth/in

Thus, the pitch diameters are dP 5 NPyPt 5 17y8.660 5 1.963 in and dG 5 52y8.660 5 6.005 in. The pitch-line velocity and transmitted force are

V 5 pdPnP

12 5 p(1.963)1800

12 5 925 ft /min

W t 5 33 000H

V 5

33 000(4) 925

5 142.7 lbf

As in Ex. 14–4, for the dynamic factor, B 5 0.8255 and A 5 59.77. Thus, Eq. (14–27) gives

Kv 5 a59.77 1 1925 59.77

b0.8255

5 1.404

The geometry factor I for helical gears requires a little work. First, the transverse pressure angle is given by Eq. (13–19) p. 684,

ft 5 tan21 atan fn

cos c b 5 tan21 a tan 20°

cos 30° b 5 22.80°

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764 Mechanical Engineering Design

The radii of the pinion and gear are rP 5 1.963y2 5 0.9815 in and rG 5 6.004y2 5 3.002 in, respectively. The addendum is a 5 1yPn 5 1y10 5 0.1, and the base-circle radii of the pinion and gear are given by Eq. (13–6), p. 672, with f 5 ft:

(rb)P 5 rP cos ft 5 0.9815 cos 22.80° 5 0.9048 in

(rb)G 5 3.002 cos 22.80° 5 2.767 in

From Eq. (14–25), the surface strength geometry factor

Z 5 2(0.9815 1 0.1)2 2 0.90482 1 2(3.004 1 0.1)2 2 2.7692

2(0.9815 1 3.004) sin 22.80°

5 0.5924 1 1.4027 2 1.544 4 5 0.4507 in

Since the first two terms are less than 1.544 4, the equation for Z stands. From Eq. (14–24) the normal circular pitch pN is

pN 5 pn cos fn 5 p

Pn cos 20° 5

10 cos 20° 5 0.2952 in

From Eq. (14–21), the load sharing ratio

mN 5 pN

0.95Z 5

0.2952 0.95(0.4507)

5 0.6895

Substituting in Eq. (14–23), the geometry factor I is

I 5 sin 22.80° cos 22.80°

2(0.6895)

3.06 3.06 1 1

5 0.195

From Fig. 14–7, geometry factors J9P 5 0.45 and J9G 5 0.54. Also from Fig. 14–8 the J-factor multipliers are 0.94 and 0.98, correcting J9P and J9G to

JP 5 0.45(0.94) 5 0.423

JG 5 0.54(0.98) 5 0.529

The load-distribution factor Km is estimated from Eq. (14–32):

Cp f 5 1.5

10(1.963) 2 0.0375 1 0.0125(1.5) 5 0.0577

with Cmc 5 1, Cpm 5 1, Cma 5 0.15 from Fig. 14–11, and Ce 5 1. Therefore, from Eq. (14–30),

Km 5 1 1 (1)[0.0577(1) 1 0.15(1)] 5 1.208

(a) Pinion tooth bending. Substituting the appropriate terms into Eq. (14–15) using Pt gives

(s)P 5 aW tKoKvKs

F KmKB

J b

P 5 142.7(1)1.404(1.043)

8.66 1.5

1.208(1)

0.423

5 3445 psi

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Spur and Helical Gears 765

Substituting the appropriate terms for the pinion into Eq. (14–41) gives

Answer (SF)P 5 aStYNy(KTKR) s

b P

5 31 350(0.977)y[1(0.85)]

3445 5 10.5

Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15) gives

(s)G 5 142.7(1)1.404(1.052)

8.66 1.5

1.208(1)

0.529 5 2779 psi

Substituting the appropriate terms for the gear into Eq. (14–41) gives

Answer (SF)G 5 28 260(0.996)y[1(0.85)]

2779 5 11.9

(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14–16) gives

(sc)P 5 Cp aWtKoKvKs

dPF Cf

I b1y2

5 2300 c142.7(1)1.404(1.043)

1.208 1.963(1.5)

0.195 d 1y2

5 48 230 psi

Substituting the appropriate terms for the pinion into Eq. (14–42) gives

Answer (SH)P 5 aScZNy(KTKR) sc

b P

5 106 400(0.948)y[1(0.85)]

48 230 5 2.46

Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is Ks. Thus,

(sc)G 5 c (Ks)G

(Ks)P d 1y2

(sc)P 5 a1.052 1.043

b1y2

48 230 5 48 440 psi

Substituting the appropriate terms for the gear into Eq. (14–42) with CH 5 1.005 gives

Answer (SH)G 5 93 500(0.973)1.005y[1(0.85)]

48 440 5 2.22

the pinion is from wear. For the gear we compare SF with S2 H, or 11.9 with 2.222 5 4.93,

so the threat is also from wear in the gear. For the meshing gearset wear controls.

It is worthwhile to compare Ex. 14–4 with Ex. 14–5. The spur and helical gear- sets were placed in nearly identical circumstances. The helical gear teeth are of greater length because of the helix and identical face widths. The pitch diameters of the helical gears are larger. The J factors and the I factor are larger, thereby reducing stresses. The result is larger factors of safety. In the design phase the gearsets in Ex. 14–4 and Ex. 14–5 can be made smaller with control of materials and relative hardnesses.

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766 Mechanical Engineering Design

Now that examples have given the AGMA parameters substance, it is time to examine some desirable (and necessary) relationships between material properties of spur gears in mesh. In bending, the AGMA equations are displayed side by side:

sP 5 aW tKoKvKs

F KmKB

J b

P sG 5 aW tKoKvKs

F KmKB

J b

(SF)P 5 aStYNy(KTKR) s

b P (SF)G 5 aStYNy(KTKR)

s b

Equating the factors of safety, substituting for stress and strength, canceling identical terms (Ks virtually equal or exactly equal), and solving for (St)G gives

(St)G 5 (St)P

(YN)P

(YN)G JP

JG (a)

The stress-cycle factor YN comes from Fig. 14–14, where for a particular hardness, YN 5 aNb. For the pinion, (YN)P 5 aNbP, and for the gear, (YN)G 5 a(NPymG)b. Substituting these into Eq. (a) and simplifying gives

(St)G 5 (St)PmbG

JG (14–44)

Normally, mG . 1 and JG . JP, so Eq. (14–44) shows that the gear can be less strong (lower Brinell hardness) than the pinion for the same safety factor.

EXAMPLE 14–6 In a set of spur gears, a 300-Brinell 18-tooth 16-pitch 20° full-depth pinion meshes with a 64-tooth gear. Both gear and pinion are of grade 1 through-hardened steel. Using b 5 20.023, what hardness can the gear have for the same factor of safety?

Solution For through-hardened grade 1 steel the pinion strength (St)P is given in Fig. 14–2:

(St)P 5 77.3(300) 1 12 800 5 35 990 psi

From Fig. 14–6 the form factors are JP 5 0.32 and JG 5 0.41. Equation (14–44) gives

(St)G 5 35 990 a64 18 b20.023

0.32 0.41

5 27 280 psi

Use the equation in Fig. 14–2 again.

Answer (HB)G 5 27 280 2 12 800

77.3 5 187 Brinell

The AGMA contact-stress equations also are displayed side by side:

(sc)P 5 Cp aWt KoKvKs

dPF Cf

I b1y2

P (sc)G 5 Cp aW t KoKvKs

dPF Cf

I b1y2

(SH)P 5 aScZNy(KTKR) sc

b P (SH)G 5 aScZNCHy(KTKR)

sc b

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Spur and Helical Gears 767

Equating the factors of safety, substituting the stress relations, and canceling identical terms including Ks gives, after solving for (Sc)G,

(Sc)G 5 (Sc)P

(ZN)P

(ZN)G a 1

CH b

G 5 (SC)PmbG a 1

CH b

where, as in the development of Eq. (14–44), (ZN)Py(ZN)G 5 mbG and the value of b for wear comes from Fig. 14–15. Since CH is so close to unity, it is usually neglected; therefore

(Sc)G 5 (Sc)PmbG (14–45)

EXAMPLE 14–7 For b 5 20.056 for a through-hardened steel, grade 1, continue Ex. 14–6 for wear.

Solution From Fig. 14–5,

(Sc)P 5 322(300) 1 29 100 5 125 700 psi

From Eq. (14–45),

(Sc)G 5 (Sc)P a64 18 b20.056

5 125 700 a64 18 b20.056

5 117 100 psi

Answer (HB)G 5 117 100 2 29 200

322 5 273 Brinell

which is slightly less than the pinion hardness of 300 Brinell.

Equations (14–44) and (14–45) apply as well to helical gears.

14–19 Design of a Gear Mesh A useful decision set for spur and helical gears includes

• Function: load, speed, reliability, life, Ko

• Unquantifiable risk: design factor nd

• Tooth system: f, c, addendum, dedendum, root fillet radius

• Gear ratio mG, Np, NG

• Quality number Qv

• Diametral pitch Pd

• Face width F

• Pinion material, core hardness, case hardness

• Gear material, core hardness, case hardness

The first item to notice is the dimensionality of the decision set. There are four design decision categories, eight different decisions if you count them separately. This is a larger number than we have encountered before. It is important to use a design strategy that is convenient in either longhand execution or computer implementation. The design

t a priori decisions

t design decisions

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768 Mechanical Engineering Design

decisions have been placed in order of importance (impact on the amount of work to be redone in iterations). The steps, after the a priori decisions have been made are

• Choose a diametral pitch.

• Examine implications on face width, pitch diameters, and material properties. If not satisfactory, return to pitch decision for change.

• Choose a pinion material and examine core and case hardness requirements. If not satisfactory, return to pitch decision and iterate until no decisions are changed.

• Choose a gear material and examine core and case hardness requirements. If not satisfactory, return to pitch decision and iterate until no decisions are changed.

With these plan steps in mind, we can consider them in more detail. First select a trial diametral pitch.

Pinion bending:

• Select a median face width for this pitch, 4pyP

• Find the range of necessary ultimate strengths

• Choose a material and a core hardness

• Find face width to meet factor of safety in bending

• Choose face width

• Check factor of safety in bending

Gear bending:

• Find necessary companion core hardness

• Choose a material and core hardness

• Check factor of safety in bending

Pinion wear:

• Find necessary Sc and attendant case hardness

• Choose a case hardness

• Check factor of safety in wear

Gear wear:

• Find companion case hardness

• Choose a case hardness

• Check factor of safety in wear

Completing this set of steps will yield a satisfactory design. Additional designs with diametral pitches adjacent to the first satisfactory design will produce several among which to choose. A figure of merit is necessary in order to choose the best. Unfortunately, a figure of merit in gear design is complex in an academic environment because material and processing costs vary. The possibility of using a process depends on the manufacturing facility if gears are made in house. After examining Ex. 14–4 and Ex. 14–5 and seeing the wide range of factors of safety, one might entertain the notion of setting all factors of safety equal.9 In

9In designing gears it makes sense to define the factor of safety in wear as (S)2 H for uncrowned teeth, so that

there is no mix-up. ANSI, in the preface to ANSI/AGMA 2001-D04 and 2101-D04, states “the use is com- pletely voluntary . . . does not preclude anyone from using . . . procedures . . . not conforming to the standards.”

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Spur and Helical Gears 769

steel gears, wear is usually controlling and (SH)P and (SH)G can be brought close to equality. The use of softer cores can bring down (SF)P and (SF)G, but there is value in keeping them higher. A tooth broken by bending fatigue not only can destroy the gear set, but can bend shafts, damage bearings, and produce inertial stresses up- and downstream in the power train, causing damage elsewhere if the gear box locks.

EXAMPLE 14–8 Design a 4:1 spur-gear reduction for a 100-hp, three-phase squirrel-cage induction motor running at 1120 rev/min. The load is smooth, providing a reliability of 0.95 at 109 revolutions of the pinion. Gearing space is meager. Use Nitralloy 135M, grade 1 material to keep the gear size small. The gears are heat-treated first then nitrided.

Solution Make the a priori decisions:

• Function: 100 hp, 1120 rev/min, R 5 0.95, N 5 109 cycles, Ko 5 1

• Design factor for unquantifiable exingencies: nd 5 2

• Tooth system: fn 5 20°

• Tooth count: NP 5 18 teeth, NG 5 72 teeth (no interference, Sec. 13–7, p. 677)

• Quality number: Qv 5 6, use grade 1 material

• Assume mB $ 1.2 in Eq. (14–40), KB 5 1

Pitch: Select a trial diametral pitch of Pd 5 4 teeth/in. Thus, dP 5 18y4 5 4.5 in and dG 5 72y4 5 18 in. From Table 14–2, YP 5 0.309, YG 5 0.4324 (interpolated). From Fig. 14–6, JP 5 0.32, JG 5 0.415.

V 5 pdP

12 5 p(4.5)1120

12 5 1319 ft/min

Wt 5 33 000H

V 5

33 000(100) 1319

5 2502 lbf

From Eqs. (14–28) and (14–27),

B 5 0.25(12 2 Qv)2y3 5 0.25(12 2 6)2y3 5 0.8255

A 5 50 1 56(1 2 0.8255) 5 59.77

Kv 5 a59.77 1 11319 59.77

b0.8255

5 1.480

From Eq. (14–38), KR 5 0.658 2 0.0759 ln (1 2 0.95) 5 0.885. From Fig. 14–14,

(YN)P 5 1.3558(109)20.0178 5 0.938

(YN)G 5 1.3558(109y4)20.0178 5 0.961

From Fig. 14–15,

(ZN)P 5 1.4488(109)20.023 5 0.900

(ZN)G 5 1.4488(109y4)20.023 5 0.929

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770 Mechanical Engineering Design

From the recommendation after Eq. (14–8), 3p # F # 5p. Try F 5 4p 5 4pyP 5 4py4 5 3.14 in. From Eq. (a), Sec. 14–10,

Ks 5 1.192 aF1Y P b0.0535

5 1.192 a3.1410.309 4

b0.0535

5 1.140

From Eqs. (14–31), (14–33) and (14–35), Cmc 5 Cpm 5 Ce 5 1. From Fig. 14–11, Cma 5 0.175 for commercial enclosed gear units. From Eq. (14–32), Fy(10dP) 53.14y [10(4.5)] 5 0.0698. Thus,

Cpf 5 0.0698 2 0.0375 1 0.0125(3.14) 5 0.0715

From Eq. (14–30),

Km 5 1 1 (1)[0.0715(1) 1 0.175(1)] 5 1.247

From Table 14–8, for steel gears, Cp 5 23001psi. From Eq. (14–23), with mG 5 4 and mN 5 1,

I 5 cos 20° sin 20°

4 4 1 1

5 0.1286

Pinion tooth bending. With the above estimates of Ks and Km from the trial diametral pitch, we check to see if the mesh width F is controlled by bending or wear considera- tions. Equating Eqs. (14–15) and (14–17), substituting ndW t for W t, and solving for the face width (F)bend necessary to resist bending fatigue, we obtain

(F)bend 5 ndWtKo Kv

Ks Pd

KmKB

JP KTKR

StYN (1)

Equating Eqs. (14–16) and (14–18), substituting ndW t for W t, and solving for the face width (F)wear necessary to resist wear fatigue, we obtain

(F)wear 5 aCpKT KR

ScZN b2

nd W t Ko

Kv Ks

KmCf

dP I (2)

From Table 14–5 the hardness range of Nitralloy 135M is Rockwell C32–36 (302–335 Brinell). Choosing a midrange hardness as attainable, using 320 Brinell. From Fig. 14–4,

St 5 86.2(320) 1 12 730 5 40 310 psi

Inserting the numerical value of St in Eq. (1) to estimate the face width gives

(F)bend 5 2(2502)(1)1.48(1.14)4 1.247(1)(1)0.885 0.32(40 310)0.938

5 3.08 in

From Table 14–6 for Nitralloy 135M, Sc 5 170 000 psi. Inserting this in Eq. (2), we find

(F)wear 5 a2300(1)(0.885) 170 000(0.900)

2(2502)1(1.48)1.14

1.247(1) 4.5(0.1286)

5 3.22 in

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Spur and Helical Gears 771

Decision Make face width 3.50 in. Correct Ks and Km:

Ks 5 1.192 a3.5010.309 4

b0.0535

5 1.147

10dP 5

3.50 10(4.5)

5 0.0778

Cp f 5 0.0778 2 0.0375 1 0.0125(3.50) 5 0.0841

Km 5 1 1 (1)[0.0841(1) 1 0.175(1)] 5 1.259

The bending stress induced by W t in bending, from Eq. (14–15), is

(s)P 5 2502(1)1.48(1.147)

4 3.50

1.259(1)

0.32 5 19 100 psi

The AGMA factor of safety in bending of the pinion, from Eq. (14–41), is

(SF)P 5 40 310(0.938)y[1(0.885)]

19 100 5 2.24

Decision Gear tooth bending. Use cast gear blank because of the 18-in pitch diameter. Use the same material, heat treatment, and nitriding. The load-induced bending stress is in the ratio of JPyJG. Then

(s)G 5 19 100

0.32 0.415

5 14 730 psi

The factor of safety of the gear in bending is

(SF)G 5 40 310(0.961)y[1(0.885)]

14 730 5 2.97

Pinion tooth wear. The contact stress, given by Eq. (14–16), is

(sc)P 5 2300 c2502(1)1.48(1.147) 1.259

4.5(3.5)

1 0.129

d 1y2

5 118 000 psi

The factor of safety from Eq. (14–42), is

(SH)P 5 170 000(0.900)y[1(0.885)]

118 000 5 1.465

By our definition of factor of safety, pinion bending is (SF)P 5 2.24, and wear is (SH)2

P 5 (1.465)2 5 2.15.

Gear tooth wear. The hardness of the gear and pinion are the same. Thus, from Fig. 14–12, CH 5 1, the contact stress on the gear is the same as the pinion, (sc)G 5 118 000 psi. The wear strength is also the same, Sc 5 170 000 psi. The factor of safety of the gear in wear is

(SH)G 5 170 000(0.929)y[1(0.885)]

118 000 5 1.51

So, for the gear in bending, (SF)G 5 2.97, and wear (SH)2 G 5 (1.51)2 5 2.29.

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772 Mechanical Engineering Design

Rim. Keep mB $ 1.2. The whole depth is ht 5 addendum 1 dedendum 5 1yPd 1 1.25yPd 5 2.25yPd 5 2.25y4 5 0.5625 in. The rim thickness tR is

tR $ mB ht 5 1.2(0.5625) 5 0.675 in

In the design of the gear blank, be sure the rim thickness exceeds 0.675 in; if it does not, review and modify this mesh design.

This design example showed a satisfactory design for a four-pitch spur-gear mesh. Material could be changed, as could pitch. There are a number of other satisfactory designs, thus a figure of merit is needed to identify the best. One can appreciate that gear design was one of the early applications of the digital computer to mechanical engineering. A design program should be interactive, presenting results of calculations, pausing for a decision by the designer, and showing the consequences of the decision, with a loop back to change a decision for the better. The program can be structured in totem-pole fashion, with the most influential decision at the top, then tumbling down, decision after decision, ending with the abil- ity to change the current decision or to begin again. Such a program would make a fine class project. Troubleshooting the coding will reinforce your knowledge, adding flexibility as well as bells and whistles in subsequent terms. Standard gears may not be the most economical design that meets the functional requirements, because no application is standard in all respects.10 Methods of design- ing custom gears are well understood and frequently used in mobile equipment to provide good weight-to-performance index. The required calculations including opti- mizations are within the capability of a personal computer.

10See H. W. Van Gerpen, C. K. Reece, and J. K. Jensen, Computer Aided Design of Custom Gears, Van Gerpen–Reece Engineering, Cedar Falls, Iowa, 1996.

PROBLEMS Problems marked with an asterisk (*) are linked to problems in other chapters, as summarized in Table 1–2 of Sec. 1–17, p. 34. Because the results will vary depending on the method used, the problems are presented by section.

Section 14–1

14–1 A steel spur pinion has a pitch of 6 teeth/in, 22 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of 1200 rev/min and transmits 15 hp to a 60-tooth gear. If the face width is 2 in, estimate the bending stress.

14–2 A steel spur pinion has a diametral pitch of 10 teeth/in, 18 teeth cut full-depth with a 20° pressure angle, and a face width of 1 in. This pinion is expected to transmit 2 hp at a speed of 600 rev/min. Determine the bending stress.

14–3 A steel spur pinion has a module of 1.25 mm, 18 teeth cut on the 20° full-depth system, and a face width of 12 mm. At a speed of 1800 rev/min, this pinion is expected to carry a steady load of 0.5 kW. Determine the bending stress.

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Spur and Helical Gears 773

14–4 A steel spur pinion has 16 teeth cut on the 20° full-depth system with a module of 8 mm and a face width of 90 mm. The pinion rotates at 150 rev/min and transmits 6 kW to the mating steel gear. What is the bending stress?

14–5 A steel spur pinion has a module of 1 mm and 16 teeth cut on the 20° full-depth system and is to carry 0.15 kW at 400 rev/min. Determine a suitable face width based on an allowable bending stress of 150 MPa.

14–6 A 20° full-depth steel spur pinion has 20 teeth and a module of 2 mm and is to transmit 0.5 kW at a speed of 200 rev/min. Find an appropriate face width if the bending stress is not to exceed 75 MPa.

14–7 A 20° full-depth steel spur pinion has a diametral pitch of 5 teeth/in and 24 teeth and transmits 6 hp at a speed of 50 rev/min. Find an appropriate face width if the allowable bending stress is 20 kpsi.

14–8 A steel spur pinion is to transmit 20 hp at a speed of 400 rev/min. The pinion is cut on the 20° full-depth system and has a diametral pitch of 4 teeth/in and 16 teeth. Find a suitable face width based on an allowable stress of 12 kpsi.

14–9 A 20° full-depth steel spur pinion with 18 teeth is to transmit 2.5 hp at a speed of 600 rev/min. Determine appropriate values for the face width and diametral pitch based on an allowable bending stress of 10 kpsi.

14–10 A 20° full-depth steel spur pinion is to transmit 1.5 kW hp at a speed of 900 rev/min. If the pinion has 18 teeth, determine suitable values for the module and face width. The bending stress should not exceed 75 MPa.

Section 14–2

14–11 A speed reducer has 20° full-depth teeth and consists of a 20-tooth steel spur pinion driving a 50-tooth cast-iron gear. The horsepower transmitted is 12 at a pinion speed of 1200 rev/min. For a diametral pitch of 8 teeth/in and a face width of 1.5 in, find the contact stress.

14–12 A gear drive consists of a 16-tooth 20° steel spur pinion and a 48-tooth cast-iron gear having a pitch of 12 teeth/in. For a power input of 1.5 hp at a pinion speed of 700 rev/min, select a face width based on an allowable contact stress of 100 kpsi.

14–13 A gearset has a module of 5 mm, a 20° pressure angle, and a 24-tooth cast-iron spur pinion driving a 48-tooth cast-iron gear. The pinion is to rotate at 50 rev/min. What horsepower input can be used with this gearset if the contact stress is limited to 690 MPa and F 5 60 mm?

14–14 A 20° 20-tooth cast-iron spur pinion having a module of 4 mm drives a 32-tooth cast-iron gear. Find the contact stress if the pinion speed is 1000 rev/min, the face width is 50 mm, and 10 kW of power is transmitted.

14–15 A steel spur pinion and gear have a diametral pitch of 12 teeth/in, milled teeth, 17 and 30 teeth, respectively, a 20° pressure angle, a face width of 7

8 in, and a pinion speed of 525 rev/min. The tooth properties are Sut 5 76 kpsi, Sy 5 42 kpsi and the Brinell hardness is 149. Use the Gerber criteria to compensate for one-way bending. For a design factor of 2.25, what is the power rating of the gearset?

14–16 A milled-teeth steel pinion and gear pair have Sut 5 113 kpsi, Sy 5 86 kpsi and a hardness at the involute surface of 262 Brinell. The diametral pitch is 3 teeth/in, the face width is 2.5 in, and the pinion speed is 870 rev/min. The tooth counts are 20 and 100. Use the Gerber criteria to compensate for one-way bending. For a design factor of 1.5, rate the gearset for power considering both bending and wear.

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774 Mechanical Engineering Design

14–17 A 20° full-depth steel spur pinion rotates at 1145 rev/min. It has a module of 6 mm, a face width of 75 mm, and 16 milled teeth. The ultimate tensile strength at the involute is 900 MPa exhibiting a Brinell hardness of 260. The gear is steel with 30 teeth and has identical material strengths. Use the Gerber criteria to compensate for one-way bending. For a design factor of 1.3 find the power rating of the gearset based on the pinion and the gear resisting bending and wear fatigue.

14–18 A steel spur pinion has a pitch of 6 teeth/in, 17 full-depth milled teeth, and a pressure angle of 20°. The pinion has an ultimate tensile strength at the involute surface of 116 kpsi, a Brinell hardness of 232, and a yield strength of 90 kpsi. Its shaft speed is 1120 rev/min, its face width is 2 in, and its mating gear has 51 teeth. Use a design factor of 2. (a) Pinion bending fatigue imposes what power limitation? Use the Gerber criteria to compen-

sate for one-way bending. (b) Pinion surface fatigue imposes what power limitation? The gear has identical strengths to

the pinion with regard to material properties. (c) Determine power limitations due to gear bending and wear. (d ) Specify the power rating for the gearset.

Section 14–3 to 14–19

14–19 A commercial enclosed gear drive consists of a 20° spur pinion having 16 teeth driving a 48-tooth gear. The pinion speed is 300 rev/min, the face width 2 in, and the diametral pitch 6 teeth/in. The gears are grade 1 steel, through-hardened at 200 Brinell, made to No. 6 quality standards, uncrowned, and are to be accurately and rigidly mounted. Assume a pinion life of 108 cycles and a reliability of 0.90. Determine the AGMA bending and contact stresses and the corresponding factors of safety if 5 hp is to be transmitted.

14–20 A 20° spur pinion with 20 teeth and a module of 2.5 mm transmits 120 W to a 36-tooth gear. The pinion speed is 100 rev/min, and the gears are grade 1, 18-mm face width, through- hardened steel at 200 Brinell, uncrowned, manufactured to a No. 6 quality standard, and considered to be of open gearing quality installation. Find the AGMA bending and contact stresses and the corresponding factors of safety for a pinion life of 108 cycles and a reli- ability of 0.95.

14–21 Repeat Prob. 14–19 using helical gears each with a 20° normal pitch angle and a helix angle of 30° and a normal diametral pitch of 6 teeth/in.

14–22 A spur gearset has 17 teeth on the pinion and 51 teeth on the gear. The pressure angle is 20° and the overload factor Ko 5 1. The diametral pitch is 6 teeth/in and the face width is 2 in. The pinion speed is 1120 rev/min and its cycle life is to be 108 revolutions at a reliability R 5 0.99. The quality number is 5. The material is a through-hardened steel, grade 1, with Brinell hardnesses of 232 core and case of both gears. For a design factor of 2, rate the gearset for these conditions using the AGMA method.

14–23 In Sec. 14–10, Eq. (a) is given for Ks based on the procedure in Ex. 14–2. Derive this equation.

14–24 A speed-reducer has 20° full-depth teeth, and the single-reduction spur-gear gearset has 22 and 60 teeth. The diametral pitch is 4 teeth/in and the face width is 31

4 in. The pinion shaft speed is 1145 rev/min. The life goal of 5-year 24-hour-per-day service is about 3(109) pinion revolu- tions. The absolute value of the pitch variation is such that the quality number is 6. The mate- rials are 4340 through-hardened grade 1 steels, heat-treated to 250 Brinell, core and case, both gears. The load is moderate shock and the power is smooth. For a reliability of 0.99, rate the speed reducer for power.

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Spur and Helical Gears 775

14–25 The speed reducer of Prob. 14–24 is to be used for an application requiring 40 hp at 1145 rev/min. For the gear and the pinion, estimate the AGMA factors of safety for bending and wear, that is, (SF)P, (SF)G, (SH)P, and (SH)G. By examining the factors of safety, identify the threat to each gear and to the mesh.

14–26 The gearset of Prob. 14–24 needs improvement of wear capacity. Toward this end the gears are nitrided so that the grade 1 materials have hardnesses as follows: The pinion core is 250 and the pinion case hardness is 390 Brinell, and the gear core hardness is 250 core and 390 case. Estimate the power rating for the new gearset.

14–27 The gearset of Prob. 14–24 has had its gear specification changed to 9310 for carburizing and surface hardening with the result that the pinion Brinell hardnesses are 285 core and 580–600 case, and the gear hardnesses are 285 core and 580–600 case. Estimate the power rating for the new gearset.

14–28 The gearset of Prob. 14–27 is going to be upgraded in material to a quality of grade 2 (9310) steel. Estimate the power rating for the new gearset.

14–29 Matters of scale always improve insight and perspective. Reduce the physical size of the gearset in Prob. 14–24 by one-half and note the result on the estimates of transmitted load W

t and power.

14–30 AGMA procedures with cast-iron gear pairs differ from those with steels because life pre- dictions are difficult; consequently (YN)P, (YN)G, (ZN)P, and (ZN)G are set to unity. The consequence of this is that the fatigue strengths of the pinion and gear materials are the same. The reliability is 0.99 and the life is 107 revolution of the pinion (KR 5 1). For lon- ger lives the reducer is derated in power. For the pinion and gear set of Prob. 14–24, use grade 40 cast iron for both gears (HB 5 201 Brinell). Rate the reducer for power with SF and SH equal to unity.

14–31 Spur-gear teeth have rolling and slipping contact (often about 8 percent slip). Spur gears tested to wear failure are reported at 108 cycles as Buckingham’s surface fatigue load-stress factor K. This factor is related to Hertzian contact strength SC by

SC 5 A 1.4K (1yE1 1 1yE2) sin f

where f is the normal pressure angle. Cast iron grade 20 gears with f 5 14 1 2 ° and 20° pressure

angle exhibit a minimum K of 81 and 112 psi, respectively. How does this compare with SC 5 0.32HB kpsi?

14–32 You’ve probably noticed that although the AGMA method is based on two equations, the details of assembling all the factors is computationally intensive. To reduce error and omissions, a computer program would be useful. Write a program to perform a power rating of an existing gearset, then use Prob. 14–24, 14–26, 14–27, 14–28, and 14–29 to test your program by com- paring the results to your longhand solutions.

14–33 In Ex. 14–5 use nitrided grade 1 steel (4140) which produces Brinell hardnesses of 250 core and 500 at the surface (case). Use the upper fatigue curves on Figs. 14–14 and 14–15. Estimate the power capacity of the mesh with factors of safety of SF 5 SH 5 1.

14–34 In Ex. 14–5 use carburized and case-hardened gears of grade 1. Carburizing and case-hardening can produce a 550 Brinell case. The core hardnesses are 200 Brinell. Estimate the power capacity of the mesh with factors of safety of SF 5 SH 5 1, using the lower fatigue curves in Figs. 14–14 and 14–15.

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776 Mechanical Engineering Design

14–35 In Ex. 14–5, use carburized and case-hardened gears of grade 2 steel. The core hardnesses are 200, and surface hardnesses are 600 Brinell. Use the lower fatigue curves of Figs. 14–14 and 14–15. Estimate the power capacity of the mesh using SF 5 SH 5 1. Compare the power capacity with the results of Prob. 14–34.

14–36* The countershaft in Prob. 3–72, p. 152, is part of a speed reducing compound gear train using 20° spur gears. A gear on the input shaft drives gear A. Gear B drives a gear on the output shaft. The input shaft runs at 2400 rev/min. Each gear reduces the speed (and thus increases the torque) by a 2 to 1 ratio. All gears are to be of the same material. Since gear B is the smallest gear, transmitting the largest load, it will likely be critical, so a preliminary analysis is to be performed on it. Use a diametral pitch of 2 teeth/in, a face-width of 4 times the circu- lar pitch, a Grade 2 steel through-hardened to a Brinell hardness of 300, and a desired life of 15 kh with a 95 percent reliability. Determine factors of safety for bending and wear.

14–37* The countershaft in Prob. 3–73, p. 152, is part of a speed reducing compound gear train using 20° spur gears. A gear on the input shaft drives gear A with a 2 to 1 speed reduction. Gear B drives a gear on the output shaft with a 5 to 1 speed reduction. The input shaft runs at 1800 rev/min. All gears are to be of the same material. Since gear B is the smallest gear, transmitting the largest load, it will likely be critical, so a preliminary analysis is to be per- formed on it. Use a module of 18.75 mm/tooth, a face-width of 4 times the circular pitch, a Grade 2 steel through-hardened to a Brinell hardness of 300, and a desired life of 12 kh with a 98 percent reliability. Determine factors of safety for bending and wear.

14–38* Build on the results of Prob. 13–40, p. 720, to find factors of safety for bending and wear for gear F. Both gears are made from Grade 2 carburized and hardened steel. Use a face-width of 4 times the circular pitch. The desired life is 12 kh with a 95 percent reliability.

14–39* Build on the results of Prob. 13–41, p. 721, to find factors of safety for bending and wear for gear C. Both gears are made from Grade 2 carburized and hardened steel. Use a face-width of 4 times the circular pitch. The desired life is 14 kh with a 98 percent reliability.

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Chapter Outline

15–1 Bevel Gearing—General 778

15–2 Bevel-Gear Stresses and Strengths 780

15–3 AGMA Equation Factors 783

15–4 Straight-Bevel Gear Analysis 795

15–5 Design of a Straight-Bevel Gear Mesh 798

15–6 Worm Gearing—AGMA Equation 801

15–7 Worm-Gear Analysis 805

15–8 Designing a Worm-Gear Mesh 809

15–9 Buckingham Wear Load 812

Bevel and Worm Gears15

777

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778 Mechanical Engineering Design

The American Gear Manufacturers Association (AGMA) has established standards for the analysis and design of the various kinds of bevel and worm gears. Chapter 14 was an introduction to the AGMA methods for spur and helical gears and contains many of the definitions of terms used in this chapter. AGMA has established similar methods for other types of gearing, which all follow the same general approach.

15–1 Bevel Gearing—General Bevel gears may be classified as follows:

• Straight bevel gears

• Spiral bevel gears

• Zerol bevel gears

• Hypoid gears

• Spiroid gears

A straight bevel gear was illustrated in Fig. 13–35, p. 701. These gears are usu- ally used for pitch-line velocities up to 1000 ft/min (5 m/s) when the noise level is not an important consideration. They are available in many stock sizes and are less expensive to produce than other bevel gears, especially in small quantities. A spiral bevel gear is shown in Fig. 15–1; the definition of the spiral angle is illustrated in Fig. 15–2. These gears are recommended for higher speeds and where the noise level is an important consideration. Spiral bevel gears are the bevel coun- terpart of the helical gear; it can be seen in Fig. 15–1 that the pitch surfaces and the nature of contact are the same as for straight bevel gears except for the differences brought about by the spiral-shaped teeth. The Zerol bevel gear is a patented gear having curved teeth but with a zero spiral angle. The axial thrust loads permissible for Zerol bevel gears are not as large as those for the spiral bevel gear, and so they are often used instead of straight bevel gears. The Zerol bevel gear is generated by the same tool used for regular spiral bevel gears. For design purposes, use the same procedure as for straight bevel gears and then simply substitute a Zerol bevel gear.

Figure 15–1 Spiral bevel gears. (Courtesy of Gleason Works, Rochester, N.Y.)

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Bevel and Worm Gears 779

It is frequently desirable, as in the case of automotive differential applications, to have gearing similar to bevel gears but with the shafts offset. Such gears are called hypoid gears, because their pitch surfaces are hyperboloids of revolution. The tooth action between such gears is a combination of rolling and sliding along a straight line and has much in common with that of worm gears. Figure 15–3 shows a pair of hypoid gears in mesh. Figure 15–4 is included to assist in the classification of spiral bevel gearing. It is seen that the hypoid gear has a relatively small shaft offset. For larger offsets, the pinion begins to resemble a tapered worm and the set is then called spiroid gearing.

Figure 15–2 Cutting spiral-gear teeth on the basic crown rack.

Basic crown rack

Cutter radius

Spiral angle

Mean radius of crown rack

Circular pitch

Face advance

Figure 15–3 Hypoid gears. (Courtesy of Gleason Works, Rochester, N.Y.)

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780 Mechanical Engineering Design

15–2 Bevel-Gear Stresses and Strengths In a typical bevel-gear mounting, Fig. 13–36, p. 702, for example, one of the gears is often mounted outboard of the bearings. This means that the shaft deflections can be more pronounced and can have a greater effect on the nature of the tooth contact. Another difficulty that occurs in predicting the stress in bevel-gear teeth is the fact that the teeth are tapered. Thus, to achieve perfect line contact passing through the cone cen- ter, the teeth should bend more at the large end than at the small end. To obtain this condition requires that the load be proportionately greater at the large end. Because of this varying load across the face of the tooth, it is desirable to have a fairly short face width. Because of the complexity of bevel, spiral bevel, Zerol bevel, hypoid, and spiroid gears, as well as the limitations of space, only a portion of the applicable standards that refer to straight-bevel gears is presented here.1 Table 15–1 gives the symbols used in ANSI/AGMA 2003-B97.

Fundamental Contact Stress Equation

sc 5 sc 5 Cp a Wt

F dP I Ko

Kv Km

Cs Cxcb1y2

(U.S. customary units)

sH 5 ZE a1000Wt

bd Z1 KA Kv KHb Zx Zxcb1y2

(SI units)

(15–1)

The first term in each equation is the AGMA symbol, whereas sc, our normal notation, is directly equivalent.

Figure 15–4 Comparison of intersecting- and offset-shaft bevel-type gearings. (From Gear Handbook by Darle W. Dudley, 1962, pp. 2–24.)

Worm

Spiroid

Hypoid

Spiral bevel

Ring gear

1Figures 15–5 to 15–13 and Tables 15–1 to 15–7 have been extracted from ANSI/AGMA 2003-B97, Rating the Pitting Resistance and Bending Strength of Generated Straight Bevel, Zerol Bevel and Spiral Bevel Gear Teeth with the permission of the publisher, the American Gear Manufacturers Association, 1001 N. Fairfax Street, Suite 500, Alexandria, VA, 22314-1587.

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Bevel and Worm Gears 781

AGMA ISO Symbol Symbol Description Units

Am Rm Mean cone distance in (mm) A0 Re Outer cone distance in (mm) CH ZW Hardness ratio factor for pitting resistance Ci Zi Inertia factor for pitting resistance CL ZNT Stress cycle factor for pitting resistance Cp ZE Elastic coefficient [lbf/in2]0.5

([N/mm2]0.5) CR ZZ Reliability factor for pitting CSF Service factor for pitting resistance CS Zx Size factor for pitting resistance Cxc Zxc Crowning factor for pitting resistance D, d de2, de1 Outer pitch diameters of gear and pinion, respectively in (mm) EG, EP E2, E1 Young’s modulus of elasticity for materials of gear and pinion, respectively lbf/in2

(N/mm2) e e Base of natural (Napierian) logarithms F b Net face width in (mm)

FeG, FeP b92, b91 Effective face widths of gear and pinion, respectively in (mm) fP Ra1 Pinion surface roughness min (mm) HBG HB2 Minimum Brinell hardness number for gear material HB HBP HB1 Minimum Brinell hardness number for pinion material HB hc Eht min Minimum total case depth at tooth middepth in (mm) he h9c Minimum effective case depth in (mm) he lim h9c lim Suggested maximum effective case depth limit at tooth middepth in (mm) I ZI Geometry factor for pitting resistance J YJ Geometry factor for bending strength JG, JP YJ2, YJ1 Geometry factor for bending strength for gear and pinion, respectively KF YF Stress correction and concentration factor Ki Yi Inertia factor for bending strength KL YNT Stress cycle factor for bending strength Km KHb Load distribution factor Ko KA Overload factor KR Yz Reliability factor for bending strength KS YX Size factor for bending strength KSF Service factor for bending strength KT Ku Temperature factor Kv Kv Dynamic factor Kx Yb Lengthwise curvature factor for bending strength met Outer transverse module (mm) mmt Mean transverse module (mm) mmn Mean normal module (mm) mNI eNI Load sharing ratio, pitting mNJ eNJ Load sharing ratio, bending N z2 Number of gear teeth NL nL Number of load cycles n z1 Number of pinion teeth nP n1 Pinion speed rev/min

Table 15–1

Symbols Used in Bevel Gear Rating Equations, ANSI/AGMA 2003-B97 Standard Source: ANSI/AGMA 2003-B97.

(Continued )

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782 Mechanical Engineering Design

P P Design power through gear pair hp (kW) Pa Pa Allowable transmitted power hp (kW) Pac Paz Allowable transmitted power for pitting resistance hp (kW) Pacu Pazu Allowable transmitted power for pitting resistance at unity service factor hp (kW) Pat Pay Allowable transmitted power for bending strength hp (kW) Patu Payu Allowable transmitted power for bending strength at unity service factor hp (kW) Pd Outer transverse diametral pitch teeth/in Pm Mean transverse diametral pitch teeth/in Pmn Mean normal diametral pitch teeth/in Qv Qv Transmission accuracy number q q Exponent used in formula for lengthwise curvature factor R, r rmpt2, rmpt1 Mean transverse pitch radii for gear and pinion, respectively in (mm) Rt, rt rmyo2, rmyo1 Mean transverse radii to point of load application for gear in (mm) and pinion, respectively rc rc 0 Cutter radius used for producing Zerol bevel and spiral bevel gears in (mm) s gc Length of the instantaneous line of contact between mating tooth surfaces in (mm) sac sH lim Allowable contact stress number lbf/in2

(N/mm2) sat sF lim Bending stress number (allowable) lbf/in2 (N/mm2) sc sH Calculated contact stress number lbf/in2 (N/mm2) sF sF Bending safety factor sH sH Contact safety factor st sF Calculated bending stress number lbf/in2 (N/mm2) swc sHP Permissible contact stress number lbf/in2 (N/mm2) swt sFP Permissible bending stress number lbf/in2 (N/mm2) TP T1 Operating pinion torque lbf in (Nm) TT uT Operating gear blank temperature °F(°C) t0 sai Normal tooth top land thickness at narrowest point in (mm) Uc Uc Core hardness coefficient for nitrided gear lbf/in2 (N/mm2) UH UH Hardening process factor for steel lbf/in2 (N/mm2) vt vet Pitch-line velocity at outer pitch circle ft/min (m/s) YKG, YKP YK2, YK1 Tooth form factors including stress-concentration factor for gear and pinion, respectively mG, mp n2, n1 Poisson’s ratio for materials of gear and pinion, respectively r0 ryo Relative radius of profile curvature at point of maximum contact stress in (mm) between mating tooth surfaces f an Normal pressure angle at pitch surface ft awt Transverse pressure angle at pitch point c bm Mean spiral angle at pitch surface cb bmb Mean base spiral angle

Table 15–1

Symbols Used in Bevel Gear Rating Equations, ANSI/AGMA 2003-B97 Standard (Continued )

AGMA ISO Symbol Symbol Description Units

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Bevel and Worm Gears 783

Permissible Contact Stress Number (Strength) Equation

swc 5 (sc)all 5 sacCLCH

SHKTCR (U.S. customary units)

sHP 5 sH lim ZNT ZW

SHKuZZ (SI units)

(15–2)

Bending Stress

st 5 Wt

F Pd KoKv

Ks Km

Kx J (U.S. customary units)

sF 5 1000Wt

b KAKv

met Yx KHb

YbYJ (SI units)

(15–3)

Permissible Bending Stress Equation

swt 5 satKL

SF KT KR (U.S. customary units)

sFP 5 sF lim YNT

SF Ku Yz (SI units)

(15–4)

15–3 AGMA Equation Factors Overload Factor Ko (KA) The overload factor makes allowance for any externally applied loads in excess of the nominal transmitted load. Table 15–2, from Appendix A of 2003-B97, is included for your guidance.

Safety Factors SH and SF

The factors of safety SH and SF as defined in 2003-B97 are adjustments to strength, not load, and consequently cannot be used as is to assess (by comparison) whether the threat is from wear fatigue or bending fatigue. Since W t is the same for the pinion and gear, the comparison of 1SH to SF allows direct comparison.

Dynamic Factor Kv

In 2003-C87 AGMA changed the definition of Kv to its reciprocal but used the same symbol. Other standards have yet to follow this move. The dynamic factor Kv makes

Character of Character of Load on Driven Machine Prime Mover Uniform Light Shock Medium Shock Heavy Shock

Uniform 1.00 1.25 1.50 1.75 or higher

Light shock 1.10 1.35 1.60 1.85 or higher

Medium shock 1.25 1.50 1.75 2.00 or higher

Heavy shock 1.50 1.75 2.00 2.25 or higher

Note: This table is for speed-decreasing drives. For speed-increasing drives, add 0.01(N/n)2 or 0.01(z2/z1) 2

to the above factors.

Table 15–2

Overload Factors Ko (KA) Source: ANSI/AGMA 2003-B97.

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784 Mechanical Engineering Design

allowance for the effect of gear-tooth quality related to speed and load, and the increase in stress that follows. AGMA uses a transmission accuracy number Qv to describe the precision with which tooth profiles are spaced along the pitch circle. Figure 15–5 shows graphically how pitch-line velocity and transmission accuracy number are related to the dynamic factor Kv. Curve fits are

Kv 5 aA 1 1vt

A bB

(U.S. customary units)

Kv 5 aA 1 1200vet

A bB

(SI units)

(15–5)

where

A 5 50 1 56(1 2 B)

B 5 0.25(12 2 Qv)2y3 (15–6)

and vt (vet) is the pitch-line velocity at outside pitch diameter, expressed in ft/min (m/s):

vt 5 p dP

nPy12 (U.S. customary units)

vet 5 5.236(1025)d1 n1 (SI units)

(15–7)

The maximum recommended pitch-line velocity is associated with the abscissa of the terminal points of the curve in Fig. 15–5:

vt max 5 [A 1 (Qv 2 3)]2 (U.S. customary units)

vet max 5 [A 1 (Qv 2 3)]2

200 (SI units)

(15–8)

where vt max and vet max are in ft/min and m/s, respectively.

Figure 15–5 Dynamic factor Kv. (Source: ANSI/AGMA 2003-B97.)

D yn

am ic

fa ct

or , K

Pitch-line velocity, vt (ft /min)

Pitch-line velocity, vet (m/s)

0 2000 4000 6000 8000 10 000

0 10 20 30 40 50

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0 Qv = 5

Qv = 7

Qv = 6

Qv = 8

Qv = 9

Qv = 10

Qv = 11

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Bevel and Worm Gears 785

Size Factor for Pitting Resistance Cs (Zx)

Cs 5 •0.5 F , 0.5 in 0.125F 1 0.4375 0.5 # F # 4.5 in 1 F . 4.5 in

(U.S. customary units)

Zx 5 •0.5 b , 12.7 mm 0.004 92b 1 0.4375 12.7 # b # 114.3 mm 1 b . 114.3 mm

(SI units)

(15–9)

Size Factor for Bending Ks (Yx)

KS 5 e0.4867 1 0.2132yPd

0.5

0.5 # Pd # 16 teeth/in Pd . 16 teeth/in

(U.S. customary units)

Yx 5 e0.5 0.4867 1 0.008 339met

met , 1.6 mm 1.6 # met # 50 mm

(SI units)

(15–10)

Load-Distribution Factor Km (KHB)

Km 5 Kmb 1 0.0036 F2 (U.S. customary units)

KHb 5 Kmb 1 5.6(1026)b2 (SI units) (15–11)

where

Kmb 5 •1.00 both members straddle-mounted 1.10 one member straddle-mounted 1.25 neither member straddle-mounted

Crowning Factor for Pitting Cxc (Zxc) The teeth of most bevel gears are crowned in the lengthwise direction during manu- facture to accommodate the deflection of the mountings.

Cxc 5 Zxc 5 e1.5 properly crowned teeth 2.0 or larger uncrowned teeth

(15–12)

Lengthwise Curvature Factor for Bending Strength Kx (YB) For straight-bevel gears,

Kx 5 Yb 5 1 (15–13)

Pitting Resistance Geometry Factor I (ZI) Figure 15–6 shows the geometry factor I (ZI) for straight-bevel gears with a 20° pres- sure angle and 90° shaft angle. Enter the figure ordinate with the number of pinion teeth, move to the number of gear-teeth contour, and read from the abscissa.

Bending Strength Geometry Factor J (YJ) Figure 15–7 shows the geometry factor J for straight-bevel gears with a 20° pressure angle and 90° shaft angle.

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786 Mechanical Engineering Design

Figure 15–6 Contact geometry factor I (ZI) for coniflex straight-bevel gears with a 20° normal pressure angle and a 90° shaft angle. (Source: ANSI/AGMA 2003-B97.)

N um

be r o

f p in

io n

te et

Geometry factor, I (ZI) 0.05 0.06 0.07 0.08 0.09

Number of gear teeth

0.10 0.11 10

50 60 70 80 90 100

Figure 15–7 Bending factor J (YJ) for coniflex straight-bevel gears with a 20° normal pressure angle and 90° shaft angle. (Source: ANSI/AGMA 2003-B97.)

N um

be r o

f t ee

th o

n ge

ar fo

r w hi

ch g

eo m

et ry

fa ct

or is

d es

ir ed

Geometry factor, J (YJ)

Number of teeth in mate

0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40 10

13 15 20 25 30 35 40 45 50 100

100

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Bevel and Worm Gears 787

Stress-Cycle Factor for Pitting Resistance CL (ZNT)

CL 5 e2 103 # NL , 104

3.4822N20.0602 L 104 # NL # 1010

ZNT 5 e2 103 # nL , 104

3.4822n20.0602 L 104 # nL # 1010

(15–14)

See Fig. 15–8 for a graphical presentation of Eqs. (15–14).

Stress-Cycle Factor for Bending Strength KL (YNT)

KL 5 µ 2.7 102 # NL , 103

6.1514N20.1192 L 103 # NL , 3(106)

1.683N20.0323 L 3(106) # NL # 1010 critical

1.3558N20.0178 L 3(106) # NL # 1010 general

(15–15)

YNT 5 µ 2.7 102 # nL , 103

6.1514n20.1192 L 103 # nL , 3(106)

1.683n20.0323 L 3(106) # nL # 1010 critical

1.3558n20.0178 L 3(106) # nL # 1010 general

See Fig. 15–9 for a plot of Eqs. (15–15).

Figure 15–8 Contact stress-cycle factor for pitting resistance CL (ZNT) for carburized case-hardened steel bevel gears. (Source: ANSI/AGMA 2003-B97.)

St re

ss -c

yc le

fa ct

or , C

L (Z

N T )

Number of load cycles, NL (nL )

104103 0.5

0.6

0.7

0.8 0.9 1.0

2.0

3.0

4.0

5.0

105 106 107 108 109 1010

CL = 3.4822 NL –0.0602

ZNT = 3.4822 nL –0.0602

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788 Mechanical Engineering Design

Hardness-Ratio Factor CH (ZW)

CH 5 1 1 B1(Nyn 2 1) B1 5 0.008 98(HBPyHBG) 2 0.008 29

ZW 5 1 1 B1(z2yz1 2 1) B1 5 0.008 98(HB1yHB2) 2 0.008 29 (15–16)

The preceding equations are valid when 1.2 # HBPyHBG # 1.7 (1.2 # HB1yHB2 # 1.7). Figure 15–10 graphically displays Eqs. (15–16). When a surface-hardened pinion (48 HRC or harder) is run with a through-hardened gear (180 # HB # 400), a work- hardening effect occurs. The CH (ZW) factor varies with pinion surface roughness fP (Ra1) and the mating-gear hardness:

CH 5 1 1 B2(450 2 HBG) B2 5 0.000 75 exp(20.0122 fP)

ZW 5 1 1 B2(450 2 HB2) B2 5 0.000 75 exp(20.52 Ra1) (15–17)

where fP (Ra1) 5 pinion surface hardness min (mm)

HBG (HB2) 5 minimum Brinell hardness of the gear

See Fig. 15–11 for carburized steel gear pairs of approximately equal hardness CH 5 ZW 5 1.

Temperature Factor KT (KU)

KT 5 e1 32°F # t # 250°F (460 1 t)y710 t . 250°F

Ku 5 e1 0°C # u # 120°C (273 1 u)y393 u . 120°C

(15–18)

St re

ss -c

yc le

fa ct

or , K

L (Y

N T )

Number of load cycles, NL (nL) 104102 103

0.5

0.6

0.7

0.8

0.9 1.0

0.5

0.6

0.7

0.8

0.9 1.0

1.5

2.0

3.0

3.5

105 106 107 108 109 1010

KL = 1.3558 NL –0.0178

YNT = 1.3558 nL –0.0178

NOTE: The choice of KL (YNT) is influenced by: Pitch-line velocity Gear material cleanliness Residual stress Material ductility and fracture toughness

KL = 1.683 NL –0.0323

YNT = 1.683 nL –0.0323

KL = 6.1514 NL –0.1192

YNT = 6.1514 nL –0.1192

Figure 15–9 Stress-cycle factor for bending strength KL (YNT) for carburized case-hardened steel bevel gears. (Source: ANSI/AGMA 2003-B97.)

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Bevel and Worm Gears 789

Reliability Factors CR (ZZ) and KR (YZ) Table 15–3 displays the reliability factors. Note that CR 5 1KR and ZZ 5 1YZ. Logarithmic interpolation equations are

YZ 5 KR 5 e0.50 2 0.25 log(1 2 R) 0.99 # R # 0.999 0.70 2 0.15 log(1 2 R) 0.90 # R , 0.99

(15–19) (15–20)

The reliability of the stress (fatigue) numbers allowable in Tables 15–4, 15–5, 15–6, and 15–7 is 0.99.

Figure 15–10 Hardness-ratio factor CH (ZW) for through-hardened pinion and gear. (Source: ANSI/AGMA 2003-B97.)

H ar

dn es

s- ra

tio fa

ct or

, C H

( Z

W )

Reduction gear ratio, N/n (z2/z1) 0 2 4 6 8 10 12 14 16 18 20

1.00

1.02

1.04

1.06

1.08

1.10

1.12

1.14

1.7

1.6

1.5

1.4

1.3

1.2

C al

cu la

te d

ha rd

ne ss

ra tio

, H

B G

H B

< 1.2

When

use CH (ZW) = 1

HBG

HBP

HB2

HB1

Figure 15–11 Hardness-ratio factor CH (ZW) for surface-hardened pinions. (Source: ANSI/AGMA 2003-B97.)

H ar

dn es

s- ra

tio fa

ct or

C H

( Z

W )

Brinell hardness of the gear HB

180 200 250 300 350 400 1.00

1.05

1.10

1.15

1.20

16 !in (0.4 !m) Surface roughness of pinion, fP (Ra1)

32 !in (0.8 !m)

63 !in (1.6 !m)

125 !in (3.2 !m)

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790 Mechanical Engineering Design

Elastic Coefficient for Pitting Resistance Cp (ZE)

Cp 5 B 1

p[(1 2 n2 P)yEP 1 (1 2 n2

G)yEG]

ZE 5 B 1

p[(1 2 n2 1)yE1 1 (1 2 n2

2)yE2]

(15–21)

Reliability Factors for Steel* Requirements of Application CR (ZZ) KR (YZ)†

Fewer than one failure in 10 000 1.22 1.50

Fewer than one failure in 1000 1.12 1.25

Fewer than one failure in 100 1.00 1.00

Fewer than one failure in 10 0.92 0.85‡

Fewer than one failure in 2 0.84 0.70§

*At the present time there are insufficient data concerning the reliability of bevel gears made from other materials. †Tooth breakage is sometimes considered a greater hazard than pitting. In such cases a greater value of KR (YZ) is selected for bending. ‡At this value plastic flow might occur rather than pitting. §From test data extrapolation.

Table 15–3

Reliability Factors Source: ANSI/AGMA 2003-B97.

Table 15–4

Allowable Contact Stress Number for Steel Gears, sac (sH lim) Source: ANSI/AGMA 2003-B97.

Minimum Allowable Contact Stress Number, Material Heat Surface* sac

(SH lim) lbf/in2 (N/mm2) Designation Treatment Hardness Grade 1 † Grade 2† Grade 3†

Steel Through-hardened‡ Fig. 15–12 Fig. 15–12 Fig. 15–12

Flame or induction 50 HRC 175 000 190 000 hardened§ (1210) (1310)

Carburized and 2003-B97 200 000 225 000 250 000 case hardened§ Table 8 (1380) (1550) (1720)

AISI 4140 Nitrided§ 84.5 HR15N 145 000 (1000)

Nitralloy 160 000 135M Nitrided§ 90.0 HR15N (1100)

*Hardness to be equivalent to that at the tooth middepth in the center of the face width. †See ANSI/AGMA 2003-B97, Tables 8 through 11, for metallurgical factors for each stress grade of steel gears. ‡These materials must be annealed or normalized as a minumum. §The allowable stress numbers indicated may be used with the case depths prescribed in 21.1, ANSI/AGMA 2003-B97.

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Bevel and Worm Gears 791

where Cp 5 elastic coefficient, 2290 1psi for steel

ZE 5 elastic coefficient, 190 2N/mm2 for steel

EP and EG 5 Young’s moduli for pinion and gear respectively, psi

E1 and E2 5 Young’s moduli for pinion and gear respectively, N/mm2

Allowable Contact Stress Tables 15–4 and 15–5 provide values of sac (sH) for steel gears and for iron gears, respectively. Figure 15–12 graphically displays allowable stress for grade 1 and 2 materials.

Material Designation

Typical Minimum Allowable Contact Heat Surface Stress Number, sac

Material ASTM ISO Treatment Hardness (SH lim) lbf/in2 (N/mm2)

Cast iron ASTM A48 ISO/DR 185 Class 30 Grade 200 As cast 175 HB 50 000 (345) Class 40 Grade 300 As cast 200 HB 65 000 (450)

Ductile ASTM A536 ISO/DIS 1083 (nodular) Grade 80-55-06 Grade 600-370-03 Quenched 180 HB 94 000 (650) iron Grade 120-90-02 Grade 800-480-02 and tempered 300 HB 135 000 (930)

Table 15–5

Allowable Contact Stress Number for Iron Gears, sac (sH lim) Source: ANSI/AGMA 2003-B97.

Minimum Bending Stress Number (Allowable), Material Heat Surface sat (SF lim) lbf/in2 (N/mm2) Designation Treatment Hardness Grade 1* Grade 2* Grade 3*

Steel Through-hardened Fig. 15–13 Fig. 15–13 Fig. 15–13

Flame or induction hardened Unhardened roots 50 HRC 15 000 (85) 13 500 (95) Hardened roots 22 500 (154)

Carburized and case 2003-B97 hardened† Table 8 30 000 (205) 35 000 (240) 40 000 (275)

AISI 4140 Nitrided†,‡ 84.5 HR15N 22 000 (150)

Nitralloy 135M Nitrided†,‡ 90.0 HR15N 24 000 (165)

*See ANSI/AGMA 2003-B97, Tables 8–11, for metallurgical factors for each stress grade of steel gears. †The allowable stress numbers indicated may be used with the case depths prescribed in 21.1, ANSI/AGMA 2003-B97. ‡The overload capacity of nitrided gears is low. Since the shape of the effective S-N curve is flat, the sensitivity to shock should be investigated before proceeding with the design.

Table 15–6

Allowable Bending Stress Numbers for Steel Gears, sat (sF lim) Source: ANSI/AGMA 2003-B97.

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792 Mechanical Engineering Design

The equations are

sac 5 341HB 1 23 620 psi grade 1 sH lim 5 2.35HB 1 162.89 MPa grade 1 sac 5 363.6HB 1 29 560 psi grade 2 sH lim 5 2.51HB 1 203.86 MPa grade 2

(15–22)

Allowable Bending Stress Numbers Tables 15–6 and 15–7 provide sat (sF lim) for steel gears and for iron gears, respec- tively. Figure 15–13 shows graphically allowable bending stress sat (sH lim) for through- hardened steels. The equations are

sat 5 44HB 1 2100 psi grade 1 sF lim 5 0.30HB 1 14.48 MPa grade 1 sat 5 48HB 1 5980 psi grade 2 sH lim 5 0.33HB 1 41.24 MPa grade 2

(15–23)

Reversed Loading AGMA recommends use of 70 percent of allowable strength in cases where tooth load is completely reversed, as in idler gears and reversing mechanisms.

Summary Figure 15–14 is a “road map” for straight-bevel gear wear relations using 2003-B97. Figure 15–15 is a similar guide for straight-bevel gear bending using 2003-B97.

Material Designation

Typical Minimum Bending Stress Number Heat Surface (Allowable), sat

Material ASTM ISO Treatment Hardness (SF lim) lbf/in2 (N/mm2)

Cast iron ASTM A48 ISO/DR 185 Class 30 Grade 200 As cast 175 HB 4500 (30) Class 40 Grade 300 As cast 200 HB 6500 (45)

Ductile ASTM A536 ISO/DIS 1083 (nodular) Grade 80-55-06 Grade 600-370-03 Quenched 180 HB 10 000 (70) iron Grade 120-90-02 Grade 800-480-02 and tempered 300 HB 13 500 (95)

Table 15–7

Allowable Bending Stress Number for Iron Gears, sat (sF lim) Source: ANSI/AGMA 2003-B97.

150 200 250 300 350 400 450 75

100

125

150

175

200

600

700

800

900

1000

1100

1200

1300

Brinell hardness HB

A llo

w ab

le c

on ta

ct s

tr es

s nu

m be

r s ac

, k ps

A llo

w ab

le c

on ta

ct s

tr es

s nu

m be

r ! H

lim , M

Maximum for grade 1 sac = 341 HB + 23 620

(!H lim = 2.35 HB + 162.89)

Maximum for grade 2 sac = 363.6 HB + 29 560

(!H lim = 2.51 HB + 203.86)

Figure 15–12 Allowable contact stress number for through-hardened steel gears, sac (sH lim). (Source: ANSI/AGMA 2003-B97.)

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Bevel and Worm Gears 793

Figure 15–13 Allowable bending stress number for through-hardened steel gears, sat (sF lim). (Source: ANSI/AGMA 2003-B97.)

150 200 250 300 350 400 450 10

100

150

200

250

300

350

Brinell hardness HB

B en

di ng

s tr

es s

nu m

be r (

al lo

w ab

le ) s

at (k

ps i)

B en

di ng

s tr

es s

nu m

be r (

al lo

w ab

le ) !

F li

m (M

Pa )

Maximum for grade 2 sat = 48 HB + 5980

(!F lim = 0.33 HB + 41.24)

Maximum for grade 1 sat = 44 HB + 2100

(!F lim = 0.30 HB + 14.48)

Figure 15–14 “Road map” summary of principal straight-bevel gear wear equations and their parameters.

STRAIGHT-BEVEL GEAR WEAR BASED ON ANSI /AGMA 2003-B97 (U.S. customary units)

Gear contact stress

Gear wear strength

Wear factor of safety

Geometry Force Analysis Strength Analysis

dP = NP Pd

dav = dP − F cos !

" = NP NG

tan−1

! = NG NP

tan−1

W t =

W r = W t tan# cos"

W a = W t tan# sin"

2T dav

W r = W t tan# cos"

W a = W t tan# sin"

W t = 2T dP

Sc = !c = Cp Ko Kv Km Cs Cxc W t

FdP I( )1/2

Swc = (!c)all = sac CL CH SH KT CR

At large end of tooth Table 15-2, p. 783

Eqs. (15-5) to (15-8), p. 784 Eq. (15-11), p. 785

Tables 15-4, 15-5, Fig. 15-12, Eq. (15-22), pp. 790–792 Fig. 15-8, Eq. (15-14), p. 787 Eqs. (15-16), (15-17), gear only, p. 788

Eq. (15-12), p. 785

Eq. (15-9), p. 785

Eqs. (15-19), (15-20), Table 15-3, pp. 789, 790 Eq. (15-18), p. 788

Fig. 15-6, p. 786 Eq. (15-21), p. 790

SH = , based on strength (!c)all

nw = , based on W t ; can be compared directly with SF

(!c)all !c( )2

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794 Mechanical Engineering Design

The standard does not mention specific steel but mentions the hardness attainable by heat treatments such as through-hardening, carburizing and case-hardening, flame- hardening, and nitriding. Through-hardening results depend on size (diametral pitch). Through-hardened materials and the corresponding Rockwell C-scale hardness at the 90 percent martensite shown in parentheses following include 1045 (50), 1060 (54), 1335 (46), 2340 (49), 3140 (49), 4047 (52), 4130 (44), 4140 (49), 4340 (49), 5145 (51), E52100 (60), 6150 (53), 8640 (50), and 9840 (49). For carburized case-hard materials the approximate core hardnesses are 1015 (22), 1025 (37), 1118 (33), 1320 (35), 2317 (30), 4320 (35), 4620 (35), 4820 (35), 6120 (35), 8620 (35), and E9310 (30). The conversion from HRC to HB (300-kg load, 10-mm ball) is

HRC 42 40 38 36 34 32 30 28 26 24 22 20 18 16 14 12 10

HB 388 375 352 331 321 301 285 269 259 248 235 223 217 207 199 192 187

Figure 15–15 “Road map” summary of principal straight-bevel gear bending equations and their parameters.

STRAIGHT-BEVEL GEAR BENDING BASED ON ANSI /AGMA 2003-B97 (U.S. customary units)

Gear bending stress

Gear bending strength

Bending factor of safety

Geometry Force Analysis Strength Analysis

dP = NP Pd

dav = dP − F cos !

! = NP NG

tan−1

! = NG NP

tan−1

W t =

W r = W t tan" cos!

W a = W t tan" sin!

2T dav

W t =

W r = W t tan" cos!

W a = W t tan" sin!

2T dP

Swt = #all = sat KL

SF KT KR

At large end of tooth

Table 15-2, p. 783 Eqs. (15-5) to (15-8), p. 784

Eq. (15-11), p. 785 Eq. (15-10), p. 785

Table 15-6 or 15-7, pp. 791, 792 Fig. 15-9, Eq. (15-15), pp. 788, 787

Fig. 15-7, p. 786

Eq. (15-13), p. 785

Eqs. (15-19), (15-20), Table 15-3, pp. 789, 790 Eq. (15-18), p. 788

SF = , based on strength #all #

nB = , based on W t , same as SF #all #

St = # = Pd Ko Kv W t

F Ks Km Kx J

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Bevel and Worm Gears 795

Most bevel-gear sets are made from carburized case-hardened steel, and the factors incorporated in 2003-B97 largely address these high-performance gears. For through- hardened gears, 2003-B97 is silent on KL and CL, and Figs. 15–8 and 15–9 should prudently be considered as approximate.

15–4 Straight-Bevel Gear Analysis

EXAMPLE 15–1 A pair of identical straight-tooth miter gears listed in a catalog has a diametral pitch of 5 at the large end, 25 teeth, a 1.10-in face width, and a 20° normal pressure angle; the gears are grade 1 steel through-hardened with a core and case hardness of 180 Brinell. The gears are uncrowned and intended for general industrial use. They have a quality number of Qv 5 7. It is likely that the application intended will require outboard mounting of the gears. Use a safety factor of 1, a 107 cycle life, and a 0.99 reliability. (a) For a speed of 600 rev/min find the power rating of this gearset based on AGMA bending strength. (b) For the same conditions as in part (a) find the power rating of this gearset based on AGMA wear strength. (c) For a reliability of 0.995, a gear life of 109 revolutions, and a safety factor of SF 5 SH 5 1.5, find the power rating for this gearset using AGMA strengths.

Solution From Figs. 15–14 and 15–15,

dP 5 NPyPd 5 25y5 5 5.000 in

vt 5 pdPnPy12 5 p(5) 600y12 5 785.4 ft /min

Overload factor: uniform-uniform loading, Table 15–2, Ko 5 1.00. Safety factor: SF 5 1, SH 5 1. Dynamic factor Kv: from Eq. (15–6),

B 5 0.25(12 2 7)2y3 5 0.731

A 5 50 1 56(1 2 0.731) 5 65.06

Kv 5 a65.06 1 2785.4 65.06

b0.731

5 1.299

From Eq. (15–8),

vt max 5 [65.06 1 (7 2 3)]2 5 4769 ft/min

vt , vt max, that is, 785.4 , 4769 ft/min, therefore Kv is valid. From Eq. (15–10),

Ks 5 0.4867 1 0.2132y5 5 0.529

From Eq. (15–11),

Kmb 5 1.25 and Km 5 1.25 1 0.0036(1.10)2 5 1.254

From Eq. (15–13), Kx 5 1. From Fig. 15–6, I 5 0.065; from Fig. 15–7, JP 5 0.216, JG 5 0.216. From Eq. (15–15),

KL 5 1.683(107)20.0323 5 0.999 96 < 1

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From Eq. (15–14),

CL 5 3.4822(107)20.0602 5 1.32

Since HBPyHBG 5 1, then from Fig. 15–10, CH 5 1. From Eqs. (15–13) and (15–18), Kx 5 1 and KT 5 1, respectively. From Eq. (15–20),

KR 5 0.70 2 0.15 log(1 2 0.99) 5 1, CR 5 1KR 5 11 5 1

(a) Bending: From Eq. (15–23),

sat 5 44(180) 1 2100 5 10 020 psi

From Eq. (15–3),

st 5 s 5 W t

F Pd KoKv

KsKm

KxJ 5

1.10 (5)(1)1.299

0.529(1.254) (1)0.216

5 18.13 W t

From Eq. (15–4),

swt 5 satKL

SFKTKR 5

10 020(1) (1)(1)(1)

5 10 020 psi

Equating st and swt,

18.13W t 5 10 020 W t 5 552.6 lbf

Answer H 5 W tvt

33 000 5

552.6(785.4) 33 000

5 13.2 hp

(b) Wear: From Fig. 15–12,

sac 5 341(180) 1 23 620 5 85 000 psi

From Eq. (15–2),

sc, all 5 sacCLCH

SHKTCR 5

85 000(1.32)(1) (1)(1)(1)

5 112 200 psi

Now Cp 5 22901psi from definitions following Eq. (15–21). From Eq. (15–9),

Cs 5 0.125(1.1) 1 0.4375 5 0.575

From Eq. (15–12), Cxc 5 2. Substituting in Eq. (15–1) gives

sc 5 Cp a W t

FdPI KoKvKmCsCxcb1y2

5 2290 c W t

1.10(5)0.065 (1)1.299(1.254)0.575(2) d 1y2

5 52422Wt

Equating sc and sc, all gives

52422W t 5 112 200, W t 5 458.1 lbf

H 5 458.1(785.4)

33 000 5 10.9 hp

796 Mechanical Engineering Design

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Bevel and Worm Gears 797

Rated power for the gearset is

Answer H 5 min(12.9, 10.9) 5 10.9 hp

KL 5 1.683(109)20.0323 5 0.8618

From Eq. (15–19),

KR 5 0.50 2 0.25 log(1 2 0.995) 5 1.075, CR 5 1KR 5 11.075 5 1.037

From Eq. (15–14),

CL 5 3.4822(109)20.0602 5 1

Bending: From Eq. (15–23) and part (a), sat 5 10 020 psi. From Eq. (15–3),

st 5 s 5 Wt

1.10 5(1)1.299

0.529(1.254) (1)0.216

5 18.13Wt

From Eq. (15–4),

swt 5 satKL

SFKTKR 5

10 020(0.8618) 1.5(1)1.075

5 5355 psi

Equating st to swt gives

18.13W t 5 5355 W t 5 295.4 lbf

H 5 295.4(785.4)

33 000 5 7.0 hp

Wear: From Eq. (15–22), and part (b), sac 5 85 000 psi. Substituting into Eq. (15–2) gives

sc,all 5 sacCLCH

SHKTCR 5

85 000(1)(1) 1.5(1)1.037

5 54 640 psi

Substituting into Eq. (15–1) gives, from part (b), sc 5 52422W t. Equating sc to sc, all gives

sc 5 sc,all 5 54 640 5 52422W t W t 5 108.6 lbf

The wear power is

H 5 108.6(785.4)

33 000 5 2.58 hp

Answer The mesh rated power is H 5 min (7.0, 2.58) 5 2.6 hp.

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798 Mechanical Engineering Design

15–5 Design of a Straight-Bevel Gear Mesh A useful decision set for straight-bevel gear design is

• Function: power, speed, mG, R

• Design factor: nd

• Tooth system

• Tooth count: NP, NG

• Pitch and face width: Pd, F

• Quality number: Qv

• Gear material, core and case hardness

• Pinion material, core and case hardness

In bevel gears the quality number is linked to the wear strength. The J factor for the gear can be smaller than for the pinion. Bending strength is not linear with face width, because added material is placed at the small end of the teeth. Consequently, face width is roughly prescribed as

F 5 min(0.3A0, 10yPd) (15–24)

where A0 is the cone distance (see Fig. 13–20, p. 682), given by

A0 5 dP

2 sin g 5

2 sin G (15–25)

t A priori decisions

t Design decisions

EXAMPLE 15–2 Design a straight-bevel gear mesh for shaft centerlines that intersect perpendicularly, to deliver 6.85 hp at 900 rev/min with a gear ratio of 3:1, temperature of 300°F, normal pressure angle of 20°, using a design factor of 2. The load is uniform-uniform. Use a pinion of 20 teeth. The material is to be AGMA grade 1 and the teeth are to be crowned. The reliability goal is 0.995 with a pinion life of 109 revolutions.

Solution First we list the a priori decisions and their immediate consequences.

Function: 6.85 hp at 900 rev/min, gear ratio mG 5 3, 300°F environment, neither gear straddle-mounted, Kmb 5 1.25 [Eq. (15–11)], R 5 0.995 at 109 revolutions of the pinion,

Eq. (15–14): (CL)G 5 3.4822(109y3)20.0602 5 1.068

(CL)P 5 3.4822(109)20.0602 5 1

Eq. (15–15): (KL)G 5 1.683(109y3)20.0323 5 0.8929

(KL)P 5 1.683(109)20.0323 5 0.8618

Eq. (15–19): KR 5 0.50 2 0.25 log(1 2 0.995) 5 1.075

CR 5 1KR 5 11.075 5 1.037

Eq. (15–18): KT 5 CT 5 (460 1 300)y710 5 1.070

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Bevel and Worm Gears 799

Design factor: nd 5 2, SF 5 2, SH 5 12 5 1.414.

Tooth system: crowned, straight-bevel gears, normal pressure angle 20°,

Eq. (15–13): Kx 5 1

Eq. (15–12): Cxc 5 1.5.

With NP 5 20 teeth, NG 5 (3)20 5 60 teeth and from Fig. 15–14,

g 5 tan21(NPyNG) 5 tan21(20y60) 5 18.43° G 5 tan21(60y20) 5 71.57°

From Figs. 15–6 and 15–7, I 5 0.0825, JP 5 0.248, and JG 5 0.202. Note that JP . JG.

Decision 1: Trial diametral pitch, Pd 5 8 teeth/in.

Eq. (15–10): Ks 5 0.4867 1 0.2132y8 5 0.5134

dP 5 NPyPd 5 20y8 5 2.5 in

dG 5 2.5(3) 5 7.5 in

vt 5 pdPnPy12 5 p(2.5)900y12 5 589.0 ft/min

W t 5 33 000 hp/vt 5 33 000(6.85)y589.0 5 383.8 lbf

Eq. (15–25): A0 5 dPy(2 sin g) 5 2.5y(2 sin 18.43°) 5 3.954 in

Eq. (15–24):

F 5 min(0.3A0, 10yPd) 5 min[0.3(3.954), 10y8] 5 min(1.186, 1.25) 5 1.186 in

Decision 2: Let F 5 1.25 in. Then,

Eq. (15–9): Cs 5 0.125(1.25) 1 0.4375 5 0.5937

Eq. (15–11): Km 5 1.25 1 0.0036(1.25)2 5 1.256

Decision 3: Let the transmission accuracy number be 6. Then, from Eq. (15–6),

B 5 0.25(12 2 6)2y3 5 0.8255

A 5 50 1 56(1 2 0.8255) 5 59.77

Eq. (15–5): Kv 5 a59.77 1 2589.0 59.77

b0.8255

5 1.325

Decision 4: Pinion and gear material and treatment. Carburize and case-harden grade ASTM 1320 to

Core 21 HRC (HB is 229 Brinell) Case 55-64 HRC (HB is 515 Brinell)

From Table 15–4, sac 5 200 000 psi and from Table 15–6, sat 5 30 000 psi.

Gear bending: From Eq. (15–3), the bending stress is

(st)G 5 Wt

F PdKoKv

KsKm

KxJG 5

383.8 1.25

8(1)1.325

0.5134(1.256) (1)0.202

5 10 390 psi

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The bending strength, from Eq. (15–4), is given by

(swt)G 5 a satKL

SFKTKR b

G 5

30 000(0.8929) 2(1.070)1.075

5 11 640 psi

The strength exceeds the stress by a factor of 11640y10390 5 1.12, giving an actual factor of safety of (SF)G 5 2(1.12) 5 2.24.

Pinion bending: The bending stress can be found from

(st)P 5 (st)G

JP 5 10 390

0.202 0.248

5 8463 psi

The bending strength, again from Eq. (15–4), is given by

(swt)P 5 a satKL

SFKTKR b

P 5

30 000(0.8618) 2(1.070)1.075

5 11 240 psi

The strength exceeds the stress by a factor of 11 240y8463 5 1.33, giving an actual factor of safety of (SF)P 5 2(1.33) 5 2.66.

Gear wear: The load-induced contact stress for the pinion and gear, from Eq. (15–1), is

sc 5 Cp a W t

FdPI KoKvKmCsCxcb1y2

5 2290 c 383.8 1.25(2.5)0.0825

(1)1.325(1.256)0.5937(1.5) d 1y2

5 107 560 psi

From Eq. (15–2) the contact strength of the gear is

(swc)G 5 asacCLCH

SHKTCR b

G 5

200 000(1.068)(1)22(1.070)1.037 5 136 120 psi

The strength exceeds the stress by a factor of 136 120y107 560 5 1.266, giving an actual factor of safety of (SH)2

G 5 1.2662(2) 5 3.21.

Pinion wear: From Eq. (15–2) the contact strength of the pinion is

(swc)P 5 a sacCLCH

SH KT CR b

P 5

200 000(1)(1)22(1.070)1.037 5 127 450 psi

The strength exceeds the stress by a factor of 127 450y107 560 5 1.185, giving an actual factor of safety of (SH)2

P 5 1.1852(2) 5 2.81. The actual factors of safety are 2.24, 2.66, 3.21, and 2.81. Making a direct com- parison of the factors, we note that the primary threat is from gear bending. We also note that the other three factors of safety are considerably higher than the target design factor. If optimization is desired, our goal would be to make changes in the design decisions that drive the factors closer to 2.

800 Mechanical Engineering Design

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Bevel and Worm Gears 801

15–6 Worm Gearing—AGMA Equation Sections 13–11 and 13–17 introduced wormgearing and its force analysis and efficiency. Here, we will present a condensed version of the AGMA recommendations for cylin- drical (single-enveloping) wormgearing2. For brevity, the equations will be shown for U.S. customary units only. Similar equations for SI units are available in the AGMA standards. Since they are essentially nonenveloping worm gears, the crossed helical gears, shown in Fig. 15–16, can be considered with other worm gearing. Crossed helical gears, and worm gears too, usually have a 90° shaft angle, though this need not be so. The relation between the shaft and helix angles is

^ 5 cP 6 cG (15–26)

where ^ is the shaft angle. The plus sign is used when both helix angles are of the same hand, and the minus sign when they are of opposite hand. The subscript P in Eq. (15–26) refers to the pinion (worm); the subscript W is used for this same purpose. The subscript G refers to the gear, also called gear wheel, worm wheel, or simply the wheel. Table 15–8 gives cylindrical worm dimensions common to worm and gear.

Figure 15–16 View of the pitch cylinders of a pair of crossed helical gears.

Pitch cylinder of B

Pitch cylinder of A

Axis of B

Axis of A

Fn 14.5° 20° 25°

Quantity Symbol NW # 2 NW # 2 NW . 2

Addendum a 0.3183px 0.3183px 0.286px

Dedendum b 0.3683px 0.3683px 0.349px

Whole depth ht 0.6866px 0.6866px 0.635px

*The table entries are for a tangential diametral pitch of the gear of Pt 5 1.

Table 15–8

Cylindrical Worm Dimensions Common to Both Worm and Gear*

2ANSI/AGMA 6034-B92, February 1992, Practice for Enclosed Cylindrical Wormgear Speed-Reducers and Gear Motors; and ANSI/AGMA 6022-C93, Dec. 1993, Design Manual for Cylindrical Wormgearing. Note: Equations (15–32) to (15–38) are contained in Annex C of 6034-B92 for informational purposes only. To comply with ANSI/AGMA 6034-B92, use the tabulations of these rating factors provided in the standard.

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802 Mechanical Engineering Design

Good proportions indicate the worm pitch diameter d falls in the range

C 0.875

3 # d #

C0.875

1.6 (15–27)

where C is the center-to-center distance. AGMA relates the allowable tangential force on the worm-gear tooth (W t)all to other parameters by

(W t )all 5 CsD0.8 m FeCmCv (15–28)

where Cs 5 materials factor

Dm 5 mean gear diameter, in

Fe 5 effective face width of the gear (actual face width, but not to exceed 0.67dm, the mean worm diameter), in

Cm 5 ratio correction factor

Cv 5 velocity factor

The friction force Wf is given by

Wf 5 f W t

cos l cos fn (15–29)

where f 5 coefficient of friction

l 5 lead angle at mean worm diameter

fn 5 normal pressure angle

The sliding velocity Vs at the mean worm diameter, in feet per minute, is

Vs 5 pnW dm

12 cos l (15–30)

where nW 5 rotative speed of the worm and dm 5 mean worm diameter. The torque at the worm gear is

TG 5 WtDm

2 (15–31)

where Dm is the mean gear diameter. The parameters in Eq. (15–28) are, quantitatively,

Cs 5 720 1 10.37C3 C # 3 in (15–32)

For sand-cast gears,

Cs 5 e1000 C . 3 Dm # 2.5 in 1190 2 477 log Dm C . 3 Dm . 2.5 in

(15–33)

For chilled-cast gears,

Cs 5 e1000 C . 3 Dm # 8 in 1412 2 456 log Dm C . 3 Dm . 8 in

(15–34)

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Bevel and Worm Gears 803

For centrifugally cast gears,

Cs 5 e1000 C . 3 Dm # 25 in 1251 2 180 log Dm C . 3 Dm . 25 in

(15–35)

The ratio correction factor Cm for gear ratio mG is given by

Cm 5 µ 0.0222m2 G 1 40mG 2 76 1 0.46 3 , mG # 20

0.010722m2 G 1 56mG 1 5145 20 , mG # 76

1.1483 2 0.006 58mG mG . 76

(15–36)

The velocity factor Cv is given by

Cv 5 •0.659 exp (20.0011Vs) Vs , 700 ft /min 13.31 V

20.571 s 700 # Vs , 3000 ft /min

65.52 V 20.774 s Vs . 3000 ft /min

(15–37)

AGMA reports the coefficient of friction f as

f 5 •0.15 Vs 5 0 0.124 exp (20.074V

0.645 s ) 0 , Vs # 10 ft /min

0.103 exp (20.110V

0.450 s ) 1 0.012 Vs . 10 ft /min

(15–38)

Now we examine some worm-gear mesh geometry. The addendum a and dedendum b are

a 5 px

p 5 0.3183px (15–39)

b 5 1.157px

p 5 0.3683px (15–40)

The full depth ht is

ht 5 µ 2.157px

p 5 0.6866px px $ 0.16 in

2.200px

p 1 0.002 5 0.7003px 1 0.002 px , 0.16 in

(15–41)

The worm outside diameter do is

do 5 d 1 2a (15–42)

The worm root diameter dr is

dr 5 d 2 2b (15–43)

The worm-gear throat diameter Dt is

Dt 5 D 1 2a (15–44)

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804 Mechanical Engineering Design

where D is the worm-gear pitch diameter. The worm-gear root diameter Dr is

Dr 5 D 2 2b (15–45)

The clearance c is

c 5 b 2 a (15–46)

The worm face width (maximum) (FW)max is

(FW) max 5 2 BaDt

2 b2

2 aD 2

2 ab2

5 222 Da (15–47)

which was simplified using Eq. (15–44). The worm-gear face width FG is

FG 5 e2 dmy3 px . 0.16 in 1.1252(do 1 2c)2 2 (do 2 4a)2 px # 0.16 in

(15–48)

The heat loss rate Hloss from the worm-gear case in ft ? lbf/min is

Hloss 5 33 000(1 2 e)Hin (15–49)

where e is efficiency, given by Eq. (13–46), p. 708, and Hin is the input horsepower from the worm. The overall coefficient h# CR for combined convective and radiative heat transfer from the worm-gear case in ft ? lbf/(min ? in2 ? °F) is

h# CR 5 µ nW

6494 1 0.13 no fan on worm shaft

3939 1 0.13 fan on worm shaft

(15–50)

The temperature of the oil sump ts is given by

ts 5 ta 1 Hloss

h# CRA 5

33 000(1 2 e)(H )in

h# CRA 1 ta (15–51)

where A is the case lateral area in in2, and ta is the ambient temperature in °F. Bypassing Eqs. (15–49), (15–50), and (15–51) one can apply the AGMA recommen- dation for minimum lateral area Amin in in2 using

Amin 5 43.20C1.7 (15–52)

Because worm teeth are inherently much stronger than worm-gear teeth, they are not considered. The teeth in worm gears are short and thick on the edges of the face; midplane they are thinner as well as curved. Buckingham3 adapted the Lewis equation for this case:

sa 5 W t

pnFey (15–53)

where pn 5 px cos l and y is the Lewis form factor related to circular pitch. For fn 5 14.5°, y 5 0.100; fn 5 20°, y 5 0.125; fn 5 25°, y 5 0.150; fn 5 30°, y 5 0.175.

3Earle Buckingham, Analytical Mechanics of Gears, McGraw-Hill, New York, 1949, p. 495.

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Bevel and Worm Gears 805

15–7 Worm-Gear Analysis Compared to other gearing systems worm-gear meshes have a much lower mechani- cal efficiency. Cooling, for the benefit of the lubricant, becomes a design constraint sometimes resulting in what appears to be an oversize gear case in light of its contents. If the heat can be dissipated by natural cooling, or simply with a fan on the wormshaft, simplicity persists. Water coils within the gear case or lubricant outpumping to an external cooler is the next level of complexity. For this reason, gear-case area is a design decision. To reduce cooling load, use multiple-thread worms. Also keep the worm pitch diameter as small as possible. Multiple-thread worms can remove the self-locking feature of many worm-gear drives. When the worm drives the gearset, the mechanical efficiency eW is given by

eW 5 cos fn 2 f tan l cos fn 1 f cot l

(15–54)

With the gear driving the gearset, the mechanical efficiency eG is given by

eG 5 cos fn 2 f cot l cos fn 1 f tan l

(15–55)

To ensure that the worm gear will drive the worm,

f stat , cos fn tan l (15–56)

where values of fstat can be found in ANSI/AGMA 6034-B92. To prevent the worm gear from driving the worm, refer to clause 9 of 6034-B92 for a discussion of self- locking in the static condition. It is important to have a way to relate the tangential component of the gear force WG

t to the tangential component of the worm force WW t , which includes the role of

friction and the angularities of fn and l. Refer to Eq. (13–45), p. 708, solved for WW t :

W t W 5 W t

G cos fn sin l 1 f cos l cos fn cos l 2 f sin l

(15–57)

In the absence of friction

W t W 5 W t

G tan l

The mechanical efficiency of most gearing is very high, which allows power in and power out to be used almost interchangeably. Worm gearsets have such poor efficiencies that we work with, and speak of, output power. The magnitude of the gear transmitted force WG

t can be related to the output horsepower H0, the application fac- tor Ka, the efficiency e, and design factor nd by

W t G 5

33 000nd H0 Ka

VGe (15–58)

We use Eq. (15–57) to obtain the corresponding worm force WW t . It follows that the

worm and gear transmitted powers in hp are

HW 5 W t

W VW

33 000 5 pdW nW W

t W

12(33 000) (15–59)

HG 5 W t

G VG

33 000 5 pdG nG W

t G

12(33 000) (15–60)

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806 Mechanical Engineering Design

From Eq. (13–44), p. 708,

Wf 5 f W t

f sin l 2 cos fn cos l (15–61)

The sliding velocity of the worm at the pitch cylinder Vs is

Vs 5 pdnW

12 cos l (15–62)

and the friction power Hf is given by

Hf 5 ZWf ZVs

33 000 hp (15–63)

Table 15–9 gives the largest lead angle lmax associated with normal pressure angle fn.

Maximum Lead Fn Angle Lmax

14.5° 16°

20° 25°

25° 35°

30° 45°

Table 15–9

Largest Lead Angle Associated with a Normal Pressure Angle fn for Worm Gearing

EXAMPLE 15–3 A single-thread steel worm rotates at 1800 rev/min, meshing with a 24-tooth worm gear transmitting 3 hp to the output shaft. The worm pitch diameter is 3 in and the tangential diametral pitch of the gear is 4 teeth/in. The normal pressure angle is 14.5°. The ambient temperature is 70°F. The application factor is 1.25 and the design factor is 1; gear face width is 2 in, lateral case area 600 in2, and the gear is chill-cast bronze. (a) Find the gear geometry. (b) Find the transmitted gear forces and the mesh efficiency. (c) Is the mesh sufficient to handle the loading? (d) Estimate the lubricant sump temperature.

Solution (a) mG 5 NGyNW 5 24y1 5 24, gear: D 5 NGyPt 5 24y4 5 6.000 in, worm: d 5 3.000 in. The axial circular pitch px is px 5 pyPt 5 py4 5 0.7854 in. C 5 (3 1 6)y2 5 4.5 in.

Eq. (15–39): a 5 pxyp 5 0.7854yp 5 0.250 in

Eq. (15–40): b 5 0.3683px 5 0.3683(0.7854) 5 0.289 in

Eq. (15–41): ht 5 0.6866px 5 0.6866(0.7854) 5 0.539 in

Eq. (15–42): do 5 3 1 2(0.250) 5 3.500 in

Eq. (15–43): dr 5 3 2 2(0.289) 5 2.422 in

Eq. (15–44): Dt 5 6 1 2(0.250) 5 6.500 in

Eq. (15–45): Dr 5 6 2 2(0.289) 5 5.422 in

Eq. (15–46): c 5 0.289 2 0.250 5 0.039 in

Eq. (15–47): (FW)max 5 212(6)0.250 5 3.464 in

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Bevel and Worm Gears 807

The tangential speeds of the worm, VW, and gear, VG, are, respectively,

VW 5 p(3)1800y12 5 1414 ft/min VG 5 p(6)1800y24

12 5 117.8 ft/min

The lead of the worm, from Eq. (13–27), p. 688, is L 5 px NW 5 0.7854(1) 5 0.7854 in. The lead angle l, from Eq. (13–28), p. 688, is

l 5 tan21 L pd

5 tan21 0.7854 p(3)

5 4.764°

The normal diametral pitch for a worm gear is the same as for a helical gear, which from Eq. (13–18), p. 684, with c 5 l is

Pn 5 Pt

cos l 5

4 cos 4.764°

5 4.014

pn 5 p

Pn 5 p

4.014 5 0.7827 in

The sliding velocity, from Eq. (15–62), is

Vs 5 pdnW

12 cos l 5 p(3)1800

12 cos 4.764° 5 1419 ft/min

(b) The coefficient of friction, from Eq. (15–38), is

f 5 0.103 exp[20.110(1419)0.450] 1 0.012 5 0.0178

The efficiency e, from Eq. (13–46), p. 708, is

Answer e 5 cos fn 2 f tan l cos fn 1 f cot l

5 cos 14.5° 2 0.0178 tan 4.764° cos 14.5° 1 0.0178 cot 4.764°

5 0.818

The designer used nd 5 1, Ka 5 1.25 and an output horsepower of H0 5 3 hp. The gear tangential force component W t

G, from Eq. (15–58), is

Answer W t G 5

33 000nd H0 Ka

VGe 5

33 000(1)3(1.25) 117.8(0.818)

5 1284 lbf

Answer The tangential force on the worm is given by Eq. (15–57):

W t W 5 W t

G cos fn sin l 1 f cos l cos fn cos l 2 f sin l

5 1284 cos 14.5° sin 4.764° 1 0.0178 cos 4.764° cos 14.5° cos 4.764° 2 0.0178 sin 4.764°

5 131 lbf

(c)

Eq. (15–34): Cs 5 1000

Eq. (15–36): Cm 5 0.010722242 1 56(24) 1 5145 5 0.823

Eq. (15–37): Cv 5 13.31(1419)20.571 5 0.211

Eq. (15–28): (W t)all 5 Cs D 0.8(Fe)G Cm Cv

5 1000(6)0.8(2)0.823(0.211) 5 1456 lbf

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808 Mechanical Engineering Design

Since W tG , (W t)all, the mesh will survive at least 25 000 h. The friction force Wf is given by Eq. (15–61):

Wf 5 f W t

f sin l 2 cos fn cos l 5

0.0178(1284) 0.0178 sin 4.764° 2 cos 14.5° cos 4.764°

5 223.7 lbf

The power dissipated in frictional work Hf is given by Eq. (15–63):

Hf 5 ZWf ZVs

33 000 5

Z223.7Z1419

33 000 5 1.02 hp

The worm and gear transmitted powers, HW and HG, are given by

HW 5 W t

W VW

33 000 5

131(1414) 33 000

5 5.61 hp HG 5 W t

G VG

33 000 5

1284(117.8) 33 000

5 4.58 hp

Answer Gear power is satisfactory. Now,

Pn 5 Ptycos l 5 4ycos 4.764° 5 4.014

pn 5 pyPn 5 py4.014 5 0.7827 in

The bending stress in a gear tooth is given by Buckingham’s adaptation of the Lewis equation, Eq. (15–53), as

(s)G 5 W t

pn FG y 5

1284 0.7827(2)(0.1)

5 8200 psi

Answer Stress in gear is satisfactory. (d )

Eq. (15–52): Amin 5 43.2C1.7 5 43.2(4.5)1.7 5 557 in2

The gear case has a lateral area of 600 in2.

Eq. (15–49): Hloss 5 33 000(1 2 e)Hin 5 33 000(1 2 0.818)5.61

5 33 690 ft ? lbf/min

Eq. (15–50): h# CR 5 nW

3939 1 0.13 5

1800 3939

1 0.13 5 0.587 ft ? lbf/(min ? in2 ? °F)

Answer Eq. (15–51): ts 5 ta 1 Hloss

h# CR A 5 70 1

33 690 0.587(600)

5 166°F

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Bevel and Worm Gears 809

15–8 Designing a Worm-Gear Mesh A usable decision set for a worm-gear mesh includes

• Function: power, speed, mG, Ka

• Design factor: nd

• Tooth system

• Materials and processes

• Number of threads on the worm: NW

• Axial pitch of worm: px

• Pitch diameter of the worm: dW

• Face width of gear: FG

• Lateral area of case: A

Reliability information for worm gearing is not well developed at this time. The use of Eq. (15–28) together with the factors Cs, Cm, and Cv, with an alloy steel case- hardened worm together with customary nonferrous worm-wheel materials, will result in lives in excess of 25 000 h. The worm-gear materials in the experience base are principally bronzes:

• Tin- and nickel-bronzes (chilled-casting produces hardest surfaces)

• Lead-bronze (high-speed applications)

• Aluminum- and silicon-bronze (heavy load, slow-speed application)

The factor Cs for bronze in the spectrum sand-cast, chilled-cast, and centrifugally cast increases in the same order. Standardization of tooth systems is not as far along as it is in other types of gearing. For the designer this represents freedom of action, but acquisition of tooling for tooth-forming is more of a problem for in-house manufacturing. When using a subcontractor the designer must be aware of what the supplier is capable of providing with on-hand tooling. Axial pitches for the worm are usually integers, and quotients of integers are common. Typical pitches are 1

4, 5

16, 3 8,

1 2,

3 4, 1, 54,

6 4,

7 4, and 2, but others are possible.

Table 15–8 shows dimensions common to both worm gear and cylindrical worm for proportions often used. Teeth frequently are stubbed when lead angles are 30° or larger. Worm-gear design is constrained by available tooling, space restrictions, shaft center-to-center distances, gear ratios needed, and the designer’s experience. ANSI/AGMA 6022-C93, Design Manual for Cylindrical Wormgearing offers the following guidance. Normal pressure angles are chosen from 14.5°, 17.5°, 20°, 22.5°, 25°, 27.5°, and 30°. The recommended minimum number of gear teeth is given in Table 15–10. The normal range of the number of threads on the worm is 1 through 10. Mean worm pitch diameter is usually chosen in the range given by Eq. (15–27). A design decision is the axial pitch of the worm. Since acceptable proportions are couched in terms of the center-to-center distance, which is not yet known, one chooses a trial axial pitch px. Having NW and a trial worm diameter d,

NG 5 mG NW Pt 5 p

px D 5

t A priori decisions

t Design decisions

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810 Mechanical Engineering Design

Then

(d)lo 5 C0.875y3 (d)hi 5 C0.875y1.6

Examine (d)lo # d # (d)hi, and refine the selection of mean worm-pitch diameter to d1 if necessary. Recompute the center-to-center distance as C 5 (d1 1 D)y2. There is even an opportunity to make C a round number. Choose C and set

d2 5 2C 2 D

Equations (15–39) through (15–48) apply to one usual set of proportions.

Fn (NG)min

14.5 40

17.5 27

20 21

22.5 17

25 14

27.5 12

30 10

Table 15–10

Minimum Number of Gear Teeth for Normal Pressure Angle fn

EXAMPLE 15–4 Design a 10-hp 11:1 worm-gear speed-reducer mesh for a lumber mill planer feed drive for 3- to 10-h daily use. A 1720-rev/min squirrel-cage induction motor drives the planer feed (Ka 5 1.25), and the ambient temperature is 70°F.

Solution Function: H0 5 10 hp, mG 5 11, nW 5 1720 rev/min. Design factor: nd 5 1.2. Materials and processes: case-hardened alloy steel worm, sand-cast bronze gear. Worm threads: double, NW 5 2, NG 5 mG NW 5 11(2) 5 22 gear teeth acceptable for fn 5 20°, according to Table 15–10. Decision 1: Choose an axial pitch of worm px 5 1.5 in. Then,

Pt 5 pypx 5 py1.5 5 2.0944

D 5 NGyPt 5 22y2.0944 5 10.504 in

Eq. (15–39): a 5 0.3183px 5 0.3183(1.5) 5 0.4775 in (addendum)

Eq. (15–40): b 5 0.3683(1.5) 5 0.5525 in (dedendum)

Eq. (15–41): ht 5 0.6866(1.5) 5 1.030 in

Decision 2: Choose a mean worm diameter d 5 2.000 in. Then

C 5 (d 1 D)y2 5 (2.000 1 10.504)y2 5 6.252 in

(d)lo 5 6.2520.875y3 5 1.657 in

(d)hi 5 6.2520.875y1.6 5 3.107 in

The range, given by Eq. (15–27), is 1.657 # d # 3.107 in, which is satisfactory. Try d 5 2.500 in. Recompute C:

C 5 (2.5 1 10.504)y2 5 6.502 in

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Bevel and Worm Gears 811

The range is now 1.715 # d # 3.216 in, which is still satisfactory. Decision: d 5 2.500 in. Then

Eq. (13–27): L 5 px NW 5 1.5(2) 5 3.000 in

Eq. (13–28):

l 5 tan21[Ly(pd)] 5 tan21[3y(p2.5)] 5 20.905° (from Table 15–9 lead angle OK)

Eq. (15–62): Vs 5 p d nW

12 cos l 5 p(2.5)1720

12 cos 20.905° 5 1205.1 ft/min

VW 5 p d nW

12 5 p(2.5)1720

12 5 1125.7 ft/min

VG 5 p D nG

12 5 p(10.504)1720y11

12 5 430.0 ft/min

Eq. (15–33): Cs 5 1190 2 477 log 10.504 5 702.8

Eq. (15–36): Cm 5 0.0222112 1 40(11) 2 76 1 0.46 5 0.772

Eq. (15–37): Cv 5 13.31(1205.1)20.571 5 0.232

Eq. (15–38): f 5 0.103 exp[20.11(1205.1)0.45] 1 0.012 5 0.0191

Eq. (15–54): eW 5 cos 20° 2 0.0191 tan 20.905° cos 20° 1 0.0191 cot 20.905°

5 0.942

(If the worm gear drives, eG 5 0.939.) To ensure nominal 10-hp output, with adjust- ments for Ka, nd, and e,

Eq. (15–57): W t W 5 1222

cos 20° sin 20.905° 1 0.0191 cos 20.905° cos 20° cos 20.905° 2 0.0191 sin 20.905°

5 495.4 lbf

Eq. (15–58): W t G 5

33 000(1.2)10(1.25) 430(0.942)

5 1222 lbf

Eq. (15–59): HW 5 p(2.5)1720(495.4)

12(33 000) 5 16.9 hp

Eq. (15–60): HG 5 p(10.504)1720y11(1222)

12(33 000) 5 15.92 hp

Eq. (15–61): Wf 5 0.0191(1222)

0.0191 sin 20.905° 2 cos 20° cos 20.905° 5 226.8 lbf

Eq. (15–63): Hf 5 Z226.8 Z1205.1

33 000 5 0.979 hp

From Eq. (15–28), with Cs 5 702.8, Cm 5 0.772, and Cv 5 0.232,

(Fe)req 5 W t

Cs D 0.8Cm

Cv 5

1222

702.8(10.504)0.80.772(0.232) 5 1.479 in

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812 Mechanical Engineering Design

Decision 3: The available range of (Fe)G is 1.479 # (Fe)G # 2dy3 or 1.479 # (Fe)G # 1.667 in. Set (Fe)G 5 1.5 in.

Eq. (15–28): W t all 5 702.8(10.504)0.81.5(0.772)0.232 5 1239 lbf

This is greater than 1222 lbf. There is a little excess capacity. The force analysis stands.

Decision 4:

Eq. (15–50): h# CR 5 nW

6494 1 0.13 5

1720 6494

1 0.13 5 0.395 ft ? lbf/(min ? in2 ? °F)

Eq. (15–49): Hloss 5 33 000(1 2 e)HW 5 33 000(1 2 0.942)16.9 5 32 347 ft ? lbf/min

The AGMA area, from Eq. (15–52), is Amin 5 43.2C1.7 5 43.2(6.502)1.7 5 1041.5 in2. A rough estimate of the lateral area for 6-in clearances:

Vertical: d 1 D 1 6 5 2.5 1 10.5 1 6 5 19 in

Width: D 1 6 5 10.5 1 6 5 16.5 in

Thickness: d 1 6 5 2.5 1 6 5 8.5 in

Area: 2(19)16.5 1 2(8.5)19 1 16.5(8.5) < 1090 in2

Expect an area of 1100 in2. Choose: Air-cooled, no fan on worm, with an ambient temperature of 70°F.

ts 5 ta 1 Hloss

h# CR A 5 70 1

32 350 0.395(1100)

5 70 1 74.5 5 144.5°F

Lubricant is safe with some margin for smaller area.

Eq. (13–18): Pn 5 Pt

cos l 5

2.094 cos 20.905°

5 2.242

pn 5 p

Pn 5 p

2.242 5 1.401 in

Gear bending stress, for reference, is

Eq. (15–53): s 5 W t

pn Fe y 5

1222 1.401(1.5)0.125

5 4652 psi

The risk is from wear, which is addressed by the AGMA method that provides (W tG)all.

15–9 Buckingham Wear Load A precursor to the AGMA method was the method of Buckingham, which identified an allowable wear load in worm gearing. Buckingham showed that the allowable gear-tooth loading for wear can be estimated from

(W t G)all 5 Kw

dG Fe (15–64)

where Kw 5 worm-gear load factor

dG 5 gear-pitch diameter

Fe 5 worm-gear effective face width

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Bevel and Worm Gears 813

Table 15–11 gives values for Kw for worm gearsets as a function of the material pair- ing and the normal pressure angle.

Material Thread Angle Fn

Worm Gear 14 1 2 ° 20° 25° 30°

Hardened steel* Chilled bronze 90 125 150 180

Hardened steel* Bronze 60 80 100 120

Steel, 250 BHN (min.) Bronze 36 50 60 72

High-test cast iron Bronze 80 115 140 165

Gray iron† Aluminum 10 12 15 18

High-test cast iron Gray iron 90 125 150 180

High-test cast iron Cast steel 22 31 37 45

High-test cast iron High-test cast iron 135 185 225 270

Steel 250 BHN (min.) Laminated phenolic 47 64 80 95

Gray iron Laminated phenolic 70 96 120 140

*Over 500 BHN surface. †For steel worms, multiply given values by 0.6.

Table 15–11

Wear Factor Kw for Worm Gearing Source: Earle Buckingham, Design of Worm and Spiral Gears, Industrial Press, New York, 1981.

For material combinations not addressed by AGMA, Buckingham’s method allows quantitative treatment.

PROBLEMS 15–1 An uncrowned straight-bevel pinion has 20 teeth, a diametral pitch of 6 teeth/in, and a transmis-

sion accuracy number of 6. Both the pinion and gear are made of through-hardened steel with a Brinell hardness of 300. The driven gear has 60 teeth. The gearset has a life goal of 109 revolutions of the pinion with a reliability of 0.999. The shaft angle is 90°, and the pinion speed is 900 rev/min. The face width is 1.25 in, and the normal pressure angle is 20°. The pinion is mounted outboard of its bearings, and the gear is straddle-mounted. Based on the AGMA bend- ing strength, what is the power rating of the gearset? Use K0 5 1 and SF 5 SH 5 1.

EXAMPLE 15–5 Estimate the allowable gear wear load (W tG)all for the gearset of Ex. 15–4 using Buckingham’s wear equation.

Solution From Table 15–11 for a hardened steel worm and a bronze bear, Kw is given as 80 for fn 5 20°. Equation (15–64) gives

(Wt G)all 5 80(10.504)1.5 5 1260 lbf

which is larger than the 1239 lbf of the AGMA method. The method of Buckingham does not have refinements of the AGMA method. [Is (W tG)all linear with gear diameter?]

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814 Mechanical Engineering Design

15–2 For the gearset and conditions of Prob. 15–1, find the power rating based on the AGMA surface durability.

15–3 An uncrowned straight-bevel pinion has 30 teeth, a diametral pitch of 6, and a transmission accuracy number of 6. The driven gear has 60 teeth. Both are made of No. 30 cast iron. The shaft angle is 90°. The face width is 1.25 in, the pinion speed is 900 rev/min, and the normal pressure angle is 20°. The pinion is mounted outboard of its bearings and the bearings of the gear straddle it. What is the power rating based on AGMA bending strength? Note: For cast iron gearsets reliability information has not yet been developed. We say that if the life is greater than 107 revolutions, then set KL 5 1, CL 5 1, CR 5 1, KR 5 1, and apply a factor of safety. Use SF 5 2 and SH 5 12.

15–4 For the gearset and conditions of Prob. 15–3, find the power rating based on AGMA surface durability.

15–5 An uncrowned straight-bevel pinion has 22 teeth, a module of 4 mm, and a transmission accuracy number of 5. The pinion and the gear are made of through-hardened steel, both having core and case hardnesses of 180 Brinell. The pinion drives the 24-tooth bevel gear. The shaft angle is 90°, the pinion speed is 1800 rev/min, the face width is 25 mm, and the normal pressure angle is 20°. Both gears have an outboard mounting. Find the power rating based on AGMA pitting resistance if the life goal is 109 revolutions of the pinion at 0.999 reliability.

15–6 For the gearset and conditions of Prob. 15–5, find the power rating for AGMA bending strength.

15–7 In straight-bevel gearing, there are some analogs to Eqs. (14–44) and (14–45) pp. 766 and 767, respectively. If we have a pinion core with a hardness of (HB)11 and we try equal power ratings, the transmitted load W t can be made equal in all four cases. It is possible to find these relations:

(a) For carburized case-hardened gear steel with core AGMA bending strength (sat)G and pin- ion core strength (sat)P, show that the relationship is

(sat)G 5 (sat) P

JG m20.0323

This allows (HB)21 to be related to (HB)11.

(b) Show that the AGMA contact strength of the gear case (sac)G can be related to the AGMA core bending strength of the pinion core (sat)P by

(sac)G 5 Cp

(CL)G CH BS2

SF (sat)P(KL)P Kx JP KT Cs Cxc

NP I Ks

If factors of safety are applied to the transmitted load Wt, then SH 5 1SF and SH 2ySF is unity.

The result allows (HB)22 to be related to (HB)11.

(sac)P 5 (sac)G m0.0602 G CH

Core Case

Pinion (HB)11 (HB)12

Gear (HB)21 (HB)22

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Bevel and Worm Gears 815

which can be established by the relations in Prob. 15–7, and see if the result matches transmitted loads W t in all four cases.

15–10 A catalog of stock bevel gears lists a power rating of 5.2 hp at 1200 rev/min pinion speed for a straight-bevel gearset consisting of a 20-tooth pinion driving a 40-tooth gear. This gear pair has a 20° normal pressure angle, a face width of 0.71 in, a diametral pitch of 10 teeth/in, and is through-hardened to 300 BHN. Assume the gears are for general industrial use, are generated to a transmission accuracy number of 5, and are uncrowned. Also assume the gears are rated for a life of 3 3 106 revolutions with a 99 percent reliability. Given these data, what do you think about the stated catalog power rating?

15–11 Apply the relations of Prob. 15–7 to Ex. 15–1 and find the Brinell case hardness of the gears for equal allowable load Wt in bending and wear. Check your work by reworking Ex. 15–1 to see if you are correct. How would you go about the heat treatment of the gears?

15–12 Your experience with Ex. 15–1 and problems based on it will enable you to write an interactive computer program for power rating of through-hardened steel gears. Test your understanding of bevel-gear analysis by noting the ease with which the coding develops. The hardness proto- col developed in Prob. 15–7 can be incorporated at the end of your code, first to display it, then as an option to loop back and see the consequences of it.

15–13 Use your experience with Prob. 15–11 and Ex. 15–2 to design an interactive computer-aided design program for straight-steel bevel gears, implementing the ANSI/AGMA 2003-B97 stan- dard. It will be helpful to follow the decision set in Sec. 15–5, allowing the return to earlier decisions for revision as the consequences of earlier decisions develop.

15–14 A single-threaded steel worm rotates at 1725 rev/min, meshing with a 56-tooth worm gear transmitting 1 hp to the output shaft. The pitch diameter of the worm is 1.50. The tangential diametral pitch of the gear is 8 teeth per inch and the normal pressure angle is 20°. The ambi- ent temperature is 70°F, the application factor is 1.25, the design factor is 1, the gear face is 0.5 in, the lateral case area is 850 in2, and the gear is sand-cast bronze. (a) Determine and evaluate the geometric properties of the gears. (b) Determine the transmitted gear forces and the mesh efficiency. (c) Is the mesh sufficient to handle the loading? (d ) Estimate the lubricant sump temperature, assuming fan-stirred air.

As in Ex. 15–4, design a cylindrical worm-gear mesh to connect a squirrel-cage induction motor to a liquid agitator. The motor speed is 1125 rev/min, and the velocity ratio is to be 10:1. The output power requirement is 25 hp. The shaft axes are 90° to each other. An overload factor Ko (see Table 15–2) makes allowance for external dynamic excursions of load from the nominal or average load W t. For this service Ko 5 1.25 is appropriate. Additionally, a design factor nd of 1.1 is to be included to address other unquantifiable risks. For Probs. 15–15 to 15–17 use the AGMA method for (W tG)all whereas for Probs. 15–18 to 15–22, use the Buckingham method. See Table 15–12.

15–15 to 15–22

15–8 Refer to your solution to Probs. 15–1 and 15–2. If the pinion core hardness is 300 Brinell, use the relations from Prob. 15–7 to determine the required hardness of the gear core and the case hardnesses of both gears to ensure equal power ratings.

15–9 Repeat Probs. 15–1 and 15–2 with the hardness protocol

Core Case

Pinion 300 373

Gear 339 345

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816 Mechanical Engineering Design

Problem Materials No. Method Worm Gear

15–15 AGMA Steel, HRC 58 Sand-cast bronze

15–16 AGMA Steel, HRC 58 Chilled-cast bronze

15–17 AGMA Steel, HRC 58 Centrifugal-cast bronze

15–18 Buckingham Steel, 500 Bhn Chilled-cast bronze

15–19 Buckingham Steel, 500 Bhn Cast bronze

15–20 Buckingham Steel, 250 Bhn Cast bronze

15–21 Buckingham High-test cast iron Cast bronze

15–22 Buckingham High-test cast iron High-test cast iron

Table 15–12

Table Supporting Problems 15–15 to 15–22

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