Spur Gears 1021 1. Introduction. 2. Friction Wheels. 3. Advantages and Disadvantages of Gear Drives. 4. Classification of Gears. 5. Terms used in Gears. 6. Condition for Constant Velocity Ratio of Gears–Law of Gearing. 7. Forms of Teeth. 8. Cycloidal Teeth. 9. Involute Teeth. 10. Comparison Between Involute and Cycloidal Gears. 11. Systems of Gear Teeth. 12. Standard Proportions of Gear Systems. 13. Interference in Involute Gears. 14. Minimum Number of Teeth on the Pinion in order to Avoid Interference. 15. Gear Materials. 16. Design Considerations for a Gear Drive. 17. Beam Strength of Gear Teeth- Lewis Equation. 18. Permissible Working Stress for Gear Teeth in Lewis Equation. 19. Dynamic Tooth Load. 20. Static Tooth Load. 21. Wear Tooth Load. 22. Causes of Gear Tooth Failure. 23. Design Procedure for Spur Gears. 24. Spur Gear Construction. 25. Design of Shaft for Spur Gears. 26. Design of Arms for Spur Gears. 28 C H A P T E R 28.1 28.1 28.1 28.1 28.1 Intr Intr Intr Intr Introduction oduction oduction oduction oduction We have discussed earlier that the slipping of a belt or rope is a common phenomenon, in the transmission of motion or power between two shafts. The effect of slipping is to reduce the velocity ratio of the system. In precision machines, in which a definite velocity ratio is of importance (as in watch mechanism), the only positive drive is by gears or toothed wheels. A gear drive is also provided, when the distance between the driver and the follower is very small. 28.2 28.2 28.2 28.2 28.2 Friction Wheels Friction Wheels Friction Wheels Friction Wheels Friction Wheels The motion and power transmitted by gears is kinematically equivalent to that transmitted by frictional wheels or discs. In order to understand how the motion can be transmitted by two toothed wheels, consider two plain circular wheels A and B mounted on shafts. The wheels have sufficient rough surfaces and press against each other as shown in Fig. 28.1.
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1021
1. Introduction.2. Friction Wheels.3. Advantages and
Disadvantages of Gear Drives.4. Classification of Gears.5. Terms used in Gears.6. Condition for Constant
Velocity Ratio of Gears–Law ofGearing.
7. Forms of Teeth.8. Cycloidal Teeth.9. Involute Teeth.10. Comparison Between Involute
and Cycloidal Gears.11. Systems of Gear Teeth.12. Standard Proportions of Gear
Systems.13. Interference in Involute Gears.14. Minimum Number of Teeth on
the Pinion in order to AvoidInterference.
15. Gear Materials.16. Design Considerations for a
Gear Drive.17. Beam Strength of Gear Teeth-
Lewis Equation.18. Permissible Working Stress for
Gear Teeth in Lewis Equation.19. Dynamic Tooth Load.20. Static Tooth Load.21. Wear Tooth Load.22. Causes of Gear Tooth Failure.23. Design Procedure for Spur
Gears.24. Spur Gear Construction.25. Design of Shaft for Spur Gears.26. Design of Arms for Spur Gears.
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28.128.128.128.128.1 IntrIntrIntrIntrIntroductionoductionoductionoductionoductionWe have discussed earlier that the slipping of a belt or
rope is a common phenomenon, in the transmission ofmotion or power between two shafts. The effect of slippingis to reduce the velocity ratio of the system. In precisionmachines, in which a definite velocity ratio is of importance(as in watch mechanism), the only positive drive is by gearsor toothed wheels. A gear drive is also provided, whenthe distance between the driver and the follower is verysmall.
28.228.228.228.228.2 Friction WheelsFriction WheelsFriction WheelsFriction WheelsFriction WheelsThe motion and power transmitted by gears is
kinematically equivalent to that transmitted by frictionalwheels or discs. In order to understand how the motion canbe transmitted by two toothed wheels, consider two plaincircular wheels A and B mounted on shafts. The wheels havesufficient rough surfaces and press against each other asshown in Fig. 28.1.
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* We know that frictional resistance, F = µ . RN
where µ = Coefficient of friction between the rubbing surfaces of the two wheels, and
RN = Normal reaction between the two rubbing surfaces.
Fig. 28.1. Friction wheels. Fig. 28.2. Gear or toothed wheel.
Let the wheel A is keyed to the rotating shaft and the wheel B to the shaft to be rotated. A littleconsideration will show that when the wheel A is rotated by a rotating shaft, it will rotate the wheel Bin the opposite direction as shown in Fig. 28.1. The wheel B will be rotated by the wheel A so long asthe tangential force exerted by the wheel A does not exceed the maximum frictional resistance betweenthe two wheels. But when the tangential force (P) exceeds the *frictional resistance (F), slipping willtake place between the two wheels.
In order to avoid the slipping, a number of projections (called teeth) as shown in Fig. 28.2 areprovided on the periphery of the wheel A which will fit into the corresponding recesses on the peripheryof the wheel B. A friction wheel with the teeth cut on it is known as gear or toothed wheel. The usuallconnection to show the toothed wheels is by their pitch circles.Note : Kinematically, the friction wheels running without slip and toothed gearing are identical. But due to thepossibility of slipping of wheels, the friction wheels can only beused for transmission of small powers.
28.328.328.328.328.3 Advantages and Disadvantages ofAdvantages and Disadvantages ofAdvantages and Disadvantages ofAdvantages and Disadvantages ofAdvantages and Disadvantages ofGear DrivesGear DrivesGear DrivesGear DrivesGear Drives
The following are the advantages and disadvantagesof the gear drive as compared to other drives, i.e. belt, ropeand chain drives :Advantages
1. It transmits exact velocity ratio.2. It may be used to transmit large power.3. It may be used for small centre distances of shafts.4. It has high efficiency.5. It has reliable service.6. It has compact layout.
Disadvantages1. Since the manufacture of gears require special
tools and equipment, therefore it is costlier thanother drives.
In bicycle gears are used to transmitmotion. Mechanical advantage canbe changed by changing gears.
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2. The error in cutting teeth may cause vibrations and noise during operation.
3. It requires suitable lubricant and reliable method of applying it, for the proper operation ofgear drives.
28.428.428.428.428.4 Classification of GearsClassification of GearsClassification of GearsClassification of GearsClassification of GearsThe gears or toothed wheels may be classified as follows :
1. According to the position of axes of the shafts. The axes of the two shafts between whichthe motion is to be transmitted, may be
(a) Parallel, (b) Intersecting, and (c) Non-intersecting and non-parallel.
The two parallel and co-planar shafts connected by the gears is shown in Fig. 28.2. These gearsare called spur gears and the arrangement is known as spur gearing. These gears have teeth parallelto the axis of the wheel as shown in Fig. 28.2. Another name given to the spur gearing is helicalgearing, in which the teeth are inclined to the axis. The single and double helical gears connectingparallel shafts are shown in Fig. 28.3 (a) and (b) respectively. The object of the double helical gear isto balance out the end thrusts that are induced in single helical gears when transmitting load. Thedouble helical gears are known as herringbone gears. A pair of spur gears are kinematically equivalentto a pair of cylindrical discs, keyed to a parallel shaft having line contact.
The two non-parallel or intersecting, but coplaner shafts connected by gears is shown inFig. 28.3 (c). These gears are called bevel gears and the arrangement is known as bevel gearing.The bevel gears, like spur gears may also have their teeth inclined to the face of the bevel, inwhich case they are known as helical bevel gears.
Fig. 28.3
The two non-intersecting and non-parallel i.e. non-coplanar shafts connected by gears is shownin Fig. 28.3 (d). These gears are called skew bevel gears or spiral gears and the arrangement isknown as skew bevel gearing or spiral gearing. This type of gearing also have a line contact, therotation of which about the axes generates the two pitch surfaces known as hyperboloids.Notes : (i) When equal bevel gears (having equal teeth) connect two shafts whose axes are mutually perpendicu-lar, then the bevel gears are known as mitres.
(ii) A hyperboloid is the solid formed by revolving a straight line about an axis (not in the same plane),such that every point on the line remains at a constant distance from the axis.
(iii) The worm gearing is essentially a form of spiral gearing in which the shafts are usually at right angles.
2. According to the peripheral velocity of the gears. The gears, according to the peripheralvelocity of the gears, may be classified as :
(a) Low velocity, (b) Medium velocity, and (c) High velocity.
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* A straight line may also be defined as a wheel of infinite radius.
Fig. 28.5. Rack and pinion.
The gears having velocity less than 3 m/s are termed as low velocity gears and gears havingvelocity between 3 and 15 m / s are known as medium velocity gears. If the velocity of gears is morethan 15 m / s, then these are called high speed gears.
3. According to the type of gearing. The gears, according to the type of gearing, may beclassified as :
(a) External gearing, (b) Internal gearing, and (c) Rack and pinion.
Fig. 28.4
In external gearing, the gears of the two shafts mesh externally with each other as shown inFig. 28.4 (a). The larger of these two wheels is called spur wheel or gear and the smaller wheel iscalled pinion. In an external gearing, the motion of the two wheels is always unlike, as shown inFig. 28.4 (a).
In internal gearing, the gears of the two shafts mesh internally with each other as shown in Fig.28.4 (b). The larger of these two wheels is called annular wheel and the smaller wheel is calledpinion. In an internal gearing, the motion of the wheels is always like as shown in Fig. 28.4 (b).
Sometimes, the gear of a shaft meshes externally andinternally with the gears in a *straight line, as shown in Fig.28.5. Such a type of gear is called rack and pinion. Thestraight line gear is called rack and the circular wheel iscalled pinion. A little consideration will show that with thehelp of a rack and pinion, we can convert linear motion intorotary motion and vice-versa as shown in Fig. 28.5.
4. According to the position of teeth on the gearsurface. The teeth on the gear surface may be
(a) Straight, (b) Inclined, and (c) Curved.
We have discussed earlier that the spur gears havestraight teeth whereas helical gears have their teeth inclinedto the wheel rim. In case of spiral gears, the teeth are curvedover the rim surface.
28.528.528.528.528.5 TTTTTerererererms used in Gearms used in Gearms used in Gearms used in Gearms used in GearsssssThe following terms, which will be mostly used in this chapter, should be clearly understood at
this stage. These terms are illustrated in Fig. 28.6.
1. Pitch circle. It is an imaginary circle which by pure rolling action, would give the samemotion as the actual gear.
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��������� � 10252. Pitch circle diameter. It is the diameter of the pitch circle. The size of the gear is usually
specified by the pitch circle diameter. It is also called as pitch diameter.
3. Pitch point. It is a common point of contact between two pitch circles.
4. Pitch surface. It is the surface of the rolling discs which the meshing gears have replaced atthe pitch circle.
5. Pressure angle or angle of obliquity. It is the angle between the common normal to two gearteeth at the point of contact and the common tangent at the pitch point. It is usually denoted by φ. The
standard pressure angles are 1214 / ° and 20°.
6. Addendum. It is the radial distance of a tooth from the pitch circle to the top of the tooth.
7. Dedendum. It is the radial distance of a tooth from the pitch circle to the bottom of the tooth.
8. Addendum circle. It is the circle drawn through the top of the teeth and is concentric with thepitch circle.
9. Dedendum circle. It is the circle drawn through the bottom of the teeth. It is also called rootcircle.Note : Root circle diameter = Pitch circle diameter × cos φ, where φ is the pressure angle.
10. Circular pitch. It is the distance measured on the circumference of the pitch circle froma point of one tooth to the corresponding point on the next tooth. It is usually denoted by pc.Mathematically,
Circular pitch, pc = π D/T
where D = Diameter of the pitch circle, and
T = Number of teeth on the wheel.
A little consideration will show that the two gears will mesh together correctly, if the two wheelshave the same circular pitch.Note : If D1 and D2 are the diameters of the two meshing gears having the teeth T1 and T2 respectively; then forthem to mesh correctly,
pc = 1 2
1 2
D D
T T
π π= or 1 1
2 2
D T
D T=
Fig. 28.6. Terms used in gears.
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11. Diametral pitch. It is the ratio of number of teeth to the pitch circle diameter in millimetres.It denoted by Pd. Mathematically,
Diametral pitch, pd =c
T
D p
π= ...π =
∵ cD
pT
where T = Number of teeth, and
D = Pitch circle diameter.
12. Module. It is the ratio of the pitch circle diameter in millimetres to the number of teeth. It isusually denoted by m. Mathematically,
Module, m = D / T
Note : The recommended series of modules in Indian Standard are 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16,20, 25, 32, 40 and 50.
The modules 1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5,5.5, 7, 9, 11, 14, 18, 22, 28, 36 and 45 are of secondchoice.
13. Clearance. It is the radial distance from the top of the tooth to the bottom of the tooth, in ameshing gear. A circle passing through the top of the meshing gear is known as clearance circle.
14. Total depth. It is the radial distance between the addendum and the dedendum circle of agear. It is equal to the sum of the addendum and dedendum.
15. Working depth. It is radial distance from the addendum circle to the clearance circle. It isequal to the sum of the addendum of the two meshing gears.
16. Tooth thickness. It is the width of the tooth measured along the pitch circle.
17. Tooth space. It is the width of space between the two adjacent teeth measured along thepitch circle.
18. Backlash. It is the difference between the tooth space and the tooth thickness, as measuredon the pitch circle.
Spur gears
��������� � 102719. Face of the tooth. It is surface of the tooth above the pitch surface.
20. Top land. It is the surface of the top of the tooth.
21. Flank of the tooth. It is the surface of the tooth below the pitch surface.
22. Face width. It is the width of the gear tooth measured parallel to its axis.
23. Profile. It is the curve formed by the face and flank of the tooth.
24. Fillet radius. It is the radius that connects the root circle to the profile of the tooth.
25. Path of contact. It is the path traced by the point of contact of two teeth from the beginningto the end of engagement.
26. Length of the path of contact. It is the length of the common normal cut-off by the addendumcircles of the wheel and pinion.
27. Arc of contact. It is the path traced by a point on the pitch circle from the beginning to theend of engagement of a given pair of teeth. The arc of contact consists of two parts, i.e.
(a) Arc of approach. It is the portion of the path of contact from the beginning of the engagementto the pitch point.
(b) Arc of recess. It is the portion of the path of contact from the pitch point to the end of theengagement of a pair of teeth.Note : The ratio of the length of arc of contact to the circular pitch is known as contact ratio i.e. number of pairsof teeth in contact.
28.628.628.628.628.6 Condition fCondition fCondition fCondition fCondition for Constant or Constant or Constant or Constant or Constant VVVVVelocity Raelocity Raelocity Raelocity Raelocity Ratio of Geartio of Geartio of Geartio of Geartio of Gears–Law of Gears–Law of Gears–Law of Gears–Law of Gears–Law of GearingingingingingConsider the portions of the two teeth, one on the wheel 1 (or pinion) and the other on the wheel
2, as shown by thick line curves in Fig. 28.7. Let the two teeth come in contact at point Q, and thewheels rotate in the directions as shown in the figure.
Let T T be the common tangent and MN be the common normal to the curves at point of contactQ. From the centres O1 and O2, draw O1M and O2N perpendicular to MN. A little consideration willshow that the point Q moves in the direction QC, when considered as a point on wheel 1, and in thedirection QD when considered as a point on wheel 2.
Let v1 and v2 be the velocities of the point Q on the wheels 1 and 2 respectively. If the teeth areto remain in contact, then the components of these velocitiesalong the common normal MN must be equal.
∴ v1 cos α = v2 cos βor (ω1 × O1Q) cos α = (ω2 × O2Q) cos β
11 1
1
( )O M
O QO Q
ω × =2
2 22
( )O N
O QO Q
ω ×
∴ ω1.O1M = ω2 . O2N
or 1
2
ωω =
2
1
O N
O M...(i)
Also from similar triangles O1MP and O2NP,
2
1
O N
O M=
2
1
O P
O P...(ii)
Combining equations (i) and (ii), we have
1
2
ωω
= 2
1
O N
O M = 2
1
O P
O P ...(iii)
We see that the angular velocity ratio is inverselyproportional to the ratio of the distance of P from the centres
Fig. 28.7. Law of gearing.
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O1 and O2, or the common normal to the two surfaces at the point of contact Q intersects the line ofcentres at point P which divides the centre distance inversely as the ratio of angular velocities.
Therefore, in order to have a constantangular velocity ratio for all positions of thewheels, P must be the fixed point (called pitchpoint) for the two wheels. In other words, thecommon normal at the point of contactbetween a pair of teeth must always passthrough the pitch point. This is fundamentalcondition which must be satisfied whiledesigning the profiles for the teeth of gearwheels. It is also known as law of gearing.
Notes : 1. The above condition is fulfilled by teethof involute form, provided that the root circles fromwhich the profiles are generated are tangential tothe common normal.
2. If the shape of one tooth profile is arbitrarychosen and another tooth is designed to satisfy theabove condition, then the second tooth is said to beconjugate to the first. The conjugate teeth are notin common use because of difficulty in manufacture and cost of production.
3. If D1 and D2 are pitch circle diameters of wheel 1 and 2 having teeth T1 and T2 respectively, thenvelocity ratio,
1
2
ωω
= 2 2 2
1 1 1
O P D T
O P D T= =
Aircraft landing gear is especially designed to absorb shock and energy when anaircraft lands, and then release gradually.
Gear trains inside a mechanical watch
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28.728.728.728.728.7 ForForForForForms of ms of ms of ms of ms of TTTTTeetheetheetheetheethWe have discussed in Art. 28.6 (Note 2) that conjugate teeth are not in common use. Therefore,
in actual practice, following are the two types of teeth commonly used.
1. Cycloidal teeth ; and 2. Involute teeth.
We shall discuss both the above mentioned types of teeth in the following articles. Both theseforms of teeth satisfy the condition as explained in Art. 28.6.
A cycloid is the curve traced by a point on the circumference of a circle which rolls withoutslipping on a fixed straight line. When a circle rolls without slipping on the outside of a fixed circle,the curve traced by a point on the circumference of a circle is known as epicycloid. On the other hand,if a circle rolls without slipping on the inside of a fixed circle, then the curve traced by a point on thecircumference of a circle is called hypocycloid.
Fig. 28.8. Construction of cycloidal teeth of a gear.
In Fig. 28.8 (a), the fixed line or pitch line of a rack is shown. When the circle C rolls withoutslipping above the pitch line in the direction as indicated in Fig. 28.8 (a), then the point P on the circletraces the epicycloid PA. This represents the face of the cycloidal tooth profile. When the circle Drolls without slipping below the pitch line, then the point P on the circle D traces hypocycloid PBwhich represents the flank of the cycloidal tooth. The profile BPA is one side of the cycloidal racktooth. Similarly, the two curves P' A' and P' B' forming the opposite side of the tooth profile are tracedby the point P' when the circles C and D roll in the opposite directions.
In the similar way, the cycloidal teeth of a gear may be constructed as shown in Fig. 28.8 (b).The circle C is rolled without slipping on the outside of the pitch circle and the point P on the circleC traces epicycloid PA, which represents the face of the cycloidal tooth. The circle D is rolled on theinside of pitch circle and the point P on the circle D traces hypocycloid PB, which represents the flankof the tooth profile. The profile BPA is one side of the cycloidal tooth. The opposite side of the toothis traced as explained above.
The construction of the two mating cycloidal teeth is shown in Fig. 28.9. A point on the circle Dwill trace the flank of the tooth T1 when circle D rolls without slipping on the inside of pitch circle ofwheel 1 and face of tooth T2 when the circle D rolls without slipping on the outside of pitch circle ofwheel 2. Similarly, a point on the circle C will trace the face of tooth T1 and flank of tooth T2. Therolling circles C and D may have unequal diameters, but if several wheels are to be interchangeable,they must have rolling circles of equal diameters.
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Fig. 28.9. Construction of two mating cycloidal teeth.
A little consideration will show that the common normalXX at the point of contact between two cycloidal teeth alwayspasses through the pitch point, which is the fundamental con-dition for a constant velocity ratio.
28.928.928.928.928.9 InInInInInvvvvvolute olute olute olute olute TTTTTeetheetheetheetheethAn involute of a circle is a plane curve generated by a
point on a tangent, which rolls on the circle without slippingor by a point on a taut string which is unwrapped from a reelas shown in Fig. 28.10 (a). In connection with toothed wheels,the circle is known as base circle. The involute is traced asfollows :
Let A be the starting point of the involute. The base circleis divided into equal number of parts e.g. AP1, P1 P2, P2 P3etc.The tangents at P1, P2, P3 etc., are drawn and the lenghtsP1A1, P2A2, P3A3 equal to the arcs AP1, AP2 and AP3 are setoff. Joining the points A, A1, A2, A3 etc., we obtain the involutecurve AR. A little consideration will show that at any instantA3, the tangent A3T to the involute is perpendicular to P3A3 andP3A3 is the normal to the involute. In other words, normal atany point of an involute is a tangent to the circle.
Now, let O1 and O2 be the fixed centres of the two basecircles as shown in Fig. 28.10(b). Let the correspondinginvolutes AB and A'B' be in contact at point Q. MQ and NQ arenormals to the involute at Q and are tangents to base circles.Since the normal for an involute at a given point is the tangentdrawn from that point to the base circle, therefore the commonnormal MN at Q is also the common tangent to the two basecircles. We see that the common normal MN intersects the lineof centres O1O2 at the fixed point P (called pitch point).Therefore the involute teeth satisfy the fundamental conditionof constant velocity ratio.
The clock built by Galelioused gears.
��������� � 1031From similar triangles O2 NP and O1 MP,
1
2
O M
O N= 1 2
2 1
O P
O P
ω=ω ...(i)
which determines the ratio of the radii of the two base circles. The radii of the base circles is given by
O1M = O1 P cos φ, and O2N = O2 P cos φwhere φ is the pressure angle or the angle of obliquity.
Also the centre distance between the base circles
= 1 2 1 21 2 cos cos cos
O M O N O M O NO P O P
++ = + =φ φ φ
Fig. 28.10. Construction of involute teeth.A little consideration will show, that if the centre distance is changed, then the radii of pitch
circles also changes. But their ratio remains unchanged, because it is equal to the ratio of the two radiiof the base circles [See equation (i)]. The common normal, at the point of contact, still passes throughthe pitch point. As a result of this, the wheel continues to work correctly*. However, the pressureangle increases with the increase in centre distance.
28.1028.1028.1028.1028.10 Comparison Between Involute and Cycloidal GearsComparison Between Involute and Cycloidal GearsComparison Between Involute and Cycloidal GearsComparison Between Involute and Cycloidal GearsComparison Between Involute and Cycloidal GearsIn actual practice, the involute gears are more commonly used as compared to cycloidal gears,
due to the following advantages :
Advantages of involute gearsFollowing are the advantages of involute gears :
1. The most important advantage of the involute gears is that the centre distance for a pair ofinvolute gears can be varied within limits without changing the velocity ratio. This is not true forcycloidal gears which requires exact centre distance to be maintained.
2. In involute gears, the pressure angle, from the start of the engagement of teeth to the end ofthe engagement, remains constant. It is necessary for smooth running and less wear of gears. But incycloidal gears, the pressure angle is maximum at the beginning of engagement, reduces to zero atpitch point, starts increasing and again becomes maximum at the end of engagement. This results inless smooth running of gears.
3. The face and flank of involute teeth are generated by a single curve whereas in cycloidalgears, double curves (i.e. epicycloid and hypocycloid) are required for the face and flank respectively.
* It is not the case with cycloidal teeth.
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Thus the involute teeth are easy to manufacture than cycloidal teeth. In involute system, the basic rackhas straight teeth and the same can be cut with simple tools.Note : The only disadvantage of the involute teeth is that the interference occurs (Refer Art. 28.13) with pinionshaving smaller number of teeth. This may be avoided by altering the heights of addendum and dedendum of themating teeth or the angle of obliquity of the teeth.
Advantages of cycloidal gearsFollowing are the advantages of cycloidal gears :
1. Since the cycloidal teeth have wider flanks, therefore the cycloidal gears are stronger thanthe involute gears for the same pitch. Due to this reason, the cycloidal teeth are preferred specially forcast teeth.
2. In cycloidal gears, the contact takes place between a convex flank and concave surface,whereas in involute gears, the convex surfaces are in contact. This condition results in less wear incycloidal gears as compared to involute gears. However the difference in wear is negligible.
3. In cycloidal gears, the interference does not occur at all. Though there are advantages ofcycloidal gears but they are outweighed by the greater simplicity and flexibility of the involute gears.
28.11 Systems of Gear 28.11 Systems of Gear 28.11 Systems of Gear 28.11 Systems of Gear 28.11 Systems of Gear TTTTTeetheetheetheetheethThe following four systems of gear teeth are commonly used in practice.
1. 1214 / ° Composite system, 2. 1
214 / ° Full depth involute system, 3. 20° Full depth involutesystem, and 4. 20° Stub involute system.
The 1214 / ° composite system is used for general purpose gears. It is stronger but has no inter-
changeability. The tooth profile of this system has cycloidal curves at the top and bottom and involutecurve at the middle portion. The teeth are produced by formed milling cutters or hobs. The tooth
profile of the 1214 / ° full depth involute system was developed for use with gear hobs for spur and
helical gears.
The tooth profile of the 20° full depth involute system may be cut by hobs. The increase of the
pressure angle from 1214 / ° to 20° results in a stronger tooth, because the tooth acting as a beam is
wider at the base. The 20° stub involute system has a strong tooth to take heavy loads.
28.1228.1228.1228.1228.12 StandarStandarStandarStandarStandard Prd Prd Prd Prd Proporoporoporoporoportions of Gear Systemstions of Gear Systemstions of Gear Systemstions of Gear Systemstions of Gear SystemsThe following table shows the standard proportions in module (m) for the four gear systems as
discussed in the previous article.
TTTTTaaaaable 28.1.ble 28.1.ble 28.1.ble 28.1.ble 28.1. Standar Standar Standar Standar Standard prd prd prd prd proporoporoporoporoportions of gear systemstions of gear systemstions of gear systemstions of gear systemstions of gear systems.....
S. No. Particulars °1214 / composite or full 20° full depth 20° stub involute
depth involute system involute system system
1. Addendum 1m 1m 0.8 m
2. Dedendum 1.25 m 1.25 m 1 m
3. Working depth 2 m 2 m 1.60 m
4. Minimum total depth 2.25 m 2.25 m 1.80 m
5. Tooth thickness 1.5708 m 1.5708 m 1.5708 m
6. Minimum clearance 0.25 m 0.25 m 0.2 m
7. Fillet radius at root 0.4 m 0.4 m 0.4 m
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28.1328.1328.1328.1328.13 InterferInterferInterferInterferInterference in Inence in Inence in Inence in Inence in Invvvvvolute Gearolute Gearolute Gearolute Gearolute GearsssssA pinion gearing with a wheel is shown in Fig.
28.11. MN is the common tangent to the base circlesand KL is the path of contact between the two matingteeth. A little consideration will show, that if the radiusof the addendum circle of pinion is increased to O1N,the point of contact L will move from L to N. Whenthis radius is further increased, the point of contact Lwill be on the inside of base circle of wheel and noton the involute profile of tooth on wheel. The tip oftooth on the pinion will then undercut the tooth on thewheel at the root and remove part of the involuteprofile of tooth on the wheel. This effect is known asinterference and occurs when the teeth are being cut.In brief, the phenomenon when the tip of a toothundercuts the root on its mating gear is known asinterference.
Fig. 28.11. Interference in involute gears.
Similarly, if the radius of the addendum circle of the wheel increases beyond O2M, then the tipof tooth on wheel will cause interference with the tooth on pinion. The points M and N are calledinterference points. Obviously interference may be avoided if the path of contact does not extendbeyond interference points. The limiting value of the radius of the addendum circle of the pinion isO1N and of the wheel is O2M.
From the above discussion, we conclude that the interference may only be avoided, if the pointof contact between the two teeth is always on the involute profiles of both the teeth. In other words,interference may only be prevented, if the addendum circles of the two mating gears cut the commontangent to the base circles between the points of tangency.
A drilling machine drilling holes for lampretaining screws
1034 � ���������������������������
Note : In order to avoid interference, the limiting value of the radius of the addendum circle of the pinion (O1 N)and of the wheel (O2 M), may be obtained as follows :
From Fig. 28.11, we see that
O1N = 2 2 2 21( ) ( ) ( ) [( ) sin ]+ = + + φbO M MN r r R
where rb = Radius of base circle of the pinion = O1P cos φ = r cos φ
Similarly O2M = 2 2 2 22( ) ( ) ( ) [( ) sin ]+ = + + φbO N MN R r R
where Rb = Radius of base circle of the wheel = O2P cos φ = R cos φ
28.1428.1428.1428.1428.14 MinimMinimMinimMinimMinimum Number of um Number of um Number of um Number of um Number of TTTTTeeth on the Pinion in Oreeth on the Pinion in Oreeth on the Pinion in Oreeth on the Pinion in Oreeth on the Pinion in Order to der to der to der to der to AAAAAvvvvvoidoidoidoidoidInterferInterferInterferInterferInterferenceenceenceenceence
We have seen in the previous article that the interference may only be avoided, if the point ofcontact between the two teeth is always on the involute profiles of both the teeth. The minimumnumber of teeth on the pinion which will mesh with any gear (also rack) without interference aregiven in the following table.
TTTTTaaaaable 28.2.ble 28.2.ble 28.2.ble 28.2.ble 28.2. Minim Minim Minim Minim Minimum number of teeth on the pinion in orum number of teeth on the pinion in orum number of teeth on the pinion in orum number of teeth on the pinion in orum number of teeth on the pinion in order to avder to avder to avder to avder to avoid interferoid interferoid interferoid interferoid interference.ence.ence.ence.ence.
S. No. Systems of gear teeth Minimum number of teeth on the pinion
1. 1214 °/ Composite 12
2. 1214 °/ Full depth involute 32
3. 20° Full depth involute 18
4. 20° Stub involute 14
The number of teeth on the pinion (TP) in order to avoid interference may be obtained from thefollowing relation :
TP = W
2
2
1 11 2 sin – 1
+ + φ
A
GG G
where AW = Fraction by which the standard addendum for the wheel should bemultiplied,
G = Gear ratio or velocity ratio = TG / TP = DG / DP,
φ = Pressure angle or angle of obliquity.
28.1528.1528.1528.1528.15 Gear MaterialsGear MaterialsGear MaterialsGear MaterialsGear MaterialsThe material used for the manufacture of gears depends upon the strength and service conditions
like wear, noise etc. The gears may be manufactured from metallic or non-metallic materials. Themetallic gears with cut teeth are commercially obtainable in cast iron, steel and bronze. The non-metallic materials like wood, rawhide, compressed paper and synthetic resins like nylon are used forgears, especially for reducing noise.
The cast iron is widely used for the manufacture of gears due to its good wearing properties,excellent machinability and ease of producing complicated shapes by casting method. The cast irongears with cut teeth may be employed, where smooth action is not important.
The steel is used for high strength gears and steel may be plain carbon steel or alloy steel.The steel gears are usually heat treated in order to combine properly the toughness and toothhardness.
��������� � 1035
The phosphor bronze is widely used for worm gears in order to reduce wear of the worms whichwill be excessive with cast iron or steel. The following table shows the properties of commonly usedgear materials.
TTTTTaaaaable 28.3.ble 28.3.ble 28.3.ble 28.3.ble 28.3. Pr Pr Pr Pr Properoperoperoperoperties of commonly used gear maties of commonly used gear maties of commonly used gear maties of commonly used gear maties of commonly used gear materterterterterialsialsialsialsials.....
Material Condition Brinell hardness Minimum tensilenumber strength (N/mm2)
28.1628.1628.1628.1628.16 Design Considerations for a Gear DriveDesign Considerations for a Gear DriveDesign Considerations for a Gear DriveDesign Considerations for a Gear DriveDesign Considerations for a Gear DriveIn the design of a gear drive, the following data is usually given :
1. The power to be transmitted.
2. The speed of the driving gear,
3. The speed of the driven gear or the velocity ratio, and
4. The centre distance.
The following requirements must be met in the design of a gear drive :
(a) The gear teeth should have sufficient strength so that they will not fail under static loadingor dynamic loading during normal running conditions.
(b) The gear teeth should have wear characteristics so that their life is satisfactory.
(c) The use of space and material should be economical.
(d) The alignment of the gears and deflections of the shafts must be considered because theyeffect on the performance of the gears.
(e) The lubrication of the gears must be satisfactory.
��������� � 1037
28.1728.1728.1728.1728.17 Beam StrBeam StrBeam StrBeam StrBeam Strength of Gear ength of Gear ength of Gear ength of Gear ength of Gear TTTTTeeth – Leeeth – Leeeth – Leeeth – Leeeth – Lewis Equawis Equawis Equawis Equawis EquationtiontiontiontionThe beam strength of gear teeth is determined from an equation (known as *Lewis equation)
and the load carrying ability of the toothed gears as determined by this equation gives satisfactoryresults. In the investigation, Lewis assumed that as the load is being transmitted from one gear toanother, it is all given and taken by one tooth, because it is not always safe to assume that the load isdistributed among several teeth. When contact begins, the load is assumed to be at the end of thedriven teeth and as contact ceases, it is at the end of the driving teeth. This may not be true when thenumber of teeth in a pair of mating gears is large, because the load may be distributed among severalteeth. But it is almost certain that at some time during the contact of teeth, the proper distribution ofload does not exist and that one tooth must transmitthe full load. In any pair of gears having unlikenumber of teeth, the gear which have the fewerteeth (i.e. pinion) will be the weaker, because thetendency toward undercutting of the teeth becomesmore pronounced in gears as the number of teethbecomes smaller.
Consider each tooth as a cantilever beamloaded by a normal load (WN) as shown in Fig.28.12. It is resolved into two components i.e.tangential component (WT) and radial component(WR) acting perpendicular and parallel to the centreline of the tooth respectively. The tangential component (WT) induces a bending stress which tends tobreak the tooth. The radial component (WR) induces a compressive stress of relatively small magnitude,therefore its effect on the tooth may be neglected. Hence, the bending stress is used as the basis fordesign calculations. The critical section or the section of maximum bending stress may be obtainedby drawing a parabola through A and tangential to the tooth curves at B and C. This parabola, asshown dotted in Fig. 28.12, outlines a beam of uniform strength, i.e. if the teeth are shaped like aparabola, it will have the same stress at all the sections. But the tooth is larger than the parabola atevery section except BC. We therefore, conclude that the section BC is the section of maximum stressor the critical section. The maximum value of the bending stress (or the permissible working stress),at the section BC is given by
σw = M.y / I ...(i)where M = Maximum bending moment at the critical section BC = WT × h,
WT = Tangential load acting at the tooth,
h = Length of the tooth,
y = Half the thickness of the tooth (t) at critical section BC = t/2,
I = Moment of inertia about the centre line of the tooth = b.t3/12,
b = Width of gear face.
Substituting the values for M, y and I in equation (i), we get
σw = T T3 2
( ) / 2 ( ) 6
. /12 .
W h t W h
b t b t
× × ×=
or WT = σw × b × t2 / 6 h
In this expression, t and h are variables depending upon the size of the tooth (i.e. the circularpitch) and its profile.
Fig. 28.12. Tooth of a gear.
* In 1892, Wilfred Lewis investigated for the strength of gear teeth. He derived an equation which is nowextensively used by industry in determining the size and proportions of the gear.
1038 � ���������������������������
Let t = x × pc , and h = k × pc ; where x and k are constants.
∴ WT =2 2 2.
6 . 6σ × × = σ × × ×c
w w cc
x p xb b p
k p kSubstituting x2 / 6k = y, another constant, we have
WT = σw . b . pc . y = σw . b . π m . y ...(∵ pc = π m)
The quantity y is known as Lewis form factor or tooth form factor and WT (which is thetangential load acting at the tooth) is called the beam strength of the tooth.
Since 2 2 2
2,
6 6 6 .( )= = × =c
cc
px t ty
k h h pp therefore in order to find the value of y, the
quantities t, h and pc may be determined analytically or measured from the drawing similarto Fig. 28.12. It may be noted that if the gear is enlarged, the distances t, h and pc will each increaseproportionately. Therefore the value of y will remain unchanged. A little consideration will showthat the value of y is independent of the size of the tooth and depends only on the number of teethon a gear and the system of teeth. The value of y in terms of the number of teeth may be expressedas follows :
y =0.684
0.124 – ,T
for 1214 °/ composite and full depth involute system.
=0.912
0.154 – ,T
for 20° full depth involute system.
=0.841
0.175 – ,T
for 20° stub system.
28.1828.1828.1828.1828.18 PPPPPererererermissible missible missible missible missible WWWWWorororororking Strking Strking Strking Strking Stress fess fess fess fess for Gear or Gear or Gear or Gear or Gear TTTTTeeth in the Leeeth in the Leeeth in the Leeeth in the Leeeth in the Lewis Equawis Equawis Equawis Equawis EquationtiontiontiontionThe permissible working stress (σw) in the Lewis equation depends upon the material for which
an allowable static stress (σo) may be determined. The allowable static stress is the stress at the
Bicycle gear mechanism switches the chain between different sized sprockets at the pedals and onthe back wheel. Going up hill, a small front and a large rear sprocket are selected to reduce the
push required for the rider. On the level, a large front and small rear. sprocket are used to prevent therider having to pedal too fast.
Going up hill On the levelIdler
sprocket
Chain
Gear cable
pulls on
mechanism
Derailleur mechanism
Tensioner
Sprocket
set
Hub
��������� � 1039elastic limit of the material. It is also called the basic stress. In order to account for the dynamiceffects which become more severe as the pitch line velocity increases, the value of σw is reduced.According to the Barth formula, the permissible working stress,
σw = σo × Cv
where σo = Allowable static stress, and
Cv = Velocity factor.
The values of the velocity factor (Cv) are given as follows :
Cv =3
,3 v+
for ordinary cut gears operating at velocities upto 12.5 m / s.
=4.5
,4.5 v+
for carefully cut gears operating at velocities upto 12.5 m/s.
=6
,6 v+ for very accurately cut and ground metallic gears
operating at velocities upto 20 m / s.
=0.75
,0.75 v+ for precision gears cut with high accuracy and
operating at velocities upto 20 m / s.
=0.75
0.25,1 v
+ + for non-metallic gears.
In the above expressions, v is the pitch line velocity in metres per second.
The following table shows the values of allowable static stresses for the different gearmaterials.
TTTTTaaaaable 28.4.ble 28.4.ble 28.4.ble 28.4.ble 28.4. VVVVValues of alloalues of alloalues of alloalues of alloalues of allowwwwwaaaaable stable stable stable stable static strtic strtic strtic strtic stressessessessess.....
Material Allowable static stress (σo) MPa or N/mm2
Cast iron, ordinary 56
Cast iron, medium grade 70
Cast iron, highest grade 105
Cast steel, untreated 140
Cast steel, heat treated 196
Forged carbon steel-case hardened 126
Forged carbon steel-untreated 140 to 210
Forged carbon steel-heat treated 210 to 245
Alloy steel-case hardened 350
Alloy steel-heat treated 455 to 472
Phosphor bronze 84
Non-metallic materials
Rawhide, fabroil 42
Bakellite, Micarta, Celoron 56
Note : The allowable static stress (σo) for steel gears is approximately one-third of the ultimate tensile stregth(σu) i.e. σo = σu / 3.
��������� � 1041where K = A factor depending upon the form of the teeth.
= 0.107, for 1214 ° full depth involute system.
= 0.111, for 20° full depth involute system.
= 0.115 for 20° stub system.
EP = Young's modulus for the material of the pinion in N/mm2.
EG = Young's modulus for the material of gear in N/mm2.
e = Tooth error action in mm.
The maximum allowable tooth error in action (e) depends upon the pitch line velocity (v) andthe class of cut of the gears. The following tables show the values of tooth errors in action (e) for thedifferent values of pitch line velocities and modules.
TTTTTaaaaable 28.6.ble 28.6.ble 28.6.ble 28.6.ble 28.6. VVVVValues of maximalues of maximalues of maximalues of maximalues of maximum alloum alloum alloum alloum allowwwwwaaaaable tooth errble tooth errble tooth errble tooth errble tooth error in action (or in action (or in action (or in action (or in action (eeeee) v) v) v) v) verererererses pitch lineses pitch lineses pitch lineses pitch lineses pitch linevvvvvelocityelocityelocityelocityelocity,,,,, f f f f for wor wor wor wor well cut commerell cut commerell cut commerell cut commerell cut commercial gearcial gearcial gearcial gearcial gearsssss.....
Pitch line Tooth error in Pitch line Tooth error in Pitch line Tooth error invelocity (v) m/s action (e) mm velocity (v) m/s action (e) mm velocity (v) m/s action (e) mm
1.25 0.0925 8.75 0.0425 16.25 0.0200
2.5 0.0800 10 0.0375 17.5 0.0175
3.75 0.0700 11.25 0.0325 20 0.0150
5 0.0600 12.5 0.0300 22.5 0.0150
6.25 0.0525 13.75 0.0250 25 and over 0.0125
7.5 0.0475 15 0.0225
TTTTTaaaaable 28.7.ble 28.7.ble 28.7.ble 28.7.ble 28.7. VVVVValues of tooth erralues of tooth erralues of tooth erralues of tooth erralues of tooth error in action (or in action (or in action (or in action (or in action (eeeee) v) v) v) v) verererererses module.ses module.ses module.ses module.ses module.
Tooth error in action (e) in mm
Module (m) in mm First class Carefully cut gears Precision gearscommercial gears
Upto 4 0.051 0.025 0.0125
5 0.055 0.028 0.015
6 0.065 0.032 0.017
7 0.071 0.035 0.0186
8 0.078 0.0386 0.0198
9 0.085 0.042 0.021
10 0.089 0.0445 0.023
12 0.097 0.0487 0.0243
14 0.104 0.052 0.028
16 0.110 0.055 0.030
18 0.114 0.058 0.032
20 0.117 0.059 0.033
1042 � ���������������������������
28.2028.2028.2028.2028.20 StaStaStaStaStatic tic tic tic tic TTTTTooth Loadooth Loadooth Loadooth Loadooth LoadThe static tooth load (also called beam strength or endurance strength of the tooth) is obtained
by Lewis formula by substituting flexural endurance limit or elastic limit stress (σe) in place ofpermissible working stress (σw).
∴ Static tooth load or beam strength of the tooth,
WS = σe.b.pc.y = σe.b.π m.y
The following table shows the values of flexural endurance limit (σe) for different materials.
TTTTTaaaaable 28.8.ble 28.8.ble 28.8.ble 28.8.ble 28.8. VVVVValues of falues of falues of falues of falues of flelelelelexural endurance limit.xural endurance limit.xural endurance limit.xural endurance limit.xural endurance limit.
Material of pinion and Brinell hardness number Flexural endurancegear (B.H.N.) limit (σe) in MPa
For safety, against tooth breakage, the static tooth load (WS) should be greater than the dynamicload (WD). Buckingham suggests the following relationship between WS and WD.
For shock loads, WS ≥ 1.5 WDNote : For steel, the flexural endurance limit (σe) may be obtained by using the following relation :
σe = 1.75 × B.H.N. (in MPa)
28.2128.2128.2128.2128.21 WWWWWear ear ear ear ear TTTTTooth Loadooth Loadooth Loadooth Loadooth LoadThe maximum load that gear teeth can carry, without premature wear, depends upon the radii of
curvature of the tooth profiles and on the elasticity and surface fatigue limits of the materials. Themaximum or the limiting load for satisfactory wear of gear teeth, is obtained by using the followingBuckingham equation, i.e.
Ww = DP.b.Q.Kwhere Ww = Maximum or limiting load for wear in newtons,
DP = Pitch circle diameter of the pinion in mm,b = Face width of the pinion in mm,Q = Ratio factor
= G
G P
22 . .,
. . 1
TV R
V R T T
× =+ + for external gears
= G
G P
22 . .,
. . – 1 –
TV R
V R T T
× = for internal gears.
V.R. = Velocity ratio = TG / TP,K = Load-stress factor (also known as material combination factor) in
N/mm2.
��������� � 1043The load stress factor depends upon the maximum fatigue limit of compressive stress, the pressure
angle and the modulus of elasticity of the materials of the gears. According to Buckingham, the loadstress factor is given by the following relation :
K =2
P G
( ) sin 1 1
1.4es
E E
σ φ +
where σes = Surface endurance limit in MPa or N/mm2,φ = Pressure angle,
EP = Young's modulus for the material of the pinion in N/mm2, andEG = Young's modulus for the material of the gear in N/mm2.
The values of surface endurance limit (σes) are given in the following table.
TTTTTaaaaable 28.9.ble 28.9.ble 28.9.ble 28.9.ble 28.9. VVVVValues of surfalues of surfalues of surfalues of surfalues of surface endurance limit.ace endurance limit.ace endurance limit.ace endurance limit.ace endurance limit.
Material of pinion Brinell hardness number Surface endurance limitand gear (B.H.N.) (σes) in N/mm2
Grey cast iron 160 630
Semi-steel 200 630
Phosphor bronze 100 630Steel 150 350
200 490240 616280 721300 770320 826350 910
400 1050
Intermediate gear wheels
Blades
Gearwheel to turnthe blades
Driving gear wheel
Clutch lever
Roller
An old model of a lawn-mower
1044 � ���������������������������
Notes : 1. The surface endurance limit for steel may be obtained from the following equation :σes = (2.8 × B.H.N. – 70) N/mm2
2. The maximum limiting wear load (Ww) must be greater than the dynamic load (WD).
28.2228.2228.2228.2228.22 Causes of Gear Causes of Gear Causes of Gear Causes of Gear Causes of Gear TTTTTooth Footh Footh Footh Footh FailurailurailurailurailureeeeeThe different modes of failure of gear teeth and their possible remedies to avoid the failure, are
as follows :1. Bending failure. Every gear tooth acts as a cantilever. If the total repetitive dynamic load
acting on the gear tooth is greater than the beam strength of the gear tooth, then the gear tooth with failin bending, i.e. the gear tooth with break.
In order to avoid such failure, the module and face width of the gear is adjusted so that the beamstrength is greater than the dynamic load.
2. Pitting. It is the surface fatigue failure which occurs due to many repetition of Hertz contactstresses. The failure occurs when the surface contact stresses are higher than the endurance limit ofthe material. The failure starts with the formation of pits which continue to grow resulting in therupture of the tooth surface.
In order to avoid the pitting, the dynamic load between the gear tooth should be less than thewear strength of the gear tooth.
3. Scoring. The excessive heat is generated when there is an excessive surface pressure, highspeed or supply of lubricant fails. It is a stick-slip phenomenon in which alternate shearing and weldingtakes place rapidly at high spots.
This type of failure can be avoided by properly designing the parameters such as speed, pressure andproper flow of the lubricant, so that the temperature at the rubbing faces is within the permissible limits.
4. Abrasive wear. The foreign particles in the lubricants such as dirt, dust or burr enter betweenthe tooth and damage the form of tooth. This type of failure can be avoided by providing filters for thelubricating oil or by using high viscosity lubricant oil which enables the formation of thicker oil filmand hence permits easy passage of such particles without damaging the gear surface.
5. Corrosive wear. The corrosion of the tooth surfaces is mainly caused due to the presence ofcorrosive elements such as additives present in the lubricating oils. In order to avoid this type of wear,proper anti-corrosive additives should be used.
28.2328.2328.2328.2328.23 Design PrDesign PrDesign PrDesign PrDesign Procedurocedurocedurocedurocedure fe fe fe fe for Spur Gearor Spur Gearor Spur Gearor Spur Gearor Spur GearsssssIn order to design spur gears, the following procedure may be followed :
1. First of all, the design tangential tooth load is obtained from the power transmitted and thepitch line velocity by using the following relation :
WT = SP
Cv
× ...(i)
where WT = Permissible tangential tooth load in newtons,P = Power transmitted in watts,
*v = Pitch line velocity in m / s ,60
D Nπ=D = Pitch circle diameter in metres,
* We know that circular pitch,
pc = π D / T = π m ...(∵ m = D / T)∴ D = m.TThus, the pitch line velocity may also be obtained by using the following relation, i.e.
v =. .. . .
60 60 60π π= = cp T ND N m T N
where m = Module in metres, andT = Number of teeth.
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��������� � 1045N = Speed in r.p.m., and
CS = Service factor.
The following table shows the values of service factor for different types of loads :
TTTTTaaaaable 28.10.ble 28.10.ble 28.10.ble 28.10.ble 28.10. VVVVValues of seralues of seralues of seralues of seralues of service fvice fvice fvice fvice factoractoractoractoractor.....
Type of serviceType of load
Intermittent or 3 hours 8-10 hours per day Continuous 24 hoursper day per day
Steady 0.8 1.00 1.25
Light shock 1.00 1.25 1.54
Medium shock 1.25 1.54 1.80
Heavy shock 1.54 1.80 2.00
Note : 1. The above values for service factor are for enclosed well lubricated gears. In case of non-enclosed andgrease lubricated gears, the values given in the above table should be divided by 0.65.
2. Apply the Lewis equation as follows :
WT = σw.b.pc.y = σw.b.π m.y
= (σo.Cv) b.π m.y ...(∵ σw = σo.Cv)
Notes : (i) The Lewis equation is applied only to the weaker of the two wheels (i.e. pinion or gear).
(ii) When both the pinion and the gear are made of the same material, then pinion is the weaker.
(iii) When the pinion and the gear are made of different materials, then the product of (σw × y) or (σo × y)is the *deciding factor. The Lewis equation is used to that wheel for which (σw × y) or (σo × y) is less.
* We see from the Lewis equation that for a pair of mating gears, the quantities like WT, b, m and Cv areconstant. Therefore (σw × y) or (σo × y) is the only deciding factor.
A bicycle with changeable gears.
1046 � ���������������������������
(iv) The product (σw × y) is called strength factor of the gear.
(v) The face width (b) may be taken as 3 pc to 4 pc (or 9.5 m to 12.5 m) for cut teeth and 2 pc to 3 pc (or 6.5 m to 9.5 m) for cast teeth.
3. Calculate the dynamic load (WD) on the tooth by using Buckingham equation, i.e.
WD = WT + WI
= TT
T
21 ( . )
21 .
v b C WW
v b C W
++
+ +In calculating the dynamic load (WD), the value of tangential load (WT) may be calculated by
neglecting the service factor (CS) i.e.
WT = P / v, where P is in watts and v in m / s.
4. Find the static tooth load (i.e. beam strength or the endurance strength of the tooth) by usingthe relation,
WS = σe.b.pc.y = σe.b.π m.y
For safety against breakage, WS should be greater than WD.
5. Finally, find the wear tooth load by using the relation,
Ww = DP.b.Q.K
The wear load (Ww) should not be less than the dynamic load (WD).
Example 28.1. The following particulars of a single reduction spur gear are given :
Gear ratio = 10 : 1; Distance between centres = 660 mm approximately; Pinion transmits 500kW at 1800 r.p.m.; Involute teeth of standard proportions (addendum = m) with pressure angle of22.5°; Permissible normal pressure between teeth = 175 N per mm of width. Find :
1. The nearest standard module if no interference is to occur;
2. The number of teeth on each wheel;
3. The necessary width of the pinion; and
4. The load on the bearings of the wheels due to power transmitted.
Solution : Given : G = TG / TP = DG / DP = 10 ; L = 660 mm ; P = 500 kW = 500 × 103 W ;NP = 1800 r.p.m. ; φ = 22.5° ; WN = 175 N/mm width
1. Nearest standard module if no interference is to occurLet m = Required module,
TP = Number of teeth on the pinion,
TG = Number of teeth on the gear,
DP = Pitch circle diameter of the pinion, and
DG = Pitch circle diameter of the gear.
We know that minimum number of teeth on the pinion in order to avoid interference,
TP = W
2
2
1 11 2 sin – 1
+ + φ
A
GG G
=2
2 1 213.3 say 14
0.151 110 1 2 sin 22.5 – 1
10 10
× = = + + °
... (∵ AW = 1 module)
∴ TG = G × TP = 10 × 14 = 140 ...(∵ TG / TP = 10)
��������� � 1047
We know that L = G GP PP
105.5
2 2 2 2
D DD DD+ = + = ...(∵ DG / DP = 10)
∴ 660 = 5.5 DP or DP = 660 / 5.5 = 120 mm
We also know that DP = m . TP
∴ m = DP / TP = 120 / 14 = 8.6 mm
Since the nearest standard value of the module is 8 mm, therefore we shall take
m = 8 mm Ans.2. Number of teeth on each wheel
We know that number of teeth on the pinion,
TP = DP / m = 120 / 8 = 15 Ans.and number of teeth on the gear,
TG = G × TP = 10 × 15 = 150 Ans.3. Necessary width of the pinion
We know that the torque acting on the pinion,
T =3
P
60 500 10 602652 N-m
2 2 1800
P
N
× × ×= =π π ×
∴ Tangential load, WT =P
265244 200 N
/ 2 0.12 / 2
T
D= = ...(∵ DP is taken in metres)
and normal load on the tooth,
WN = T 44 20047 840 N
cos cos 22.5
W = =φ °
Since the normal pressure between teeth is 175 N per mm of width, therefore necessary width ofthe pinion,
b =47 840
273.4 mm175
= AAns.
4. Load on the bearings of the wheels
We know that the radial load on the bearings due to the power transmitted.WR = WN . sin φ = 47 840 × sin 22.5° = 18 308 N = 18.308 kN Ans.
Example 28.2. A bronze spur pinion rotating at 600 r.p.m. drives a cast iron spur gear at atransmission ratio of 4 : 1. The allowable static stresses for the bronze pinion and cast iron gear are84 MPa and 105 MPa respectively.
The pinion has 16 standard 20° full depth involute teeth of module 8 mm. The face width ofboth the gears is 90 mm. Find the power that can be transmitted from the standpoint of strength.
Solution. Given : NP = 600 r.p.m. ; V.R. = TG / TP = 4 ; σOP = 84 MPa = 84 N / mm2 ;σOG = 105 MPa = 105 N/mm2 ; TP = 16 ; m = 8 mm ; b = 90 mm
We know that pitch circle diameter of the pinion,
DP = m.TP = 8 × 16 = 128 mm = 0.128 m
∴ Pitch line velocity,
v = P P. 0.128 6004.02 m/s
60 60
D Nπ π × ×= =
Since the pitch line velocity (v) is less than 12.5 m/s, therefore velocity factor,
Cv =3 3
0.4273 3 4.02v
= =+ +
1048 � ���������������������������
We know that for 20° full depth involute teeth, tooth form factor for the pinion,
yP =P
0.912 0.9120.154 – 0.154 – 0.097
16T= =
and tooth form factor for the gear,
yG =G
0.912 0.9120.154 – 0.154 – 0.14
4 16T= =
×... (∵ TG / TP = 4)
∴ σOP × yP = 84 × 0.097 = 8.148
and σOG × yG = 105 × 0.14 = 14.7
Since (σOP × yP) is less than (σOG × yG), therefore the pinion is weaker. Now using the Lewisequation for the pinion, we have tangential load on the tooth (or beam strength of the tooth),
= WT × v = 7870 × 4.02 = 31 640 W = 31.64 kW Ans.Example 28.3. A pair of straight teeth spur gears is to transmit 20 kW when the pinion rotates
at 300 r.p.m. The velocity ratio is 1 : 3. The allowable static stresses for the pinion and gear materialsare 120 MPa and 100 MPa respectively.
The pinion has 15 teeth and its face width is 14 times the module. Determine : 1. module;2. face width; and 3. pitch circle diameters of both the pinion and the gear from the standpoint ofstrength only, taking into consideration the effect of the dynamic loading.
Assuming steady load conditions and 8-10 hours of service per day, the service factor (CS) fromTable 28.10 is given by
CS = 1
We know that design tangential tooth load,
WT =3
S20 10 84 746
1 N0.236
PC
v m m
×× = × =
and velocity factor, Cv =3 3
3 3 0.236v m=
+ +
��������� � 1049We know that tooth form factor for the pinion,
yP =P
0.912 0.9120.154 – 0.154 –
15T=
= 0.154 – 0.0608 = 0.0932and tooth form factor for the gear,
yG =G
0.912 0.9120.154 – 0.154 –
3 15T=
×= 0.154 – 0.203 = 0.1337 ... (∵ TG = 3TP)
∴ σOP × yP = 120 × 0.0932 = 11.184
and σOG × yG = 100 × 0.1337 = 13.37
Since (σOP × yP) is less than (σOG × yG), therefore the pinion is weaker. Now using the Lewisequation to the pinion, we have
WT = σwP.b.π m.yP = (σOP × Cv) b.π m . yP
∴84 746
m=
23 1476120 14 0.0932
3 0.236 3 0.236
mm m
m m × π × = + +
or 3 + 0.236 m = 0.0174 m3
Solving this equation by hit and trial method, we find thatm = 6.4 mm
The standard module is 8 mm. Therefore let us takem = 8 mm Ans.
2. Face widthWe know that the face width,
b = 14 m = 14 × 8 = 112 mm Ans.
Kitchen Gear :Kitchen Gear :Kitchen Gear :Kitchen Gear :Kitchen Gear : This 1863 fruit and vegetable peeling machine uses a rack and pinion to drive spurgears that turn an apple against a cutting blade. As the handle is pushed round the semi-circular
base, the peel is removed from the apple in a single sweep.
Pinion
HandleSpurgearCutting
blade
Semi-c i r c u l a rrack
1050 � ���������������������������
3. Pitch circle diameter of the pinion and gearWe know that pitch circle diameter of the pinion,
DP = m.TP = 8 × 15 = 120 mm Ans.and pitch circle diameter of the gear,
DG = m.TG = 8 × 45 = 360 mm Ans. ... (∵ TG = 3 TP)
Example 28.4. A gear drive is required to transmit a maximum power of 22.5 kW. The velocityratio is 1:2 and r.p.m. of the pinion is 200. The approximate centre distance between the shafts maybe taken as 600 mm. The teeth has 20° stub involute profiles. The static stress for the gear material(which is cast iron) may be taken as 60 MPa and face width as 10 times the module. Find the module,face width and number of teeth on each gear.
Check the design for dynamic and wear loads. The deformation or dynamic factor in theBuckingham equation may be taken as 80 and the material combination factor for the wear as 1.4.
Solution. Given : P = 22.5 kW = 22 500 W ; V.R.= DG/DP = 2 ; NP = 200 r.p.m. ; L = 600 mm ;σOP = σOG = 60 MPa = 60 N/mm2 ; b = 10 m ; C = 80 ; K = 1.4
ModuleLet m = Module in mm,
DP = Pitch circle diameter of the pinion, and
DG = Pitch circle diameter of the gear.
We know that centre distance between the shafts (L),
600 = GP P PP
21.5
2 2 2 2
DD D DD+ = + = ... (∵ DG = V.R. × DP)
Arm of a material handler In addition to gears, hydraulic rams as shown above, play important role intransmitting force and energy.
��������� � 1051∴ DP = 600 / 1.5 = 400 mm = 0.4 m
and DG = 2 DP = 2 × 400 = 800 mm = 0.8 m
Since both the gears are made of the same material, therefore pinion is the weaker. Thus thedesign will be based upon the pinion.
We know that pitch line velocity of the pinion,
v = P P. 0.4 2004.2 m / s
60 60
π π × ×= =D N
Since v is less than 12 m / s, therefore velocity factor,
Cv =3 3
0.4173 3 4.2v
= =+ +
We know that number of teeth on the pinion,
TP = DP / m = 400 / m
∴ Tooth form factor for the pinion,
yP =P
0.841 0.8410.175 – 0.175 –
400
m
T
×= ... (For 20° stub system)
= 0.175 – 0.0021 m ...(i)Assuming steady load conditions and 8–10 hours of service per day, the service factor (CS)
from Table 28.10 is given by
CS = 1
We know that design tangential tooth load,
WT = S22 500
1 5357 N4.2
PC
v× = × =
We also know that tangential tooth load (WT),
5357 = σwP . b. π m.yP =(σOP × Cv) b.π m.yP
= (60 × 0.417) 10 m × π m (0.175 – 0.0021 m)
= 137.6 m2 – 1.65 m3
Solving this equation by hit and trial method, we find that
m = 0.65 say 8 mm Ans.Face width
We know that face width,
b = 10 m = 10 × 8 = 80 mm Ans.Number of teeth on the gears
We know that number of teeth on the pinion,
TP = DP / m = 400 / 8 = 50 Ans.and number of teeth on the gear,
TG = DG / m = 800 / 8 = 100 Ans.Checking the gears for dynamic and wear load
We know that the dynamic load,
WD = TT
T
21 ( . )
21 .
v b C WW
v b C W
+++ +
=21 4.2 (80 80 5357)
535721 4.2 80 80 5357
× × ++× + × +
1052 � ���������������������������
=61.037 10
5357 5357 5273 10 630 N196.63
×+ = + =
From equation (i), we find that tooth form factor for the pinion,
From Table 28.8, we find that flexural endurance limit (σe) for cast iron is 84 MPa or 84 N/mm2.
∴ Static tooth load or endurance strength of the tooth,
WS = σe . b. π m. yP = 84 × 80 × π × 8 × 0.1582 = 26 722 N
We know that ratio factor,
Q =2 . . 2 2
1.33. . 1 2 1
V R
V R
× ×= =+ +
∴ Maximum or limiting load for wear,
Ww = DP . b. Q . K = 400 × 80 × 1.33 × 1.4 = 59 584 NSince both WS and Ww are greater than WD, therefore the design is safe.
Example 28.5. A pair of straight teeth spur gears, having 20° involute full depth teeth is totransmit 12 kW at 300 r.p.m. of the pinion. The speed ratio is 3 : 1. The allowable static stresses forgear of cast iron and pinion of steel are 60 MPa and 105 MPa respectively. Assume the following:
Number of teeth of pinion = 16; Face width = 14 times module; Velocity factor (Cv ) = 4.5
4.5 + v,
v being the pitch line velocity in m / s; and tooth form factor (y) = –0.912
0.154No. of teeth
Determine the module, face width and pitch diameter of gears. Check the gears for wear; givenσes = 600 MPa; EP = 200 kN/mm2 and EG = 100 kN/mm2. Sketch the gears.
Solving this equation by hit and trial method, we find thatm = 5.6 say 6 mm Ans.
Face widthWe know that face width,
b = 14 m = 14 × 6 = 84 mm Ans.Pitch diameter of gears
We know that pitch diameter of the pinion,DP = m.TP = 6 × 16 = 96 mm Ans.
and pitch diameter of the gear,DG = m.TG = 6 × 48 = 288 mm Ans. ...(∵ TG = 3 TP)
This is a close-up photo (magnified 200 times) of a micromotor’s gear cogs. Micromotorshave been developed for use in space missions and microsurgery.
1054 � ���������������������������
Checking the gears for wear
We know that the ratio factor,
Q =2 . . 2 3
1.5. . 1 3 1
V R
V R
× ×= =+ +
and load stress factor, K =2
P G
( ) sin 1 1
1.4es
E E
σ φ +
=2
3 3
(600) sin 20 1 1
1.4 200 10 100 10
° + × × = 0.44 + 0.88 = 1.32 N/mm2
We know that the maximum or limiting load for wear,
Ww = DP.b.Q.K = 96 × 84 × 1.5 × 1.32 = 15 967 N
and tangential load on the tooth (or beam strength of the tooth),
WT =3 347.8 10 47.8 10
7967 N6m
× ×= =
Since the maximum wear load is much more than the tangential load on the tooth, therefore thedesign is satisfactory from the standpoint of wear. Ans.
Example 28.6. A reciprocating compressor is to be connected to an electric motor with thehelp of spur gears. The distance between the shafts is to be 500 mm. The speed of the electric motoris 900 r.p.m. and the speed of the compressor shaft is desired to be 200 r.p.m. The torque, to betransmitted is 5000 N-m. Taking starting torque as 25% more than the normal torque, determine :1. Module and face width of the gears using 20 degrees stub teeth, and 2. Number of teeth and pitchcircle diameter of each gear. Assume suitable values of velocity factor and Lewis factor.
Solution. Given : L = 500 mm ; NM = 900 r.p.m. ; NC = 200 r.p.m. ; T = 5000 N-m ; Tmax = 1.25 T1. Module and face width of the gears
Let m = Module in mm, andb = Face width in mm.
Since the starting torque is 25% more than the normal torque, therefore the maximum torque,Tmax = 1.25 T = 1.25 × 5000 = 6250 N-m = 6250 × 103 N-mm
We know that velocity ratio,
V.R. = M
C
9004.5
200
N
N= =
Let DP = Pitch circle diameter of the pinion on the motor shaft, andDG = Pitch circle diameter of the gear on the compressor shaft.
We know that distance between the shafts (L),
500 = GP
2 2
DD + or DP + DG = 500 × 2 = 1000 ...(i)
and velocity ratio, V.R. = G
P
4.5D
D= or DG = 4.5 DP ...(ii)
Substituting the value of DG in equation (i), we haveDP + 4.5 DP = 1000 or DP = 1000 / 5.5 = 182 mm
and DG = 4.5 DP = 4.5 × 182 = 820 mm = 0.82 mWe know that pitch line velocity of the drive,
v = G C. 0.82 2008.6 m / s
60 60
π π × ×= =D N
��������� � 1055∴ Velocity factor,
Cv =3 3
0.263 3 8.6v
= =+ +
...(∵ v is less than 12.5 m/s)
Let us assume than motor pinion is made of forged steel and the compressor gear of cast steel.Since the allowable static stress for the cast steel is less than the forged steel, therefore the designshould be based upon the gear. Let us take the allowable static stress for the gear material as
σOG = 140 MPa = 140 N/mm2
We know that for 20° stub teeth, Lewis factor for the gear,
yG =G G
0.841 0.8410.175 – 0.175 –
×= m
T DG
G... = ∵
DT
m
=0.841
0.175 – 0.175 – 0.001820
mm=
and maximum tangential force on the gear,
WT =3
G
2 2 6250 1015 244 N
820maxT
D
× ×= =
We also know that maximum tangential force on the gear,
WT = σwG.b.π m.yG = (σOG × Cν) b × π m × yG ...(∵ σwG = σOG. Cν)
15 244 = (140 × 0.26) × 10 m × π m (0.175 – 0.001 m)
= 200 m2 – 1.144 m3 ...(Assuming b = 10 m)
Solving this equation by hit and trial method, we find that
m = 8.95 say 10 mm Ans.and b = 10 m = 10 × 10 = 100 mm Ans.2. Number of teeth and pitch circle diameter of each gear
We know that number of teeth on the pinion,
TP = P 18218.2
10
D
m= =
TG = G 82082
10
D
m= =
1056 � ���������������������������
In order to have the exact velocity ratio of 4.5, we shall takeTP = 18 and TG = 81 Ans.
∴ Pitch circle diameter of the pinion,DP = m × TP = 10 × 18 = 180 mm Ans.
and pitch circle diameter of the gear,DG = m × TG = 10 × 81 = 810 mm Ans.
28.2428.2428.2428.2428.24 Spur Gear ConstructionSpur Gear ConstructionSpur Gear ConstructionSpur Gear ConstructionSpur Gear ConstructionThe gear construction may have different designs depending upon the size and its application.
When the dedendum circle diameter is slightly greater than the shaft diameter, then the pinion teethare cut integral with the shaft as shown in Fig. 28.13 (a). If the pitch circle diameter of the pinion isless than or equal to 14.75 m + 60 mm (where m is the module in mm), then the pinion is made solidwith uniform thickness equal to the face width, as shown in Fig. 28.13 (b). Small gears upto 250 mmpitch circle diameter are built with a web, which joins the hub and the rim. The web thickness isgenerally equal to half the circular pitch or it may be taken as 1.6 m to 1.9 m, where m is the module.The web may be made solid as shown in Fig. 28.13 (c) or may have recesses in order to reduce itsweight.
Fig. 28.13. Construction of spur gears.
Fig. 28.14. Gear with arms.
��������� � 1057Large gears are provided with arms to join the hub and the rim, as shown in Fig. 28.14. The
number of arms depends upon the pitch circle diameter of the gear. The number of arms may beselected from the following table.
TTTTTaaaaable 28.11.ble 28.11.ble 28.11.ble 28.11.ble 28.11. Number of ar Number of ar Number of ar Number of ar Number of arms fms fms fms fms for the gearor the gearor the gearor the gearor the gearsssss.....
S. No. Pitch circle diameter Number of arms
1. Up to 0.5 m 4 or 5
2. 0.5 – 1.5 m 6
3. 1.5 – 2.0 m 8
4. Above 2.0 m 10
The cross-section of the arms is most often elliptical, but other sections as shown in Fig. 28.15may also be used.
Fig. 28.15. Cross-section of the arms.
The hub diameter is kept as 1.8 times the shaft diameter for steel gears, twice the shaft diameterfor cast iron gears and 1.65 times the shaft diameter for forged steel gears used for light service. Thelength of the hub is kept as 1.25 times the shaft diameter for light service and should not be less thanthe face width of the gear.
The thickness of the gear rim should be as small as possible, but to facilitate casting and toavoid sharp changes of section, the minimum thickness of the rim is generally kept as half of thecircular pitch (or it may be taken as 1.6 m to 1.9 m, where m is the module). The thickness of rim (tR)may also be calculated by using the following relation, i.e.
tR =T
mn
where T = Number of teeth, and
n = Number of arms.
The rim should be provided with a circumferential rib of thickness equal to the rim thickness.
Krishna
Highlight
1058 � ���������������������������
28.2528.2528.2528.2528.25 Design of Shaft for Spur GearsDesign of Shaft for Spur GearsDesign of Shaft for Spur GearsDesign of Shaft for Spur GearsDesign of Shaft for Spur GearsIn order to find the diameter of shaft for spur gears,
the following procedure may be followed.
1. First of all, find the normal load (WN), actingbetween the tooth surfaces. It is given by
WN = WT / cos φwhere WT = Tangential load, and
φ = Pressure angle.
A thrust parallel and equal to WN will act at thegear centre as shown in Fig. 28.16.
2. The weight of the gear is given byWG = 0.001 18 TG.b.m2 (in N)
where TG = No. of teeth on the gear,b = Face width in mm, andm = Module in mm.
3. Now the resultant load acting on the gear,
WR = 2 2N G N G( ) ( ) 2 cosW W W W+ + × φ
4. If the gear is overhung on the shaft, then bending moment on the shaft due to the resultantload,
M = WR × xwhere x = Overhang i.e. the distance between the centre of gear and the centre
of bearing.5. Since the shaft is under the combined effect of torsion and bending, therefore we shall
determine the equivalent torque. We know that equivalent torque,
Te = 2 2M T+where T = Twisting moment = WT × DG / 2
6. Now the diameter of the gear shaft (d ) is determined by using the following relation, i.e.
Te =16
d 3π × τ ×
where τ = Shear stress for the material of the gear shaft.Note : Proceeding in the similar way as discussed above, we may calculate the diameter of the pinion shaft.
28.2628.2628.2628.2628.26 Design of Design of Design of Design of Design of ArArArArArms fms fms fms fms for Spur Gearor Spur Gearor Spur Gearor Spur Gearor Spur GearsssssThe cross-section of the arms is calculated by assuming them as a cantilever beam fixed at the
hub and loaded at the pitch circle. It is also assumed that the load is equally distributed to all the arms.It may be noted that the arms are designed for the stalling load. The stalling load is a load that willdevelop the maximum stress in the arms and in the teeth. This happens at zero velocity, when thedrive just starts operating.
The stalling load may be taken as the design tangential load divided by the velocity factor.
Let WS = TDesign tangential loadStalling load =
Velocity factor v
W
C= ,
DG = Pitch circle diameter of the gear,
n = Number of arms, and
σb = Allowable bending stress for the material of the arms.
Fig. 28.16. Load acting on the gear.
��������� � 1059Now, maximum bending moment on each arm,
M = S G S G/ 2
2
W D W D
n n
× ×=
and the section modulus of arms for elliptical cross-section,
Z =2
1 1( )
32
a bπ
where a1 = Major axis, and b1 = Minor axis.
The major axis is usually taken as twice the minor axis. Now, using the relation, σb = M / Z,we can calculate the dimensions a1 and b1 for the gear arm at the hub end.Note : The arms are usually tapered towards the rim about 1/16 per unit length of the arm (or radius of the gear).
∴ Major axis of the section at the rim end
= G G
1 1 1 11 1
– Taper = – Length of the arm = – –16 16 2 32
× × =D Da a a a
Example 28.7. A motor shaft rotating at 1500 r.p.m. has to transmit 15 kW to a low speed shaft
with a speed reduction of 3:1. The teeth are °1214 / involute with 25 teeth on the pinion. Both the
pinion and gear are made of steel with a maximum safe stress of 200 MPa. A safe stress of 40 MPamay be taken for the shaft on which the gear is mounted and for the key.
Design a spur gear drive to suit the above conditions. Also sketch the spur gear drive. Assumestarting torque to be 25% higher than the running torque.
Solution : Given : NP = 1500 r.p.m. ; P = 15 kW = 15 × 103 W ; V.R. = TG/TP = 3 ; φ = 1214 °/ ;
Design for spur gearsSince the starting torque is 25% higher than the running torque, therefore the spur gears should
be designed for power,
P1 = 1.25 P = 1.25 × 15 × 103 = 18 750 W
We know that the gear reduction ratio (TG / TP) is 3. Therefore the number of teeth on the gear,
TG = 3 TP = 3 × 25 = 75Let us assume that the module (m) for the pinion and gear is 6 mm.
∴ Pitch circle diameter of the pinion,
DP = m.TP = 6 × 25 = 150 mm = 0.15 m
and pitch circle diameter of the gear,
DG = m. TG = 6 × 75 = 450 mm
We know that pitch line velocity,
v = P P. 0.15 150011.8 m / s
60 60
π π × ×= =D N
Assuming steady load conditions and 8–10 hours of service per day, the service factor (CS)from Table 28.10 is given by
CS = 1
∴ Design tangential tooth load,
WT = 1S
18 7501 1590 N
11.8
PC
v× = × =
We know that for ordinary cut gears and operating at velocities upto 12.5 m/s, the velocityfactor,
Cv = 3 3
0.2033 3 11.8v
= =+ +
1060 � ���������������������������
Since both the pinion and the gear are made ofthe same material, therefore the pinion is the weaker.
We know that for 1214 °/ involute teeth, tooth form
factor for the pinion,
yP = P
0.684 0.6840.124 – 0.124 – 0.0966
25T= =
Let b =Face width for both the pinion and gear.
We know that the design tangential tooth load(WT),
1590 = σwP.b.π m.yP = (σOP.Cv) b.π m.yP
= (200 × 0.203) b × π × 6 × 0.0966 = 74 b
∴ b = 1590 / 74 = 21.5 mm
In actual practice, the face width (b) is taken as9.5 m to 12.5 m, but in certain cases, due to spacelimitations, it may be taken as 6 m. Therefore let ustake the face width,
b = 6 m = 6 × 6 = 36 mm Ans.From Table 28.1, the other proportions, for the
pinion and the gear having 1214 °/ involute teeth, are as
follows :
Addendum = 1 m = 6 mm Ans.Dedendum = 1.25 m = 1.25 × 6 = 7.5 mm Ans. Working depth = 2 m = 2 × 6 = 12 mm Ans.Minimum total depth = 2.25 m = 2.25 × 6 = 13.5 mm Ans.Tooth thickness = 1.5708 m = 1.5708 × 6 = 9.4248 mm Ans.Minimum clearance = 0.25 m = 0.25 × 6 = 1.5 mm Ans.
Design for the pinion shaft
We know that the normal load acting between the tooth surfaces,
= 2 2 12(1643) (38) 2 1643 38 cos 14 / °+ + × × × = 1680 N
Assuming that the pinion is overhung on the shaft and taking overhang as 100 mm, therefore
Bending moment on the shaft due to the resultant load,
M = WR × 100 = 1680 × 100 = 168 00 N-mm
* Since the weight of the pinion (WP) is very small as compared to the normal load (WN), therefore it may beneglected. Thus the resultant load acting on the pinion (WR) may be taken equal to WN.
This mathematical machine calleddifference engine, assembled in 1832,
∴ (dP)3 = 206 × 103 / 7.855 = 26.2 × 103 or dP = 29.7 say 30 mm Ans.
We know that the diameter of the pinion hub
= 1.8 dP = 1.8 × 30 = 54 mm Ans.
and length of the hub = 1.25 dP = 1.25 × 30 = 37.5 mm
Since the length of the hub should not be less than that of the face width i.e. 36 mm, therefore letus take length of the hub as 36 mm. Ans.Note : Since the pitch circle diameter of the pinion is 150 mm, therefore the pinion should be provided with aweb and not arms. Let us take thickness of the web as 1.8 m, where m is the module.
∴ Thickness of the web = 1.8 m = 1.8 × 6 = 10.8 mm Ans.Design for the gear shaft
We have calculated above that the normal load acting between the tooth surfaces,
WN = 1643 N
We know that weight of the gear,
WG = 0.001 18 TG.b.m2 = 0.001 18 × 45 × 36 × 62 = 69 N
Assuming that the gear is overhung on the shaft and taking the overhang as 100 mm, thereforebending moment on the shaft due to the resultant load,
M = WR × 100 = 1710 × 100 = 171 000 N-mm
and twisting moment on the shaft,
T = GT
4501590 357 750 N-mm
2 2
DW × = × =
∴ Equivalent twisting moment,
Te = 2 2 2 2 3(171 000) (357 750) 396 10 N-mmM T+ = + = ×Let dG = Diameter of the gear shaft.
We know that equivalent twisting moment (Te),
396 × 103 =3 3 3
G G G( ) 40 ( ) 7.855 ( )16 16
d d dπ π× τ = × =
∴ (dG)3 = 396 × 103 / 7.855 = 50.4 × 103 or dG = 37 say 40 mm Ans.
1062 � ���������������������������
We know that diameter of the gear hub
= 1.8 dG = 1.8 × 40 = 72 mm Ans.and length of the hub = 1.25 dG = 1.25 × 40 = 50 mm Ans.Design for the gear arms
Since the pitch circle diameter of the gear is 450 mm, therefore the gear should be providedwith four arms. Let us assume the cross-section of the arms as elliptical with major axis (a1) equal totwice the minor axis (b1).
∴ Section modulus of arms,
Z =2 2
31 1 1 11
( ) ( )0.05 ( )
32 32 2
a b a aa
π π= × = ... (∵ b1 = a1/2)
Since the arms are designed for the stalling load and stalling load is taken as the design tangentialload divided by the velocity factor, therefore stalling load,
WS = T 15907830 N
0.203v
W
C= = ... (∵ Cv = 0.203)
∴ Maximum bending moment on each arm,
M = S G 7830 450440 440 N-mm
2 4 2
W D
n× = × =
We know that bending stress (σb),
42 =6
3 31 1
440 440 9 10
0.05 ( ) ( )
M
Z a a
×= = ... (Taking σb = 42 N/mm2)
∴ (a1)3 = 9 × 106 / 42 = 0.214 × 106 or a1 = 60 mm Ans.and b1 = a1 / 2 = 60 / 2 = 30 mm Ans.These dimensions refer to the hub end. Since the arms are tapered towards the rim and the taper
is 1 / 16 per unit length of the arm (or radius of the gear), thereforeMajor axis of the arm at the rim end,
a2 = G1 1
1– Taper = –
16 2
Da a ×
=1 450
60 – 46 mm16 2
× = Ans.
and minor axis of the arm at the rim end,
b2 =Major axis 46
23 mm2 2
= = Ans.
Design for the rimThe thickness of the rim for the pinion (tRP) may be taken as 1.6 m to 1.9 m, where m is the
module. Let us take thickness of the rim for the pinion,tRP = 1.6 m = 1.6 × 6 = 9.6 say 10 mm Ans.
The thickness of the rim for the gear (tRG) may be obtained by using the relation,
tRG = G 456 20 mm
4
Tm
n= = Ans.
EEEEEXEXEXEXEXERRRRRCISECISECISECISECISESSSSS
1. Calculate the power that can be transmitted safely by a pair of spur gears with the data given below.Calculate also the bending stresses induced in the two wheels when the pair transmits this power.
Number of teeth in the pinion = 20
Number of teeth in the gear = 80
��������� � 1063Module = 4 mm
Width of teeth = 60 mm
Tooth profile = 20° involute
Allowable bending strength of the material
= 200 MPa, for pinion
= 160 MPa, for gear
Speed of the pinion = 400 r.p.m.
Service factor = 0.8
Lewis form factor =0.912
0.154 –T
Velocity factor =3
3 v+ [Ans. 13.978 kW ; 102.4 MPa ; 77.34 MPa]
2. A spur gear made of bronze drives a mid steel pinion with angular velocity ratio of 123 / : 1. The
pressure angle is 1214 °/ . It transmits 5 kW at 1800 r.p.m. of pinion. Considering only strength, design
the smallest diameter gears and find also necessary face width. The number of teeth should not be lessthan 15 teeth on either gear. The elastic strength of bronze may be taken as 84 MPa and of steel as 105
MPa. Lewis factor for 1214 °/ pressure angle may be taken as
y =0.684
0.124 –No. of teeth
[Ans. m = 3 mm ; b = 35 mm ; DP = 48 mm ; DG = 168 mm]3. A pair of 20° full-depth involute tooth spur gears is to transmit 30 kW at a speed of 250 r.p.m. of the
pinion. The velocity ratio is 1 : 4. The pinion is made of cast steel having an allowable static stress, σo= 100 MPa, while the gear is made of cast iron having allowable static stress, σo = 55 MPa.
The pinion has 20 teeth and its face width is 12.5 times the module. Determine the module, face widthand pitch diameters of both the pinion and gear from the standpoint of strength only taking velocityfactor into consideration. The tooth form factor is given by the expression
y =0.912
0.154 –No. of teeth
and velocity factor is given by
Cv =3
,3 + v where v is the peripheral speed of the gear in m/s.
[Ans. m = 20 mm ; b = 250 mm ; DP = 400 mm ; DG = 1600 mm]4. A micarta pinion rotating at 1200 r.p.m. is to transmit 1 kW to a cast iron gear at a speed of 192 r.p.m.
Assuming a starting overload of 20% and using 20° full depth involute teeth, determine the module,number of teeth on the pinion and gear and face width. Take allowable static strength for micarta as 40MPa and for cast iron as 53 MPa. Check the pair in wear.
5. A 15 kW and 1200 r.p.m. motor drives a compressor at 300 r.p.m. through a pair of spur gears having20° stub teeth. The centre to centre distance between the shafts is 400 mm. The motor pinion is madeof forged steel having an allowable static stress as 210 MPa, while the gear is made of cast steelhaving allowable static stress as 140 MPa. Assuming that the drive operates 8 to 10 hours per dayunder light shock conditions, find from the standpoint of strength,
1. Module; 2. Face width and 3. Number of teeth and pitch circle diameter of each gear.
Check the gears thus designed from the consideration of wear. The surface endurance limit may betaken as 700 MPa.[Ans. m = 6 mm ; b = 60 mm ; TP = 24 ; TG = 96 ; DP = 144mm ; DG = 576 mm]
6. A two stage reduction drive is to be designed to transmit 2 kW; the input speed being 960 r.p.m. andoverall reduction ratio being 9. The drive consists of straight tooth spur gears only, the shafts beingspaced 200 mm apart, the input and output shafts being co-axial.
1064 � ���������������������������
(a) Draw a layout of a suitable system to meet the above specifications, indicating the speeds of allrotating components.
(b) Calculate the module, pitch diameter, number of teeth, blank diameter and face width of thegears for medium heavy duty conditions, the gears being of medium grades of accuracy.
(c) Draw to scale one of the gears and specify on the drawing the calculated dimensions and otherdata complete in every respect for manufacturing purposes.
7. A motor shaft rotating at 1440 r.p.m. has to transmit 15 kW to a low speed shaft rotating at 500 r.p.m.The teeth are 20° involute with 25 teeth on the pinion. Both the pinion and gear are made of cast ironwith a maximum safe stress of 56 MPa. A safe stress of 35 MPa may be taken for the shaft on whichthe gear is mounted. Design and sketch the spur gear drive to suit the above conditions. The startingtorque may be assumed as 1.25 times the running torque.
8. Design and draw a spur gear drive transmitting 30 kW at 400 r.p.m. to another shaft running approxi-mately at 100 r.p.m. The load is steady and continuous. The materials for the pinion and gear are caststeel and cast iron respectively. Take module as 10 mm. Also check the design for dynamic load andwear.
3 v+; σe = 84 MPa ; e = 0.023 mm ; σes = 630 MPa ; EP = 210 kN/mm2 ; EG = 100 kN/mm2]
9. Design a spur gear drive required to transmit 45 kW at a pinion speed of 800 r.p.m. The velocity ratiois 3.5 : 1. The teeth are 20° full-depth involute with 18 teeth on the pinion. Both the pinion and gearare made of steel with a maximum safe static stress of 180 MPa. Assume a safe stress of 40 MPa forthe material of the shaft and key.
10. Design a pair of spur gears with stub teeth to transmit 55 kW from a 175 mm pinion running at 2500r.p.m. to a gear running at 1500 r.p.m. Both the gears are made of steel having B.H.N. 260. Approximatethe pitch by means of Lewis equation and then adjust the dimensions to keep within the limits set bythe dynamic load and wear equation.
QQQQQUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS
1. Write a short note on gear drives giving their merits and demerits.
2. How are the gears classified and what are the various terms used in spur gear terminology ?3. Mention four important types of gears and discuss their applications, the materials used for them and
their construction.4. What condition must be satisfied in order that a pair of spur gears may have a constant velcoity ratio?5. State the two most important reasons for adopting involute curves for a gear tooth profile.6. Explain the phenomenon of interference in involute gears. What are the conditions to be satisfied in
order to avoid interference ?7. Explain the different causes of gear tooth failures and suggest possible remedies to avoid such fail-
ures.8. Write the expressions for static, limiting wear load and dynamic load for spur gears and explain the
various terms used there in.9. Discuss the design procedure of spur gears.
10. How the shaft and arms for spur gears are designed ?
OBJECTOBJECTOBJECTOBJECTOBJECTIVEIVEIVEIVEIVE TTTTT YPYPYPYPYPE E E E E QQQQQUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS
1. The gears are termed as medium velocity gears, if their peripheral velocity is
(a) 1–3 m / s (b) 3–15 m / s(c) 15–30 m / s (d) 30–50 m / s
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��������� � 10652. The size of gear is usually specified by
3. A spur gear with pitch circle diameter D has number of teeth T. The module m is defined as(a) m = d / T (b) m = T / D(c) m = π D / T (d) m = D.T
4. In a rack and pinion arrangement, the rack has teeth of ........... shape.(a) square (b) trepazoidal
5. The radial distance from the ............ to the clearance circle is called working depth.(a) addendum circle (b) dedendum circle
6. The product of the diametral pitch and circular pitch is equal to(a) 1 (b) 1/π(c) π (d) π × No. of teeth
7. The backlash for spur gears depends upon(a) module (b) pitch line velocity(c) tooth profile (d) both (a) and (b)
8. The contact ratio for gears is(a) zero (b) less than one(c) greater than one (d) none of these
9. If the centre distance of the mating gears having involute teeth is increased, then the pressure angle(a) increases (b) decreases(c) remains unchanged (d) none of these
10. The form factor of a spur gear tooth depends upon(a) circular pitch only (b) pressure angle only(c) number of teeth and circular pitch (d) number of teeth and the system of teeth
11. Lewis equation in spur gears is used to find the(a) tensile stress in bending (b) shear stress(c) compressive stress in bending (d) fatigue stress
12. The minimum number of teeth on the pinion in order to avoid interference for 20° stub system is(a) 12 (b) 14(c) 18 (d) 32
13. The allowable static stress for steel gears is approximately ......... of the ultimate tensile stress.(a) one-fourth (b) one-third(c) one-half (d) double
14. Lewis equation in spur gears is applied(a) only to the pinion (b) only to the gear(c) to stronger of the pinion or gear (d) to weaker of the pinion or gear
15. The static tooth load should be ........... the dynamic load.(a) less than (b) greater than
(c) equal to
ANSWEANSWEANSWEANSWEANSWERRRRRSSSSS
1. (b) 2. (b) 3. (a) 4. (b) 5. (a)
6. (c) 7. (d) 8. (c) 9. (a) 10. (d)
11. (c) 12. (b) 13. (b) 14. (d) 15. (b)
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1066 � �������������� �����������
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1066
1. Introduction.2. Terms used in Helical Gears.3. Face Width of Helical Gears.4. Formative or Equivalent
Number of Teeth for HelicalGears.
5. Proportions for HelicalGears.
6. Strength of Helical Gears.
29�
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29.129.129.129.129.1 IntrIntrIntrIntrIntroductionoductionoductionoductionoductionA helical gear has teeth in form of helix around the
gear. Two such gears may be used to connect two parallelshafts in place of spur gears. The helixes may be righthanded on one gear and left handed on the other. The pitchsurfaces are cylindrical as in spur gearing, but the teethinstead of being parallel to the axis, wind around thecylinders helically like screw threads. The teeth of helicalgears with parallel axis have line contact, as in spur gearing.This provides gradual engagement and continuous contactof the engaging teeth. Hence helical gears give smooth drivewith a high efficiency of transmission.
We have already discussed in Art. 28.4 that the helicalgears may be of single helical type or double helical type.In case of single helical gears there is some axial thrustbetween the teeth, which is a disadvantage. In order toeliminate this axial thrust, double helical gears (i.e.
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����� �������� � 1067
Fig. 29.1. Helical gear(nomenclature).
herringbone gears) are used. It is equivalent to two single helical gears, in which equal and oppositethrusts are provided on each gear and the resulting axial thrust is zero.
29.229.229.229.229.2 TTTTTerererererms used in Helical Gearms used in Helical Gearms used in Helical Gearms used in Helical Gearms used in Helical GearsssssThe following terms in connection with helical gears, as shown in
Fig. 29.1, are important from the subject point of view.1. Helix angle. It is a constant angle made by the helices with the
axis of rotation.2. Axial pitch. It is the distance, parallel to the axis, between similar
faces of adjacent teeth. It is the same as circular pitch and is thereforedenoted by pc. The axial pitch may also be defined as the circular pitchin the plane of rotation or the diametral plane.
3. Normal pitch. It is the distance between similar faces of adjacentteeth along a helix on the pitch cylinders normal to the teeth. It is denotedby pN. The normal pitch may also be defined as the circular pitch in the normal plane which is a planeperpendicular to the teeth. Mathematically, normal pitch,
pN = pc cos αNote : If the gears are cut by standard hobs, then the pitch (or module) and the pressure angle of the hob willapply in the normal plane. On the other hand, if the gears are cut by the Fellows gear-shaper method, the pitchand pressure angle of the cutter will apply to the plane of rotation. The relation between the normal pressureangle (φN) in the normal plane and the pressure angle (φ) in the diametral plane (or plane of rotation) is given by
tan φN = tan φ × cos α
29.329.329.329.329.3 FFFFFace ace ace ace ace WWWWWidth of Helical Gearidth of Helical Gearidth of Helical Gearidth of Helical Gearidth of Helical GearsssssIn order to have more than one pair of teeth in contact, the tooth displacement (i.e. the ad-
vancement of one end of tooth over the other end) or overlap should be atleast equal to the axialpitch, such that
Overlap = pc = b tan α ...(i)The normal tooth load (WN) has two components ; one is tangential component (WT) and the
other axial component (WA), as shown in Fig. 29.2. The axial or end thrust is given byWA = WN sin α = WT tan α ...(ii)
From equation (i), we see that as the helix angle increases, then thetooth overlap increases. But at the same time, the end thrust as given byequation (ii), also increases, which is undesirable. It is usually recom-mended that the overlap should be 15 percent of the circular pitch.
∴ Overlap = b tan α = 1.15 pc
or b =1.15 1.15
tan tancp m× π
=α α
... (∵ pc = π m)
where b = Minimum face width, andm = Module.
Notes : 1. The maximum face width may be taken as 12.5 m to 20 m, where m isthe module. In terms of pinion diameter (DP), the face width should be 1.5 DP to2 DP, although 2.5 DP may be used.
2. In case of double helical or herringbone gears, the minimum face widthis given by
b =2.3 2.3
tan tancp m× π=
α αThe maximum face width ranges from 20 m to 30 m.
Fig. 29.2. Face width ofhelical gear.
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1068 � �������������� �����������
3. In single helical gears, the helix angle ranges from 20° to 35°, while for double helical gears, it may bemade upto 45°.
29.429.429.429.429.4 ForForForForFormamamamamativtivtivtivtive or Equive or Equive or Equive or Equive or Equivalent Number of alent Number of alent Number of alent Number of alent Number of TTTTTeeth feeth feeth feeth feeth for Helical Gearor Helical Gearor Helical Gearor Helical Gearor Helical GearsssssThe formative or equivalent number of teeth for a helical gear may be defined as the number of
teeth that can be generated on the surface of a cylinder having a radius equal to the radius of curvatureat a point at the tip of the minor axis of an ellipse obtained by taking a section of the gear in the normalplane. Mathematically, formative or equivalent number of teeth on a helical gear,
TE = T / cos3 αwhere T = Actual number of teeth on a helical gear, and
α = Helix angle.
29.529.529.529.529.5 PrPrPrPrProporoporoporoporoportions ftions ftions ftions ftions for Helical Gearor Helical Gearor Helical Gearor Helical Gearor Helical GearsssssThough the proportions for helical gears are not standardised, yet the following are recommended
by American Gear Manufacturer's Association (AGMA).
Pressure angle in the plane of rotation,
φ = 15° to 25°
Helix angle, α = 20° to 45°
Addendum = 0.8 m (Maximum)
Dedendum = 1 m (Minimum)
Minimum total depth = 1.8 m
Minimum clearance = 0.2 m
Thickness of tooth = 1.5708 m
In helical gears, the teeth are inclined to the axis of the gear.
����� �������� � 1069
29.629.629.629.629.6 StrStrStrStrStrength of Helical Gearength of Helical Gearength of Helical Gearength of Helical Gearength of Helical GearsssssIn helical gears, the contact between mating teeth is gradual, starting at one end and moving
along the teeth so that at any instant the line of contact runs diagonally across the teeth. Therefore inorder to find the strength of helical gears, a modified Lewis equation is used. It is given by
WT = (σo × Cv) b.π m.y'
where WT = Tangential tooth load,
σo = Allowable static stress,
Cv = Velocity factor,
b = Face width,
m = Module, and
y' = Tooth form factor or Lewis factor corresponding to the formativeor virtual or equivalent number of teeth.
Notes : 1. The value of velocity factor (Cv) may be taken as follows :
Cv =6
,6 v+ for peripheral velocities from 5 m / s to 10 m / s.
=15
,15 v+
for peripheral velocities from 10 m / s to 20 m / s.
=0.75
,0.75 v+
for peripheral velocities greater than 20 m / s.
= 0.75
0.25,1 v
++
for non-metallic gears.
2. The dynamic tooth load on the helical gears is given by
WD = WT + 2
T
2T
21 ( . cos ) cos
21 . cos
v b C W
v b C W
α + α
+ α +where v, b and C have usual meanings as discussed in spur gears.
3. The static tooth load or endurance strength of the tooth is given by
WS = σe.b.π m.y'
4. The maximum or limiting wear tooth load for helical gears is given by
Ww =P
2
. . .
cos
D b Q K
αwhere DP, b, Q and K have usual meanings as discussed in spur gears.
In this case, K =2
N
P G
( ) sin 1 1
1.4es
E E
σ φ +
where φN = Normal pressure angle.
Example 29.1. A pair of helical gears are to transmit 15 kW. The teeth are 20° stub in diametralplane and have a helix angle of 45°. The pinion runs at 10 000 r.p.m. and has 80 mm pitch diameter.The gear has 320 mm pitch diameter. If the gears are made of cast steel having allowable staticstrength of 100 MPa; determine a suitable module and face width from static strength considerationsand check the gears for wear, given σes = 618 MPa.
Since both the pinion and gear are made of the same material (i.e. cast steel), therefore thepinion is weaker. Thus the design will be based upon the pinion.
We know that the torque transmitted by the pinion,
T =3
P
60 15 10 6014.32 N-m
2 2 10000
P
N
× × ×= =
π π ×∴ *Tangential tooth load on the pinion,
WT =P
14.32358 N
/ 2 0.08 / 2
T
D= =
We know that number of teeth on the pinion,
TP = DP / m = 80 / m
and formative or equivalent number of teeth for the pinion,
TE =P3 3 3
80 / 80 / 226.4
cos cos 45 (0.707)
T m m
m= = =
α °∴ Tooth form factor for the pinion for 20° stub teeth,
y'P =E
0.841 0.8410.175 0.175 0.175 0.0037
226.4 /m
T m− = − = −
We know that peripheral velocity,
v = P P. 0.08 1000042 m / s
60 60
D Nπ π × ×= =
∴ Velocity factor,
Cv =0.75 0.75
0.1040.75 0.75 42v
= =+ +
...(∵ v is greater than 20 m/s)
Since the maximum face width (b) for helical gears may be taken as 12.5 m to 20 m, where m isthe module, therefore let us take
b = 12.5 m
We know that the tangential tooth load (WT),
358 = (σOP . Cv) b.π m.y'P= (100 × 0.104) 12.5 m × π m (0.175 – 0.0037 m)
= 409 m2 (0.175 – 0.0037 m) = 72 m2 – 1.5 m3
Solving this expression by hit and trial method, we find that
m = 2.3 say 2.5 mm Ans.and face width, b = 12.5 m = 12.5 × 2.5 = 31.25 say 32 mm Ans.Checking the gears for wear
We know that velocity ratio,
V.R. = G
P
3204
80
D
D= =
∴ Ratio factor,
Q =2 . . 2 4
1.6. . 1 4 1
V R
V R
× ×= =+ +
We know that tan φN = tan φ cos α = tan 20° × cos 45° = 0.2573
∴ φN = 14.4°
* The tangential tooth load on the pinion may also be obtained by using the relation,
WT = P P., where (in m / s)
60P D N
vv
π=
����� �������� � 1071
Since both the gears are made of the same material (i.e. cast steel), therefore let us takeEP = EG = 200 kN/mm2 = 200 × 103 N/mm2
∴ Load stress factor,
K =2
N
P G
( ) sin 1 1
1.4es
E E
σ φ +
=2
23 3
(618) sin 14.4 1 10.678 N/mm
1.4 200 10 200 10
° + = × × We know that the maximum or limiting load for wear,
Ww = P2 2
. . . 80 32 1.6 0.6785554 N
cos cos 45
D b Q K × × ×= =α °
Since the maximum load for wear is much more than the tangential load on the tooth, thereforethe design is satisfactory from consideration of wear.
Example 29.2. A helical cast steel gear with 30° helix angle has to transmit 35 kW at 1500r.p.m. If the gear has 24 teeth, determine the necessary module, pitch diameter and face width for 20°full depth teeth. The static stress for cast steel may be taken as 56 MPa. The width of face may betaken as 3 times the normal pitch. What would be the end thrust on the gear? The tooth factor for 20°
full depth involute gear may be taken as E
.. ,
0 9120 154
T− where TE represents the equivalent number
of teeth.
Solution. Given : α = 30° ; P = 35 kW = 35 × 103 W ; N = 1500 r.p.m. ; TG = 24 ; φ = 20° ;σo = 56 MPa = 56 N/mm2 ; b = 3 × Normal pitch = 3 PN
ModuleLet m = Module in mm, and
DG = Pitch circle diameter of the gear in mm.
We know that torque transmitted by the gear,
T =3
360 35 10 60223 N-m 223 10 N-mm
2 2 1500
P
N
× × ×= = = ×π π ×
The picture shows double helical gears which are also called herringbone gears.
= (σo × Cv) 3 × pc cos α × π m × y' ...(∵ pN = pc cos α)
= (σo × Cv) 3 π m cos α × π m × y' ...(∵ pc = π m)
∴18 600 15
56 3 cos 30 0.12915 1.885
m mm m
= π × ° × π × +
=22780
15 1.885
m
m+or 279 000 + 35 061 m = 2780 m3
Solving this equation by hit and trial method, we find that
m = 5.5 say 6 mm Ans.Pitch diameter of the gear
We know that the pitch diameter of the gear,
DG = m × TG = 6 × 24 = 144 mm Ans.Face width
It is given that the face width,
b = 3 pN = 3 pc cos α = 3 × π m cos α= 3 × π × 6 cos 30° = 48.98 say 50 mm Ans.
End thrust on the gearWe know that end thrust or axial load on the gear,
WA =18 600 18 600
tan tan 30 0.5776TW
mα = × ° = × =1790 N Ans.
����� �������� � 1073Example 29.3. Design a pair of helical gears for transmitting 22 kW. The speed of the driver
gear is 1800 r.p.m. and that of driven gear is 600 r.p.m. The helix angle is 30° and profile iscorresponding to 20° full depth system. The driver gear has 24 teeth. Both the gears are made of caststeel with allowable static stress as 50 MPa. Assume the face width parallel to axis as 4 times thecircular pitch and the overhang for each gear as 150 mm. The allowable shear stress for the shaftmaterial may be taken as 50 MPa. The form factor may be taken as 0.154 – 0.912 / TE, where TE is
the equivalent number of teeth. The velocity factor may be taken as ,350
350 v+ where v is pitch line
velocity in m / min. The gears are required to be designed only against bending failure of the teethunder dynamic condition.
Solution. Given : P = 22 kW = 22 × 103 W ; NP = 1800 r.p.m.; NG = 600 r.p.m. ; α = 30° ;φ = 20° ; TP = 24 ; σo = 50 MPa = 50 N/mm2 ; b = 4 pc t ; Overhang = 150 mm ; τ = 50 MPa = 50 N/mm2
Design for the pinion and gearWe know that the torque transmitted by the pinion,
T =3
P
60 22 10 60116.7 N-m 116 700 N-mm
2 2 1800
P
N
× × ×= = =π π ×
Since both the pinion and gear are made of the same material (i.e. cast steel), therefore thepinion is weaker. Thus the design will be based upon the pinion. We know that formative or equivalentnumber of teeth,
TE = P3 3 3
24 2437
cos cos 30 (0.866)
T= = =
α °
∴ Form factor, y' =E
0.912 0.9120.154 0.154 0.129
37T− = − =
Gears inside a car
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First of all let us find the module of teeth.
Let m = Module in mm, and
DP = Pitch circle diameter of the pinion in mm.
We know that the tangential tooth load on the pinion,
WT =P P P
2 2
/ 2
T T T
D D m T= =
×...(∵ DP = m.TP)
=2 116 700 9725
N24m m
×=
×
and peripheral velocity, v = π DP.NP = π m.TP.NP
= π m × 24 × 1800 = 135 735 m mm / min = 135.735 m m / min
∴ Velocity factor, Cv =350 350
350 350 135.735v m=
+ +We also know that the tangential tooth load on the pinion,
WT = (σo.Cv) b.π m.y' = (σo.Cv) 4 pc × π m × y' ... (∵ b = 4 pc)
= (σo.Cv) 4 × π m × π m × y' ... (∵ pc = π m)
∴ 2
2 29725 350 89 12650 4 0.129
350 135.735 350 135.735
mm
m m m = × π × = + +
3.4 × 106 + 1.32 × 106 m = 89 126 m3
Solving this expression by hit and trial method, we find that
m = 4.75 mm say 6 mm Ans.
We know that face width,
b = 4 pc = 4 π m = 4 π × 6 = 75.4 say 76 mm Ans.and pitch circle diameter of the pinion,
DP = m × TP = 6 × 24 = 144 mm Ans.Since the velocity ratio is 1800 / 600 = 3, therefore number of teeth on the gear,
TG = 3 TP = 3 × 24 = 72
and pitch circle diameter of the gear,
DG = m × TG = 6 × 72 = 432 mm Ans.
Helical gears.
����� �������� � 1075Design for the pinion shaft
Let dP = Diameter of the pinion shaft.
We know that the tangential load on the pinion,
WT =9725 9725
1621 N6m
= =
and the axial load of the pinion,
WA = WT tan α = 1621 tan 30°
= 1621 × 0.577 = 935 N
Since the overhang for each gear is 150 mm, therefore bending moment on the pinion shaft dueto the tangential load,
M1 = WT × Overhang = 1621 × 150 = 243 150 N-mm
and bending moment on the pinion shaft due to the axial load,
M2 = PA
144935 67 320 N-mm
2 2
DW × = × =
Since the bending moment due to the tangential load (i.e. M1) and bending moment due to theaxial load (i.e. M2) are at right angles, therefore resultant bending moment on the pinion shaft,
The pinion shaft is also subjected to a torque T = 116 700 N-mm, therefore equivalent twistingmoment,
Te = 2 2 2 2(252 293) (116 700) 277 975 N-mmM T+ = + =We know that equivalent twisting moment (Te),
277 975 =3 3 3
P P P( ) 50 ( ) 9.82 ( )16 16
d d dπ π× τ × =
∴ (dP)3 = 277 975 / 9.82 = 28 307 or dP = 30.5 say 35 mm Ans.Let us now check for the principal shear stress.We know that the shear stress induced,
τ = 23 3
P
16 16 27797533 N/mm 33 MPa
( ) (35)eT
d
×= = =π π
and direct stress due to axial load,
σ =A 2
2 2P
9350.97 N/mm 0.97 MPa
( ) (35)4 4
W
d= = =
π π
Helical gears
1076 � �������������� �����������
∴ Principal shear stress,
= 2 2 2 21 14 (0.97) 4(33) 33 MPa
2 2 σ + τ = + =
Since the principal shear stress is less than the permissible shear stress of 50 MPa, therefore thedesign is satisfactory.
We know that the diameter of the pinion hub
= 1.8 dP = 1.8 × 35 = 63 mm Ans.and length of the hub = 1.25 dP = 1.25 × 35 = 43.75 say 44 mm
Since the length of the hub should not be less than the face width, therefore let us take length ofthe hub as 76 mm. Ans.
Note : Since the pitch circle diameter of the pinion is 144 mm, therefore the pinion should be provided with aweb. Let us take the thickness of the web as 1.8 m, where m is the module.
∴ Thickness of the web = 1.8 m = 1.8 × 6 = 10.8 say 12 mm Ans.
Design for the gear shaft
Let dG = Diameter of the gear shaft.
We have already calculated that the tangential load,
We also know that equivalent twisting moment (Te),
472 000 = 3 3 3
G G G( ) 50 ( ) 9.82 ( )16 16
d d dπ π× τ × = × =
∴ (dG)3 = 472 000 / 9.82 = 48 065 or dG = 36.3 say 40 mm Ans.Let us now check for the principal shear stress.
We know that the shear stress induced,
τ = 23 3
G
16 16 472 00037.6 N/mm 37.6 MPa
( ) (40)eT
d
×= = =π π
����� �������� � 1077and direct stress due to axial load,
σ =A 2
2 2G
9350.744 N/mm 0.744 MPa
( ) (40)4 4
W
d= = =
π π
∴ Principal shear stress
=2 2 2 21 1
4 (0.744) 4 (37.6) 37.6 MPa2 2
σ + τ = + = Since the principal shear stress is less than the permissible shear stress of 50 MPa, therefore the
design is satisfactory.
We know that the diameter of the gear hub
= 1.8 dG = 1.8 × 40 = 72 mm Ans.and length of the hub = 1.25 dG = 1.25 × 40 = 50 mm
We shall take the length of the hub equal to the face width, i.e. 76 mm. Ans.Since the pitch circle diameter of the gear is 432 mm, therefore the gear should be provided
with four arms. The arms are designed in the similar way as discussed for spur gears.Design for the gear arms
Let us assume that the cross-section of the arms is elliptical with major axis (a1) equal to twicethe minor axis (b1). These dimensions refer to hub end.
∴ Section modulus of arms,
Z =2 3
31 1 11
( ) ( )0.05 ( )
32 64
b a aa
π π= = 11 2
ab
= ∵
Since the arms are designed for the stalling load and it is taken as the design tangential loaddivided by the velocity factor, therefore
Stalling load, WS = T 350 135.7351621
350v
W m
C
+ =
= 1621350 135.735 6
5393 N350
+ × = ∴ Maximum bending moment on each arm,
M = S G 5393 432291 222 N-mm
2 4 2
W D
n× = × =
We know that bending stress (σb),
42 =3
3 31 1
291 222 5824 10
0.05 ( ) ( )
M
Z a a
×= = ... (Taking σb = 42 N/mm2)
∴ (a1)3 = 5824 × 103 / 42 = 138.7 × 103 or a1 = 51.7 say 54 mm Ans.
and b1 = a1 / 2 = 54 / 2 = 27 mm Ans.Since the arms are tapered towards the rim and the taper is 1/16 mm per mm length of the arm
(or radius of the gear), thereforeMajor axis of the arm at the rim end,
a2 = a1 – Taper = a1 G1
16 2
D− ×
=1 432
54 40 mm16 2
− × = Ans.
and minor axis of the arm at the rim end,b2 = a2 / 2 = 40 / 2 = 20 mm Ans.
1078 � �������������� �����������
Design for the rimThe thickness of the rim for the pinion may be taken as 1.6 m to 1.9 m, where m is the module.
Let us take thickness of the rim for pinion,
tRP = 1.6 m = 1.6 × 6 = 9.6 say 10 mm Ans.The thickness of the rim for the gear (tRG) is given by
tRG = m G 726 25.4 say 26 mm
4
T
n= = Ans.
EEEEEXEXEXEXEXERRRRRCISECISECISECISECISESSSSS
1. A helical cast steel gear with 30° helix angle has to transmit 35kW at 2000 r.p.m. If the gear has 25 teeth, find the necessarymodule, pitch diameters and face width for 20° full depthinvolute teeth. The static stress for cast steel may be taken as100 MPa. The face width may be taken as 3 times the normalpitch. The tooth form factor is given by the expression
y' = 0.154 – 0.912/TE , where TE represents the equivalent num-
ber of teeth. The velocity factor is given by Cv = 6
,6 v+ where
v is the peripheral speed of the gear in m/s. [Ans. 6 mm ; 150 mm ; 50 mm]
2. A pair of helical gears with 30° helix angle is used to transmit15 kW at 10 000 r.p.m. of the pinion. The velocity ratio is 4 : 1.Both the gears are to be made of hardened steel of static strength 100 N/mm2. The gears are 20° stuband the pinion is to have 24 teeth. The face width may be taken as 14 times the module. Find themodule and face width from the standpoint of strength and check the gears forwear. [Ans. 2 mm ; 28 mm]
Gears inside a car engine.
����� �������� � 10793. A pair of helical gears consist of a 20 teeth pinion meshing with a 100 teeth gear. The pinion rotates at
720 r.p.m. The normal pressure angle is 20° while the helix angle is 25°. The face width is 40 mm andthe normal module is 4 mm. The pinion as well as gear are made of steel having ultimate strength of600 MPa and heat treated to a surface hardness of 300 B.H.N. The service factor and factor of safetyare 1.5 and 2 respectively. Assume that the velocity factor accounts for the dynamic load and calculatethe power transmitting capacity of the gears. [Ans. 8.6 kW]
4. A single stage helical gear reducer is to receive power from a 1440 r.p.m., 25 kW induction motor. Thegear tooth profile is involute full depth with 20° normal pressure angle. The helix angle is 23°,number of teeth on pinion is 20 and the gear ratio is 3. Both the gears are made of steel with allowablebeam stress of 90 MPa and hardness 250 B.H.N.
(a) Design the gears for 20% overload carrying capacity from standpoint of bending strength andwear.
(b) If the incremental dynamic load of 8 kN is estimated in tangential plane, what will be the safepower transmitted by the pair at the same speed?
QQQQQUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS
1. What is a herringbone gear? Where they are used?
2. Explain the following terms used in helical gears :
(a) Helix angle; (b) normal pitch; and
(c) axial pitch.
3. Define formative or virtual number of teeth on a helical gear. Derive the expression used to obtain itsvalue.
4. Write the expressions for static strength, limiting wear load and dynamic load for helical gears andexplain the various terms used therein.
OBJECTOBJECTOBJECTOBJECTOBJECTIVEIVEIVEIVEIVE TTTTT YPYPYPYPYPE E E E E QQQQQUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS
1. If T is the actual number of teeth on a helical gear and φ is the helix angle for the teeth, the formativenumber of teeth is written as
(a) T sec3 φ (b) T sec2 φ(c) T/sec3φ (d) T cosec φ
2. In helical gears, the distance between similar faces of adjacent teeth along a helix on the pitch cylin-ders normal to the teeth, is called
(a) normal pitch (b) axial pitch(c) diametral pitch (d) module
3. In helical gears, the right hand helices on one gear will mesh ......... helices on the other gear.
(a) right hand (b) left hand
4. The helix angle for single helical gears ranges from
(a) 10° to 15° (b) 15° to 20°
(c) 20° to 35° (d) 35° to 50°
5. The helix angle for double helical gears may be made up to
(a) 45° (b) 60°
(c) 75° (d) 90°
ANSWEANSWEANSWEANSWEANSWERRRRRSSSSS
1. (a) 2. (a) 3. (b) 4. (c) 5. (a)
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1080 � �������������� �����������
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1080
1. Introduction.2. Classification of Bevel
Gears.3. Terms used in Bevel Gears.4. Determination of Pitch
Angle for Bevel Gears.5. Proportions for Bevel Gears.6. Formative or Equivalent
Number of Teeth for BevelG e a r s — T r e d g o l d ' sApproximation.
7. Strength of Bevel Gears.8. Forces Acting on a Bevel
Gear.9. Design of a Shaft for Bevel
Gears.
30�
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�
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30.130.130.130.130.1 IntrIntrIntrIntrIntroductionoductionoductionoductionoductionThe bevel gears are used for transmitting power at a
constant velocity ratio between two shafts whose axesintersect at a certain angle. The pitch surfaces for the bevelgear are frustums of cones. The two pairs of cones in contactis shown in Fig. 30.1. The elements of the cones, as shownin Fig. 30.1 (a), intersect at the point of intersection of theaxis of rotation. Since the radii of both the gears areproportional to their distances from the apex, therefore thecones may roll together without sliding. In Fig. 30.1 (b),the elements of both cones do not intersect at the point ofshaft intersection. Consequently, there may be pure rollingat only one point of contact and there must be tangentialsliding at all other points of contact. Therefore, these cones,cannot be used as pitch surfaces because it is impossible tohave positive driving and sliding in the same direction atthe same time. We, thus, conclude that the elements of bevel
����������� � 1081gear pitch cones and shaft axes must intersect at the same point.
Fig. 30.1. Pitch surface for bevel gears.
30.230.230.230.230.2 Classification of Bevel GearsClassification of Bevel GearsClassification of Bevel GearsClassification of Bevel GearsClassification of Bevel GearsThe bevel gears may be classified into the following types, depending upon the angles between
the shafts and the pitch surfaces.
1. Mitre gears. When equal bevel gears (having equal teeth and equal pitch angles) connect twoshafts whose axes intersect at right angle, as shown in Fig. 30.2 (a), then they are known as mitre gears.
2. Angular bevel gears. When the bevel gears connect two shafts whose axes intersect at anangle other than a right angle, then they are known as angular bevel gears.
The bevel gear is used to change the axis of rotational motion. By using gears of differingnumbers of teeth, the speed of rotation can also be changed.
1082 � �������������� �����������
3. Crown bevel gears. When the bevel gears connect two shafts whose axes intersect at an anglegreater than a right angle and one of the bevel gears has a pitch angle of 90º, then it is known as acrown gear. The crown gear corresponds to a rack in spur gearing, as shown in Fig. 30.2 (b).
Fig. 30.2. Classification of bevel gears.
4. Internal bevel gears. When the teeth on the bevel gear are cut on the inside of the pitch cone,then they are known as internal bevel gears.Note : The bevel gears may have straight or spiral teeth. It may be assumed, unless otherwise stated, that thebevel gear has straight teeth and the axes of the shafts intersect at right angle.
30.330.330.330.330.3 TTTTTerererererms used in Bems used in Bems used in Bems used in Bems used in Bevvvvvel Gearel Gearel Gearel Gearel Gearsssss
Fig. 30.3. Terms used in bevel gears.
A sectional view of two bevel gears in mesh is shown in Fig. 30.3. The following terms inconnection with bevel gears are important from the subject point of view :
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����������� � 10831. Pitch cone. It is a cone containing the pitch elements of the teeth.
2. Cone centre. It is the apex of the pitch cone. It may be defined as that point where the axes oftwo mating gears intersect each other.
3. Pitch angle. It is the angle made by the pitch line with the axis of the shaft. It is denoted by ‘θP’.
4. Cone distance. It is the length of the pitch cone element. It is also called as a pitch coneradius. It is denoted by ‘OP’. Mathematically, cone distance or pitch cone radius,
OP = GP
P P1 P2
/ 2/ 2Pitch radiussin � ��� ���
DD= =θ θ
5. Addendum angle. It is the angle subtended by the addendum of the tooth at the cone centre.It is denoted by ‘α’ Mathematically, addendum angle,
α = tan–1 a
OP
where a = Addendum, and OP = Cone distance.
6. Dedendum angle. It is the angle subtended by the dedendum of the tooth at the cone centre.It is denoted by ‘β’. Mathematically, dedendum angle,
β = tan–1 d
OP
where d = Dedendum, and OP = Cone distance.
7. Face angle. It is the angle subtended by the face of the tooth at the cone centre. It is denotedby ‘φ’. The face angle is equal to the pitch angle plus addendum angle.
8. Root angle. It is the angle subtended by the root of the tooth at the cone centre. It is denotedby ‘θR’. It is equal to the pitch angle minus dedendum angle.
9. Back (or normal) cone. It is an imaginary cone, perpendicular to the pitch cone at the end ofthe tooth.
10. Back cone distance. It is the length of the back cone. It is denoted by ‘RB’. It is also calledback cone radius.
11. Backing. It is the distance of the pitch point (P) from the back of the boss, parallel to thepitch point of the gear. It is denoted by ‘B’.
12. Crown height. It is the distance of the crown point (C) from the cone centre (O), parallel tothe axis of the gear. It is denoted by ‘HC’.
13. Mounting height. It is the distance of the back of the boss from the cone centre. It isdenoted by ‘HM’.
14. Pitch diameter. It is the diameter of the largest pitch circle.
15. Outside or addendum cone diameter. It is the maximum diameter of the teeth of the gear.It is equal to the diameter of the blank from which the gear can be cut. Mathematically, outsidediameter,
DO = DP + 2α cos θP
where DP = Pitch circle diameter,
α = Addendum, and
θP = Pitch angle.
16. Inside or dedendum cone diameter. The inside or the dedendum cone diameter is given by
Dd = DP – 2d cos θP
where Dd = Inside diameter, and
d = Dedendum.
1084 � �������������� �����������
30.430.430.430.430.4 DeterDeterDeterDeterDeterminaminaminaminamination of Pitch tion of Pitch tion of Pitch tion of Pitch tion of Pitch Angle fAngle fAngle fAngle fAngle forororororBevel GearsBevel GearsBevel GearsBevel GearsBevel Gears
Consider a pair of bevel gears in mesh, asshown in Fig. 30.3.
Let θP1 = Pitch angle for the pinion,
θP2 = Pitch angle for the gear,
θS = Angle between the two shaft axes,
DP = Pitch diameter of the pinion,
DG = Pitch diameter of the gear, and
V.R. = Velocity ratio = G G P
P P G
D T ND T N
= =
From Fig. 30.3, we find that
θS = θP1 + θP2 or θP2 = θS – θP1
∴ sin θP2 = sin (θS – θP1) = sin θS . cos θP1 – cos θS . sin θP1 ...(i)We know that cone distance,
OP = GP
P1 P2
/ 2/ 2sin sin
DD=
θ θ or GP2
P1 P
sinsin
D
D
θ=
θ = V.R.
∴ sin θP2 = V.R. × sin θP1 ...(ii)From equations (i) and (ii), we have
V.R. × sin θP1 = sin θS . cos θP1 – cos θS . sin θP1
Dividing throughout by cos θP1 we get
V.R. tan θP1 = sin θS – cos θS. tan θP1
or tan θP1 = S
S
sinV.R cos
θ+ θ
∴ θP1 = tan–1 S
S
sinV.R cos
θ + θ
...(iii)
Similarly, we can find that
tan θP2 = S
S
sin1
cosV.R
θ
+ θ
∴ θP2 = tan–1 S
S
sin1
cosV.R
θ + θ
...(iv)
Note : When the angle between the shaft axes is 90º i.e. θS = 90º, then equations (iii) and (iv) may be written as
θP1 = tan–1 1
V.R
= tan–1 P
G
DD
= tan–1 P
G
TT
= tan–1 G
P
NN
and θP2 = tan–1 (V.R.) = tan–1 G
P
DD
= tan–1
G
P
T
T
= tan–1 P
G
TN
30.530.530.530.530.5 PrPrPrPrProporoporoporoporoportions ftions ftions ftions ftions for Beor Beor Beor Beor Bevvvvvel Gearel Gearel Gearel Gearel GearThe proportions for the bevel gears may be taken as follows :
1. Addendum, a = 1 m
Mitre gears
����������� � 10852. Dedendum, d = 1.2 m
3. Clearance = 0.2 m
4. Working depth = 2 m
5. Thickness of tooth = 1.5708 m
where m is the module.
Note : Since the bevel gears are not interchangeable, therefore these are designed in pairs.
30.630.630.630.630.6 For For For For Formamamamamativtivtivtivtive or Equive or Equive or Equive or Equive or Equivalent Number of alent Number of alent Number of alent Number of alent Number of TTTTTeeth feeth feeth feeth feeth for Beor Beor Beor Beor Bevvvvvel Gearel Gearel Gearel Gearel Gears – s – s – s – s – TTTTTrrrrredgold’edgold’edgold’edgold’edgold’sssssApprApprApprApprApproooooximaximaximaximaximationtiontiontiontion
We have already discussed that the involute teeth for a spur gear may be generated by the edgeof a plane as it rolls on a base cylinder. A similar analysis for a bevel gear will show that a true sectionof the resulting involute lies on the surface of a sphere. But it is not possible to represent on a planesurface the exact profile of a bevel gear tooth lying on the surface of a sphere. Therefore, it isimportant to approximate the bevel gear tooth profiles as accurately as possible. The approximation(known as Tredgold’s approximation) is based upon the fact that a cone tangent to the sphere at thepitch point will closely approximate the surface of the sphere for a short distance either side of thepitch point, as shown in Fig. 30.4 (a). The cone (known as back cone) may be developed as a planesurface and spur gear teeth corresponding to the pitch and pressure angle of the bevel gear and theradius of the developed cone can be drawn. This procedure is shown in Fig. 30.4 (b).
Fig. 30.4
Let θP = Pitch angle or half of the cone angle,
R = Pitch circle radius of the bevel pinion or gear, and
RB = Back cone distance or equivalent pitch circle radius of spur pinionor gear.
Now from Fig. 30.4 (b), we find thatRB = R sec θP
We know that the equivalent (or formative) number of teeth,
TE = B2 Rm
... Pitch circle diameter
Number of teeth =Module
∵
= PP
2 secsec
RT
mθ
= θ
where T = Actual number of teeth on the gear.
1086 � �������������� �����������
Notes : 1. The action of bevel gears will be same as that of equivalent spur gears.
2. Since the equivalent number of teeth is always greater than the actual number of teeth, therefore agiven pair of bevel gears will have a larger contact ratio. Thus, they will run more smoothly than a pair of spurgears with the same number of teeth.
30.730.730.730.730.7 StrStrStrStrStrength of Beength of Beength of Beength of Beength of Bevvvvvel Gearel Gearel Gearel Gearel GearsssssThe strength of a bevel gear tooth is obtained in a similar way as discussed in the previous
articles. The modified form of the Lewis equation for the tangential tooth load is given as follows:
WT = (σo × Cv) b.π m.y' L b
L−
where σo = Allowable static stress,
Cv = Velocity factor,
=3
3 v+ , for teeth cut by form cutters,
=6
6 v+ , for teeth generated with precision machines,
v = Peripheral speed in m / s,
b = Face width,
m = Module,
y' = Tooth form factor (or Lewis factor) for the equivalent number ofteeth,
L = Slant height of pitch cone (or cone distance),
=2 2
G P
2 2D D +
Hypoid bevel gears in a car differential
InputPinion
Driveshaft
Ring gear
����������� � 1087DG = Pitch diameter of the gear, and
DP = Pitch diameter of the pinion.
Notes : 1. The factor L b
L
− may be called as bevel factor.
2. For satisfactory operation of the bevel gears, the face width should be from 6.3 m to 9.5 m, where m isthe module. Also the ratio L / b should not exceed 3. For this, the number of teeth in the pinion must not less than
2
48
1 ( . .)V R+ , where V.R. is the required velocity ratio.
3. The dynamic load for bevel gears may be obtained in the similar manner as discussed for spur gears.
4. The static tooth load or endurance strength of the tooth for bevel gears is given by
WS = σe.b.π m.y' –L b
L
The value of flexural endurance limit (σe) may be taken from Table 28.8, in spur gears.
5. The maximum or limiting load for wear for bevel gears is given by
Ww =P
P1
. . .cos
D b Q Kθ
where DP, b, Q and K have usual meanings as discussed in spur gears except that Q is based on formative orequivalent number of teeth, such that
Q =EG
EG EP
2 TT T+
30.830.830.830.830.8 ForForForForForces ces ces ces ces Acting on a BeActing on a BeActing on a BeActing on a BeActing on a Bevvvvvel Gearel Gearel Gearel Gearel Gear
Consider a bevel gear and pinion in mesh as shown in Fig. 30.5. The normal force (WN) on thetooth is perpendicular to the tooth profile and thus makes an angle equal to the pressure angle (φ) tothe pitch circle. Thus normal force can be resolved into two components, one is the tangential component(WT) and the other is the radial component (WR). The tangential component (i.e. the tangential toothload) produces the bearing reactions while the radial component produces end thrust in the shafts.The magnitude of the tangential and radial components is as follows :
WT = WN cos φ, and WR = WN sin φ = WT tan φ ...(i)
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1088 � �������������� �����������
These forces are considered to act at the mean radius (Rm). From the geometry of the Fig. 30.5,we find that
Rm = PP1sin
2 2 2Db b
L LL
− θ = − ... P
P1/ 2
sinD
L θ = ∵
Now the radial force (WR) acting at the mean radius may be further resolved into twocomponents, WRH and WRV, in the axial and radial directions as shown in Fig. 30.5. Therefore theaxial force acting on the pinion shaft,
WRH = WR sin θP1 = WT tan φ . sin θP1 ...[From equation (i)]
and the radial force acting on the pinion shaft,
WRV = WR cos θP1 = WT tan φ. cos θP1
Fig. 30.5. Forces acting on a bevel gear.
A little consideration will show that the axial force on the pinion shaft is equal to the radial forceon the gear shaft but their directions are opposite. Similarly, the radial force on the pinion shaft isequal to the axial force on the gear shaft, but act in opposite directions.
30.930.930.930.930.9 Design of a Shaft for Bevel GearsDesign of a Shaft for Bevel GearsDesign of a Shaft for Bevel GearsDesign of a Shaft for Bevel GearsDesign of a Shaft for Bevel GearsIn designing a pinion shaft, the following procedure may be adopted :
1. First of all, find the torque acting on the pinion. It is given by
T =P
602P
N×π N-m
where P = Power transmitted in watts, and
NP = Speed of the pinion in r.p.m.
2. Find the tangential force (WT) acting at the mean radius (Rm) of the pinion. We know thatWT = T / Rm
3. Now find the axial and radial forces (i.e. WRH and WRV) acting on the pinion shaft asdiscussed above.
4. Find resultant bending moment on the pinion shaft as follows :
The bending moment due to WRH and WRV is given by
M1 = WRV × Overhang – WRH × Rm
and bending moment due to WT,
M2 = WT × Overhang
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����������� � 1089∴ Resultant bending moment,
M = 2 21 2( ) ( )M M+
5. Since the shaft is subjected to twisting moment (T ) and resultant bending moment (M),therefore equivalent twisting moment,
Te = 2 2M T+6. Now the diameter of the pinion shaft may be obtained by using the torsion equation. We
know that
Te =16π
× τ (dP)3
where dP = Diameter of the pinion shaft, and
τ = Shear stress for the material of the pinion shaft.
7. The same procedure may be adopted to find the diameter of the gear shaft.
Example 30.1. A 35 kW motor running at 1200 r.p.m. drives a compressor at 780 r.p.m.through a 90° bevel gearing arrangement. The pinion has 30 teeth. The pressure angle of teeth is
1214 / ° . The wheels are capable of withstanding a dynamic stress,
σw = 140 280
280 v +
MPa, where v is the pitch line speed in m / min.
The form factor for teeth may be taken as
0.124 – E
0.686T
, where TE is the number of teeth
equivalent of a spur gear.
The face width may be taken as 14
of the
slant height of pitch cone. Determine for thepinion, the module pitch, face width, addendum,dedendum, outside diameter and slant height.
Solution : Given : P = 35 kW = 35 × 103
W ; NP = 1200 r.p.m. ; NG = 780 r.p.m. ; θS = 90º ;TP = 30 ; φ = 14 1/2º ; b = L / 4
Module and face width for the pinionLet m = Module in mm,
b = Face width in mm
= L / 4, and ...(Given)
DP = Pitch circle diameter of the pinion.
We know that velocity ratio,
V.R. =P
G
12001.538
780NN
= =
∴ Number of teeth on the gear,
TG = V.R. × TP = 1.538 × 30 = 46
Since the shafts are at right angles, therefore pitch angle for the pinion,
θP1 = tan–1 1 11 1tan tan (0.65)
. . 1.538V R− − = =
= 33º
and pitch angle for the gear,
θP2 = 90º – 33º = 57º
High performance 2- and 3 -way bevel gearboxes
1090 � �������������� �����������
We know that formative number of teeth for pinion,TEP = TP.sec θP1 = 30 × sec 33º = 35.8
and formative number of teeth for the gear,TEG = TG.sec θP2 = 46 × sec 57º = 84.4
Tooth form factor for the pinion
y'P = 0.124 – EP
0.686 0.6860.124 0.105
35.5T= − =
and tooth form factor for the gear,
y'G = 0.124 – EG
0.686 0.6860.124 0.116
84.4T= − =
Since the allowable static stress (σo) for both the pinion and gear is same (i.e. 140 MPa orN/mm2) and y'P is less than y'G, therefore the pinion is weaker. Thus the design should be based uponthe pinion.
We know that the torque on the pinion,
T =3
P
60 35 10 602 2 1200P
N× × ×=π π × = 278.5 N-m = 278 500 N-mm
∴ Tangential load on the pinion,
WT =P P
2 2 2 278 500 18 567. 30
T TD m T m m
×= = =× N
We know that pitch line velocity,
v = P P P P. . . 30 12001000 1000 1000D N m T N mπ π π × ×= = m / min
= 113.1 m m / min∴ Allowable working stress,
σw = 140 280 280
140280 280 113.1v m
= + + MPa or N / mm2
We know that length of the pitch cone element or slant height of the pitch cone,
L = P P
P1 P1
302 sin 2 sin 2 sin 33º
D m T m× ×= =θ θ = 27.54 m mm
Since the face width (b) is 1/4th of the slant height of the pitch cone, therefore
b =27.54
4 4L m= = 6.885 m mm
We know that tangential load on the pinion,
WT = (σOP × Cv) b.π m.y'P L b
L−
= σw.b.π m.y′P L b
L−
... (∵ σw = σOP × Cv)
or 18 567
m = 140
280280 113.1m
+
6.885 m × π m × 0.105 27.54 6.885
27.54m m
m−
=266 780
280 113.1m
m+
or 280 + 113.1 m = 66 780 m2 × 18 567
m = 3.6 m3
Solving this expression by hit and trial method, we find that
����������� � 1091m = 6.6 say 8 mm Ans.
and face width, b = 6.885 m = 6.885 × 8 = 55 mm Ans.Addendum and dedendum for the pinion
We know that addendum,
a = 1 m = 1 × 8 = 8 mm Ans.and dedendum, d = 1.2 m = 1.2 × 8 = 9.6 mm Ans.Outside diameter for the pinion
We know that outside diameter for the pinion,
DO = DP + 2 a cos θP1 = m.TP + 2 a cos θP1 ... (∵ DP = m . TP)
= 8 × 30 + 2 × 8 cos 33º = 253.4 mm Ans.Slant height
We know that slant height of the pitch cone,
L = 27.54 m = 27.54 × 8 = 220.3 mm Ans.Example 30.2. A pair of cast iron bevel gears connect two shafts at right angles. The pitch
diameters of the pinion and gear are 80 mm and 100 mm respectively. The tooth profiles of the gearsare of 14 1/2º composite form. The allowable static stress for both the gears is 55 MPa. If the piniontransmits 2.75 kW at 1100 r.p.m., find the module and number of teeth on each gear from the stand-point of strength and check the design from the standpoint of wear. Take surface endurance limit as630 MPa and modulus of elasticity for cast iron as 84 kN/mm2.
Solution. Given : θS = 90º ; DP = 80 mm = 0.08 m ; DG = 100 mm = 0.1 m ; φ = 1214 ° ;
Since the shafts are at right angles, therefore pitch angle for the pinion,
θP1 = tan–1 1. .V R
= tan–1 P
G
DD
= tan–1 80
100
= 38.66º
and pitch angle for the gear,
θP2 = 90º – 38.66º = 51.34º
We know that formative number of teeth for pinion,
TEP = TP . sec θP1 = 80m
× sec 38.66º = 102.4
m ... (∵ TP = DP / m )
and formative number of teeth on the gear,
TEG = TG . sec θP2 = 100m
× sec 51.34º = 160m
... (∵ TG = DG / m)
Since both the gears are made of the same material, therefore pinion is the weaker. Thus thedesign should be based upon the pinion.
We know that tooth form factor for the pinion having 14 1/2º composite teeth,
y'P = 0.124 – EP
0.684T
= 0.124 – 0.684
102.4m×
= 0.124 – 0.006 68 m
and pitch line velocity,
v = P P. 0.08 110060 60
D Nπ π × ×= = 4.6 m/s
1092 � �������������� �����������
Taking velocity factor,
Cv =6 6
6 6 4.6v=
+ + = 0.566
We know that length of the pitch cone element or slant height of the pitch cone,
*L =2 2
G P
2 2D D +
= 2 2
100 802 2
+ = 64 mm
Assuming the face width (b) as 1/3rd of the slant height of the pitch cone (L), therefore
b = L / 3 = 64 / 3 = 21.3 say 22 mm
We know that torque on the pinion,
T =P
60 2750 602 2 1100P
N× ×=
π × π × = 23.87 N-m = 23 870 N-mm
∴ Tangential load on the pinion,
WT =P
23 870/ 2 80/ 2
TD
= = 597 N
We also know that tangential load on the pinion,
WT = (σOP × Cv) b × π m × y'P L b
L−
or 597 = (55 × 0.566) 22 × π m (0.124 – 0.00 668 m) 64 22
64−
= 1412 m (0.124 – 0.006 68 m)
= 175 m – 9.43 m2
Solving this expression by hit and trial method, wefind that
m = 4.5 say 5 mm Ans.Number of teeth on each gear
We know that number of teeth on the pinion,
TP = D P / m = 80 / 5 = 16 Ans.and number of teeth on the gear,
TG = DG / m = 100 / 5 = 20 Ans.Checking the gears for wear
We know that the load-stress factor,
K = 2
P G
( ) sin 1 11.4
es
E Eσ φ +
= 2 1
23 3
(630) sin 14 / 1 11.4 84 10 84 10
°+
× × = 1.687
and ratio factor,Q = EG
EG EP
2 2 160 /160 / 102.4 /
T mT T m m
×=+ + = 1.22
* The length of the pitch cone element (L) may also obtained by using the relation
L = DP / 2 sin θ P1
The bevel gear turbine
����������� � 1093∴ Maximum or limiting load for wear,
Ww = P
1
. . . 80 22 1.22 1.687cos cos 38.66ºP
D b Q K × × ×=θ = 4640 N
Since the maximum load for wear is much more than the tangential load (WT), therefore thedesign is satisfactory from the consideration of wear. Ans.
Example 30.3. A pair of bevel gears connect two shafts at right angles and transmits 9 kW.Determine the required module and gear diameters for the following specifications :
Particulars Pinion Gear
Number of teeth 21 60
Material Semi-steel Grey cast ironBrinell hardness number 200 160
Since WS is less that WD, therefore the design is not satisfactory from the standpoint of dynamicload. We have already discussed in spur gears (Art. 28.20) that WS ≥ 1.25 WD for steady loads. Fora satisfactory design against dynamic load, let us take the precision gears having tooth error in action(e = 0.015 mm) for a module of 5 mm.
∴ Deformation or dynamic factor,
C =
3 3
0.107 0.0151 1
210 10 84 10
×
+× ×
= 96 N/mm
and dynamic load on the gear,
WD = 21 6.6 (54 96 1364)
136421 6.6 54 96 1364
× × ++× + × +
= 5498 N
From above we see that by taking precision gears, WS is greater than WD, therefore the design issatisfactory, from the standpoint of dynamic load.
Check for wear loadFrom Table 28.9, we find that for a gear of grey cast iron having B.H.N. = 160, the surface
endurance limit is,
σes = 630 MPa = 630 N/mm2
∴ Load-stress factor,
K =2
P G
( ) sin 1 11.4
es
E Eσ φ +
=
2 12
3 3
(630) sin14 1 11.4 210 10 84 10
°+
× × = 1.18 N/mm2
and ratio factor, Q = EG
EG EP
2 2 181.5181.5 22.26
TT T
×=+ + = 1.78
1096 � �������������� �����������
We know that maximum or limiting load for wear,
Ww = DP.b.Q.K = 105 × 54 × 1.78 × 1.18 = 11 910 N
Since Ww is greater then WD, therefore the design is satisfactory from the standpoint of wear.
Example 30.4. A pair of 20º full depthinvolute teeth bevel gears connect two shaftsat right angles having velocity ratio 3 : 1.The gear is made of cast steel havingallowable static stress as 70 MPa and thepinion is of steel with allowable static stressas 100 MPa. The pinion transmits 37.5 kWat 750 r.p.m. Determine : 1. Module and facewidth; 2. Pitch diameters; and 3. Pinionshaft diameter.
Assume tooth form factor,
y = 0.154 – E
.0 912T
, where TE is the
formative number of teeth, width = 1/3 rdthe length of pitch cone, and pinion shaftoverhangs by 150 mm.
Solution. Given : φ = 20º ; θS = 90º ;V.R. = 3 ; σOG = 70 MPa = 70 N/mm2 ;σOP = 100 MPa = 100 N/mm2 ; P = 37.5 kW = 37 500 W ; NP = 750 r.p.m. ; b = L / 3 ; Overhang = 150 mm
Module and face widthLet m = Module in mm,
b = Face width in mm = L / 3, ...(Given)
DG = Pitch circle diameter of the gear in mm.Since the shafts are at right angles, therefore pitch angle for the pinion,
θP1 = tan–1 1. .V R
= tan–1 13
= 18.43º
and pitch angle for the gear,θP2 = θS – θP1 = 90º – 18.43º = 71.57º
Assuming number of teeth on the pinion (TP) as 20, therefore number of teeth on the gear,
TG = V.R. × TP = 3 × 20 = 60 ... G P( . . / )V R T T=∵We know that formative number of teeth for the pinion,
TEP = TP . sec θP1 = 20 × sec 18.43º = 21.08and formative number of teeth for the gear,
TEG = TG . sec θP2 = 60 sec 71.57º = 189.8We know that tooth form factor for the pinion,
Since the shaft is subjected to twisting moment (T ) and bending moment (M ), thereforeequivalent twisting moment,
����������� � 1099
Te = 2 2+M T
2 2(893 000) (477 400)= +
= 1013 × 103 N-mmWe also know that equivalent twisting moment (Te),
1013 × 103 = 16π
× τ (dP)3
= 16π
× 45 (dP)3 = 8.84 (dP)3
... (Taking τ = 45 N/mm2)
∴ (dP)3 = 1013 × 103 / 8.84 = 114.6 × 103
or dP = 48.6 say 50 mm Ans.
EEEEEXEXEXEXEXERRRRRCISECISECISECISECISESSSSS
1. A pair of straight bevel gears is required to transmit 10 kW at 500 r.p.m. from the motor shaft toanother shaft at 250 r.p.m. The pinion has 24 teeth. The pressure angle is 20°. If the shaft axes are atright angles to each other, find the module, face width, addendum, outside diameter and slant height.The gears are capable of withstanding a static stress of 60 MPa. The tooth form factor may be taken as
0.154 – 0.912/TE, where TE is the equivalent number of teeth. Assume velocity factor as 4.5
4.5 + v ,
where v the pitch line speed in m/s. The face width may be taken as 14
of the slant height of the pitch
cone. [Ans. m = 8 mm ; b = 54 mm ; a = 8 mm ; DO = 206.3 mm ; L = 214.4 mm]
2. A 90º bevel gearing arrangement is to be employed to transmit 4 kW at 600 r.p.m. from the drivingshaft to another shaft at 200 r.p.m. The pinion has 30 teeth. The pinion is made of cast steel having astatic stress of 80 MPa and the gear is made of cast iron with a static stress of 55 MPa. The toothprofiles of the gears are of 141/2º composite form. The tooth form factor may be taken as
y' = 0.124 – 0.684 / TE, where TE is the formative number of teeth and velocity factor, Cv = 3
3 + v ,
where v is the pitch line speed in m/s.
The face width may be taken as 1/3 rd of the slant height of the pitch cone. Determine the module, facewidth and pitch diameters for the pinion and gears, from the standpoint of strength and check thedesign from the standpoint of wear. Take surface endurance limit as 630 MPa and modulus ofelasticity for the material of gears is EP = 200 kN/mm2 and EG = 80 kN/mm2.
[Ans. m = 4 mm ; b = 64 mm ; DP = 120 mm ; DG = 360 mm]
3. A pair of bevel gears is required to transmit 11 kW at 500 r.p.m. from the motor shaft to another shaft,the speed reduction being 3 : 1. The shafts are inclined at 60º. The pinion is to have 24 teeth with apressure angle of 20º and is to be made of cast steel having a static stress of 80 MPa. The gear is to bemade of cast iron with a static stress of 55 MPa. The tooth form factor may be taken asy = 0.154 – 0.912/TE, where TE is formative number of teeth. The velocity factor may be taken as
33 + v , where v is the pitch line velocity in m/s. The face width may be taken as 1/4 th of the slant
height of the pitch cone. The mid-plane of the gear is 100 mm from the left hand bearing and 125 mmfrom the right hand bearing. The gear shaft is to be made of colled-rolled steel for which the allowabletensile stress may be taken as 80 MPa. Design the gears and the gear shaft.
Bevel gears
1100 � �������������� �����������
QQQQQUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS
1. How the bevel gears are classified ? Explain with neat sketches.
2. Sketch neatly the working drawing of bevel gears in mesh.
3. For bevel gears, define the following :
(i) Cone distance; (ii) Pitch angle; (iii) Face angle; (iv) Root angle; (v) Back cone distance; and
(vi) Crown height.
4. What is Tredgold's approximation about the formative number of teeth on bevel gear?
5. What are the various forces acting on a bevel gear ?
6. Write the procedure for the design of a shaft for bevel gears.
OBJECTOBJECTOBJECTOBJECTOBJECTIVEIVEIVEIVEIVE TTTTT YPYPYPYPYPE E E E E QQQQQUEUEUEUEUESTSTSTSTSTIONSIONSIONSIONSIONS
1. When bevel gears having equal teeth and equal pitch angles connect two shafts whose axes intersectat right angle, then they are known as
4. If b denotes the face width and L denotes the cone distance, then the bevel factor is written as
(a) b / L (b) b / 2L
(c) 1 – 2 b.L (d) 1 – b / L
5. For a bevel gear having the pitch angle θ, the ratio of formative number of teeth (TE) to actual numberof teeth (T) is
(a)1
sin θ (b)1
cos θ
(c)1
tan θ (d) sin θ cos θ
ANSWEANSWEANSWEANSWEANSWERRRRRSSSSS
1. (d) 2. (b) 3. (c) 4. (d) 5. (b)
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��������� � 1101
���������
1101
31
�
�
�
�
�
1. Introduction2. Types of Worms3. Types of Worm Gears.4. Terms used in Worm
Gearing.5. Proportions for Worms .6. Proportions for Worm Gears.7. Efficiency of Worm Gearing.8. Strength of Worm Gear
Teeth .9. Wear Tooth Load for Worm
Gear. 10. Thermal Rating of Worm
Gearing.11. Forces Acting on Worm
Gears.12. Design of Worm Gearing. 31.131.131.131.131.1 IntrIntrIntrIntrIntroductionoductionoductionoductionoduction
The worm gears are widely used for transmittingpower at high velocity ratios between non-intersectingshafts that are generally, but not necessarily, at right angles.It can give velocity ratios as high as 300 : 1 or more in asingle step in a minimum of space, but it has a lowerefficiency. The worm gearing is mostly used as a speedreducer, which consists of worm and a worm wheel orgear. The worm (which is the driving member) is usuallyof a cylindrical form having threads of the same shape asthat of an involute rack. The threads of the worm may beleft handed or right handed and single or multiple threads.The worm wheel or gear (which is the driven member) issimilar to a helical gear with a face curved to conform tothe shape of the worm. The worm is generally made ofsteel while the worm gear is made of bronze or cast ironfor light service.
1102 � ���������������������������
The worm gearing is classified as non-interchangeable, because a worm wheel cut with a hob ofone diameter will not operate satisfactorily with a worm of different diameter, even if the thread pitchis same.
31.231.231.231.231.2 TTTTTypes of ypes of ypes of ypes of ypes of WWWWWororororormsmsmsmsmsThe following are the two types of worms :
1. Cylindrical or straight worm, and
2. Cone or double enveloping worm.
The cylindrical or straight worm, as shown in Fig. 31.1 (a), is most commonly used. The shapeof the thread is involute helicoid of pressure angle 14 ½° for single and double threaded worms and20° for triple and quadruple threaded worms. The worm threads are cut by a straight sided millingcutter having its diameter not less than the outside diameter of worm or greater than 1.25 times theoutside diameter of worm.
The cone or double enveloping worm, as shown in Fig. 31.1 (b), is used to some extent, but itrequires extremely accurate alignment.
Fig. 31.1. Types of worms.
31.331.331.331.331.3 TTTTTypes of ypes of ypes of ypes of ypes of WWWWWorororororm Gearm Gearm Gearm Gearm GearsssssThe following three types of worm gears are important from the subject point of view :
1. Straight face worm gear, as shown in Fig. 31.2 (a),
2. Hobbed straight face worm gear, as shown in Fig. 31.2 (b), and
3. Concave face worm gear, as shown in Fig. 31.2 (c).
Fig. 31.2. Types of worms gears.
The straight face worm gear is like a helical gear in which the straight teeth are cut with a formcutter. Since it has only point contact with the worm thread, therefore it is used for light service.
The hobbed straight face worm gear is also used for light service but its teeth are cut with ahob, after which the outer surface is turned.
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��������� � 1103
The concave face worm gear is the accepted standard form and is used for all heavy service andgeneral industrial uses. The teeth of this gear are cut with a hob of the same pitch diameter as themating worm to increase the contact area.
31.431.431.431.431.4 TTTTTerererererms used in ms used in ms used in ms used in ms used in WWWWWorororororm Gearm Gearm Gearm Gearm GearingingingingingThe worm and worm gear in mesh is shown in Fig. 31.3.
The following terms, in connection with the worm gearing, are important from the subject pointof view :
1. Axial pitch. It is also known as linear pitch of a worm. It is the distance measured axially(i.e. parallel to the axis of worm) from a point on one thread to the corresponding point on theadjacent thread on the worm, as shown in Fig. 31.3. It may be noted that the axial pitch (pa) of a wormis equal to the circular pitch ( pc ) of the mating worm gear, when the shafts are at right angles.
Fig. 31.3 . Worm and Worm gear.
Worm gear is used mostly where the power source operates at a high speed and output is at a slowspeed with high torque. It is also used in some cars and trucks.
1104 � ���������������������������
2. Lead. It is the linear distance through which a point on a thread moves ahead in onerevolution of the worm. For single start threads, lead is equal to the axial pitch, but for multiple startthreads, lead is equal to the product of axial pitch and number of starts. Mathematically,
Lead, l = pa . n
where pa = Axial pitch ; and n = Number of starts.
3. Lead angle. It is the angle between the tangent to the thread helix on the pitch cylinder andthe plane normal to the axis of the worm. It is denoted by λ.
A little consideration will show that if one completeturn of a worm thread be imagined to be unwound fromthe body of the worm, it will form an inclined plane whosebase is equal to the pitch circumference of the worm andaltitude equal to lead of the worm, as shown in Fig. 31.4.
From the geometry of the figure, we find that
tan λ =Lead of the worm
Pitch circumference of the worm
=W W
.ap nl
D D=
π π...(� l = pa . n)
=W W W
. . .cp n m n m n
D D D
π= =π π ...(� pa = pc ; and pc = π m)
where m = Module, and
DW = Pitch circle diameter of worm.
The lead angle (λ) may vary from 9° to 45°. It has been shown by F.A. Halsey that a lead angleless than 9° results in rapid wear and the safe value of λ is 12½°.
Fig. 31.4. Development of a helix thread.
Model of sun and planet gears.
INPUTSpline to AcceptMotor Shaft
Housing OD Designed to meetRAM Bore Dia, and Share MotorCoolant Supply
OUTPUT- External Spline toSpindle
Ratio Detection SwitchesHydraulic or Pneumatic SpeedChange Actuator
Round Housing With O-ringSeated Cooling Jacket
Motor Flange
Hollow Through Bore forDrawbar Integration
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For a compact design, the lead angle may be determined by the following relation, i.e.
tan λ =1/ 3
G
W
,N
N
where NG is the speed of the worm gear and NW is the speed of the worm.
4. Tooth pressure angle. It is measured in a plane containing the axis of the worm and is equalto one-half the thread profile angle as shown in Fig. 31.3.
The following table shows the recommended values of lead angle (λ) and tooth pressureangle (φ).
TTTTTaaaaable 31.1.ble 31.1.ble 31.1.ble 31.1.ble 31.1. Recommended v Recommended v Recommended v Recommended v Recommended values of lead angle and pralues of lead angle and pralues of lead angle and pralues of lead angle and pralues of lead angle and pressuressuressuressuressure angle.e angle.e angle.e angle.e angle.
For automotive applications, thepressure angle of 30° is recommendedto obtain a high efficiency and to per-mit overhauling.
5. Normal pitch. It is the distancemeasured along the normal to the threadsbetween two corresponding points ontwo adjacent threads of the worm.Mathematically,
Normal pitch, pN = pa.cos λNote. The term normal pitch is used for aworm having single start threads. In case of aworm having multiple start threads, the termnormal lead (lN) is used, such that
lN = l . cos λ6. Helix angle. It is the angle
between the tangent to the thread helix on the pitch cylinder and the axis of the worm. It is denoted byαW, in Fig. 31.3. The worm helix angle is the complement of worm lead angle, i.e.
αW + λ = 90°
It may be noted that the helix angle on the worm is generally quite large and that on the wormgear is very small. Thus, it is usual to specify the lead angle (λ) on the worm and helix angle (αG) onthe worm gear. These two angles are equal for a 90° shaft angle.
7. Velocity ratio. It is the ratio of the speed of worm (NW) in r.p.m. to the speed of the worm gear(NG) in r.p.m. Mathematically, velocity ratio,
V.R. = W
G
N
NLet l = Lead of the worm, and
DG = Pitch circle diameter of the worm gear.
We know that linear velocity of the worm,
vW = W.
60
l N
Worm gear teeth generation on gear hobbing machine.
1106 � ���������������������������
and linear velocity of the worm gear,
vG = G G
60
D Nπ
Since the linear velocity of the worm and worm gear are equal, therefore
W.
60
l N= G G W G
G
.or
60
D N N D
N l
π π=
We know that pitch circle diameter of the worm gear,
DG = m . TG
where m is the module and TG is the number of teeth on the worm gear.
∴ V.R. = W G G
G
.N D m T
N l l
π π= =
= G G G. .
.c a
a
p T p T T
l p n n= = ... ( �pc = π m = pa ; and l = pa . n)
where n = Number of starts of the worm.
From above, we see that velocity ratio may also be defined as the ratio of number of teeth on theworm gear to the number of starts of the worm.
The following table shows the number of starts to be used on the worm for the different velocityratios :
TTTTTaaaaable 31.2.ble 31.2.ble 31.2.ble 31.2.ble 31.2. Number of star Number of star Number of star Number of star Number of starts to be used on the wts to be used on the wts to be used on the wts to be used on the wts to be used on the worororororm fm fm fm fm for difor difor difor difor differferferferferent vent vent vent vent velocity raelocity raelocity raelocity raelocity ratiostiostiostiostios.
Velocity ratio (V.R.) 36 and above 12 to 36 8 to 12 6 to 12 4 to 10
Number of starts orthreads on the worm Single Double Triple Quadruple Sextuple(n = Tw)
31.531.531.531.531.5 PrPrPrPrProporoporoporoporoportions ftions ftions ftions ftions for or or or or WWWWWororororormsmsmsmsmsThe following table shows the various porportions for worms in terms of the axial or circular
S. No. Particulars Single and double Triple and quadruplethreaded worms threaded worms
1. Normal pressure angle (φ) 14½° 20°
2. Pitch circle diameter for 2.35 pc + 10 mm 2.35 pc + 10 mmworms integral with the shaft
3. Pitch circle diameter for 2.4 pc + 28 mm 2.4 pc + 28 mmworms bored to fit over the shaft
4. Maximum bore for shaft pc + 13.5 mm pc + 13.5 mm
5. Hub diameter 1.66 pc + 25 mm 1.726 pc + 25 mm
6. Face length (LW) pc (4.5 + 0.02 TW) pc (4.5 + 0.02 TW)
7. Depth of tooth (h) 0.686 pc 0.623 pc
8. Addendum (a) 0.318 pc 0.286 pc
Notes: 1. The pitch circle diameter of the worm (DW) in terms of the centre distance between the shafts (x) maybe taken as follows :
DW =0.875( )
1.416
x... (when x is in mm)
��������� � 1107
2. The pitch circle diameter of the worm (DW ) may also be taken as
DW = 3 pc, where pc is the axial or circular pitch.
3. The face length (or length of the threaded portion) of the worm should be increased by 25 to 30 mm forthe feed marks produced by the vibrating grinding wheel as it leaves the thread root.
31.631.631.631.631.6 Pr Pr Pr Pr Proporoporoporoporoportions ftions ftions ftions ftions for or or or or WWWWWorororororm Gearm Gearm Gearm Gearm GearThe following table shows the various proportions for worm gears in terms of circular pitch
S. No. Particulars Single and double threads Triple and quadruple threads
1. Normal pressure angle (φ) 14½° 20°
2. Outside diameter (DOG) DG + 1.0135 pc DG + 0.8903 pc
3. Throat diameter (DT) DG + 0.636 pc DG + 0.572 pc
4. Face width (b) 2.38 pc + 6.5 mm 2.15 pc + 5 mm
5. Radius of gear face (Rf) 0.882 pc + 14 mm 0.914 pc + 14 mm
6. Radius of gear rim (Rr) 2.2 pc + 14 mm 2.1 pc + 14 mm
31.731.731.731.731.7 EfEfEfEfEfffffficiencicienciciencicienciciency of y of y of y of y of WWWWWorororororm Gearm Gearm Gearm Gearm GearingingingingingThe efficiency of worm gearing may be defined as the ratio of work done by the worm gear to
the work done by the worm.
Mathematically, the efficiency of worm gearing is given by
η =tan (cos tan )
cos tan
λ φ− µ λφ λ + µ
...(i)
where φ = Normal pressure angle,
µ = Coefficient of friction, and
λ = Lead angle.
The efficiency is maximum, when
tan λ = 21 + µ − µIn order to find the approximate value of
the efficiency, assuming square threads, thefollowing relation may be used :
Efficiency, η =tan (1 – tan )
tan
λ µ λλ + µ
1 tan
1 / tan
− µ λ=+ µ λ
1
tan
tan ( )
λ=λ + φ
...(Substituting in equation (i), φ = 0, forsquare threads)
where φ1 = Angle of friction, such that tan φ1 = µ. A gear-cutting machine is used to cut gears.
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The coefficient of friction varies with the speed, reaching a minimum value of 0.015 at a
rubbing speed .
cosW W
rD N
vπ = λ
between 100 and 165 m/min. For a speed below 10 m/min, take
µ = 0.015. The following empirical relations may be used to find the value of µ, i.e.
µ = 0.25
0.275,
( )rv for rubbing speeds between 12 and 180 m/min
= 0.02518000
rv+ for rubbing speed more than 180 m/min
Note : If the efficiency of worm gearing is lessthan 50%, then the worm gearing is said to beself locking, i.e. it cannot be driven by applyinga torque to the wheel. This property of selflocking is desirable in some applications suchas hoisting machinery.
Example 31.1. A triple threadedworm has teeth of 6 mm module and pitchcircle diameter of 50 mm. If the worm gearhas 30 teeth of 14½° and the coefficient offriction of the worm gearing is 0.05, find1. the lead angle of the worm, 2. velocityratio, 3. centre distance, and 4. efficiencyof the worm gearing.
Solution. Given : n = 3 ; m = 6 ;DW = 50 mm ; TG = 30 ; φ = 14.5° ;µ = 0.05.
1. Lead angle of the wormLet λ = Lead angle of the worm.
We know that tan λ =W
. 6 30.36
50
m n
D
×= =
∴ λ = tan–1 (0.36) = 19.8° Ans.2. Velocity ratio
We know that velocity ratio,V.R. = TG / n = 30 / 3 = 10 Ans.
3. Centre distanceWe know that pitch circle diameter of the worm gear
DG = m.TG = 6 × 30 = 180 mm∴ Centre distance,
x = W G 50 180115 mm
2 2
D D+ += = Ans.
4. Efficiency of the worm gearingWe know that efficiency of the worm gearing.
η =tan (cos tan )
cos . tan
λ φ − µ λφ λ + µ
=tan 19.8 (cos 14.5 0.05 tan 19.8 )
cos 14.5 tan 19.8 0.05
° ° − × °° × ° +
=0.36 (0.9681 0.05 0.36) 0.342
0.858 or 85.8%0.9681 0.36 0.05 0.3985
− × = =× +
Ans.
Hardened and ground worm shaft and worm wheelpair
Krishna
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Krishna
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Krishna
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��������� � 1109Note : The approximate value of the efficiency assuming square threads is
η =1 – tan 1 0.05 0.36 0.982
0.86 or 86%1 / tan 1 0.05 / 0.36 1.139
µ λ − ×= = =+ µ λ + Ans.
31.831.831.831.831.8 Str Str Str Str Strength of ength of ength of ength of ength of WWWWWorororororm Gear m Gear m Gear m Gear m Gear TTTTTeetheetheetheetheethIn finding the tooth size and strength, it is safe to assume that the teeth of worm gear are always
weaker than the threads of the worm. In worm gearing, two or more teeth are usually in contact, butdue to uncertainty of load distribution among themselves it is assumed that the load is transmitted byone tooth only. We know that according to Lewis equation,
WT = (σo . Cv) b. π m . ywhere WT = Permissible tangential tooth load or beam strength of gear tooth,
σo = Allowable static stress,Cv = Velocity factor,b = Face width,m = Module, andy = Tooth form factor or Lewis factor.
Notes : 1. The velocity factor is given by
Cv =6
,6 v+
where v is the peripheral velocity of the worm gear in m/s.
2. The tooth form factor or Lewis factor (y) may be obtained in the similar manner as discussed in spurgears (Art. 28.17), i.e.
y =G
0.6840.124 ,
T− for 14½° involute teeth.
=G
0.9120.154 ,
T− for 20° involute teeth.
3. The dynamic tooth load on the worm gear is given by
WD =T
T6
6v
W vW
C
+ = where WT = Actual tangential load on the tooth.
The dynamic load need not to be calculated because it isnot so severe due to the sliding action between the worm andworm gear.
4. The static tooth load or endurance strength of the tooth(WS) may also be obtained in the similar manner as discussedin spur gears (Art. 28.20), i.e.
WS = σe.b π m.y
where σe = Flexural endurance limit. Itsvalue may be taken as 84 MPafor cast iron and 168 MPa forphosphor bronze gears.
31.9 31.9 31.9 31.9 31.9 WWWWWear ear ear ear ear TTTTTooth Load footh Load footh Load footh Load footh Load for or or or or WWWWWorororororm Gearm Gearm Gearm Gearm GearThe limiting or maximum load for wear (WW) is
given by
WW = DG . b . K
where DG = Pitch circle diameterof the worm gear, Worm gear assembly.
1110 � ���������������������������
b = Face width of the worm gear, and
K = Load stress factor (also known as material combination factor).
The load stress factor depends upon the combination of materials used for the worm and wormgear. The following table shows the values of load stress factor for different combination of worm andworm gear materials.
TTTTTaaaaable 31.5.ble 31.5.ble 31.5.ble 31.5.ble 31.5. VVVVValues of load stralues of load stralues of load stralues of load stralues of load stress fess fess fess fess factor (actor (actor (actor (actor (KKKKK ).).).).).
Material
S.No. Load stress factor (K)Worm Worm gear N/mm2
1. Steel (B.H.N. 250) Phosphor bronze 0.415
2. Hardened steel Cast iron 0.345
3. Hardened steel Phosphor bronze 0.550
4. Hardened steel Chilled phosphor bronze 0.830
5. Hardened steel Antimony bronze 0.830
6. Cast iron Phosphor bronze 1.035
Note : The value of K given in the above table are suitable for lead angles upto 10°. For lead angles between 10°and 25°, the values of K should be increased by 25 per cent and for lead angles greater than 25°, increase thevalue of K by 50 per cent.
31.1031.1031.1031.1031.10 TherTherTherTherThermal Ramal Ramal Ramal Ramal Rating of ting of ting of ting of ting of WWWWWorororororm Gearm Gearm Gearm Gearm GearingingingingingIn the worm gearing, the heat generated due to the work lost in friction must be dissipated in
order to avoid over heating of the drive and lubricating oil. The quantity of heat generated (Qg) isgiven by
Qg = Power lost in friction in watts = P (1 – η) ...(i)where P = Power transmitted in watts, and
η = Efficiency of the worm gearing.
The heat generated must be dissipated through the lubricating oil to the gear box housing andthen to the atmosphere. The heat dissipating capacity depends upon the following factors :
1. Area of the housing (A),
2. Temperature difference between the housing surface and surrounding air (t2 – t1), and
3. Conductivity of the material (K).
Mathematically, the heat dissipating capacity,
Qd = A (t2 – t1) K ...(ii)From equations (i) and (ii), we can find the temperature difference (t2 – t1). The average value
of K may be taken as 378 W/m2/°C.Notes : 1. The maximum temperature (t2 – t1) should not exceed 27 to 38°C.
2. The maximum temperature of the lubricant should not exceed 60°C.
3. According to AGMA recommendations, the limiting input power of a plain worm gear unit from thestandpoint of heat dissipation, for worm gear speeds upto 2000 r.p.m., may be checked from the followingrelation, i.e.
P =1.73650
. . 5
x
V R +where P = Permissible input power in kW,
x = Centre distance in metres, andV.R. = Velocity ratio or transmission ratio.
��������� � 1111
31.1131.1131.1131.1131.11 ForForForForForces ces ces ces ces Acting on Acting on Acting on Acting on Acting on WWWWWorororororm Gearm Gearm Gearm Gearm GearsssssWhen the worm gearing is transmitting power, the forces acting on the worm are similar to those
on a power screw. Fig. 31.5 shows the forces acting on the worm. It may be noted that the forces ona worm gear are equal in magnitude to that of worm, but opposite in direction to those shown inFig. 31.5.
Fig. 31.5. Forces acting on worm teeth.
The various forces acting on the worm may be determined as follows :
1. Tangential force on the worm,
WT =W
2 Torque on worm
Pitch circle diameter of worm ( )D
×
= Axial force or thrust on the worm gear
The tangential force (WT) on the worm produces a twisting moment of magnitude (WT × DW / 2)and bends the worm in the horizontal plane.
2. Axial force or thrust on the worm,
WA = WT / tan λ = Tangential force on the worm gear
=G
2 Torque on the worm gear
Pitch circle diameter of worm gear( )D
×
The axial force on the worm tends to move the worm axially, induces an axial load on thebearings and bends the worm in a vertical plane with a bending moment of magnitude (WA × DW / 2).
3. Radial or separating force on the worm,
WR = WA . tan φ = Radial or separating force on the worm gear
The radial or separating force tends to force the worm and worm gear out of mesh. This forcealso bends the worm in the vertical plane.
Example 31.2. A worm drive transmits 15 kW at 2000 r.p.m. to a machine carriage at 75 r.p.m.The worm is triple threaded and has 65 mm pitch diameter. The worm gear has 90 teeth of 6 mmmodule. The tooth form is to be 20° full depth involute. The coefficient of friction between the matingteeth may be taken as 0.10. Calculate : 1. tangential force acting on the worm ; 2. axial thrust andseparating force on worm; and 3. efficiency of the worm drive.
Solution. Given : P = 15 kW = 15 × 103 W ; NW = 2000 r.p.m. ; NG = 75 r.p.m. ; n = 3 ;DW = 65 mm ; TG = 90 ; m = 6 mm ; φ = 20° ; µ = 0.10
1. Tangential force acting on the wormWe know that the torque transmitted by the worm
=3
W
60 15 10 6071.6 N-m 71 600 N-mm
2 2 2000
P
N
× × ×= = =π π ×
1112 � ���������������������������
∴ Tangential force acting on the worm,
WT = Torque on worm 71 600
2203 NRadius of worm 65/ 2
= = Ans.
2. Axial thrust and separating force on wormLet λ = Lead angle.
We know that tan λ =W
. 6 30.277
65
m n
D
×= =
or λ = tan–1 (0.277) = 15.5°∴ Axial thrust on the worm,
WA = WT / tan λ = 2203 / 0.277 = 7953 N Ans.and separating force on the worm
WR = WA . tan φ = 7953 × tan 20° = 7953 × 0.364 = 2895 N Ans.3. Efficiency of the worm drive
We know that efficiency of the worm drive,
η =tan (cos . tan )
cos .tan
λ φ − µ λφ λ + µ
=tan 15.5 (cos 20 0.10 tan 15.5 )
cos 20 tan 15.5 0.10
° ° − × °° × ° +
= 0.277 (0.9397 0.10 0.277) 0.25260.701 or 70.1%
0.9397 0.277 0.10 0.3603
− × = =× +
Ans.
31.1231.1231.1231.1231.12 Design of Design of Design of Design of Design of WWWWWorororororm Gearm Gearm Gearm Gearm GearingingingingingIn designing a worm and worm gear, the quantities like the power transmitted, speed, velocity
ratio and the centre distance between the shafts are usually given and the quantities such as leadangle, lead and number of threads on the worm are to be determined. In order to determine thesatisfactory combination of lead angle, lead and centre distance, the following method may be used:
From Fig. 31.6 we find that the centre distance,
x = W G
2
D D+
Fig. 31.6. Worm and worm gear.
Worm gear boxes are noted for reliablepower transmission.
��������� � 1113
The centre distance may be expressed in terms of the axial lead (l), lead angle (λ) and velocityratio (V.R.), as follows :
x = (cot . .)2
lV Rλ +
πIn terms of normal lead (lN = l cos λ), the above expression may be written as :
x =N 1 . .
2 sin cos
l V R + π λ λ
orN
1 1 . .
2 sin cos
x V R
l = + π λ λ ...(i)
Since the velocity ratio (V.R.) is usually given, therefore the equation (i) contains three variablesi.e. x, lN and λ. The right hand side of the above expression may be calculated for various values ofvelocity ratios and the curves are plotted as shown in Fig. 31.7. The lowest point on each of the curvesgives the lead angle which corresponds to the minimum value of x / lN. This minimum value repre-sents the minimum centre distance that can be used with a given lead or inversely the maximum leadthat can be used with a given centre distance. Now by using Table 31.2 and standard modules, we candetermine the combination of lead angle, lead, centre distance and diameters for the given designspecifications.
Fig. 31.7. Worm gear design curves.
Note : The lowest point on the curve may be determined mathematically by differentiating the equation (i) withrespect to λ and equating to zero, i.e.
3 3
2 2
( . .) sin cos
sin . cos
V R λ − λλ λ
= 0 or V.R. = cot3 λ
Example 31.3. Design 20° involute worm and gear to transmit 10 kW with worm rotating at1400 r.p.m. and to obtain a speed reduction of 12 : 1. The distance between the shafts is 225 mm.
Solution. Given : φ = 20° ; P = 10 kW = 10 000 W ; NW = 1400 r.p.m. ; V.R.= 12 ; x = 225 mm
The worm and gear is designed as discussed below :
1. Design of wormLet lN = Normal lead, and
λ = Lead angle.
1114 � ���������������������������
We have discussed in Art. 31.12 that the value of x / lN will be minimum corresponding to
cot3 λ = V.R. = 12 or cot λ = 2.29
∴ λ = 23.6°
We know thatN
x
l=
1 1 . .
2 sin cos
V R + π λ λ
N
225
l=
1 1 12 1(2.5 13.1) 25
2 sin 23.6 cos 23.6 2 + = + = π ° ° π
∴ lN = 225 / 2.5 = 90 mm
and axial lead, l = lN / cos λ = 90 / cos 23.6° = 98.2 mm
From Table 31.2, we find that for a velocity ratio of 12, the number of starts or threads on theworm,
n = TW = 4
∴ Axial pitch of the threads on the worm,
pa = l / 4 = 98.2 / 4 = 24.55 mm
∴ m = pa / π = 24.55 / π = 7.8 mm
Let us take the standard value of module, m = 8 mm
∴ Axial pitch of the threads on the worm,
pa = π m = p × 8 = 25.136 mm Ans.Axial lead of the threads on the worm,
l = pa . n = 25.136 × 4 = 100.544 mm Ans.and normal lead of the threads on the worm,
lN = l cos λ = 100.544 cos 23.6° = 92 mm Ans.We know that the centre distance,
x = N 1 . . 92 1 12
2 sin cos 2 sin 23.6 cos 23.6
l V R + = + π λ λ π ° ° = 14.64 (2.5 + 13.1) = 230 mm Ans.
Let DW = Pitch circle diameter of the worm.
We know that tan λ =W
l
Dπ
∴ DW =100.544
73.24 mmtan tan 23.6
l = =π λ π °
Ans.
Worm gear of a steering mechanism in an automobile.
��������� � 1115
Since the velocity ratio is 12 and the worm has quadruple threads (i.e. n = TW = 4), thereforenumber of teeth on the worm gear,
TG = 12 × 4 = 48
From Table 31.3, we find that the face length of the worm or the length of threaded portion is
LW = pc (4.5 + 0.02 TW)
= 25.136 (4.5 + 0.02 × 4) = 115 mm ...(� pc = pa)
This length should be increased by 25 to 30 mm for the feed marks produced by the vibratinggrinding wheel as it leaves the thread root. Therefore let us take
LW = 140 mm Ans.We know that depth of tooth,
h = 0.623 pc = 0.623 × 25.136 = 15.66 mm Ans....(From Table 31.3)
and addendum, a = 0.286 pc = 0.286 × 25.136 = 7.2 mm Ans.∴ Outside diameter of worm,
DOW = DW + 2a = 73.24 + 2 × 7.2 = 87.64 mm Ans.2. Design of worm gear
We know that pitch circle diameter of the worm gear,
DG = m . TG = 8 × 48 = 384 mm = 0.384 m Ans.From Table 31.4, we find that outside diameter of worm gear,
DOG = DG + 0.8903 pc = 384 + 0.8903 × 25.136 = 406.4 mm Ans.Throat diameter,
DT = DG + 0.572 pc = 384 + 0.572 × 25.136 = 398.4 mm Ans.and face width, b = 2.15 pc + 5 mm = 2.15 × 25.136 + 5 = 59 mm Ans.
Let us now check the designed worm gearing from the standpoint of tangential load, dynamicload, static load or endurance strength, wear load and heat dissipation.
(a) Check for the tangential loadLet NG = Speed of the worm gear in r.p.m.
We know that velocity ratio of the drive,
V.R. =W
GG
1400or 116.7 r.p.m
. . 12WN N
NN V R
= = =
∴ Torque transmitted,
T =G
60 10 000 60818.2 N-m
2 2 116.7
P
N
× ×= =π π ×
and tangential load acting on the gear,
WT =G
2 Torque 2 818.24260 N
0.384D
× ×= =
We know that pitch line or peripheral velocity of the worm gear,
v = G G. . 0.384 116.72.35 m/s
60 60
D Nπ π × ×= =
∴ Velocity factor,
Cv =6 6
0.726 6 2.35v
= =+ +
1116 � ���������������������������
and tooth form factor for 20° involute teeth,
y =G
0.912 0.9120.154 0.154 0.135
48T− = − =
Since the worm gear is generally made of phosphor bronze, therefore taking the allowable staticstress for phosphor bronze, σo = 84 MPa or N/mm2.
We know that the designed tangential load,WT = (σo . Cv) b. π m . y = (84 × 0.72) 59 × π × 8 × 0.135 N
= 12 110 N
Since this is more than the tangential load acting on the gear (i.e. 4260 N), therefore the designis safe from the standpoint of tangential load.
(b) Check for dynamic loadWe know that the dynamic load,
WD = WT / v = 12 110 / 0.72 = 16 820 N
Since this is more than WT = 4260 N, therefore the design is safe from the standpoint ofdynamic load.
(c) Check for static load or endurance strengthWe know that the flexural endurance limit for phosphor bronze is
σe = 168 MPa or N/mm2
∴ Static load or endurance strength,
WS = σe . b. π m . y = 168 × 59 × π × 8 × 0.135 = 33 635 N
Since this is much more than WT = 4260 N, therefore the design is safe from the standpoint ofstatic load or endurance strength.
(d) Check for wearAssuming the material for worm as hardened steel, therefore from Table 31.5, we find that for
hardened steel worm and phosphor bronze worm gear, the value of load stress factor,K = 0.55 N/mm2
Gears are usually enclosed in boxes to protect them from environmentalpollution and provide them proper lubrication.
��������� � 1117
∴ Limiting or maximum load for wear,WW = DG . b . K = 384 × 59 × 0.55 = 12 461 N
Since this is more than WT = 4260 N, therefore the design is safe from the standpoint of wear.(e) Check for heat dissipation
First of all, let us find the efficiency of the worm gearing (η).We know that rubbing velocity,
vr =W W. 0.07324 1400
351.6 m / mincos cos 23.6
D Nπ π × ×= =λ °
...(DW is taken in metres)
∴ Coefficient of friction,
µ =351.6
0.025 0.025 0.044518 000 18 000
rv+ = + =
...(∴ vr is greater than 180 m/min)
and angle of friction, φ1 = tan–1 µ = tan–1 (0.0445) = 2.548°
We know that efficiency,
η =1
tan tan 23.6 0.43690.89 or 89%
tan ( ) tan (23.6 2.548) 0.4909
λ °= = =λ + φ +
Assuming 25 per cent overload, heat generated,Qg = 1.25 P (1 – η) = 1.25 × 10 000 (1 – 0.89) = 1375 W
We know that projected area of the worm,
AW = 2 2 2W( ) (73.24) 4214 mm
4 4D
π π= =
and projected area of the worm gear,
AG = 2 2 2G( ) (384) 115 827 mm
4 4D
π π= =
∴ Total projected area of worm and worm gear,A = AW + AG = 4214 + 115 827 = 120 041 mm2
= 120 041 × 10–6 m2
We know that heat dissipating capacity,Qd = A (t2 – t1) K = 120 041 × 10–6 (t2 – t1) 378 = 45.4 (t2 – t1)
The heat generated must be dissipated in order to avoid over heating of the drive, thereforeequating Qg = Qd, we have
t2 – t1 = 1375 / 45.4 = 30.3°CSince this temperature difference (t2 – t1) is within safe limits of 27 to 38°C, therefore the design
is safe from the standpoint of heat.3. Design of worm shaft
Let dW = Diameter of worm shaft.We know that torque acting on the worm gear shaft,
Tgear =G
1.25 60 1.25 10000 601023 N-m
2 2 116.7
P
N
× × ×= =π π ×
= 1023 × 103 N-mm ...(Taking 25% overload)
∴ Torque acting on the worm shaft,
Tworm = 31023
96 N-m 96 10 N-mm. . 12 0.89
gearT
V R= = = ×
× η ×
1118 � ���������������������������
We know that tangential force on the worm,
WT = Axial force on the worm gear
= 3
W
2 2 96 102622 N
73.24wormT
D
× × ×= =
Axial force on the worm,
WA = Tangential force on the worm gear
=3
G
2 2 1023 105328 N
384gearT
D
× × ×= =
and radial or separating force on the worm
WR = Radial or separating force on the worm gear
= WA . tan φ = 5328 × tan 20° = 1940 N
Let us take the distance between the bearings of the worm shaft (x1) equal to the diameter of theworm gear (DG), i.e.
x1 = DG = 384 mm
∴ Bending moment due to the radial force (WR ) in the vertical plane
= R 1 1940 384186240 N-mm
4 4
W x× ×= =
and bending moment due to axial force (WA) in the vertical plane
= A W 5328 73.2497556 N-mm
4 4
W D× ×= =
∴ Total bending moment in the vertical plane,
M1 = 186 240 + 97 556 = 283 796 N-mm
We know that bending moment due to tangential force (WT) in the horizontal plane,
or dG = 48.8 say 50 mm Ans.Let us now check the maximum shear stress induced.
We know that actual shear stress,
τ =6
23 3
G
16 16 1.14 1046.4 N/mm 46.4 MPa
( ) (50)
egT
d
× ×= = =π π
and direct compressive stress on the shaft due to the axial force,
σc =3
2 2G
Axial force 26221.33 N/mm 1.33 MPa
( ) (50)4 4
d= = =
π π
∴ Maximum shear stress,
τmax = 2 2 2 21 1( ) 4 (1.33) 4 (46.4) 46.4MPa
2 2cσ + τ = + =
Since the maximum shearstress induced is less than 50 MPa(assumed), therefore the design forworm gear shaft is satisfactory.
Example 31.4. A speedreducer unit is to be designed foran input of 1.1 kW with atransmission ratio 27. The speed ofthe hardened steel worm is 1440r.p.m. The worm wheel is to be madeof phosphor bronze. The tooth formis to be 20° involute.
Solution. Given : P = 1.1 kW= 1100 W ; V.R. = 27 ; NW = 1440r.p.m. ; φ = 20°
A speed reducer unit (i.e.,worm and worm gear) may bedesigned as discussed below. Sun and Planet gears.
Speed changeshift axis
Bearing housingoutput belt pulley
Slide dogclutch
Output sungear
Motorflange
Input sungear
Planetgears
Oil collector
Krishna
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��������� � 1121
Since the centre distance between the shafts is not known, therefore let us assume that for thissize unit, the centre distance (x) = 100 mm.
We know that pitch circle diameter of the worm,
DW =0.875 0.875( ) (100)
39.7 say 40 mm1.416 1.416
x = =
∴ Pitch circle diameter of the worm gear,
DG = 2x – DW = 2 × 100 – 40 = 160 mm
From Table 31.2, we find that for the transmission ratio of 27, we shall use double startworms.
∴ Number of teeth on the worm gear,
TG = 2 × 27 = 54
We know that the axial pitch of the threads on the worm (pa) is equal to circular pitch of teeth onthe worm gear ( pc).
∴ pa = G
G
1609.3 mm
54cD
pT
π π ×= = =
and module, m = 9.3
2.963 say 3 mmcp= =
π π∴ Actual circular pitch,
pc = π m = π × 3 = 9.426 mm
Actual pitch circle diameter of the worm gear,
DG = G. 9.426 54162 mmcp T ×= =
π πAns.
and actual pitch circle diameter of the worm,
DW = 2x – DG = 2 × 100 – 162 = 38 mm Ans.The face width of the worm gear (b) may be taken as 0.73 times the pitch circle diameter of
worm (DW).
∴ b = 0.73 DW = 0.73 × 38 = 27.7 say 28 mm
Let us now check the design from the standpoint of tangential load, dynamic load, static load orendurance strength, wear load and heat dissipation.
1. Check for the tangential load
Let NG = Speed of the worm gear in r.p.m.
We know that velocity ratio of the drive,
V.R. =W W
GG
1440or 53.3 r.p.m
. . 27
N NN
N V R= = =
∴ Peripheral velocity of the worm gear,
v =G G. 0.162 53.3
0.452 m/s60 60
D Nπ π × ×= =
... (DG is taken in metres)
and velocity factor, Cv = 6 6
0.936 6 0.452v
= =+ +
We know that for 20° involute teeth, the tooth form factor,
y =G
0.912 0.9120.154 0.154 0.137
54T− = − =
1122 � ���������������������������
From Table 31.4, we find that allowable static stress for phosphor bronze is
σo = 84 MPa or N/mm2
∴ Tangential load transmitted,
WT = (σo . Cv) b . π m . y = (84 × 0.93) 28 × π × 3 × 0.137 N
= 2825 N
and power transmitted due to the tangential load,
P = WT × v = 2825 × 0.452 = 1277 W = 1.277 kW
Since this power is more than the given power to be transmitted (1.1 kW), therefore the designis safe from the standpoint of tangential load.
2. Check for the dynamic load
We know that the dynamic load,
WD = WT / Cv = 2825 / 0.93 = 3038 N
and power transmitted due to the dynamic laod,
P = WD × v = 3038 × 0.452 = 1373 W = 1.373 kW
Since this power is more than the given power to be transmitted, therefore the design is safefrom the standpoint of dynamic load.
3. Check for the static load or endurance strength
From Table 31.8, we find that the flexural endurance limit for phosphor bronze is
σe = 168 MPa or N/mm2
∴ Static load or endurance strength,
WS = σe . b. π m . y = 168 × 28 × π × 3 × 0.137 = 6075 N
and power transmitted due to the static load,
P = WS × v = 6075 × 0.452 = 2746 W = 2.746 kW
Since this power is more than the power to be transmitted (1.1 kW), therefore the design is safefrom the standpoint of static load.
4. Check for the wear loadFrom Table 31.5, we find that the load stress factor for hardened steel worm and phosphor
bronze worm gear is
K = 0.55 N/mm2
∴ Limiting or maximum load for wear,
WW = DG . b . K = 162 × 28 × 0.55 = 2495 N
and power transmitted due to the wear load,
P = WW × v = 2495 × 0.452 = 1128 W = 1.128 kW
Since this power is more than the given power to be transmitted (1.1 kW), therefore the designis safe from the standpoint of wear.
5. Check for the heat dissipationWe know that permissible input power,
P =1.7 1.73650 ( ) 3650 (0.1)
2.27 kW. 5 27 5
x
V R= =
+ +... (x is taken in metres)
Since this power is more than the given power to be transmitted (1.1 kW), therefore the designis safe from the standpoint of heat dissipation.
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1. A double threaded worm drive is required for power transmission between two shafts having theiraxes at right angles to each other. The worm has 14½° involute teeth. The centre distance is approxi-mately 200 mm. If the axial pitch of the worm is 30 mm and lead angle is 23°, find 1. lead; 2. pitchcircle diameters of worm and worm gear; 3. helix angle of the worm; and 4. efficiency of the drive ifthe coefficient of friction is 0.05. [Ans. 60 mm ; 45 mm ; 355 mm ; 67° ; 87.4%]
2. A double threaded worm drive has an axial pitch of 25 mm and a pitch circle diameter of 70 mm. Thetorque on the worm gear shaft is 1400 N-m. The pitch circle diameter of the worm gear is 250 mm andthe tooth pressure angle is 25°. Find : 1. tangential force on the worm gear, 2. torque on the wormshaft, 3. separating force on the worm, 4. velocity ratio, and 5. efficiency of the drive, if the coefficientof friction between the worm thread and gear teeth is 0.04.
[Ans. 11.2 kN ; 88.97 N-m ; 5220 N ; 82.9%]
3. Design a speed reducer unit of worm and worm wheel for an input of 1 kW with a transmission ratioof 25. The speed of the worm is 1600 r.p.m. The worm is made of hardened steel and wheel ofphosphor bronze for which the material combination factor is 0.7 N/mm2. The static stress for thewheel material is 56 MPa. The worm is made of double start and the centre distance between the axesof the worm and wheel is 120 mm. The tooth form is to be 14½° involute. Check the design forstrength, wear and heat dissipation.
4. Design worm and gear speed reducer to transmit 22 kW at a speed of 1440 r.p.m. The desired velocityratio is 24 : 1. An efficiency of atleast 85% is desired. Assume that the worm is made of hardened steeland the gear of phosphor bronze.
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1. Discuss, with neat sketches, the various types of worms and worm gears.
2. Define the following terms used in worm gearing :
(a) Lead; (b) Lead angle; (c) Normal pitch; and (d) Helix angle.
3. What are the various forces acting on worm and worm gears ?
4. Write the expression for centre distance in terms of axial lead, lead angle and velocity ratio.
The worm in its place. One can also see the two cubic worm bearing blocks and thebig gear.
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1. The worm gears are widely used for transmitting power at ........... velocity ratios between non-inter-secting shafts.
(a) high (b) low
2. In worm gears, the angle between the tangent to the thread helix on the pitch cylinder and the planenormal to the axis of worm is called
(a) pressure angle (b) lead angle
(c) helix angle (c) friction angle
3. The normal lead, in a worm having multiple start threads, is given by
(a) lN = l / cos λ (b) lN = l . cos λ(c) lN = l (d) lN = l tan
where lN = Normal lead,
l = Lead, and
λ = Lead angle.
4. The number of starts on the worm for a velocity ratio of 40 should be
(a) single (b) double
(c) triple (d) quadruple
5. The axial thrust on the worm (WA) is given by
(a) WA = WT . tan φ (b) WA = WT / tan φ(c) WA = WT . tan λ (d) WA = WT / tan λ
