Kom Unit 6 - (revised) (1).docx

Unit 6 - GEARS Sub – Topic 1 – SPUR GEARS Learning Objectives : By the end of studying this topic, the student must be able to understand :  The main advantage of gear drives over the other forms of drives is that this is a positive drive, and thereby transmits a constant velocity ratio.

 A friction wheel with Teeth cut on it is termed as a Toothed Wheel (or) Gear.

 Classification of gears such as Spur gears, Helical, Spiral, Bevel, and Worm gears.  Gear tooth terminology of Spur gears.  Different forms of gear teeth such as Involute and Cycloidal profiles.  Law of Gearing, and Condition for constant velocity ratio of toothed wheels.  How to compute the velocity of sliding of gear teeth, path of contact, and arc of contact.

 What is ‘Interference’ in Involute gears, and how to avoid / reduce it.  Interference between rack and pinion.  What is ‘Under – cutting’, and why it is done. Introduction : The transmission of motion from one shaft to another can be achieved by Belt drives, Rope drives, Chain drives, or Gear drives. The gear drives are especially used to connect two shafts which are at a short distance apart, while the other types of drives are suitable to 1

connect the shafts which are at large distance apart. The main advantage of gear drives over the other forms of drives is that this is a positive drive, and thereby transmits a constant velocity ratio. Of course the manufacture of gears is quite complex, as compared to the other types of drives, because it involves careful cutting of the gear tooth profiles accurately on special purpose machines. Consider two plain cylindrical wheels 1 and 2, with sufficient rough surfaces – pressed against each other. The wheel 1 is keyed to a shaft, which is rotating at some speed. The wheel 2 is keyed to another shaft, which is to be rotated. Then the wheel 2 will be rotated in a direction opposite to that of 1 -- so long as -- the tangential force exerted by the wheel 1 does not exceed the maximum frictional resistance between the wheels. (Refer Fig.1)

Fig.1 When the tangential force exceeds the maximum frictional resistance between the wheels, slipping occurs between the wheels. To avoid slipping  a number of projections, called TEETH are provided on the periphery of the wheel 1, which will fit into the corresponding recesses on the periphery of the wheel 2 Thus, a friction wheel → with Teeth cut on it → is termed as a Toothed Wheel (or) GEAR. (Refer Fig.2)

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Fig.2 Classification of Gears : (1) According to the position of axes of shafts : (a) Parallel → Spur gears, and Helical gears. [Refer Fig.3 (a)] (b) Intersecting → Bevel gears → Straight, and Helical gears. [Fig.3(b)]

Fig.3(a) 3

Fig.3(a)

Fig.3(b) 4

(c) Non – Intersecting, and Non – parallel → Skew Bevel gears (or) Crossed Helical gears (or) Spiral gears. [Refer Fig.3(c)]

Fig.3(c) (d) Worm gears:[Refer Fig.3(d)] → For connecting non – intersecting (skew) Shafts. (a) Two Parallel and Coplanar shafts are connected by  Spur Gears. In these gears, the gear teeth are parallel to the axis of shafts, whereas in Helical gears, the gear teeth are inclined to the axis of the gear wheel. Helical gears  are again classified into two types : (i)

Single Helical gears.

(ii)

Double Helical gears (or) Herringbone gears [Fig.3(a)]

(b) Two intersecting – Non – parallel – but Coplanar shafts are connected by Bevel gears of either straight teeth (Spur Bevel gears) or of inclined teeth (Helical Bevel gears)  [Refer Fig.3(b)]. (c) Two non – intersecting, non – parallel, and non – coplanar shafts are connected by Skew Bevel gears or Spiral gears. [Refer Fig.3(c)]. These gears have curved teeth -- and -- this type of gearing will have line contact. Rotation of the line of contact about the axes of

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the mating gears  generates the two pitch surfaces -- which are hyperboloids. [A hyperboloid is a solid formed by revolving a straight line about an axis (which is not in the same plane), such that every point on the line remains at a constant distance from the axis]. (d) Worm gears are used to connect two non – intersecting (skew) shafts. They are normally used when high velocity reduction is required between the driver and driven gears. The pinion (or Worm) has a small number of teeth (threads) – usually 1 to 4 – and its mating gear is called the Worm wheel. The worm gears are used where high loads are to be transmitted. (2) According to the type of gearing : (a) External Gearing :

In this type of gearing, the larger wheel is

referred to as the (Gear) Wheel, and the smaller wheel is called Pinion. The direction of rotation of the gear wheels are opposite to each other. (b) Internal Gearing :

In this type of gearing, the larger wheel is

referred to as the Annular wheel (or) Annulus, and the smaller wheel is called Pinion. The direction of rotation of the gear wheels are the same. (c) Rack – and – Pinion :

Here, the pitch circle of the gear wheel is

a straight line (or) the radius of curvature of the pitch circle is infinite. Such a gear wheel is called Rack. The smaller wheel, as usual, called Pinion. This type of gearing is used to convert the linear motion into rotary motion -- or -- vice – versa. Gear Tooth Terminology [Refer Fig.4(a) and (b)]

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Fig.4 (1) Pitch Circle : An imaginary circle -- which, by pure rolling motion -Would give the same motion as the actual gear. (2) Pitch circle diameter (or) Pitch diameter (d) :

The size of gear is

usually specified by the Pitch diameter (d) in mm. (3) Pitch Point : The common point of contact between the pitch circles of two mating gears. 7

(4) Pitch Surface :

This is the surface of the rolling discs, which

were replaced by the mating gears, at the pitch circle. (5) Pitch Line :

The line of contact of the pitch surfaces of two mating

gears. (6) Pressure angle (or) Angle of Obliquity (φ) :

The angle between

the common normal to the two mating gear teeth at the point of contact and the common tangent at the pitch point. The standard values for φ are 14.50 and 200. (7) Addendum :

The radial distance of a gear tooth from the pitch

circle to the top of the tooth. (8) Dedendum :

The radial distance of a gear tooth from the pitch

circle to the bottom of the tooth. (9) Addendum Circle : The circle drawn through the top of the gear teeth -- and -- which is concentric with the pitch circle. (10) Dedendum Circle (or) Root Circle : The circle drawn through the bottom of the gear teeth -- and -- which is concentric with the pitch circle. Root Circle diameter = Pitch circle diameter x cos φ. (11) Circular Pitch (pc) :

The distance measured on the circumference

of a pitch circle -- from a point of one gear tooth -- to a corresponding point on the next tooth. pc =

d T

, where ‘d’ is the pitch diameter and ‘T’ is number of

teeth on the gear wheel. Note :

The two meshing gear wheels should have the same circular pitch for correct meshing.

∴ pc =

 d1  d 2 d T d N  1 = 1 ; Also  π d1 N1 = π d2N2  1 = 2 = T1 d2 T2 d 2 N1 T2

(12) Diametral Pitch (pd) :

This is the ratio of the number of teeth to 8

the pitch cir cle diameter in mm. T d   ( pc = ∴ pd = = ) d pc T (13) Module (m) :

This is the inverse of diametral pitch  ∴ m =

d . T

(14) Clearance : The radial distance from the bottom of the tooth in one gear to the top of the meshing tooth in the mating gear. (15) Clearance Circle : A circle passing through the top of the teeth in the mating gear. (16) Total Depth : The radial distance between the Addendum circle and Dedendum circle of a gear. ∴ Total Depth = Addendum + Dedendum. (17) Working Depth :

The radial distance between the Addendum circle

and Clearance circle. ∴ Working Depth = Sum of addend of the two meshing gears. (18) Tooth Thickness : The width of a gear tooth measured along its pitch circle. (19) Tooth Space :

The width of space between two adjacent gear

teeth measured along its pitch circle. (20) Face of Tooth : The surface of a gear tooth above the pitch surface. (21) Top Land : The surface on the top of a gear tooth. (22) Flank of Tooth :The surface of a gear tooth below the pitch surface. (23) Face Width : The width of a gear tooth measured parallel to its axis. (24) Profile :

The curve formed by the face and flank of a gear tooth.

(25) Fillet Radius :

The radius connecting the Root circle to the tooth

profile. (26) Path of Contact : The curve traced by the point of contact of the mating teeth -- from the beginning to the end of their engagement. (27) Arc of Contact :

The path followed by a point on either pitch line

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of the mating gear teeth -- from the beginning to the end of their engagement. The first part of this -- from the beginning of engagement of the mating gear teeth until the two teeth are in contact at the pitch point  is termed as  the Arc of Approach. The second part -- from the pitch point until the end of engagement of the mating gear teeth  is termed as  the Arc of Recess. Forms of Gear Teeth 1. Cycloidal teeth. 2. Involute teeth. 1. Cycloidal teeth :

[Refer Fig.5(a), (b), and (c)]

A Cycloid is the locus of a point on the circumference of a circle -- which rolls without slipping -- along a straight line, as shown in Fig.5(a). An Epicycloid is the locus of a point on the circumference of a circle -- which rolls without slipping -- on the outside of a circular arc. A Hypocloid is the locus of a point on the circumference of a circle -- which rolls without slipping -- on the inside of a circular arc. An important aspect, which makes the cycloid suitable for the profile of a gear tooth, is that the line joining the tracing point A to the point of contact P between the rolling circle and the straight / curved line is normal to the cycloid. Cycloidal Profile :

[Refer Fig.5(b)]

Let two circles X and Y -- with centers at O and Q respectively --

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Fig.5(a)

Fig.5(b) 11

be in contact at the point P. A smaller circle Z with center at R is also in contact with the circle X at P. Now, if all the three circles X, Y, and Z turn about their centers -- and roll together without slipping -- then a point A on the circle Z will trace a hypocycloid on X, and an epicycloid on Y. Thus, if the circle X turns clockwise through an angle α, then Y and Z will turn clockwise through angles β and γ respectively. Now, if we consider three points A, a, and a1 initially coinciding with the point P (at the beginning of motion), then  the arcs PA, Pa, and Pa1 will be of equal length -- and the tracing point A on the circle Z will describe the portion aA of hypocycloid on X, and the portion a1A of epicycloid on Y. Thus, A is the point of contact of the cycloids, and AP is the common normal. To transmit continuous motion from X to Y, a number of similar pairs of cycloids are to spaced around the circumferences of X and Y  so that the contact between the next pair begins before the contact

between the previous pair ends. Therefore, for maintain contact beyond P, a 2nd generating circle Z1 in contact with the circle Y is considered. And the tracing point B on Z1 traces the epicycloid portion Bb on X, and the hypocycloid portion Bb1 on Y. Construction of the Cycloidal tooth profile :

[Refer Fig.5(c)]

 A point on the circle D will trace the flank of the gear tooth T1, when D rolls without slipping on the inside of the pitch circle of gear wheel 1.  A point on the circle D will trace the face of the gear tooth T2, when D rolls without slipping on the outside of the pitch circle of gear wheel 2.  Similarly, a point on the circle C will trace the flank of the gear tooth T2, when C rolls without slipping on the inside of the pitch circle of gear wheel 2. 12

Fig.5(c)  And a point on the circle C will trace the face of the gear tooth T1, when C rolls without slipping on the outside of the pitch circle of gear wheel 1.  The rolling circles C and D need not be of equal diameter.  The common normal at the point of contact between the mating cycloidal teeth always passes through the pitch point P. Therefore, the Law of Gearing is satisfied by the profiles of the mating cycloidal teeth, and hence a constant velocity ratio can be obtained. Alternate method of constructing the Cycloidal Tooth Profile : [Fig.5(d)]

Fig.5(d) 13

 The circle A, rolling on the outside of the pitch circle of a gear tooth, generates an epicycloids, which forms the face portion of the tooth profile.  The circle A rolls, without slipping, to the right side.  The circle B, which rolls without slipping to the left side, on the inside of the pitch circle generates a hypocloid.  This hypocloid forms the flank profile of the cycloidal gear tooth. Construction of the Involute tooth Profile :

(Refer Fig.6)

Fig.6 Definition of Involute : The locus of a point on a straight line -- which rolls without slipping -- on the circumference of a circle -- or -- the locus of a point on a cord, which is held tout, and unwound from a cylinder. Constructing the Involute Profile :

(Refer Fig.6)

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 Divide the base circle into a number of equal parts -- starting from the point A -- like A1, 12, 23, ... , etc.,  Draw tangents to the base circle at the points 1, 2, 3, ... , etc.,  Set off lengths 1a, 2b, 3c, .... , etc., (from the points 1, 2, 3, ... , etc.,) equal to the arc lengths of 1A, 2A, 3A, ... , etc., along the tangents.  Join A, a, b, c, ... , etc., by a smooth curve.  This is the required Involute curve. On the involute curve, if a tangent is drawn at any point, say d to the curve, it will be perpendicular to the corresponding generating line, i.e., 4d. Therefore, the line 4d will become normal to the Involute. As a corollary, the normal to the involute at any point will be tangent to the base circle. Law of Gearing -- Condition for Constant Velocity Ratio of Toothed Wheels :

(Refer Fig.7)

Fig.7 15

 Consider the two portions 1 and 2 -- one on the gear wheel 1, and the other on the gear wheel 2.  The portions 1 and 2 are in contact at the point Q.  The gear wheels 1 and 2 are rotating in opposite sense to each other -- about the fixed centers O1 and O2 -- and with angular velocities ω1 and ω2 -- respectively.  N1N2 is the common normal to both the curves 1 and 2, drawn at the point of contact Q.  The line O1O2 -- which is the line of centers -- cuts N1N2 at the point P.  Further, the lines O1M and O2N are drawn such that they are at right angles to the common normal N1N2. Now  Q is a point common to both the wheels. ∴ When Q is considered as a point on the wheel 1  it moves in the direction QC  which is perpendicular to O1Q. But  when Q is considered as a point on the wheel 2  it moves in the direction QD  which is perpendicular to O2Q. Thus  v1  is the velocity of Q as a point on the wheel 1 -- and -v2  is the velocity of Q as a point on the wheel 2. Now  if the teeth on the wheels are to remain in contact → then, The component of v1 along MN = The component of v2 along MN. i.e., v1 cos α = v2 cos β -- (or) -- (ω1  O1Q) cos α = (ω2  O2Q) cos β. ∴ ω1  O1Q 

O1M O N O N ω = ω2  O2Q  2  ω1  O1M = ω2  O2N  1 = 2 -- (1) O1Q O 2Q ω 2 O1M

Now  from the similar triangles O1MP and O2NP  ∴

O2 N O2 P = -- (2) O1M O1P

ω1 O 2 N O 2 P = = ------------------------------- (3) ω 2 O1M O1P

Thus, the angular velocities of wheels 1 and 2 are in a ratio which is 16

inversely proportional to the ratio of distances of the point ‘P’ from their centers O1 and O2 -- (or) -- The common normal to the two surfaces at the point of contact Q intersects the line of centers O 1O2 at a point P, which divides the line O1O2 inversely as the ratio of their angular velocities. ω  ∴ To have a constant velocity ratio  1  for all positions of the wheels,  ω2 

P must be a fixed point for both the wheels. This fixed point P is called the Pitch point. In other words  the common normal at the point of contact between a pair of teeth must always pass through the Pitch point  This is the fundamental condition to be satisfied while designing the profiles for the teeth of gear wheels. This is called ‘The Law of Gearing’. Note : (1) If the shape of one gear tooth profile is chosen arbitrarily, and the profile of the mating gear tooth is designed to satisfy the law of gearing  then the 2nd gear tooth is said to be Conjugate to the 1st gear tooth.

However, conjugate teeth are not common  because of the difficulty in manufacturing them, hence higher cost of production. (2) If the pitch circle diameters of the mating gear wheels are d1 and d2  d2  N1 ω1 O2 P  2  d 2 T2  then      . N 2 ω2 O1P  d1  d1 T1   2 Velocity of Sliding of Gear Teeth :  Sliding between a pair of gear teeth in contact at Q occurs along the common tangent TT to the teeth profile curves .  The velocity of one gear tooth relative to its mating gear tooth along the common tangent at their point of contact is called the 17

Velocity of Sliding. Now  The velocity of point Q along TT → as a point on the gerar wheel 1 → is represented by EC. From the similar triangles QEC and O1MQ 

EC QC v   1  ω1 . MQ O1Q O1Q

∴ EC = ω1 .MQ Similarly  the velocity of point Q along TT→ as a point on the gear wheel 2 → is represented by ED. Again, from the similar triangles QED and O2NQ 

v ED QD   2  ω2 . QN O2 Q O2 Q

∴ ED = ω 2 .QN Now  let vs = Velocity of sliding at Q. ∴ vs = (ED – EC) = ( ω 2 .QN - ω1 .MQ) = ω 2 (QP + PN) – ω1(MP – QP) = (ω1 + ω 2 ).QP + ( ω 2 .PN - ω1.MP). But  From the similar triangles O2NP and O1MP  ∴ ω1 .MP = ω 2 .PN



ω1 O2 P PN .   ω2 O1P MP

∴ vs = (ω1 + ω 2 ).QP

Thus, vs is directly proportional to distance of the point of contact Q from the pitch point P. Centre distance between the mating Gear wheels :

(Refer Fig.8)

 Let O1 and O2  be the fixed centres of base circles of the two mating gear wheels -- and -- AB and A′B′ be the involute profiles of the corresponding gear teeth.  Q is the point of contact between AB and A′B′ at a given instant.  QM and QN are the tangents drawn at Q to the two base circles.  TT is the common tangent to the two involute profiles AB and A′B′.  ∴ QN is normal to the involute profile A′B′ -- and -- hence normal to the common tangent TT. 18

 Similarly  QM is normal to the involute profile AB -- and -hence normal to the common tangent TT.

Fig.8  Thus  both QM and QN are normal to the same line TT -hence MQN is a single (straight) line, and it is the common normal to the two involute profiles AB and

A′B′.

 ∴ The normal to an involute at a given point is the tangent drawn from that point to the base circle.  Thus  the common normal MN at Q to the involute profiles is also common tangent to the corresponding base circles.  Further, it can be seen that the common normal MN drawn at Q intersects the line of centres O1O2 at the fixed point P -- which is the Pitch point.

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 ∴ The involute profile teeth satisfy the fundamental law of gearing -- hence a constant velocity ratio can be obtained. Now  from the similar triangles O2NP and O1MP 

O1M O1P ω2 = =  O2 N O2P ω1

(from the law of gearing). ∴ Radii of the base circles  i.e., O1M and O2N are given by : O1M = O1P cos φ -- and -- O2N = O2P cos φ  where φ is the Pressure angle (or) Angle of Obliquity. Further  the centre distance O1O2= (O1P+O2P) = Length of Path of Contact :

O1M O2 N (O1M+O2 N)   cos cos cos

(Refer Fig.9)

Fig.9 Consider a pinion rotating clockwise driving a gear wheel. The contact between the two gears begins at K (i.e., at the point where the addendum circle of the gear wheel meets the common normal MN), and 20

ends at the point L (i.e., at the point where the addendum circle of the pinion meets the common normal). MN  is the common normal drawn through the points of contact K / L -- and -- it is also the common tangent to the base circles. ∴ Length of the Path of Contact = KL = (KP + PL) KP = Path of Approach -- and -- PL = Path of Recess. Now  From the ∟ triangle O2KN  KN =

 O K -O N  2

2

2

2

=

 O K -O P 2

2

2

2

R

cos2   =

2 A

- R 2 cos2   .

And -- PN = O2P sin φ = R sin φ . ∴ Path of Approach = KP = (KN – PN) =

R

2 A

- R 2 cos2   - R sin φ .

Similarly  From the ∟ triangle O1ML  ML =

O L - O M  = O L - O P 2

1

2

2

1

1

2

1

r

cos2   =

2

A

- r 2 cos2  

.

And -- MP = O1P sin φ = r sin φ . ∴ Path of Recess = PL = (ML – MP) = ∴ Path of Contact = KL = (KP + PL) =

 r - r cos   - r sin φ .  R - R cos   +  r - r 2

2

2

A

2

2

2

2

A

A

2

cos2  

- (R + r) sin φ. Length of Arc of Contact :

(Refer Fig.10 and Fig.11))

By definition  Arc of Contact is the path traced by a point on the pitch circle from the beginning to the end of engagement of a pair of gear teeth -- or --

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Fig.10 – Arc of Contact the distance travelled by a point on either pitch circle of the mating gear wheels during the period of contact of a pair of gear teeth. GH, JK, and DL  are the positions of the same involute at different instants of time. GH  The position of involute at the beginning of engagement. JK  The position of involute at the pitch point. DL  The position of involute at the end of engagement . P′PP′′  The Arc of Contact = Arc of Approach + Arc of Recess. (P′P) (PP′′) Now  let the time to traverse the Arc of Approach = ta -- and -the time to traverse the Arc of Recess = tr . Then  Arc of Approach (P′P) = (Tangential velocity of P)  ta = (ω.r).ta t t = ω  (r cos φ)  a = (Tangential velocity of H)  a  ( AH = AF) cos cos

Arc(HK)  Arc(FK)  Arc(FH) = . cos  cos Now  Arc FK = Path FP  (the point P is on the generator FP which =

rolls on the base circle FHK to generate the involute PK). Similarly  Arc FH = Path FC. ∴ Arc of Approach =

(FP - FC) CP Path of Approach   cos cos cos 22

.

Now  Arc of Recess = PP′′ = (Tangential velocity of P)  tr = (ω.r).tr t t  ( AH = AF) = ω  (r cos φ)  r = (Tangential velocity of K)  r cos  cos 

tr Arc(KL)  Arc(FL)  Arc(FK) = = cos  cos  cos  (FD - FP) PD Path of Re cess = .   cos  cos  cos  CP PD CD ∴ Arc of Contact = P′PP′′ = (P′P + PP′′) = + = cos  cos cos  = (Tangential velocity of K) x

= Alternate approach :

Path of Contact . cos

(Refer Fig.11)

Fig.11

The Arc of Contact is the arc EPF (or) arc GPH. Now  consider the arc GPH  This is divided into {arc GP + arc PH} Arc GP  is the Arc of approach ;Arc PH  is the Arc of Recess.

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The angle subtended by the arc GP at the centre O1 of gear wheel 1 is termed as the Angle of Approach -- and -- the angle subtended by the arc PH at the centre O2 of gear wheel 2 is the Angle of Recess. Arc of Approach GP = Arc of recess PH =

Path of Approach KP KP     cos   = . GP cos cos  

Path of Re cess PL = . cos  cos 

∴ Length of the Arc of Contact = (Arc GP + Arc PH) = = Interference in Involute Gears :

KP PL + cos cos 

Path of Contact (KP + PL) KL = = . cos  cos  cos (Refer Fig.12)

MN  is common tangent to the two base circles. KL  Path of contact between the two mating gear teeth.

Fig.12 24

 Now  if the radius of addendum circle of the pinion is increased to O1N  then, the point of contact L moves to N.  If the radius is further increased  then L will move on to the inside of the base circle of the wheel  but the point of contact will not lie on the involute profile of the gear tooth on wheel.  This results in  the tip of the gear tooth on pinion will undercut (or dig out the flank of the wheel tooth) the gear tooth on the wheel at its root -- thereby removing part of the involute profile of the gear tooth on the wheel.  This effect  is termed as Interference between the mating gear teeth. Definition : The phenomenon of the tip of a gear tooth undercutting the root of its mating gear tooth  is called ‘Interference’.  Similarly  if the radius of addendum circle of wheel is increased beyond O2M  then, the tip of the gear tooth on wheel causes interference with the tooth on pinion.  ∴ The points M and N  are called ‘Interference Points’.  Thus  Interference is avoided  if the path of contact does not extend beyond the Interference points.  ∴The limiting value of radius of the addendum circle of pinion = O1N. -&- the limiting value of radius of the addendum circle of wheel = O2M.  ∴ To avoid Interference  the point of contact between the two mating gear teeth should always lie on the involute profiles of both the gear teeth.  In other words  the addendum circles of the two mating gears should the common tangent (to the two base circles) between the points of tangency (or) the interference points.

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Effect of changing the centre distance on the Velocity ratio in Involute Gears :

(Refer Fig.13)

Fig.13

 Suppose the centre of rotation of one of the gear wheels is changed, say from O1 to O1′.  ∴ The point of contact between the mating gear teeth at a given instant will shift from Q to Q′.  Draw common normal M′N′ from Q′ tangential to the modified base circles.  The point of intersection of M′N′ with O1′O2 is P′  which is the new pitch point.  Now  from the similar triangles O1MP and O2NP 

O1M O1P . = O2 N O2 P

 Similarly  from the similar triangles O1′ M′ P′ and O2 N′ P′  O1' M' O1' P' . = O2 N' O2 P '

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 But  O1M = O1′ M′ -- and -- O2N = O2 N′  (the LHS and RHS in either equation are the radii of the same base circle). O1P O1' P'  Thus, if the distance between the centres of  ∴ = O2 P O2 P '

rotation of the mating gear wheels is changed -- within limits -the Velocity ratio remains unchanged  However, the pressur angle changes from φ to φ′.  Thus  the possibility to change the centre distance -- without destroying the correct tooth action (i.e., without violating the law of gearing) -- is an important advantage with involute profile of gear teeth. Important Characteristics of Involute profile gear teeth : 1. The contact between a pair of mating gear teeth starts when the tip of the tooth of driven wheel touches the flank of the tooth of driving wheel. The contact ends when the tip of the tooth on the driving wheel touches the flank of the tooth on driven wheel. 2. The points of contact between a pair of mating gear teeth at every instant lie on the line of action -- which is the common tangent to the base circles of the two mating gear wheels. 3. If the direction of rotation of gear wheels is reversed, the points of contact lie on the other common tangent to the base circles. 4. The initial contact between a pair of mating gear teeth occurs at the point where the addendum circle of driven wheel intersects the line of action. The final contact occurs at the point where the addendum circle of driving wheel intersects the line of action. To find the Number of pairs of gear teeth in contact at any given instant of time : Arc of contact = Length of the pitch circle traversed by a point on it during the mating of a pair of gear teeth. 27

∴ All the teeth lying in between the arc of contact P′ PP′′ on a gear wheel will be meshing with the teeth on the other wheel (Refer Fig.10). ∴ Number of gear teeth within the arc P′ PP′′ = n = ∴n=

Path of Contact 1  cos  pc

Arc length PP . Circular pitch

.

For continuous transmission of motion at any given instant of time  at least one tooth of one gear wheel should be in contact with another tooth on the mating gear wheel. ∴ n is always > 1. If n lies between 1 and 2  the number of pairs of gear teeth in contact at any given instant of time will be < 1 -- and -- > 2. Similarly  if n lies between 2 and 3  the number of pairs of gear teeth in contact at any given instant of time will be < 2 -- and -- > 3, .... and so on. e.g., if n = 1.6  it means that  one pair of teeth is always in contact between the mating gear wheels -- and -- two pairs are in contact for 60% of the time during their engagement. To find the Minimum number of teeth to avoid Interference :

(Fig.12)

Maximum value of the addendum radius of gear wheel = O2M. Now  from the ∟ triangle O2NM  O2M2 = O2N2 + MN2 = O2N2 + (MP + PN)2 = (R cos φ)2 + (r sin φ + R sin φ)2 = R2 cos2 φ + r2 sin2 φ + R2 sin2 φ + 2 r R sin2 φ = R2 (cos2 φ + sin2 φ) + r2 sin2 φ + 2 r R sin2 φ = R2 + (r2 + 2 r R) sin2 φ = R2 [1 +

  r 2 2r   1 2 2 2 2 (r + 2 r R) sin φ] = R 1   2   .sin   2 R R  R 

 r r   = R2 1  .  2 .sin 2    ∴ O2M = R .   R R 

28

 r r  2  1  .  2 .sin    R R    

∴ Maximum value of Addendum of the Wheel = (O2M – O2P)      r r  2  =  R  1  .  2  .sin     R  = R.      R R   .Now

   r  r    2    1  .  2  .sin     1     R  R   

 Suppose the adopted value for the addendum of the wheel is

equal to  (aw  module), where aw is addendum coefficient (aw), then  the addendum of wheel should be < (aw)max  to avoid Interference .

   r  r    2  i.e.,   1  .  2  .sin     1 ≥ (aw  m)     R  R    d D 2r 2R mT mt   R  ;r  But  m =   . t T t T 2 2

mT    t  t     2 ∴ .   1  .   2  .sin     1 ≥ (aw  m) 2    T  T         T    1  1  2  1  .  2 .sin  .       1 2    G  G     

∴T ≥

Tmin =

-- (or) --

≥ aw  where G =

2 aw    1  1       1  .  2  .sin 2     1     G  G    2 aw    1  1       1  .  2  .sin 2     1     G  G   

T = Gear Ratio. t

 In the limit 



where Tmin is the

Minimum number of teeth on the Gear wheel to avoid Interference. Similarly  The Interference = tmin =

minimum

number

of

teeth

on

pinion

Tmin . G

Interference between Rack and Pinion :

29

(Refer Fig.14)

to

avoid

Fig.14 A Rack is a gear wheel of infinite pitch circle radius. Now  let PN be the line of action  The engagement of rack with the pinion tooth begins at the point L. To avoid Interference  Maximum addendum of the Rack = GN. Now  let the adopted value of addendum of the rack = ar.m  where ar is the Addendum coefficient of the rack. Now  from the ∟ triangle GPN, GN = PN sin φ = (r sin φ).sin φ = r sin2 φ.

mt .sin2 φ. 2 To avoid Interference  GN ≥ ar.m -- or -- ar.m < GN. ∴ GN =

i.e.,

2ar 2ar mt  ∴ tmin = . sin2 φ ≥ ar.m -- or -- t ≥ . 2 sin φ sin 2 φ 2

tmin is the minimum number of teeth on pinion to avoid Interference. Under – Cutting :

[Refer Fig.15(a) and (b)]

30

Fig.15(a)

Fig.15(b) The situation where a part of the addendum of a pinion falls inside the base circle is shown in Fig.15(a). The profile of the tooth inside the base circle is radial. If the addendum of the mating gear exceeds its limiting value, then it interferes with the dedendum of the pinion, resulting in the two mating gears getting locked. However, if a cutting rack with similar teeth is used to cut the teeth in pinion [see Fig.15(b)], then it will remove that portion of the pinion tooth which would have interfered with the gear {as shown earlier in Fig.15(a)}. A gear in which the material is removed like this  to avoid interference, is said to be undercut, and this process is termed as Undercutting.

When

a

gear

meshes

with

such

undercut

pinion,

interference does not occur. However, undercutting is not required if the gear / pinion teeth are designed to avoid interference. Methods of Eliminating (or) Reducing Interference : 31

1. The part of the flank of pinion tooth which lies within the base circle, and the part of the face of gear tooth which engages with it may be made cycloidal, instead of involute profile in shape. Effect :

The modified involute profile shape or the composite

profile shape (which is partly involute and partly cycloidal) for the meshing gear teeth has the following disadvantages : (a) Difficulty in manufacture of the profile shape. (b) For the cycloidal portions of profiles of the mating gear teeth, the path of contact includes parts of the rolling circles  hence correct gearing is obtained only if the centre distance between the mating gear wheels is maintained

exactly  thus the

advantage with regard to changing the centre distance (albeit within limits) which lies with the involute profile is lost thereby. 2. The addenda of the teeth on the gear wheel and pinion may be modified  the addendum of the wheel is reduced by the amount necessary to avoid interference -- and -- that of the pinion is correspondingly increased. Effect :

(Refer Fig.12)  To avoid interference, the limiting

value of addendum of the wheel is GM -- and -- that of the pinion is HN. Further, HN > GM  Therefore, if the addendum of wheel is equal to the addendum of the pinion (as per the standard practice), since GM < HN, the addendum of the gear wheel would become the deciding factor for interference to occur or otherwise. Therefore, to avoid interference  it will be necessary to reduce the addendum of wheel -- and -- increase the addendum of pinion by the corresponding amount  so as to retain the same total working depth. 3. The centre distance between the two mating gear wheels (having teeth of involute profile) may be made larger than the standard 32

centre distance  This results in increasing the pressure angle, and thereby avoiding interference. Effect :

For the wheels of involute profile gear teeth  the

centre distance between the mating gear wheels may be changed (within limits) -- without destroying the correctness of tooth action. This has the following advantages : (a) The radii of addendum circle and dedendum circle of pinion can be increased  so as to increase the total working depth. (b) The profile on one of the sides of pinion teeth is shifted without altering its shape -- until -- it touches the next wheel tooth profile  in order to eliminate the backlash (i.e., improper or incorrect meshing), which results from changing the centre distance between the mating gear wheels. The above two aspects together -- in effect -- help to increase the centre distance beyond the standard value for a pair of gear wheels. Angle of Action (δ) :

The angle turned by a gear from the start of

engagement to the end of engagement between a pair of teeth, i.e., the angles subtended by the arcs of contact of the wheel (or pinion) at the centre of the wheel (or pinion) respectively. ∴ δ = (α + β)  where α = Angle of approach -- & -- β = Angle of Recess. Now  δ for wheel ≠ δ for pinion.

Arc of contact  3600 Arc of contact  3600  δ= . Pitch circle circumference d Contact Ratio :

If γ is the Pitch angle (i.e., the angle subtended by the

circular pitch at the centre of the wheel)  then the Contact ratio is :

Angle of Action δ 3600  pc  γ = Contact Ratio = = . Pitch angle γ d Worked Examples 33

1. Two mating involute gears, each having 200 pressure angle are in mesh with each other. The number of teeth on pinion is 20, and the gear ratio is 2. Module of the gears is 12 mm, and the pinion rotates at 250 rpm clockwise. If the addendum on each wheel is such that the paths of approach and recess are half of the maximum possible length, find the velocity of the point of contact along the surface of each mating gear tooth at a given instant when the tip of a tooth on the pinion is in contact with the wheel. Also find the following : (a) The velocity of sliding at this instant. (b) The maximum velocity of sliding during approach and recess. (c) The lengths of path of contact and arc of contact. (d) The addenda of pinion and gear. (e) The number of pairs of teeth in contact, and the angle of rotation of the pinion when any one pair of gear teeth is in contact. (f) The normal force between the teeth, assuming that  the pinion transmits 15 HP, there are two pairs of teeth in contact, and the total force is divided equally between the two pairs, and friction is neglected. Solution :

(Refer Fig.1-1)

E is the point of contact of the tip of the tooth on the pinion. ∠ PAE = α ;

∠ EBJ = β.

ω1 = Angular velocity of pinion;

ω2 = Angular velocity of gear wheel.

∴ At the point of contact E  Linear velocity of pinion = ω1.AE - and Linear velocity of gear wheel = ω2.BE. For the two mating gear surfaces to be in contact with each other, the velocity components along the common normal must be equal. i.e., v1 cos (α+φ) = v2 cos β. Now  from the triangle AEP,

EP AE AE AE = = = . sin α sin (APE) sin (90+φ) cos φ 34

Fig.1-1

Path of Recess r = A  where rA is the addendum circle radius sin α cos φ of pinion. T Given that : t = 20; and Gear ratio = G = = 2; ∴ T = 40; t Pressure angle = φ = 200. n×t 250  20 n T  Speed of pinion =n= 250 rpm    2  ∴ N= = 125 rpm. N t T 40 2 n 2  250 2 N 2  125   ∴ ω1 = = 26.2 rad/s; ω2 = = 13.1 rad/s. 60 60 60 60 D d Module of the gears = m = 12 mm = =  ∴ D = 480 mm; d = 240 mm. T t ∴ R = 240 mm; r = 120 mm. Now  the paths of approach and recess will be maximum  when interference is just avoided. ∴ Maximum value of path of approach = r sin φ = 120  sin 200 = 41 mm. i.e.,

∴ The path of approach =

R

2 A

- R 2 cos2   - R sin φ = 20.5 mm. 35

R

(or)

2 A

- R 2 cos2   = 20.5 + R sin φ = 20.5 + 240  sin 200 = 20.5 + 82.1 =

102.6 mm. ∴  R A 2 - R 2 cos 2 φ  = (102.6)2 = 10526.8  ∴ RA2 = 10526.8 + R2 cos2 φ RA2 = 10526.8 + (240)2  cos2 200 = 10526.8 + 50862.1 = 61388.9 mm2. ∴ RA = 247.8 mm = 24.78 cm. Similarly  Maximum value of path of recess = R sin φ = 240  sin 200 = 82.1 mm. ∴ The path of recess =

r

∴

A



2

r

A

2

- r 2 cos2   - r sin φ = 41.05 mm = 4.105 cm

- r 2 cos2   = 41.05 + r sin φ = 41.05 + 120  sin 200 = 82.1.



∴ rA 2 - r 2 cos 2  = (82.1)2 = 6739.2  ∴ rA2 = 6739.2 + r2 cos2 φ = 6739.2 + (120)2  cos2 200 = 6739.2 + 12715.5 = 19454.72  ∴ rA= 139.48 mm  14 cm. ∴

Path of Recess 14 4.105 r  ∴ sin α = 0.2755  α = 160. = A  = 0 sin α cos 20 sin α cos φ

Similarly  from the triangle PEB, ∴

R BP EP = = . sin(φ-β) sin(90+β) cosβ

41.05 EP sin(φ-β) (sin φ.cos β-cos φ.sin β)  = 0.171 =  240 R cosβ cosβ (sin 200.cos β-cos 200 .sin β) = = (sin 200 -cos 200 .tanβ) cos β = (0.342 – 0.934  tan β).

∴ tan β = (0.342 – 0.171) ÷ 0.934 = 0.183  β = 10.40. Now  the velocity of the point of contact along the surface of the pinion tooth, i.e., perpendicular to the common normal = v1 sin (α+φ) = ω1.r. sin (α+φ) = 26.2  120  sin (16+20)0 = 1848 cm/s = 18.48 m/s.

36

Similarly  the velocity of the point of contact along the surface of the wheel tooth, i.e., perpendicular to the common normal = v2 sin β =ω2.R. sin β = 13.1  240  sin 10.40 = 568 cm/s = 5.68 m/s. (a) The velocity of sliding at this instant = [v1 sin (α+φ) - v2 sin β] = (18.48 – 5.68) = 12.8 m/s. (b) Maximum velocity of sliding = (ω1 + ω2)  Maximum distance of the point of contact from the pitch point ∴ Maximum velocity of sliding during approach =

r sin φ  (ω1 +ω2 ) 2

120  sin 200 (26.2  13.1) = 806.5 mm/s = 0.81 m/s. = 2 R sin φ Maximum velocity of sliding during recess =  (ω1 +ω2 ) 2 240  sin 200 (26.2  13.1) = 1613 mm/s = 1.61 m/s. = 2

(c) Length of Path of Contact =

r sin φ R sin φ + 2 2

120  sin 200 240  sin 200 = + = (20.52 + 41.04) = 61.56 mm = 6.2 cm. 2 2 6.2 Path of contact Path of contact Length of Arc of Contact = = = 0 0.934 cos φ cos 20 = 6.6 cm.

(d) Addendum of pinion = (rA – r) = (14 – 12) = 2 cm. Addendum of gear wheel = (RA – R) = (24.78 – 24) = 0.78 cm. (e) Number of pairs of teeth in contact = n = =

Arc of contact pc

Arc of contact 6.6 = = 1.75. d     24       t   20 

The angle of rotation of the pinion when any one pair of gear teeth

Arc of contact  3600 is in contact = Angle of action = δ = = 31.50. d 37

15×4500 HP×4500 = = 43 kg-m. 2πN 2π  250 T 43 ∴ Tangential force on pinion = ft* = p = = 358 kg. 0.12 r

(f) Torque on the pinion = Tp =

ft* is divided equally between the two pairs of teeth. 179 f t * 358 f ∴ ft = = 179 kg  ∴ fn = t = = 192 kg.  cos 200 2 2 cos 

2. Two spur gear wheels have 30 teeth each of involute profile. The circular pitch is 25 mm, and pressure angle is 200.. Determine the addendum of the wheels if the arc of contact is twice the circular pitch. Solution : Given that : ∴d=D=

t = T = 30;

25  30



pc =

πd  D  = 25 mm; φ = 200. t T

= 238.7 mm; ∴ r = R = 119.4 mm;

Arc of contact = 2  25 = 50 mm. Path of contact = ( Ra 2  R2 cos2  )  (ra 2  r 2 cos2  )  ( R  r )sin  = ( Ra 2  (119.4)2 cos2 200 )  (ra 2  (119.4)2 cos2 200 )  (2 119.4)sin 200 Since r = R  ra = Ra  ∴ Path of contact = 2  ( Ra 2  (119.4)2 cos2 200 )  2  (119.4)sin 200 = 2  ( Ra 2 14256.4  0.883)  238.8  0.342 = 2  ( Ra 2 12588.4)  81.67 mm. ∴ Arc of contact =

2 Path of contact 2  ( Ra  12588.4)  81.67  = 50 mm.  cos  cos 200

38

∴ 2  ( Ra 2  12588.4)  81.67 = 50  cos 200 = 50  0.94 = 46.98. ( Ra 2 12588.4) = (46.98 + 81.67)÷2 = 64.33  (Ra2 – 12588.4) = (64.33)2 =

4138  Ra2 = 4138 + 12588.4 = 16726.4  ∴ Ra = 129.33 mm. ∴ R + Addendum = 119.4 + Addendum = 129.33  ∴ Addendum on each gear wheel = (129.33 – 119.4) = 9.93 mm. 3. Two 200 pressure angle involute gears in mesh have a module of 10 mm, and the addendum is equal to one module. The larger gear has 50 teeth, and the pinion has 13 teeth. Does interference occur? If it occurs, to what value should the pressure angle be changed to avoid interference? Solution : Given that : φ = 200; m = T = 50;

d D = = 10 mm; Addendum = 1 module = 10 mm; t T

t = 13;

To avoid Interference  aw (max) =

mT 2

  tt     1    2  sin 2    1       T  T

where aw is the ‘Maximum value of Addendum of the gear wheel’. ∴ aw (max) =

10  50   13  13     1    2  sin 2 200   1 = 250   1  0.5876  0.117  1   2   50  50   

= 8.445 mm ˂ m (=10 mm)  ∴ Interference shall occur. To avoid Interference  aw (max) shall be ≥ m. ∴ aw (max) =

10  50   13  13     1    2  sin 2  '  1 = 10 2   50  50   

 250   1  0.5876 sin2    1 = 10   1  0.5876 sin2    1 =0.04       1  0.5876 sin2    = 1.04  (1+0.5876 sin2 φ′) = (1.04)2 = 1.082  

∴ 0.5876 sin2 φ′ = 0.082  sin2 φ′ = 0.1389  sin φ′ = 0.37265  φ′ = 21.880. ∴ To avoid Interference  The pressure angle must be changed to 21.880. 39

4. A rack is being driven by an 18 – teeth pinion of involute profile and 120 mm pitch circle diameter. The addendum of both pinion and rack is 6 mm. (i) Find the least pressure angle required to avoid interference. (ii) With this pressure angle, find the length of arc of contact. (iii) Also Find the minimum number of teeth in contact at a time. Solution : Given that : t = 18;

d = 120 mm;

Addendum of pinion / rack = 6 mm.

120 6 = 6.67 mm  ∴ Addendum coefficient= ar = = 0.9. 6.67 18 2a 2  0.9 2a To avoid Interference  t ≥ 2r  (or) sin2 φ ≥ r = = 0.1. 18 t sin  d t

(i) Module =m = =

∴ sin φ = 0.316  ∴ φ = 18.4350. (ii) Path of contact =

Addendum of rack  cos 

r

a

2

 r 2 cos 2    r sin 

Now  ra = (r + Addendum) = 60 + 6 = 66 mm. ∴ Path of contact =

6  cos(18.4350 )

 66

2

 602  cos2 18.4350   60  sin18.4350

= 6.325 + 33.407 – 18.974 = 20.76 mm. ∴ Arc of contact =

20.76 Path of contact = = 21.88 mm. cos18.4350 cos 

(iii) Number of pairs of teeth in contact at a time= n = ∴n=

Arc of contact 21.88 = m Circular pitch

21.88 = 1.045  1 pair of teeth (or) 2 teeth.   6.67

Summary A friction wheel with Teeth cut on it is termed as a Toothed Wheel

(or) Gear. The main advantage of gear drives over the other

forms of drives is that this is a positive drive, and thereby transmits a constant velocity ratio. Gears are classified as Spur gears, Helical,

Spiral, Bevel, and Worm gears. There are two forms of gear teeth known as the Involute and Cycloidal profiles. Expressions for the 40

law of gearing for constant velocity ratio of toothed wheels, the velocity of sliding of gear teeth, path of contact, and arc of contact for involute

profile gear teeth are derived, from first principles. The

phenomenon of

‘Interference’ in Involute gears, and Interference

between involute rack and pinion are explained, and the methods of avoiding / reducing it have been discussed. What is ‘Under – cutting’, and why it is done. Further, what is ‘Under – cutting’, and why it is done, is also explained. QUIZ 1. The main advantage of gear drives over the other forms of drives is (a) it forms a higher pair between the mating gear teeth (b) it is a modified form of friction discs (c) it is a positive drive, transmitting a constant velocity ratio (d) it does not give rise to any thrust force

(c)

2. The purpose of cutting projections (teeth) on two mating friction discs is (a) To avoid slipping

(b) to avoid creep

(c) to reduce interference

(d) to avoid under-cutting

(a)

3. Which of the following gears are used to connect two non–intersecting, and non – parallel shafts? (a) Helical gears (b) Spiral gears

(c) Bevel gears

(d) Spur gears (b)

4. A helical gear which is made of a pair of helical gears secured together, one with right – hand helix and the other with left – hand helix, is known as (a) skew spur gear

(b) skew bevel gear

(c) mitre gear

(d) herringbone gear

(d)

5. A gear wheel, which has the radius of curvature of its pitch circle as infinite, is called a 41

(a) mitre gear

(b) pinion

(c) skew gear

(d) rack

(d)

6. The size of a gear is usually specified by its (a) pitch circle diameter

(b) number of teeth

(c) module pitch

(d) diametral pitch

(a)

7. Which of the following relations is correct for a gear tooth of involute profile? (φ = pressure angle) (a) Pitch Circle diameter = Root circle diameter x cos φ. (b) Root Circle diameter = Pitch circle diameter x cos φ. (c) Pitch Circle diameter = Root circle diameter x sin φ. (d) Root Circle diameter = Pitch circle diameter x sin φ.

(b)

8. With the usual notations of gear tooth terminology, which of the following relations is not correct? (a) pc = (c) m =

d T T d

(b) Total Depth = Addendum + Dedendum (d) Arc of contact =

Path of contact cos 

(c)

9. Which of the following is not a reason as to why cycloidal profile is not commonly used for gear teeth? (a) Pressure angle is constant throughout the engagement (b) Requires exact centre distance (c) Operation is less smooth than in involute gears (d) Difficult to manufacture as compared to the involute gears

(a)

10. If the shape of one gear tooth profile is chosen arbitrarily, and the profile of the mating gear tooth is designed to satisfy the law of gearing, such gear teeth are known as (a) skew teeth

(b) conjugate teeth

(c) involute teeth

(d) heeringbone teeth

(b)

11. The velocity of one gear tooth relative to its mating gear tooth along the common tangent at their point of contact is called the 42

(a) velocity of sliding

(b) pitch line velocity

(c) velocity of rubbing

(d) tangential velocity

(a)

12. The common normal at the point of contact to the mating involute profiles is also common tangent to the corresponding (a) addendum circles

(b) pitch circles

(c) base circles

(d) root circles

(c)

13. The distance travelled by a point on either pitch circle of the mating gear wheels during the period of contact of a pair of gear teeth is termed as the (a) path of contact

(b) arc of contact

(c) path of approach

(d) arc of approach

(b)

14. The tip of the gear tooth on pinion digging out the flank of the gear tooth on the wheel at its root, thereby removing part of the involute profile of the gear tooth on the wheel, is known as (a) under – cutting

(b) conjugation

(c) change of pressure angle

(d) interference

(d)

15. Which of the following is not a correct method of eliminating the interference between mating gear teeth? (a) making the profiles of mating parts of the gear teeth cycloidal (b) modifying the addenda of the teeth on the gear wheel and pinion (c) modifying the centre distance between the two mating gear wheels (d) modifying the pressure angle of the mating gear wheels

(d)

GLOSSARY Gear :

A friction wheel with teeth cut on it.

Spur gears :

Gears whose teeth are parallel to the axis of shafts, and are used to connect two parallel and coplanar shafts.

Helical gears :

Gears whose teeth are inclined to the axis of the gear wheel, and are used to connect two parallel and coplanar shafts. 43

Double Helical gears : A pair of helical gears secured together, one with right – hand helix and the other with left – hand helix, and the teeth of the two rows separated by a groove for tool run – out. Herringbone gears :

A double helical gear without a groove between the two rows of teeth.

Bevel gears :

The gears used to connect two intersecting, non–parallel, but coplanar shafts. If the gear teeth are straight (parallel to the shaft axes), they are called Spur bevel gears, and if the teeth are inclined to the shaft axes, they are called Helical bevel gears.

Spiral gears :

The gears which have curved teeth, and used to connect two non – intersecting, non – parallel, and non – coplanar shafts. They are also known as Skew Bevel gears.

Worm and Worm gears : These are used to connect two non – intersecting (skew) shafts. The pinion (or Worm) has a small number of teeth (threads) – usually 1 to 4 – and its mating gear is called the Worm wheel. Annular wheel (or) Annulus : The larger of the two mating gear wheels with internally cut gear teeth. The smaller wheel is called Pinion, and the directions of rotation of both wheels are the same. Rack – and – Pinion :

The gear wheel whose radius of curvature of the pitch circle is infinite is called the Rack, and the smaller wheel is called Pinion.

Pressure angle (or) Angle of Obliquity (φ) : The angle between the common normal to the two mating gear teeth at the point of contact and the common tangent at the pitch point. d Module (m) : This is the inverse of diametral pitch  ∴ m = . T 44

Path of Contact : The curve traced by the point of contact of the mating teeth -- from the beginning to the end of their engagement. Arc of Contact : The path followed by a point on either pitch line of the mating gear teeth -- from the beginning to the end of their engagement. Cycloidal gear teeth : The part of the flank of pinion tooth which lies within the base circle, and the part of the face of gear tooth which engages with it may be made cycloidal in shape. Involute gear teeth :

Gear teeth having involute profile -- which may

be defined as the locus of a point on a straight line -which rolls without slipping -- on the circumference of a circle. Interference :

The phenomenon of the tip of a gear tooth undercutting the root of its mating gear tooth.

Undercutting :

If a cutting rack with similar teeth is used to cut the teeth in pinion, then it will remove that portion of the pinion tooth which would have interfered with the gear. A gear in which the material is removed like this to avoid interference, is said to be undercut, and this process is termed as Undercutting.

Unit 6 – Sub – Topic 2 Helical Gears, Spiral Gears, Bevel Gears, Worm and Worm Gears

Learning Objectives : By the end of studying this topic, the student must be able to understand :  The helical gears are used to transmit power between two parallel shafts or two non – parallel and non–intersecting shafts.

45

 When helical gears are used to connect two non – parallel and non – intersecting shafts, they are termed as Crossed (or skew) helical gears or Spiral gears.  For helical gears connecting parallel shafts  the contact between a pair of mating gear teeth is a line contact through the face width of the gear teeth. But  in the case of crossed helical gears which are used to connect two shafts whose axes are not parallel to each other (i.e., skew shafts)  the contact between a pair of mating gear teeth will be a point contact only.  ∴ While the helical gears for parallel shafts are considered stronger than spur gears, the crossed helical gears (or spiral gear gears) for connecting skew shafts are not preferred to transmit heavy loads.  Parallel helical gears have smoother action  hence less noise and vibration than spur gears.  Helical gears are more suitable than spur gears for high speed applications.  The cutting of gear teeth in helical gears is more difficult than in spur gears.  Helical gears are subjected to radial and axial (end) thrust loads, whereas the spur gears impose radial loads only.  A worm and worm gear are used when high speed reduction is required between skew shafts.  Usually, the Worm and Worm gears are used to connect two skew shafts whose axes are perpendicular to each other.  A worm can be a single, double, or triple start worm, if one, two, or three threads respectively are traversed on the worm for the advancement of one tooth on the worm wheel.  Bevel gears are used to transmit drive between two shafts whose axes are intersecting – but – coplanar. A helical gear has teeth in the form of a helix around the gear. On two mating helical gears, if the helix on one gear wheel is right – handed, that on the other gear wheel may be left–handed [Refer Fig.3(a)] or right – handed. The helical gears are used to transmit power between two parallel shafts or two non – parallel and non–intersecting shafts. When used to connect two parallel shafts, they are called Helical gears, 46

and when used to connect two non – parallel and non – intersecting shafts, they are termed as Crossed (or skew) helical gears or Spiral gears. Helical and Spiral gears :  In helical and spiral gears  the gear teeth may be right – handed or left – handed.  (Refer Fig.16)  Here, the gear 1 is a right – handed helical gear while the gear 2 is left – handed  And when the two mating gears are mounted on shafts with parallel axes, the helix angles of the gears must be equal  Thus, ψ1 = ψ2 = ψ.

Fig.16  [Refer Fig.17(a)]  In this, the same two helical gears are shown, when viewed from above. Now, if ψ2 is slightly reduced -- and -it is still desired that the two gears mesh tangentially with each other  then, the axis of the gear 2 must be rotated through some angle θ  as shown in Fig.17(b)  such that θ = (ψ1 - ψ2).  Now  if ψ2 is made equal to zero  the gear 2 will become a

47

Fig.17 (a) & (b)

Fig.17 (c) & (d) straight spur gear [Refer Fig.17(c)] -- and -- the angle between the shaft axes θ = ψ1.  But  if ψ2 is made negative  [Refer Fig.17(d)]  i.e., the teeth of gear 2 are made of the same hand helix as that of gear 1  then, the angle between the shaft axes θ = (ψ1) – (- ψ2) = (ψ1 + ψ2). 48

Fig.17 (e)  For helical gears connecting parallel shafts  the contact between a pair of mating gear teeth is a line contact through the face width of the gear teeth. But  in the case of crossed helical gears which are used to connect two shafts whose axes are not parallel to each other (i.e., skew shafts)  the contact between a pair of mating gear teeth will be a point contact only  ∴ While the helical gears for parallel shafts are considered stronger than spur gears, the crossed helical gears (or spiral gear gears) for connecting skew shafts are not preferred to transmit heavy loads. Velocity ratio of Helical gears :

[Refer Fig.17(e)]

 The pitch line velocities v1 and v2 of the gears 1 and 2 act in the directions, as shown in Fig.17(e).  v12 represents the sliding velocity of the teeth of gear 1 relative to that of gear 2 -- in magnitude and direction -- parallel to the common tangent t – t.  However  the normal components of v1 and v2 (perpendicular to t – t) must be equal. vn = v1 cos ψ1 = v2 cos ψ2  ∴

cos ψ1 v2 = . cos ψ 2 v1 49

Now  Velocity Ratio = V.R. = ∴ VR =

(v / d ) (d / v ) (v / r ) ω2 = 2 2 = 2 2 = 1 2 . (v1 / r1 ) ω1 ( d 2 / v1 ) (v1 / d1 )

(m / cos ψ1 ) T1 cos ψ1 m T cos ψ1 T d1 cos ψ1 . = 1 1. = n = 1. . . (mn / cos ψ 2 ) T2 cos ψ 2 m2 T2 cos ψ 2 T2 d 2 cos ψ 2

Thus  VR =

ω2 N T = 2 = 1. T2 ω1 N1

Advantages of Helical gears over Spur gears :  Helical gears connecting two parallel shafts have line contact, which runs diagonally across the face of the gear teeth.  Parallel helical gears have smoother action  hence less noise and vibration than spur gears  because the contact between gear teeth is gradual, beginning at one end of the tooth – and – progressing across the tooth surface gradually  whereas in spur gears, the contact occurs simultaneously over the entire face width of the gear tooth.  Pick – up of load by the gear tooth occurs gradually in helical gears  resulting in smooth engagement and quiet operation  even at high speeds – whereas in spur gears, the application of load is sudden  hence results in impact conditions  thereby generating noise, especially under high speed conditions.  More suitable than spur gears for high speed applications.  Contact ratio is higher than in spur gears. Disadvantages of Helical gears over Spur gears :  Helical gear teeth are cut in the form of a helix on the pitch cylinder between meshing gears, whereas the spur gear teeth are cut parallel to the axis of the shaft  hence the gear – cutting operation is more difficult in helical gears.  Helical gears are subjected to radial and axial (end) thrust loads, whereas the spur gears impose radial loads only. 50

Helical Gear Terminology :

(Refer Fig.18)

Fig.18 Helix angle (ψ) : The angle between a line drawn through one of the gear teeth and the centre line of the shaft on which the gear is mounted. It varies from 150 to 300. Normal Circular pitch (pn) : The distance between corresponding points on adjacent teeth  measured in a plane normal to the helix  Thus, it 51

is the perpendicular distance between two adjacent teeth. For two mating Helical gears  the normal circular pitch must be the same. ∴ pn = p cos ψ ;

Further  p = π.m ;

pn = π.mn ; mn = m cos ψ.

∴ pn = π m cos ψ, where p = Transverse circular pitch, m = module =

d , and T

mn = Normal Module = The module measured in a plane normal to the helix. Transverse Circular pitch (p) :

The distance between corresponding

points on adjacent teeth  measured in a plane perpendicular to the axis of the shaft on which the gear is mounted  ∴ p =

Normal Diametral pitch (pnd) : plane normal to the helix  pnd = Transverse Diametral pitch (ptd) :

d = π m. T

Fig.19 The diametral pitch measured in a p . cos ψ

The diametral pitch measured in the

plane of rotation, i.e., normal to the axis of rotation of the shaft  52

ptd =

T 1   ∴ ptd  p = pnd  pn = π. d m

Transverse Pressure angle (ψt) :

The pressure angle measured in the

transverse plane (or) plane of rotation. Normal Pressure angle (ψn) : The pressure angle measured in the normal plane (or) the plane perpendicular to the teeth. Axial Pitch (px) :

The distance measured in a plane parallel to the shaft

axis between corresponding points on adjacent gear teeth  px = pt.cot ψ. Lead :

The distance measured parallel to the shaft axis  to represent

the distance advanced by each gear tooth per revolution. Lead angle :

The acute angle between the tangent to the helix and a

plane perpendicular to the axis of the cylinder. Pitch Diameter (d) : The diameter of the pitch cylinder  d =

T.mn . cos ψ

Center distance between Helical gears (C) : 1 1 (d1  d 2 )  (m1T1  m2T2 ) . 2 2  m 1 m ∴ C =  n T1  n T2   For helical gears connecting parallel shafts 2  cos ψ1 cos ψ2 

C = (r1 + r2) =

 ψ1 = ψ 2 = ψ 

∴C=

Forces on Helical gears :

mn (T1  T2 ) . 2 cos ψ

(Refer Fig.20)

 The force exerted by a helical gear on its mating gear  acts normal to the contacting surfaces (if friction is neglected).  However  a normal force in the case of helical gears has three components  Tangential component, Radial component, and a third component parallel to the axis of the shaft of gear  known as the Axial (or) Thrust force component.  Fig.20  shows the normal force and its components  acting on a driven (helical) gear  which are exerted on it by the driving helical gear. 53

Fig.20 Let

Fnt = Total normal force ; Fr = Radial force ; φ = Pressure angle ;

Ft = Tangential force ;

Fa = Axial force ;

Fn = Normal force in the plane of Ft and Fa ; φn = Normal pressure angle ;

ψ = Helix angle.

Then  Fn = Fnt cos φn -- and -- Fr = Fnt sin φn -- and -Ft = Fn cos ψ -- and -- Fa = Fn sin ψ. Efficiency of Helical and Spiral gears :

(Refer Fig.21)

Two mating spiral gears are shown separately in Fig.21(a) and Fig.21(b) along with the force acting on them.  Gear 1 is driver and gear 2 is the driven gear. Let

Ft1 = Tangential force acting on the gear wheel 1, Ft2 = Tangential force acting on the gear wheel 2, Fn1 = Fn2 = Fn = Normal reaction force between the two surfaces in contact.

54

Fig.21  The direction of the sliding velocity of gear 2 relative to that of gear 1 (v21) is shown in Fig.21(c).  The friction force μ.Fn2 acts in the opposite direction.  Fn2 and μ.Fn2 combine into one reaction force F2 inclined at an angle φ with the normal reaction  where φ is the angle of friction. Now  Ft1 = F1 cos (ψ1 - φ) -- and -- Ft2 = F2 cos (ψ2 + φ). But  Fn1 = Fn2 = Fn -- and -- μ.Fn1 = μ.Fn2 -- and -- F1 = F2 = F. ∴ Ft1 = F cos (ψ1 - φ) -- and -- Ft2 = F cos (ψ2 + φ). And  Input torque = Ft1  v1 -- and -- Output torque = Ft2  v2 . 55

∴ Efficiency = η = ∴η= =

Ft 2  v2 cos ψ1 v .  But 2 = cos ψ 2 Ft1  v1 v1

2 cos ψ1 cos(ψ 2   ) cos(ψ1  ψ 2   )  cos(ψ1  ψ 2   ) cos(ψ 2   ) cos ψ1 . = = 2 cos ψ 2 cos(ψ1   ) cos(ψ 2  ψ1   )  cos(ψ 2  ψ1   ) cos(ψ1   ) cos ψ 2

cos(θ   )  cos(ψ1  ψ 2   ) cos(θ   )  cos(ψ1  ψ 2   ) = . cos(θ   )  cos[(ψ1  ψ 2   )] cos(θ   )  cos(ψ1  ψ 2   )

∴ η will be maximum if the common term in the numerator and denominator, i.e., cos(ψ1  ψ2   ) is maximum  i.e., equal to 1. ∴ (ψ1  ψ2   ) = 0  ψ1 = (ψ2 + φ)  Now, since θ = (ψ1 + ψ2), ψ1 = [(θ - ψ1) + φ]  2 ψ1 = (θ + φ) -- or -- ψ1 = ∴ ηmax =

cos(θ  φ)  1 cos(θ - φ)  1

(θ  φ) . 2

-------------------------------------------------------------- (1) Worm and Worm Gearing

 A worm and worm gear are used when high speed reduction is required between skew shafts.  Of course, we may use spiral gears for this purpose  but then, a small driver and large follower are required. Further  the load transmitted through these gears is limited.  ∴ To transmit higher loads -- than with the spiral gears -- Worm and Worm gears can be used.  This transmission of higher tooth loads is possible with worm and worm gears  because the contact between the worm threads (teeth) and the worm wheel teeth is a line contact.  Usually, the Worm and Worm gears are used to connect two skew shafts whose axes are perpendicular to each other.  However  the high sliding velocities give rise to high heating of the worm  thereby restricting its use.  A worm can be a single, double, or triple start worm, if one, two, or three threads respectively are traversed on the worm for the advancement of one tooth on the worm wheel. 56

Terminology of Worm gears :

(Refer Fig.22)

Fig.22 Axial pitch (px) : The distance between corresponding points on adjacent teeth, measured parallel to the axis of the worm. Lead (L) : The distance by which a helix advances along the axis of the gear for one turn around.

57

In a single – start worm, the axial pitch is equal to the Lead, in a double – start worm, it is one – half of the lead, and in a triple – start worm, it is one – third of the lead, and so on. Lead angle (λ) : The angle at which the teeth are inclined to the normal to the axis of rotation. Thus  the lead angle is the compliment of the helix angle  i.e., (ψ + λ) = 900. Let  1 denote the worm -- and 2 denote the worm wheel. As the shaft axes of the worm and worm wheel are at 900 to each other  (ψ1 + ψ2) = 900  ∴ (90 – λ1) + ψ2 = 90  ∴ λ1 = ψ2.

i.e., The lead angle of the worm = Helix angle of the worm wheel. Also  pn of worm = pn of worm wheel. i.e., px1 cos λ1 = p2 cos ψ2  But, λ1 = ψ2  ∴ px1 = p2. i.e., Axial pitch of the worm = circular pitch of the worm wheel. Velocity ratio : Assume that  a worm rotates through one revolution about its axis. ∴ The angle turned by the worm = 2π. Now  The lead of the worm is equal to the axial distance advanced by a thread in one revolution of the worm. Also  The lead is the distance moved by the pitch circle of the worm wheel. ∴ The angle turned by the worm wheel during the time of one revolution of the worm =

2L L (or) . d2 R2

∴ Velocity Ratio = VR =

(2L / d 2 ) Angle turned by the worm wheel L = = . 2  d2 Angle turned by the worm

Centre Distance between the shaft axes of Worm and Worm Wheel : Centre distance = C =

T2   m2 cos ψ2 1  cos ψ2 mn  T1 . .T1  T2       2  cos ψ1 cos ψ2  2 cos ψ2  cos ψ1 

But  ψ2 = λ1 -- and -- ψ1 = (90 – λ1). 58

∴C=

  m m2  cos 1 m  cos 1 . .T1  T2  = 2 . T1 cot 1  T2 . .T1  T2  = 2 .  2 2  cos (90 - 1 ) 2  sin 1  





Efficiency of Worm Gears : Efficiency of Worm Gears = η =

Ft 2  v2 cos ψ1 v , where v1 and  But 2 = cos ψ 2 Ft1  v1 v1

v2 are the pitch line velocities of the worm and worm wheel respectively. ∴η=

cos(ψ 2   ) cos ψ1 .  (Refer Fig. 21) cos(ψ1   ) cos ψ 2 cos(1   ) cos(1   ) sin 1 cos(1   ) cos(90  1 ) = = . .tan 1 . cos[90  1   ] cos 1 sin(1   ) cos[90  (1   )] cos 1

= ∴η=

tan 1 . tan(1   )

Now  ηmax =

cos(θ  φ)  1  From Eqn. (1)  where θ = (ψ1 + ψ2). cos(θ - φ)  1

But, since θ = 900 in Worm gears  ηmax =

cos(90  φ)  1 (1  sin φ) = . (1  sin φ) cos(90 - φ)  1

If the worm wheel is the driver, then  it can be deduced that : η=

(1  sin φ) tan(1  φ) -- and -- ηmax = . tan 1 (1  sin φ)

BEVEL GEARS Bevel gears are used to transmit drive between two shafts whose axes are intersecting – but – coplanar. Terminology of Bevel Gears :

(Refer Fig.23)

Let γg and γp  be the pitch angles of the gear and pinion respectively. rg and rp  be the pitch radii of the gear and pinion respectively. The pitch cones for the two mating external bevel gears are shown in Fig.23(b). Now  sin γg=

rg OP

=

rg ( rp /sin  p )

=

rg rp

sin(θ -  g ) (or) sin γg =

rg rp

(sin θ cos  g - cos θ sin  g )

Dividing both sides of the above eqn. by cos γg  we get : tan γg =

rg rp

(sin θ - cos θ sin  g ) -- or --

rp rg

tan  g = (sin θ - cos θ sin  g ) 

59

Legend :

x  Root cone ;

y  Pitch cone ; z  Face cone ;

β  Root angle ; φ  Pitch angle ;

ψ  Face angle ;

angle ; ∠ d  Dedendum angle ; f  Face width ; ∠ d  Dedendum ;

CD  Centre Distance ;

∠ a  Addendum a  Addendum ;

BCR  Back Cone Radius ;

Fig.23 tan γg =

sin θ

 Now, since vp = ωg.rg = ωp.rp 

 rp    cos θ   rg  sin θ ∴ tan γg = -- and -- Similarly  tan γp =  ωg   cos θ    ω p  

rp rg

=

ωg ωp

.

sin θ  ωp   cos θ     ωg 

.

Worked Examples 1. The centre distance between two meshing spiral gears is 150 mm, and The angle between the shaft axes is 600. The gear ratio is 2, and the normal circular pitch is 10 mm. The driven gear has a helix angle of 250. Determine the (i) number of teeth on each wheel, (ii) exact centre distance, and (iii) efficiency, if the friction angle is 40. Solution : 60

Given that :

T = 2; t

pn = 10 mm; θ = 600; G =

C = 150 mm;

ψ2 = 250.

Friction angle = φ = 40. (i) Assuming that both the meshing gears are of the same hand helix  θ = (ψ1 + ψ2) = (ψ1 + 25) = 60  ψ1 = (60 – 25) = 350. pn = π.mn = 10 mm  mn = 1 m

10 = 3.1831 mm; 

 1 3.1831

And since

T = 2  T = 2t. t

 C=  n T1  n T2  =   2t   t  =  7.7717  3.5122  = 11.28 0 0 2 cos 25 2  cos ψ1 cos ψ2  2  cos 35  2

∴t=

m

2C (2 150)  = 26.6  27; 11.28 11.28

(ii) Exact centre distance = ∴ Cexact = (iii) η =

t

3.1831

t

∴ T = 2t = 54.

 1  3.1831 mn 1  mn 3.1831  T1  T2  =   54   27   0 0 cos 25 2  cos ψ1 cos ψ2  2  cos 35 

1 (209.84 + 94.83) = 152.33 mm. 2

cos(θ   )  cos(ψ1  ψ 2   ) cos(60  4)  cos(35  25  4) = = 0.92 (or) 92 %. cos(θ   )  cos(ψ1  ψ 2   ) cos(60  4)  cos(35  25  4)

2. Two left – handed helical gears connect two shafts 700 apart. The larger gear has 50 teeth, the smaller gear has 20, and the normal module is 4 mm. If the centre distance between the gears is 0.167 m, find the helix angles of the two gears. Solution : Given that : θ = 700;

T1 = 20;

T2 = 50;

mn = 4 mm;

C = 167 mm;

 1 4   40 mn 1  mn 4 100   20  T1    50  =  T2  =   = 167 mm. 2  cos ψ1 cos ψ2 cos ψ2  2  cos ψ1   cos ψ1 cos ψ2 

C= 

 40  100   = 167 mm.  cos ψ1 cos(70  ψ1 ) 

θ = (ψ1 + ψ2) = 700  ψ2 = (70 – ψ1)  C = 

As the exact solution of the above equation is beyond the scope of the reader, we try to obtain an approximate solution by hit and trial method, as shown below :

61

 40  1   100 2.5    = 167 mm  40   = 167 mm   cos ψ1 cos(70  ψ1 )   cos ψ1 cos(70  ψ1 ) 

C= 

 1  167 2.5 = 4.175.    = 40  cos ψ1 cos(70  ψ1 ) 

ψ1

1 cos ψ1

2.5 cos(70  ψ1 )

 1  2.5     cos ψ1 cos(70  ψ1 ) 

25

1.103

3.536

4.639

26

1.113

3.475

4.588

27

1.122

3.418

4.540

28

1.133

3.364

4.497

29

1.143

3.313

4.456

30

1.155

3.264

4.419

35

1.221

3.052

4.273

40

1.305

2.887

4.192

41

1.325

2.858

4.183

42

1.346

2.831

4.177

43

1.367

2.806

4.173

42.5

1.356

2.818

4.1745

(degrees)

∴ ψ1 = 42.50 ;

ψ2 = (70 – 42.5) = 27.50.

∴ The helix angles are :

42.50 and 27.50.

3. Two meshing bevel gears are mounted on two intersecting shafts, the 62

angle between the shafts being 480. The velocity ratio of the gears is 2.4. Determine the pitch angles of gears. Solution: Given that :

θ = 480; Velocity ratio = VR = 2.4.

Since VR ˃ 1, the gear wheel is the driver. Let

γg and γp  be the pitch angles of gear and pinion respectively.

∴ tan γg =

sin θ  ωg   cos θ     ωp 

=

sin 480  1   cos 480    2.4 



0.743 = 0.6841. (0.417  0.669)

∴ γg = 34.390. tan γp =

sin θ  ωp   cos θ     ωg 

=

sin 480

 2.4  cos 48  0



0.743 = 0.2421. (2.4  0.669)

∴ γp = 13.610. 4. A three – start worm has a pitch diameter of 80 mm, and a pitch of 20 mm. It rotates at 600 rpm, and drives a 40 – teeth worm gear. If the coefficient of friction is 0.05, determine the (i) helix angle of the worm; (ii) speed of the worm gear; (iii) centre distance; (iv) efficiency; and (v) maximum efficiency. Solution : Given that :

N1 = 600 rpm; T2 = 40;

p1 = 20 mm;

d1 = 80 mm;

μ = tan φ = 0.05  φ = 2.860. (i) Since it is a three – start worm  Lead = L = 3 p1 = 60 mm. If λ1 is the lead angle of the worm  tan λ1 = =

60 = 0.2387  λ1 = 13.430;   80

Lead = pitch circumference

∴ Helix angle = ψ1 = (90 – λ1) = 76.570.

(ii) Pitch of the worm wheel = Axial pitch of the worm = 20 mm. ∴ p2 =

 d2 T2

 20 =

 d2 20  40  d2 = = 254.65 mm.  40

63

VR =

60 ω2 N 2 L = = = 0.075  ∴ N2 = 0.075  600 = 45 rpm.  ω1 N1  d 2   254.65

Alternately 

3 N 2 T1 T = 45 rpm.  N2 = N1  1 = 600   40 N1 T2 T2

(iii)Centre distance=C=









p m2 20 . T1 cot 1  T2 = 2 . T1 cot 1  T2 = 3cot 13.430  40   2 2 2

∴ C = 167.3 mm. (iv) η =

0.2388 tan13.430 tan 1 = = = 0.8172 (or) 81.72 %. 0 0 0.2922 tan(1   ) tan(13.43  2.86 )

(v) ηmax =

(1  sin 2.860 ) (1  sin  ) = = 0.905 (or) 90.5 %. (1  sin  ) (1  sin 2.860 )

SUMMARY In Spur gears, the gear teeth are cut on the pitch cylinder of the gear parallel to the axis of rotation of the gear wheel. Therefore, they can be used only to connect shafts whose axes of rotation are parallel. But in Helical gears, the gear teeth are cut at an angle of inclination to the axis of rotation of the gear wheel. This angle is termed as the helix angle. When helical gears are used to connect two non – parallel and non – intersecting shafts, they are termed as Crossed (or skew) helical gears or Spiral gears. When the transmission of drive required is between two parallel shafts, helical gears are considered to be stronger than spur gears. Helical gears, when used to connect two parallel axes shafts, they will have smoother action, hence less noise and vibration than spur gears. So, helical

gears

are

more

suitable

than

spur

gears

for

high

speed

applications. However, the cutting of gear teeth in helical gears is more difficult than in spur gears. Further, helical gears are subjected to radial and axial (end) thrust loads, whereas the spur gears impose radial loads only. A worm and worm gear are used when high speed reduction is required between skew shafts. Usually, the Worm and Worm gears are

64

used to connect two skew shafts whose axes are perpendicular to each other. Bevel gears are used to transmit drive between two shafts whose axes are intersecting – but – coplanar. QUIZ 1. Which of the following gears are used to connect two non –parallel and non – intersecting shafts? (a) Spur gears (b) Helical gears (c) Spiral gears

(d) Bevel gears

(c)

2. Which of the following statements is not true? (a) helical gears are considered stronger than spur gears for connecting parallel shafts (b) spiral gears are preferred to transmit heavy loads between

skew

shafts

(c) helical gears are more suitable than spur gears for high speed applications (d) A worm and worm gear are used when high speed reduction is required between skew shafts

(b)

3. Usually, the Worm and Worm gears are used to connect two skew shafts whose axes are (a) perpendicular to each other

(b) parallel to each other

(c) inclined at any angle other than 900 (d) short in length

(a)

4. In a worm and worm wheel, if the worm is a three- start worm, then (a) three threads are traversed on the worm for the advancement of one tooth on the worm wheel

(b) three threads are traversed on the worm for the advancement of three teeth on the worm wheel

(c) one thread is traversed on the worm for the advancement of three teeth on the worm wheel

(d) One – third of a thread is traversed on the worm for the advancement of three teeth on the worm wheel 65

(a)

5. Bevel gears are used to transmit drive between two shafts whose axes are (a) intersecting, and non – coplanar.

(b) non – intersecting, but coplanar

(b) non–intersecting, and non – coplanar (d) intersecting, but, coplanar (d) 6. When two helical gears with helix angles ψ1 and ψ2 are used to connect two parallel shafts, then it is essential that (a) ψ1 = (- ψ2)

(b) ψ1 =

1 ψ2

(c)

ψ1 =1 ψ2

(d) ψ1 = 90 – ψ2

(c)

7. When two helical gears, both of same hand helix, and with helix angles ψ1 and ψ2 are used to connect two parallel shafts, and if ψ2 is slightly less than ψ1, and, it is still desired that the two gears mesh tangentially with each other, then the axis of the gear 2 must be rotated through some angle θ such that θ = (a) (ψ1 - ψ2)

(b) (ψ1 + ψ2)

(c) (900 - ψ2)

(d) (900 – ψ1)

(b)

8. When two helical gears, one of right hand helix and the other with left hand helix, with helix angles ψ1 and ψ2 are used to connect two parallel shafts, and if ψ2 is slightly less than ψ1, and, it is still desired that the two gears mesh tangentially with each other, then the axis of the gear 2 must be rotated through some angle θ such that θ = (a) (ψ1 - ψ2)

(b) (ψ1 + ψ2)

(c) (900 - ψ2)

(d) (900 – ψ1)

(a)

9. When two helical gears, with helix angles ψ1 and ψ2 are used to connect two shafts, and if ψ2 is made equal to zero, then the angle between the shaft axes is equal to (a) Zero

(b) ψ1

(c) (900 - ψ1)

(d) 900

(b)

10. In the case of crossed helical gears which are used to connect two shafts whose axes are not parallel to each other (i.e., skew shafts), the contact between a pair of mating gear teeth will be a (a) straight line contact

(b) curved line contact

(c) point contact

(d) skew line contact

11. Which of the following gears are not preferred for connecting skew 66

(c)

shafts while transmitting heavy loads? (a) Bevel gears

(b) Helical gears

(c) Worm and worm gears

(d) Spiral gears

(d)

12. Parallel helical gears have smoother action, hence less noise and vibration than spur gears, because (a) they have line contact between the mating gear teeth (b) the contact between mating gear teeth is gradual (c) the helical hears are normally used for low speed applications (d) the contact ratio in helical gears is higher than in spur gears

(b)

13. Which of the following is higher in helical gears than in spur gears? (a) path of contact

(b) arc of contact

(c) angle of action

(d) contact ratio

(d)

14. In the design of helical gears, an imaginary spur gear is considered in a plane normal to the gear teeth. This gear is called the (a) design spur gear

(b) design spur gear

(c) virtual spur gear

(d) formative helical gear

(c)

15. In helical gears, the angle between a line drawn through one of the gear teeth and the centre line of the shaft on which the gear is mounted is termed as the (a) helix angle

(b) angle of contact

(c) angle of action

(d) pitch angle

(a)

16. In helical gears, the distance between corresponding points on adjacent teeth, measured in a plane perpendicular to the axis of the shaft on which the gear is mounted, is called the (a) normal circular pitch

(b) transverse diametral pitch

(c) normal diametral pitch

(d) transverse circular pitch

(d)

17. In helical gears, the acute angle between the tangent to the helix and a plane perpendicular to the axis of the cylinder, is termed as the (a) helix angle

(b) lead angle 67

(c) pitch angle

(d) angle of action

(b)

18. When two helical gears with helix angles ψ1 and ψ2, and pressure angle φ, are used to transmit motion between two shafts, the efficiency will be maximum when (a) sin(ψ1  ψ2   ) is maximum

(b) cot(ψ1  ψ2   ) is maximum

(c) cos(ψ1  ψ2   ) is maximum

(d) tan(ψ1  ψ2   ) is maximum

(c)

19. Usually, a worm and worm gear are used when high speed reduction is required between skew shafts. However, if we want to use spiral gears for this purpose, then (a) a small driver and large follower are required (b) a large driver and small follower are required (c) a large driver and larger follower are required (d) a small driver and smaller follower are required

(a)

20. The use of worm and worm gear is restricted because (a) they can be used only to connect skew shafts whose axes are at 900 to each other (b) the contact between the worm threads and the worm wheel teeth is a point contact (c) the transmission of higher tooth loads is not possible with worm and worm gears (d) the high sliding velocities give rise to high heating of the worm(d) 21. The ---------------------------- of the worm is equal to the helix angle of the worm wheel. (a) pitch angle

(b) lead angle

(c) thread angle

(d) angle of action

(b)

22. In the case of worm and worm wheel, the maximum efficiency will be equal to (φ being the pressure angle) (a)

(1  cot φ) (1  cot φ)

(b) 68

(1  sin φ) (1  cos φ)

(c)

(1  sin φ) (1  sin φ)

(d)

(1  tan φ) (1  tan φ)

(c)

23. In bevel gears, if γg and γp are the pitch angles of the gear and pinion respectively, ωg and ωp are their respective angular velocities, and θ is the angle between the shafts on which the gears are mounted, then γg will be equal to (a) tan-1

(c) tan-1

sin θ

(b) tan-1

 ωg   cos θ     ωp 

cosθ

(d) tan-1

 ωg   sin θ     ωp 

sin θ  ωp   cos θ     ωg 

cosθ  ωp   sin θ     ωg 

(a)

24. In bevel gears, if γg and γp are the pitch angles of the gear and pinion respectively, ωg and ωp are their respective angular velocities, and θ is the angle between the shafts on which the gears are mounted, then γp will be equal to (a) tan-1

(c) tan-1

sin θ

(b) tan-1

 ωg   cos θ     ωp 

cosθ

(d) tan-1

 ωg   sin θ     ωp 

sin θ  ωp   cos θ     ωg  cosθ  ωp   sin θ     ωg 

(b)

GLOSSARY Helical gear :

A gear which has teeth cut in the form of a helix around the gear. In these gears, the teeth are inclined to the axis of the gear.

Spiral gear :

This is a crossed helical gear, i.e., the mating helical gears are mounted on shafts whose axes are not parallel, but inclined with each other.

Bevel gear :

These gears are used to connect two intersecting, but 69

coplanar shafts. Kinematically bevel gears are equivalent to rolling cones. Worm and worm gear :

In these gears, there are screw threads on the worm, and gear teeth cut on the mating worm wheel.

Helix angle (ψ) : The angle between a line drawn through one of the gear teeth and the centre line of the shaft on which the gear is mounted. Lead of helical gear :

This is the distance measured parallel to the shaft axis to represent the distance advanced by each gear tooth per revolution.

Lead (L) in worm gears :

The distance by which a helix advances along the axis of the gear for one turn around. In a single – start worm, the axial pitch is equal to the Lead, in a double – start worm, it is one – half of the lead, and in a triple – start worm, it is one – third of the lead, and so on.

Lead angle (λ) in worm gears :

The angle at which the teeth are inclined to the normal to the axis of rotation. Thus, the lead angle is the compliment of the helix angle  i.e., (ψ + λ) = 900.

xxxxx

70