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WEEK 12MECHANISMS

• References• (METU, Department of Mechanical Engineering)

Text Book: “Mechanisms” Web Page: http://www.me.metu.edu.tr/people/eres/ME301/Index.ht

• Analitik Çözümlü Örneklerle Mekanizma Tekniği, Prof. Dr. Mustafa SABUNCU, 2004

• Mekanizma Tekniği, Prof. Dr. Eres SÖYLEMEZ, 2007

• Theory of Machines and Mechanisms, J.J. Uicker, G.R.Pennock ve J.E. Shigley, 2003

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Gears are an extremely common component used in many machines. The Figure illustrates the drive mechanism for the paper feed rollers of a scanner. In this application, an electric motor drives a small gear that drives larger gears to turn the feed rollers. The feed rollers then draw the document into the machine’s scanning device.

Feed rollers for a scanner.

GEARS

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TYPES OF GEARS

Spur gears are simplest and, hence, the most common type of gear. The teeth of a spur gear are parallel to the axis of rotation. Spur gears are used to transmit motion between parallel shafts, which encompasses the majority of applications. A pair of mating spur gears is illustrated in the Figure

A rack is a special case of spur gear where the teeth of the rack are not formed around a circle, but laid flat. The rack can be perceived as a spur gear with aninfinitely large diameter.When the rack mates with a spur gear, translating motion is produced. A mating rack and gear are illustrated in the Figure.

Rack and pinion pair

Internal gear and pinion meshing.External gear and

pinion meshing.

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Helical gears are similar to, and can be used in the same applications as, spur gears. The difference is that the teeth of a helical gear are inclined to the axis of rotation. The angle of inclination is termed the helix angle, ϕ . This angle provides amore gradual engagement of the teeth during meshing and produces less impact and noise. Because of this smoother action, helical gears are preferred in high-speed applications. However, the helix angle produces thrust forces and bendingcouples, which are not generated in spur gears. A pair of mating helical gears is illustrated in the Figure.

Helical gears.

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Herringbone gears are used in the same applications as spur gears and helical gears. In fact, they are also referred to as double helical gears. The herringbone gear appears as two opposite-hand helical gears butted against each other.This complex configuration counterbalances the thrust force of a helical gear. A herringbone gear is shown in the Figure.

Herringbone gear

Bevel gears have teeth formed on a conical surface and are used to transmit motion between nonparallel shafts. Although most of their applications involve connecting perpendicular shafts, bevel gears can also be used in applications that require shaft angles that are both larger and smaller than 90°. As bevel gears mesh, their cones have a common apex. However, the actual cone angle of each gear depends on the gear ratio of the mating gears. Therefore bevel gears are designed as a set, and replacing one gear to alter the gear ratio is not possible. A pair of mating bevelgears is illustrated in the Figure.

Bevel gear

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A worm and worm gear is used to transmit motion between nonparallel and nonintersecting shafts. The worm has one tooth that is formed in a spiral around a pitch cylinder. This one tooth is also referred to as the thread because it resembles a screw thread. Similar to the helical gear, the spiral pitch of the worm generates an axial force that must be supported. In most applications, the worm drives the wormgear to produce great speed reductions. Generally, a worm gear drive is not reversible. That is, the worm gear cannot drive the worm. A mating worm and worm gear are shown in the Figure

Worm gears.

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Spur gear toothGear tooth features.

The pitch diameter, d, of a gear is simply the diameter of the pitch circle. Because the kinematics of a spur gear are identical to an analogous friction roller, the pitch diameter is a widely referenced gear parameter.

The number of teeth, T, is simply the total number of teeth on the gear.

The module, m, is a commonly referenced gear parameter in the SI unit system, themodule has units of millimeters. The module is also a relative measure of tooth size. It is defined as the ratio of pitch diameter to the number of teeth in a gear:

millimeter millimeter=dm , m : , d :T

The circular pitch, p (or cp) can be calculated from the number of teeth and the pitch diameter of a gear. The governing equation is π

=dpT

The kinematic function of gears is to transfer rotational motion from one shaft to another. Since these shafts may be parallel, perpendicular, or at any other angle with respect to each other, gears designed for any of these cases take different forms and have different names: spur, helical, bevel, worm, etc.

GEAR KINEMATICS

13 13 2 2 223

12 12 3 3 3

ω= = = − = − = −ω

n r d TRn r d T

12ω13ω

2r3r

2d 3d2 3

Two gears can only mesh if they have the same ”module”

( ) ( )

2 3

12 2 13 3

12 2 13 3

12 2 13 3

12 2 2 13 3 3

2 3 12 2 13 3

260

=

ω = − ωω = − ω

= −π

ω=

= − =

= ⇒ = −

t tV V.r ( ) .r.d ( ) .d

n .d ( )n .d , n : rpmn ,rad / sn

n . m .T ( )n . m .T , d m.T*****m m n .T ( )n .T

3T

2T

Gear Ratio23 23=R N Speed Ratio

Gear Ratio is equal to the speed ratio for simple gear trains

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Annular (internal) gear and pinion meshing.Vp

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3d

2d

13ω12ω

3T

2T

( ) ( )

2 3

12 2 13 3

12 2 13 3

12 2 13 3

12 2 2 13 3 3

2 3 12 2 13 3

260

=

ω = ωω = ω

=π

ω=

= =

= ⇒ =

t tV V.r .r.d .d

n .d n .d , n : rpmn ,rad / sn

n . m .T n . m .T , d m.Tm m n .T n .T

13 13 2 2 223

12 12 3 3 3

ω= = = = =ω

n r d TRn r d T

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12

13

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BELT AND CHAIN DRIVES

The primary function of a belt or chain drive is identical to that of a gear drive. All three of these mechanisms are used to transfer power between rotating shafts. However, the use of gears becomes impractical when the distance between the shafts is large. Both belt and chain drives offer the flexibility of efficient operation at large and small center distances.

Belt drive geometry.

Chain drive geometry

1 21 2

1 2 2

2 1 1

2 2= ω =ω

ω= =

ω

bd dV . .

r dr d

pulley

1 21 2

1 2 2

2 1 1

2 2= ω =ω

ω= =

ω

cd dV . .

r dr d

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BELT DRIVE:Cross drive

Vb

Vb

Grinding wheel Chipper/shredder

Examples

( )1 21 2

1 2 2

2 1 1

2 2= ω = − ω

ω= =

ω

bd dV . .

r dr d

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Example: A gear train is shown in the Figure The gears have the following properties:

Gear 2: T2 = 12 teeth and m2 = 1Gear 3: d3 = 48 mmGear 4: T4 = 18 teethGear 5: d5 = 48 mm and m5 =2Gear 6: d6 = 24 mm and m6=1.5Gear 7: T7 = 48 teeth

1 1 1 1

2

7

3

4

5

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Determine the rotational velocityof gear 7 as gear 2 drives at 1800 rpm counterclockwise.

[ ]12 2 13 3 12 2 13 3 3 3

14 4 15 5 15 15

16 6 17 7 17 17

1800 12 48 450450 36 48 337 5

337 5 24 72 112 5

= − = − ⇒ = − ⇒ = −

= − ⇒ − = − ⇒ == − ⇒ = − ⇒ = −

n .d ( )n .d or n .T ( )n .T . .N N rpm,CWn .d ( )n .d . n . n , rpm,CCWn .d ( )n .d , . n . n , rpm,CW

12 1800=n rpm,CCW

( )1

1

Product of driving gear tooth numbers1 k=number of external meshesProduct of driven gear tooth numbers

= = − kjij

i

nR ,

n

( )317 2 4 6 1727 17

12 3 5 7

12 18 161 112 51800 48 24 48

= = − = − ⇒ = −n T .T .T n . .R , n . rpm,(CW )n T .T .T . .

OR

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PLANETARY GEAR TRAINS

Planetary gear trains allow obtaining high transmission ratios in a compactdesign, which makes them suitable for applications in, for example, machine tools, hoists, and automatic transmissions.

3

4

5

2

5

20

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PLANETARY GEAR TRAINS

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VPi = VPj = VA + VP/A

VPj = VPi = ω1j rj

VA = ω1k(rj±ri) (+ if external, - if internal mesh)

VP/A=± ω1iri (- if external, + if internal mesh)

Where ri and rj are the radii of the gears. Then:

ω1j rj= ω1k(rj±ri) ± ω1iri

1 1

1 1

ω −ω± =

ω −ωj ki

j i k

rr

± = ± =± =i i iij

j j j

r d T R , Gear Ratior d T

Gear Ratio is not equal to the speed ratio

Let us assume that the angular speed is positive if it is counter clockwise. Since the velocities considered are along the same direction (but, may be in opposite sense), the vector velocity equation can be treated as a scalar equation. The velocities can be expressed in terms of angular speeds and the gear radii as:

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Note that the definition of the gear ratio is the same for the planetary gear trains and for simple gear trains. For external gearing Rij= -Ti/Tj and for internal gearing Rij =+Ti/Tj. It is also interesting to note that the terms (ω1j - ω1k) and (ω1i - ω1k ) are really the angular speeds of the sun and planet gears relative to the arm, e.g.:

ωkj = (ω1j - ω1k) (angular speed of link j with respect to link k)ωki = (ω1i - ω1k) (angular speed of link i with respect to link k)

Therefore:

1 1

1 1

ω ω −ω= =ω ω −ω

kj j kij

ki i k

R

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A simple epicyclic gear train

its schematic diagram.

12 13 15 134 4

2 14 13 5 14 13

ω −ω ω −ω− = =

ω −ω ω −ωT T, ,T T

1 1

1 1

ω −ω± = ± =

ω −ωj ki i

j j i k

r Tr T

Example:

15 132

5 12 13

ω −ω= −

ω −ωTT

1513 12 15

20040200 100 25080 100 200

ω −ω = ω = ⇒ = − ⇒ω =

−rpm, rpm rpm(CCW )

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Example:

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The below figure illustrates a reverted planetary gear train. Gear 2 is fastened to its shaft and is driven at 250 rev/min in a clockwise direction. Gears 4 and 5 are planet gears that are joined but are free to turn on the shaft carried by the arm. Gear 6 is stationary. Find the speed and direction of rotation of the arm.

Example:

View Direction

1 1

1 1

ω −ω± = ± =

ω −ωj ki i

j j i k

r Tr T

12 13 5 16 134

2 14 13 6 15 13

ω −ω ω −ω− = − =

ω −ω ω −ωTT , ,

T T

15 14ω =ω

13 13

14 13 15 13

250 030 1620 34

− −ω −ω− = − =

ω −ω ω −ω, ,

( )12 16250 0ω = − ω =rpm, CW ;

( )13 14 15114 28 128 57ω = ω =ω = −. rpm, CCW ; . rpm(CW )

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As another example, consider the planetary gear train shown below.Since ,

1 1

1 1

ω ω −ω= =ω ω −ω

kj j kij

ki i k

R 1 1

1 1

ω ω −ω= =ω ω −ω

kj j kpj

kp p k

R

1 1

1 1

ω −ω= =

ω −ωij p k

ippj i k

RR

R1= jp

pj

RR

( )1= = −′

k i jip ij jp

j p

T .TR R .R

T .T

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An gear train with two planet gears

R47= (T4/T5).(T5’/T6).(T6/T7)

Example:

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If there is more than one arm in the planetary gear train, the gear train is called compound planetary gear train. In such a case the above equations must be written for each simple planetary gear train. In Compound gear trains one must first identify gears that have a moving axis of rotation. These gears are the planet gears. Next, one must identify the links that are connected to the planet by a revolute joint. This link is the arm of the planet considered. Then for each arm found, the planetary gear relation must be written.

( )1= = −′

k i jip ij jp

j p

T .TR R .R

T .T

Example: Determine the output speed and the direction of rotation when the input speed is 3000 rpm

Link 3 does not have a fixed axis of rotation. Therefore it is a planet. Link 2 is connected to link 3 by a revolute joint. Therefore it is the arm. For this planetary system:

For the remaining portion of the gear train all the gears have a fixed axis of rotation (they are connected by a revolute joint to link 1) therefore, simple gear train relation yields:

for n12=3000 rpm, n16=-59.8 rpm (in opposite direction of the input speed).

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Example:

R14= (T1/T3).(T3’/T4)R16= (T1/T5).(T5’/T6)

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Example: A speed changer is shown. The input speed is 1800 rpm. Determine the output speed and state its relative direction of rotation with respect to the input link.

In the first portion of the gear train (links 2, 3, 4 and 6) the gears have a fixed axis of rotation. Therefore

In the upper part of the gear train link 5 does not have a fixed axis of rotation. Therefore link 5 is the planet and link 4 connected to 5 by a revolute joint is the arm. Therefore for this planetary gear train:

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GEAR TRAINS WITH BEVEL GEARS

The direction of rotation must be determined by noting that the velocity of the two cones at the point of contact (P) will be the same. If link 2 is rotating CW when viewed from point O, link 3 will rotate CCW when viewed from point O. A convenient form of determining the direction of rotation of the gears is to consider an arrow representing the direction of rotation for each gear. If the arrow is piercing into the paper, one will see the end of the arrow which will look like a " +" sign. If it is piercing out of the paper, its tip will be seen and it will look like a "." sign. In the side view of a gear, the arrow showing the rotation of the gear can be represented by the above two signs. At the point of contact of two gears, the arrow will either pierce into (+) or out of (.) the paper and on the other sides of the gears it will be the opposite. Hence, the direction of rotation of each gear can be determined using this convention.

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Example: Consider a compound gear train shown. Determine R24 and the direction of rotation of link 4 with respect to link 2.

Consider a compound gear train shown. Determine R24 and the direction of rotation of link 4 with respect to link 2.

View Direction

In order to determine the direction of rotation of link 4 relative to link 2, consider link 2 rotating in a certain direction i.e. let us assume clockwise rotation when viewed from the right.

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BEVEL PLANETARY GEAR TRAINS

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Example: Consider the planetary gear train shown. We can write the gear ratio between links 1 and 4, when the arm (link 2) is considered fixed as:

The sign of the gear ratio is determined with the convention described before (considering the arm as fixed).

n11 = 0, and it is obtained:

Where (-) sign states that link 2 (arm) and link 4 rotate in opposite direction relative to each other.

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Example: The gear train shown below is known as "Humpage's bevel gear differential" link 2 is the input and link 4 is the output. We would like to determine the speed ratio N24 Considering the gear train 2, 3 and 1, link 3 is the planet and link 5 is the arm. Hence:

Since n11=0 :

Considering the gear train 1, 3 and 4, link 3 is the planet and link 5 is the arm. Hence we can also write:

Since n11=0 :

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Example

R43= (T4/T6).(T6’/T3)

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1- Find the angular velocities of the gears and arm.

1

11

4142

2

D3=40 mmD41=80 mmz42=20 teethm5=2 mmDK1=80 mmDK2=40 mmN3=2400 rpmNK2=800 rpm

1

View Direction

CCW

K1

35

×K2

BeltView

DirectionCCW

z=T

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2- Find the angular velocities of the gears and arm.

1

1 1

8 7

9

106

m5=2 mmm2=4 mmz3=40 teethz4=60 teethr6=90 mmm10=2 mmz10=30 teeth

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1

1

5 2600 rpm 1800 rpm

View Direction

CCW

View Direction

CCW

z=T

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3- Find the angular velocities of the gears and arms. (r2-arm = reffective)

1

1 1

79

8

1600 d/d CW

r5=160 mmr2-arm =120 mm

z6=80 teethz7=40 teethz9=160 teeth

1600 d/d CCW

3

45

2

6

View Direction View Direction

View Direction

z=T

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4-Find the angular velocities of the gears and arm