Gear Trains

GEAR TRAINS

MENG 364 MACHINE DYNAMICS

Prof. Hamza Diken

September 2011

Gear Trains

• A gear train is a set or system of gears arranged to transfer torque from one part of a mechanical system to another.

• Gear trains may consist of:• Driving gears - attached to the input shaft• Driven gears - attached to the output shaft• Idler gears - interposed between the driving and driven

gear in order to maintain the direction of the output shaft the same as the input shaft or to increase the distance between the drive and driven gears. A compound gear train refers to two or more gears used to transmit motion.

Types of gear trains include:

• Simple gear train

• Compound gear train

• Epicyclic gear train

• Reverted gear train

Simple gear train: having only one gear on each axis.

Compound gear train: has two or more gears on one or more axes.

Reverted gear train: the first and last gears are on the same axis.

Epicyclic gear train: one of the gear center moves

Gear train composed of bevel crossed-helical and spur gears

• The gear ratio is the relationship between the numbers of teeth on two gears that are meshed or two sprockets connected with a common roller chain, or the circumferences of two pulleys connected with a drive belt.

mND teeth ofnumber module diameter pitch

22Vc

bb

aa

mNmN

bbaa NN

• The input gear or driver gear in a gear train is the gear directly connected to the motor or other power source. Thus the driver is the gear that transmits power to the other gears in the gear train. In a simple two-gear system, the second gear is called the output gear or driven gear. In a gear train consisting of more than two gears, the final gear (the gear connected to a wheel axle or other rotating mechanical component) is the output gear.

TP speed torquepower

o

i

i

o

ooii

T

T

TTP

ueinput torq

queoutput tor advantage mechanical

• gear ratio (gr) = (number of teeth on output or driven gear)/(number of teeth on input or driver gear)

i

o

N

Ngr

o

i

i

o

N

Nsr

Speed ratio =angular velocity of output shaft or driven gear/ angular velocity of input shaft or driver gear.

ratiogearratiospeed

1

• Mating gear teeth acting against each other must be “conjugate” in shape to ensure that the angular velocity ratio remains constant. The gear shape or profile that is in universal use is the “involute”.

The involute gear profile is the most commonly used system for gearing today. In an involute gear, the profiles of the teeth are involutes of a circle. (The involute of a circle is the spiraling curve traced by the end of an imaginary taut string unwinding itself from that stationary circle called the base circle.)

• Spur gear teeth are straight and parallel to the axis of rotation. Shafts of spur gears in contact are also parallel. When motion is to be transmitted between shafts that are not parallel, use ca be made of worm, bevel, hypoid or crossed-helical gears.

Spur gear

Rack and pinion gearing Worm gearInternal gear

Bevel gearHelical gear

Hypoid gear

• Pitch circle is an imaginary circle such that pitch circles of mating gears are tangent to each other. The circle we draw to show a gear is the pitch circle of that gear.

• The pinion is the smaller of two mating gears. The pinion generally has 9 or more teeth. The ratio of the number of teeth between a gear and its pinion is kept generally equal to or less than 6 in good engineering practice.

• Module m is the ratio of the diameter D of the pitch circle to the number of teeth N.

m=D/N

• Module is expressed in mm/tooth.

• Common modules have values from 0.3 to 10 mm/tooth.(these values are standard)

• Diametral pitch is the inverse of the module

P=N/D

• The pressure line is the line of action of the force between the contacting teeth of gears in mesh.

• The pressure angle is the smaller angle formed by the pressure line and the normal to the line joining the centers of two gears in mesh.

• Commonly used pressure angles are 20, 22.5 and 25 degrees.

• A rack is a spur gear of infinite radius. It is in the form of a straight gear path.

• An internal gear, also called ring gear or annular gear, has teeth cut on its inside diameter.

• An external gear has teeth cut on its outer diameter.

• The pitch of a screw thread is the distance a nut moves axially when the nut makes one turn. For a screw and nut to mesh together, they must have the same pitch. For two gears to mesh together, they must have the same module.

• Two or more gears form a gear train when they are used to transmit motion from one shaft to another. Belts, chains and screws are many times used in conjunction with gear trains. There are two types of gear trains;

• In the ordinary gear train, axes of all gears are fixed relative to the frame.

• In the epicyclic or planetary gear train the axis of at least one gear moves in a circular path.

ExampleInput speed is given. Find output speed of the gear train. Arrows indicate direction of

rotation.

eedd

ddcc

cb

bbaa

NN

NN

NN

a

b

a

e

ca

b

a

d

c

e

dc

d

c

e

dd

e

de

N

N

N

N

N

N

N

N

N

N

N

N

N

N

N

N

a

b

c

e

b

a

e

c

a

e

N

N

N

N

srgr

N

N

N

Nsr

1

• From arrows on the gear train, we deduce that the output is in the opposite direction to the input. Gear D is an idler gear in this train. Idler gears do not affect the speed ratio.

• We can generalize the process of computation for ordinary gear trains to other linear mechanical systems such as belt and chain drives. In the case of belt drives diameters of the pulleys are used in the relationships. For chain drives one can use either the number of teeth on the sprockets or the sprocket diameter itself.

example

Belt drive

example

You are to design an ordinary gear train that will reduce an input speed of 1370 rpm cw to 52 rpm ccw at output. Use gears of module 5 such that the pinions are not less than 50 mm in diameter and the gear diameters are not more than 200 mm.

Total reduction ratio is

This ratio is too high for a single pair of gears since it is greater than 6. If two identical pairs of gears are used the reduction ratio will be

This ratio is acceptable. Supposing that the pinion diameter is selected as 50 mm, the corresponding gear diameter will be

Which is greater than 200 mm

3.2652

13701

0379.01370

52

o

i

i

o

srgr

sr

21 grgrgr1.53.26

3.262

gr

gr

If three identical pairs of gears are used the reduction ratio will be

If we assume number teeth on the pinion is 12 and module m=5 than pinion diameter is

Gear diameter and gear teeth number is

Which is acceptable

3.263 gr 397.23.26 3

1

gr

36312N

mm 180360

g

grN

grDD

p

pg

Possible layout of gearbox

Output speed is

Since the desired output speed is 52 rpm, it would be convenient to modify one of the gears

Example

In the binding press shown in the figure, input motion (shaft A) is transmitted by bevel gears 1 and 2 and then spur gears 3 and 4 to screw 5. gears 2 and 3 are integral, and gear 4 is permitted to slide up and down along gear 3. Screw 5 is single threaded and has pitch of 8 mm. Screw 6 is machined into a bore in screw 5 single threaded and has a pitch of 4 mm. Screw 5 is right handed (it will move up if gear 4 is rotated ccw looking from top) and screw 6 is left handed. Pressure plate B can move up and down but cannot rotate. We wish to find the number of turns shaft A must make to lower the plate by 25 mm.

For every revolution of screw 5, plate B moves 8 + 4 = 12 mm up or down.

EXAMPLE

A typical three speed automotive transmission of the manual type is shown in figure. For an engine speed of 5000 rpm, compute the speed of the output shaft J for the cases of A) Direct drive (A to F) B) second gear (C and D coupling) C) first gear (E and F couples) D) reverse gear (G, H and F couples).

A) For direct drive shaft J is shifted toward A such that the clutch between the input shaft and shaft J is engaged, thus making J to rotate at the same speed as A.

B) In the second gear position C meshes with D so that

C) Gear E meshes with F in the first gear position such that

D) In reverse gear G meshes via idler H with F so that the direction is reversed

EXAMPLE

We wish to design a two speed gear box for a new drill, for which the input speed is 1500 rpm. The desired output speeds are 450 and 280 rpm, approximately. Strength considerations dictate a module of m=5 mm. It is also known that Na=15 and Nb=40 teeth.

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bababa

NNNNNN

m

LNN

mNNmNmNDDL

also constant, is which 2

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22

Epicyclic Gear Trains

• Epicyclic gearing or planetary gearing is a gear system consisting of one or more outer gears, or planet gears, revolving about a central, or sun gear. Typically, the planet gears are mounted on a movable arm or carrier which itself may rotate relative to the sun gear. Epicyclic gearing systems also incorporate the use of an outer ring gear or annulus, which meshes with the planet gears.

• Sun: The central gear

• Planet carrier: Holds one or more peripheral planet gears, all of the same size, meshed with the sun gear

• Annulus or ring gear: An outer ring with inward-facing teeth that mesh with the planet gear or gears

Figure 1. An epicyclic gear train and its associated terminology

• Epicyclic gear train typically comprises a sun gear S with fixed center, a planet gear P or gears the center(s) of which orbit in a circular path and an arm or planet carrier “a” that connects the center of the planet gear to the center of the sun gear. A ring gear R sometimes provided. The sun gear may be omitted at that times. More than one planet gear is used in practice for reasons of stress distribution and balancing. More than one set of planetary trains is usually provided for achieving a variety of speed.

Once the epicyclic elements of a simple epicyclic train have been identified, analysis is done as follows:

• Lock the train (freeze the elements so that no relative movement is possible) and rotate the entire train one positive turn.

• Hold (fix) the arm, and rotate the stationary (fixed gear by one negative turn. Note down the number of rotations of relevant gears.

• Add the resultant revolutions.

example

The number of teeth as follows Ns=40, Np=20, Nr=80. Find the angular velocity of the sun gear S if the arm rotates ccw at 50 rad/s. Notice that ring gear is fixed.

solution

step arm ring planet sun

1. Gears locked

+1 +1 +1 +1

2. Arm fixed 0 -1 -80/20 = -4 +(80/20)(20/40) = +2

3. Result +1 0 -3 +3

Thus every revolution of the arm, the planet P rotates three times in the opposite direction, and the sun gear S rotates three revolutions in the same direction as the arm. The ring gear does not rotate. Since the arm rotates at 50 rad/s ccw, it is concluded that the sun gear must turn in the same direction with speed of 150 rad/s.

example

The sun gear A of the train runs at a speed of 1200 rpm. Determine the speed of the output shaft F.

solution

step A C Arm BE

1 +1 +1 +1

2 (142/60)(60/22) = +6.45 -1 0

3 +7.45 0 +1

Thus input gear A must turn 7.45 revolutions for the arm to make one revolution.

Second train yields similar result thus the speed of shaft F is

• In case there is more than one input into the epicyclic train (multi input gear train) the principle of superposition is employed. Each input is analyzed singly, while keeping the other inputs fixed. Total output is obtained by summing up all outputs.

example

Automotive differential shown in figure which is an epicyclic device found in every car. Given that the speed of the left wheel is 100 rpm and of the right wheel 135 rpm, find the speed of gear A which is coupled to the torque shaft leading to the engine.

solution

Step C E arm and B D

1 +1 +1 +1 +1

2 No need (16/11)(11/16)=+1 0 -1

3 No need +2 +1 0

Fix the left wheel

step D E Arm and B C

1 +1 +1 +1 +1

2 (16/11)(11/16)=+1 -1 0 No need

3 +2 0 +1 No need

Fix the right wheel

• The velocity of gear A is found as

example

Input-1 rotates ccw at 120 rad/s and input-2 rotates cw at 360 rad/s. Find the magnitude and direction of the output speed for the teeth numbers indicated.

• Note that gears A and B are not epicyclic, although C is a sun gear with planet D. From the drawing it should be understood that gears B and C and also D and E are integral. We first fix the input-1. This fixes the arm and the train becomes ordinary.

step C D and E F arm

1 +1 +1 +1 +1

2 -1 Not needed (48/24)(36/108)=+0.67 0

3 0 Not needed +1.67 +1

Fixing input-2, we form the following table