Gear Trains

Gear Train A combination of gears used to transmit motion from one shaft to another. Why combination?: to obtain large speed reduction within a small space. Main types: • Simple gear train: one gear per shaft • Compound gear train: two or more gears per shaft, but for the first, last • Reverted gear train: input and output gears are coaxial. • Planetary/epicyclic gear train: one with a relative motion of the axes.

Simple Gear Train

Gear Ratio

+ : Internal gearing - : External gearing

Simple Gear Train

-

Intermediate gears

Input :Output motion

Odd

Even

Alike

Opposite

Intermediate gears are called idlers in SGT as they do not affect the speed ratio of the gear train. They are primarily used to; (a) Connect gears with large center distance (b) To obtain the desired direction for the driven

Compound Gear Train The compound gears are rigidly fixed to the shaft, hence have the same speed.

-

-

(-1)pairs

-

-

Reverted Gear Train

Epicyclic Gear Train The axes of the shafts over which the gears are mounted may move relative to the fixed axis. Gear A and the arm C have a common axis at O1 about which they can rotate. The gear B meshes with gear A, and has its axis on the arm at O2 about which it can rotate.

If the arm is fixed: Simple gear train

If the gear A is fixed and the arm is rotated about O1 then the gear B is forced to rotate upon and around gear A. Such a motion is called epicyclic (epi: upon; cyclic:around)

EGTs are useful for transmitting high speed ratios with gears of moderate size……… ….backgear of lathe; Differential gears of automobiles, wrist watches…

Epicyclic Gear Train

Planetory Gear Train Annulus (Internal T) Planet (Around Sun

Sun (Center)

Epicyclic Gear Train Speed ratio: Algebraic method; Tabular method

The motion of each element of the EGT relative to the arm is written in the form of equations These equations are solved using two conditions: (i) Some element is fixed, implying zero velocity (ii) Motion of some other element is specified.

One caution: You’ll need to associate +/- signs to clockwise and anticlockwise directions

Anticlockwise: + Clockwise: -

Epicyclic Gear Train: Problem-1

- No. of teeth are as shown in the diagram -A is fixed -B and C are carried on an arm which revolves at 100 rpm about the axis of A or D. # Find the number of teeth of C and the speed & sense of rotation of C?

Epicyclic Gear Train: Problem-1 tA= 150; tB=25; tD=40; NC=?

NA=0; Narm =100 CL =-100

Geometry:

dA/2 = dB + dC + dD/2 tA/2 = tB + tC + tD/2 75=25+tC+20

tC=30

Engagement of AB: Internal implies + (NB-Narm) / (NA-Narm) = tA/tB NB- (-100) = (150/25) * (0-(-100))

NB=500

Engagement of BC: External implies (NB-Narm) / (NC-Narm) = -tC/tB NC- (-100) = (-25/30) * (500-(-100))

NC=-600 = 600 rpm clockwise

Epicyclic Gear Train: Problem-2

# Find the speed and direction of wheel D when wheel A is fixed and arm F makes 200rpm clockwise.

Epicyclic Gear Train: Problem-2 tA= 80; tC=72; tD=48 NA=0; Narm =200 CL =-200 Note: NB=NC due to compound shaft; Geometry: rA + rB = rC + rD tA + tB = tC + tD tB = 40 Engagement of AB: External implies -

(NB-Narm) / (NA-Narm) = -tA/tB NB- (-200) = (-80/40) * (0-(-200))

Engagement of CD: External implies (ND-Narm) / (NC-Narm) = -tC/tD Note NC= NB ND = 400 = 400 rpm CCL

NB=-600

600 rpm CL

ND?

Sun-Planet Epicyclic Gear Train

(R)

(A) (P) (S)

Torques in Epicyclic Gear Train When a geared system transmits power, torques are transmitted from one element to another.

Torques in Epicyclic Gear Train

When the rotating parts of an epicyclic gear train have uniform speeds, i.e., zero angular acceleration, then the gear train is kept in equilibrium by three externally applied torques. The net (external) torque applied must be zero.

There are no losses in power transmission, i.e., there are no internal friction losses at the bearings and at the contact surfaces. In other words, the net energy dissipated by the gear train is zero.

Torques in Epicyclic Gear Train: Problem 1

Determine the torque necessary to keep the internal gear stationary

Torques in Epicyclic Gear Train: Problem 2 Shown is a compound epicyclic gear train in which two gears S1 and S2 are integral with the Input shaft B. The arm A2 is integral with the output shaft, C. The planet gear P2 revolves on a pin attached to the arm A2, and meshes with the Sun gear S2 and internal gear I2 that is co-axial with the input shaft.

Arm A1

The planet gear P1 meshes with the fixed internal gear I1 and sun gear S1. The planet gear P1 revolves on a pin fixed to internal gear I2. The number of teeth are as stated in the figure. The input power on the shaft B is 10.47 kW at 1000 rpm (CCL). Find: •The speed and torque at the output shaft •The torque required to hold the internal gear I1.

I1

P1

S1

A1 I2

P2

S2

Torques in Epicyclic Gear Train: Problem 2 The input power on the shaft B is 10.47 kW at 1000 rpm (CCL). Find: •The speed and torque at the output shaft •The torque required to hold the internal gear I1.

Arm A1

Inp:B Teeth (T) Speed (N)

1000 CCL

S1

P1

I1

26

31

88

1000 CCL

0

Rad(S1) + Dia (P1) = Rad(P1) (TS1)/2 + TP1 = (TP1)/2

A1

I2

P2

S2

83

26

31 1000 CCL

Rad(S2) + Dia (P2) = Rad(P2) (TS2)/2 + TP2 = (TP2)/2

A2/C

Torques in Epicyclic Gear Train: Problem 2 The input power on the shaft B is 10.47 kW at 1000 rpm (CCL). Find: •The speed and torque at the output shaft •The torque required to hold the internal gear I1.

I1

P1

S1

Torques in Epicyclic Gear Train: Problem 2 The input power on the shaft B is 10.47 kW at 1000 rpm (CCL). Find: •The speed and torque at the output shaft •The torque required to hold the internal gear I1.

I2

P2

S2

Torques in Epicyclic Gear Train: Problem 2 The input power on the shaft B is 10.47 kW at 1000 rpm (CCL). Find: •The speed and torque at the output shaft •The torque required to hold the internal gear I1.

Epicyclic Gear Train with Bevel Gear

Bevel gears are used for high speed reduction with fewer gears. The method applicable for EGT with spur gears is also applicable for EGT with bevel gears with some change: For those gears whose axes are inclined to the main axis, the terms clockwise and anticlockwise are not applicable, because direction of rotation for different gears ought to be with respect to the same axis.

Epicyclic Gear Train with Bevel Gear: Problem 1

The figure shows an epicyclic bevel GT (Humpage’s reduction gear) in which the gear B is connected to the input shaft, and gear F is connected to the output shaft. The Arm (A), carrying the compound wheels D & E, turns freely on the output shaft. If the input speed is 1000 rpm CCL (seen from the right), determine the speed of the output shaft, when: I: gear C is fixed II: gear C is rotated at 10 rpm CCL.

Epicyclic Gear Train with Bevel Gear: Problem 1 NB = 1000 rpm CCL, determine NF, when: I: gear C is fixed II: gear C is rotated at 10 rpm CCL.

1 B

2 D

C

E

F

3

Epicyclic Gear Train with Bevel Gear: Problem 1 NB = 1000 rpm CCL, determine NF, when: I: gear C is fixed II: gear C is rotated at 10 rpm CCL.

1 B

2 D

C

E

F

3