1..INTRODUCTION

Any combination of gear wheels by means of which motion is transmitted from one shaft to another

shaft is called a gear-train. In case of epicyclic gear-trains the axes of the shafts on which the gears are

mounted may move relative to a fixed.

The Epicyclic Gear Trains are useful for transmitting high velocity ratios with gears of moderate size

in a comparatively lesser space. The epicyclic gear trains are used in the back gear of lathe, differential gears

of the automobiles, hoists, pulley blocks, wrist watches etc.

Objectives of Project:

To measure Epicyclic Gear ratio between input shaft and output shaft.

To measure Input torque, Holding torque and Output torque.

1

2. THEORY

Any combination of gear wheels by means of which motion is transmitted from one shaft to another

shaft is called a gear-train. In case of epicyclic gear-trains the

axes of the shafts on which the gears are mounted may move

relative to a fixed axis.

1) External type epicyclic Gear-Train –

It consists of Bearing Blocks supporting input

shaft and output shaft, through the epicyclic gear-train.

The input or driving shaft carries on the arm A, PIN on

which the compound wheel B-C is free to revolve.

Wheel C meshes with the fixed wheel E and wheel B meshes with a wheel D keyed to the driven

shaft.

2) Internal Type Epicyclic Gear-Train –

It consists of a SUN gear mounted on input shaft. Two planet gears, and gears on both side meshes

with SUN gear and which also meshes with the internal tooth of the angular gear. Two planet gears are

mounted on the pins which are fitted into both ends of the arm. 0utput shaft is connected to the arm (A) on

which a drum is fixed.

Tooth Loads and Torques in Epicyclic Gear Trains –

If the parts of an epicyclic gear-train are all moving at uniform speeds, so that no angular

accelerations are involved, the algebraic sum of all the external torques applied to the train must be zero, or

(T) = 0 ............................ (1)

There are at least three external torques for every train , and in many cases there are only three. These are,

Ti the input torque on the driving member, Arm.

To the resisting, or load torque on the driven member.

Th the holding , or braking torque on the fixed member.

If there is no acceleration,

Ti + To + Th = 0 ................... (2)

2

An epicyclic speed-reduction gear is shown in Fig. The driving shaft carries on the arm , a pin, on which the

compound wheel is free to revolve.

3. System Development

3

System development includes

Design of machine components

Fabrication and assembly

Cost of project

Working

3.1 Design of Machine Component:

Design of “Epicyclic Gear Train Apparatus” consists of design of following systems-

Design of Shaft

Design of Bearing

Design of Stand

Thermal analysis of pulley

3.1.1 Design of shaft

To design shaft from gearbox outlet to last end between which two bearing and one pulley is present.

Selected motor-1 HP, 1500 rpm.

3.1.1.1 Design shaft for maximum power transmission

Consider the speed ratio= 3.85

i.e. n1n2

=3.85

n2=15003.85

n2=389

Speed of the shaft=389 rpm.

Consider material of shaft=40C8

Sut=600 N/mm2

Syt=380 N/mm2

G=79300 N/mm2

P=2∗π∗n∗T

60 (P=746 w, n=389 rpm)

4

746= 2∗π∗389∗T

60

T = 746∗60

2∗π∗389

T=18.313 Nm ………(1)

Consider the material of belt is Polyamide,

µ=0.5

Angle of lap=θ=π=3.1412 rad

Let T1=tension in tight side

T2= tension in slack side

T 1T 2

=eµθ(µ=Coefficient of friction)

T 1T 2

=e0.5∗3.1412

T 1T 2

=4.8

T1=4.8 T2

Torque transmitted by pulley,

T=(T1-T2)*r ............(2)

Consider, a standard diameter of pulley=d=200mm

T=(T1-T2)*0.1

18.313=(T1-T2)*0.1

183.13=T1-T2

183.13=4.8*T2-T2

183.13=3.8*T2

T2=48.192 N and T1=4.8*T2=231.32 N

5

F=T1+T2

F=231.32+48.192

F=279.51 N

The force F is act at the midpoint of shaft , therefore the reaction at bearing is exactly half of the force and

opposite to that of direction of force.

Ra=Rb=139.756N

The bending moment at the midpoint of the shaft i.e. at pulley is

M=F*0.18

M=25.156*103 Nmm

The load is applied gradually by hand wheel,

Kt=1.5, Kb=1

The Equivalent Torque ,

Te=√(Kb∗M)2+(Kt∗T )2

Te=√(25.156)2+(1.5∗18.313)2

Te=37.246*103 N-mm6

3.1.1.2 Design of shaft by ASME code,

𝞽max=0.18*Sut

𝞽max =0.18*600

𝞽max =108N/mm2

Also, 𝞽max =16∗Teπ∗d3

d3= 16∗37.247∗103

π∗108

d=12.065mm

3.1.1.3 Design for torsional rigidity

θ=0.25/m length

θ=T∗LG∗J

θ=32∗T∗LG∗π∗d4

θ=32∗18.313∗1000

79300∗π∗d4

d=9.848mm

By considering large value, d=12.065mm

By Selecting next standard value, d=15mm.

3.1.2 Design of Bearing

Due to gearing, axial thrust is produced and speed is low. So we select the hydrodynamic bearing

with grease as lubricant.

Let, d=15mm

Assuming l/d=1

Let, w=139.756N

P=w

l∗d7

P=139.756

d2

P=0.62 N/mmm2

Pmax=2N/mm2

Standard value for gear reducer (V.B.Bhandari-625)

P

Pmax =0.31

From l/d=1 and P

Pmax =0.31 we can find the remaining values for bearing,

€=0.8

ho/c=0.2

s=0.0446

Φ=36.24

(r/c)*f=1.7

Q/(rcnsl)=4.62

Qs/Q=0.842

3.1.3 Design of stand:

Design of stand consist of two parts i.e. design of horizontal member and design of vertical

member.

3.1.3.1 Design of horizontal member

8

For Design of horizontal member we have to consider the forces acting on the belt and support

reactions. From these we find cross section of beam.

Beam material is selected as CI,

Syt =300N/mm2

FOS=3

Considering the beam is rectangular in cross section and width is three times thickness.

(b=3*t).

Fig: force acting on beam

𝞼max=3003 N/mm2

=100 N/mm2

𝞽all =0.5 Syt

3 N/mm2

= 50 N/mm2

But 𝞽all =SAYbI N/mm2 where, S=force actin on beam

A= Area of cross section

Y=distance of CG from NA

b= Width

I= moment of inertia

9

109.3 N170.277 N

231.32N48.192N

A=b×t/2=1.5×t2

y=t/4

Ixx=(1/12)×b×t3

𝞽all=(231.02×1.5×.25×t3) / (3×0.25×t4)

=150

Solving this we get,

t=1.514mm .

This is very less therefore considering next standard value of t=5mm.

we get, b=3t

=3×5 =15mm.

3.1.3.2 Design for vertical member

Selecting 0.5” hollow pipe for design and checking the condition for buckling we can find maximum

load that can be applied.

Maximum load applied=π × π × E ×I

¿×≤¿¿

E=200GPa

We get the value of maximum force P=23190 N

But we are going to apply the force of 231.32N only hence design is suitable.

10

3.1.4 Thermal analysis of Pulley

We are selecting the pulley of 200 mm diameter.

The weight of pulley is equal to 3 kg.

ΔT=E

Cp∗m

ΔT=746

500∗3

ΔT=0.5ºc

3.2 Fabrication and Assembly:

Fabrication of components consists of detailed procedure followed, sequence of operation performed

during manufacturing.

3.2.1 Frame:

11

3.3 Cost of Project

Sr. no. Name Of Component Rate Quantity Total1 Texmo Bear motor 1HP

(DC motor)

8500 1 8500

2 Epicyclic Gear Box 1HP

flange holding

9000 1 9000

3 Spring Balance 10 kg 560 3 16804 Dimmer 4DLF 2600 1 26005 RPM Digital indicator 975 1 9756 AD Coverter bridge

rectifier

350 1 350

7 Flanged Coupling 220 4 8808 SKF Bearing 370 2 7409 Pulley (20 dia) 450 1 45010 Wheel 280 1 28011 3 Way toggle switch 80 1 8012 2 way toggle switch 80 1 80

12

13 Standard voltmeter 220 1 22014 Standard Ammeter 220 1 22015 Plug pin 40 1 4016 Glass Fuse 20 1 2017 1.5 X 1.5 Sheet 3 mm

thick

120 6 720

18 Pipe 15 14 foot 21019 Rectangular Bar

1.25”X0.25

60 4 Foot 240

20 Wire G and S 20

15

8 foot

6 foot

160

9021 Shaft 2 foot 850

22 Rope 12 mm 30 2 mtr 6023 Nut Bolt

8 mm

6 mm

4mm

8

6

5

18

8

4

144

48

2024 Washer 30 1/2 1525 1.5 Steel pipe X Heavy 32 8 foot 25626 C Channel C3*4.1 70 18 foot 126027 Screw 1.5 cm, 1 foot 120 3 360

Total 3054

3.4 Working of ‘EPICYCLIC GEAR TRAIN APPARATUS”:EGTA consists of three sub-systems and their working is as follows-

l. To measure epicyclic gear ratio

2. Measure Input torque i.e. Ti. Output Torque To & Holding torque Th.

3. To find efficiency of gearbox.

Ti + To + Th = 0.

Holding Torque Th = - (To + Ti )

13

INTERNAL TYPE EPICYCLIC GEAR-TRAIN (MOTORISED) –

PREQUATIONS

1) Check the experimental set-up.

2) Bring dimmer to zero position before start.

3) Give supply to Motor from control panel.

4) Adjust the RPM of Input shaft to some fix value.

5) Apply holding torque, just to hold the drum. This must be done carefully.

6) Apply Load to the output shaft.

EXPERIMENT NO. 1 –

To Study and verify the speed ratio of Epicyclic gear train (Motorised).

a) Rotate the motor coupling by hand and see that for what no. of revolutions of Motor

shaft, the output shaft rotates through one revolution. (Hold the gear box in position)

b) Start the motor (By properly connecting the control panel) and slowly increase the

speed and note the output shaft speed. Take two such speeds and calculate the speed

ratio and verify.

Gear Ratio =

EXPERIMENT N0. 2 –

To study and verify the Torque relationship,

Ti + Th + To = 0…………………….(1)

14

Input Speed

Output Speed

where,

Ti = Input Torque

Th = Holding Torque

To = Output Torque

a) Thp = Practical holding Torque

This can be calculated as below

Th = S1 x L

S1 - reading of Tension on spring balance.

L - distance of spring balance from centre of gear box=0.2m

b) To = Output Torque

This can be calculated as below

To = (S2 – S3) x R

R= Radius of the loading drum=0.095m

S2 – S3 = readings of Tension on spring balances.

c) Net kinematic energy dissipated by gear train must be zero

Tiωi + To ωo + Thωh = 0…………………….(2)

Here,

ωi= Angular velocity at i/p

ωo= Angular velocity at o/p

ωh= Angular velocity of fixed member

As fix member is held between suspended tension meters ωh=0

15

So,

Tiωi + To ωo = 0

Ti = - To×( ωo / ωi) …………………….(3)

From (1),

Th = -(Ti + To )

From (3),

Th = -(- To×( ωo / ωi) + To )

Th = To×( ωo / ωi) - To

Th = To{( ωo / ωi) - 1}

From equation for angular velocity,

Th = To{( No / Ni) - 1}

This is theoretical holding torque.

d) Efficiency of gear train

= Practical holdingTorquetheoreticalholding torque .

= ( Tha / Tht)

We find that Ti + Th = To

Ti+To+Th=0

PROCEDURE

Take following steps to verify the Torque relationship,

1) Put on the spring balances on gear unit and output shaft

pulley.

16

2) Connect the control panel to motor.

3) Start the motor and measure input & output shaft speeds.

4) Note down the readings of Input & Output RPM and

readings of spring balance.

4. SYSTEM ANALYSIS

ANALYSIS FOR LOAD IN ANSYS

RESULT FROM ANSYS

NODE FX FY

1 511.54 70.954

2 -594.50 85.444

3 597.62 79.690

4 -730.90 -25.192

5 -253.81 -39.472

6 236.76 -29.184

7 744.86 -11.173

8 -510.74 85.756

9 -633.26 -43.851

10 -199.46 -12.881

11 201.92 -34.671

12 629.96 -17.50917

OBSERVATION TABLE

RESULT TABLE

Sr.no Speed Ratio Output torque T2(kgm)

Holding torque theo. Ttheo(kgm)

Holding torque actualTactual(kgm)

Efficiency Η

1 5.73 0.11875 -0.09804 -0.09 0.9182 5.734 0.1615 -0.1333 -0.12 0.9003 5.717 0.209 -0.1724 -0.15 0.8704 5.753 0.24225 -0.20 -0.17 0.850

18

Sr.no Spring Balance Reading

S1

Tension meter 1 reading

S2

Tension meter 2 reading

S3

I/P RPMN1

O/P RPMN2

1 0.45 2 0.75 1244 2172 0.6 2.5 0.8 1202 209.73 0.75 3 0.8 1150 201.14 0.85 3.5 0.95 1120 194.75 1.05 4 1.05 1075 1886 1.1 4.5 1.21 1050 183.87 1.25 5 1.25 1010 175.88 1.4 5.5 1.3 985.8 172.69 1.6 5.75 1.35 980 170.910 1.65 6 1.4 975 17011 1.255 4.75 1.1 1023 179.512 1.45 5.25 1.15 995.3 173.513 1.2 4.25 0.95 1082 18614 1 3.75 0.85 1114 194.115 0.85 3.25 0.8 1144 200

5 5.717 0.28025 -0.2312 -0.21 0.9086 5.714 0.31255 -0.2578 -0.22 0.8537 5.7438 0.35625 -0.2942 -0.25 0.8498 5.711 0.399 -0.3291 -0.28 0.8519 5.733 0.418 -0.3451 -0.32 0.92710 5.734 0.437 -0.3608 -0.33 0.91511 5.7 0.34675 -0.2859 -0.25 0.87812 5.737 0.3895 -0.3216 -0.29 0.90213 5.817 0.3135 -0.2596 -0.24 0.92414 5.74 0.2755 -0.2275 -0.2 0.87915 5.72 0.2327 -0.1921 -0.17 0.885

5. CONCLUSIONS

5.1 Conclusions:

We have studied different types of vibration such as forced lateral vibration, torsional vibration,

damped and undamped vibration.

We have calculated various parameters related epicyclic gear train such as speed ratio of epicyclic

gear box, holding torque, efficiency at different load and speed conditions.

5.2 Future Scope

This machine is operated manually. So errors can occur due to human mistakes. So to minimize these

errors machine can be made automatic so that we can get the result in digital form and thus human errors can

be eliminated.

19