7/29/2019 Design of Transmission Systems 1 Key

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Key

03M603 Design of Transmission System

UNIT - I

1.a. Toothed belts No slippage more transmission efficiency power transmission at

constant angular velocity ratio no need for initial tension and belt tensioning devices.

- 2 marks

1.b. Determination of Nominal pitch length - 4 marks

C

dDdDCL

4

)(

2

)(2

2+

++=

= 2903 mm

Selection of various modification factors. - 3 marks

Correction factor for length =Fc = 1.05

Service factor = Fa =1.3

Arc of contact =00 60180

C

dD= 157.30 or 2.74 radians

Calculation of maximum power capacity - 3 marks

KW = SSd

Se

)1032.18.5079.0( 2409.0 = 2.759.

Determination of number belts - 4 marks

dc

ab

FFKW

FPn

= = 4 belts.

Calculation of actual centre distance - 4 marks

875.46684

=

+=

dDLA

5.153128/)(2

== dDB

05.9172 =+= BAACActual centre distance = 917.05 mm.

OR

2.a. Regular lay rope In this type, wires are twisted in one direction to form strands and

the strands are twisted in the other direction to form the rope. They are untwisting and

they are accepted as standard. - 2 marks

Long lay rope In this type, wires and strands are twisted in the same direction. They

have greater resistance to abrasive wear and failure due to fatigue. - 2 marks

2.b. Determination of the transmission ratioi= 970/330 = 2.94 - 1 mark

Selection of teeth on the driver sprocket

Z1 = 25 (for i= 2 to 3) &

Number of teeth on the driven sprocket

Z2 = iZ1 = 74 - 1 mark

Determination of optimum centre distance & pitch - 1 mark

a = 500 mm ; pmax = a/30 = 16.6 mm

pmin = a/50 = 10 mm; p = 15.875 mm is chosen.

Selection of chain & chain number - 1 mark

Duplex 10A-2/DR50

Determination of total load on the driving side of the chain - 1 markPT = Pt + Pc + Ps = 1876.4 N

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Determination of Design Load = Ts Pk - 2marks

ks = k1 . k2 . k3 . k4 . k5 . k6 = 1.6

Design load = 1.6*1876.4 = 3002 N

Determination of Factor of safety - 1 mark

Breaking load / Design load = 14.79Check for induced Bearing stress on Rollers - 2 marks

Induced stress = 97.19=

=A

kP st N/mm2 < 4.22][ = N/mm2

Determination of Corrected Chain Length - 2 marks

pppa

zzzzaI /

222

2

1221

+

++=

= 116 links.

= p116 = 1.842 m.Determination of Exact Centre Distance - 2 marks

p

Mee

a

+

= 482

= 512.91 mmDetermination of Sprocket diameters - 2 marks

Pitch circle diameter of smaller sprocket,)/180sin( 1

1z

pd = = 126.66 mm

Outside diameter of smaller sprocket, d01 = d1 + 0.8dr= 134.788 mm

Pitch circle diameter of larger sprocket,)/180sin( 2

2z

pd = = 374.034 mm

Outside diameter of larger sprocket d02 = d2 + 0.8 dr= 382.162 mm

UNIT - II3.a. To account for uneven distribution of tooth load along the face width of the tooth.

- 2 marks

3.b. 1) Selection of material: - 1 mark

C45 steel

Core hardness < 350 BHN

2720mm

Nu = 2360 mm

Ny =

2) Calculation of design stresses - 2 marks

[ ] 14.1

=

nK

Kblb Where 50)(25.01 ++= yu

= 119.5 2mm

N

[ ] 2740. mmNKHRCC clRc ==

3) Centre distance calculation - 4 marks

[ ]

[ ]3

22

.

74.0)1(

+

i

MEia t

c

mma 108

2

7/29/2019 Design of Transmission Systems 1 Key

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4) Calculation of module - 4 marks

mmzz

am 6.3

2

21

=+

=

Adopt m = 4, Revise a

( )

2

21zzm

a

+= =120 mm

5) Calculation of Gear geometries - 2 marks

Pinion Diameter, d = 80 mm

Wheel diameter, D = 160 mm

Number of teeth in pinion, Z1 = 20

Number of teeth in Wheel, Z2 = 40

6) Revise Design Torque [ ]tM - 2marks

[ ] NmMkkkM tdot 130... ==6) Check for induced stresses - 3 marks

( )[ ][ ]b

t

bamby

Mi