Chain Drive Calculations

CHAIN DRIVE SELECTION: (For Cart Travelling with Single Motor)

INPUT DATA :

Power to be Transmitted in H.P.(N) = Power to be Transmitted in kW. = Required Transmission Ratio (Z2/Z1) = Recommended No. of Teeth on Driver Sprocket (Z 1): (Ref. Design Data Book Page No. 7.74) Recommended No. of Teeth for Required Transmission Ratio = But, Where Space is a Problem We can Select, Minimum No. of Teeth on Driver Sprocket : We Consider, No. of Teeth on Driver Sprocket (Z 1) = Therefore, No. of Teeth on Driven Sprocket (Z 2) =

Maximum Spped of Rotation for selected Teeth in RPM = (Ref. Design Data Book Page No. 7.74)

Center Distance (a) : We know that, Center Distance (a) = (30 to 50) x Pitch of Sprocket We Select Duplex Chain having Chain having following specifications, Chain with 1 " Pitch (Class 16A-2) Here, Selected Pitch of Sprocket (p) = Therefore, Center Distance (a) =

Minimum Center Distance (amin): (Ref. Design Data Book Page No. 7.74) For required transmission ratio, Minimum centre distance is given as, amin = a' + (30 to 50 ) mm Where, a' = (D1 + D2)/ 2

Where, 0

D1 =Tip Diameter of Driver Sprocket = p/ (tan(180 /Z1)) + 0.6 p Therefore , Tip Diamter of Driver Sprocket (D 1) =

0

D2 =Tip Diameter of Driven Sprocket = p/ (tan(180 /Z2)) + 0.6 p Therefore , Tip Diamter of Driven Sprocket (D2) =

Therefore, a' = (D1 + D2)/ 2 = Therefore, Minimum Center Distance (amin) = a' + (30 to 50 ) mm

Maximum Center Distance (amax): Amax = (80xp) in mm

Relation Between Center Distance & Length of Chain:

Length of continuous chain in multiples of pitches ( i.e. approximate number of links) is given by, 2

Lp = (2 x ap ) + (Z1+Z2)/2 + [{(Z2- Z1) / 2 X 3.1415} ]/ ap Where, ap = Approximate Approximate Centre in multiples of pitches = ao / p Where, ao = Initially assumed centre distance in mm =


Where, 0


0


Therefore, a' = (D1 + D2)/ 2 = Therefore, Minimum Center Distance (amin) = a' + (30 to 50 ) mm


Relation Between Center Distance & Length of Chain:

Length of continuous chain in multiples of pitches ( i.e. approximate number of links) is given by, 2

Lp = (2 x ap ) + (Z1+Z2)/2 + [{(Z2- Z1) / 2 X 3.1415} ]/ ap Where, ap = Approximate Approximate Centre in multiples of pitches = ao / p Where, ao = Initially assumed centre distance in mm =

P = Pitch of Chain in mm = Therefore, Length of continuous chain in multiples of pitches (Lp) =

Final Centre Distance corrected to even number of pitches (a) : 2

1/2

a = [(e+(e – 8m) ) / 4] x p Where, e = Lp – (Z2+ Z1) / 2 = 2

M = {(Z2- Z1) / 2 X 3.1415}

We know that, Length of Chain (L) = Lp x p Where, Lp = Length of continuous chain in multiples of pitches ( i.e. approximate number of links) P = Pitch of Chain in mm = Therefore , Length of Chain (L) =

POWER TRANSMITTED BY CHAIN WITH BREAKING LOAD:

Power Transmitted by the Chain on the basis of Breaking Load (P) in H.P. Is given by, P = (Q X V) /(75 X n X Ks) (Ref. Design Data Book Page No. 7.77) Where, Q = Breaking Load in kgf (Ref. Design Data Book Page No. 7.72 ) For Chain with 1 " Pitch (Class 16A-2)

Q=

V = Linear Velocity of the Driver Sprocket in m/s = (3.142 X D1 X N )/ 60

Where, D1 - Pitch Circle Diameter of Driver Sprocket in mm D1= p / sin(180º/Z1) p = Pitch of Chain in mm Therefore , Pitch Circle Diamter of Driver Sprocket (D1) = Pitch Circle Diamter of Driver Sprocket (D1) = RPM of the Driver Sprocket i.e. RPM of Gearbox Output Shaft (N) = Linear Velocity of the Driver Sprocket in m/s (V) =

n - Factor of Safety (Ref. Design Data Book Page No. 7.76 ) For Pitch 25.4 mm & 92 RPM,

n=

Ks - Service Factor Ks = K1 X K2 X K3 X K4 X K5 X K6 K1 = Load Factor (For Variable Load with mild shocks) : K2 = Distance Regulation Factor (For Adjustable Supports) K3 = Factor For Centre Distance of Sprocket ( For A < 25p) 0

K4 = Factor for Position of Sprocket (Inclination of the line joining centers of sprocket to horizontal > 60 ) K5 = Factor for Lubrication (For Periodic) K6 = Rating Factor (Single Shift of 8 hours a day) Therefore,

Ks =

Therefore, Power Transmitted by the Chain on the basis of Breaking Load (P) is given as, Power Transmitted by the Chain on the basis of Breaking Load (P) =

Therefore, Selected Chain is Safe Safe for for Carrying the Load without Breaking.

POWER TRANSMITTED ON THE BASIS OF BEARING STRESS:

Power transmitted on the basis of allowable bearing stress is given by, Pb = (b x A x v) / (75 x ks) Where, 2

b = Allowable bearing stress in kgf/cm (Ref. Design Data Book) = 2

A = Projected bearing area in cm (Ref. Design Data Book) = v = Chain Velocity in m/s = ks = Service Factor (Ref. Design Data Book, From Calculations) = Therefore, Power transmitted on the basis of allowable bearing stress (Pb) = Power transmitted on the basis of allowable bearing stress (Pb) =

Therefore, Selected Chain is Safe for Carrying the Load without Bearing failure.

CHECK FOR ACTUAL FACTOR OF SAFETY: Actual factor of Safety (n) is given as, n=Q/F Where, Q = Breaking Load of chain in kgf (Ref. Design Data Book) = F = Resultant Load in kgf = Pt + Pc + Ps Where,

Pt = Tangential force due to power transmission in kgf = (75 x N)/ v = Where, N = Actual Power to be transmitted in H.P. = v = Chain Velocity (m/s) =

Therefore, Tangential force due to power transmission (Pt) =

2

Pc = Centrifugal Tension in kgf = (w x v )/g

Where, w = Weight per metre of chain in kgf (Ref. Design Data Book) =

v = Chain Velocity (m/s) = 2

g = Acceleration due to gravity(m/s ) =

Therefore, Centrifugal Tension (Pc) =

Ps = Tension due to sagging of chain in kgf = k x w x a

Where, k = Coefficent of Sag for Vertical position chain drive (Ref. Design Data Book) = w = Weight per metre of chain in kgf (Ref. Design Data Book) = a = Centre Distance in metre =

Therefore, Tension due to sagging of chain in kgf (Ps) =

Therefore, Resultant Load in kgf (F) =

Therefore, Actual factor of Safety (n) =

LOAD ON SHAFT: Load on Shaft due to Chain Dr ive in kgf is given by, Qo = k1 x Pt Where, k1 = Load Factor ( Position of Drive is Vertical & Shock Load) Ref. to Design Data Book,

k1 =

Pt = Tangential force due to power transmission in kgf = (75 x N) / v

Where, v = Linear Velocity of the Driver Sprocket in m/s = N = Actual Power to be transmitted in H. P. = Therefore, Tangential force due to power transmission in kgf (Pt) =

Therefore, Load on Shaft due to Chain Drive (Qo) =

3

H.P.

2.238

kW

1

30-27

7 13 13

1300

25.4

mm

762

to

1270

mm

118.29501

mm

118.29501

mm

118.29501

mm

148.29501

mm

2032

mm

25.4

mm

0

25.4

mm mm

11400

kgf

25.4

mm

106.1

mm

0.1061

m

92 0.51

m/s

7.8

1.25 1 1.25 1.251 1.5 1 2.93

3.4

H.P.

315

kgf/cm

4.2

cm

0.5112682

m /s

2

2.9320313

3.0759465

H.P.

2.2946561

kW

11400

kgf

3

H.P.

0.5112682

m /s

440.1

kgf

5.4026504

kgf

2

0.5112682

m /s

9.81

m/s

0.1439579

kgf

2

1 5.4026504

kgf

0.60

m

3.2415902

kgf

443.46767

kgf

25.706496

1.15

0.5112682

m /s

3.0

H.P.

440.1

kgf

506.1

kgf

CHAIN DRIVE SELECTION: (For Pallet Picking Drive Assembly) INPUT DATA : Power to be Transmitted in H.P.(N) =

2

Power to be Transmitted in kW. =

1.492

Required Transmission Ratio (Z2/Z1) =

1

Recommended No. of Teeth on Driver Sprocket (Z 1): (Ref. Design Data Book Page No. 7.74) Recommended No. of Teeth for Required Transmission Ratio =

30-27

But, Where Space is a Problem We can Select, Minimum No. of Teeth on Driver Sprocket :

7

We Consider, No. of Teeth on Driver Sprocket (Z 1) =

17

Therefore, No. of Teeth on Driven Sprocket (Z 2) =

17

Maximum Spped of Rotation for selected Teeth in RPM =

1800

(Ref. Design Data Book Page No. 7.74)

Center Distance (a) : We know that, Center Distance (a) = (30 to 50) x Pitch of Sprocket We Select Duplex Chain having following specifications, Chain with 5/8 " Pitch (Class 12A-2) Here, Selected Pitch of Sprocket (p) =

15.875

Therefore, Center Distance (a) =

476.25


Where,

0


94.45131

0


Therefore, a' = (D1 + D2)/ 2 =

94.45131

94.45131

Therefore, Minimum Center Distance (amin) = a' + (30 to 50 ) mm

124.4513


1270

Relation Between Center Distance & Length of Chain: Length of continuous chain in multiples of pitches ( i.e. approximate number of links) is given by, 2

Lp = (2 x ap ) + (Z1+Z2)/2 + [{(Z2- Z1) / 2 X 3.1415} ]/ ap Where, ap = Approximate Centre in multiples of pitches = ao / p Where, ao = Initially assumed centre distance in mm = P = Pitch of Chain in mm =

15.875

Therefore, Length of continuous chain in multiples of pitches (Lp) =

Final Centre Distance corrected to even number of pitches (a) : 2

1/2

a = [(e+(e – 8m) ) / 4] x p Where, e = Lp – (Z2+ Z1) / 2 = 2

M = {(Z2- Z1) / 2 X 3.1415}

0

We know that, Length of Chain (L) = Lp x p Where, Lp = Length of continuous chain in multiples of pitches ( i.e. approximate number of links) P = Pitch of Chain in mm =

15.875

Therefore , Length of Chain (L) =

POWER TRANSMITTED BY CHAIN WITH BREAKING LOAD: Power Transmitted by the Chain on the basis of Breaking Load (P) in H.P. Is given by, P = (Q X V) /(75 X n X Ks) (Ref. Design Data Book Page No. 7.77) Where, Q = Breaking Load in kgf (Ref. Design Data Book Page No. 7.72 ) For Chain with 5/8" Pitch (Class 12A-2)

Q=

6360

V = Linear Velocity of the Driver Sprocket in m/s = (3.142 X D1 X N )/ 60 Where, D1 - Pitch Circle Diameter of Driver Sprocket in mm D1 = p / sin(180º/Z1) p = Pitch of Chain in mm

15.875

Therefore , Pitch Circle Diamter of Driver Sprocket (D1) =

86.4

Pitch Circle Diamter of Driver Sprocket (D1) =

0.0864

RPM of the Driver Sprocket i.e. RPM of Gearbox Output Shaft (N) = Linear Velocity of the Driver Sprocket in m/s (V) =

47 0.21


n=

7

Ks - Service Factor Ks = K1 X K2 X K3 X K4 X K5 X K6 K1 = Load Factor (For Constant Load) :

1

K2 = Distance Regulation Factor (For Adjustable Supports)

1

K3

= Factor For Centre Distance of Sprocket {For lp / (Z1+Z2 ) = 1.5 or ap=30 to 50 p} 1

K4 = Factor for Position of Sprocket (Inclination 0 to 60 Degree)

1

K5 = Factor for Lubrication (For Periodic)

1.5

K6 = Rating Factor (Single Shift of 8 hours a day)

1

Therefore, Therefore,

Ks =

1.50

Power Transmitted by the Chain on the basis of Breaking Load (P) is given as, Power Transmitted by the Chain on the basis of Breaking Load (P) =

1.7

Therefore, Selected Chain is Not Safe for Carrying the Load without Breaking.

POWER TRANSMITTED ON THE BASIS OF BEARING STRESS: Power transmitted on the basis of allowable bearing stress is given by, Pb = (b x A x v) / (75 x ks) Where, 2

b = Allowable bearing stress in kgf/cm (Ref. Design Data Book) =

315

A = Projected bearing area in cm (Ref. Design Data Book) =

2

2.1

v = Chain Velocity in m/s =

0.21261

ks = Service Factor (Ref. Design Data Book, From Calculations) =

1.5

Therefore, Power transmitted on the basis of allowable bearing stress (Pb) =

1.250147

Power transmitted on the basis of allowable bearing stress (Pb) =

0.93261

Therefore, Selected Chain is Not Safe for Carrying the Load without Bearing failure.

CHECK FOR ACTUAL FACTOR OF SAFETY: Actual factor of Safety (n) is given as, n=Q/F Where, Q = Breaking Load of chain in kgf (Ref. Design Data Book) =

6360

F = Resultant Load in kgf = Pt + Pc + Ps Where,

Pt = Tangential force due to power transmission in kgf = (75 x N)/ v = Where,

N = Actual Power to be transmitted in H.P. =

2

v = Chain Velocity (m/s) =

0.21261


705.5

2



2.96


0.21261 2



9.81 0.013639


Where, k = Coefficent of Sag for Vertical position chain drive (Ref. Design Data Book) =

1

w = Weight per metre of chain in kgf (Ref. Design Data Book) =

2.96

a = Centre Distance in metre =

0.60


1.776


707.3065


8.991859

LOAD ON SHAFT: Load on Shaft due to Chain Dr ive in kgf is given by, Qo = k1 x Pt Where, k1 = Load Factor ( Position of Drive is Vertical & Shock Load) Ref. to Design Data Book,

k1 =

1.15


Where, v = Linear Velocity of the Driver Sprocket in m/s =

0.21261

N = Actual Power to be transmitted in H. P. =

2.0

Therefore, Tangential force due to power transmission in kgf (Pt) =

705.5

Therefore, Load on Shaft due to Chain Drive (Qo) =

811.3

H.P. kW

mm to

793.75

mm

mm

mm mm

mm

mm

mm

mm

kgf

mm mm m

m/s

H.P.

kgf/cm cm

2

m/s

H.P. kW

kgf

H.P. m/s kgf

2

kgf m/s m/s

2

kgf

kgf m kgf

kgf

m/s H.P. kgf

kgf

Sheet3

CHAIN DRIVE SELECTION: (When Servo Motor is used is used for Cart Travelling for Accuracy in Stopping) SPROCKET 1.25” PITCH

INPUT DATA : Rotary motion of Gearbox output shaft is converted into linear motion of cart by using sprocket – chain drive. Power to be Transmitted in H.P.(N) =

3


2.238

Required Transmission Ratio (Z 2/Z1) =

1


30-27


7


20


1300


We Select Simplex Chain having following specifications, Chain with 1.25 " Pitch (Class 20A-1) Here, Selected Pitch of Sprocket (p) =

31.75

0


219.5176213


31.750

Page 23

Sheet3 Therefore , Length of Chain (L) =

POWER TRANSMITTED BY CHAIN WITH BREAKING LOAD: Power Transmitted by the Chain on the basis of Breaking Load (P) in H.P. Is given by, P = (Q X V) /(75 X n X Ks) (Ref. Design Data Book Page No. 7.77) Where, Q = Breaking Load in kgf (Ref. Design Data Book Page No. 7.72 ) For Chain with 1.25 " Pitch (Class 20A-1)

Q=

8850

V = Linear Velocity of the Driver Sprocket in m/s = (3.142 X D1 X N )/ 60 Where, D1 - Pitch Circle Diameter of Driver Sprocket in mm º

D1= p / sin(180 /Z1) p = Pitch of Chain in mm

31.75


203.0


0.2030

RPM of the Driver Sprocket i.e. RPM of Gearbox Output Shaft (N) =

92

Linear Velocity of the Driver Sprocket in m/s (V) =

0.98


n=

7.8

Ks - Service Factor Ks = K1 X K2 X K3 X K4 X K5 X K6 K1 = Load Factor (For Variable Load with heavy shocks) :

1.5


1

K3 = Factor For Centre Distance of Sprocket ( For A < 25p)

1.25


1


1.5


1

Therefore,

Ks =

Page 24

2.81

Sheet3 Therefore, Power Transmitted by the Chain on the basis of Breaking Load (P) is given as, Power Transmitted by the Chain on the basis of Breaking Load (P) =

5.3

Therefore, Selected Chain is Safe for Carrying the Load without breaking.



315


2

2.62


0.977682029


2.8125


3.825213528


2.853609292

Therefore, Selected Chain is Safe for Carrying the Load without Bearing failure.


8850


Pt = Tangential force due to power transmission in kgf = (75 x N)/ v = Where, N = Actual Power to be transmitted in H.P. =

3


0.977682029

Page 25

Sheet3


230.1

2



3.87


0.977682029 2

9.81



0.377083233



1


3.87


0.60


2.322


232.8352556


38.00970766

LOAD ON SHAFT: Load on Shaft due to Chain Drive in kgf is given by, Qo = k1 x Pt Where, k1 = Load Factor ( Position of Drive is Vertical & Shock Load) Ref. to Design Data Book,

k1 =

1.15



0.977682029


3.0


Page 26

230.1

Sheet3 Therefore, Load on Shaft due to Chain Drive (Qo) =

264.7

Page 27

Sheet3

H.P. kW

mm

mm

mm

Page 28

Sheet3 mm

kgf

mm mm m

m/s

Page 29

Sheet3

H.P.

kgf/cm cm

2

2

m/s

H.P. kW

kgf

H.P. m/s

Page 30

Sheet3 kgf

kgf m/s m/s

2

kgf

kgf m kgf

kgf

m/s H.P. kgf

Page 31

Sheet3

kgf

Page 32

Sheet4

CHAIN DRIVE SELECTION: (When Servo Motor is used is used for Cart Travelling for Accuracy in Stopping) SPROCKET 1” PITCH INPUT DATA : Rotary motion of Gearbox output shaft is converted into linear motion of cart by using sprocket – chain driv Power to be Transmitted in H.P.(N) =

3


2.238

Required Transmission Ratio (Z 2/Z1) =

1


30-27


7


25


1300


We Select Simplex Chain having following specifications, Chain with 1" Pitch (Class 20A-1) Here, Selected Pitch of Sprocket (p) =

25.4

0


216.307696


25.400

Page 33

Sheet4 Therefore , Length of Chain (L) =

POWER TRANSMITTED BY CHAIN WITH BREAKING LOAD: Power Transmitted by the Chain on the basis of Breaking Load (P) in H.P. Is given by, P = (Q X V) /(75 X n X Ks) (Ref. Design Data Book Page No. 7.77) Where, Q = Breaking Load in kgf (Ref. Design Data Book Page No. 7.72 ) For Chain with 1" Pitch (Class 16A-1)

Q=

5700

V = Linear Velocity of the Driver Sprocket in m/s = (3.142 X D1 X N )/ 60 Where, D1 - Pitch Circle Diameter of Driver Sprocket in mm D1= p / sin(180º/Z1) p = Pitch of Chain in mm

25.4


202.666


0.2027

RPM of the Driver Sprocket i.e. RPM of Gearbox Output Shaft (N) =

92

Linear Velocity of the Driver Sprocket in m/s (V) =

0.98


n=

7.8

Ks - Service Factor Ks = K1 X K2 X K3 X K4 X K5 X K6 K1 = Load Factor (For Variable Load with heavy shocks) :

1.5


1

K3 = Factor For Centre Distance of Sprocket ( For A < 25p)

1.25


1


1.5


1

Therefore,

Ks =

Page 34

2.81

Sheet4 Therefore, Power Transmitted by the Chain on the basis of Breaking Load (P) is given as, Power Transmitted by the Chain on the basis of Breaking Load (P) =

3.4

Therefore, Selected Chain is Safe for Carrying the Load without breaking.



315


2

1.79


0.97623383


2.8125


2.60953811


1.94671543

Therefore, Selected Chain is Not Safe for Carrying the Load without Bearing failure.


5700


Pt = Tangential force due to power transmission in kgf = (75 x N)/ v = Where, N = Actual Power to be transmitted in H.P. =

3


0.97623383

Page 35

Sheet4


230.5

2



2.62


0.97623383 2

9.81



0.25453059



1


2.62


0.60


1.572


232.304099


24.5368033

LOAD ON SHAFT: Load on Shaft due to Chain Drive in kgf is given by, Qo = k1 x Pt Where, k1 = Load Factor ( Position of Drive is Vertical & Shock Load) Ref. to Design Data Book,

k1 =

1.15



0.97623383


3.0


Page 36

230.5

Sheet4 Therefore, Load on Shaft due to Chain Drive (Qo) =

265.0

Page 37

Sheet4

. H.P. kW

mm

mm

mm

Page 38

Sheet4 mm

kgf

mm mm m

m/s

Page 39

Sheet4

H.P.

kgf/cm cm

2

2

m/s

H.P. kW

kgf

H.P. m/s

Page 40

Sheet4 kgf

kgf m/s m/s

2

kgf

kgf m kgf

kgf

m/s H.P. kgf

Page 41

Chain Drive Calculations

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