chain drive calculations

8/12/2019 Chain Drive Calculations

1/42

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 (Z1):

(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 (Z1) =

Therefore, No. of Teeth on Driven Sprocket (Z2) =

Maximum Spped of Rotation for selected Teeth in RPM =


Center Distance (a) :

We know that, Center Distance (a) = (30 to 50) x Pitch of Sprocket

We Select Duplex Chainhaving following specifications,

Chain with 1 " Pitch (Class 16A-2)

Here, Selected Pitch of Sprocket (p) =

Therefore, Center Distance (a) =


2/42


3/42

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) :

a = [(e+(e2 8m)

1/2) / 4] x p

Where,

e = Lp (Z2+ Z1) / 2 =

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

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)


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 D1X N )/ 60


4/42

Where,

D1- Pitch Circle Diameter of Driver Sprocket in mm

D1= p / sin(180/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)

K4 = Factor for Position of Sprocket (Inclination of the line joining centers of sprocket to horizontal > 600)

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 Safefor Carrying the Load without Breaking.

POWER TRANSMITTED ON THE BASIS OF BEARING STRESS:


5/42

Power transmitted on the basis of allowable bearing stress is given by,

Pb = (b x A x v) / (75 x ks)

Where,

b = Allowable bearing stress in kgf/cm

2

(Ref. Design Data Book) =

A = Projected bearing area in cm2

(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 Safefor 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) =

Pc = Centrifugal Tension in kgf = (w x v2)/g

Where,

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


6/42

v = Chain Velocity (m/s) =

g = Acceleration due to gravity(m/s2) =

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 Drive in kgf is given by,

Qo = k1x 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) =


7/42

3 H.P.

2.238 kW

1

30-27

7

13

13

1300

25.4 mm

762 to 1270 mm


8/42

118.29501 mm

118.29501 mm

118.29501 mm

148.29501 mm

2032 mm


9/42

25.4 mm

0

25.4 mm

mm

11400 kgf


10/42

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.


11/42

315kgf/cm

2

4.2 cm2

0.5112682 m/s

2.9320313

3.0759465 H.P.

2.2946561 kW

11400 kgf

3 H.P.

0.5112682 m/s

440.1 kgf

5.4026504 kgf


12/42

0.5112682 m/s

9.81 m/s2

0.1439579 kgf

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


13/42


(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 for Required Transmission Ratio = 30-27


Minimum No. of Teeth on Driver Sprocket : 7

We Consider, No. of Teeth on Driver Sprocket (Z1) = 17

Therefore, No. of Teeth on Driven Sprocket (Z2) = 17

Maximum Spped of Rotation for selected Teeth in RPM = 1800


Center Distance (a) :

We know that, Center Distance (a) = (30 to 50) x Pitch of Sprocket

We Select Duplex Chainhaving following specifications,

Chain with 5/8 " Pitch (Class 12A-2)

Here, Selected Pitch of Sprocket (p) = 15.875

Therefore, Center Distance (a) = 476.25

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,


14/42

D1=Tip Diameter of Driver Sprocket = p/ (tan(1800/Z1)) + 0.6 p

Therefore , Tip Diamter of Driver Sprocket (D1) = 94.45131

D2=Tip Diameter of Driven Sprocket = p/ (tan(1800/Z2)) + 0.6 p

Therefore , Tip Diamter of Driven Sprocket (D2) = 94.45131

Therefore, a' = (D1+ D2)/ 2 = 94.45131

Therefore, Minimum Center Distance (amin) = a' + (30 to 50 ) mm 124.4513

Maximum Center Distance (amax):

Amax= (80xp) in mm 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,

Lp = (2 x ap) + (Z1+Z2)/2 + [{(Z2- Z1) / 2 X 3.1415}2]/ 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) :

a = [(e+(e2 8m)

1/2) / 4] x p

Where,

e = Lp (Z2+ Z1) / 2 =

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

2

0

We know that,





15/42




P = (Q X V) /(75 X n X Ks)


Where,



For Chain with 5/8" Pitch (Class 12A-2) Q = 6360


Where,


D1 = p / sin(180/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) = 47

Linear Velocity of the Driver Sprocket in m/s (V) = 0.21


For Pitch 15.875 mm & 47 RPM, n = 7

Ks - Service Factor


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, Ks = 1.50

Therefore,


16/42


Power Transmitted by the Chain on the basis of Breaking Load (P) = 1.7

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



Pb = (b x A x v) / (75 x ks)

Where,

b = Allowable bearing stress in kgf/cm2

(Ref. Design Data Book) = 315

A = Projected bearing area in cm2

(Ref. Design Data Book) = 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


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



n = Q / F

Where,

Q = Breaking Load of chain in kgf (Ref. Design Data Book) = 6360


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

Therefore, Tangential force due to power transmission (Pt) = 705.5


17/42


Where,

w = Weight per metre of chain in kgf (Ref. Design Data Book) = 2.96


g = Acceleration due to gravity(m/s2) = 9.81

Therefore, Centrifugal Tension (Pc) = 0.013639


Where,

k = Coefficent of Sag for Vertical position chain drive (Ref. Design Data Book) = 1


a = Centre Distance in metre = 0.60

Therefore, Tension due to sagging of chain in kgf (Ps) = 1.776

Therefore,Resultant Load in kgf (F) = 707.3065

Therefore, Actual factor of Safety (n) = 8.991859

LOAD ON SHAFT:


Qo = k1x 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


18/42

H.P.

kW

mm

to 793.75


19/42

mm

mm

mm

mm

mm

mm

mm


20/42

mm

kgf

mm

mm

m

m/s


21/42

H.P.

kgf/cm2

cm2

m/s

H.P.

kW

kgf

H.P.

m/s

kgf


22/42

kgf

m/s

m/s2

kgf

kgf

m

kgf

kgf

m/s

H.P.kgf

kgf


23/42

Sheet3


(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.












We Select Simplex Chainhaving following specifications,

Chain with 1.25 " Pitch (Class 20A-1)




We know that,




Page 23


24/42

Sheet3




P = (Q X V) /(75 X n X Ks)


Where,



For Chain with 1.25 " Pitch (Class 20A-1) Q = 8850


Where,


D1= p / sin(180/Z1)







For Pitch 31.75 mm & 92 RPM, n = 7.8

Ks - Service Factor


K1 = Load Factor (For Variable Load with heavy shocks) : 1.5


K3 = Factor For Centre Distance of Sprocket ( For A < 25p) 1.25

K4 = Factor for Position of Sprocket (Inclination 0 to 60 Degree) 1




Page 24


25/42

Sheet3

Therefore,



Therefore, Selected Chain is Safefor Carrying the Load without breaking.



Pb = (b x A x v) / (75 x ks)

Where,



A = Projected bearing area in cm

2

(Ref. Design Data Book) =2.62



Therefore,



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



n = Q / F

Where,



Where,


Where,



Page 25


26/42

Sheet3



Where,






Where,







LOAD ON SHAFT:


Qo = k1x Pt

Where,




Where,


N = Actual Power to be transmitted in H. P. = 3.0

Therefore, Tangential force due to power transmission in kgf (Pt) = 230.1

Page 26


27/42

Sheet3

Therefore,


Page 27


28/42

Sheet3

H.P.

kW

mm

mm

mm

Page 28


29/42

Sheet3

mm

kgf

mm

mm

m

m/s

Page 29


30/42

Sheet3

H.P.

kgf/cm2

cm

2

m/s

H.P.

kW

kgf

H.P.

m/s

Page 30


31/42

Sheet3

kgf

kgf

m/s

m/s2

kgf

kgf

m

kgf

kgf

m/s

H.P.

kgf

Page 31


32/42

Sheet3

kgf

Page 32


33/42

Sheet4


(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












We Select Simplex Chainhaving following specifications,

Chain with 1" Pitch (Class 20A-1)




We know that,




Page 33


34/42

Sheet4




P = (Q X V) /(75 X n X Ks)


Where,



For Chain with 1" Pitch (Class 16A-1) Q = 5700


Where,


D1= p / sin(180/Z1)







For Pitch 25.4 mm & 92 RPM, n = 7.8

Ks - Service Factor


K1 = Load Factor (For Variable Load with heavy shocks) : 1.5


K3 = Factor For Centre Distance of Sprocket ( For A < 25p) 1.25

K4 = Factor for Position of Sprocket (Inclination 0 to 60 Degree) 1




Page 34


35/42

Sheet4

Therefore,



Therefore, Selected Chain is Safefor Carrying the Load without breaking.



Pb = (b x A x v) / (75 x ks)

Where,



A = Projected bearing area in cm

2

(Ref. Design Data Book) =1.79



Therefore,



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



n = Q / F

Where,



Where,


Where,



Page 35


36/42

Sheet4



Where,






Where,







LOAD ON SHAFT:


Qo = k1x Pt

Where,




Where,


N = Actual Power to be transmitted in H. P. = 3.0

Therefore, Tangential force due to power transmission in kgf (Pt) = 230.5

Page 36


37/42

Sheet4

Therefore,


Page 37


38/42

Sheet4

.

H.P.

kW

mm

mm

mm

Page 38


39/42

Sheet4

mm

kgf

mm

mm

m

m/s

Page 39


40/42

Sheet4

H.P.

kgf/cm2

cm

2

m/s

H.P.

kW

kgf

H.P.

m/s

Page 40


41/42

Sheet4

kgf

kgf

m/s

m/s2

kgf

kgf

m

kgf

kgf

m/s

H.P.

kgf

Page 41


42/42

Sheet4

kgf

chain drive calculations

Documents