chain drive calculations
TRANSCRIPT
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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 =
(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 Chainhaving following specifications,
Chain with 1 " Pitch (Class 16A-2)
Here, Selected Pitch of Sprocket (p) =
Therefore, Center Distance (a) =
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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)
(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 D1X N )/ 60
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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:
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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) =
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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) =
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3 H.P.
2.238 kW
1
30-27
7
13
13
1300
25.4 mm
762 to 1270 mm
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118.29501 mm
118.29501 mm
118.29501 mm
148.29501 mm
2032 mm
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25.4 mm
0
25.4 mm
mm
11400 kgf
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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.
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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
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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
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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 (Z1):
(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 (Z1) = 17
Therefore, No. of Teeth on Driven Sprocket (Z2) = 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 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,
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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,
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
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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 D1X N )/ 60
Where,
D1- Pitch Circle Diameter of Driver Sprocket in mm
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
n - Factor of Safety (Ref. Design Data Book Page No. 7.76 )
For Pitch 15.875 mm & 47 RPM, 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, Ks = 1.50
Therefore,
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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 Safefor 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,
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
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
Therefore, Tangential force due to power transmission (Pt) = 705.5
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Pc = Centrifugal Tension in kgf = (w x v2)/g
Where,
w = Weight per metre of chain in kgf (Ref. Design Data Book) = 2.96
v = Chain Velocity (m/s) = 0.21261
g = Acceleration due to gravity(m/s2) = 9.81
Therefore, Centrifugal Tension (Pc) = 0.013639
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) = 1
w = Weight per metre of chain in kgf (Ref. Design Data Book) = 2.96
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:
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 = 1.15
Pt = Tangential force due to power transmission in kgf =(75 x N) / v
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
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H.P.
kW
mm
to 793.75
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mm
mm
mm
mm
mm
mm
mm
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mm
kgf
mm
mm
m
m/s
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H.P.
kgf/cm2
cm2
m/s
H.P.
kW
kgf
H.P.
m/s
kgf
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kgf
m/s
m/s2
kgf
kgf
m
kgf
kgf
m/s
H.P.kgf
kgf
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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
Power to be Transmitted in kW. = 2.238
Required Transmission Ratio (Z2/Z1) = 1
Recommended No. of Teeth on Driver Sprocket (Z1):
(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 (Z1) = 20
Maximum Spped of Rotation for selected Teeth in RPM = 1300
(Ref. Design Data Book Page No. 7.74)
We Select Simplex Chainhaving following specifications,
Chain with 1.25 " Pitch (Class 20A-1)
Here, Selected Pitch of Sprocket (p) = 31.75
D1=Tip Diameter of Driver Sprocket = p/ (tan(1800/Z1)) + 0.6 p
Therefore , Tip Diamter of Driver Sprocket (D1) = 219.5176213
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 = 31.750
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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 D1X N )/ 60
Where,
D1- Pitch Circle Diameter of Driver Sprocket in mm
D1= p / sin(180/Z1)
p = Pitch of Chain in mm 31.75
Therefore , Pitch Circle Diamter of Driver Sprocket (D1) = 203.0
Pitch Circle Diamter of Driver Sprocket (D1) = 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 - Factor of Safety (Ref. Design Data Book Page No. 7.76 )
For Pitch 31.75 mm & 92 RPM, 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
K2 = Distance Regulation Factor (For Adjustable Supports) 1
K3 = Factor For Centre Distance of Sprocket ( For A < 25p) 1.25
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 = 2.81
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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 Safefor 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,
b = Allowable bearing stress in kgf/cm2
(Ref. Design Data Book) = 315
A = Projected bearing area in cm
2
(Ref. Design Data Book) =2.62
v = Chain Velocity in m/s = 0.977682029
ks = Service Factor (Ref. Design Data Book, From Calculations) = 2.8125
Therefore,
Power transmitted on the basis of allowable bearing stress (Pb) = 3.825213528
Power transmitted on the basis of allowable bearing stress (Pb) = 2.853609292
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) = 8850
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. = 3
v = Chain Velocity (m/s) = 0.977682029
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Sheet3
Therefore, Tangential force due to power transmission (Pt) = 230.1
Pc = Centrifugal Tension in kgf = (w x v2)/g
Where,
w = Weight per metre of chain in kgf (Ref. Design Data Book) = 3.87
v = Chain Velocity (m/s) = 0.977682029
g = Acceleration due to gravity(m/s2) = 9.81
Therefore, Centrifugal Tension (Pc) = 0.377083233
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) = 1
w = Weight per metre of chain in kgf (Ref. Design Data Book) = 3.87
a = Centre Distance in metre = 0.60
Therefore, Tension due to sagging of chain in kgf (Ps) = 2.322
Therefore,Resultant Load in kgf (F) = 232.8352556
Therefore, Actual factor of Safety (n) = 38.00970766
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 = 1.15
Pt = Tangential force due to power transmission in kgf =(75 x N) / v
Where,
v = Linear Velocity of the Driver Sprocket in m/s = 0.977682029
N = Actual Power to be transmitted in H. P. = 3.0
Therefore, Tangential force due to power transmission in kgf (Pt) = 230.1
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Sheet3
Therefore,
Load on Shaft due to Chain Drive (Qo) = 264.7
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Sheet3
H.P.
kW
mm
mm
mm
Page 28
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Sheet3
mm
kgf
mm
mm
m
m/s
Page 29
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Sheet3
H.P.
kgf/cm2
cm
2
m/s
H.P.
kW
kgf
H.P.
m/s
Page 30
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Sheet3
kgf
kgf
m/s
m/s2
kgf
kgf
m
kgf
kgf
m/s
H.P.
kgf
Page 31
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Sheet3
kgf
Page 32
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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
Power to be Transmitted in kW. = 2.238
Required Transmission Ratio (Z2/Z1) = 1
Recommended No. of Teeth on Driver Sprocket (Z1):
(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 (Z1) = 25
Maximum Spped of Rotation for selected Teeth in RPM = 1300
(Ref. Design Data Book Page No. 7.74)
We Select Simplex Chainhaving following specifications,
Chain with 1" Pitch (Class 20A-1)
Here, Selected Pitch of Sprocket (p) = 25.4
D1=Tip Diameter of Driver Sprocket = p/ (tan(1800/Z1)) + 0.6 p
Therefore , Tip Diamter of Driver Sprocket (D1) = 216.307696
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 = 25.400
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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 D1X N )/ 60
Where,
D1- Pitch Circle Diameter of Driver Sprocket in mm
D1= p / sin(180/Z1)
p = Pitch of Chain in mm 25.4
Therefore , Pitch Circle Diamter of Driver Sprocket (D1) = 202.666
Pitch Circle Diamter of Driver Sprocket (D1) = 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 - Factor of Safety (Ref. Design Data Book Page No. 7.76 )
For Pitch 25.4 mm & 92 RPM, 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
K2 = Distance Regulation Factor (For Adjustable Supports) 1
K3 = Factor For Centre Distance of Sprocket ( For A < 25p) 1.25
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 = 2.81
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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 Safefor 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,
b = Allowable bearing stress in kgf/cm2
(Ref. Design Data Book) = 315
A = Projected bearing area in cm
2
(Ref. Design Data Book) =1.79
v = Chain Velocity in m/s = 0.97623383
ks = Service Factor (Ref. Design Data Book, From Calculations) = 2.8125
Therefore,
Power transmitted on the basis of allowable bearing stress (Pb) = 2.60953811
Power transmitted on the basis of allowable bearing stress (Pb) = 1.94671543
Therefore, Selected Chain is NotSafefor 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) = 5700
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. = 3
v = Chain Velocity (m/s) = 0.97623383
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Therefore, Tangential force due to power transmission (Pt) = 230.5
Pc = Centrifugal Tension in kgf = (w x v2)/g
Where,
w = Weight per metre of chain in kgf (Ref. Design Data Book) = 2.62
v = Chain Velocity (m/s) = 0.97623383
g = Acceleration due to gravity(m/s2) = 9.81
Therefore, Centrifugal Tension (Pc) = 0.25453059
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) = 1
w = Weight per metre of chain in kgf (Ref. Design Data Book) = 2.62
a = Centre Distance in metre = 0.60
Therefore, Tension due to sagging of chain in kgf (Ps) = 1.572
Therefore,Resultant Load in kgf (F) = 232.304099
Therefore, Actual factor of Safety (n) = 24.5368033
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 = 1.15
Pt = Tangential force due to power transmission in kgf =(75 x N) / v
Where,
v = Linear Velocity of the Driver Sprocket in m/s = 0.97623383
N = Actual Power to be transmitted in H. P. = 3.0
Therefore, Tangential force due to power transmission in kgf (Pt) = 230.5
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Therefore,
Load on Shaft due to Chain Drive (Qo) = 265.0
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.
H.P.
kW
mm
mm
mm
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mm
kgf
mm
mm
m
m/s
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H.P.
kgf/cm2
cm
2
m/s
H.P.
kW
kgf
H.P.
m/s
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kgf
kgf
m/s
m/s2
kgf
kgf
m
kgf
kgf
m/s
H.P.
kgf
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kgf