U2_S5_Selection of chain drives.pptx

U2-S5-Selection of Chain Drives
- Dr. M. Satthiyaraju

Contents
• Chain drives
• Applications
• Design procedure for chain drives
• Selection of chain drives

Chain:
A chain can be deﬁned as a series of links
connected by pin joints.

Chain Drives
 A chain drive is another major type of
ﬂexible drive used to transmit power over
comparatively long Centre distances.
 A chain drive consists of an endless chain
wrapped around two sprockets

Advantages of chain drives compared
with BELT and GEAR drives
 Chain drives can be used for long as well as short Centre distances.
 a number of shafts can be driven in the same or opposite direction by
means of the chain from a single driving sprocket
 Chain drives have small overall dimensions than belt drives, resulting in
compact unit.
 Chain drive is a positive drive.
 The efﬁciency of chain drives is high.
 Chain does not require initial tension
 Easy replace and Not affected by atmospheric conditions

Disadvantages of chain drives compared
with BELT and GEAR drives
 Chain drives operate without full lubricant ﬁ lm between the joints unlike
gears.
 Chain drives are not suitable for non-parallel shafts.
 Chain drive is unsuitable where precise motion is required due to
polygonal effect
 Require housing
 Compared with belt drives, chain drives require precise alignment of
shafts.
 Chain drives generate noise.

Chain Drives
Applications:
 Conveyor systems,
 Automobiles
 Motor cycles
 Bicycles and many other similar applications
When it is suitable?
 Desirable at low to moderate speeds,
 High torque applications and
 Usually at lower speed stage of a power transmission system

CHAIN DRIVES:
Chain drive:
A chain drive consists
1. A driving sprocket
2. A driven sprocket
3. A chain loop

CHAIN DRIVES:
Sprocket:
The sprocket is a toothed wheel with a special proﬁle for the teeth.

Structure of a Roller Chain:
• Single-strand chain drive Double-strand chain drive
1. Inner link plate, 2. Outer link plate, 3. Pin, 4. Bushing, 5. Roller

Materials and Methods
• The inner and outer link plates are made of medium
carbon steels
• These link plates are blanked from cold-rolled sheets
and hardened to 50 HRC.
• The pins, bushes and rollers are made of case
carburising alloy steels and hardened to 50 HRC

Chain Strands:
 Single
 Double
 Triple and
 Quadruple strands

Chain drive working conditions
 Velocity ratios less than 10 : 1
 Chain velocities of up to 25 m/s
 Recommended to transmit power up to 100 kW.

Chains Classification
 With respect to their purpose;
(i) Load lifting chains
(ii) Hauling chains
(iii) Power transmission chains

PITCH
 The pitch (p) of the chain is the linear distance between the axes of
adjacent rollers.
 Roller chains are standardized and manufactured on the basis of the pitch.

Designation
 The roller chains are designated on the basis of ‘pitch’
 Designation: Example : 08B or 16A
First : Number = The pitch of this chain is (08/16) inch or
(08/16) x (25.4) mm, i.e., 12.7 mm.
Second : Letter (A or B)
A: American Standard ANSI series
B: British standard series
08B-2: In this 2 indicates double strand

Kinematic Analysis
Chain drive geometry:

GEOMETRIC RELATIONSHIPS
Here,
D - is the pitch circle diameter of the sprocket
α - is called the pitch angle.
Pitch Circle Diameter (PCD):
The pitch circle diameter of the sprocket is deﬁned as the diameter of an
imaginary circle that passes through the Centres of link pins as the chain is
wrapped on the sprocket.

GEOMETRIC RELATIONSHIPS
Pitch Circle Diameter (D); 𝐷 =
𝑝
𝑆𝑖𝑛 (
α
2
)
𝐷 =
𝑝
𝑆𝑖𝑛 (
180
𝑍
)
Where, Z – No of teeth on a sprocket
D - is the pitch circle diameter of the sprocket
α - is called the pitch angle.

Velocity Analysis of a Chain Drive
Transmission Ratio (i):
Where,
Z1 - No. of teeth on sprocket pinion (Driver)
Z2 - No. of teeth on sprocket wheel (Driven)
N1 - Speed of rotation of pinion in rpm
N2 - Speed of rotation of wheel in rpm

Polygonal Effect
 Sprocket with less number of teeth can affect the smooth running of a
chain drive. This unsmooth running condition is termed as chordal action of
the chain.
 When a chain roller approaches the sprocket and has just seated, it has
radius Rc, also known as chordal radius.
 When the roller passes through half of the pitch angle α/2, the roller has
radius R which is the pitch radius. R > Rc

Polygonal Effect
Maximum linear velocity is,
Minimum linear velocity is,
The variation in velocity is,

Number of teeth:
 As the number of teeth (z) increases to ∞ , cos (180/z) or cos (180/∞),
i.e., cos (0°) will approach unity and (vmax. – vmin.) will become zero.
 For smooth operation at moderate and high speeds, it is considered a good
practice to use a driving sprocket with at least 17 teeth.
 From durability and noise considerations, the minimum number of teeth on
the driving sprocket should be 19 or 21.

Failure criteria in the Chain Drive:
 Wear: The wear of the chain is caused by the articulation of pins in the bushings.
The wear results in elongation of the chain, or in other words, the chain pitch is
increased. This makes the chain ‘ride out’ on the sprocket teeth, resulting in a faulty
engagement
 Fatigue: The chain link is, therefore, subjected to one complete cycle of ﬂuctuating
stresses during every revolution of the sprocket wheel. This results in a fatigue
failure of side link plates.
 Impact: The engagement of rollers with the teeth of the sprocket results in impact.
 Galling: is a stick-slip phenomenon between the pin and the bushing. When the
chain tension is high, welds are formed at the high spots of the contacting area.

SILENT CHAIN or inverted-tooth chain
 It consists of a series of links formed from laminated steel plates.
 Each plate has two teeth with a space between them to accommodate the
mating tooth of the sprocket.
 The sprocket teeth have a trapezoidal proﬁle.

Design procedure for the selection of
chain drives
 Determining the transmission ratio (i)
i =
𝒁𝟐
𝒁𝟏
=
𝒏𝟏
𝒏𝟐
Z1 - No. of teeth on sprocket pinion (Driver)
Z2 - No. of teeth on sprocket wheel (Driven)
n1 - Speed of rotation of pinion in rpm.
n2 - Speed of rotation of wheel in rpm
1- Pinion
2- Wheel

chain drives
 Select the no. of teeth on sprocket pinion (Z1)
PSG Data book P. No: 7.74
 Calculate the No. of teeth on sprocket wheel Z2
Select odd no of teeth for uniform wear

chain drives
 Calculate of pitch p
Refer databook Page No. 7.74
a = Centre distance
= (30-50)p
Where p – pitch value
pmax = ?
pmin = ?
Note: Choose any standard pitch in between pmax and pmin
(databook Page No. 7.71, 7.72 and 7.73)

chain drives
 Select chain No., according to the selected standard
pitch.
(PSG Data book page No. 7.71, 7.72 and 7.73)
Take the following values according to the selected chain
No.
Simplex - R
Duplex - DR
Triplex - TR
(1) Bearing area in cm2 (A)
(2) Weight per meter in kgf (W)
(3) Breaking load in kgf (Q)

chain drives
 Refer PSG databook, Page No. 7.77. Checking of breaking
load Q in kgf
 Power transmitted on the basis of Breaking load
N=
𝑸 𝒗
𝟏𝟎𝟐 𝒏 𝑲𝒔
in kW
Where, N - given power in kW.
Q - induced breaking load (to be calculated)
v - chain velocity in m/sec

chain drives
 Chain velocity
v =
𝒁 𝒑 𝑵
𝟔𝟎 ×𝟏𝟎𝟎𝟎
in m/sec
p - pitch in mm
N1 - Speed of pinion in rpm
Z - no. of teeth on pinion
n - factor of safety allowable
(Refer databook Page No. 7.77)

chain drives
The value of n depends on
1. speed of rotation of pinion
2. pitch in mm ks
3. Service factor
Refer PSG databook Page No. 7.76 and 7.77
Ks = k1 × k2 × k3 × k4 × k5 × k6
Note: If conditions are not given, select Ks = 1

chain drives
 Q < [Q] then design is safe.
 If Q > [Q] then change the chain no. and read the values
of A, W and [Q] and check again Q breaking load.
Check the Actual factor of safety.
Refer PSG data book Page No. 7.78.
[n] = Q/ ΣP
[n] = actual Fos > n (allowable factor of safety)
Q = Breaking load of chain in kgf

chain drives
 Q = Breaking load of chain in kgf
 ΣP = Pt + Pc + Ps
Pt - Tangential force due to power transmission
Pt - Centrifugal tension
Ps - Tension due to sagging of chain

chain drives
 Pt - Tangential force due to power transmission
𝑃𝑡 =
102 𝑁
𝑣
in kgf
 Pc - Centrifugal tension
𝑃𝑐 =
𝑊 𝑣2
𝑔
in kgf
 Ps - Tension due to sagging of chain
Ps = K W a
Where, K - Coefficient of sag (refer PSG Data book page No. 7.78) value
of K depends on position of chain drive
W - Weight/m length of chain kgf
a - Centre distance in meter

chain drives
To Find ‘a’: Centre distance
1. To find length of chain Refer PSG Data book Page No. 7.75
𝑙𝑝 = 2𝑎𝑝 +
𝑍1+ 𝑍2
2
+
𝑍1− 𝑍2
2𝜋
2
𝑎𝑝
Actual length of chain ; l = lp + p
where ap = ao/p
ap - Approximate centre distance
ao - Initially assumed centre distance
p - Pitch

chain drives
To find exact centre distance (a)
The value of ‘m’ can be read directly from the PSG data book. Page
No.7.76
The value of ‘m’ depends on (Z2-Z1) value. Calculate (Z2- Z1) and take the
corresponding value of ‘m’.

chain drives
Checking of bearing stress:
Refer PSG Data book, Page No. 7.77
The allowable bearing pressure [σ] depends on 1. Pitch 2. Chain velocity
Read the value of [σ] in kgf/mm2 and convert in kgf /cm2
[kgf/mm2 x 102 kgf/cm2]
Power transmitted on the basis of allowable bearing stress;

chain drives
where N - given power in kW
[σ] - Induced bearing stress in kgf / cm2
A - bearing Area cm2
v - Chain speed m/sec
Ks - service factor
Calculate induced stress in kgf/cm2
Then the design is satisfactory.

chain drives
 Diameters of pinion and wheel d and D
Refer PSG data book Page No. 7.78
𝑑 =
𝑝
𝑆𝑖𝑛 (
180
𝑍1
)
𝐷 =
𝑝
𝑆𝑖𝑛 (
180
𝑍2
)
d - dia. of smaller sprocket
D - dia. of larger sprocket

DESIGN OF CHAIN DRIVES
Design a chain drive to operate a compressor from a 15
kW electric motor at 900 rpm; The compressor is to run
at a speed of 300 rpm; The minimum Centre distance
should be 550 mm.

 Solution:
 Given Data:
Power of Motor (P) = 15 kW
Speed of motor (N1) = 900 rpm
Speed of the compressor N2 = 300 rpm
Minimum centre distance ao = 550 mm;

 Calculation of transmission ratio
Transmission ratio = Z2/Z1 = N1/N2
i = Transmission ratio = = N1/N2 = 900/300 = 3
From PSG Design data book, refer Page No. 7.74.
For i = 2 to 3 ; Z1 = 25 to 27 i = 3 to 4 ; Z1 = 23 to 25
Take Z1 = 23 to 27 (select any odd no. of teeth)
Select Z1 = 27 teeth (no. of teeth on sprocket pinion)
Z2 = i Z1 = 3 x 27 = 81 = 82 (no. of teeth on sprocket wheel)

 Calculation of pitch
From PSG design data book, P.No.7.74
Optimum centre distance (a) = 30 to 50 p
where a - approximate centre distance
pmax = 550/30 = 18.33 mm
pmin = 550/50 = 11 mm
Select standard pitch from PSG data book,
Take any standard pitch between 11 to 18.33 mm
Select Pitch; p = 15.875 mm

 Selection of Chain No.
Select roller chain from PSG Data book Page No. 7.72
The available chain No. are 10A and 10B
Select 10A-2 Duplex Chain
Pitch ; p = 15.875 mm
Corresponding to chain No. selected, take the values of
A - Bearing area = 1.4 cm2
W - Weight per m length = 1.78 kgf
Q - Breaking load = 4440 kgf

 Calculate Power transmitted based on breaking load:
From data book, Pg. No. 7.77
From the above equation calculate ‘Q’ breaking load by considering,
N = given power
N = 15 kW
=
27 ×15.875 ×900
60 ×1000
= 6.429 m/sec

n - Minimum value of factor of safety
Ks = 1 ; Z1 = 15 to 30 from PSG data book, Pg. No. 7.77,
Select n = 11 for a pitch 15.875 mm and n1 < 1000 rpm
Since the specific conditions are not given in the problem,
assume K1 = K2 = K3 = K4 = K5 = K6 = 1
Ks = 1
15 =
𝑄 ×6.429
102 ×11 ×1
; Q =
15 ×102 ×11
16.429
= 2617.825 kgf

Breaking load Q = 2617.825 kgf which is less than the selected chain
Breaking load (4440 kgf)
The selection of chain no. is satisfactory based on breaking load.
 To find (a) Calculation of Length of chain
(b) Final centre distance.
(a) Length of continuous chain in multiples of pitches:
𝑍1 + 𝑍2
2
+
𝑍1 − 𝑍2
2𝜋
2
𝑎𝑝
where ap = ao /p = 550/15.875
ap = 34.64 mm

ao - initially assumed centre distance in mm = 550 mm
p - pitch = 15.875 mm
No. of teeth on sprocket wheel Z2 = 82;
No. of teeth on sprocket pinion Z1 = 27
Refer PSG data book Page No. 7.76,
Read the value of ‘m’ directly from PSG data book Pg.No. 7.63
(Z2 - Z1) = 82 - 27 = 55;

the value of m = 76.6
𝑙𝑝 = 2 34.64 +
27+82
2
+
76.6
34.64
= 125.99 = 126 mm (approximated to 126)
Length of chain 𝑙 = 𝑙p . p = 126 x 15.875 = 2000.25 mm
Take, 𝑙 = 2000 mm
(b) Final centre distance

e = 126 −
27+82
2
= 71.5 mm

Final Centre distance
𝑎 =
71.5 + 71.52 − 8 × 76.6
4
× 15.875
= 550 mm
 Check the actual factor of safety
From PSG data book, Refer P.No. 7.78
Actual factor of safety [n] =
𝑄
Σ𝑃
Q - Breaking load of the chain = 4440 kg ΣP = Pt + Ps + Pc

Pt - Tangential force due to power Transmission:
Pt =
102 N
𝑣
Pt =
102 ×15
6.429
= 237.98 kgf
Pc - Centrifugal Tension:
Pc =
𝑊 𝑣2
𝑔
=
1.78 × 6.4292
9.81
= 7.499 𝑘𝑔𝑓
W = 1.78 kgf
Ps - tension due to slagging = kWa metres

Ps - tension due to slagging
Ps = k W a in metres
= 6 × 1.78 ×
550
1000
= 5.874 kg
K - Coefficient of Sag from PSG data book 7.78
= 6 (Horizontal)
a = 550 mm = 0.55 m
ΣP = 237.98 +7.499 + 5.874
= 251.353 kgf

[n] =
𝑄
Σ𝑃
=
4440
251.353
= 17.66 > 11
which is greater than allowable factor of safety
∴ The design is safe.
Step 8: Checking of Allowable Bearing Stress
From PSG data book, Pg.No. 7.77
The allowable bearing stress σ = 2.24 kgf/mm2
(for a pitch of 15.875 and speed < 1000 rpm)
Refer PSG data book, from Pg. no. 7.77

Power transmitted on the basis of allowable bearing stress
15 =
𝜎 × 1.4 × 6.429
102 × 1
σ = 169.98 kgf/cm2
σ = 1.69 kgf/mm2
σ = [σ]
< 2.24 kgf/mm2
Therefore the design is safe.

Pitch dia. of small sprocket:
𝑑 =
𝑝
𝑆𝑖𝑛 (
180
𝑍1
)
=
15.875
𝑆𝑖𝑛 (
180
27
)
= 136.74 mm
Pitch dia. of large sprocket
𝐷 =
𝑝
𝑆𝑖𝑛 (
180
𝑍2
)
=
15.875
𝑆𝑖𝑛 (
180
82
)
= 414.46 mm

A 15 kW squirrel cage motor with a speed of, 1250 rpm,
is driving a centrifugal pump at 550 rpm. The
centrifugal pump is located at 700 mm from the motor.
Design a chain drive.

Solution:
Given Data:
Power of motor N = 15 kW
Speed of motor N1 = 1250 rpm
Speed of centrifugal pump N2 = 550 rpm
Minimum centre distance a = 700 mm

 Calculation of transmission ratio
STEP:1 Transmission ratio = Z2/Z1 = N1/N2
i = Transmission ratio = = N1/N2 = 1250/550 = 2.27
STEP:2 From PSG Design data book, refer Page No. 7.74.
For i = 2 to 3 ; Z1 = 25 to 27 i = 3 to 4 ; Z1 = 23 to 25
Take Z1 = 23 to 27 (select any odd no. of teeth)
Select Z1 = 27 teeth (no. of teeth on sprocket pinion)
Z2 = i Z1 = 2.27 x 27 = 62 = 62 (no. of teeth on sprocket wheel)

STEP 3: Calculation of pitch
From PSG design data book, P.No.7.74
Optimum centre distance (a) = 30 to 50 p
where a - approximate centre distance
pmax = 7000/30 = 23.33 mm
pmin = 700/50 = 14 mm
Select standard pitch from PSG data book, Pg.No. 7.72
Take any standard pitch between 14 to 23.33 mm
Select Pitch; p = 15.875 mm

STEP 4: Selection of Chain No.
Select roller chain from PSG Data book Page No. 7.72
The available chain No. are 10A and 10B
Select 10A-2 Duplex Chain
Pitch ; p = 15.875 mm
Corresponding to chain No. selected, take the values of
A - Bearing area = 1.4 cm2
W - Weight per m length = 1.78 kgf
Q - Breaking load = 4440 kgf

STEP 5: Calculate Power transmitted based on breaking load:
From data book, Pg. No. 7.77
From the above equation calculate ‘Q’ breaking load by considering,
N = given power
N = 15 kW
=
27 ×15.875 ×1250
60 ×1000
= 8.929 m/sec

n - Minimum value of factor of safety
Ks = 1 ; Z1 = 15 to 30 from PSG data book, Pg. No. 7.77,
Select n = 13.2 for a pitch 15.875 mm and n1 < 1600 rpm
Since the specific conditions are not given in the problem,
assume K1 = K2 = K3 = K4 = K5 = K6 = 1
Ks = 1
15 =
𝑄 ×6.429
102 ×11 ×1
; Q =
15 ×102 ×11
16.429
= 2261.66 kgf

Breaking load Q = 2617.825 kgf which is less than the selected chain
Breaking load (4440 kgf)
The selection of chain no. is satisfactory based on breaking load.
STEP:6 To find (a) Calculation of Length of chain
(b) Final centre distance.
(a) Length of continuous chain in multiples of pitches: from data book, Pg.No. 7.7
𝑍1 + 𝑍2
2
+
𝑍2 − 𝑍1
2𝜋
2
𝑎𝑝
where ap = ao /p = 700/15.875
ap = 44.09 mm

ao - initially assumed centre distance in mm = 700 mm
p - pitch = 15.875 mm
No. of teeth on sprocket wheel Z2 = 62;
No. of teeth on sprocket pinion Z1 = 27
Refer PSG data book Page No. 7.76,
Read the value of ‘m’ directly from PSG data book Pg.No. 7.63
(Z2 - Z1) = 62 - 27 = 35;

the value of m = 31
𝑙𝑝 = 2 44.09 +
27+62
2
+
31
34.64
= 133.38
Length of chain 𝑙 = 𝑙p . p = 133.38 x 15.875 = 2117.46 mm

e = 133.38 −
27+62
2
= 88.8 mm

Final Centre distance
𝑎 =
88.8 + 88.82 − (8 × 31)
4
× 15.875
= 699.90 mm
a = 700 mm

STEP: 7 Check the actual factor of safety
From P.S.G. Data book, Pg.No. 7.78
Actual factor of safety [n] =
𝑄
Σ𝑃
where
Q - Breaking load of the chain = 4440 kg
Σ𝑃 = Pt + Ps + Pc

Pt - Tangential force due to power Transmission:
Pt =
102 N
𝑣
Pt =
102 ×15
8.929
= 171.35 kgf
Pc - Centrifugal Tension:
Pc =
𝑊 𝑣2
𝑔
where W - Weight per meter length.
From P.S.G. Data book, Pg.No. 7.72, corresponding to p =15.875 and chain
number. W = 1.78 kgf/m length

Pc =
1.78 × 8.9292
9.8
= 14.46 kgf
Ps - tension due to slagging
Ps = k W a in metres
From P.S.G. Data book, Pg.No. 7.78; k - coefficient for sag. k = 6 for horizontal
position of chain drive. and a - is the centre distance W is weight per meter
length
= 6 × 1.78 ×
700
1000
= 7.476 kgf
ΣP = 237.98 +7.499 + 5.874
= 251.353 kgf

[n] =
𝑄
Σ𝑃
=
4440
193.286
= 22.97 > 13.2
which is greater than allowable factor of safety
∴ The design is safe.
Step 8: Checking of Allowable Bearing Stress
From P.S.G. data book, Pg.No. 7.77, corresponding to speed of rotation of
sprocket < 1600 rpm, and pitch 15.875 mm.
The allowable bearing stress σ = 1.85 kgf/mm2
(for a pitch of 15.875 and speed < 1600 rpm)
Refer PSG data book, from Pg. no. 7.77

Power transmitted on the basis of allowable bearing stress
15 =
𝜎 × 1.4 × 8.929
102 × 1
σ = 122.39 kgf/cm2
σ = 1.22 kgf/mm2
σ < [σ]
< 1.85 kgf/mm2
Therefore, the design is safe.

STEP:9:
Pitch dia. of small sprocket:
𝑑 =
𝑝
𝑆𝑖𝑛 (
180
𝑍1
)
=
15.87
𝑆𝑖𝑛 (
180
27
)
= 136.74 mm
Pitch dia. of large sprocket:
𝐷 =
𝑝
𝑆𝑖𝑛 (
180
𝑍2
)
=
15.87
𝑆𝑖𝑛 (
180
62
)
= 313.43 mm

U2_S5_Selection of chain drives.pptx

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