MACHINE DESIGN LAB - Indian Institute of Technologyshibayan/MCC 15204 Machine...Table of Contents...
Transcript of MACHINE DESIGN LAB - Indian Institute of Technologyshibayan/MCC 15204 Machine...Table of Contents...
MANUAL ON
MACHINE DESIGN LAB
(Course Code No: MCC 15204)
Department of Mechanical Engineering
INDIAN INSTITUTE OF TECHNOLOGY
(INDIAN SCHOOL OF MINES), DHANBAD
GENERAL LABORATORY RULES FOR MACHINE
DESIGN:
β’ Revise and be prepared with all the basics of machine design theories taught
in theory class by instructor. Students are advised to bring class notes of
Machine Design.
β’ Read the procedure as set forth in the lab manual before you begin any design.
β’ It is compulsory to bring machine design data book in laboratory.
β’ For numerical calculations students have to bring scientific calculator without
fail.
β’ Use of mobile phones other than design purpose is strictly prohibited in the
laboratory.
β’ Drawing of the machine components is to prepared with AutoCAD.
β’ Final report of a design is to be submitted in the subsequent Lab class.
Table of Contents
Sl. No. List of design of machine element Page No.
1 Design of Journal Bearing 1-3
2 Design of Spur Gear 4-9
3 Design of Helical Gear 10-13
4 Design of Bevel Gear 14-16
5 Design of Worm Gear 17-19
6 Design of Flywheel 20-25
7 Selection of Rolling Element Bearing from Manufacturerβs Catalogue 26-29
8 Selection of V Belt Drive 30-33
9 Selection of Chain Drive 34-35
Note:
1. At least eight design problems are to be completed in the semester from the
above list.
2. Two more designs of machine components are to be carried out as designed
and set by the instructor as per the scope of the syllabus.
1
1. Journal Bearing
Objective: Design of journal bearing
Theory: Journal bearings are mechanical components used to support shafts of a machine.
Journal bearings are therefore designed to carry radial loads. The load carrying capacity is
developed due to the generation of pressure by the fluid film formed in the clearance space
between the bearing and journal. The shape of the clearance space is shown in Figure 1. The
expression for film thickness can be easily determined from the geometry of the film shape.
Figure 1. Journal bearing geometry
Material: Cast iron, alloy steel, bronze, aluminium etc.
Design equations:
β’ Eccentricity (e) = distance ObOj, where Ob= axis of bearing, Oj= axis of journal
β’ Radial clearance (c) = Rb- R , where Rb = radius of bearing , R= radius of journal
β’ Eccentricity ratio (Ι) = π
π
β’ Sommerfield number (S) = (π
π)
2 Β΅π
π, where Β΅ = viscosity of lubricant, π = shaft
speed
2
β’ Sliding velocity (U) = ΟR , Ο =2ΟN
β’ Film thickness (h) = AB which is the film thickness at B at angle ΞΈ from line of
action
h = Rb - π
π πππ sin[π β π ππβ1 (
π
π π ππ π)]
For simplicity assumed π
π << 1, so then
h = c(1+ Ι cos ΞΈ)
L= length of bearing
β’ Infinitely long bearing approximation (L/2R) > 2,
p = (6Β΅ππ Ι
π2 ) (2+ Ι cosΞΈ)π ππ π
(2+ Ι2)(1+ Ι cosΞΈ)2
Wπ = 12Β΅ππΏ (π
π)
2 Ι2
(2+ Ι2)(1βΙ2)
Wπ‘ = 6Β΅ππΏ (π
π)
2 πΙ
(2+ Ι2)(1βΙ2)1
2β
W= 6Β΅ππΏ (π
π)
2 Ι[π2βΙ2(π2β4)]1
2β
(2+ Ι2)(1βΙ2)
Π€= π‘ππβ1 (Wπ‘
Wπ) = (
πβ1βΙ2
2Ι)
π= (2+ Ι2)(1βΙ2)
6πΙ(
1
(π2(1βΙ2))+4Ι2)
π= π
π (
Ι sinΠ€
2+
2π2π
(1βΙ2)1
2β)
where, P = Pressure
Wπ = Radial load
Wπ‘ = Normal load
W = Total load carrying capacity
Π€ = Attitude angle
π = Coefficient of friction
β’ Infinitely short bearing approximation (L/2 R) < 1
p = 3Β΅π
π π2 (πΏ2
4β π¦2)
Ι sin π
(1+Ι cos π )2
Wπ = βΒ΅ππΏ3
π2
Ι2
(1βΙ2)2
Wπ‘ = Β΅ππΏ3
4π2
ΟΙ
(1βΙ2)3
2β
W= Β΅ππΏ3
4π2
Ι
(1βΙ2)2[(π2(1 β Ι2) + 16Ι2]
12β
Π€= π‘ππβ1 (Wπ‘
Wπ) = (
πβ1βΙ2
4Ι)
π= (1βΙ2)
2
ΟΙ[(π2(1βΙ2)+16Ι2]1
2β(
2π
πΏ)
2
π= π
π (
2π2π
(1βΙ2)1
2β)
ππΏ = ΙULc
F = Β΅ππΏπ
π
2π
(1βΙ2)1
2β
where, P = Pressure
Wπ = Radial load
Wπ‘ = Normal load
W = Total load carrying capacity
Π€ = Attitude angle
π = Coefficient of friction
ππΏ = Leakage flow rate
F = Friction force
3
β’ Finite journal bearing design: For finite journal bearing, solution of Raimondi and
Boyd method can be used. Dimensionless performance parameters are available in the
form of Charts and tables. It can be used for solving problems.
Power loss (ππ‘) = 2ΟRNF
Temperature rise (ΞT) = ππ‘
π½πππππΏ
where J = mechanical equivalent of
heat
ππ = specific heat
Refer machine design data book for required data if needed.
Problems:
Problem 1: Write a computer program for design of journal bearing.
Problem 2: A full journal bearing of width 20 cm with a journal of diameter 10 cm has diametric
clearance of 100 micro meters. The journal rotates at 1200 rpm. The absolute viscosity of
lubricant at 20Β° C is 0.04 Pas. For an eccentricity ratio of 0.6, determine the minimum film
thickness, load carrying capacity, attitude angle, Sommerfeld number, friction factor. Mass
density, and specific heat of the oil at constant pressure may be taken as 900 kg/m3 and 2.0
J/g/0K, respectively.
Problem 3: A journal bearing is operating under following operating conditions: Journal
diameter = 20 cm, bearing length = 10 cm and journal speed = 600 r.p.m. Clearance ratio may
be chosen between 0.5, 1, 1.5, and 2.0 mm/m. Select a clearance ratio and determine load
carrying capacity, oil flow rate, power loss, and temperature rise of lubricant while the viscosity
of the oil at 38Β°C is 100cS and at 100Β°C is 12cS. Specific gravity of the oil is 0.9. The bearing
is designed to run at an eccentricity ratio 0.6.
4
2. Spur Gear
Objective: Design of spur gear
Theory: In spur gears, the teeth are cut parallel to the axis of the shaft. As the teeth are parallel
to the axis of the shaft, spur gears are used only when the shafts are parallel. The profile of the
gear tooth is in the shape of an involute curve and it remains identical along the entire width of
the gear wheel. Spur gears impose radial loads on the shafts. Spur gear nomenclature and a pair
of spur gear is shown in Figure 1 and Figure 2 respectively.
Figure 1. Gear nomenclature
Figure 2. Spur gear
Material: Steel, hardened teeth, cast iron, bronze, stainless steel, aluminium etc.
5
Design equations/ data:
β’ Module of gear
P = π§
π
m = π
π§
d = mz
where ,d = pitch circle diameter (mm),
z = number of teeth on the gear
P = Diametral Pitch
m = Module (mm)
β’ Recommended series of module (mm)
Choice 1 1 1.25 1.5 2 2.5 3 4
5 6 8 10 12 16 20
Choice 2 1.125 1.375 1.75 2.25 2.75 3.5 4.5
`5.5 7 9 11 14 18 -
Choice 3 3.25 3.75 6.5 - - - -
β’ Gear Ratio and transmission ratio
i = ππ
ππ =
π§π
π§π
where, np = speed of pinion,
ng = speed of wheel,
zp = number of teeth on the pinion,
zg = number of teeth on the wheel
β’ Basic Relationship
P = π§
π
p = ππ
π§
PΓp = Ο
a = π(π§π+π§π)
2
where p = Circular Pitch (mm),
d = pitch circle diameter (mm),
z = number of teeth on the gear
P = Diametral Pitch
m = Module (mm)
a = centre-to-centre distance (mm)
zp = number of teeth on the pinion,
zg = number of teeth on the wheel
β’ Standard proportions of gear tooth for 20Β° full depth involute system
Dimension Notation Proportion
Addendum ha ha = m
Dedendum hf hf = 1.25m
Clearance c c = 0.25m
Working depth hk hk = 2 m
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Whole depth h H = 2.25 m
Tooth thickness s s = 1.5708m
Tooth space β¦. 1.5708m
Fillet radius β¦. 0.4m
β’ Spur gears - Component of tooth force
Mt = 60Γ106(ππ)
2πππ
Pt = 2ππ‘
ππ
Pr = Pt tanΞ±
dp = mzp
where Mt = Torque transmitted by gear (N-mm)
np = speed of pinion (rpm)
Pt = Tangential component of resultant tooth force (N)
Pr = Radial component of resultant tooth force (N)
Ξ± = Pressure angle (Β°)
dp = pitch circle diameter of pinion (mm)
β’ Minimum number of teeth
Pressure angle (Ξ±) 14.5Β° 20Β° 25Β°
zmin (theoritical) 32 17 11
zmin (practical) 27 14 9
Note: The minimum number of teeth to avoid interference is given by, zmin = 2
sin2 Ξ±
β’ Face width of tooth
Optimum range of face width : (8m) < b < (12m) or (b=10m)
β’ Beam Strength of gear tooth (Lewisβ equation)
Beam strength (Sb) indicates the maximum value of tangential force that the tooth can
transmit without failure:
ππ = ππππππ
ππ = ππ = (1
3) ππ’π‘
Sb= beam strength of gear tooth (N)
Οb= permissible bending stress (MPa)
Y= Lewis form factor based on virtual number of teeth
Se= endurance limit (MPa)
Sut=ultimate tensile strength (MPa)
β’ Wear strength of gear tooth (Buckinghamβs Equation)
Q = π§π
π§π+π§π (for external gear)
Q = 2π§π
π§πβπ§π (for internal gear)
Sw = b.Q.dp.K
K = ππ
2π πππΌπππ πΌ (1
πΈπ + 1/πΈπ)
1.4
Q = Ratio factor
π§π = number of teeth on pinion
π§π = number of teeth on wheel
Sw = wear strength of the gear tooth (N)
dp = pitch circle diameter of pinion (mm)
K = load-stress factor (MPa)
7
Οc = Surface endurance strength of the material
(MPa)
Ξ± = pressure angle
Ep = modulus of elasticity of pinion materials
(MPa)
Eg = modulus of elasticity of wheel materials
(MPa)
β’ For steel gears with 20Β° pressure angle
K = 0.16 (π΅π»π
100)
2
where BHN = Surface hardness of gears
(Brinell hardness number)
According to G. Niemann: Οc = 0.27(BHN) kgf/mm2 = 0.27 (9.81) (BHN) N/mm2
β’ Values of Modulus of elasticity and poissonβs ratio for gear materials
Material Modulus of elasticity E (MPa) Poissonβs ratio
Steel 206000 0.3
Cast Steel 202000 0.3
Spheroidal cast iron 173000 0.3
Cast tin bronze 103000 0.3
Tin bronze 113000 0.3
Grey cast iron 118000 0.3
β’ Effective load on gear tooth
Tangential force due to rated torque or rated power (Pt)
Mt = 60Γ106(ππ)
2πππ
Pt = 2ππ‘
ππ
Pt = Tangential force due to rated torque (N)
Mt = rated torque (N-mm)
Kw = power transmitted by gears (kW)
np = speed of pinion (rpm)
dp = pitch circle diameter of pinion (mm)
Effective load on gear tooth (Peff) β Preliminary gear design
Peff = CSππ‘
πΆπ£
Peff = effective load on gear tooth (N)
CS = Service factor
Cv = Velocity factor
Effective load on gear tooth (Peff) β Final gear design
Peff = (CSPt+Pd) Pd = incremental dynamic load (N)
(Buckinghamβs equation)
β’ Service factor for speed reduction gearboxes (Cs)
Working characteristic of
driving machine
Working characteristic of driven machine
Uniform Moderate shock Heavy shock
8
Uniform 1.00 1.25 1.75
Light shock 1.25 1.50 2.00
Medium shock 1.5 1.75 2.25
Note : For Electric motors, Cs = starting torque
rated torque
β’ Velocity factor (Cv)
For ordinary and commercially cut gears made with form cutters and with (v<10 m/s)
Cv = 3
3+π£
For accurately hobbed and generated gears with (v<20 m/s)
Cv = 6
6+π£
For precision gears with shaving, grinding, and lapping operations and with (v>20 m/s)
Cv = 5.6
5.6+βπ£
π£ = πππππ
60Γ103 where, π£ = pitch line velocity (m/s)
Pd = 21π£(πΆππ+ππ‘)
21π£+β(πΆππ+ππ‘ )
Pd = dynamic load or incremental dynamic load (N)
π£ = pitch line velocity (m/s)
C = deformation factor (MPa or N/mm2)
e = sum of errors between two meshing teeth (mm)
b = face width of tooth (mm)
Pt = tangential force due to rated torque (N)
β’ Deformation factor (C)
Deformation factor C depends upon moduli of elasticity of materials for pinion and
gear and the form of tooth or pressure angle
C = π
[1
πΈπ+
1
πΈπ]
k = constant depending upon the form of tooth
Ep = Modulus of elasticity of pinion material (MPa or N/mm2)
Eg = Modulus of elasticity of wheel material (MPa or N/mm2)
The values of k for various tooth forms are as follows:
k = 0.107 (for 14.5Β° full depth teeth)
k = 0.111 (for 20Β° full depth teeth)
k = 0.107 (for 20Β° stub teeth)
β’ Values of deformation factor C
Materials 14.5Β° full
depth teeth
20Β° full depth
teeth 20Β° stub teeth
Pinion Material Gear Material
Grey C.I. Grey C.I. 5500 5700 5900
9
Steel Grey C.I. 7600 7900 8100
Steel Steel 11000 11400 11900
Refer machine design data book for required data if needed.
Problems:
Problem 1: Write a computer program for design of spur gear.
Problem 2: It is required to design a pair of spur gears with 20Β° full-depth involute teeth based
on the Lewis equation. The velocity factor is to be used to account for dynamic load. The pinion
shaft is connected to a 10kW, 1440 rpm motor. The starting torque of the motor is 150% of the
rated torque. The speed reduction is 4:1. The pinion as well as the gear is made plain carbon
steel 40C8 (Sut = 600 N/mm2). The factor of safety can be taken as 1.5. Design the gears, specify
their dimensions and suggest suitable surface hardness for the gears.
Problem 3: A pair of spur gears with 20Β° full-depth involute teeth consists of a 20 teeth pinion
meshing with a 41 teeth gear. The module is 3 mm while the face width is 40 mm. The material
for pinion as well as gear is steel with an ultimate tensile strength of 600 N/mm2. The gears are
heat treated to a surface hardness of 400 BHN. The pinion rotates at 1450 rpm and the service
factor for the application is 1.75. Assume that velocity factor accounts for the dynamic load
and the factor of safety is 1.5. Determine the rated power that the gears can transit.
Problem 4: A pair of spur gears consists of a 24 teeth pinion, rotating at 1000 rpm and
transmitting power to a 48 teeth gear. The module is 6 mm, while the face width is 60 mm.
Both gears are made of steel with an ultimate tensile strength of 450 N/mm2. They are heat
treated to a surface hardness of 250 BHN. Assume that velocity factor accounts for the dynamic
load. Calculate (i) beam strength; (ii) wear strength; and (iii) the rated power that the gears can
transmit, if service factor and the factor of safety are 1.5 and 2, respectively.
10
3. Helical Gear
Objective: Design of helical gear.
Theory: A pair of helical gears is shown in Figure 1. The teeth of these gears are cut at an
angle with the axis of the shaft. Helical gears have an involute profile similar to that of spur
gears. However, this involute profile is in a plane, which is perpendicular to the tooth element.
The magnitude of the helix angle of pinion and gear is same; however, the hand of the helix is
opposite. A right-hand pinion meshes with a left-hand gear and vice versa. Helical gears impose
radial and thrust loads on shafts.
Figure 1. Helical gear
Material: Cast iron, alloy steel, bronze, aluminum etc.
Design equations:
No. of teeth on pinion (Zp), No. of teeth on gear (Zg), Helix angle (Ο), normal pressure angle
(Ξ±n), normal module (m), Rotational speed (N)
β’ Basic equations of helical gears
pn = pπππ π
Pn = P/πππ π
mn = ππππ π
pa = p/π‘πππ
Cos Ο = tanΞ±n/tanΞ± Ξ±n
d = zmn/cos Ο
a = mπ(π§π + π§π)
2
where p= transverse circular pitch(mm)
pn= normal circular pitch(mm)
Ο = Helix angle
Pn= normal diametric pitch(mm)
P = tranverse diametric pitch(mm)
mn= normal module(mm)
m = transverse module(mm)
pa = axial pitch(mm)
Ξ±n = normal pressure angle (0)
11
π =π€π
π€π=
π§π
π§π
Ξ±=transverse pressure angle (0)
d=pitch circle diameter(mm)
z= number of teeth
a=center to center distance(mm)
i=speed ratio
π§π = number of teeth on pinion
π§π = number of teeth on wheel
Recommended series of normal module (mn) in mm is: 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8 and 10.
β’ Standard proportions of helical gears
ππ = ππ [π§
πππ π+ 2]
ππ = ππ [π§
πππ πβ 2.5]
where, Addendum(ha)=mn
Dedendum(hf)=1.25mn
Clearance(c)=0.25mn
da=addendum circle diameter(mm)
df=dedendum circle diameter (mm)
β’ Components of tooth force
ππ‘ =60π₯106(ππ)
2πππ
ππ‘ =2ππ‘
ππ
ππ = ππ‘ [π‘πππΌπ
πππ π]
ππ = ππ‘π‘πππ
ππ =π§πππ
πππ π
Suffix p is used for pinion
Mt = torque transmitted by gears (N-mm)
(kW) = power transmitted by gears
np= speed of pinion (rpm)
Pt= tangential component of resultant tooth force (N)
Pr= radial component of resultant tooth force (N)
Pa= axial or thrust component of resultant tooth force (N)
dp= pitch circle diameter of pinion (mm)
Ξ±n= normal pressure angle (0)
β’ Design by Lewis and Buckinghamβs Equations
Beam Strength of gear tooth (Lewisβ equation)
Beam strength (Sb) indicates the maximum value of tangential force that the tooth can
transmit without failure
ππ = ππππππ Sb= beam strength of gear tooth (N)
Οb= permissible bending stress (MPa)
Y= Lewis form factor based on virtual number
of teeth (zβ)
12
ππ = ππ = (1
3) ππ’π‘
Se= endurance limit (MPa)
Sut=ultimate tensile strength (MPa)
β’ Wear strength of gear tooth (Buckinghamβs equation)
Wear strength (Sw) indicates the maximum value of tangential force that the tooth can
transmit without pitting failure.
ππ€ =πππππΎ
πππ 2π
Sw= wear strength of thr gear tooth (N)
Q= ratio factor
dp= pitch circle diameter of pinion (mm)
K= load-stress factor (MPa)
π =2π§π
π§π+π§π (for external gears)
π =2π§π
π§πβπ§π (for internal gears)
zp= number of teeth on pinion
zg= number of teeth on wheel
πΎ =ππ
2π πππΌππππ πΌπ(1πΈπ
β + 1πΈπ
β )
1.4
Οc= surface endurance strength of the
material(MPa)
πΌπ= normal pressure angle(0)
πΈπ= modulus of elasticity of pinion material
(MPa)
πΈπ= modulus of elasticity of wheel material
(MPa)
For steel gears with 200 pressure angle,
πΎ β 0.16 (π΅π»π
100)
2
BHN= surface hardness of gears (Brinell
Hardness Number)
According to G. Niemann,
Οc=0.27(BHN) kgf/mm2 = 0.27(9.81)(BHN) N/mm2
β’ Effective Load on gear tooth
Tangential force due to rated torque or rated power (Pt)
ππ‘ =60 β 106(ππ)
2πππ
ππ‘ =2ππ‘
ππ
Pt= tangential force due to rated troque (N)
Mt = rated torque(N-mm)
(kW) = power transmitted by gears
np= speed of pinion (rpm)
dp= pitch circle diameter of pinion (mm)
Effective load on gear tooth (Peff) β Preliminary gear design
ππππ =πΆπ ππ‘
πΆπ£
πΆπ£ =5.6
5.6 + βπ£
Peff= effective load on gear tooth (N)
Cs= service factor
Cv= velocity factor
v= pitch line velocity (m/s)
13
Effective load on gear tooth (Peff) β Final gear design
ππππ = (πΆπ ππ‘ + ππ) Pd= incremental dynamic load (N) (Buckinghamβs
equation)
Refer machine design data book for required data if needed.
Problems:
Problem 1: Write a computer program for design of helical gear.
Problem 2: A pair of parallel helical gear consists of an 18 teeth pinion meshing with a 45 teeth
gear. 7.5kW power at 2000 rpm is supplied to the pinion through its shaft. The normal module
is 6mm, while the normal pressure angle is 200. The helix angle is 230. Determine the tangential,
radial and axial components of the resultant tooth force between the meshing teeth.
Problem 3: A pair of parallel helical gears consists of 20 teeth pinion meshing with a 100 teeth
gear. The pinion rotates at 720 rpm. The normal pressure angle is 200, while the helix angle is
250. The face width is 40 mm and the normal module is 4 mm. The pinion as well as the gear
is made of steel 40C8 (Sut=600N/mm2) and heat treated to a surface hardness of 300 BHN. The
service factor and the factor of safety are 1.5 and 2 respectively. Assume that the velocity factor
accounts for the dynamic tool load and calculate the power transmitting capacity of the gears.
14
4. Bevel Gear
Objective: Design of bevel gear
Theory: Bevel gears, as shown in Figure 1, have the shape of a truncated cone. The size of the
gear tooth, including the thickness and height, decreases towards the apex of the cone. Bevel
gears are normally used for shafts, which are at right angles to each other. This, however, is
not a rigid condition and the angle can be slightly more or less than 90 degrees. The tooth of
the bevel gears can be cut straight or spiral. Bevel gears impose radial and thrust loads on the
shafts.
Figure 1. Bevel gear
Material: Cast steel, Plain carbon steels, Alloy steels etc.
Design Equations:
β’ Pitch angle formulas
Pinion (π³π) = tanβ1 sin π³
ππ +cos π³
VR= Velocity Ratio
π = the angle between the shafts.
Gear (π³π) = tanβ1sin π³
1ππ + cos π³
β’ Components of tooth force
15
ππ‘ =60 β 106(ππ)
2πππ
ππ‘ =ππ‘
ππ
ππ = ππ‘ π‘ππ πΌ πππ πΎ
ππ = ππ‘ππ‘ π‘ππ πΌ π ππ πΎ
ππ = [π·π
2β
π sin πΎ
2]
Suffix p is used for pinion
Mt = torque transmitted by gears (N-mm)
(kW) = power transmitted by gears
np= speed of pinion (rpm)
Pt= tangential component of resultant tooth force (N)
Pr= radial component of resultant tooth force (N)
Pa= axial or thrust component of resultant tooth force (N)
ππ= radius of pinion at the midpoint along the face width (mm)
πΌ = pressure angle (0)
π·π= pitch circle diameter at large end of the tooth (mm)
β’ Design by Lewis Equations
Beam Strength of gear tooth (Lewisβ equation)
Beam strength (Sb) indicates the maximum value of tangential force that the tooth can transmit
without failure
ππ = πππππ [1 βπ
π΄0]
π€βπππ π΄0 = βπ·π
2
4+
π·π2
4 π =
1
4π΄0
Sb= beam strength of gear tooth (N)
Οb= permissible bending stress (MPa)
Y= Lewis form factor based on formative
number of teeth
m= module at the large end of the tooth (mm)
ππ = ππ = (1
3) ππ’π‘
Se= endurance limit (MPa)
Sut=ultimate tensile strength (MPa)
β’ Standard proportions of bevel gears
Addendum,(a) = 1 m; Dedendum (d) = 1.2 m Clearance= 0.2 m Working depth= 2 m Thickness
of tooth = 1.57 m
β’ Equivalent number of Teeth for Bevel Gears- Tredgoldβs Approximation:
ππ = π sec π³π ππ π
β’ Outside diameter for the Bevel Gears (π«π) = π·π ππ π + 2π πππ π³π ππ π
β’ The relation between π¨π, π«π, ππ is: sin π³π =π·π/2
π΄0
β’ Wear strength of gear tooth (Buckinghamβs equation)
Wear strength (Sw) indicates the maximum value of tangential force that the tooth can
transmit without pitting failure.
16
ππ€ = .75πππ·ππΎ
πππ 2πΎ
Sw= wear strength of thr gear tooth (N)
Q= ratio factor
Dp= pitch circle diameter of at large end of
the tooth (mm)
K= load-stress factor (MPa)
π =2π§π
π§π + π§π tan πΎ
zp= number of teeth on pinion
zg= number of teeth on wheel
πΎ =ππ
2π πππΌ cos πΌ (1πΈπ
β + 1πΈπ
β )
1.4
Οc= surface endurance strength of the
material(MPa)
πΌπ= normal pressure angle(0)
πΈπ= modulus of elasticity of pinion material
(MPa)
πΈπ= modulus of elasticity of wheel material
(MPa)
For steel gears with 200 pressure angle,
πΎ β 0.16 (π΅π»π
100)
2
BHN= surface hardness of gears (Brinell
Hardness Number)
According to G. Niemann,
Οc=0.27(BHN)kgf/mm2 = 0.27(9.81)(BHN) N/mm2
Refer machine design data book for required data if needed.
Problems:
Problem 1: Write a computer program for design of bevel gear.
Problem 2: A pair of straight bevel gears is required to transmit 10 kW at 500 rpm from motor shaft
to another shaft at 250 rpm. The pinion has 24 teeth. The pressure angle is 20Β°. If the shaft axes are
at right angles to each other find the module, face width, addendum, outside diameter and slant
height. The gears are capable of withstanding a static stress of 60 MPa. The tooth form factor may
be taken as . 154 β.684
ππΈ , where ππΈ is equivalent number of teeth. Assume velocity factor πΆπ£ =
4.5
4.5+π£ where π£ the pitch line speed in m/s.
Problem 3: A 35 kW motor running at 1200 rpm drives a compressor at 780 rpm. The arrangement
is made through a 90α΅ bevel gearing arrangement. The pressure angle of the teeth is 14.5α΅. The
allowable static stress for both pinion and gear are 85 MPa. Determine the number of teeth on the
gear, face width, addendum, dedendum outside diameter and slant height, thickness of tooth. Also
check your design from the standpoint of wear. Take surface endurance limit as 690 MPa and
Modulus of Elasticity as 89kN/ mm2.
17
5. Worm Gear
Objective: Design of worm gear
Theory: The worm gears, as shown in Figure 1, consist of a worm and a worm wheel. The
worm is in the form of a threaded screw, which meshes with the matching wheel. The threads
on the worm can be single or multi-start and usually have a small lead. Worm gear drives are
used for shafts, the axes of which do not intersect and are perpendicular to each other. The
worm imposes high thrust load, while the worm wheel imposes high radial load on the shafts.
Worm gear drives are characterized by high speed reduction ratio. High gear ratios like 200:1
can be achieved.
Figure 1. Worm gear
Figure 2. Nomenclature of worm gear
18
Material: For worm- Cast iron, alloy steel, cast steel, carbon steel etc.
For worm wheel- Aluminium, brass, copper, plastic etc.
Design equations/ data:
π =π1
π
π = ππ₯π§1
ππ₯ = Οm
π2 = ππ§2
tan Ο =π
ππ1 =
π§1
π
Ο + Ο = π
2
a= 1
2 (π1+π2) =
π
2 (π+π§2)
i = π§2
π§1
where, π1= pitch circle diameter of worm (mm)
π= lead of worm (mm)
ππ₯= axial pitch of worm (mm)
π2= pitch circle diameter of worm wheel
(mm)
Ο = lead angle of worm (o)
Ο = helix angle of worm (o)
a = centre to centre distance between worm
and worm wheel (mm)
i = speed ratio
β’ Dimensions of worm and worm wheel
βπ1= m
βπ1= (2.2πππ Ο β 1)π
π = 0.2ππππ Ο
ππ1= m (q+2)
ππ1= (π + 2 β 4.4πππ Ο)π
where, βπ1= addendum of worm (mm)
βπ1= dedendum of worm (mm)
π = clearance (mm)
ππ1= outside diameter of the worm (mm)
βπ1= root diameter of the worm (mm)
β’ Dimensions of worm wheel
βπ2= (2πππ Ο β 1)π
βπ2= (0.2πππ Ο + 1)π
ππ2= (π§2+ 4πππ Ο β 2)m
ππ2= (π§2 β 2 β 0.4πππ Ο)π
πΉ= 2mβ(π + 1)
ππ = (ππ1 + 2π)π ππβ1 [πΉ
(ππ1 + 2π)]
where, βπ2= addendum at throat of worm wheel
(mm)
βπ2= dedendum in medium plane (mm)
ππ2= throat diameter of the worm wheel (mm)
ππ2= root diameter of the worm wheel (mm)
F = effective face width of worm wheel (mm)
ππ= length of root of worm wheel teeth (mm)
β’ Forces acting on worm and worm wheel
(π1)π‘ =2ππ‘
π1 (π1)π = axial component on the worm (N)
(π1)π‘ = tangential component on the worm (N)
19
(π1)π = (π1)π‘
(cos πΌπππ Ο β Β΅π ππΟ)
(cos πΌπ ππΟ + Β΅πππ Ο)
(π1)π = (π1)π‘
π πππΌ
(cos πΌπ ππΟ + Β΅πππ Ο)
(π2)π‘ = (π1)π
(π2)π = (π1)π‘
(π2)π = (π1)π
(π1)π = radial component on the worm (N)
(π2)π = axial component on the worm wheel (N)
(π2)π‘ = tangential component on the worm wheel
(N)
(π2)π = radial component on the worm wheel (N)
ππ‘ = torque transmitted by gears (N-mm)
Ξ± = normal pressure angle (o)
Β΅ = coefficient of friction
β’ Friction in worm gears
π£π = ππ1π1
60000πππ Ο
Ξ· = (cos πΌβΒ΅π‘ππΟ)
(cos πΌ+Β΅πππ‘Ο)
π£π = rubbing velocity (m/s)
π1= speed (rpm)
Ξ· =efficiency of worm gear drive
Refer machine design data book for required data if needed.
Problems:
Problem 1: Write a computer program for design of worm gear.
Problem 2: A pair of worm gears is designated as 2/54/10/5. Calculate (i) the centre distance;
(ii) the speed reduction; (iii) the dimensions of the worm; and (iv) the dimensions of the worm
wheel.
Problem 3: A pair of worm and worm wheel is designated as 2/52/10/4, 10 kW power at 720
rpm is supplied to the worm shaft. The coefficient of friction is 0.04 and the pressure angle is
20Β°. Calculate the tangential, axial and radial components of the resultant gear tooth force
acting on the worm wheel.
20
6. Flywheel
Objective: Design of flywheel
Theory: A flywheel is a heavy rotating body that acts as a reservoir of energy. The energy is
stored in the flywheel in the form of kinetic energy. The flywheel acts as an energy bank
between the source of power and the driven machinery. Depending upon the source of power
and type of driven machinery, there are two distinct applications of the flywheel.
Figure 1. Flywheel
Materials:
Material Mass density (kg/m3) (π)
Grey cast iron
FG 150 7050
FG 200 7100
FG 220 7150
FG 260 7200
FG 300 7250
Steels
Carbon steels 7800
Design equation/data:
β’ Torque analysis
Maximum fluctuation of speed = Οmax -
Οmin
Cs= Οmax β Οmin
Ο
Ο = Οmax + Οmin
2
Οmax = Maximum speed during a cycle
Οmin = Minimum speed during a cycle
Cs= Coefficient of fluctuation of speed
U0 = Change in kinetic energy
21
U0 = IΟ2Cs
Cs = 2 (nmax β nmin )
(nmax + nmin )
Ο is the average or mean angular
velocity of the flywheel
n is the speed in rpm
Table 1: Coefficients of fluctuations of speed
Type of Equipment ππ¬
Punching, shearing and forming presses 0.200
Compressor (belt driven) 0.120
Compressor (gear driven) 0.020
Machine tools 0.025
Reciprocating pumps 0.040
Geared drives 0.020
Internal combustion engines 0.030
D.C. generators (direct drive) 0.010
A.C. generators (direct drive) 0.005
β’ Coefficient of fluctuation of energy
The intercepted areas between the torque developed by the engine and the mean
torque line, taken in order from one end are -a1, +a2, -a3, +a4, -a5, +a6, and -a7
respectively.
Figure 2. Turning Moment Diagram
It is assumed that the energy stored in the flywheel is U at the point A. Therefore,
Energy at B = U β a1
Energy at C = U β a1 + a2
22
Energy at D = U β a1 + a2 β a3
Energy at E = U β a1 + a2 β a3 + a4
Energy at F = U β a1 + a2 β a3 + a4 β a5
Energy at G = U β a1 + a2 β a3 + a4 β a5 + a6
Energy at H = U β a1 + a2 β a3 + a4 β a5 + a6 β a7 = U
Maximum fluctuation of energy = maximum kinetic energy β minimum kinetic
energy
Coefficient of fluctuation of energy (Ce) = Maximum fluctuations of energy
work done per cycle
The work done per cycle is given by,
Work done/cycle = area below mean torque line from 00 to 3600
Work done/cycle = (2Ο)Tm (for two stroke engine)
Work done/cycle = (4Ο)Tm (for four stroke engine)
Table 2: Coefficient of fluctuations of energy
Type of engine Ce
Single-cylinder, double-acting steam engine 0.21
Cross-compound steam engine 0.096
Single-cylinder, four-stroke petrol engine 1.93
Four-cylinder, four-stroke petrol engine 0.066
Six-cylinder, four-stroke petrol engine 0.031
β’ Solid disk flywheel
Figure 3. Solid Disk Flywheel
23
I = mR
2
2
m = ΟR2tΟ
I = Ο
2ΟtR4
I = mass moment of inertia of disk (kg-m2)
m = mass of disk (kg)
R = outer radius of disk (m)
t = thickness of disk (m)
Ο = mass density of flywheel material
(kg/m3)
Two principal stresses in the rotating disk is given by,
Οt = Οv2
106 (ΞΌ+3
8) [1 β (
3ΞΌ+1
ΞΌ+3) (
r
R)
2
]
Οr = Οv2
106 (ΞΌ+3
8) [1 β (
r
R)
2
]
where, Οt = tangential stress at radius r
(N/mm2)
Οr = radial stress at radius r (N/mm2)
ΞΌ = Poissonβs ratio (0.3 for steel and 0.27
for cast iron)
v = peripheral velocity (m/s) (RΟ)
The maximum tangential stress and
maximum radial stress are equal and both
occur at (r = 0)
Therefore,
(Οt)max. = (Οr)max. = Οv2
106 (ΞΌ+3
8)
β’ Rimmed flywheel
Ir = KI = π0πΎ
π2πΆπ
[since, I = π0
π2πΆπ ]
where, Ir =Moment of Inertia of the rim is
given
Ir = moment of inertia of rim (kg-m2)
I = required moment of inertia (kg-m2)
K = 1, When entire moment of Inertia is due
to rim alone
K = 0.9, When the rim contributes 90% of the
required moment of inertia
If thickness of rim is usually very small compared with the mean radius. Then the
radius of gyration of the rim is equal to the mean radius.
Ir = mrR2 mr = mass of the rim (kg)
24
R = mean radius of the rim (m
Flywheels are usually made of grey cast iron, for which the limiting mean rim
velocity is 30 m/s. When the velocity exceeds this limit, there is a possibility of
bursting due to centrifugal force, resulting in an explosion.
v = πR
R <30
π
ππ‘ = π1
π΄1
π1 = 2
3[
(1000)ππ£2
πΆ]
m = btπ
ππ‘π = (1000)ππ£2
ππ‘[1 β
cos β
3πΆ sin πΌΒ±
2(1000)π
πΆπ‘(
1
πΌβ
cos β
sin πΌ)]
where, π1 = Tensile force in each spoke (N)
π΄1 = Cross-sectional area of spoke (mm2)
ππ‘ = Tensile stress in spokes (N/mm2)
m = mass of the rim per mm of
circumference (kg/mm)
v = velocity at the mean radius (m/s)
C = constant
π = mass density of flywheel material
(kg/mm3)
b = width of the rim (mm)
t = thickness of the rim (mm)
ππ‘π= tensile stress in the rim (N/mm2)
R = mean radius of rim (m)
2πΌ = angle between two consecutive spokes
(rad)
β = angle from centre line between spokes to
the section where the stress is being
calculated (rad)
β’ Expression for Constant C:
For 4 spokes: 2πΌ = π
2
C = [72960 π 2
π‘2 + 0.643 + π΄
π΄1]
For 6 spokes: 2πΌ = π
3
C = [20280 π 2
π‘2 + 0.957 + π΄
π΄1]
For 8 spokes: 2πΌ = π
4
C = [9120 π 2
π‘2+ 1.274 +
π΄
π΄1]
where, A = cross-sectional area of rim (mm2)
A1 = cross-sectional area of spokes (mm2)
Refer machine design data book for required data if needed.
25
Problems:
Problem 1: Write a computer program to design a flywheel.
Problem 2: The turning moment diagram of a multi-cylinder engine is drawn with a scale of (1
mm = 10) on the abscissa and (1 mm = 250 N-m) on the ordinate. The intercepted areas between
the torque developed by the engine and the mean resisting torque of the machine, taken in order
from one end are β 350, + 800, - 600, + 900, -550, +450 and -650 mm2. The engine is running
at a mean speed of 750 rpm and the coefficient of speed fluctuations is limited to 0.02. A
rimmed flywheel made of grey cast iron FG 200 (π = 7100 kg/m3) is provided. The spokes,
hub and shaft are assumed to contribute 10% of the required moment of inertia. The rim has
rectangular cross-section and the ratio of width to thickness is 1.5. Determine the dimensions
of the rim.
Problem 3: A machine is driven by a motor, which exerts a constant torque. The resisting torque
of the machine increases uniformly from 500 N-m to 1500 N-m through a 3600 rotation of the
driving shaft and drops suddenly to 500 N-m again at the beginning of the next revolution. The
mean angular velocity of the machine is 30 rad/s and the coefficient of speed fluctuations is
0.2. A solid circular steel disk, 25 mm thick, is used as flywheel. The mass density of steel is
7800 kg/m3 while Poissonβs ratio is 0.3. Calculate the outer radius of the flywheel disk and the
maximum stresses induced in it.
26
7. Rolling Element Bearing
Objective: Selection of Rolling Element Bearing from Manufacturerβs Catalogue
Theory: Bearing is mechanical device that permits relative motion between two parts, such as
the shaft and the housing, with minimum friction. Bearings are classified in different ways the
most important criterion to classify the bearings is the type of friction between the shaft and
the bearing surface. Depending on type of friction, bearings are classified into two main group
Sliding contact bearing and Rolling contact bearing.
Rolling contact bearing: Rolling contact bearing are also called antifriction bearings or simply
ball bearing. Rolling element such as balls or rollers, are introduced between the surfaces that
are in in relative motion. Figure 1 shows rolling contact bearing. Rolling contact bearing are
used in following applications: Machine tool spindle, Automobile front and rear axle, Gear
boxes, Small size electric motors and Rope etc.
Figure 1. Rolling Contact Bearing
The types of rolling contact bearing, which are frequently used are:
(i) Deep Groove Ball bearing (ii) Cylindrical roller bearing (iii) Angular contact bearing (iv)
Self-aligning bearing (v)Taper roller bearing (vi) Thrust ball bearing. Different types of
rolling contact bearing are shown in Figure 2.
27
Figure 2. Types of Rolling Contact Bearing
Materials: Chrome Steel - SAE 52100, Stainless Steels, Stainless Steel Bearingsβ ACD34
/KS440 / X65Cr13 etc.
Selection of Rolling Element Bearing:
The information given here should serve to indicate which are the most important of the
following points to be considered when selecting bearing type and thus facilitate an
appropriate choice.
β’ Cylindrical & Needle roller β pure radial load.
β’ Thrust (cylindrical roller, ball), four point angular contact ball bearings β pure axial load.
β’ Taper roller, spherical roller, angular contact ball bearings β combined Load.
β’ Cylindrical roller, angular contact ball bearingβ high speed.
β’ Deep groove, angular contact, and cylindrical roller bearing β high running accuracy.
Design equations/data:
β’ Equivalent dynamic load:
π = πππΉπ + ππΉπ
π = equivalent dynamic load (N)
πΉπ = radial load acting on bearing (N),
πΉπ = axial or thrust load acting on bearing
π = race- rotation factor
π = radial factor,
π= thrust factor
28
β’ Load life equation:
πΏ10 = (πΆ
π)
π
where, πΏ10 = rating bearing life (in
million revolutions)
C = dynamic load capacity (N)
p = 3 (for ball bearings)
p = 10/3 (for roller bearing)
β’ Life in hours:
πΏ10 =60ππΏ10β
106
where, πΏ10β = rated bearing life (hours)
π = speed of rotation (rpm)
β’ Cyclic loads and speed:
ππ = β[π1π1
3+π2π23+β―
π1+π2+β―
3]
ππ = β[β ππ3
β π]
3
where N = π1 + π2 + β― ππ
ππ = equivalent dynamic load for
complete work cycle (N)
π1, π2 β¦ ππ = dynamic load during first,
second, .... nth element of work cycle
π1, π2 β¦ ππ= number of revolutions
completed by first, second.... nth element
of work cycle
N = life of complete work cycle (rev)
β’ Cyclic loads and speeds (continuous variation of load):
ππ = [1
πβ« π3 ππ]1 3β
β’ Bearing with probability of survival other than 90%:
π = πβ(πΏ πβ )π, where π = reliability (in fraction), L = corresponding life (in million of
revolution), π and π = constants (π = 6.84 and π = 1.17)
(πΏ
πΏ10) = [
logπ(1
π )
logπ(1
π 90)]1 πβ , where πΏ10 = life corresponding to a reliability of 90% or
π 90, π 90 = 0.9
β’ System reliability: π π = (π )π, where π = number of bearings in the system (each
having the same Reliability π ), π π = reliability of the complete system
Refer machine design data book for required data if needed.
Problems:
Problem 1: Write a computer program for selection of ball bearing from SKF/FAG
manufacturer catalogue.
Problem 2: A ball bearing operates on the following work cycle:
29
Element
No
Radial load
(N)
Speed
(rpm)
Element time
(%)
1 3000 720 30
2 7000 1440 50
3 5000 900 20
The dynamic load capacity of the bearing is16.6 kN. Calculate (i) the average speed of rotation;
(ii) the equivalent radial load; and (iii) the bearing life.
Problem 3: A ball bearing is subjected to a radial force of 2500 N and an axial force of 1000
N. The dynamic load carrying capacity of the bearing is 7350 N. The values of X and Y factors
are 0.56 and 1.6 respectively. The shaft is rotating at 720 rpm. Calculate the life of the bearing.
Problem 3: The gear-reduction unit shown has a gear that is press fit onto a cylindrical sleeve
that rotates around a stationary shaft. The helical gear transmits an axial thrust load T of 1kN
as shown in the figure. Tangential and radial loads (not shown) are also transmitted through
the gear, introducing radial ground reaction forces at the bearings of 3.5kN for bearing A and
2.5 kN for bearing B. The desired life for each bearing is 90 kh at a speed of 150 rev/min with
a 90 percent reliability. The first iteration of the shaft design indicates approximate diameters
of 28mm at A and 25mm at B. Select suitable tapered roller bearings from TIMKEN catalogue.
30
8. V Belt Drive
Objective: Selection of V Belt Drive
Theory: Belts are used to transmit power between two shafts by means of friction. A belt drive
consists of three elementsβdriving and driven pulleys and an endless belt, which envelopes
them. Belt drives are mainly used in electric motors, automobiles, machine tools and
conveyors. Depending upon the shape of the cross-section, belts are classified as flat belts and
V-belts. Flat belts have a narrow rectangular cross-section, while V-belts have a trapezoidal
cross-section.
Among flexible machine elements, perhaps V-belt drives have widest industrial application.
These belts have trapezoidal cross section and do not have any joints. Therefore, these belts
are manufactured only for certain standard lengths. To accommodate these belts the pulleys
have V shaped grooves which makes them relatively costlier. V belt can have transmission
ratio up to 1:15 and belt slip is very small.
Material: V belts are made of polyester fabric and cords with rubber reinforcement as shown
in
Figure 1.
Figure 1. Cross section of V belts
Polyester fabrics which are located on horizontal lines near the center of gravity of the belt
cross section carry the central load. The surrounding layer of rubber transmit force from cords
to side walls. Cover is polychloroprene impregnated elastic cover.
Design equation / data :
β’ Designation of V belt
To specify a V belt, give the belt-section letter, followed by the inside circumference
in inches. e.g. B42, C60
31
Inside length + X = Pitch Length
β’ V- belt Equation
V-belts have additional friction grip due to the presence of wedge. Therefore,
modification is needed in the equation for belt tension. The equation is modified as,
πΉ1 β π π£2
πΉ2 β ππ£2= πππΌ/π ππππ½
β’ Equivalent smaller pulley diameter (de): In a belt drive, both the pulleys are not
identical, hence to consider severity of flexing, equivalent smaller pulley diameter is
calculated based on speed ratio. The power rating of V-belt is then estimated based on
the equivalent smaller pulley diameter
de = small diameter correction factor x d = Fb x d, kW power rating of different sections
of belts are also given in tabular form by manufacturer catalogue for different pitch
diameter of the pulley and different pulley speeds for an angle of wrap of 180o.
β’ V- belt design factors
Service Factor - A belt drive is designed based on the design power, which is the
modified required power. The modification factor is called the service factor. The
service factor depends on hours of running, type of shock load expected and nature of
duty.
Design Power= service factor x Rated power = Fa x P
β’ Modification of kW rating:
Power rating of a typical V-belt section requires modification, since, the ratings are
given for the conditions other than operating conditions. The factors are as follows:
β’ Angle of wrap correction factor: The power rating of V-belts are based on angle of
wrap, Ξ± =180ΞΏ Hence, Angle of wrap correction factor is incorporated when Ξ± is not
equal to 180o .
β’ Selection of V- belt
Selection of V belt depends on two concepts
1. Design Power (Total power considering correction factor)
2. Belt power Rating (Power transmitting capacity of one belt)
32
Step 1. Determine the correction factor according to service Fa and calculate the design
power as below
π·ππ πππ πππ€ππ = π‘ππππ πππ‘π‘ππ(πππ‘ππ) πππ€ππ π Γ πΉπ
Step 2. Select the type of belt according to the design power and speed(rpm) of faster
pulley (smaller pulley) then determine the recommended pitch diameter of the smaller
pulley that depends upon the cross section of the belt. Calculate the pitch diameter of
bigger pulley by following relationship
π·
π=
π
π
Step 3. Determine the pitch length of the of the belt.
Step 4. Compare the above value of L with the preferred pitch length. In case of
nonstandard value , the nearest value should be taken. And now on the basis of standard
value of L corrected center distance can be calculated by the formula given in step 3.
Step 5. Determine the correction factors for belt pitch length
Step 6. Calculate the arc of contact for the smaller pulley by following relation and
determine the correction factor for arc of contact. It is not advisable to use arc of contact
less than 1200 for V-belt drive.
ππ = 180π β 2 sinβ1 (π· β π
2πΆ)
Step 7. Determine the power rating by the given relationship or it may be taken from
table depending upon the type of cross-section. It depends upon speed of smaller pulley,
pitch diameter of the smaller pulley and the speed ratio.
Step 8. Calculate the modified power rating of the belt;
ππππππππ πππ€ππ π ππ‘πππ = πππ€ππ πππ‘πππ ππ π ππππ‘ (πΎπ) Γ πΉπ Γ πΉπ
Step 9. The number of belt is obtained by the following relationship
ππ’ππππ ππ ππππ‘ = π·ππ πππ πππ€ππ
ππππππππ πππ€ππ π ππ‘πππ=
π Γ πΉπ
πΎπ Γ πΉπ Γ πΉπ
Refer machine design data book for required data if needed.
33
Problems:
Problem 1: Write a computer program for selection of V Belt Drive.
Problem 2: It is required to select a V-belt drive to connect a 20-kW, 1440 rpm motor to a
compressor running at 480 rpm for 15 hours per day. Space is available for a centre distance
of approximately 1.2 m. Determine (i) the specifications of the belt; (ii) diameters of motor and
compressor pulleys;(iii) the correct centre distance; and(iv) the number of belts.
Problem3: The following data is given for an open-type V-belt drive:
diameter of driving pulley = 200 mm
diameter of driven pulley = 600 mm
groove angle for sheaves = 34Β°
mass of belt = 0.5 kg/m
maximum permissible tension in belt = 500 N
coefficient of friction = 0.2
contact angle for smaller pulley = 157Β°
speed of smaller pulley = 1440 rpm
power to be transmitted = 10 kW
How many V-belts should be used, assuming each belt takes its proportional part of the load?
34
9. Chain Drive
Objective: Selection of Chain Drive
Theory: A chain drive consists of an endless chain wrapped around two sprockets as shown in
Figure 1. A chain can be defined as a series of links connected by pin joints. The sprocket is a
toothed wheel with a special profile for the teeth. The chain drive is intermediate between belt
and gear drives. It has some features of belt drives and some of gear drives.
Figure 1. Chain Drive
Material: Alloy steel, stainless steel etc.
Design Equation/Data:
β’ Pitch Angle
πΌ =360
π§
where Ξ± = pitch angle (Degrees)
z = Number of teeth on sprocket
β’ Pitch circle diameter of sprocket
π· =π
sin(180
π )
where D = pitch circle diameter of
sprocket (mm)
P = pitch of chain (mm)
β’ Velocity Ratio
π =π1
π2
=π§2
π§1
where I =Velocity ratio
n1, n2 = speeds of rotation of driving and driven shafts (rpm)
z1, z2 = number of teeth on driving and driven sprockets
β’ Average velocity of chain
v =ΟDn
60 Γ 103=
zpn
60 Γ 103
where v =Average velocity of chain
(m/s)
n = Speed of rotation of sprocket (rpm)
35
β’ Length of chain
L = Ln Γ p
Ln = 2 (a
p) + (
z1 + z2
2) + (
z2 β z1
2Ο)
2
Γ (p
a)
a =p
4{[Ln β (
z1+z2
2)] +
β[Ln β (z1+z2
2)]
2
β 8 (z2βz1
2Ο)
2
}
L=Length of chain (mm)
Ln = Number of links in chain
a=Centre distance between axes of
driving and driven sprockets (mm)
π§1 = Number of teeth on the smaller sprocket
π§2 = Number of teeth on the larger sprocket
Refer machine design data book for required data if needed.
Problems:
Problem 1: Write a computer program for selection of Chain Drive.
Problem 2: It is required to design a chain drive with a duplex chain to connect a 15 kW, 1400
rpm electric motor to a transmission shaft running at 350 rpm. The operation involves moderate
shocks. (i) Specify the number of teeth on the driving and driven sprockets. (ii) Select a proper
roller chain. (iii) Calculate the pitch circle diameters of the driving and driven sprockets. (iv)
Determine the number of chain links. (v) Specify the correct centre distance. During
preliminary stages, the centre distance can be assumed to be 40 times the pitch of the chain.
Problem 3: It is required to design a chain drive to connect 5 kW, 1400 rpm electric motor to a
drilling machine. The speed reduction is 3:1. The centre distance should be approximately 500
mm. (i) Select a proper roller chain for the drive. (ii) Determine the number of chain links. (iii)
Specify the correct centre distance between the axes of sprockets.