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:

• 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.

• Read the procedure as set forth in the lab manual before you begin any design.

• It is compulsory to bring machine design data book in laboratory.

• For numerical calculations students have to bring scientific calculator without

fail.

• Use of mobile phones other than design purpose is strictly prohibited in the

laboratory.

• Drawing of the machine components is to prepared with AutoCAD.

• 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 Manufacturer’s 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:

• Eccentricity (e) = distance ObOj, where Ob= axis of bearing, Oj= axis of journal

• Radial clearance (c) = Rb- R , where Rb = radius of bearing , R= radius of journal

• Eccentricity ratio (ɛ) = 𝑒

𝑐

• Sommerfield number (S) = (𝑅

𝑐)

2 µ𝑁

𝑝, where µ = viscosity of lubricant, 𝑁 = shaft

speed

2

• Sliding velocity (U) = ωR , ω =2πN

• Film thickness (h) = AB which is the film thickness at B at angle θ from line of

action

h = Rb - 𝑅

𝑠𝑖𝑛𝜃 sin[𝜃 − 𝑠𝑖𝑛−1 (

𝑒

𝑅𝑠𝑖𝑛 𝜃)]

For simplicity assumed 𝑒

𝑅 << 1, so then

h = c(1+ ɛ cos θ)

L= length of bearing

• Infinitely long bearing approximation (L/2R) > 2,

p = (6µ𝑈𝑅ɛ

𝑐2 ) (2+ ɛ cosθ)𝑠𝑖𝑛 𝜃

(2+ ɛ2)(1+ ɛ cosθ)2

W𝑟 = 12µ𝑈𝐿 (𝑅

𝑐)

2 ɛ2

(2+ ɛ2)(1−ɛ2)

W𝑡 = 6µ𝑈𝐿 (𝑅

𝑐)

2 𝜋ɛ

(2+ ɛ2)(1−ɛ2)1

2⁄

W= 6µ𝑈𝐿 (𝑅

𝑐)

2 ɛ[𝜋2−ɛ2(𝜋2−4)]1

2⁄

(2+ ɛ2)(1−ɛ2)

Ф= 𝑡𝑎𝑛−1 (W𝑡

W𝑟) = (

𝜋√1−ɛ2

2ɛ)

𝑆= (2+ ɛ2)(1−ɛ2)

6𝜋ɛ(

1

(𝜋2(1−ɛ2))+4ɛ2)

𝑓= 𝑐

𝑅(

ɛ sinФ

2+

2𝜋2𝑆

(1−ɛ2)1

2⁄)

where, P = Pressure

W𝑟 = Radial load

W𝑡 = Normal load

W = Total load carrying capacity

Ф = Attitude angle

𝑓 = Coefficient of friction

• Infinitely short bearing approximation (L/2 R) < 1

p = 3µ𝑈

𝑅𝑐2 (𝐿2

4− 𝑦2)

ɛ sin 𝜃

(1+ɛ cos 𝜃 )2

W𝑟 = −µ𝑈𝐿3

𝑐2

ɛ2

(1−ɛ2)2

W𝑡 = µ𝑈𝐿3

4𝑐2

πɛ

(1−ɛ2)3

2⁄

W= µ𝑈𝐿3

4𝑐2

ɛ

(1−ɛ2)2[(𝜋2(1 − ɛ2) + 16ɛ2]

12⁄

Ф= 𝑡𝑎𝑛−1 (W𝑡

W𝑟) = (

𝜋√1−ɛ2

4ɛ)

𝑆= (1−ɛ2)

2

πɛ[(𝜋2(1−ɛ2)+16ɛ2]1

2⁄(

2𝑅

𝐿)

2

𝑓= 𝑐

𝑅(

2𝜋2𝑆

(1−ɛ2)1

2⁄)

𝑄𝐿 = ɛULc

F = µ𝑈𝐿𝑅

𝑐

2𝜋

(1−ɛ2)1

2⁄

where, P = Pressure

W𝑟 = Radial load

W𝑡 = Normal load

W = Total load carrying capacity

Ф = Attitude angle

𝑓 = Coefficient of friction

𝑄𝐿 = Leakage flow rate

F = Friction force

3

• 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 (𝑃𝑡) = 2πRNF

Temperature rise (ΔT) = 𝑃𝑡

𝐽𝜌𝑐𝑝𝑄𝐿

where J = mechanical equivalent of

heat

𝑐𝑝 = 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 20° 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 38°C is 100cS and at 100°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:

• Module of gear

P = 𝑧

𝑑

m = 𝑑

𝑧

d = mz

where ,d = pitch circle diameter (mm),

z = number of teeth on the gear

P = Diametral Pitch

m = Module (mm)

• 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 - - - -

• Gear Ratio and transmission ratio

i = 𝑛𝑝

𝑛𝑔 =

𝑧𝑔

𝑧𝑝

where, np = speed of pinion,

ng = speed of wheel,

zp = number of teeth on the pinion,

zg = number of teeth on the wheel

• Basic Relationship

P = 𝑧

𝑑

p = 𝜋𝑑

𝑧

P×p = π

a = 𝑚(𝑧𝑝+𝑧𝑔)

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

• Standard proportions of gear tooth for 20° 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

6

Whole depth h H = 2.25 m

Tooth thickness s s = 1.5708m

Tooth space …. 1.5708m

Fillet radius …. 0.4m

• Spur gears - Component of tooth force

Mt = 60×106(𝑘𝑊)

2𝜋𝑛𝑝

Pt = 2𝑀𝑡

𝑑𝑝

Pr = Pt tanα

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)

α = Pressure angle (°)

dp = pitch circle diameter of pinion (mm)

• Minimum number of teeth

Pressure angle (α) 14.5° 20° 25°

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 α

• Face width of tooth

Optimum range of face width : (8m) < b < (12m) or (b=10m)

• Beam Strength of gear tooth (Lewis’ equation)

Beam strength (Sb) indicates the maximum value of tangential force that the tooth can

transmit without failure:

𝑆𝑏 = 𝑚𝑛𝑏𝜎𝑏𝑌

𝜎𝑏 = 𝑆𝑒 = (1

3) 𝑆𝑢𝑡

Sb= beam strength of gear tooth (N)

σb= permissible bending stress (MPa)

Y= Lewis form factor based on virtual number of teeth

Se= endurance limit (MPa)

Sut=ultimate tensile strength (MPa)

• Wear strength of gear tooth (Buckingham’s Equation)

Q = 𝑧𝑔

𝑧𝑔+𝑧𝑝 (for external gear)

Q = 2𝑧𝑔

𝑧𝑔−𝑧𝑝 (for internal gear)

Sw = b.Q.dp.K

K = 𝜎𝑐

2𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼 (1

𝐸𝑝 + 1/𝐸𝑔)

1.4

Q = Ratio factor

𝑧𝑝 = number of teeth on pinion

𝑧𝑔 = number of teeth on wheel

Sw = wear strength of the gear tooth (N)


K = load-stress factor (MPa)

7

σc = Surface endurance strength of the material

(MPa)

α = pressure angle

Ep = modulus of elasticity of pinion materials

(MPa)

Eg = modulus of elasticity of wheel materials

(MPa)

• For steel gears with 20° pressure angle

K = 0.16 (𝐵𝐻𝑁

100)

2

where BHN = Surface hardness of gears

(Brinell hardness number)

According to G. Niemann: σc = 0.27(BHN) kgf/mm2 = 0.27 (9.81) (BHN) N/mm2

• Values of Modulus of elasticity and poisson’s ratio for gear materials

Material Modulus of elasticity E (MPa) Poisson’s 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

• Effective load on gear tooth

Tangential force due to rated torque or rated power (Pt)

Mt = 60×106(𝑘𝑊)

2𝜋𝑛𝑝

Pt = 2𝑀𝑡

𝑑𝑝

Pt = Tangential force due to rated torque (N)

Mt = rated torque (N-mm)

Kw = power transmitted by gears (kW)

np = speed of pinion (rpm)


Effective load on gear tooth (Peff) – Preliminary gear design

Peff = CS𝑃𝑡

𝐶𝑣

Peff = effective load on gear tooth (N)

CS = Service factor

Cv = Velocity factor

Effective load on gear tooth (Peff) – Final gear design

Peff = (CSPt+Pd) Pd = incremental dynamic load (N)

(Buckingham’s equation)

• 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

• Velocity factor (Cv)

For ordinary and commercially cut gears made with form cutters and with (v<10 m/s)

Cv = 3

3+𝑣

For accurately hobbed and generated gears with (v<20 m/s)

Cv = 6

6+𝑣

For precision gears with shaving, grinding, and lapping operations and with (v>20 m/s)

Cv = 5.6

5.6+√𝑣

𝑣 = 𝜋𝑑𝑝𝑛𝑝

60×103 where, 𝑣 = pitch line velocity (m/s)

Pd = 21𝑣(𝐶𝑒𝑏+𝑃𝑡)

21𝑣+√(𝐶𝑒𝑏+𝑃𝑡 )

Pd = dynamic load or incremental dynamic load (N)

𝑣 = 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)

• 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 = 𝑘

[1

𝐸𝑝+

1

𝐸𝑔]

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 14.5° full depth teeth)

k = 0.111 (for 20° full depth teeth)

k = 0.107 (for 20° stub teeth)

• Values of deformation factor C

Materials 14.5° full

depth teeth

20° full depth

teeth 20° 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


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 20° 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 20° 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 (ψ), normal pressure angle

(αn), normal module (m), Rotational speed (N)

• Basic equations of helical gears

pn = p𝑐𝑜𝑠𝜓

Pn = P/𝑐𝑜𝑠𝜓

mn = 𝑚𝑐𝑜𝑠𝜓

pa = p/𝑡𝑎𝑛𝜓

Cos ψ = tanαn/tanα αn

d = zmn/cos ψ

a = m𝑛(𝑧𝑝 + 𝑧𝑔)

2

where p= transverse circular pitch(mm)

pn= normal circular pitch(mm)

ψ = Helix angle

Pn= normal diametric pitch(mm)

P = tranverse diametric pitch(mm)

mn= normal module(mm)

m = transverse module(mm)

pa = axial pitch(mm)

αn = normal pressure angle (0)

11

𝑖 =𝑤𝑝

𝑤𝑔=

𝑧𝑔

𝑧𝑝

α=transverse pressure angle (0)

d=pitch circle diameter(mm)

z= number of teeth

a=center to center distance(mm)

i=speed ratio

𝑧𝑝 = number of teeth on pinion

𝑧𝑔 = 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.

• Standard proportions of helical gears

𝑑𝑎 = 𝑚𝑛 [𝑧

𝑐𝑜𝑠 𝜓+ 2]

𝑑𝑓 = 𝑚𝑛 [𝑧

𝑐𝑜𝑠𝜓− 2.5]

where, Addendum(ha)=mn

Dedendum(hf)=1.25mn

Clearance(c)=0.25mn

da=addendum circle diameter(mm)

df=dedendum circle diameter (mm)

• Components of tooth force

𝑀𝑡 =60𝑥106(𝑘𝑊)

2𝜋𝑛𝑝

𝑃𝑡 =2𝑀𝑡

𝑑𝑝

𝑃𝑟 = 𝑃𝑡 [𝑡𝑎𝑛𝛼𝑛

𝑐𝑜𝑠𝜓]

𝑃𝑎 = 𝑃𝑡𝑡𝑎𝑛𝜓

𝑑𝑝 =𝑧𝑝𝑚𝑛

𝑐𝑜𝑠𝜓

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)

αn= normal pressure angle (0)

• Design by Lewis and Buckingham’s Equations

Beam Strength of gear tooth (Lewis’ equation)

Beam strength (Sb) indicates the maximum value of tangential force that the tooth can

transmit without failure

𝑆𝑏 = 𝑚𝑛𝑏𝜎𝑏𝑌 Sb= beam strength of gear tooth (N)


Y= Lewis form factor based on virtual number

of teeth (z’)

12

𝜎𝑏 = 𝑆𝑒 = (1

3) 𝑆𝑢𝑡



• Wear strength of gear tooth (Buckingham’s equation)

Wear strength (Sw) indicates the maximum value of tangential force that the tooth can

transmit without pitting failure.

𝑆𝑤 =𝑏𝑄𝑑𝑝𝐾

𝑐𝑜𝑠2𝜓

Sw= wear strength of thr gear tooth (N)

Q= ratio factor


K= load-stress factor (MPa)

𝑄 =2𝑧𝑔

𝑧𝑔+𝑧𝑝 (for external gears)

𝑄 =2𝑧𝑔

𝑧𝑔−𝑧𝑝 (for internal gears)

zp= number of teeth on pinion

zg= number of teeth on wheel

𝐾 =𝜎𝑐

2𝑠𝑖𝑛𝛼𝑛𝑐𝑜𝑠𝛼𝑛(1𝐸𝑝

⁄ + 1𝐸𝑔

⁄ )

1.4

σc= surface endurance strength of the

material(MPa)

𝛼𝑛= normal pressure angle(0)

𝐸𝑝= modulus of elasticity of pinion material

(MPa)

𝐸𝑔= modulus of elasticity of wheel material

(MPa)

For steel gears with 200 pressure angle,

𝐾 ≈ 0.16 (𝐵𝐻𝑁

100)

2

BHN= surface hardness of gears (Brinell

Hardness Number)

According to G. Niemann,

σc=0.27(BHN) kgf/mm2 = 0.27(9.81)(BHN) N/mm2

• Effective Load on gear tooth

Tangential force due to rated torque or rated power (Pt)

𝑀𝑡 =60 ∗ 106(𝑘𝑊)

2𝜋𝑛𝑝

𝑃𝑡 =2𝑀𝑡

𝑑𝑝

Pt= tangential force due to rated troque (N)

Mt = rated torque(N-mm)




Effective load on gear tooth (Peff) – Preliminary gear design

𝑃𝑒𝑓𝑓 =𝐶𝑠𝑃𝑡

𝐶𝑣

𝐶𝑣 =5.6

5.6 + √𝑣

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) – Final gear design

𝑃𝑒𝑓𝑓 = (𝐶𝑠𝑃𝑡 + 𝑃𝑑) Pd= incremental dynamic load (N) (Buckingham’s

equation)


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:

• Pitch angle formulas

Pinion (𝛳𝑝) = tan−1 sin 𝛳

𝑉𝑅+cos 𝛳

VR= Velocity Ratio

𝜃 = the angle between the shafts.

Gear (𝛳𝑔) = tan−1sin 𝛳

1𝑉𝑅 + cos 𝛳

• Components of tooth force

15

𝑀𝑡 =60 ∗ 106(𝑘𝑊)

2𝜋𝑛𝑝

𝑃𝑡 =𝑀𝑡

𝑟𝑚

𝑃𝑟 = 𝑃𝑡 𝑡𝑎𝑛 𝛼 𝑐𝑜𝑠𝛾

𝑃𝑎 = 𝑃𝑡𝑃𝑡 𝑡𝑎𝑛 𝛼 𝑠𝑖𝑛 𝛾

𝑟𝑚 = [𝐷𝑝

2−

𝑏 sin 𝛾

2]

Suffix p is used for pinion

Mt = torque transmitted by gears (N-mm)



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)

𝑟𝑚= radius of pinion at the midpoint along the face width (mm)

𝛼 = pressure angle (0)

𝐷𝑝= pitch circle diameter at large end of the tooth (mm)

• Design by Lewis Equations

Beam Strength of gear tooth (Lewis’ equation)

Beam strength (Sb) indicates the maximum value of tangential force that the tooth can transmit

without failure

𝑆𝑏 = 𝑚𝑏𝜎𝑏𝑌 [1 −𝑏

𝐴0]

𝑤ℎ𝑒𝑟𝑒 𝐴0 = √𝐷𝑝

2

4+

𝐷𝑔2

4 𝑏 =

1

4𝐴0

Sb= beam strength of gear tooth (N)


Y= Lewis form factor based on formative

number of teeth

m= module at the large end of the tooth (mm)

𝜎𝑏 = 𝑆𝑒 = (1

3) 𝑆𝑢𝑡



• 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

• Equivalent number of Teeth for Bevel Gears- Tredgold’s Approximation:

𝑇𝑒 = 𝑇 sec 𝛳𝑝 𝑜𝑟 𝑔

• Outside diameter for the Bevel Gears (𝑫𝟎) = 𝐷𝑝 𝑜𝑟 𝑔 + 2𝑎 𝑐𝑜𝑠 𝛳𝑝 𝑜𝑟 𝑔

• The relation between 𝑨𝟎, 𝑫𝒑, 𝜭𝒑 is: sin 𝛳𝑝 =𝐷𝑝/2

𝐴0

• Wear strength of gear tooth (Buckingham’s equation)

Wear strength (Sw) indicates the maximum value of tangential force that the tooth can

transmit without pitting failure.

16

𝑆𝑤 = .75𝑏𝑄𝐷𝑝𝐾

𝑐𝑜𝑠2𝛾

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)

𝑄 =2𝑧𝑔

𝑧𝑔 + 𝑧𝑝 tan 𝛾

zp= number of teeth on pinion

zg= number of teeth on wheel

𝐾 =𝜎𝑐

2𝑠𝑖𝑛𝛼 cos 𝛼 (1𝐸𝑝

⁄ + 1𝐸𝑔

⁄ )

1.4

σc= surface endurance strength of the

material(MPa)

𝛼𝑛= normal pressure angle(0)

𝐸𝑝= modulus of elasticity of pinion material

(MPa)

𝐸𝑔= modulus of elasticity of wheel material

(MPa)

For steel gears with 200 pressure angle,

𝐾 ≈ 0.16 (𝐵𝐻𝑁

100)

2

BHN= surface hardness of gears (Brinell

Hardness Number)

According to G. Niemann,

σc=0.27(BHN)kgf/mm2 = 0.27(9.81)(BHN) N/mm2


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 20°. 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 −.684

𝑇𝐸 , where 𝑇𝐸 is equivalent number of teeth. Assume velocity factor 𝐶𝑣 =

4.5

4.5+𝑣 where 𝑣 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 90ᵒ bevel gearing arrangement. The pressure angle of the teeth is 14.5ᵒ. 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:

𝑞 =𝑑1

𝑚

𝑙 = 𝑝𝑥𝑧1

𝑝𝑥 = πm

𝑑2 = 𝑚𝑧2

tan ϒ =𝑙

𝜋𝑑1 =

𝑧1

𝑞

ϒ + ψ = 𝜋

2

a= 1

2 (𝑑1+𝑑2) =

𝑚

2 (𝑞+𝑧2)

i = 𝑧2

𝑧1

where, 𝑑1= pitch circle diameter of worm (mm)

𝑙= lead of worm (mm)

𝑝𝑥= axial pitch of worm (mm)

𝑑2= pitch circle diameter of worm wheel

(mm)

ϒ = lead angle of worm (o)

ψ = helix angle of worm (o)

a = centre to centre distance between worm

and worm wheel (mm)

i = speed ratio

• Dimensions of worm and worm wheel

ℎ𝑎1= m

ℎ𝑓1= (2.2𝑐𝑜𝑠ϒ − 1)𝑚

𝑐 = 0.2𝑚𝑐𝑜𝑠ϒ

𝑑𝑎1= m (q+2)

𝑑𝑓1= (𝑞 + 2 − 4.4𝑐𝑜𝑠ϒ)𝑚

where, ℎ𝑎1= addendum of worm (mm)

ℎ𝑓1= dedendum of worm (mm)

𝑐 = clearance (mm)

𝑑𝑎1= outside diameter of the worm (mm)

ℎ𝑓1= root diameter of the worm (mm)

• Dimensions of worm wheel

ℎ𝑎2= (2𝑐𝑜𝑠ϒ − 1)𝑚

ℎ𝑓2= (0.2𝑐𝑜𝑠ϒ + 1)𝑚

𝑑𝑎2= (𝑧2+ 4𝑐𝑜𝑠ϒ − 2)m

𝑑𝑓2= (𝑧2 − 2 − 0.4𝑐𝑜𝑠ϒ)𝑚

𝐹= 2m√(𝑞 + 1)

𝑙𝑟 = (𝑑𝑎1 + 2𝑐)𝑠𝑖𝑛−1 [𝐹

(𝑑𝑎1 + 2𝑐)]

where, ℎ𝑎2= addendum at throat of worm wheel

(mm)

ℎ𝑓2= dedendum in medium plane (mm)

𝑑𝑎2= throat diameter of the worm wheel (mm)

𝑑𝑓2= root diameter of the worm wheel (mm)

F = effective face width of worm wheel (mm)

𝑙𝑟= length of root of worm wheel teeth (mm)

• Forces acting on worm and worm wheel

(𝑃1)𝑡 =2𝑀𝑡

𝑑1 (𝑃1)𝑎 = axial component on the worm (N)

(𝑃1)𝑡 = tangential component on the worm (N)

19

(𝑃1)𝑎 = (𝑃1)𝑡

(cos 𝛼𝑐𝑜𝑠ϒ − µ𝑠𝑖𝑛ϒ)

(cos 𝛼𝑠𝑖𝑛ϒ + µ𝑐𝑜𝑠ϒ)

(𝑃1)𝑟 = (𝑃1)𝑡

𝑠𝑖𝑛𝛼

(cos 𝛼𝑠𝑖𝑛ϒ + µ𝑐𝑜𝑠ϒ)

(𝑃2)𝑡 = (𝑃1)𝑎

(𝑃2)𝑎 = (𝑃1)𝑡

(𝑃2)𝑟 = (𝑃1)𝑟

(𝑃1)𝑟 = radial component on the worm (N)

(𝑃2)𝑎 = axial component on the worm wheel (N)

(𝑃2)𝑡 = tangential component on the worm wheel

(N)

(𝑃2)𝑟 = radial component on the worm wheel (N)

𝑀𝑡 = torque transmitted by gears (N-mm)

α = normal pressure angle (o)

µ = coefficient of friction

• Friction in worm gears

𝑣𝑠= 𝜋𝑑1𝑛1

60000𝑐𝑜𝑠ϒ

η = (cos 𝛼−µ𝑡𝑎𝑛ϒ)

(cos 𝛼+µ𝑐𝑜𝑡ϒ)

𝑣𝑠= rubbing velocity (m/s)

𝑛1= speed (rpm)

η =efficiency of worm gear drive


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

20°. 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) (𝛒)

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:

• Torque analysis

Maximum fluctuation of speed = ωmax -

ωmin

Cs= ωmax − ωmin

ω

ω = ωmax + ωmin

2

ωmax = Maximum speed during a cycle

ωmin = Minimum speed during a cycle

Cs= Coefficient of fluctuation of speed

U0 = Change in kinetic energy

21

U0 = Iω2Cs

Cs = 2 (nmax − nmin )

(nmax + nmin )

ω 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 𝐂𝐬

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

• 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 – a1

Energy at C = U – a1 + a2

22

Energy at D = U – a1 + a2 – a3

Energy at E = U – a1 + a2 – a3 + a4

Energy at F = U – a1 + a2 – a3 + a4 – a5

Energy at G = U – a1 + a2 – a3 + a4 – a5 + a6

Energy at H = U – a1 + a2 – a3 + a4 – a5 + a6 – a7 = U

Maximum fluctuation of energy = maximum kinetic energy – 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 = (2π)Tm (for two stroke engine)

Work done/cycle = (4π)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

• Solid disk flywheel

Figure 3. Solid Disk Flywheel

23

I = mR

2

2

m = πR2tρ

I = π

2ρtR4

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)

ρ = mass density of flywheel material

(kg/m3)

Two principal stresses in the rotating disk is given by,

σt = ρv2

106 (μ+3

8) [1 − (

3μ+1

μ+3) (

r

R)

2

]

σr = ρv2

106 (μ+3

8) [1 − (

r

R)

2

]

where, σt = tangential stress at radius r

(N/mm2)

σr = radial stress at radius r (N/mm2)

μ = Poisson’s ratio (0.3 for steel and 0.27

for cast iron)

v = peripheral velocity (m/s) (Rω)

The maximum tangential stress and

maximum radial stress are equal and both

occur at (r = 0)

Therefore,

(σt)max. = (σr)max. = ρv2

106 (μ+3

8)

• Rimmed flywheel

Ir = KI = 𝑈0𝐾

𝜔2𝐶𝑠

[since, I = 𝑈0

𝜔2𝐶𝑠]

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 = 𝜔R

R <30

𝜔

𝜎𝑡 = 𝑃1

𝐴1

𝑃1 = 2

3[

(1000)𝑚𝑣2

𝐶]

m = bt𝜌

𝜎𝑡𝑟 = (1000)𝑚𝑣2

𝑏𝑡[1 −

cos ∅

3𝐶 sin 𝛼±

2(1000)𝑅

𝐶𝑡(

1

𝛼−

cos ∅

sin 𝛼)]

where, 𝑃1 = Tensile force in each spoke (N)

𝐴1 = Cross-sectional area of spoke (mm2)

𝜎𝑡 = 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

𝜌 = mass density of flywheel material

(kg/mm3)

b = width of the rim (mm)

t = thickness of the rim (mm)

𝜎𝑡𝑟= tensile stress in the rim (N/mm2)

R = mean radius of rim (m)

2𝛼 = angle between two consecutive spokes

(rad)

∅ = angle from centre line between spokes to

the section where the stress is being

calculated (rad)

• Expression for Constant C:

For 4 spokes: 2𝛼 = 𝜋

2

C = [72960 𝑅2

𝑡2 + 0.643 + 𝐴

𝐴1]


3

C = [20280 𝑅2

𝑡2 + 0.957 + 𝐴

𝐴1]


4

C = [9120 𝑅2

𝑡2+ 1.274 +

𝐴

𝐴1]

where, A = cross-sectional area of rim (mm2)

A1 = cross-sectional area of spokes (mm2)


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 – 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 (𝜌 = 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 Poisson’s 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 Manufacturer’s 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 Bearings– 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.

• Cylindrical & Needle roller – pure radial load.

• Thrust (cylindrical roller, ball), four point angular contact ball bearings – pure axial load.

• Taper roller, spherical roller, angular contact ball bearings – combined Load.

• Cylindrical roller, angular contact ball bearing– high speed.

• Deep groove, angular contact, and cylindrical roller bearing – high running accuracy.

Design equations/data:

• Equivalent dynamic load:

𝑃 = 𝑋𝑉𝐹𝑟 + 𝑌𝐹𝑎

𝑃 = equivalent dynamic load (N)

𝐹𝑟 = radial load acting on bearing (N),

𝐹𝑎 = axial or thrust load acting on bearing

𝑉 = race- rotation factor

𝑋 = radial factor,

𝑌= thrust factor

28

• Load life equation:

𝐿10 = (𝐶

𝑃)

𝑝

where, 𝐿10 = rating bearing life (in

million revolutions)

C = dynamic load capacity (N)

p = 3 (for ball bearings)

p = 10/3 (for roller bearing)

• Life in hours:

𝐿10 =60𝑛𝐿10ℎ

106

where, 𝐿10ℎ = rated bearing life (hours)

𝑛 = speed of rotation (rpm)

• Cyclic loads and speed:

𝑃𝑒 = √[𝑁1𝑃1

3+𝑁2𝑃23+⋯

𝑁1+𝑁2+⋯

3]

𝑃𝑒 = √[∑ 𝑁𝑃3

∑ 𝑁]

3

where N = 𝑁1 + 𝑁2 + ⋯ 𝑁𝑛

𝑃𝑒 = equivalent dynamic load for

complete work cycle (N)

𝑃1, 𝑃2 … 𝑃𝑛 = dynamic load during first,

second, .... nth element of work cycle

𝑁1, 𝑁2 … 𝑁𝑛= number of revolutions

completed by first, second.... nth element

of work cycle

N = life of complete work cycle (rev)

• Cyclic loads and speeds (continuous variation of load):

𝑃𝑒 = [1

𝑁∫ 𝑃3 𝑑𝑁]1 3⁄

• Bearing with probability of survival other than 90%:

𝑅 = 𝑒−(𝐿 𝑎⁄ )𝑏, where 𝑅 = reliability (in fraction), L = corresponding life (in million of

revolution), 𝑎 and 𝑏 = constants (𝑎 = 6.84 and 𝑏 = 1.17)

(𝐿

𝐿10) = [

log𝑒(1

𝑅)

log𝑒(1

𝑅90)]1 𝑏⁄ , where 𝐿10 = life corresponding to a reliability of 90% or

𝑅90, 𝑅90 = 0.9

• System reliability: 𝑅𝑆 = (𝑅)𝑁, where 𝑁 = number of bearings in the system (each

having the same Reliability 𝑅), 𝑅𝑆 = reliability of the complete system


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 elements—driving 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 :

• 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

• 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,

𝐹1 − 𝑚 𝑣2

𝐹2 − 𝑚𝑣2= 𝑒𝑓𝛼/𝑠𝑖𝑛𝑒𝛽

• 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.

• 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

• 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:

• Angle of wrap correction factor: The power rating of V-belts are based on angle of

wrap, α =180ο Hence, Angle of wrap correction factor is incorporated when α is not

equal to 180o .

• 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

𝐷𝑒𝑠𝑖𝑔𝑛 𝑃𝑜𝑤𝑒𝑟 = 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑(𝑟𝑎𝑡𝑒𝑑) 𝑝𝑜𝑤𝑒𝑟 𝑃 × 𝐹𝑎

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

𝐷

𝑑=

𝑛

𝑁

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.

𝜃𝑑 = 180𝑜 − 2 sin−1 (𝐷 − 𝑑

2𝐶)

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;

𝑀𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝑃𝑜𝑤𝑒𝑟 𝑅𝑎𝑡𝑖𝑛𝑔 = 𝑝𝑜𝑤𝑒𝑟 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝑎 𝑏𝑒𝑙𝑡 (𝐾𝑊) × 𝐹𝑐 × 𝐹𝑑

Step 9. The number of belt is obtained by the following relationship

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑒𝑙𝑡 = 𝐷𝑒𝑠𝑖𝑔𝑛 𝑃𝑜𝑤𝑒𝑟

𝑀𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝑃𝑜𝑤𝑒𝑟 𝑅𝑎𝑡𝑖𝑛𝑔=

𝑃 × 𝐹𝑎

𝐾𝑊 × 𝐹𝑐 × 𝐹𝑑


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 = 34°

mass of belt = 0.5 kg/m

maximum permissible tension in belt = 500 N

coefficient of friction = 0.2

contact angle for smaller pulley = 157°

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:

• Pitch Angle

𝛼 =360

𝑧

where α = pitch angle (Degrees)

z = Number of teeth on sprocket

• Pitch circle diameter of sprocket

𝐷 =𝑃

sin(180

𝑍 )

where D = pitch circle diameter of

sprocket (mm)

P = pitch of chain (mm)

• Velocity Ratio

𝑖 =𝑛1

𝑛2

=𝑧2

𝑧1

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

• Average velocity of chain

v =πDn

60 × 103=

zpn

60 × 103

where v =Average velocity of chain

(m/s)

n = Speed of rotation of sprocket (rpm)

35

• Length of chain

L = Ln × p

Ln = 2 (a

p) + (

z1 + z2

2) + (

z2 − z1

2π)

2

× (p

a)

a =p

4{[Ln − (

z1+z2

2)] +

√[Ln − (z1+z2

2)]

2

− 8 (z2−z1

2π)

2

}

L=Length of chain (mm)

Ln = Number of links in chain

a=Centre distance between axes of

driving and driven sprockets (mm)

𝑧1 = Number of teeth on the smaller sprocket

𝑧2 = Number of teeth on the larger sprocket


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.

MACHINE DESIGN LAB - Indian Institute of Technologyshibayan/MCC 15204 Machine...Table of Contents...

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