Vel Tech High Tech Dr. RR & Dr. SR Engineering College

Department of Mechanical Engineering

Design of Machine Elements (ME6503) course material

Class: V- Sem. Mechanical Prepared By: Dr P. Nagasankar

ME8594 DESIGN OF MACHINE ELEMENTS L T P C

3 0 0 3

OBJECTIVES

• To familiarize the various steps involved in the Design Process

• To understand the principles involved in evaluating the shape and dimensions of a component to satisfy

functional and strength requirements.

• To learn to use standard practices and standard data

• To learn to use catalogues and standard machine components

(Use of P S G Design Data Book is permitted)

UNIT I STEADY STRESSES AND VARIABLE STRESSES IN MACHINE MEMBERS 10

Introduction to the design process - factors influencing machine design, selection of materials based on

mechanical properties - Preferred numbers, fits and tolerances – Direct, Bending and torsional stress

equations – Impact and shock loading – calculation of principle stresses for various load combinations,

eccentric loading – curved beams – crane hook and ‘C’ frame- Factor of safety -theories of failure – Design

based on strength and stiffness – stress concentration – Design for variable loading.

UNIT II SHAFTS AND COUPLINGS 8

Design of solid and hollow shafts based on strength, rigidity and critical speed – Keys, keyways and splines

- Rigid and flexible couplings.

UNIT III TEMPORARY AND PERMANENT JOINTS 9

Threaded fasteners - Bolted joints including eccentric loading, Knuckle joints, Cotter joints – Welded joints,

riveted joints for structures - theory of bonded joints.

UNIT IV ENERGY STORING ELEMENTS AND ENGINE COMPONENTS 9

Various types of springs, optimization of helical springs - rubber springs - Flywheels considering

stresses in rims and arms for engines and punching machines- Connecting Rods and crank shafts.

UNIT V BEARINGS 9

Sliding contact and rolling contact bearings - Hydrodynamic journal bearings, Somerfield Number,

Raimondi and Boyd graphs, -- Selection of Rolling Contact bearings.

TOTAL: 45 PERIODS

OUTCOMES:

Upon completion of this course, the students can able to successfully design machine Components

UNIT I STEADY STRESSES AND VARIABLE STRESSES IN MACHINE

MEMBERS

Stress

Stress, σ = P/A N/mm2 PSGDB -7.1

where P= Force or load acting on a body and A= cross sectional area of the body

Strain

PSGDB -7.1

Tensile Stress and Strain

Let P = Axial tensile force acting on the body,

A = Cross-sectional area of the body,

l = Original length, and

δl= Increase in length.

Tensile stress, σ = P/A and tensile strain, εt= δl/ l

Young's Modulus or Modulus of Elasticity

Hooke's law* states that when a material is loaded within elastic limit, the stress is

directly proportional to strain, i.e.

PSGDB -7.1

Poisson's Ratio

Torsional Shear Stress

PSGDB -7.1

where τ = Torsional shear stress induced at the outer surface of the shaft or maximum

shear stress, N/mm2

r = Radius of the shaft, mm

T = Torque or twisting moment, N-mm

J = Second moment of area of the section about its polar axis or polar moment of inertia,

mm4

C = Modulus of rigidity for the shaft material, N/mm2

l = Length of the shaft, mm and θ = Angle of twist in radians on a length l

Impact Stress

N/mm2

BENDING STRESS IN STRAIGHT BEAMS

Consider a straight beam subjected to a bending moment M as shown in figure. The

following assumption are usually made delivering the bending formula

(i) The material of the beam is perfectly homogeneous and isotropic.

(ii) The material of the beam obeys Hooks law.

(iii) The Young’s modulus is same in tension and compression.

(iv) The loads are applied in the plane of bending.

The bending equation is given by

PSGDB -7.1

M- Bending moment acting at the given section N mm

-bending stress, N/mm2

I-moment of inertia of the cross section about the neutral axis mm4

y- Distance from the neutral axis to the extreme fiber mm

E- Young’s modulus of the material of the beam N/mm2

R- Radius of curvature of the beam mm

BENDING STRESS IN CURVED BEAMS

Consider a curved beam subjected to a bending moment M as shown in figure. The

general expression for bending stress (b)in a curved beam at any fibre at a distance y

from the neutral axis is

N/mm2 PSGDB -6.2

M- Bending moment acting at the given section about the centroidal axis N-mm

A- Area of cross-section mm2

e- Radius of curvature of the neutral axis mm

R- Radius of curvature of the centroidal axis, mm

Rn- radius of curvature of the neutral axis, mm

y- Distance from the neutral axis to the fiber under consideration. It is positive for the

distances towards the center of curvature and negative for the distance away from the

center of curvature.

STRESS CONCENTRATION

σmax = maximum stress/normal stress, Kt = Stress concentration factor, (it depends

upon material and geometry of the part.)

Methods of Reducing Stress Concentration

• Avoiding sharp corners

• Providing fillets

• Use of multiple holes instead of single hole.

• Undercutting the shoulder part

PRINCIPLE STRESSES AND PRINCIPLE PLANES

The planes which have no shear stress are known as principle planes (τ=0)

The normal stresses acting on the principle planes are known as principle stresses.

Two principle stresses are

(N/mm2) PSGDB -7.2

Maximum shear stress

N/mm2 PSGDB -7.2

APPLICATION OF PRINCIPLE STRESSES AND PRINCIPLE PLANES

Maximum tensile stress

N/mm2 PSGDB -7.2

Maximum compressive stress

N/mm2 PSGDB -7.2

Maximum shear stress

N/mm2 PSGDB -7.2

THEORIES OF FAILURE

According to the important theories the failure takes place when a certain limiting

value is reached by one of the following

(i) Maximum principal stress (or) maximum normal stress (or) Ranking theory

Failure occurs when the maximum normal stress is equal to the tensile yield

strength.

PSGDB -7.3

This theory is based on failure in tensile or compression and ignores the

possibility of failure due to shearing stress, ductile material mostly fail by

shearing. So this theory is used for brittle material.

(ii) Maximum shear theory (or) Guest’s theory (or) Coloumb theory

Failure occurs when the maximum shear stress developed in the machine

member becomes equal to the maximum shear stress at yielding in a tensile test.

PSGDB -7.3

This theory is mostly used for ductile materials.

(iii) Maximum strain theory (or) st Venant’s theory

Failure occurs when the maximum strain in the member equal in the tensile

yield strain.

PSGDB -7.3

(iv) Maximum strain energy theory

Failure is induced in the member when the strain energy stored per unit volume

of the member becomes equal to the strain energy per unit volume at the yield

point.

PSGDB -7.3

(v) Distortion energy theory (Von mises henky theory)

PSGDB -7.3

UNIT – I (PART –A)

1. Describe the material properties of hardness, stiffness and resilience. (NOV 13)

Hardness- It embraces many different properties such as resistance to wear, scratching,

deformation and machinability, Stiffness is the ability of a material to resist elastic deflection.

Resilience- It is the property of the material to absorb energy and to resist shock and impact loads.

2. What are the methods used to improve fatigue strength? (NOV 13)

a) Maximum load to be applied on a machine member should be far less than its endurance limit.

b) working temperature should be properly maintained.

3. What is ‘Adaptive design? Where is it used? Give examples. (NOV 12)

Adaptive design is the process in which the design parameters of existing machine-member or any

engineering devices is slightly modified and a new member is formed without changing much the

existing structure but the efficiency and model can be improved to suit for new trend. Various

models of bi-cycles, watches, etc

4. State the difference between straight beams and curved beams. (NOV 12)

In straight beam the neutral axis coincide with the centroidal axis.

In curved beam the neutral axis does not coincide with the centroidal axis.

5. List the various steps involved in the design of a machine elements.(MAY 12)

Recognition of need, synthesis, force analysis, material selection, design of elements,

modification, detailed drawing and manufacturing.

6. Compare the characteristics of Gerber curve Soderberg and Goodman lines.(MAY 12)

Gerber curve is a parabolic curve, used for brittle material, Soderberg line is used for finding stress for

ductile material and Goodman line is for brittle material

7. What are the factors to be considered for deciding the magnitude of factor of safety?(MAY13)

Type of product and effects of its failure

8. List out the methods of reducing stress concentration factor. (MAY 13)

i) Methods for reducing stress concentration factor in cylindrical members with shoulders,

ii) Methods for reducing stress concentration factor in cylindrical members with threads,

iii) Methods for reducing stress concentration factor in cylindrical members with holes.

9. Which theory of failure is suitable for the design of brittle materials? (DEC 15)

Rankine’s theory or Maximum principal stress theory.

10. What are the common materials used in mechanical engineering design? (DEC 15)

Cast iron, steel, wrought iron, copper, aluminium, carbon steel, alloy steel

11) Define creep.

When a part is subjected to a constant stress at high temperature for a long period of time,

it will undergo a slow and permanent deformation called creep.

12) Give some methods of reducing stress concentration. (Nov 2012)

i. Avoiding sharp corners. ii. Providing fillets.

iii. Use of multiple holes instead of single hole iv. Undercutting the shoulder parts.

13) What is an S-N Curve?

An S- N curve has fatigue stress on Y axis and number of loading cycles in X

axis. It is used to find the fatigue stress value corresponding to a given number of cycles

14) Differentiate between repeated stress and reversed stress.

Repeated stress refers to a stress varying from zero to a maximum value of same nature.

Reversed stress of cyclic stress varies from one value of tension to the same value of

compression.

15) Distinguish between brittle fracture and ductile fracture.

In brittle fracture, crack growth is up to a small depth of the material. In ductile fracture

large amount of plastic deformation is present to a higher depth

UNIT – I (PART-B)

1.) Describe the design process

(i) Recognition of need

It explains why the new product has to be developed; why the existing product has to be improved;

how it satisfies the human needs. Here the needs are examined thoroughly and identified.

(ii) Definition of problem

After identifying the needs, the problem should be defined clearly in terms of specifications

required. Examples: #Dimensions required, #Stress required, #Materials required, #Speed, feed required,

#Hardness required, #Power required, etc.

(iii) Synthesis

After defining the problem, the relationship between various parameters is to be considered. After

considering all parameters, the products are to be modeled (i.e., dimensioned/ shaped).

(iv) Analysis and optimization

Analysis tells us to find whether the design satisfy the requirements (or) solve the problem.(It can be

done through the mechanical software like ANSYS).

By optimization technique, we can check whether the problem has got optimal solution. If it is not

optimized, the problem has to be iterated again and again, i.e., the synthesis part has to be changed again

and again until we get optimal solution.(Performance).

(v) Evaluation and presentation

In evaluation part, the new/improved system has to be proved, i.e., the system has to be operated in

real condition or at least it has to be tested in laboratory.In presentation part, the improved design has to

be communicated properly to all i.e., people who are going to produce or buy the product .

2) Describe the factors influencing machine design

1. Type of load and stresses caused by the load. The load, on a machine component, may be in

acting several ways due to which the internal stresses are set up. The various types of load and

stresses (like static, variable, impact/shock) may be considered.

2. Motion of the parts or kinematics of the machine. The operation of any machine depends upon

(a) rectilinear motion like unidirectional and reciprocating motions. (b) Curvilinear motion like

rotary, oscillatory & SHM.(c) Constant/variable velocity or acceleration.

3. Selection of materials. It is essential that a designer should have thorough knowledge of the

properties of the materials and their behaviour under working conditions.

4. Form and size of the parts. The smallest practicable cross-section may be used, but it may be

checked that the stresses induced in the designed cross-section are reasonably safe.

5. Frictional resistance and lubrication. There is always loss of power due to frictional resistance

Therefore, all frictional surfaces should be identified and lubricated.

6. Convenient and economical features. .The operating features like the starting, and stopping

levers should be located on the basis of convenient. If parts are to be changed or replaced, easy

access should be provided. The economical operation of a machine should be carefully studied.

7. Use of standard parts. The standard parts which are interchangeable should be used wherever and

whenever possible; parts for which patterns are already in existence such as gears, pulleys, bearings,

screws, nuts and pins. We can avoid the delay caused by manufacturing these parts.

8. Safety of operation. Some machines are dangerous to operate, especially those which are speeded

up should be provided with safety devices to avoid accidents.

9. Workshop facilities. A design engineer should provide all the facilities in the workshop, in order to

avoid the necessity of having work done in some other workshop.

10. Number of articles to be manufactured. The number of articles to be manufactured in a machine

should be clearly stated by the design engineer to avoid overhead expenses

11. Cost of construction. The cost of construction of an article is the most important consideration

involved in design.

12. Assembling. Every machine or structure must be assembled as a unit before it can function. Large

units must often be assembled, tested and erected in suitable location.

3. An unknown weight falls through 10 mm on a collar rigidly attached to the lower end of vertical bar

3m long and 600mm2cross section. If the maximum instantaneous extension is 2 mm, what is the

corresponding value of stress and the unknown weight?.Take E= 2 x105 N/mm2.

4. A shaft, as shown in Fig., is subjected to a bending load of 3kN, pure torque of 1000 N-m and an axial

pulling force of 15 kN. Calculate the stresses at A and B

5. Determine the thickness of a 120 mm wide uniform plate for safe continuous operation if the plate is

to be subjected to a tensile load that has a maximum value of 250 kN and a minimum value of 100 kN.

The properties of the plate material are as follows: Endurance limit stress = 225 MPa, and Yield point

stress = 300 MPa. The factor of safety based on yield point may be taken as 1.5.

6. Determine the diameter of a circular rod made of ductile material with a fatigue strength (complete stress

reversal), σe = 265 MPa and a tensile yield strength of 350 MPa. The member is subjected to a varying axial

load from Wmin= – 300 kN to Wmax= 700 kN and has a stress concentration factor=1.8. Take FOS = 2.

7. A circular bar of 500 mm length is supported freely at its two ends. It is acted upon by a central concentrated

cyclic load having a minimum value of 20 kN and a maximum value of 50 kN. Determine the diameter of bar by

taking a factor of safety of 1.5, size effect of 0.85, surface finish factor of 0.9. The material properties of bar is

given by: ultimate strength of 650 MPa, yield strength of 500 MPa and endurance strength of 350MPa.

8. A 50 mm diameter shaft is made from carbon steel having ultimate tensile strength of 630MPa. It is subjected

to a torque which fluctuates between 2000 Nm to -800 Nm. Using Soderberg method, calculate the factor of

safety. Assume suitable values for any other data needed.

9. A Simple supported beam has a concentrated load at the centre which fluctuates from a value of P to 4 P. The

span of the beam is 500 mm and its cross-section is circular with a diameter of 60 mm. Taking for the beam

material an ultimate stress of 700 MPa, a yield stress of 500 MPa, endurance limit of 330 MPa for reversed

bending, and a factor of safety of 1.3, calculate the maximum value of P. Take a size factor of 0.85 and a surface

finish factor of 0.9.

e = rc - rn = 100 - 91.024 = 8.976mm

Section A-B is subjected to a combination of direct load and bending, due to the eccentricity of force.

Stress due to direct load will be,

y = rn – r = 91.0 24 – r

Mb = 22X103X1 00 = 2.2X106 N-mm

11. Determine the value of “t” in the cross section of a curved beam as shown in fig such that t he normal

stress due to bending at the extreme fibers are numerically equal.

Given data;

Inner radius = ri=150mm, Outer radius, ro=150+40+100=290mm

Solution;

From Figure Ci + CO = 40 + 100 = 140mm… (1)

Since the normal stresses due to bending at the extreme fiber are numerically equal, we have,

(i) eCi=

= 0 .51724Co…………… (2)

Radius of neutral axis, rn=

rn =197.727 mm

ai = 40mm; bi = 100mm; b2 =t;

ao = 0; bo = 0; ri = 150m m; ro = 290mm;

= 4000+100𝑡

23.639+0.423𝑡

i.e., 4674.069+83.6 1t = 4000+100t;t = 41.126mm.

UNIT II -DESIGN OF SHAFTS AND COUPLINGS

SHAFT

A shaft is a rotating machine element which is used to transmit power from one

place to other place. Carbon steels of grade 40C8, 45C8, 50C4, 50C12 are normally used

as shaft materials.

Material properties

➢ It should have high strength

➢ It should have good machinability.

➢ It should have low notch sensitivity factor.

➢ It should have good heat treatment properties.

➢ It should have high wear resistance.

Stresses in shaft

Following stresses are induced in the shaft.

1. Shear stress due to transmission of torque

2. Bending stress due to forces acting upon machine elements like gears, pulleys etc.

3. Stresses due to combined torsional and bending loads.

DESIGN OF SHAFTS

The shaft may be designed on the basis of 1. Strength; 2. Rigidity and stiffness

In designing shaft on the basis of strength, the following cases are considered

1. Shafts subjected to twisting moment only

2. Shaft subjected to bending moment only

3. Shaft subjected to combined twisting moment and bending moment

4. Shaft subjected to fluctuating loads

1. SHAFTS SUBJECTED TO TWISTING MOMENT ONLY

PSGDB 7.21

T1- Tension in the tight side N

T2- Tension in the slack side N

R- Radius of the pulley mm

2. SHAFT SUBJECTED TO BENDING MOMENT ONLY

PSGDB 7.21

2. SHAFT SUBJECTED TO COMBINED TWISTING MOMENT AND

BENDINGMOMENT

When the shaft is subjected to combined twisting moment ad bending moment then

the shaft must be designed on the basic of two moments simultaneously

For solid shaft

PSGDB 7.21

KEY

A key is a piece of mild steel inserted between the shaft and hub or boss of the pulley

to connect these together in order to prevent relative motion between them.

TYPES OF KEYS

1. Sunk key,

2. Saddle key,

3. Tangent key,

4. Round key

5. Splines

SUNK KEYS The sunk keys are provided half in the keyway of the shaft and half in

the keyway of the hub or boss of the pulley.

TYPES OF SUNK KEYS

1. Rectangular sunk key

2. Square sunk key -The only difference from the rectangular sunk key is the width

and thickness is equal w=t=d/2

3. Parallel sunk key -The parallel sunk key may be of rectangular or square cross

section. The cross section is uniform in width and thickness throughout length.

4. Gib head key- A gib head key is similar to a square or rectangular key but it has a

head at one end, generally at the larger end of the taper sunk key. The gib head is used

for driving the key while assembling or disassembling.

5. Feather key -Feather key is used where it is necessary to slide a keyed gear, pulley

assembly along the shaft. Keys are tight fitted or screwed on the shaft.

6. Woodruff key- A woodruff key is used to transmit small amount of torque in

automotive and machine tool industries. The keyway in the shaft is milled in a curved

shape whereas the keyway in the hub is usually straight. The main advantage of this

key is that it will align itself in the keyway.

SADDLE KEYS

The saddle keys are of the following two types:

1. Flat saddle key, and 2. Hollow saddle key

Force acting on sunk key

Strength of a Sunk Key

A key connecting the shaft and hub

Let T = Torque transmitted by the shaft, N-mm

F = Tangential force N; d = Diameter of shaft, mm

l = Length of key, mm; w = Width of key.mm

t = Thickness of key, mm and

τ and σc= shear and crushing stress for material of key, N/mm2

F = Area resisting shearing × Shear stress = l x w x τ

Therefore, Torque transmitted by the shaft,

Considering crushing of the key, the tangential crushing force acting at the

circumference of the shaft, i.e., F= area resisting crushing X crushing stress

Therefore, torque transmitted by the shaft

When the key material is same as that of the shaft, then τ = τ1. So, l = 1.571 d.

DESIGN OF COUPLING

TYPES OF SHAFT COUPLINGS

1. Rigid coupling -

It is used to connect two shafts which are perfectly aligned. The types are

➢ Sleeve (or) muff coupling

➢ Clamp(or) split muff (or) compression coupling

➢ Flange coupling

2. Flexible coupling

It is used to connect two shafts having lateral and angular misalignments. They are:

➢ Bushed pin type coupling

➢ Universal coupling

➢ Oldham coupling

SLEEVE (or) MUFF COUPLING

It is made of cast iron. It consists of a hollow cylinder whose inner diameter is that

same as that of the shaft. It is fitted over the ends of two shafts by means of a gib head

key. The power transmitted from one shaft to two other shafts by a key and a sleeve.

Outer diameter of sleeve D=2d+13mm

Length of sleeve L=3.5d; d- diameter of shaft

DESIGN OF MUFF COUPILNG

1. Design for sleeve

The sleeve is designed by considering it as a hollow shaft

2. Design for key

The length of coupling key is at least equal to the length of the sleeve. The

coupling key is usually made into two parts so that the length of key in each shaft

after that the induced shearing and crushing stresses may be checked.

CLAMP (or) COMPRESSION COUPLING

In this case the muff or sleeve is made into two halves are bolted together. The halves of

the muff are made of cast iron. The shaft end is made to abut each other and a single key is

fitted directly in the keyway of both the shaft. Both the halves are held together by means

of mild steel bolts and nuts. The number of bolt may be two or four or six.

Diameter of muff D=2d+13mm, Length of muff or sleeve L=3.5d

DESIGN OF CLAMP (or) COMPRESSION COUPILNG

1. Design for sleeve

It is similar to “Design for sleeve” of muff coupling.

2. Design for key

It is similar to “Design for key” of muff coupling

3. Design of clamping bolts

Force exerted by each bolt

Force exerted by each side of the shaft

Torque transmitted by the coupling

Where T-torque transmitted by the shaft, N-mm

d- diameter of shaft, mm; db- root or effective dia of bolt, mm; n- Number of bolt

σ- Permissible stress for bolt , N/mm2

µ-co-efficient of friction between the muff and shaft; L- Length of muff , mm

FLANGE COUPLING

A flange coupling usually applied to a coupling having two separate cast iron flanges.

Each flange is mounted on the shaft and keyed to it. The faces are turned up at right

angle to the axis of the shaft. One of the flanges has a projected portion and the other

flange has a corresponding recess. This helps to bring the shaft into line and to maintain

alignment. The two flanges are coupled together by means of bolt and nuts.

1. Design for hub

The hub is designed by considering it as a hollow shaft

D=2*d; Length of hub L=1.5d

2. Design for key

The key is designed with equal properties and then checked for shearing and

crushing stress. The length of key is taken equal to the length of hub

3. Design for flange

tf- thickness of flange(d/2)

4. Design for bolt

The bolts are subjected to shear stress due to torque transmitted. The number of

bolts (n) depends upon the diameter of shaft and pitch circle diameter is taken

D1=3d

Torque transmitted for Shear is

where d1- diameter of bolt, Torque transmitted for crushing is

`

Bushed pin flexible coupling

Let l = Length of bush in the flange, mm

D2 = Diameter of bush, mm

pb = Bearing pressure on the bush or pin, N/mm2

n = Number of pins, and

D1 = Diameter of pitch circle of the pins.mm

We know that bearing load acting on each pin,

W = pb× d2 × l

Then, Total bearing load on the bush or pins

=W x n = pb x d2 x l x n

And the torque transmitted by the coupling,

Direct shear stress due to pure torsion in the coupling halves

Maximum bending of the pin

Bending stress

The value of maximum principal stress varies from

28 to 42MPa

UNIT– II (PART - A)

1. What is the main use of woodruff keys? (NOV 13)

Woodruff key is used in machine tool and automobile construction.

2. Why a hollow shaft has greater strength and stiffness than solid shaft of equal weight?

(NOV 12)

The two reasons for preferring hollow shaft over solid shaft are,

a) For same weight of shaft, hollow shaft can transmit 1.5 times the torque transmitted by

solid shaft.

b) For a particular power transmission, hollow shaft requires minimum weight.

3. Under what circumstances flexible couplings are used? (NOV 12)

It is used to connect two shafts having both lateral and angular misalignment.

4. Classify keys with its applications .(MAY 12)

Sunk key, Feather key, Woodruff key, Splines, Peg key, Pin key and Cone key.

5. Write the advantages of the knuckle joints.(MAY 12)

The advantages of the knuckle joint are to connect two rods which are under the action of tensile

loads and may be readily disconnected for adjustments.

6. State Castiglione’s theorem. (MAY 13)

The partial derivative of the strain energy with respect to any force gives the deflection

corresponding to that force. It is used to determine the deflection at a point where no force is acting.

7. What is the difference between rigid and flexible coupling? (MAY 13)

Rigid coupling is used to connect two shafts which are perfectly aligned, Flexible coupling is used

to connect two shafts having both lateral and angular misalignment.

8. What is meant by design of a shaft based on rigidity? (DEC 15)

The shaft that permits an excessive angular displacement may contribute to vibrations that affect the

gear action and cause premature bearing wear failure. In addition, the particular applications such as

machine tools require that the spindles be especially rigid.

9. What are the possible modes of failure of the pin (bolt) in a flexible coupling? (DEC 15)

The pin (bolt) may fail due to plain shear load in the flexible coupling.

10. A shaft of 70mm long is subjected to shear stress of 40 MPa and has an angle of twist

equal to 0.017 radian. Determine the diameter of the shaft. Take G = 80GPa (NOV 13)

l = 70 mm, τ = 40 N/mm2, θ = 0.017 radian, G = 80000 N/mm2, τ/R = Gθ/l, 40/R = (80000 X

0.017)/70, (40 X 70)/(80000 X 0.017) = R = 2.05, D = 4.1 mm.

11. Define the term critical speed

The speed, at which the shaft runs so that the additional deflection of the shaft from the axis of

rotation becomes infinite, is known as critical or whirling speed.

12. What are splines?

The keys are made integral with the shaft which fits in the keyways broached in the hub.

Such shafts are known as splined shafts. These shafts usually have four, six, ten or sixteen splines.

The splined shafts are relatively stronger than shafts having a single key way.

13. What are the purposes in machinery for which couplings are used?

1. To provide the connection of shafts of units those are manufactured separately such as

motor and generator and to provide for disconnection for repairs or alterations.

2. To provide misalignment of the shafts or to introduce mechanical flexibility.

3. To reduce the transmission of shock from one shaft to another.

4. To introduce protection against over load

14. List out the requirements of a shaft coupling?

The requirements of a shaft coupling are,

a) It should be easy to connect or disconnect. b).It should transmit the full power of the shaft.

c). It should hold the shafts in perfect alignment. d).It should have no projecting parts.

15. What are the main functions of the knuckle joints?

a)It is used to transmit axial load from one machine element to other.

b) Small angular movement is possible between the rods.

UNIT – II (PART-B)

1. A shaft is supported by two bearings placed 1 m apart. A 600 mm diameter pulley is mounted at a distance

of 300 mm to the right of left hand bearing and this drives a pulley directly below it with the help of belt

having maximum tension of 2.25 kN. Another pulley 400 mm diameter is placed 200 mm to the left of

right hand bearing and is driven with the help of electric motor and belt, which is placed horizontally to

the right. The angle of contact for both the pulleys is 180° and μ = 0.24. Determine the suitable diameter

for a solid shaft, allowing working stress of 63MPa in tension and 42MPa in shear for the material of

shaft. Assume that the torque on one pulley is equal to that on the other pulley.

Fig. 2.1

The vertical and horizontal load diagram at C and D is shown in Figs.2.1(c) and (d) respectively.

The bending moment diagram for vertical diagram is shown in the Fig. 2.1 (e)

The bending moment diagram for horizontal diagram is shown in the Fig.2.1. (f)

The resultant bending moment diagram is shown in the Fig.2.1 (g).We see that the bending moment is

maximum at D, therefore, Maximum B.M, M = MD = 887 874 N-mm Ans.

2. A steel solid shaft transmitting 15 kW at 200 r.p.m. is supported on two bearings 750 mm apart and has two

gears keyed to it. The pinion having 30 teeth of 5 mm module is located 100 mm to the left of the right hand

bearing and delivers power horizontally to the right. The gear having 100 teeth of 5 mm module is located 150

mm to the right of the left hand bearing and receives power in a vertical direction from below. Using an

allowable stress of 54MPa in shear, determine the diameter of the shaft.

The space and torque diagrams of the shaft are shown in Figs. 2.2 (a) and (b) respectively.

The vertical and horizontal load diagram at C and D is shown in the Figs.2.2. (c) and (d) respectively.

Fig. 2.2

The bending moment diagram for vertical diagram is shown in the Fig. 2.2 (e)

The bending moment diagram for horizontal diagram is shown in the Fig. 2.2 (f)

The resultant bending moment diagram is shown in the Fig.2.2 (g).We see that the bending moment is

maximum at D, therefore,

3. A steel spindle transmits 4 kW at 800 rpm. The angular deflection should not exceed 0.25° per meter of

the spindle. If the modulus of rigidity for the material of the spindle is 84 GPa, find the diameter of the

spindle and the shear stress induced in the spindle.

4. Compare the weight, strength and stiffness of a hollow shaft of the same external diameter as that of

solid shaft. The inside diameter of the hollow shaft being half the external diameter, both the shafts have

the same material and length.

5. Design and make a neat dimensioned sketch of a muff coupling which is used to connect two steel shafts

transmitting 40 kW at 350 r.p.m. The material for the shafts and key is plain carbon steel for which allowable

shear and crushing stresses may be taken as 40 MPa and 80 MPa respectively. The material for the muff is

cast iron for which the allowable shear stress may be assumed as 15MPa.

Given: P = 40 kW = 40 × 103 W; N = 350 r.p.m.; τs = 40 MPa = 40 N/mm2; σcs = 80 MPa = 80 N/mm2;

σc = 15 MPa = 15N/mm2.

2. Design for sleeve

We know that outer diameter of the muff, D = 2d + 13 mm = 2 × 55 + 13 = 123 say 125 mm Ans.

and length of the muff, L = 3.5 d = 3.5 × 55 = 192.5 say 195 mm Ans.

Let us now check the induced shear stress in the muff. Let τc be the induced shear stress in the

muff which is made of cast iron. Since the muff is considered to be a hollow shaft, therefore the torque

transmitted ((T),

Since the induced shear stress in the muff (cast iron) is less than the permissible shear stress of

15 N/mm2, therefore the design of muff is safe.

3. Design for key

From Design data Book, we find that for a shaft of 55 mm diameter, Width of key, w = 18 mm Ans.

Since the crushing stress for the key material is twice the shearing stress, therefore a square

key may be used. Then, Thickness of key, t = w = 18 mm Ans.

We know that length of key in each shaft, l = L / 2 = 195 / 2 = 97.5 mm Ans.

Let us now check the induced shear and crushing stresses in the key. First of all, let us consider

shearing of the key. We know that torque transmitted (T),

Since the induced shear and crushing stresses are less than the permissible stresses, therefore

the design of key is safe.

6. Design a cast iron protective type flange coupling to transmit 15 kW at 900 rpm from an electric motor to

a compressor. The service factor may be assumed as 1.35. The following permissible stresses may be used :

Shear stress for shaft, bolt and key material = 40 MPa Crushing stress for bolt and key = 80 MPa

Shear stress for cast iron = 8 MPa; Draw a neat sketch of the coupling.

Fig.2.3 (a) Unprotected Flange coupling Fig.2.3. (b) Protected Flange coupling

Given: P = 15 kW = 15 × 103 W; N = 900 r.p.m. ; Service factor = 1.35 ; τs = τb = τk

= 40 MPa = 40 N/mm2 ;σcb = σck = 80 MPa = 80 N/mm2 ; τc = 8 MPa = 8 N/mm2. The protective

type flange coupling is designed as discussed below:

1. Design for hub

First of all, let us find the diameter of the shaft (d). We know that the torque transmitted by the shaft,

Since the service factor is 1.35, therefore the maximum torque transmitted by the shaft,

Tmax = 1.35 × 159.13 = 215 N-m = 215 × 103 N-mm

We know that the torque transmitted by the shaft (T),

We know that outer diameter of the hub,

D = 2d = 2 × 35 = 70 mm Ans.

And length of hub, L = 1.5 d = 1.5 × 35 = 52.5 mm Ans.

Let us now check the induced shear stress for the hub material which is cast iron. Considering the

hub as a hollow shaft, we know that the maximum torque transmitted (Tmax).

Then, τc = 215 × 103/63 147 = 3.4 N/mm2 = 3.4 MPa

Since the induced shear stress for the hub material (i.e. cast iron) is less than the permissible value

of 8 MPa, therefore the design of hub is safe.

2. Design for key

Since the crushing stress for the key material is twice its shear stress (i.e. σck = 2τk ), therefore

a square key may be used. From DDB, we find that for a shaft of 35mm diameter, Width of key,

w = 12 mm Ans.

And thickness of key, t = w = 12 mm Ans.

The length of key (l) is taken equal to the length of hub.

Then, l = L = 52.5 mm Ans.

Let us now check the induced stresses in the key by considering it in shearing and crushing. Considering

the key in shearing, we know that the maximum torque transmitted (Tmax),

Then, τk = 215 × 103/11 025 = 19.5 N/mm2 = 19.5 MPa

Considering the key in crushing, we know that the maximum torque transmitted (Tmax),

Σck = 215 × 103/ 5512.5 = 39 N/mm2 = 39 MPa.

Since the induced shear and crushing stresses in the key are less than the permissible stresses,

therefore the design for key is safe.

3. Design for flange

The thickness of flange (tf) is taken as 0.5 d.

Then, tf = 0.5 d = 0.5 × 35 = 17.5 mm Ans.

Let us now check the induced shearing stress in the flange by considering the flange at the

junction of the hub in shear.

We know that the maximum torque transmitted (Tmax),

τc = 215 × 103/134 713 = 1.6 N/mm2 = 1.6 MPa

Since the induced shear stress in the flange is less than 8MPa, therefore the design of flange is safe.

4. Design for bolts

Let d1 = Nominal diameter of bolts. Since the diameter of the shaft is 35mm, therefore let us take the

number of bolts, n=3 and pitch circle diameter of bolts,

D1 = 3d = 3 × 35 = 105 mm

The bolts are subjected to shear stress due to the torque transmitted. We know that the

maximum torque transmitted (Tmax),

(d1)2 = 215 × 103/4950 = 43.43 or d1 = 6.6 mm

Assuming coarse threads, the nearest standard size of bolt is M8. Ans. Other proportions of

the flange are taken as follows:

Outer diameter of the flange, D2 = 4 d = 4 × 35 = 140 mm Ans.

Thickness of the protective circumferential flange, tp = 0.25 d = 0.25 × 35 = 8.75 say 10 mm Ans.

7) Design a bushed-pin type of flexible coupling to connect a pump shaft to a motor shaft transmitting

32 kW at 960 rpm. The overall torque is 20 percent more than mean torque. The material properties are

as follows:

(a) The allowable shear and crushing stress for shaft and key material is 40MPa and 80MPa

respectively.

(b) The allowable shear stress for cast iron is 15 MPa.

(c) The allowable bearing pressure for rubber bush is 0.8 N/mm2.

(d)The material of the pin is same as that of shaft and key. Draw neat sketch of the coupling.

Solution. P = 32 kW = 32 × 103 W; N = 960 r.p.m. ; Tmax = 1.2 Tmean ; τs = τk = 40 MPa = 40 N/mm2 ;

σcs = σck = 80 MPa = 80 N/mm2 ; τc = 15 MPa = 15 N/mm2 ; pb = 0.8 N/mm2.

1. Design for pins and rubber bush

In order to allow for the bending stress induced due to the compressibility of the rubber bush, the

diameter of the pin (d1) may be taken as 20 mm. Ans.

The length of the pin of least diameter i.e. d1=20 mm is threaded and secured in the right hand

coupling half by a standard nut and washer. The enlarged portion of the pin which is in the left hand

coupling half is made of 24mm diameter. On the enlarged portion, a brass bush of thickness 2 mm is

pressed. A brass bush carries a rubber bush. Assume the thickness of rubber bush as 6mm.

So, Overall diameter of rubber bush,

d2 = 24 + 2 × 2 + 2 × 6 =40mm Ans.

and diameter of the pitch circle of the pins,

Let l = Length of the bush in the flange.

We know that the bearing load acting on each pin, W = pb × d2

× l = 0.8 × 40 × l = 32 l N

And the maximum torque transmitted by the coupling (Tmax),

l = 382 × 103/12 672 = 30.1 say 32 mm

And W = 32 l = 32 × 32 = 1024N

So, direct stress due to pure torsion in the coupling halves,

Since the pin and the rubber bush are not rigidly held in the left hand flange, therefore the tangential

load (W) at the enlarged portion will exert a bending action on the pin. Assuming a uniform distribution

of load (W) along the bush, the maximum bending moment on the pin,

Since the max. principal stress and max. shear stress are within limits, the design is safe.

2. Design for hub D = 2d =2x 40 = 80 mm And length of hub, L = 1.5 d = 1.5 × 40 = 60 mm

Let us now check the induced shear stress for the hub material which is cast iron. Considering the hub as a

hollow shaft, we know that the maximum torque transmitted (Tmax),

τc = 382 × 103 / 94.26 × 103 = 4.05 N/mm2 = 4.05 MPa

Since the induced shear stress for the hub material (i.e. cast iron) is less than the permissible

value of 15 MPa, therefore the design of hub is safe.

3. Design for key

Since the crushing stress for the key material is twice its shear stress (i.e. σck = 2 τk ), therefore a square

key may be used. From Table 13.1, we find that for a shaft of 40 mm diameter,

Width of key, w =14mm Ans.

And thickness of key, t=w=14mm Ans.

The length of key (L) is taken equal to the length of hub, i.e.

L = 1.5 d = 1.5 × 40 = 60 mm

Let us now check the induced stresses in the key by considering it in shearing and crushing.

Considering the key in shearing, we know that the maximum torque transmitted (Tmax),

Considering the key in crushing, we know that the maximum torque transmitted (Tmax),

σck = 382 × 103/8400 = 45.48 N/mm2 = 45.48 MPa

Since the induced shear and crushing stress in the key are less than the permissible stresses of

40MPa and 80MPa respectively, therefore the design for key is safe.

4. Design for flange

The thickness of flange ( tf ) is taken as 0.5 d. tf = 0.5 d = 0.5 × 40 = 20 mm

Let us now check the induced shear stress in the flange by considering the flange at the

junction of the hub in shear, we know that the maximum torque transmitted (Tmax),

τc =382 × 103 / 201 × 103 = 1.9 N/mm2 = 1.9 MPa

Since the induced shear stress in the flange of cast iron is less than 15 MPa, therefore the

design of flange is safe.

UNIT-III

DESIGN OF TEMPORARY AND PERMANENT JOINTS

SCREWED JOINTS

ECCENTRIC LOAD ACTING PERPENDICULAR TO THE AXIS OF BOLTS

Direct shear load Ws=W/n, where n is number of bolts

Maximum tensile load on bolt 3 or 4

Wt2= Wt= WLL2/2[(L1)2+(L2)2]

Equivalent tensile load Wte=[Wt+√((Wt)2+4(Ws)2)]/2

Equivalent shear load Wse= [√((Wt)2 +4(Ws)2)]/2

ECCENTRIC LOAD ON A BRACKET WITH CIRCULAR BASE

Maximum load Wt=(2WL/n)[(R+rcos(180/n))/2R2+r2]

Let R = Radius of the column flange; r = Radius of the bolt pitch circle,

w = Load per bolt per unit distance from the tilting edge

L = Distance of the load from the tilting edge

L1, L2, L3, and L4 = Distance of bolt centres from the tilting edge A

KNUCKLE JOINT

A knuckle joint is used to connect two rods which are under the action of tensile

loads. It consists of mainly three elements a fork or double eye rod, a single eye rod and

knuckle pin. Its use may be found in the link of a cycle chain, tie rod joint for roof truss.

Dimension of various parts of the knuckle joint

d- diameter of rod ,mm; d1- diameter of pin outer dia of eye, mm

d2=2d diameter of knuckle pin head and collar, mm

d3=1.5d thickness of single eye or rod end, mm; t=1.25d thickness of fork, mm

t1=0.75d thickness of pin head, mm; t2=0.5d, mm

1. Failure of the solid rod in tension

Load (P) = πd2σt/4 N/mm2

d- diameter of the rod

2. Failure of the knuckle pin in shear

Load(P)=2πd12τ/4

3. Failure of the single eye or rod end in tension

P=(d2-d1)tσt

4. Failure of the single eye or rod end in shearing

P=(d2-d1)tτ

5. Failure of the single eye or rod end in crushing

P=d1tσc

6. Failure of the forked end in tension

P=(d2-d1)2t1σt

7. Failure of the forked end in shear

P=(d2-d1)2t1τ

8. Failure of the forked end in crushing

P=d12t1σc

COTTER JOINTS:

Types of Cotter Joints

Following are the three commonly used cotter joints to connect two rods by a cotter:

1. Socket and spigot cotter joint 2. Sleeve and cotter joint, 3. Gib and cotter joint.

Socket and Spigot Cotter Joint

Design of Socket and Spigot Cotter Joint

d = Diameter of the rods, mm; d1 = Outside diameter of socket, mm

d2 = Diameter of spigot or inside diameter of socket, mm

d3 = Outside diameter of spigot collar, mm; t1= Thickness of spigot collar, mm

d4 = Diameter of socket collar, mm; c = Thickness of socket collar, mm

b = Mean width of cotter, mm; t = Thickness of cotter, mm; l = Length of cotter, mm

a = Distance from the end of the slot to the end of rod, mm

σt= Permissible tensile stress for the rods material, N/mm2

τ= Permissible shear stress for the cotter material, N/mm2

σc= Permissible crushing stress for the cotter material. N/mm2

1. Failure of the rods in tension

2. Failure of spigot in tension across the weakest section (or slot)

3. Failure of the rod or cotter in crushing

4. Failure of the socket in tension across the slot

5. Failure of cotter in shear

6. Failure of the socket collar in crushing

7. Failure of socket end in shearing

8. Failure of rod end in shear

9. Failure of spigot collar in crushing

10. Failure of the spigot collar in shearing

11. Failure of cotter in bending

Bending stress induced in the cotter,

12. The length of cotter (l) in taken as 4 d

13. The taper in cotter should not exceed 1 in 24 .in case the greater taper is required,

then a locking device must be provided.

14. The draw of cotter is generally taken as 2 to 3 mm.

WELDED JOINTS

Strength of transverse filler welded joints

t- Throat thickness, mm

s- Size of weld, mm

l – Length of weld, mm

t- s x sin45o = 0.707s,mm

P= throat area x allowable tensile stress = 0.707 s x l x σt , N PSGDB 11.3

STRENGTH OF PARALLEL FILLET WELDED JOINTS

A=0.707s x l PSGDB 11.3

P= throat area x allowable shear stress = 0.707s x l x τ

Shear strength of the joint for double parallel fillet weld

P=1.414s x l x τ, N PSGDB 11.3

CIRCULAR FILLET WELD SUBJECTED TO TORSION

PSGDB 11.3

CIRCULAR FILLET WELD SUBJECTED TO BENDING MOMENT

LONG FILLET WELD SUBJECTED TO TORSION

T=torque acting on the vertical plane, N

L=length of weld, mm; s=size of weld, mm ; t=throat thickness, mm

J=polar moment of inertia of the weld section= txl3/6 (of both sides weld)

Shear stress, τ= 3T/tl2

τmax= 4.242 T/sl2 PSGDB 11.3

STRENGTH OF BUTT JOINTS

Tensile strength of the butt joint (single v or square butt joint)

P=tlσt -----where l= length of weld, mm

P= (t1+t2) lσt, N

t1=throat thickness at the top, mm; t2=throat thickness at the bottom, mm

Axially loaded unsymmetrical welded sections

la=length of weld at the top, mm; lb=length of weld at the bottom, mm

l=total length of weld = la+lb, mm; P=axial load, N

a = distance of top weld from gravity axis, mm; b=distance of bottom weld from gravity

axis, mm; f=resistance offered by the weld per unit length

la= lb/(a+b); lb= la/(a+b)

ECCENTRICALLY LOADED WELDED JOINTS

Maximum normal stress, σtmax= σb/2+ (√ (σb) 2+ 4τ2)/2; τmax = (√(σb)2+ 4τ2)/2

where σb= bending stress; τ= shear stress, N/mm2

(i) Direct shear due to the shear force (P)

(ii) Bending stress due to bending moment P*e

Area of throat (A)=throat thickness*length of weld

=t*l*2 = 1.414*s*l PSGDB 11.3

Shear stress in the weld

Bending stress

UNIT– III (PART –A)

1. State the two types of eccentric welded connections. (NOV 13)

Linear fillet and Circular fillet weld joint

2. What is a Gib? Why is it provided in a cotter joint? (NOV 13)

Gib is a piece having same thickness, and provides a larger bearing surface for the cotter to slide

on, due to the increased holding power.

3. Define the term self-locking of power screws. (NOV 12)

The torque required to lower the load will be positive, indicating that an effort is applied to lower

the load. Such screws are termed as self-locking power screws.

4. What are the possible modes of failure of riveted joint? (NOV 12)

1. Tearing at the edge of plate near the rivet hole,

2. Tearing of the plate at the section weakened by holes,

3. Shearing rivets and crushing of rivets.

5. List the advantages and disadvantages of a threaded joint. (MAY 12) Advantages-

Screw joints are highly reliable in operation, simple manufacture, made in any convenient shape

and size, Disadvantages - stress concentration is available, self-loosening properties and air-tight

joints cannot be maintained.

6. Define efficiency of a riveted joint. (MAY 12)

It is defined as ratio between least strength of joint against various failures to the solid strength of

plate per pitch length.

7. Why are through bolts not preferred in assembly applications? (MAY 13)

Through bolts do not require tapping of threads in the parts that are fastened. However they are

inconvenient in assembly operation.

8. Which factor causing residual stress in welded joint? How are they relieved? (MAY 13)

Homogeneity of the weld metal, thermal stresses in welds, changes of physical properties due to

high rate of cooling. They are relieved by load is distributed uniformly along entire length and stress

spread uniformly over entire section.

9. What are the different applications of screwed fasteners?

The different applications of screwed fasteners are

a. For readily connecting & disconnecting machine parts without damage

b. The parts can be rigidly connected c. Used for transmitting power.

10. State the two types of eccentric welded connections.

i. Welded connections subjected to moment in a plane of the weld.

ii. Welded connections subjected to moment in a plane normal to the plane of the weld.

11. What are the advantages of screwed fasteners? (Nov 2012)

The advantages of screwed fasteners are (a). They are highly reliable in operation

b. They are convenient to assemble & disassemble c. A wide range of screws can be used

for various operating conditions d. They are relatively cheap to produce.

12. Define pitch.

Pitch is defined as the distance from a point on one thread to the corresponding on the

adjacent thread in the same axis plane.

13. What kind of failures will occur in a riveted joint and how can they be rectified?

Four types of failure will occur in a riveted joint which are

1. Tearing the edge of plate near the rivet hole.

2. Tearing of the plate at the section weakened by the holes.

3. Shearing of rivets and Crushing of rivets

14. What is meant by temporary joint and permanent joint? Give examples.

If the machine parts can be dismantled, without damaging or destroying the connecting

elements for maintenance or repair, then that joint is called as temporary joint such as screw joint.

When the machine parts cannot be dismantled without destroying the connecting elements these

joints are called as permanent joints. For example riveted joints, welded joints and so on.

15. State three conditions where tap bolts are used.

1. One of the parts is thick enough to accommodate a threaded hole.

2. It has sufficient strength to ensure durable threads.

3. There is no place to accommodate the nut.

UNIT – III (PART-B)

1. A cast iron cylinder head is fastened to a cylinder of 500 mm bore with 8 stud bolts. The maximum

pressure inside the cylinder is 2 MPa. The stiffness of part is thrice the stiffness of the bolt. What

should be the initial tightening load so that the point is leak proof at maximum pressure? Also choose

a suitable bolt for the above application.

Given D = 500mm n = 8, p = 2 N/mm2, qp = 3 qb

Solution

PT =2 x p x 𝜋

4 D2

=392699.06 N

Since this load is shared by 8 bolts, load/bolt = 392699/8 = 49087.38 N

Pmax = Pi (qb + qp/ qp) = Pi (qb + 3 qb / 3 qb) = 36 824 N = 37000 N

Total load on the bolt

Pb = Pi + P (qb / qb + qp)

= Pi + P (1 / (1 + (qp/ qb)))

= 37000 + 49087 (1/1+3)

= 49271.75 N

Using

Ac = ( 60𝑃𝑏

𝜎𝑦 ) 2/3, d < 45 𝑚𝑚

= ( 60 𝑥 49271

300𝜎𝑦 ) 2/3 = 459.64 mm2

From PSG page 5.42, for 561 mm2 of stress area [i.e., nearest to Ac value of 459.64 mm2], the bolt

chosen is M30.

2. A bracket, as shown in Fig.3.1, supports a load of 30 kN. Determine the size of bolts, if the maximum

allowable tensile stress in the bolt material is 60 MPa. The distances are: L1 = 80 mm, L2 = 250 mm,

and L = 500mm.

Fig 3.1

From DDB (coarse series), we find that the standard core diameter of the bolt is 28.706 mm

and the corresponding size of the bolt is M 33. Ans.

3. For supporting the travelling crane in a workshop, the brackets are fixed on steel columns as shown

in Fig. 3.2. The maximum load that comes on the bracket is 12 kN acting vertically at a distance of 400

mm from the face of the column. The vertical face of the bracket is secured to a column by four bolts, in

two rows (two in each row) at a distance of 50 mm from the lower edge of the bracket. Determine the

size of the bolts if the permissible value of the tensile stress for the bolt material is 84 MPa. Also find the

cross-section of the arm of the bracket which is rectangular.

Fig.3.2

Since the load W will try to tilt the bracket in the clockwise direction about the lower edge, therefore

the bolts will be subjected to tensile load due to turning moment. The maximum loaded bolts are 3 and 4

(See Fig.3.2), because they lie at the greatest distance from the tilting edge A–A (i.e. lower edge). We know

that maximum tensile load carried by bolts 3 and 4,

Since the bolts are subjected to shear load as well as tensile load, therefore equivalent tensile load,

4. Design a knuckle joint to transmit 150 kN as shown in Fig. 3.3. The design stresses may be taken as 75 MPa in

tension, 60 MPa in shear and 150 MPa in compression.

Fig. 3.3

5. A plate 100 mm wide and 12.5 mm thick is to be welded to another plate by means of parallel fillet welds.

The plates are subjected to a load of 50 kN. Find the length of the weld so that the max. stress does not

exceed 56 MPa. Consider the joint first under static loading and then under fatigue loading.

6. A plate 75 mm wide and 12.5 mm thick is joined with another plate by a single transverse weld and a

double parallel fillet weld as shown in Fig. 3.4.The maximum tensile and shear stresses are 70 MPa and

56 MPa respectively. Find the length of each parallel fillet weld, if the joint is subjected to both static and

fatigue loading.

Fig.3.4

7. A welded joint as shown in Fig. 3.5, is subjected to an eccentric load of 2 kN. Find the size of

weld, if the maximum shear stress in the weld is 25MPa.

Fig.3.5

8. A bracket carrying a load of 15 kN is to be welded as shown in Fig. 3.6. Find the size of weld

required if the allowable shear stress is not to exceed 80 MPa.

Fig. 3.6

UNIT-4

ENERGY STORING ELEMENTS AND ENGINE COMPONENTS

TYPES OF SPRINGS

o Helical springs

o Conical and volute spring

o Torsion spring

o Laminated or leaf spring

o Disc or Belleville spring

o Special purpose spring

TERMS USED IN COMPRESSION SPRING

Solid length

When the compression spring is compressed until the coils come in contact with

each other the spring is said to be solid. The solid length of a spring is the product of total

number of coils and the diameter of the wire.

LS=n’*d, mm; where n’- total number of coils, d- diameter of the wire ,mm

Free length

It is the length of the spring in the free or unloaded condition. It is equal to the solid

length plus the maximum deflection or compression of the spring and the clearance

between the adjacent coils.

LF=n’*d+δmax+0.15 δmax, mm PSGDB 7.101

Spring index

It is defined as the ratio of the mean diameter of the coil to the diameter of the coil

to the diameter of the wire. C=D/d PSGDB 7.100

D- mean diameter of coil ,mm: d- diameter of wire ,mm

Spring rate

It is defined as the load required per unit deflection of the spring.

q=P/y ; where P- applied load in N and y- deflection of the spring in mm

Pitch

The pitch of the coil is defined as the axial distance between adjacent coil in

uncompressed state.

Pitch length=free length/(n’-1), mm

n- number of active turns

p-pitch of coils d-diameter of spring, mm

STRESSES IN HELICAL SPRING

Consider a compression spring made of circular wire and subjected to an axial load W

Ks- Shear stress factor

When the springs are subjected to static loading, the effect of wire curvature may

be neglected. Because yielding of material will relive the stresses. In order to consider the

effect of both direct shear as well as curvature of the wire, Wahl’s stress factor is

introduced.

PSGDB 7.100

DEFLECTION OF HELICAL SPRING OF CIRCULAR WIRE

PSGDB 7.100

STIFFNESS OF SPRING (or) SPRING RATE

PSGDB 7.100

LEAF SPRING

The laminated or leaf spring consists of a number of flat plates of varying lengths

held together by means of clamps and bolts. These are mostly used in automobiles.

2L1- length of span (or) overall length of spring

L-width of band (or) distance between centres of U bolt, ineffective length of spring

effective length of spring 2L=2L1-L

PSGDB 7.104

PSGDB 7.104

DESIGN OF FLYWHEEL

PSGDB 7.120

PSGDB 7.120

UNIT– IV (PART – A)

1 What type of external forces act on connecting rod? (NOV 12)

Gas force and inertia bending force.

2. What is the purpose of flywheel that is used in an IC engine? (NOV 13, DEC 15)

The excess power than requirement, produced in power stroke is absorbed by the flywheel and

this absorbed power is released during the operation of other working strokes in which no power

is developed.

3. Define free length of spring. (NOV 14)

It is the length of spring when it is free from load, and equal to the total values of solid length plus

maximum allowable compression plus required clearance between all the coils for cushioning

action as in the case of compression spring.

4. How does the function of flywheel differ from that of governor? (NOV 12)

Flywheel function is to absorb the excess power and governor function is regulate speed of the

engine.

5. What is the objective of the nipping of the leaf spring? (MAY 12)

The objective of nipping employed in leaf spring in which large radius of curvature is provided to

the full length leaves and small radius of curvature is provided to the graduated leaves, thus

maintaining a gap between these leaves.

6. Write the advantages of Belleville spring. (MAY 12)

They are made in the form of a cone disc to carry a high compressive force. In order to improve

their load carrying capacity, they may be stacked up together. The major stresses are tensile and

compressive. Advantages of Belleville spring is where high loads over small deflections are

needed in compact spaces.

7. Define spring index. (MAY 13)

it is defined as the ratio of mean diameter to the coil to wire diameter.

8. What is Wahl factor and why is it required? (MAY 13)

The factor which is used to employ the stress due to curvature may be neglected for static loading,

wire curvature effect will give better result.

9. Define spring rate. (MAY 14)

It is defined as the load required per unit deflection of the spring.

10. What type of spring is used to maintain an effective contact between a cam and a

reciprocating roller or flat faced follower? (DEC 15)

Composite helical springs.

11. Define: Leaf springs

A leaf spring consists of flat bars of varying lengths clamped together and supported at

both ends, thus acting as a simply supported beam.

12. What is buckling of springs?

The helical compression spring behaves like a column and buckles at a comparative small

load when the length of the spring is more than 4 times the mean coil diameter.

13. Explain about surge in springs?

When one end of the spring is resting on a rigid support and the other end is loaded suddenly,

all the coils of spring does not deflect equally, because some time is required for the propagation

of stress along the wire. Thus a wave of compression propagates to the fixed end from where it is

reflected back to the deflected end this wave passes through the spring indefinitely. If the time

interval between the load application and that of the wave to propagate are equal, then resonance

will occur. This will result in very high stresses and cause failure. This phenomenon is called surge.

The methods used for eliminating surge are

a. By using dampers on the centre coil so that the wave propagation dies out

b. By using springs having high natural frequency.

14. Write notes on the master leaf & graduated leaf?

The longest leaf of the spring is known as main leaf or master leaf has its ends in the form

of an eye through which bolts are passed to secure the spring. The leaf of the spring other than

master leaf is called the graduated leaves.

15. What is meant by nip in leaf springs?

By giving greater radius of curvature to the full length leaves than the graduated leaves,

before the leaves are assembled to form a spring thus a gap or clearance will be left between the

leaves. This initial gap is called nip.

UNIT – IV (PART-B)

1. A helical spring is made from a wire of 6 mm diameter and has outside diameter of 75 mm. If the permissible shear stress is 350 MPa and modulus of rigidity 84 kN/mm, find the axial load which the spring can carry and the deflection per active turn.

2. Design a spring for a balance to measure 0 N to 1000 N over a scale of length 80 mm. The spring

is to be enclosed in a casing of 25 mm diameter. The approximate number of turns is 30. The

modulus of rigidity is 85kN/mm2. Also calculate the maximum shear stress induced.

3. Design a helical compression spring for a maximum load of 1000 N for a deflection of 25 mm

using the value of spring index as 5. The maximum permissible shear stress for spring wire is

420 MPa and modulus of rigidity is 84kN/mm2. Take Wahl’s factor, K= 4𝐶−1

4𝑐−4+

0.615

𝐶.

4. Design a close coiled helical compression spring for a service load ranging from 2250 N to 2750 N.

The axial deflection of the spring for the load range is 6 mm. Assume a spring index of 5. The

permissible shear stress intensity is 420 MPa and modulus of rigidity, G = 84 kN/mm2. Neglect the

effect of stress concentration. Draw a fully dimensioned sketch of the spring, showing details of the

finish of the end coils.

5. Design a leaf spring for the following specifications:

Total load = 140 kN; Number of springs supporting the load = 4 ; Maximum number of leaves

= 10; Span of the spring = 1000 mm; Permissible deflection = 80 mm. Take Young’s modulus, E

= 200 kN/mm2 and allowable stress in spring material as 600MPa.

6. A truck spring has a 12 number of leaves, two of which are full length leaves. The spring supports are

1.05 m apart and the central band is 85 mm wide. The central load is to be 5.4 kN with a permissible

stress of 280 MPa. Determine the thickness and width of the steel spring leaves. The ratio of the total

depth to the width of the spring is 3. Also determine the deflection of the spring.

7. Describe concentric or composite spring

A concentric spring is used to obtain greater spring force within a given space and insure the operation

of a mechanism in the event of failure of one of the springs. So, these springs may have two or more springs

and have the same free lengths as shown in Fig. 4.1(a) and are compressed equally. Such springs are used in

automobile clutches; valve springs in aircraft, heavy duty diesel engines and rail-road car suspension systems.

Sometimes these springs are used to obtain a spring force which does not increase in a direct relation to the

deflection but increases faster. Such springs are made of different lengths as shown in Fig.4.1 (b). The shorter

spring begins to act only after the longer spring is compressed to a certain amount. These springs are used in

governors of variable speed engines to take care of the variable centrifugal force. The adjacent coils of the

concentric spring are wound in opposite directions to eliminate any tendency to bind. If the same material is

used, the concentric springs are designed for the same stress. In order to get the same stress factor (K), it is

desirable to have the same spring index (C).

Assuming that both the springs are made of same material, then the maximum shear stress induced in both the

springs is approximately same, i.e.

When stress factor, K1 = K2, then,

When both the springs are compressed until the adjacent coils meet, then the solid length of both the

springs is equal, i.e.

n1.d1 = n2.d2

The equation (ii) may be written as

Now dividing equation (iii) by equation (i), we have

i.e. the springs should be designed in such a way that the spring index for both the springs is same. From

equations (i) and (iv), we have

From Fig. 4.1 (a), we find that the radial clearance between the two springs,

Usually, the radial clearance between the two springs is taken as

------ (vi)

From equation (iv), we find that

D1= C.d1, and D2= C.d2

Substituting the values of D1 and D2 in equation (vi), we have

UNIT-V

BEARINGS

Design of Bearing

❖ Sliding contact bearing

❖ Rolling contact bearing

Sliding contact bearing

a) Hydrodynamic journal bearing

b) Spilt bearing or Plummer block

c) Thrust bearing

i) Foot step or pivot bearing

ii) Collar bearing

Rolling contact bearing

a) Ball bearing

b) Roller bearing

DESIGN OF ROLLING CONTACT BEARING

Bearing life

𝐿

𝐿`10

= [𝑙𝑛 (

1𝑝

)

𝑙𝑛 (1

𝑝10)

]

1𝑏

L-required life of bearing in million revolutions

L`10-calculated life of selected bearing, for the given load, for 90% survival

p- Probability of survival

p10- probability of survival for 90%

b is a constant =1.17 for a median life=5L10

=1.34for a median life=4.08L10 (deep groove ball bearing)

Equivalent bearing load

P= (XFr+YFa) S

where

X-radial factor: Y-thrust factor

S-service factor (refer PSGDB 4.2TO4.4)

Load life relation ship

The relationship between the dynamic load carrying capacity C, the equivalent

dynamic load P and the bearing life is given by

L=(𝐶

𝑃)

𝑏

Where b= constant = 3 for ball bearing, 10/3 for roller bearing (refer PSGDB 4.6, 4.7)

The relationship between life in million revolutions and life in working hours is given by

L=60𝑛𝐿ℎ

106

where

Lh= bearing life in hours; n= speed of rotation in rpm

PROCEDURE FOR THE SELECTION OF BEARINGS

1. Calculate the radial and axial forces acting on the bearing.

2. Calculate the shaft diameter.

3. Determine the radial load factor(X) and thrust load factor(Y) from the manufacturer`s

catalogue. The value of X and Y for ball and roller bearing are given in PSGDB 4.4

4. Calculate the equivalent dynamic load from the equation

P= (XFr + YFa)S (PSGDB 4.2)

5. Decide the expected life of the bearing. Convert the expected life in hours in millions of

revolutions.

6. Calculate the dynamic load capacity from the equation.

L=(𝐶

𝑃)

𝑏

7. Check whether the selected bearing has the required dynamic load capacity. If yes, the

selected bearing is suitable for this purpose. Otherwise, select another bearing from the

next series and go back to step3 and continue.

Bearing for cyclic loads-cubic mean load

Cubic mean load, Fm=[𝐹1

3𝑛1𝑡1+𝐹23𝑛2𝑡2+⋯+𝐹𝑛

3𝑛𝑛𝑡𝑛

𝛴𝑛]

1

3,N

where F1, F2,…Fn are the equivalent loads at corresponding speed and time fraction of

cycle.

DESIGN OF SLIDING CONTACT BEARINGS - JOURNAL BEARING

A sliding contact bearing that supports a load in a radial direction is known as journal

bearing.

TERMS USED IN HYDRODYNAMIC JOURNAL BEARING

1. Diameter clearance: It the different between the dia. of journal and the bearing. c= D-d

2. Radial clearance: it is the different between the radial of the bearing and the journal.

c1=R-r=D-d/2

3. Diameteral clearance ratio: It is the ratio of the diameteral clearance to the diameter of

the journal. =c/d=(D-d)/d

4. Eccentricity: It is the radial distance between the centre of the bearing and the displaced

centre of the bearing under load.

5. Minimum oil film thickness: It is the minimum distance between the bearing and the

journal under complete lubrication condition. It is denoted by ho and occurs at the line of

centres.

COEFFICENT OF FRICTION FOR JOURNAL BEARINGS

PSGDB 7.34

µ- Coefficient of friction; Z- Absolute viscosity of oil in Ns/m2

N- Speed of journal in rpm; p- bearing pressure on the projected bearing area N/mm2

d- Diameter of journal in mm; l- length of bearing in mm; c- diameter clearance in mm

k- Factor to correct for end leakage.

CRITICAL PRESSURE OF THE JOURNAL BEARING

PSGDB 7.34

SOMMERFELD NUMBER

PSGDB 7.34

HEAT GENERATED IN A JOURNAL BEARING

PSGDB 7.34

Heat dissipated by the bearing

PSGDB 7.34

C- heat dissipation coefficient ,W/mm2oC

A- projected area of the bearing,mm2

tb- temperature of the bearing surface, oC

ta- temperature of the surrounding air, oC

DESIGN PROCEDURE FOR JOURNAL BEARING

1. Determine the length of bearing.

2. Calculate the bearing pressure.

3. Select the type of oil used.

4. Determine the amount of heat generated.

5. Determine the amount of heat rejected

UNIT–V (PART –A)

1. What is meant by life of anti-friction bearings? (NOV 13)

The life of anti-friction bearing may be defined as the number of revolutions which the bearing runs

before the first evidence of fatigue develops in the material of one of the rings or any of the rolling

elements.

2. What are the essential requirements in an end face seal? (NOV 13)

All face-type mechanical seals consist of a primary seal, secondary seal, and mechanical loading

device. All face seals require a secondary seal, which can be some type of elastomeric compound

or gasket material. The secondary seal closes leak paths between the rotary face and the shaft as

well as between the housing/gland and the stationary face.

3. In hydrodynamic bearing, what are the factors which influence the formation wedge fluid

film? (NOV 14)

Viscosity of oil, Bearing load, minimum oil film thickness and eccentricity.

4. What are the factors used for formation of oil film in hydrodynamic bearings? (NOV 12)

i) A continuous supply of oil, ii) A relative motion between the two surfaces in a direction

approximately tangential to the surfaces, iii) the ability of one of the surfaces to take up a small

inclination to the other surface.

5. What is the limitation of McKee’s equation? (MAY 12)

The limitation of McKee’s equation is design of journal bearing depends on frictional coefficient

and bearing modulus.

6. Distinguish between hydrostatic and hydrodynamic bearings.(MAY 12)

Hydrostatic bearings, in which the lubricant may be sometimes be air or water is supplied with a

sufficient pressure to separate the rotating surfaces as similar to thick lubrication Hydrodynamic

bearings are in which high viscous lubricant is supplied which separates the rotating surfaces and

prevented metal to metal contact.

7. Differentiate sliding contact and rolling contact bearings. (MAY 13)

Sliding contact bearing Rolling contact bearing

i) Starting and running friction is high Low starting and running friction

ii) High resistance to shock loading Low resistance to shock loading

iii) Service is not reliable Service is reliable

iv) Overall dimensions are huge Overall dimensions are small

8. What are the types of sliding contact bearings? (MAY 13)

i) Full journal bearing, ii) Partial journal bearing, iii) Fitted journal bearing

9. What is meant by square journal bearing? (DEC 15)

If the length of journal is equal to its diameter it is called as square journal bearing.

10. Give an example for anti-friction bearing. (DEC 15)

Ball bearing, cylindrical roller bearing, Needle roller bearing, Tapered roller bearing

11. What are the required properties of bearing materials? (Nov 2012)

Bearing material should have the following properties. i. High compressive strength ii. Low

coefficient of friction iii. High thermal conductivity. High resistance to corrosion v. Sufficient

fatigue strength vi. It should be soft with a low modulus of elasticity vii. Bearing materials should

not get weld easily to the journal material.

12. Define Rolling Contact Bearings

In rolling contact bearings, the contact between the bearing surfaces is rolling instead of sliding

as in sliding contact bearings. The ordinary sliding bearing starts from rest with practically metal-

to-metal contact and has a high coefficient of friction. It is an outstanding advantage of a rolling

contact bearing over a sliding bearing that it has a low starting friction. Due to this low friction

offered by rolling contact bearings, these are called antifriction bearings.

13).How do you express the life of a bearing? What is an average or median life?

The life of an individual ball (or roller) bearing may be defined as the number of revolutions

(or hours at some given constant speed) which the bearing runs before the first evidence of fatigue

develops in the material of one of the rings or any of the rolling elements. The rating life of a group

of apparently identical ball or roller bearings is defined as the number of revolutions (or hours at

some given constant speed) that 90 per cent of a group of bearings will complete or exceed before

the first evidence of fatigue develops (i.e. only 10 per cent of a group of bearings fail due to fatigue).

14. Define the term Reliability of a Bearing?

The reliability (R) is defined as the ratio of the number of bearings which have successfully

completed L million revolutions to the total number of bearings under test. Sometimes, it becomes

necessary to select a bearing having a reliability of more than 90%.

15. Define and where Self-aligning bearing is used?

These bearings permit shaft deflections within 2-3 degrees. It may be noted that normal clearance

in a ball bearing are too small to accommodate any appreciable misalignment of the shaft relative

to the housing. If the unit is assembled with shaft misalignment present, then the bearing will be

subjected to a load that may be in excess of the design value and premature failure may occur.

UNIT V

PROBLEMS

1. Design a journal bearing for a centrifugal pump from the following data:

Load on the journal = 20 kN

Speed of the journal = 900 rpm

Type of oil = SAE 10

Absolute viscosity at 55 º C = 0.017 kg/m-s

Ambient temperature oil = 15.5 º C

Maximum bearing pressure = 1.5 MPa

Rise of temperature of oil = 10 º C

Heat dissipation coefficient = 1232 W/m2/ º C

Solution

Step 1. Length of the journal

Assume diameter of shaft = 100mm

For centrifugal pump;

𝑙

𝑑 = 1 to 2

take 𝑙

𝑑 = 1.6

l = 1.6 x 100 =160 mm

Step 2. Check for bearing pressure

𝑝 =𝑊

𝑙×𝑑 =

20 x 1000

160×100 =1.25 N/mm2

Therefore, p = 1.25 N/mm2 < 1.5 N/mm2, hence the design is safe.

Step 3. Bearing characteristics number on Sommerfold number

𝑍𝑁

𝑝=

0.017 𝑥 900

1.25= 12.24

Actual minimum value of bearing modulus is given by 𝑍𝑁

𝑝= 28 --Sommerfold number;

The minimum value of of bearing modulus is given by 3𝑘 =𝑍𝑁

𝑝

3𝑘 = 28 ; k= 28/3 = 9.33

Now 𝑍𝑁

𝑝= 12.24 is greater than k= 9.33, hence the bearing will operate under dynamic condition.

Step 4. Clearance ratio

For centrifugal pump, 𝑐

𝑑= 0.0013

Step 5. Co efficient of friction

µ=33

108 (𝑍𝑁

𝑝)

𝑑

𝑐+ 𝑘, Assume 𝑘 = 0.002

µ=33

108(12.24)

1

0.0013+ 0.002

µ=0.0051

Step 6 Heat generated

Qg = µWV = 0.0051X 20x 103 x𝜋𝑥0.1𝑥900

60 = 480.7 watts

Step 7 Heat dissipated

Qd =C l d (tb-ta)

tb-ta =1/2(55-15.5) = 19.75

Qd = 1232 x 0.16 x 0.1 x19.75 = 389.3 watts

Step 8 The amount of artificial cooling required

We see that the heat generated is greater than the heat dissipated which indicates that the bearing is

warming up. Therefore, either the bearing should be redesigned by taking t0 = 63°C or the bearing should

be cooled artificially.

We know that

The amount of artificial cooling required = Heat generated – Heat dissipated

= Qg – Qd = 480.7 – 389.3 = 91.4 W

Step 9 Mass of lubricating oil required for artificial cooling

Let m = Mass of the lubricating oil required for artificial cooling in kg / s.

We know that the heat taken away by the oil,

Qt = m.S.t = m × 1900 × 10 = 19 000 m W

Equating this to the amount of artificial cooling required, we have

19000 m = 91.4

m = 91.4 / 19 000

= 0.0048 kg / s

= 0.288 kg / min

2. Design a self-aligning ball bearing for a radial load of 7000 N and a thrust load of 2100 N. The desired

life of the bearing is 160 million of revolutions at 300 rpm. Assume uniform and steady load.

Given :

Fr = 7000 N

Fa = 2100 N

L = 160 × 106 rev

N = 300 r.p.m.

Solution.

From Table 27.4, we find that for a self-aligning ball bearing, the values of radial factor (X thrust factor

(Y) for Fa / Fr = 2100 / 7000 = 0.3, are as follows : X = 0.65 and Y = 3.5

From PSGDB, we find that for uniform and steady load, the service factor, S for ball bearings is 1.

therefore dynamic equivalent load,

P= (X Fr + Y Fa) S

= (0.65 × 7000 + 3.5 × 2100) 1

= 11900 N

We know that the basic dynamic load rating,

C = P ( 𝐿

106 )1

𝑘

= 11900 ( 160 𝑥 106

106 )1

3

= 64600 N = 64.6 kN

From PSGDB, let us select bearing number 219 having C = 65.5 kN

3. A 150 mm diameter shaft supporting a load of 10 kN has a speed of 1500 r.p.m. The shaft runs in a bearing

whose length is 1.5 times the shaft diameter. If the diametral clearance of the bearing is 0.15 mm and the

absolute viscosity of the oil at the operating temperature is 0.011 kg/m-s, find the power wasted in friction.

Given :

d = 150 mm = 0.15 m

W = 10 kN = 10000 N

N = 1500 r.p.m.

l = 1.5 d

c = 0.15 mm

Z = 0.011 kg/m-s

Solution.

We know that length of bearing,

l = 1.5 d

= 1.5 × 150

= 225 mm

Bearing pressure,

p = 𝑊

𝐴

= 𝑊

l x d

=10000

(225 x150)

= 0.296 N/mm2

We know that coefficient of friction,

µ=33

108 (𝑍𝑁

𝑝)

𝑑

𝑐+ 𝑘, Assume 𝑘 = 0.002

µ=33

108 (0.011 𝑥 1500

0,296)

150

0.15+ 0.002

µ=0.02

Rubbing velocity, V = πdN/60

= π x 0.15 x 1500 / 60

= 11.78 m/ s

We know that heat generated due to friction,

Qg = μ.W.V

= 0.02 × 10 000 × 11.78

= 2356 W

Power wasted in friction

Qg = 2356 W = 2.356 kW

4. A single row angular contact ball bearing number 310 is used for an axial flow compressor. The bearing

is to carry a radial load of 2500 N and an axial or thrust load of 1500 N. Assuming light shock load, determine

the rating life of the bearing.

Given:

Fr = 2500 N

Fa = 1500 N

Solution.

From PSGDB, we find that for single row angular contact ball bearing, the values of radial factor

(X) and thrust factor (Y ) for Fa / Fr = 1500 / 2500 = 0.6 are

X = 1 and Y = 0

From PSGDB, we find that for light shock load, the service factor (S) is 1.5. Therefore the design

dynamic equivalent load should be taken as

P = (X Fr + Y Fa) S

= (1 × 2500 + 0 × 1500) 1.5

= 3750 N

From PSGDB, we find that for a single row angular contact ball bearing number 310, the basic

dynamic capacity,

C = 53 kN

= 53000 N

We know that rating life of the bearing in revolutions,

L = (C/P) k x 106

= (53000/3750)3 x 106

= 2823 × 106 rev

5. The thrust of propeller shaft in a marine engine is taken up by a number of collars integral with the shaft

which is 300 mm is diameter. The thrust on the shaft is 200 kN and the speed is 75 r.p.m. Taking μ constant

and equal to 0.05 and assuming the bearing pressure as uniform and equal to 0.3 N/mm2, find : 1. Number

of collars required, 2.Power lost in friction, and 3. Heat generated at the bearing in kJ/min.

Given

d = 300 mm

r = 150 mm

W = 200 kN = 200 × 103 N;

N = 75 rpm ;

μ = 0.05 ;

p = 0.3 N/mm2

Solution.

1. Number of collars required

Let n = Number of collars required.

Since the outer diameter of the collar (D) is taken as 1.4 to 1.8 times the diameter of shaft (d),

therefore

let us take D = 1.4 d

= 1.4 × 300

= 420 mm

R = 210 mm

We know that the bearing pressure (p),

0.3 = W / nπ (R2 – r2)

= 200 x 103 / nπ (2102 – 1502)

n = 2.947 / 0.3

= 9.8 say 10

2. Power lost in friction

We know that total frictional torque,

𝑇 =2

3µ𝑊 (

𝑅3−𝑟3

𝑅2−𝑟2)

𝑇 =2

3𝑥0.05𝑥 200 𝑥 103 (

2003−1503

2002−1502)

=1817 × 103 N-mm

= 1817 N-m

Power lost in friction,

P = 2πN T / 60

= 2 π x 75 x 1817 / 60

= 14270 W

= 14.27 kW