74878611 MachineDesign I

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MACHINE DESIGN AND

DRAWING-I


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B.TECH. DEGREE COURSE

SCHEME AND SYLLABUS

(2002-03 ADMISSIONONWARDS)

MAHATMA GANDHI UNIVERSITY

KOTTAYAM,KERALA

MACHINE DESIGN AND DRAWING - IM 705 2+0+2

Module 1

Definitions - Design principles common engineering materials selection

and their properties general steps in design design criteria types of

failures - types of cyclic loading.

Stresses in Machine parts tension, compression and shear elastic

constants-working stress-factor of safety-bending and torsion-combined

stresses-stress concentration-fatigue-endurance limit-fatigue diagram-fatigue

factors-theories of failure-Goodman and Soderberg lines

Detachable joints-socket and spigot cotter joint, knuckle joint pins, keys,

splines -set screws, threaded fasteners and power screws Shaft coupling

sleeve coupling split muff coupling flange coupling protected type

flange coupling thick and thin cylinders

Riveted joints: Lap joint Butt joint failures of riveted joint strength ofriveted joint efficiency of riveted joint design of longitudinal butt joint for

boiler design of circumferential lap joint for boiler joints of uniform

strength Lozange joint eccentrically loaded riveted joint.

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Module 2

Springs Classification and uses of springs design of helical springs

effect of end turns energy absorbed deflection design for fluctuating

loads vibration in springs buckling of spring materials

Shafts Torsion and bending of shafts hollow shafts design of shafts for

strength an deflection effect of keyways transverse vibration and critical

speed of shafts

Design of IC engine parts connecting rod piston flywheel

Welded joints: Lap joint Butt joint weld symbols parallel and transverse

fillet welds strength of welded joints axially loaded welded joints

eccentrically loaded welded joints.

References

1. Mechanical Engg. Design Joseph Shigley

2. Machine Design Mubeen

3. Machine Design Black

4. Machine Design R. K. Jain

5. Machine Design an integral approach Norton, Pearson

6. Machine Design data hand book Lingayah Vol I.

7. Elements of Machine Design Pandya & Shah

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MACHINE DESIGN AND DRAWING

MODULE I

Lecturer 1:

Design Principles

A machine may be defined as a combination of stationary and moving parts

constructed for the useful purpose of generating, transforming or utilizing

mechanical energy. Machines can be classified in to:

1. Machines for generating mechanical energy: Converts some form of energy

(electrical, heat, hydraulics etc.) into mechanical work. eg: steam engines, IC

engines, water turbines etc.

2. Machines for transforming mechanical energy: known as converting

machines. These types of machines transform mechanical energy into

another form of energy. eg: Electric generators, hydraulic pumps etc.

3. Machines for utilizing mechanical energy: These machines receive

mechanical energy and deliver and utilize it as such in the performance of

useful work. eg: lathe, m/c tools etc.

A m/c element or part is a separate part of machine, either integral or consist

of several small pieces which are rigidly joined together by riveting, welding etc. The

m/c element can be classified into

1. General purpose elements. eg: nuts, bolts, key, axles, shafts.

2. Special purpose elements: These m/c elements are employed only with a

particular type of machine. eg: Piston, Connecting rods, Cam shafts etc.

Further sub divided into.

3. Fasteners: which connects or join the parts of a machine.

4. Elements of Rotary motion drive: These elements transmit power: Eg: belt,

rope, chain, gears, shafts etc.

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It is engineers tasks to define, calculate, motions, forces and changes in

energy in order to determine the sizes, shapes, and materials needed for each the

interrelated parts in the machine.

Basic Requirements for Machine Elements and Machine

Cost

High o/p and efficiency

Strength

Stiffness or rigidity

Wear resistant

Light weight and Mini dimensions

Reliability

Durability

Economy of performance

Accessibility

Processability

General Steps in Design

Market survey

Define specification of product

Feasibility study

Creative Design synthesis

Preliminary Design and Development

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Detailed Design

Prototype

Design

Design for product

ASSIGNMENT 1

1. Classification of Engg. materials [P.C. Sharma]

2. Mechanical properties of Engg. Material. [V.B. Bhandari, 10 pts]

Material Properties

Mechanical properties of a material generally determined through destructive

testing samples under controlled loading conditions.

Tensile test

This is one of the simplest and be basic test and determines values of number

of parameters concerned with mechanical properties. A materials like strength,

ductility and toughness. The information which can be obtained from the tests are:

i) Proportional Limit

ii) Elastic limit

iii) Modulus of Elasticity

iv) Yield strength

A typical tensile specimen is shown in fig. 1.1.

The tensile bar is mechanical from the material to be tested in one of the

several standard diameters do and gauge lengths l0. The gauge length is an

arbitrary length defined along the small-diameter portion of the specimen by two

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indentations so that its increase can be measured during the test. The larger dia

sections are threaded for insertion into a tensile test machine which is capable of

applying either controlled loads or controlled deflections to the end of the bars; and

gauge length portion in mirror polished to eliminate stress concentration from surface

defects. The bar is stretched slowly in tension until it breaks, while the load and

distance across the gauge length are monitored.

STRESS-STRAIN DIAGRAMS

Stress ( ) is defined as the load per unit area

0

p

A

=

p Applied load at any instant.

A0 Original cross-section area of the specimen.

Strain is the change in length per unit length and is calculated from

0 =0

l l

l l0 Original gauge length at any load P.

The results of tensile test are expressed by means of stress-strain

relationships and plotted in the form of a graph.

I. It is observed from the diagram that stress-strain relationship is linear from O

to P. OP is a straight line and after P, the curve begins to deviate from straight line.

Point P is the proportional limit below which stress is proportional to strain; as

expressed by Hookes law.

E

=

Where E defines the slope of stress-strain curve up to proportional limit called

Youngs Modulus OR Modulus of Elasticity of the material.

E=tan =PX

OX

=

E is the measure of the stiffness of the material in its elastic range and has the

units of the of stress.

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For most ductile materials, the modulus of elasticity in compression is the

same as in tension.

II. ELASTIC LIMIT:

Even if the specimen is stressed beyond point P and up to E, it will regain its

initial size and shape when load is removed. This shows that the material is the

elastic stage up to point E. Therefore E is called elastic limit. Elastic limit can be

defined as the maximum stress without any permanent deformation. Point P and E

are typically close together that they are often considered as the same.

III YIELD STRENGTH

When the specimen is stressed beyond point E, plastic deformation occurs

and material starts yielding. During this stage, it is not possible to recover the initial

size and shape of the specimen on the removal of the load. From the diagram, beyond

point E, the strain increases at a faster rate up to point y1. In the case of mild steel, it

is observed that there is small reduction in load and the curve drops down to point y2,

immediately after yielding starts. The points y1 and y2 are called upper and lower

yield points respectively. For many materials, y1 and y2 are close to each other and in

such a case, two points are considered as same and denoted by y. The stress

corresponding to yield point y is called yield strength. Yield strength is defined as

the maximum stress at which a marked increase in elongation occurs without

increase in load.

Many varieties of steel, especially heat-treated steels, aluminium and cold-

drawn steels do not have a well defined yield point on the stress strain diagram.

This type of material yields gradually after passing through elastic limit E. If the

loading is stopped at point Y, at a stress level slightly higher than elastic limit E, and

specimen is unloaded and readings taken, the curve would follow the dotted line and

a permanent at a plastic deformation will exist.

The strain corresponding to this permanent deformation is indicated by OA.

For such materials which do not exhibit a well defined yield point, the yield strength

is defined as the stress corresponding to a permanent set of 0.2% of gauge length. In

such a case the yield strength is determined by offset method. A distance OA equal

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to 0.002 mm/mm strain is marked on X-axis. A line is constructed from pt A parallel

to st. line portion OP of the stress-strain curve. The pt of intersection of this line and

the stress-strain curve is called Y. The corresponding stress is called 0.2% yield

strength. The term proof load G proof strength are frequently used in the design of

fasteners. The proof strength is similar to yield strength. It is determined by offset

method, however offset in this case is 0.001 mm/mm. 0.1% proof strength denoted

by the symbol Rp 0.1.

Ultimate tensile strength

After the yield point Y2, plastic deformation of the specimen increases. The

material become stronger due to strain hardening and higher and higher load is

required to deform the material. Finally the load and corresponding stress reaches a

maximum value, given by pt u. The stress corresponding to pt U is called ultimate

strength. The ultimate tensile strength is the maximum stress that can be reached in

tension test. For ductile material the diameter of the specimen begins to decrease

rapidly beyond maximum load point U. There is localized reduction in cross-section

area called necking. As the test progresses, the C.A. at the neck decreases rapidly

and fracture taken place at cross-section of the neck. The fracture point is shown in

fig. (F).

Ultimate tensile strength is considered as failure criterion in brittle material.

vi) Percentage Elongation:

Ductility is measured by percentage elongation and is given by0

0

100

l l

l

vii) Percentage reduction in area.

Ratio of decrease in C.A. of the specimen after fracture to original C.A.

Percentage reduction in area =0

0

A A100

A

A0 Original C.A. of specimen.

A Final C.A.

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SIMPLE STRESSES IN MACHINE PARTS

Stress: Force per unit area

Stress ( ) = P/A

P = Force or load acting on the body

A = Cross sectional area

Units:

In S.I. unit; stress is usually expressed as (Pa)

1 Pa = 1 N/m2

Strain: Deformation per unit length

Strain

=l

l l change in length, l original length

Youngs Modulus:

Stress is directly proportional to strain (with in elastic limit)

= E

E = / =A

l

l

E Constant of proportionality: Youngs Modulus unit : GPa : GN/m2 or

kN/mm2.

PROBLEMS

A coil chain of a crane required to carry a max. load of 50 KN is shown in fig. A

Find diameter of link stock, if the permissible tensile stress in the link material is not

to exceed 75 MPa?

p = 50 KN = 50 103 N

t = 75 MPa = 75 106 N/m2 = 75 N/mm2

d = ?

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A =2d

4

= 0.7854 d2

t =P

A =

3

2

50 10

0.7854 d

32 50 10d 850

0.7854 75

= =

d 29.13 30 mm =

2. A M.S. bar of 12 mm dia: is subjected to an axial load of 50 KN in tension.

Find magnitude of induced stress?

( )

3

t 2

P 50 10

A / 12

= =

= 442.09 N/mm2 = 442.09 MPa.

3. If the length of bar in 1 m (dia 12 mm) and the modulus of elasticity of

material of bar is 2 105 MPa, find elongation of bar.

E

=

P / A PEdl / Ad

= = =

l

l l

( )

3

2 5

P 50 10 1 1000d

AE / 4 12 2 10

= =

ll = 2.21 mm

SHEAR STRESS AND SHEAR STRAIN

When the external force acting on a component tends to slide the adjacent

planes w.r.t each other, the resulting stress on these planes are called direct shearstresses.

shear stress

A Cross sectional area (mm2)

r Shear strain (radians)

= G.r.

G Modulus of rigidity (N/mm2)

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The relationship between the modulus of elasticity, the modulus of rigidity

and Poissons ratio is given by

E = 2G [1+ ]

- Poissons ratio

= strain in lateral dim.strain in axial dim.

The permissible shear stress is given by

( )0y

S

fs = Sy0 = yield strength in shear (N/mm2)

STRESSES DUE TO BENDING MOMENT

A st. beam is subjected to bending moment Mb as in fig.

bb

M y

I = b - bending stress at a distance of y from neutral axis (N/mm 2)

For an irregular cross section.

Mb - Applied Bending Moment

I Moment of inertia of C.O. abt NA (mm4)

I g =2y . dA

The parallel axis theorem:

21I 1 I g Ay = +

STRESSES DUE TO TORSIONAL MOMENT

The internal stresses, which are induced to resist the action of twist, are

called torsional shear stress.

tM .

J

= - Torsional shear stress (N/mm2)

Mt applied torque (N-mm)

J Polar moment of inertia (mm4)

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The angle of twist is given by

tM

JG =

l - angle of twist (radians)

l length of shaft (mm)

Power transmitted

t6

2 nMkW

60 10

=

1. Two plates subjected to a tensile force of 50 kN are fixed together by means

of three rivets (as in fig.). The plates and rivets are made of plain carbon steed 10C4

with a tensile strength of 250 N/mm2. The yield strength in shear is 50% of tensile

yield strength, and factor of safety is 2.5. Neglecting stress concentration

determines:

(i) The diameter of rivets

(ii) Thickness of plate

yield strength in shear is 50% of yield strength in tension.

To find diameter of Rivets

( )

Sys

fs = ( )

( )2

0

0.5 Syt 0.5 25050 N / mm

fs 2.5

= = =

3

32

P 50 10

A / 4 d

= =

3

250 1050

3 / 4 d=

2 50 1000 4d50 3

=

= 424.413

d = 20.68 = 22 mm

To find plate thickness:

2Syt 250

t 100 N / mmfs 2.5 = = =

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( ) ( )

P 50 1000 50 1000t

A 200 3d t 200 66 t

= = =

50 1000t 3.73 mm 4 mm134 100

= = =

2. A suspension link, used in bridge is shown in fig. The plates and the pins are

made of plain carbon steel (Syt = 400 N/mm2) and factor of safety is 5. The

maximum load in link is 100 kN. The ratio of width of the link plate to its thickness

(b/t) can be taken as 5. Calculate.

(i) Thickness and width of link plate

(ii) Dia. of knuckle pin

(iii) Width of the link plate at centre line of pin

(iv) Crushing stress on pin.

2t

Syt 40080N/mm

fs 5 = = =

2Sys 0.5Syt 0.5 400 40N/mm

fs fm 5

= = = =

LINK PLATE

P tA

=

3 5

2

100 10 100 1000 10t

(2t b) t 2t 5t 10t

= = =

52 10

t 12580 10= =

t = 11.18 = 12 mm

b = 5 t = 60 mm

1. A link shown in fig. is made of gray cast iron FG 150. It transmits a pull p of

10 KN. Assume that the link has square cross section (b = h) and using for 5,

determine the dimensions of the cross section of the link?

Solution:

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Given

Made of Grey cast iron (FG 150)

According to Indian Standard Specification (IS: 210-1993) the grey cast iron isdesignated by alphabets FG, followed by a figure indicating the minimum tensile

strength in MPa or N/mm2. FG150 means C.I. with 150 MPa or 150 N/mm2.

[C.I. is a brittle material]

- Tensile strength will be 100 200 MPa.

- Compressive strength = 400 to 1000 MPa.

- Shear strength = 120 MPa

Load P = 10 KN = 10 103 N

Square cross section A = b h

[allowable stress]( )

2Sut 150t 30N / mmfs 5

= = =

3P 10 10t

A b h

= =

310 10

30b h

=

b = h

32 10 10h

30

= = 333.33

h = 18.26 mm = 20 mm

The cross section of link is 20 20 mm

Points to be remembered:

- The dimensions of simple m/c parts are determined in the basis of pure

tensile stress, pure compressive stress, direct shear stress, bending stress or torsional

shear stress. The analysis is simple but approximate, because number of factors such

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as principal stresses, stress concentration reversal of stresses is neglected. Therefore

a higher FOS up to 5 can be taken into....

- It is incorrect to take allowable stress as data are design.

- The max. shear stress theory; proposes the yield strength in shear is 50% of

yield strength is tension.

Sys = 0.5 Syt

=( )

Sys

fs

2. The forces exerted by the levers of the pump on a rocking shaft are shown in

fig. The rocking shaft does not transmit torque. It is made of plain carbon steel 30 C 8

[Syt 400 N/mm2] and factor of safety is 5. Calculate diameter of the shaft.

Taking Moment about A:

20 200 + 30 800 RB 1050 = 0

RB 1050 = 28000

RB = 26.67 KN

y = 0

A BR R 20 30+ = +

AR 26.67 50+ =

AR 23.33 kN=

FACTOR OF SAFETY - FOS

While designing a component, it is necessary to ensure sufficient reserve strength in

the case of an accident. It is ensured by taking a suitable factor of safety (fs)

FOS can be defined as: [ all = allowable stress]

s

Failure Stressf

Allowable Stress= OR

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Failure loadfs

Working load=

The allowable stress is the stress value which is used in design to determine the

dimensions of the component. It is considered as a stress which the designer, expects

will not exceed under normal operating conditions.

For ductile material, the allowable stress ( all) is given by,

( )all

Syt

fs =

For brittle material:

( )all

Sut

fg = Syt = yield tensile strength, Sut = ultimate tensile stress

The FOS ensures against

- Uncertainty in the magnitude of external force acting on the component.

- Variations in the properties of materials like yield strength or ultimate

strength.

- Variations in the dimensions of the component due to imperfect

workmanship.

The magnitude of FOS depends upon following conditions:

1. Effect of failure:

Failure of the ball bearing in gear box.

Failure of valve in pressure vessel

2. Types of Load

- When external force acting on the m/c element is static - FOS is low.

- Impact load FOS is high

3. Degree of Accuracy in force analysis.

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When the force acting on the m/c element is precisely determined low FOS

can be selected. Where as higher FOS is considered when the m/c component is

subjected to a force whose magnitude or direction is uncertain and unpredictable.

4. Material of Component

When the component is made of homogenous ductile material, like steel,

yield strength is the criterion of feature. FOS is small in such cases. Cast Iron

component has non-homogenous structure and a higher FOS based on ultimate

strength is chosen.

5. Reliability of the component:

FOS & Reliability

- Defense

- Power stations

6. Cost of Components:

FOS & Cost

7. Testing of Machine element

Low FOS when m/c comp. are tested under actual condition of service and

operation.

8. Service conditions:

Higher FOS when m/c element is likely to operate in corrosive atmosphere

or high temp. environment.

9. Quality of Manufacture:

Quality is inversely proportional to FOS.

Lecture - 5

TYPES OF CYCLIC LOADING

Machine components are subjected to external force or load. The external

load acting on the component is either static or dynamic. The dynamic load is

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further classified into cyclic and impact loads. Static load is one: as a load which

does not vary in magnitude or direction with respect to time, after it has been applied.

Dynamic load is a load which varies in magnitude and direction w.r.t time,

after it has been applied.

There are two types of dynamic loads.

Cyclic load

Impact load

Cyclic Loads:

This is a load, which when applied, varies in magnitude in a repetitive cyclicmanner; either completely reversing itself from tension to compression or oscillating

about some mean value. In this case, the pattern of load variation w.r.to time is

repeated again and again.

Examples of cyclic loads are

- Force induced in gear teeth

- Loads induced in a rotating shaft subjected to B.M.

There are three types of mathematical models are of cyclic loads.

- Fluctuating OR alternating load

- Repeated loads

- Reversed loads

Stress time relationship corresponding to these three types of loads are given

below.

(a) Fluctuating stresses

(b) Repeated stresses

(c) Reversed stresses

The fluctuating or alternative load varies in a sinusoidal manner with respect

to time. It has some mean value as well as amplitude value. It fluctuates between

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two limits maximum and minimum load. The load can be tensile compressive or

partially tensile.

The repeated load varies in a sinusoidal manner w.r.to time but varies from

zero to some maximum value. The minimum value (load) is zero in this case and

therefore, amplitude load and mean loads are equal.

The reversed load varies in a sinusoidal manner w.r.to time, but it has zero

mean load. In this case, half portion of the cycle consists of tensile load and

remaining half of compressive load. There is complete reversal from tension to

compression between these two halves and therefore, mean load is zero.

max - Maximum stress

min - Minimum stress

m - Mean stress

a - Stress Amplitude

m - ( max + min)

a - ( max min)In the analysis of fluctuating stresses, tensile stress is considered as +ve, while

compressive stresses are ve.

STRESS CONCENTRATION

In design of m/c elements, following fundamental equations are used.

Pt =

A

bM yb =I

tM r=J

These equations are called elementary equations. These equations are based

on a number of assumptions:

1. Stress is proportional to strain.

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2. Modulus of elasticity is same in tension and in compression.

3. No high intensity contact pressure at the regions of the contact of support and

loading.

4. Machine members have constant section.

5. No abrupt change in section.

However, in practice, discontinuities and abrupt changes in cross-section are

unavoidable due to certain features of the component such as oil holes and grooves,

keyways and splines, screw threads etc. Hence it cannot be assumed that cross-

section of the machine component is uniform. Under these circumstances, the

elementary equations do not give correct result.

In order to visualize the effect of discontinuities and abrupt changes in cross

section on distribution of stresses, a concept called FLOW ANALOGY is used.

Fig. (a) shows a bar of uniform cross section subjected to axial tensile force. In flow

analogy concept, the force is visualized as flowing through the bar. Each flow line in

the figure represents a certain amount of force. These flow lines are called Force

flow lines. Since bar has uniform cross section, flow lines are uniformly spaced. Fig.

(b) shows an identical bar but with notch cut on its circumference. If we consider a

cross section of this bar away from the notch, the flow lines are uniformly spaced

showing normal distribution of tensile stress. As the line approaches the notch they

are bent in order to pass through restricted openings. The bending of force flow lines

indicates weakening of the material. The effect of stress concentration is proportional

to bending of flow lines. When force flow lines are bent, the load carrying capacity is

reduced and material becomes weak. Let us consider a flat plate Fig. (c) of uniform

thickness subjected to axial tensile force. At the section xx, there is sudden change in

height.

At both ends flow lines are parallel indicating uniform distribution of stresses. At the

right end, they are close together indicating higher magnitude of stresses. At the left

end they are spaced comparatively away from each other, indicating lower

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magnitude of stress. When these lines from left and right side join at section at xx,

they are bent indicating weakening of material.

There are two terms normal stresses and localized stresses. Normal stresses

are shown at two ends with uniform distribution. The localized stresses are restricted

to local regions of the component such as sections of discontinuity.

(i) Normal stresses: are stresses in the machine component at a section away

from discontinuity or abrupt change of cross section. Localized stresses are stresses

in local regions of the component such as the sections of discontinuity or sections of

change of cross section.

(ii) Normal stresses are determined by elementary equations. It is not possible to

use these formulae for localized stresses.

(iii) The normal stresses are comparatively of small magnitude. The localized

stresses in the vicinity of discontinuity are frequently of large magnitude. This may

give rise to a crack.

(iv) Failure rarely occurs in region of normal stresses. The region of localized

stresses is more vulnerable to fatigue failure.

Let us consider a plate with a small circular hole as in figure (d).

The distribution of stresses near the hole can be observed by keeping a model

of the plate made of epoxy resin in circular polariscope. The localized stresses in the

neighbourhood of the hole are far greater than the stress obtained by elementary

equation.

Stress concentration is defined as the localization of high stresses due to

irregularities or abrupt changes of cross section. In order to consider the effect of

stress concentration and find out localized stresses stress concentration factor is

used. It is denoted by Kt.

Highest value of actual stress near discontinuityKt

Normal stress obtained by elementary equation for mini. cross section=

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max max

0 0

Kt

= =

0

,0

stresses determined by elementary equip.

max , max Localized stresses at discontinuities.

The causes of stress concentration are:

(i) Variation of properties of materials

(ii) Local application

(iii) Abrupt change in section

(iv) Discontinuities in the component

(iv) Machining scratches

Methods to reduce stress concentration

(i) Additional notches and holes in tension member.

Eg. It is observed that a single notch results in a high degree of stress

concentration. The severity of stress concentration can be reduced by three methods:

(a) Use of multiple notches

(b) Drilling additional holes

(c) Removal of undesired material

The method of removal of undesired material is called Principle of

Minimization of Material.

STRESS CONCENTRATION FACTORS

Case 1

Stress concentration factor for a rectangular plate with a transverse hole

loaded in tension:

0

P P

A (w d)t = =

t plate thickness.

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Case 2

The values of stress concentration factor for a flat plate with a shoulder fillet

subjected to tensile or compressive force are determined from,

0

P

d t =

Case 3

Stress concentration factor for a round shaft with shoulder fillet subjected to

tensile force, bending moment and torsional moment are taken from Data book; The

nominal stresses in these three cases are as follows:

(a) Tensile force,0

2

P

d4

=

(b) Bending Moment, b0M y

I =

4

dI64

=

dy

2=

(c) Torsional Moment, t0M r

J =

4dJ

32

=

dy

2=

There are a number of geometric shapes and conditions of loading. A

separate chart for the stress concentration factor should be used for each

combination.

The effect of stress concentration depends upon the material of component.

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PROBLEMS

Q1. A flat plate subjected to a tensile force of 5KN is shown in fig. The plate

material is grey cast iron FG200 and factor of safety is 2.5. Determine the

thickness of the plate.

Ans:

The stresses are critical at two sections.

At Fillet Section:

0 Pd t

= 500030 t

= (1)

D 451.5

d 30= =

r 50.167

d 30= =

From Data Book:

Kt = 1.8

Kt =max

0

max t 0

5000K . 1.8

30 t = = 2

300N / mm

t

=

(A)

At the Hole Section:

20

P 5000 5000N / mm

(w d) t (30 15) t 15 t = = =

To find out Max. stress ( max )

d 150.5

w 30= =

From Data Book:

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Kt = 2.15

2max t 0

2.16 5000 720K . N / mm

15 t t

= = =

(B)

From (A) and (B), Maximum stress is induced at hole section. Hence equating it with

permissible stress,

utmax

s

S

f =

PRINCIPAL STRESSES

Lot of mechanical components are available which is subjected to several

types of loads simultaneously. Stresses can be classified mainly into two groups:

Normal and shear stress.

There is a particular system of notation available.

The normal stresses are denoted by x, y and z in x, y, z direction.

Tensile stresses are considered as positive.

Compressive stresses are considered as ve.

Fig.

xy The subscript x indicates that the shear stress is acting on the area,

which is perpendicular to the x-axis; y indicates in the y-direction.

xy = yx

Fig.

This fig. shows stresses acting on an oblique plane. The normal to the plane

makes an angle with x-axis. and are normal and shear stress associated with

this plane.

x y x yxycos 2 sin 2

2 2

+ = + +

x yxysin 2 sin 2

2

= +

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27/75

xy

x y

2tan 2

=

On giving Max. value

( )2

2x y x y1 xy

2 2

+ = + +

( )2

2x y x y2 xy

2 2

+ = +

( )2

2x ymax xy

2 = +

Application of principal stresses in designing Machine Parts

- There are many cases in practice, in which m/c members are subjected to combined

stresses due to simultaneous action of either tensile or compressive stresses

combined with shear stress.

Eg: propeller shafts, c-frames etc.

Maximum tensile stress:

( ) ( )2 2t

tt max

14

2 2

= + +

Maximum compressive stress:

( ) ( )2 2c

cc max

14

2 2

= + +

Maximum shear stress:

( )2 2

max t

14

2

= +

A hollow shaft of 40 mm outer diameter and 25 mm inner diameter is subjected to a

twisting moment of 120 N m, simultaneously, it is subjected to an axial thrust of 10

kN and a bending moment of 80 N m. Calculate the maximum compressive and

shear stress?

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Step-1: Write given data:

d0 = 40 mm

di = 25 mm

mT = 120 N m

= 120 N m = 120 103 N mm

Mb = 80 N m = 80 103 N mm

Step-2: Find cross sectional area of shaft

A = ( ) ( )2 2

0 1d d

4

= ( ) ( )2 2

40 254

= 765.783 mm2

Step-3: Compressive stress due to axial thrust:

32

0

P 10 1013.058 N / mm

A 765.783

= = =

Step-4: To find bending stress:

1. Find section modulus:

z =( ) ( )

4 4

0 1

0

d d

32 d

=( ) ( )

4 440 25

32 40

= 5325 mm3

2. Bending stress due to bending moment:

b =M

z=

380 10

5325

= 15.02 M Pa

Step 5: Find total compressive stress

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c = c b +

= 15.02 + 13.05

= 28.07 N/mm2

Step 6: Find shear stress

Mt =( ) ( )

4 4

0 1

0

d d

16 d

= 11.27 N/mm2

Step 7: Maximum compressive stress

( )c max = ( )2 2c

c

14 32.035 MPa

2 2

+ + =

Step 8: Maximum shearing stress:

max = ( )2 2

c

14

2

+

= ( ) ( )

2 21

28.07 11.272

+

= 18 M Pa

MOHRS CIRCLE

One of the most effective means to determine the principal stress and the principal

shear stress.

It is the graphical representation of stresses.

Steps:

The normal stresses x y, and the principle stresses 1 2, are plotted in the

abscissa. The tensile stress considered as positive, is plotted to the right of the origin

and compressive stresses are considered as negative, to its left.

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The shear stresses xy yx, and principal shear stress max are plotted on the

ordinate. A pair of shear stresses is considered as positive if they tend to rotate the

element clockwise and negative if they tend to rotate in anticlockwise.

The Mohrs circle is constructed by the following method:

(i) Plot the following

OA = x

OC = y

AB = xy

CD = yxx y

xy

tan2

=

FATIGUE FAILURE

It has been observed that materials fail under fluctuating stresses, at a stress

magnitude which is lower than the ultimate tensile strength of material. The

magnitude of stress causing fatigue failure decreases on the number of stress cycle

increases. This phenomenon of decreased resistance of the materials to fluctuating

stresses is called fatigue. The fatigue failure begins with a crack at some points in

the material. The crack is more likely to occur in following regions:

(i) Regions of discontinuity, such as oil holes, keyways, screw thread

(ii) Regions of irregularities in machining operations such as scratches on the

surface stamp mark, inspection mark etc.

(iii) Internal cracks due to defects in materials like blow holes.

These regions are subjected to stress concentration due to the crack. The

crack spreads due to fluctuating stresses, under the cross- section of the component is

20 reduced that the remaining portion is subjected to sudden fracture.

(i) Regions indicating skew growth of crack with a fine fibrous appearance.

(ii) Region of sudden fracture with a course granular appearance.

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In the case of failure under static load, there is always sufficient plastic

deformation prior to failure which gives warning well in advance.

ENDURANCE LIMIT

The fatigue or endurance limit for a material is defined as the maximum

value of completely reversing stress that the standard specimen can sustain for an

unlimited number of cycles with out fatigue failure. The endurance limit is denoted

by 1Se is considered as a criterion of failure under fluctuating.

ENDURANCE LIMIT APPROXIMATE ESTIMATION

The laboratory method for determining the endurance strength of materials

although more precise, is laborious and time consuming.

When the laboratory method for determining endurance strength of material

is not available then another as explained below is used

Se - Endurance limit stress for a rotating beam specimen subjected to reversed

bending stress (N/mm2)

Se Endurance limit stress for a particular mechanical component subjected to

reversed bending stress (N/mm2)

The approximate relationship between the endurance strength and ultimate

strength (tensile) is given as:

utSe 0.5 S = For steels

utSe 0.4 S = For cast iron and cast steels

While designing a mechanical component, it is unrealistic to expect that itsendurance limit will match the values obtained in the laboratory standard test and

controlled condition.

Therefore the laboratory derived endurance strength needs a correction. Thus

the endurance strength of a mechanical part can be found out by:

a b c d eSe k k k k S =

ka - Surface finish factor

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kb Size factor

kc Reliability factor

kd Modifying factor to accurent for stress concentration.

The rotating beam specimen has a highly published surface which is free

from scratches. When the surface finish is poor the endurance strength is reduced

because of scratches. The variation in surface finish between the specimen and

component is accounted by surface furnish factor (ka).

The size factor kb depends upon the size of the cross section as the size of the

component increases, the surface area also increases, resulting in greater number of

surface defects.

DIAMETER (d) mm K b

d 7.5 1.00

7.5 d 50 0.85

d > 50 0.75

FOR CIRCULAR CROSS SECTION

(d) - dia. of cross section

For non-circular cross section d depth

ENDURANCE LIMIT

The fatigue or endurance limit for a material is defined as the maximum value

of completely reversing stress that the standard specimen can sustain for an

unlimited number of cycles without fatigue failure.

The endurance limit, denoted as Se is considered as the criterion of failure

under fluctuating stress.

In the laboratory, the endurance limit is determined by means of a rotating

beam machine.

The rotating beam machine is developed by R.R. Moore.

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The test specimen is of standard size and has a highly polished surface. It is

rotated by an electric motor and numbers of revolutions before the

appearance of the first crack are recorded on a revolution counter.

Self aligning ball bearings are used to ensure that only radial loads are

applied to the specimen.

Test specimen is subjected to pure bending moment, and magnitude of

bending stress is adjusted by means of weights.

To determine endurance limit stress of a material large numbers of tests are

carried out.

SODERBERG AND GOODMAN DIAGRAMS

When a component is subjected to fluctuating stresses the stresses are

resolved in to two components, mean stress m and stress amplitude a. The

fatigue diagram for the general case is shown in fig.

In this case mean stress is plotted on the abscissa, with tensile stress to right

of the origin and compressive stress to its left. The stress amplitude is plotted

on the ordinate.

The magnitude of m and a depends upon the magnitude of force acting on

the component.

When stress amplitude a is zero, the load is purely static and criterion of

failure is Sut or Syt.

These limits for tension as well as for compression are plotted on the

abscissa.

When the mean stress m is zero, the stress is completely reversing and

criterion of failure is endurance strength Se, which plotted on the ordinate.

When the component is subjected to both types of stresses m and a, the

actual failure occurs at different scattered points as in the figure.

A 2t line joining Se on the ordinate to Sut on the abscissa is called the

Goodman line.

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While a line joining Se on the ordinate to Syt on the abscissa is called

soderberg line.

The Goodman line or Soderberg line are used as criteria of failure when the

component is subjected to mean stress as well as stress amplitude.

MODIFIED GOODMAN DIAGRAM

The components which are subjected to fluctuating stresses are designed by

constructing the modified Goodman diagram. For the purpose of design the

problems are classified into two groups.

Components subjected to axial or bending stress due to fluctuating force or

bending moment.

Components subjected to fluctuating torsional shear stress.

Modified Goodman diagram for axial or bending stress

In this diagram, the yield strength Syt is plotted on both the axes abscissa

and ordinate, and a line is drawn to given these two points to define failure by

yielding.

Another line is constructed to join Se on the ordinate with Sut on the abscissa,

which is the Goodman line.

There will be a point of intersection of two lines. The area under curve

represents the region of safety for components subjected to fluctuating loads.

From Fig.

tan =a

m

=a a

m m

P / A P

P / A P=

tan =a

m

P

P

Similarly it can be proved

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tan =( )

( )

b a

b m

M

M

The point of intersection of lines AB and OE is X. The point X indicates the

dividing line between the safe region and region of failure.

The co-ordinates of the point X (Sm, Sa), represents limiting values of stress which

are used to calculate the dimensions of the component.

( )a

A

S

fs = and

( )m

m

S

fs =

FLUCTUATING TORSIONAL SHEAR STRESS

The torsional yield strength is plotted on both the axes and a line is

constructed to join these two points.

A line is drawn through Sse on the ordinate and parallel to abscissa.

The pt of intersection of two lines is B.

Area under the curve represents regions of safety.

While solving the problem, a line OE with slope tan is constructed.

( )

( )

ta a

m t m

Mtan

M

= =

saa

S

fa =

( )sm

ms

S

f =

THEORIES OF FAILURE

The mechanical properties are generally obtained by the simple tension test.

The simple tension test gives information about the tensile yield strength, the

ultimate tensile strength and percentage elongation.

The relationship between the strength of a mechanical component subjected

to a complex state of stresses and mechanical properties of simple tension test is

obtained by theories of simple tension test is obtained by theories of failures. With

the help of these theories, the data obtained in the simple tension test that can be used

35


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to determine the dimensions of the component, irrespective of the nature of stresses

induced in the complex loads.

MAXIMUM NORMAL STRESS THEORY

This criterion of failure is accredited to British engineer W.J.M. Rankine

(1850).

The theory states that the failure of the mechanical component subjected to

bi-axial or tri-axial stresses occurs when the maximum normal stress reaches the

yield or ultimate strength of the material.

If 1, 2 and 3 are three principal stresses at a point on the component

and

1 > 2 > 3

Then according to this theory, the failure occurs whenever

1 = Syt OR 2 = Sut whichever is applicable.

The theory considers only the maximum principal stresses and disregards the

influence of other two principal stresses.

For tensile stresses

1 =yt

s

S

f

1 =ut

s

S

f

For compressive stresses:

1 =( )

yc

s

S

f

1 = ( )uc

s

S

f

For bi-axial stresses:

Syc = Syt [Assumption]

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[Ref. Fig. (A)]

Maximum Shear Stress Theory

This criterion of failure is accredited to CA Coulomb H. Tresca and J.J.Guest. This theory states that the failure of a mechanical component is subjected to

bi-axial or tri-axial stresses occurs when the maximum shear stress at any point in the

component become equal to the maximum shear stress in the standard specimen of

the simple tension test, when yielding starts. The stress in the specimen and

correspond Mohrs circle diagram.

From Fig.

max 1/ 2 =

When the specimen starts yielding ( )1 ytS =

ytmax

S

2 =

yt ytS 0.5S =

If 1, 2 and 3 are the three principle stresses at a point on the

component, the shear stress on three different planes is given by

1 212

2

= 2 323

2

= 1 331

2

=

The largest of these stresses equal to max (or) Syt/2

1 2

2

=

( )

ytS

2 fs

1 2 =( )

ytS

fs

yt2 3S

2 fs

= yt1 3

S

fs =

The square represents the region of safety. If a point with co-ordinates

( 1, 2) falls outside this square then it indicates the failure condition.

This theory not recommended for ductile material.

DISTORTION ENERGY THEORY

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This theory was advanced by M.T. Huber in Poland 1904 and independently

by R. Von Mises in Germany (1913) and H. Henky (1925). It is known as Huber van

Mises and Henkys theory. This theory states that the failure of the mechanical

component subjected to bi-axial or tri-axial stresses occurs when the S.E. of

distortion per unit vol. at any point in the component become equal to the S.E. of

distortion per unit vol. in a standard tension test specimen when yielding starts.

A unit cube subjected to the three principal stress 1, 2 and 3 as shown

in fig.

The total S.E. U of the cube is given by:

1 1 2 2 3 31 1 1U2 2 2

= + + .......... (I)

Where 1 2 3, , are the strains in respective direction.

Also we have,

( )1 1 2 31

E = +

( )2 2 1 31E

= + ............. (II)

( )3 3 1 21

E = +

The total S.E. U is resolved in two components Ur and Ud.

U = Ur + Ud............ (III)

Correspondingly stresses are resolved into two components:

1 1 1d v = +

2 2d v = + ........ (IV)

3 3d v = +

The components 1 2 3d, d, d cause distortion of the cube while v causes

a volumetric change. Since 1 2 3d, d, d do not change vol. of the cube.

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1 2 3d d d 0 + + = ........... (V)

[ ]1 1 2 31

d d d dE

= +

[ ]2 2 1 31

d d d dE

= + ............. (VI)

[ ]3 3 1 21

d d d dE

= +

Sub. IV in V.

( ) [ ]1 2 31 2 d d d 0 + + =

1 2 0

1 2 3d d d + + = 0 .......... (VII)

VII in IV.

( )v 1 2 31

3 = + + ....... (VIII)

The S.E. Ur corresponding to the change of vol. for the cube is given by

( )v v v3

U2

= ....... (IX)

( )v v v v1

= +

( ) v1 2

E

.........(X)

X in IX

Ur =( ) 2v3 1 2

2E

......... (XI)

VIII in XI Uv =( ) ( )

2

1 2 31 2

6E

+ + ...... (XII)

Ud = U Ur.... (XIII)

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Putting XII in XIII

( ) ( )2

1 2 3d

1 2U u

6

+ + =

( ) ( )2 2 21 2 3 1 2 2 3 1 31

U 22E

= + + + +

( ) ( ) ( )2 22

1 2 2 3 3 1

1Ud

6E

+ = + + ......... (XIV)

Simple tension test:

1 = Syt

2 = 3 = 0

2d yt

1U S

3E

+

........ (XV)

From XIV and XIII, the criterion of failure for distortion energy theory is expressed

as:

( ) ( ) ( )2 222yt 1 2 2 3 3 12S = + + ................. (XIV)

( ) ( ) ( )2 222

yt 1 2 2 3 3 1

1S

2 = + + ........ (XVII)

( ) ( ) ( )2 22

yt 1 2 2 3 3 1S = + + .......... (XVIII)

For bi-axial stress

( )yt 2 2

1 1 2 2

S

f s= +

POWER SCREW

A mechanical device meant for converting rotary motion into translational

motion for transmitting power. Main applications of power screws are:

(i) Raise the load, eg: Screw Jack

(ii) Accurate motion in machine operation

40


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(iii) To clamp work piece

(iv) To load a specimen

The main advantage of power screws are their large load carrying capacitywith small overall dimensions. Power screws are simple to design, easy to

manufacture and give smooth and noiseless service.

FORMS OF THREADS:

There are four types of threads used in power screws: They are acme, I.S.O.

metric, trapezoidal, and buttress.

Selection guide lines:

1. The efficiency of square thread is more than that of other types of threads.

2. Square threads are difficult to manufacture.

3. The strength of a screw depends upon the thread thickness at the core

diameter.

4. The wear of the thread surface becomes a serious problem in application like

the lead screw of the lathe.

5. Buttress thread can transmit power and motion only in one direction. While

square and trapezoidal threads can transmit force and motion in both

direction.

FORCE ANALYSIS

The major dimensions of the power screws are shown in figure.

d Nominal or outer dia (mm)

dc core or inner dia (mm)

dm mean diameter(mm)

Selection guide lines:

When square threads are used for the screw, it can be treated as an inclined

plane wrapped helically round a cylinder. The helix angle of the thread is given

by:

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1tan

dm =

W, is the load which is raised or lowered by rotating the screw by means of

an imaginary force P acting at the mean radius. There are two different cases

depending upon the load being raised or lowered.

For an equilibrium of horizontal forces,

NP cos Nsin= + ...... (A)

For an equilibrium of vertical forces,

N cos Nsin = ........ (B)

Dividing expression (A) by (B).

( )

( )

cos sinP

cos sin

+ =

Dividing the numerator and denominator of the right hand side by cos .

( )

( )

tanP

1 tan

+ =

(C)

The coefficient of friction is expressed as

= tan

Where is the friction angle

( )P w tan= +

The torque required to raise the load.

t

PdmM

2=

( )tWdm

M tan2

= +

Equilibrium horizontal and vertical forces:

P N cos Nsin= ......... (A)

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W N cos Nsin= + ......... (B)

Dividing (A) by (B)

( )w cos sinPcos sin =

+

Dividing the numerator and denominator of R.H.S. bycos .

( )w tanP

1 tan

=

+

KNUCKLE JOINT

Knuckle join is used to connect two rods whose axes are either coincide orintersect and lie in one plane.

The construction of this joint permits limited relative angular movement

between rods in this plane about the axis of the pin.

The rods connected by knuckle joint are subjected to tensile force.

APPLICATION

Joint between the links of suspension bridge.

Joint in valve mechanism of reciprocating engine.

Fulcrum for levers.

Joint between tie bars in roof stress.

Knuckle joint is unsuitable to connect two rotating shaft which transmit torque.

ADVANTAGES

The joint is simple to design and manufacture.

The assembly or dismantling of parts of knuckle joint is quick and simple.

For the purpose of stress analysis of knuckle joint following assumptions are made:

i) The rods are subjected to axial tensile force.

ii) The effect of stress concentration due to holes is neglected.

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45/75

b 3

32 P b a

d 2 4 3

= +

6. Tensile Failure of Eye

( )t

0

P

b d d =

7. Shear failure of eye

The eye is subjected to double shear. The area of each of the two planes

resulting shear failure is [ ]0b d d / 2

( )0

P

2 b d d / 2 =

( )0

P

b d d=

8. Tensile Failure of Fork:

( )

t

0

p

2a d d

=

9. Shear failure of Fork:

( )0

P

2a d d =

a = 0.75 n

b = 1.25 D

d1 = 1.5 d

x = 10 mm

COTTER JOINT

Cotter joint is used to connect two co-axial rods, which are subjected to either

axial tensile or compressive force.

It is used to connect rod on one side with some machine parts like cross head

or base plate on the other side.

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The principle of wedge action is used in cotter joints.

A cotter is a wedge shaped piece of made of steel plate.

The joint is tightened and adjusted by means of wedge action of cotter.

Notations:

P = tensile force acting on the rod (N)

d = diameter of each rod (mm)

D1 = outside diameter of socket (mm)

d1 = dia. of spigot or inside dia of socket (mm)

d2 = Diameter of spigot collar (mm)

D = Diameter of socket collar (mm)

t1 = thickness of spigot collar (mm)

l = Axial distance from slot to end of socket collar (mm)

B = Mean width of cotter (mm)

t = thickness of cotter (mm)

L = Length of cotter (mm)

l1 = Distance from end of slot to end of spigot on rod B (mm)

In order to design the cotter joint and find out the above dimensions, failure

in different parts and at different cross sections are considered.

1. Tensile Failure of Rods

Each rod of diameter d is subjected to tensile force P.

DESIGN OF SQUARE AND FLAT KEYS:

The design of square and flat keys is based two criteria.

1. Failure due to either shear stress or compressive stress.

The forces acting on the key is shown in fig.

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47/75

The force p acts as a resisting couple preventing the key to roll in the

keyway. It is assumed that the force p is tangential to shaft diameter.

t tM 2M

p d / 2 d= = .......... (I)

The shear failure is given by

p

b =

l...... (II)

From (I) and II

t2M

db =

l

The compressive stress is given by

c

p 2p

h / 2 h = =

l l....... (III)

From I and III

tc

4M

dh

=l

SPLINES:

Splines are keys which are made integral with the shaft. They are used when

there is a relative axial motion between the shaft and hub.

The torque transmitting capacity of splines

Mt = pm ARm

The area A is given by

A = (D d) n

km =D d

4

+

( )2 2t m m1

M p D d8

= l

MUFF COUPLING:

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48/75

Design procedure

1. Calculate the diameter of each shaft by following equations

6

t60 10 pM

2 n =

Or t

316M

d =

2. Calculate the dimensions of sleeve by following equations.

D = 2d + 13

L = 3.5 d

also check the torsional shear stress

tM .r =

4 4D d

32

=

Dr

2=

3. Determine the standard cross section of flat key from Data book.

L

2=l

t2M

db =

l

tc

4M

dh =

l

Shafts and keys are made of plain carbon steel. The sleeve is usually made

of grey cast iron of grade FG 200.

CLAMP COUPLING

1. Calculate the diameter of each shaft by following eqn.

6

t

60 10 PM

2zn

= t3

16M

d =

2. Calculate the main dimensions of sleeve halves;

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D = 2.5 d, L = 3.5 d

3. Determine standard cross section of flat key.L

2=l

t2M

db =

l

tc

4M

dh =

l

4. Calculate the diameter of clamping bolt

t1

2MP

fdn

= and 21 1 tz

p d

4

=

FLANGE COUPLING

1. Shaft Diameter: Calculate the shaft diameter using following equations.

6

t

60 10 pM

2 n

=

t

3

16M

d =

2. Dimensions of Flange.

dh = 2d, lh = 1.5 d, D = 3d, t = 0.5 d, t1 = 0.5 d, t1 = 0.25 d, dr = 1.5 d,

D0 = (4d + 2t1)

Torsional shear stress in hub can be calculated by considering it as hollow

shaft subjected to torsional moment Mt.

tM r

J = ( )

4 4hz d d

J32

=

hdr2

=

The flange at the jun. of hub is under shear while transmitting the torsional moment

Mt.

2t h

1M d t

2=

3. Diameter of Bolts:

N = 3 for d < 40

N = 40 for 40 d < 100 mm

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N = 6 for 100 d < 180 mm

2 t1

8Md

DN=

4. Dimensions of key:

Standard dimensions of key can be obtained from table 4.1 Data book. The

length of key in each shaft is lh.

h =l l

t2M

db =

l

tc

4M

dh =

l

FLEXIBLE COUPLING

1. Shaft diameter: Calculate shaft dia. by using following equation.

6

t

60 10 pM

2 n

=

t

3

16M

d =

2. Dimensions of flanges: Calculate dimensions of flanges by following

empirical equations.

dh = 2d, lh = 1.5 d, D = 3d to 4d, t = 0.5 d, t1 = 0.25 d

The torsional shear stress in the hub can be calculated by considering it as

hollow shaft subjected to torsional moment Mt.

tM .r

J = ( )

4 4hd d

J32

=

hd

r 2=

3. Diameter of pins:

The number of pins is usually 4 or 6. The diameter of pins is calculated by

1

0.5 dd

N=

Determine the shear stress

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t21

8M

d DN =

4. Dimensions of pins: Calculate the outer diameter of the rubber bush.

2t b

1M D DN

2=

lb = Db.

5. Dimensions of the key

l = ln

t2Mdb = l

tc

4M

dh =

l

THIN CYLINDERS

A cylinder is considered to be thin when the ratio of its inner diameter to the

wall thickness is more than 1.5. There are two principal stresses in their cylinders.

i i tD p 2 t=

i it

p D

2t =

Considering equilibrium of forces in longitudinal direction.

( )2i i t iP D D t4

=

t = i iP D

4t

It is seen that circumferential stress t is twice the longitudinal stress.

i i

t

P Dt

2=

THICK CYLINDER

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When the ratio of inner diameter of cylinder to the wall thickness is less than

1.5, the cylinder is said to be thick walled.

i) In thick cylinder, the tangential stresst

has highest magnitude at inner

surface of the cylinder.

ii) The radial stress r is neglected in this cylinder.

( ) ( )t r r r 2 dr 2 r dr d 2r + + + =

Neglecting the term ( )rdr d+

( )t r rd

r 0dr + + = ............... (I)

It is further an aimed that the axial stress t is uniformly distributed over the

cylinder wall thickness.

tr

E E E

= + l

l ........... (II)

r t

E

=

l

l

Where E is the modulus of elasticity and is the poisons ratio.

r t 12c =

adding I and III

( )r r 1d

2 r 2cdn

+ =

Multiplying both sides of the above equation by r.

( )2r r 1d

2 .r r 2c r dr

+ = ............ (IV)

Since ( ) ( )2 2r r rd d

r 2r r dr dr

= +

( )2 r 1d

r 2c r

dr

=

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53/75

Integrating w.r. to r.

2 2r 1 2r c r c = +

2r 1 2

cc

r = +

Value of constants c1 and c2 are evaluated from

r = pi when iD

ri

=

r = 0 when 0D

r2

=

( )

21 2

1 2 20

p Dc

D D=

l

( )

2 21 2 0

2 2 20 2

p D Dc

4 D D=

( )

220i 1

r 22 20 1

Dp D1

4rD D

=

RIVETED JOINTS

Rivet Materials

Rivets are made of Wrought iron on soft steel for most uses, but where

corrosive resistance or light weight is requirement, rivets of copper or aluminium is

used.

Kinds of riveted joints: Joints may be single, double triple. If the rivets in the

rows are in line crosswise the joint is said to be chain riveted. While if the rivets in

adjacent rows offset by one half the centre to centre distance, then it is said to be zig-

zag riveted.

Failure of Riveted joints: These joints may fail in no. of ways, they are single

shear, double shear, failure by tearing of plate. Failure of margin, crushing of rivet or

hole. Failure by tearing due to bending of plate.

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Design of a Riveted joint

1. Plates may tear in the line of mini section.

The line of mini section is the line through the centre of hole.

Area of plate along this line = (p d) t

Plate renaissance to lesion = (p d) t ft

3. Plate and rivets may crush or fail in compression.

Resistance to crushing = dt fc z.

4. The rivets may shear

Resistance to single shear = /4 d2 fs z.

Resistance to double shear = /4 d2 fs kz.

Efficiency of joint:

Tearing efficiently =( ) tp d t f p d

pt ft p

=

cc

dt f z

pt ft =

( )e

1 2 2 22 2 3 a

p nF =

+ + +

l

l l l l

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MODULE - 2

SPRINGS

Helical springs Stress Equations

There are two basic equations for the design of helical springs: They are (a)

Load-stress (b) Load-deflection.

A helical spring is made of round wire.

Spring Index: (c) which is defined as

DC

d

= D Mean coil diameter, d diameter of wire.

In practice, the value of spring index varies from 6 to 12.

When the wire of helical spring is uncoiled and straightened, it takes the

shape of a bar of diameter d and length DN.

N No. of active coils.

The torsional shear stress in the bar is given by

t1 3

16M

d =

Mt = F .D/2 From Fig.

1 3

8FD

d =

............... (I)

The direct shear stress in the bar is given by:

2 2

F F

A / 4 d = =

2

4P

d=

...... (II)

Maximum shear stress induced in the wire

1 2 = +

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3 2

8FD 4F

d d= +

3

8FD d

1 2Dd

= +

3

8FD 0.5 d1

Dd

= + ......... (III)

A shear stress correction factor ks is defined as

s

0.5k 1

c= + ........ (IV)

Substituting this value in eqn. (III)

s 3

8FDk

d

= eqn. 11.1.a. pg. 139

When the bar is bent in the form of a helical coil, the length of inside fibre is less

than the length of outer fibre. This results in stress concentration at the inside fibre

of coil.

The equation for resultant stress is given by

s 3

8FDk

d

=

where k is called Wahl factor:

4c 1 0.615k

4c 4 e

= + eqn. 11.2a

The Wahl correction factor k consists of two factors correction factor ks for directshear stress and correction factor kc for curvature effect.

k = ks kc.......... (A)

value of k can be obtained from pg. 156 Fig. 11.1

cs

kk

k = .......... (B)

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ENERGY STORED IN HELICAL SPRINGS

The springs are used to store energy.

Assuming that the load applied is gradually, the energy stored in the spring.

1U . F .

2= ........ (I)

We know

3

8FD

d

=

3

dF8DK

= ............. (A)

To find deflection:

The work done by the axial force F is converted into strain energy and stored

in the spring.

Strain energy U = Work done by F

= avg. torque angular displacement

U = tM

2

= tM

Gl

Mt = F . D / 2

l = DN

=4d

32

U =2 3

4

4F D N

Gd

According to Cartiglianos theorem, the displacement corresponding to force

P is obtained by partially differentiating strain energy w.r. to that force F

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( ) 2 3

4

u 4P D N

p p Gd

= =

=3

48FD N

Gd- eqn. 11.5a (D.B.) Pg. 139

- axial deflection of spring.

N No. of active coils.

Energy stored in the spring.

1U F

2=

3 3

4

1 d 8FD N

2 8KD Gd

=

21 FD N

2 KGd

= From DB 11.8

The stiffness of the spring (k) is defined as the force required to produce unit

deflection.

Fk=

4

3

Gdk

8D N=

Spring in series:

1 2

1 1 1

k k k

= + since1 2

= +

Spring in parallel

k = k1 + k2 since F = F1 + F2

BUCKLING OF SPRING MATERIALS

It has been found that the free length of the spring (LF) is more than four

times the mean or pitch diameter (D), then the spring behaves like a column and may

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fail by buckling at a comparatively low load. The critical axial load (For) that causes

buckling may be calculated using following relations;

Fcr= k kB LF

k spring rate or stiffness of the springF

=

kB Buckling factor depending up on the (k1 in data book] ration LF/D.

LF Free length of the spring. (l0 in dB)

value of kB can be obtained from Figure 11.3. Pg. 157

DESIGN AGAINST STATIC LOADING

1. Material for the spring.

From Table. 11.2 Or 11.8

2. The style of end coils:

From table 11.7. Different types of spring coil materials.

i (in data book) or N no. of active coils can be found out from this step.

(iii) The spring index (c)

- For industrial application (c) varies from 8 to 10. A spring Index of 8

is considered as good value.

- The spring index (c) for springs in values and dutches is taken as 5.

- The spring index should never be less than 3.

The factor of safety based on torsional yield strength (Ssy) is 1.5 for springs

subjected to static forces.

The permissible shear stress d is given by.

syd

S

1.5 =

Syt = 0.75 Sut Syt yield strength in tension

Ssy = 0.557 Syt Sut ultimate strength

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d ut0.3 S =

The Indian Standard 4454-1975 recommends.

d ut0.5 S =

We know s 28FC

kd

=

i.e, 21

d

DESIGN AGAINST FLUCTUATING LOADS

The springs subjected to fluctuating stresses are designed on the basis of two

criteria.

(1) Design for infinite life

(2) Design for finite life.

Let us consider a spring subjected to a fluctuating force, which changes its

magnitude from Pmax to Pmin in the load cycle. The mean force Pm and the force

amplitude Pa are given by

( )m max min1

P P P2

= +

( )a max min1

P P P2

= .......... (I)

The mean torsional shear stress is given by

mm s 3

P .D

k d

= ......... (II)

where ks is the correction factor or direct shear stress.

For torsional shear stress amplitude a, it is necessary to consider the effect

of stress concentration.

aa c s 3

P . Dk k

d

=

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aa 3

P .Dk

d

= ........ (III)

- A spring is never subjected to a completely reversed load, changing its

magnitude from tension to compression and passing through zero w.r. to time.

- A helical spring is subjected to purely compressive forces.

- In general, the spring wires are subjected to pulsating shear stress, which vary

from zero to seS

seS - Endurance limit in shear

For cold drawn steel wires;

se utS 0.21 S =

sy utS 0.42 S=

For oil hardened and tempered steel wires.

se utS 0.22 S =

sy utS 0.45 S=

Where Sut is the ultimate tensile strength. The failure diagram for the spring

is shown in fig.

Point A with co-ordinates [ ]se seS , S indicates the failure point of spring

wire.

Pt. B on the abscissa indicates failure under static condition.

When the stress reaches the torsional yield strength (Ssy)

AB is called line of failure.

sySOD

fs=

Line DC is parallel to BA . Any point on line CD such as X represents a stress

situation with the same factor of safety.

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Consider similar triangles XFD and AEB.

XF AE

FD EB=

a se

sy sym

S

S S Se

fs

=

SHAFT SUBJECTED TO COMBINED TWISTING AND BENDING

MOMENT.

When the shaft is subjected to combined twisting moment and bending

moment, then the shaft must be designed on the basis of two moments

simultaneously varies theories have been suggested to accent for the elastic failure of

the material when they are subjected to various types of combined stresses.

Following two theories are important.

1. Maximum shear stress theory. It is used for ductile materials such as milk

steel.

2. Maximum normal stress theory or Rankines theory.

- shear stress induced due to twisting moment.

b - Bending stress (tensile or compressive).

Induced due to bending moment.

According to maximum shear stress theory:

( )2 2

max b

14

2

= +

Substituting

bb 3

32M

d =

t3

16M

d =

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2 2

b tmax 3 3

32M 16M14

2 d d

= +

2 2b t3

16 M Md

= +

The expression 2 2b tM M+ is known as equivalent twisting moment. (Mte).

From this diameter of shaft can be evaluated. According to maximum normal

stress theory; the maximum normal stress in the shaft

( )2 2

b bb max

1 14

2 2

= + +

Substituteb

b 3

32M

d =

t3

16M

d =

( )2 2

b b tb 3 3 3max

32M 32M 16M1 14

2 2d d d

= + +

( )2 2b b t332 1

M M M2d

= + +

( ) 3 2 2b b b tmax1

d M M M32 2

= + +

The expression ( )2 2b b t1 M M M2

+ + is known as equivalent bending moment. (Mbe)

[Check: equations in data book 3.49, 3.4b, 3.5 a]

SHAFT SUBJECTED TO FLUCTUATING LOADS

In practice, the shafts are subjected to fluctuating torques and bending

moment. In order to design such shafts like line shafts and countershafts, the

combined shock and fatigue factors must be taken into account for computed

twisting and bending moment.

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Thus for a shaft subjected to combined bending and torsion, the equivalent

twisting moment (Mte).

( ) ( )2 2

te m b t tM C .M C M= +

and Equivalent bending moment.

( ) ( )2 2

be m b m b t t

1M C M C M C M

2

= + +

Cm - combined shock and fatigue factor for bending.

Ct combined shock and fatigue factor for torsion.

Values of Cm and Ct can be obtained from TABLE 3.1 constants for ASME

codes Page. 47.

* Check eqns. 3.69, 3.66, 37a, 3.7b. Pg: 43

Shafts subjected to axial load in addition to combined Torsion and bending loads.

When the shaft is subjected to an axial load (F) in addition to torsion and

bending loads as in propeller shafts of ships and shafts for driving worm gear, then

the stress due to axial load must be added to bending stress ( b).

bM

I y

=

b

M.y

I =

4

M.d/2

/ 64 d=

b3

32M

d=

Stress due to axial loads

F

A=

2 2

F 4F

/ 4 d d

= =

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For hollow shaft d is replaced by d.

2 2 2 20 1 0 1

F 4F

/ 4 d d d d= =

2

2 i0

0

4F

dd 1

d

=

2 20

4F

d 1 k=

Resultant stress for solid shaft:

b1 3 2

32M 4F

d d = +

b3

32 F dM

8d

= +

For hollow shafts:

( )( )20

1 2 40

Fd 1 k 32M

8d 1 k

+ = +

1 - Resultant stress

In case of long shafts slender shafts subjected to compressive loads, a factor

known as column factor ( ) must be introduced to take column effect.

Stress due to the compressive load

c 2

4F

d

=

D

2 20

4F

d (1 k )

=

D

The value of column factor ( ) for compressive loads may be obtained from,

following relation.

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Column factorL

K

1

1 0.0044( ) =

When

L

115K< , above equation is used.

WhenL

115K

>

2Ly K

2

( )

E

=

D

DESIGN OF SHAFT ON THE BASIS OF RIGIDITY

1. Torsional rigidity: is important in the case of Cam shaft of an IC Engine

where the timing of the valves would not be affected. The permissible amount of

twist should not exceed 0.25o per metre length of such shaft.

For line shafts or transmission shafts, deflection 2.5 to 3 degree/meter length

may be used as limiting value.

T G.

L

=

Torsional deflection or angular deflection.

2. Lateral Rigidity:

It is important for maintaining proper bearing clearances and for correct gear

teeth alignment.

When shaft is of varying cross section, then the lateral deflection may be

determined from the fundamental equation for the elastic curve

2d y2

dx

M

EI=

DESIGN OF CONNECTING ROD

Connecting Rods:

It is an intermediate link between the piston and the crank shaft of an IC

engine. The basic purpose of it is to transmit motion and force from piston to the

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crank pin. It consists of a long shank, a small and a big end. The cross section of

shank may be rectangular, circular tubular, I-section or H-section. Length of

connecting rod (l) depends up on the ratio l r , where r is the radius of crank.

Design of Connecting Rod:

In designing a connecting rod, the following dimensions are required;

1. Dimensions of cross-section of the connecting rod:

A connecting rod is to subjected to alternating direct compressive and

tensile forces; Hence the cross section of the connecting rod is designed as a strut and

the Rankines formulae is used.

A connecting rod as in figure-A, is subjected to an axial load W may buckle with X-

axis as neutral axis or Y-axis as neutral axis. The connecting rod is considered like

both ends hinged for buckling about X-axis and both ends fixed for buckling about

Y-axis.

A = Cross sectional area of the connecting.

l = Length of connecting rod.

c = Compressive yield stress

WB = Buckling load

Ixx, Iyy = Moment of Inertia of section about X-axis and Y-axis

respectively.

Kxx, Kyy = Radius of gyration of the section about X-axis and Y-axisrespectively.

Rankines formulae

WB about X-axis =

c2

xx

.A

l1 a

K

+

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WB about Y-axis =

c2

yy

.A

l1 a

2K

+

a = 1/7500 for mild steel

= 1/9000 for Wrought Iron.

= 1/1600 for Cast Iron

In order to have a connecting rod equally strong in buckling about both the axis, the

buckling loads must be equal.

c2

xx

.Al

1 aK

+

=

c2

yy

.Al

1 a2K

+

2

xx

l

K

=

2

yy

l

2K

2xxK =

2yy4K

Ixx = 4Iyy

This shows that connecting rod is four times strong in buckling about Y-axis than

about X-axis. In actual practice, Ixx is kept slightly less than 4Iyy. It is usually taken

between 3 and 3.5.

The most suitable section for the connecting rod is I-section with the proportions as

in figure.

Thickness of the flange and web = t

Width of the section = B = 4t

Depth or Height of the section = H = 5t

Area of the section = 2(4t*t) + 3 txt

= 11 t2

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Moment of Inertia of the section about X-axis

3 3 4xx

1 419I 4t(5t) 3t(3t) t

12 12 = =

3 3 4yy

1 1 131I 2 t (4t) (3t)t t

12 12 12

= + =

xx

yy

I3.2

I=

After determining the proportions for I-sections of the connecting rod, its dimensions

are determined by considering the buckling of the rod about X-axis and applying the

Rankines formulae.

Buckling load

cB 2

xx

.AW

L1 a

K

=

+

OR

WB = Max. gas force Factor of safety

FOS may be taken as 5 to 6.

2. DIMENSIONS OF THE CRANK PIN AT THE BIG END AND THE

PISTON PIN AT THE SMALL END:

Since dimensions of the crank pin at the big end and the piston pin at the

small end are limited. The crank pin at the big end has removable precision bearing

shells of brass or bronze or steel with a thin lining of bearing metal; on the inner

surface of the shell.

The allowable bearing pressure on the crankpin depends upon many factors such as

material of the bearing, viscosity of lubricating coil, method of lubrication and the

space limitation.

The value of bearing pressure may be taken as 7 N/mm2 to 12 N/mm2

depending up on the material and method of lubrication used.

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The piston pin bearing is usually a phosphor bronze bush of about 3mm

thickness and the allowable bearing pressure may be taken as 10.5 N/mm2 15

N/mm2.

Since the maximum load to be carried by the crank pin and piston pin bearing is the

maximum force in the connecting rod (Fc). Therefore dimensions for these two pins

are determined for the maximum force in the connecting rod (F c) which is taken

equal to maximum force on piston due to gas pressure (FL) neglecting inertia forces.

Maximum gas force

FL =2D P

4

.................. (i)

D Cylinder bore or piston dia in mm

P Maximum gas pressure in N/mm2.

To find out dimensions of crankpin and piston pin;

dc Dia of the crankpin

lc Length of crankpin

Pbc Allowable bearing pressure in N/mm2.

dp, lp, Pbp Corresponding values for the piston pins.

Load on crankpin:

= dc.lc.Pbc .................. (ii)

Load on piston pin:

= dp.lp.Pbp .................. (iii)

Equating (i) and (ii).

FL = dc.lc.Pbc

Take lc = 1.25 dc to 1.5 dc

Equating (i) and (iii)

FL = dp.lp.Pbp (lp = 1.5 dp to 2 dp)

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3. SIZE OF BOLTS FOR SECURING BIG END CAP:

Inertia force on reciprocating parts;

21 R

rF m .w .r 1l

= +

mR Man of the reciprocating part

W Angular speed of the engine rad/s

r Radius of crank in meters

l Length of connecting rod.

The both may be made of high carbon steel or nickel alloy steel.

Factor of safety may be taken as 6.

Force on the bolts;

=2

cb b(d ) t n4

D

dcb Core dia of bolt in mm.

Equating the inertia force to force on bolt;

2I cb bF .(d ) . t.n

4=

D

bn No. of bolts

From this expression dcb is obtained

cb

b

dd

0.84=

bd Nominal or major dia.

4. Thickness of the big end cap:

Thickness of big end cap (tc) may be determined by treating the cap as a

beam freely supported at the cap bolt centres and loaded by the inertia forces at the

top dead centre on the exhaust stroke. This load i assumed to be act in between the

uniformly distributed load and centrally concentrated load.

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Max. bending moment acting on the cap

Ic

F xM

6

=

PISTON MATERIALS

Cast iron is the most popular material used for the construction of piston

for low speed engines. Mostly closed grain pearlitic CI is used.

Limitation of CI is high weight density, which in turn increases the

inertia force.

Modern high speed engine uses aluminium.

DESIGN OF PISTON:

1. Piston Crown or Piston Head

Simplest piston crown is a flat type.

The selection of piston crown depends up on the requirement of volume

for the combustion chamber and valve arrangement.

The thickness of the piston crown can be calculated by using eqn. assuming the

crown to be a flat plate of uniform thickness and fixed at the edges;

0.5

maxPt 0.43Dt

=

Besides this, the piston crown has to carry the heat that is generated during

combustion of fuel.

2

c e

D qt 1600k(T T )=

2. PISTON RINGS:

Generally two types of piston rings are used

Compression rings

Oil rings

The radical thickness of cast iron piston ring may be computed by

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nrad

3Pt D=

Minimum depth of piston ring or axial thickness (h) can be found out by equation

Dh

10i=

The distance from top to the first groove tg from [equ. 10.30 pg 363]

PISTON BARREL

This the cylindrical portion of the piston.

Maximum thickness can be found out using equation [18.20 page 362]

t3 = 0.03 D + b + 4.5

PISTON SKIRT

The portion below the ring section is known as piston skirt.

The length of piston skirt should be such that the bearing pressure on the

piston barrel due to side thrust does not exceed 0.25 N/mm2.

Max. gas load on the piston

2DP P

4=

D

Max. side thrust on cylinder; [R]

2

P10

DR 0.1P

4= =

D[Usually side thrust [R] is taken as 1/10 of gas load].

R = Bear

74878611 MachineDesign I

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Transcript of 74878611 MachineDesign I