Design of Machine Elements (ME30109) - Amazon AWS

BY: Dr. Rahul KumarAssistant Professor

(Dept. of Mechanical Engineering)

C. V. Raman Global University, Bhubaneswar

Design of Machine Elements

(ME30109)

Overview

Design of Machine Elements 2

The main objective of the course is to build a strongfoundation and to acquaint the students with the:

1 Fundamentals and the phases of machine design.

2 Understanding on the process of material selection and design.

3 Use of design data books & various codes of practice.

4 Concepts of strength design related to various machine elements.

5 Design methodology for simple joints, couplings, fasteners, leversand springs under static loading.

Overview


CO1 Explain basic principles in the design of machine elements

& apply them effectively from material selection to design

analysis using design data book to interpret standardized

data.

CO2 Design and analyze permanent and temporary joints under

various loading conditions.

CO3 Design and analyze power transmission shafts and couplings with

different geometrical features under various loading conditions.

CO4 Explain various thread forms and design power screws.

CO5 Identify types of springs & levers and perform design analysis.

Overview


Unit 1• Fundamentals of Machine Design

Unit 2• Design of Temporary and Permanent Joints

Unit 3• Design of Shaft, Keys and Couplings

Unit 4• Design of Power Screws and Threaded joints

Unit 5• Design of Mechanical Springs and levers

Topics to be covered:

Stress and deflection equations forhelical compression Springs, Typesof ends, Design of helicalcompression and tension springs,Helical torsion Spring, Surge insprings, Design of semi-ellipticalleaf springs

Overview


Unit 5 • Springs

Topics to be covered (by self):

References:

“Mechanical Engineerng Design”, J.E. Shigley, C.R. Mischke, R.G. Budynas andK.J. Nisbett, Tata McGraw-Hill , 8th Edition, 2008

“Design of Machine Elements”, V.B. Bhandari, Tata McGraw-Hill, 3rd Edition, 2010.T3. “A Text Book of Machine Design”,

R.S. Khurmi and J.K. Gupta, S. Chand Publication, 14th Edition, 2005.

Design Hand Book”, S.M. Jalaluddin; Anuradha Agencies Publications.

Overview


Unit 4 • Power Screws and Threaded joints

Unit 5


Unit 5 • Springs

STYLES OF END

Unit 5


Unit 5 • Springs

STRESS AND DEFLECTION EQUATIONS

There are two basic equations for the design of helical springs, viz.,

load-stress equation and load-deflection equation.

The torsional moment

The torsional shear stress in

the bar is

Unit 5


Unit 5 • Springs


When the equivalent bar is bent in the form of helical coil, there are

additional stresses on account of following two factors:

There is direct or transverse shear stress in the spring wire.

When the bar is bent in the form of coil, the length of the inside

fibre is less than the length of the outside fibre. This results in stress

concentration at the inside fibre of the coil.

The resultant stress consists of superimposition of torsional shear

stress, direct shear stress and additional stresses due to the curvature of

the coil.

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Unit 5 • Springs


Pure Torsional

Stress

Direct Shear Stress

Combined Pure Torsional, Direct Shear

Stress and Stress due to curvature

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Unit 5 • Springs


The combined effect of these two factors

The direct shear stress in the bar is

The resultant shear stress in the spring wire is

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Unit 5 • Springs


The shear stress correction factor (Ks) is defined as

Substituting the above equation in the expression

AM Wahl derived the equation for resultant stress, which includes

torsional shear stress, direct shear stress and stress concentration due

to curvature

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Unit 5 • Springs


The angle of twist for the equivalent bar

The axial deflection of the spring, for small values of angle of twist is

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Unit 5 • Springs


The rate of spring (k) is given by

DESIGN OF HELICAL SPRINGS

STEP 1: For the given application, estimate the maximum spring force

(P) and the corresponding required deflection (d) of the spring. In some

cases, maximum spring force (P) and stiffness k, which is (P/d), are

specified

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Unit 5 • Springs


STEP 2: Select a suitable spring material and find out ultimate tensile

strength (Sut) from the data.

Calculate the permissible shear stress for the spring wire by following

relationship:

STEP 3: Assume a suitable value for the spring index (C).

For industrial applications, the spring index varies from 8 to 10. A

spring index of 8 is considered as a good value. The spring index for

springs in valves and clutches is 5.

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Unit 5 • Springs


STEP 4: Calculate the Wahl factor

STEP 5: Determine wire diameter (d)

STEP 6: Determine mean coil diameter (D)

STEP 7: Determine the number of active coils (N)

STEP 8: Decide the style of ends for the spring depending upon the

configuration of the application. Determine the number of inactive coils.

Adding active and inactive coils, find out the total number of coils (Nt)

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Unit 5 • Springs


STEP 9: Determine the solid length of the spring

STEP 10: Determine the actual deflection of the spring

STEP 11: Assume a gap of 0.5 to 2 mm between adjacent coils, when

the spring is under the action of maximum load. The total axial gap

between coils is given by,

Total gap = (Nt – 1) * gap between two adjacent coils

STEP 12: Determine the free length of the spring

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Unit 5 • Springs


STEP 13: Determine the pitch of the coil

STEP 14: Determine the rate of spring

Question1 : It is required to design a helical compression spring subjected

to a maximum force of 1250 N. The deflection of the spring corresponding

to the maximum force should be approximately 30 mm. The spring index

can be taken as 6. The spring is made of patented and cold-drawn steel wire.

The ultimate tensile strength and modulus of rigidity of the spring material

are 1090 and 81370 MPa respectively. The permissible shear stress for the

spring wire should be taken as 50% of the ultimate tensile strength.

Unit 5


Unit 5 • Springs


Question 2: A helical compression spring, made of circular wire, is

subjected to an axial force, which varies from 2.5 kN to 3.5 kN. Over this

range of force, the deflection of the spring should be approximately 5 mm.

The spring index can be taken as 5. The spring has square and ground ends.

The spring is made of patented and cold-drawn steel wire with ultimate

tensile strength of 1050 MPa and modulus of rigidity of 81370 MPa. The

permissible shear stress for the spring wire should be taken as 50% of the

ultimate tensile strength.

Unit 5


Unit 5 • Springs

DESIGN OF LEAF SPRINGS

Multi-leaf springs are widely used for the suspension of cars, trucks and

railway wagons. A multi-leaf spring consists of a series of flat plates,

usually of semi-elliptical shape

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Unit 5 • Springs


Unit 5


Unit 5 • Springs


Graduated-length Leaves as Triangular Plate

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Unit 5 • Springs


Extra full-length leaves asrectangular plate

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Unit 5 • Springs


Since the deflection of full-length leaves is equal to the deflection of

graduated- length leaves

NOTE: It is seen from the above equations that bending stresses in full-

length leaves are 50% more than those in graduated-length leaves.

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Unit 5 • Springs


The deflection at the end of the spring is determined from

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Unit 5 • Springs

NIPPING OF LEAF SPRINGS

As discussed in the previous section, the stresses in extra full-length

leaves are 50% more than the stresses in graduated-length leaves. One

of the methods of equalising the stresses in different leaves is to pre-

stress the spring. The pre-stressing is achieved by bending the leaves

to different radii of curvature, before they are assembled with the

centre clip.

Unit 5


Unit 5 • Springs


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Unit 5 • Springs


Under the maximum force P, the deflection of graduated-length leaves

will exceed the deflection of extra full-length leaves by an amount

equal to the initial nip C

The initial pre-load Pi required to close the gap C between the extra

full-length leaves and graduated length leaves is determined by

considering the initial deflection of leaves

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Unit 5 • Springs


The resultant stress in the extra

full-length leaves is obtained by

superimposing the stresses due to

initial pre-load (Pi) and the

external force (P)

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Unit 5 • Springs


Question 3: A semi-elliptic leaf spring used for automobile suspension

consists of three extra full-length leaves and 15 graduated-length leaves,

including the master leaf. The centre-to-centre distance between two eyes

of the spring is 1 m. The maximum force that can act on the spring is 75

kN. For each leaf, the ratio of width to thickness is 9:1. The modulus of

elasticity of the leaf material is 207000 MPa. The leaves are pre-stressed

in such a way that when the force is maximum, the stresses induced in all

leaves are same and equal to 450 MPa. Determine:

(i) the width and thickness of the leaves;

(ii) the initial nip; and

(iii) the initial pre-load required to close the gap C between extra full-

length leaves and graduated-length leaves

Unit 5


Unit 5 • Springs

SURGE OF SPRINGS

When the natural frequency of vibrations of the spring coincides with

the frequency of external periodic force, which acts on it, resonance

occurs. In this state, the spring is subjected to a wave of successive

compressions of coils that travels from one end to the other and back.

This type of vibratory motion is called ‘surge’ of spring. Surge is found

in valve springs, which are subjected to periodic force.

The natural frequency of helical compression springs held between two

parallel plates is given by

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Unit 5 • Springs

SURGE OF SPRINGS

The natural frequency of helical compression springs with one end on

the flat plate and the other end free, supporting the external force is

given by

Surge in springs is avoided by the following methods:

(i) The spring is designed in such a way that the natural frequency of the

spring is 15 to 20 times the frequency of excitation of the external force

Unit 5


Unit 5 • Springs

SURGE OF SPRINGS

Surge in springs is avoided by the following methods:

(ii) The spring is provided with friction dampers on central coils. This

prevents propagation of surge wave.

(iii) A spring made of stranded wire reduces the surge.

Any suggestions and Queries?????????

Design of Machine Elements (ME30109) - Amazon AWS

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