Page 1


Machine Design For

Mechanical Engineering By

www.thegateacademy.com


Syllabus

Machine Design

Syllabus for Machine Design Design for static and dynamic loading; failure theories; fatigue strength and the S-N diagram; principles of the design of machine elements such as bolted, riveted and welded joints, shafts, spur gears, rolling and sliding contact bearings, brakes and clutches.

Analysis of GATE Papers (Machine Design) Year

Percentage of marks

2013

6.00

2012

8.00

2011

3.00

2010

6.00

2009

6.00

2008

9.33

2007

11.33

2006

5.33

2005

3.33

Overall Percentage

6.48%

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Content

Machine Design

CONTENTS

#1.

#2.

#3.

Chapters

Page No.

Design for static Loading

1 - 18

      

1 1-7 8 - 13 14- 15 15 - 16 17 17 - 18

Introduction Theories of Failure Solved Examples Assigment 1 Assigment 2 Answer Keys Explanations

Design for Dynamic Loading

19 - 43

       

19 19 – 20 21 – 31 32 – 36 37 – 38 38 – 39 40 40 – 43

Introduction Stress Concentration Fatigue and Endurance Limit Solved Examples Assignment 1 Assignment 2 Answer Keys Explanations

Design of Joints

44-69

       

44 44 – 51 51 – 57 58 – 63 64 – 65 65 – 66 67 67 – 69

Introduction Riveted Joints Bolted/Screw Joints Solved Examples Assignment 1 Assignment 2 Answer Keys Explanations

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Content

#4.

#5.

#6.

#7.

Machine Design

Design of Shaft and Shaft Components

70-89

       

70 70 – 72 72 – 78 79 – 83 84 85 – 86 87 87 – 89

Introduction Shaft design for stress Shaft components Solved Example Assignment 1 Assignment 2 Answer Keys Explanations

Design of Bearing

90-112

        

Introduction Rolling contact Bearings Bearing Life Sliding contact/Journal Bearing Solved Example Assignment 1 Assignment 2 Answer Keys Explanations

90 90 – 96 96 – 99 99 – 103 104 – 106 107 – 108 108 – 109 110 110 – 112

Design of Brakes and Clutches

113-135

       

Introduction Brake Design Clutch Design Solved Example Assignment 1 Assignment 2 Answer Keys Explanations

113 113– 118 119 – 124 125 – 127 128 – 129 129 – 131 132 132 – 135

Design of Spur Gears

136-150

  

136 136 – 138 138

Introduction Gear Nomenclature Spur Gear: Theory of Machines

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Content

      

Lewis Equation – Beam strength of gear teeth Permissible working stress Solved Example Assignment 1 Assignment 2 Answer Keys Explanations

Machine Design

138 – 139 139 – 141 142 – 145 146 147 148 148 – 150

Module Test

151-161

Test Questions

151– 156

Answer Keys

157

Explanations

157 – 161

Reference Books

162

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Chapter 1

Machine Design

CHAPTER 1 Design for Static Loading Introduction A static load is a stationary force or couple applied to a member. To be stationary, the force or couple must be unchanging in magnitude, point or points of application and direction. A static load can produce axial tension or compression, a shear load, a bending load, a torsional load, or any combination of these. To be considered static, the load cannot change in any manner. In most testing of those properties of materials that relate to the stress-strain diagram, the load is applied gradually, to give sufficient time for the strain to fully develop. Furthermore, the specimen is tested to destruction, and so the stresses are applied only once. Testing of this kind is applicable, to what are known as static conditions; such conditions closely approximate the actual conditions to which many structural and machine members are subjected. Another important term in design is “failure”. The definition of failure varies depending upon the component and its application. Failure can mean a part has separated into two or more pieces; has become permanently distorted, thus ruining its geometry; has had its reliability downgraded; or has had its function compromised, whatever the reason.

Theories of Failure Events such as distortion, permanent set, cracking and rupturing are among the ways that a machine element fails. In uni-axial tension test the failure mechanisms is simple as elongations are largest in the axial direction, so strains can be measured and stresses inferred up to “failure.” The “failure” conclusion becomes challenging when the loading is bi-axial or tri-axial. Unfortunately, there is no universal theory of failure for the general case of material properties and stress state. Instead, over the years several hypotheses have been formulated and tested, leading to today’s accepted practices. These “practices” as known as theories of failure and are used to analyse the failure of materials. Structural metal behaviour is typically classified as being ductile or brittle, although under special situations, a material normally considered ductile can fail in a brittle manner Ductile materials are normally classified such that εf ≥ 0.05 and have an identifiable yield strength that is often the same in compression as in tension (Syt = Syc = Sy ). Brittle materials, εf < 0.05, do not exhibit identifiable yield strength, and are typically classified by ultimate tensile and compressive strengths, Sut and Suc, respectively (where Suc is given as a positive quantity). The generally accepted theories are: Ductile Materials (yield criteria)   

Maximum shear stress (MSS) Distortion energy (DE) Octahedral shear stress theory

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Chapter 1

Machine Design

Maximum normal stress (MNS)

Maximum-Shear-Stress Theory for Ductile Materials The maximum-shear-stress theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension test specimen of the same material when that specimen begins to yield. The MSS theory is also referred to as the Tresca or Guest theory. Many theories are postulated on the basis of the consequences seen from tensile tests. As a strip of a ductile material is subjected to tension, slip lines (called Lüder lines) form at approximately 45° with the axis of the strip. These slip lines are the beginning of yield, and when loaded to fracture, fracture lines are also seen at angles approximately 45° with the axis of tension. Since the shear stress is maximum at 45° from the axis of tension, it can be considered as the mechanism of failure.MSS theory is an acceptable but conservative predictor of failure; and can be applied to many cases where over design is not a problem. Recall that for simple tensile stress, σ = P/A. And the maximum shear stress occurs on a surface 45° from the tensile surface with a magnitude of = σ/2. So the maximum shear stress at yield is = /2. For a general state of stress, three principal stresses can be determined and ordered such that σ ≥ σ ≥ σ . The maximum shear stress is then = σ σ )/2 Thus, for a general state of stress, the maximum-shear-stress theory predicts yielding when =

σ

σ

σ

σ ≥

Note that this implies that the yield strength in shear is given by = 0.5 Which as we will see later is about 15 percent low (conservative) For design purposes, equation can be modified to incorporate a factor of safety, n. Thus =

n

or σ

σ =

n

Distortion-Energy Theory for Ductile Materials (The von Mises or von Mises – Hencky theory) The distortion-energy theory predicts that yielding occurs when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same material. The distortion-energy (DE) theory originated from the observation that ductile materials stressed hydrostatically exhibited yield strengths greatly in excess of the values given by the simple tension test. Therefore it was postulated that yielding was not a simple tensile or compressive phenomenon at all but, rather that it was related somehow to the angular distortion of the stressed element. To develop the theory, note in Fig. a, the unit volume subjected to any three-dimensional stress state designated by the stresses σ1, σ2, and σ3. The stress state shown in Fig. b is one of hydrostatic tension due to the stresses σav acting in each of the same principal directions as in Fig. a. The formula for σav is simply σ

=

(a)

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Chapter 1

Machine Design

Thus the element in Fig. b undergoes pure volume change, that is no angular distortion. If we regard σav as a component of σ1, σ2 and σ3, then this component can be subtracted from them, resulting in the stress state shown in Fig. c. This element is subjected to pure angular distortion that is, no volume change.

=

) (a)

)

)

Element with triaxial stresses; this element undergoes both volume change and angular distortion. Element under hydrostatic tension undergoes only volume change Element has angular distortion without volume change.

(b) (c)

The strain energy per unit volume for simple tension is u = the strain energy per unit volume is u = [ σ principal strain gives = [

σ. For the elements of figure (a) σ ]. Substituting equation for the

σ

)]

)

The strain energy for producing only volume change u can be obtained by substituting , , and in equation (b). The result is σ E

u =

)

for

.

)

If we now substitute the square of equation (a) in equation (c) and simplify the expression we get =

[

]

)

Then the distortion energy is obtained by subtracting equation (d) from equation (b). This gives u =u

u =

E

[

σ

σ )

σ

σ )

σ

σ )

]

)

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Chapter 1

Machine Design

Maximum-Normal-Stress Theory for Brittle Materials The maximum-normal-stress (MNS) theory states that failure occurs whenever one of the three principal stresses equals or exceeds the strength. Again we arrange the principal stresses for a general stress state in the ordered form σ ≥ σ ≥ σ . This theory then predicts that failure occurs whenever σ ≥

or σ

(p)

where Sut and Suc are the ultimate tensile and compressive strengths respectively, given as positive quantities. For plane stress, with the principal stresses with σ ≥ σ , Eq. (p) can be written as σ ≥

or σ

(q)

which is plotted in Fig. a. As before, the failure criteria equations can be converted to design equations. We can consider two sets of equations for load lines where σ ≥ σ as σ

σ 0 σ

σ

)

)

Note that the distortion energy is zero if σ = σ = σ . For the simple tensile test, at yield σ = is u =

v E

and σ = σ = 0 and the Eq. (e) the distortion energy

f)

So for the general of stress given by Eq. (e) yield is predicated if Eq. (e) equals or exceeds Eq. (f) this gives THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  info@thegateacademy.com © Copyright reserved. Web: www.thegateacademy.com Page 4


Chapter 1

[

σ )

σ

σ

σ )

σ )

σ

Machine Design

/

]

g)

If we had a simple case of tension σ, then yield would occur when σ ≥ . Thus, the left of Eq. (g) can be thought of as a single, equivalent or effective stress for the entire general state of stress given by σ , σ , and σ . This effective stress is usually called the von Mises stress, σ named after Dr. R. von Mises, who contributed to the theory. Thus Eq. (g) for yield, can be written as σ ≥ where the von Mises stress is σ =[

σ

σ )

)

σ )

σ

σ )

σ

/

]

)

For plane stress, let σ and σ be the two nonzero principal stresses. Then from Eq. (i) we get )

σ =

.

)

Octahedral- Shear – Stress Theory Octahedral-shear-stress theory is an extension of Distortion-Energy theory and it is based on the assumption that critical quantity is the shearing stress on the octahedral plane. Consider an isolated element in which the normal stresses on each surface are equal to the hydrostatic stress σ . There are eight surfaces symmetric to the principal directions that contain this stress. This forms an octahedron as shown in Fig. The shear stresses on these surfaces are equal and are called the octahedral shear stresses (Fig. has only one of the octahedral surfaces labeled). Through coordinate transformations the octahedral shear stress is given by = [ σ

σ )

σ

σ )

σ

σ ) ]

m)

σ

σ

σ σ

Octahedral surfaces

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Machine design : Mechanical Engineering, THE GATE ACADEMY  

THE GATE ACADEMY's GATE Correspondence Materials consist of complete GATE syllabus in the form of booklets with theory, solved examples, mod...

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