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Mechanics of Aeronautical Solids Materials and Structures

Mechanics of Aeronautical Solids Materials and

Structures

Christophe Bouvet

First published 2017 in Great Britain and the United States by ISTE Ltd and John Wiley amp Sons Inc

Apart from any fair dealing for the purposes of research or private study or criticism or review as permitted under the Copyright Designs and Patents Act 1988 this publication may only be reproduced stored or transmitted in any form or by any means with the prior permission in writing of the publishers or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address

ISTE Ltd John Wiley amp Sons Inc 27-37 St Georgersquos Road 111 River Street London SW19 4EU Hoboken NJ 07030 UK USA

wwwistecouk wwwwileycom

copy ISTE Ltd 2017 The rights of Christophe Bouvet to be identified as the author of this work have been asserted by him in accordance with the Copyright Designs and Patents Act 1988

Library of Congress Control Number 2016962010 British Library Cataloguing-in-Publication Data A CIP record for this book is available from the British Library ISBN 978-1-78630-115-4

Contents

Foreword ix

Preface xi

Introduction xiii

Chapter 1 Stress 1

11 Notion of stress 1 111 External forces 1 112 Internal cohesive forces 2 113 Normal stress shear stress 2

12 Properties of the stress vector 3 121 Boundary conditions 3 122 Torsor of internal forces 5 123 Reciprocal actions 8 124 Cauchy reciprocal theorem 9

13 Stress matrix 11 131 Notation 11 132 Invariants of the stress tensor 13 133 Relation between the stress matrix and the stress vector 15 134 Principal stresses and principal directions 18

14 Equilibrium equation 21 15 Mohrrsquos circle 23

Chapter 2 Strain 27

21 Notion of strain 27 211 Displacement vector 27

vi Mechanics of Aeronautical Solids Materials and Structures

212 Unit strain 28 213 Angular distortion 30

22 Strain matrix 33 221 Definition of the strain matrix 33 222 Principal strains and principal directions 37 223 Volume expansion 39 224 Invariants of strain tensor 40 225 Compatibility condition 40

23 Strain measurement strain gage 41

Chapter 3 Behavior Law 43

31 A few definitions 43 32 Tension test 43

321 Brittle materials 44 322 Ductile materials 45 323 Particular cases 46

33 Shear test 46 331 Brittle materials 47 332 Ductile materials 48

34 General rule 48 341 Linear elasticity 48

35 Anisotropic materials example of a composite 53 351 Elasticity 53

36 Thermoelasticity 54

Chapter 4 Resolution Methods 59

41 Assessment 59 42 Displacement method 61 43 Stress method 61 44 Finite element method 62

Chapter 5 Work-energy Theorem Principle of Finite Element Method 63

51 Work-energy theorem 63 511 Hypotheses 63 512 Strain energy 64 513 Work of external forces 65 514 Strain energy 66 515 Energy minimization Ritz method 68

Contents vii

52 Finite element method 69 521 General principle of finite element method 69 522 Example of the three-node triangular element 74

53 Application triangle with plate finite element using Catia 80

Chapter 6 Sizing Criteria of an Aeronautical Structure 83

61 Introduction 83 62 Experimental determination of a sizing criterion 85 63 Normal stress or principal stress criterion brittle material 87 64 Stress or maximum shear energy criterion ductile material 91

641 Tresca criterion 91 642 Von Mises criterion 93 643 Rupture of a ductile material 96

65 Maximum shear criterion with friction compression of brittle materials 99 66 Anisotropic criterion example of the composite 105

Chapter 7 Plasticity 109

71 Introduction 109 72 Plastic instability necking true stress and true strain 111 73 Plastic behavior law RambergndashOsgood law 116 74 Example of an elasticndashplastic calculation plate with open hole in tension 118

Chapter 8 Physics of Aeronautical Structure Materials 127

81 Introduction 127 82 Aluminum 2024 130 83 Carbonepoxy composite T300914 135 84 Polymers 140

Chapter 9 Exercises 151

91 Rosette analysis 151 92 Pure shear 154 93 Compression of an elastic solid 154 94 Gravity dam 155 95 Shear modulus 156 96 Modulus of a composite 157 97 Torsional cylinder 158 98 Plastic compression 160 99 Bi-material beam tension 162

viii Mechanics of Aeronautical Solids Materials and Structures

910 Beam thermal expansion 164 911 Cube under shear stress 165 912 Spherical reservoir under pressure 166 913 Plastic bending 169 914 Disc under radial tension 171 915 Bending beam resolution by the Ritz method 173 916 Stress concentration in open hole 174 917 Bending beam 178

Chapter 10 Solutions to Exercises 183

101 Rosette analysis 183 102 Pure shear 191 103 Compression of an elastic solid 192 104 Gravity dam 196 105 Shear modulus 201 106 Modulus of a composite 203 107 Torsional cylinder 206 108 Plastic compression 212 109 Bi-material beam tension 215 1010 Beam thermal expansion 225 1011 Cube under shear stress 231 1012 Spherical reservoir under pressure 235 1013 Plastic bending 240 1014 Disc under radial tension 245 1015 Bending beam resolution by the Ritz method 252 1016 Stress concentration in open hole 256 1017 Bending beam 259

Appendix 273

Bibliography 279

Index 281

Foreword

This book follows a long-standing tradition of mechanical engineering tuition which is already a century old and comes from the Toulouse mechanical engineering scientific community now merged into the Institut Cleacutement Ader (wwwinstitut-clement-aderorg) Just as all of its illustrious predecessors this book is very timely and illustrates the specificity and originality of the approaches that we have developed which have both a high scientific standing and a quasi-permanent connection with the aeronautical industry This publication provides the reader with the necessary knowledge and techniques to calculate structures and decompartmentalize disciplines and fields The aeronautical engineer will find all of the helpful information he or she needs within these pages the basis of continuum mechanics the finite element method and knowledge of materials metals and composites both within linear and non-linear fields The information is presented in an extremely clear and educational manner The reader may draw on an impressive series of exercises with detailed corrections something which is not so commonly found

Bruno CASTANIEacute INSA Toulouse

Institut Cleacutement Adler

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Mechanics of Aeronautical Solids Materials and Structures

Mechanics of Aeronautical Solids Materials and

Structures

Christophe Bouvet

First published 2017 in Great Britain and the United States by ISTE Ltd and John Wiley amp Sons Inc

Apart from any fair dealing for the purposes of research or private study or criticism or review as permitted under the Copyright Designs and Patents Act 1988 this publication may only be reproduced stored or transmitted in any form or by any means with the prior permission in writing of the publishers or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address

ISTE Ltd John Wiley amp Sons Inc 27-37 St Georgersquos Road 111 River Street London SW19 4EU Hoboken NJ 07030 UK USA

wwwistecouk wwwwileycom

copy ISTE Ltd 2017 The rights of Christophe Bouvet to be identified as the author of this work have been asserted by him in accordance with the Copyright Designs and Patents Act 1988

Library of Congress Control Number 2016962010 British Library Cataloguing-in-Publication Data A CIP record for this book is available from the British Library ISBN 978-1-78630-115-4

Contents

Foreword ix

Preface xi

Introduction xiii

Chapter 1 Stress 1

11 Notion of stress 1 111 External forces 1 112 Internal cohesive forces 2 113 Normal stress shear stress 2

12 Properties of the stress vector 3 121 Boundary conditions 3 122 Torsor of internal forces 5 123 Reciprocal actions 8 124 Cauchy reciprocal theorem 9

13 Stress matrix 11 131 Notation 11 132 Invariants of the stress tensor 13 133 Relation between the stress matrix and the stress vector 15 134 Principal stresses and principal directions 18

14 Equilibrium equation 21 15 Mohrrsquos circle 23

Chapter 2 Strain 27

21 Notion of strain 27 211 Displacement vector 27

vi Mechanics of Aeronautical Solids Materials and Structures

212 Unit strain 28 213 Angular distortion 30

22 Strain matrix 33 221 Definition of the strain matrix 33 222 Principal strains and principal directions 37 223 Volume expansion 39 224 Invariants of strain tensor 40 225 Compatibility condition 40

23 Strain measurement strain gage 41

Chapter 3 Behavior Law 43

31 A few definitions 43 32 Tension test 43

321 Brittle materials 44 322 Ductile materials 45 323 Particular cases 46

33 Shear test 46 331 Brittle materials 47 332 Ductile materials 48

34 General rule 48 341 Linear elasticity 48

35 Anisotropic materials example of a composite 53 351 Elasticity 53

36 Thermoelasticity 54

Chapter 4 Resolution Methods 59

41 Assessment 59 42 Displacement method 61 43 Stress method 61 44 Finite element method 62

Chapter 5 Work-energy Theorem Principle of Finite Element Method 63

51 Work-energy theorem 63 511 Hypotheses 63 512 Strain energy 64 513 Work of external forces 65 514 Strain energy 66 515 Energy minimization Ritz method 68

Contents vii

52 Finite element method 69 521 General principle of finite element method 69 522 Example of the three-node triangular element 74

53 Application triangle with plate finite element using Catia 80

Chapter 6 Sizing Criteria of an Aeronautical Structure 83

61 Introduction 83 62 Experimental determination of a sizing criterion 85 63 Normal stress or principal stress criterion brittle material 87 64 Stress or maximum shear energy criterion ductile material 91

641 Tresca criterion 91 642 Von Mises criterion 93 643 Rupture of a ductile material 96

65 Maximum shear criterion with friction compression of brittle materials 99 66 Anisotropic criterion example of the composite 105

Chapter 7 Plasticity 109

71 Introduction 109 72 Plastic instability necking true stress and true strain 111 73 Plastic behavior law RambergndashOsgood law 116 74 Example of an elasticndashplastic calculation plate with open hole in tension 118

Chapter 8 Physics of Aeronautical Structure Materials 127

81 Introduction 127 82 Aluminum 2024 130 83 Carbonepoxy composite T300914 135 84 Polymers 140

Chapter 9 Exercises 151

91 Rosette analysis 151 92 Pure shear 154 93 Compression of an elastic solid 154 94 Gravity dam 155 95 Shear modulus 156 96 Modulus of a composite 157 97 Torsional cylinder 158 98 Plastic compression 160 99 Bi-material beam tension 162

viii Mechanics of Aeronautical Solids Materials and Structures

910 Beam thermal expansion 164 911 Cube under shear stress 165 912 Spherical reservoir under pressure 166 913 Plastic bending 169 914 Disc under radial tension 171 915 Bending beam resolution by the Ritz method 173 916 Stress concentration in open hole 174 917 Bending beam 178

Chapter 10 Solutions to Exercises 183

101 Rosette analysis 183 102 Pure shear 191 103 Compression of an elastic solid 192 104 Gravity dam 196 105 Shear modulus 201 106 Modulus of a composite 203 107 Torsional cylinder 206 108 Plastic compression 212 109 Bi-material beam tension 215 1010 Beam thermal expansion 225 1011 Cube under shear stress 231 1012 Spherical reservoir under pressure 235 1013 Plastic bending 240 1014 Disc under radial tension 245 1015 Bending beam resolution by the Ritz method 252 1016 Stress concentration in open hole 256 1017 Bending beam 259

Appendix 273

Bibliography 279

Index 281

Foreword

This book follows a long-standing tradition of mechanical engineering tuition which is already a century old and comes from the Toulouse mechanical engineering scientific community now merged into the Institut Cleacutement Ader (wwwinstitut-clement-aderorg) Just as all of its illustrious predecessors this book is very timely and illustrates the specificity and originality of the approaches that we have developed which have both a high scientific standing and a quasi-permanent connection with the aeronautical industry This publication provides the reader with the necessary knowledge and techniques to calculate structures and decompartmentalize disciplines and fields The aeronautical engineer will find all of the helpful information he or she needs within these pages the basis of continuum mechanics the finite element method and knowledge of materials metals and composites both within linear and non-linear fields The information is presented in an extremely clear and educational manner The reader may draw on an impressive series of exercises with detailed corrections something which is not so commonly found

Bruno CASTANIEacute INSA Toulouse

Institut Cleacutement Adler

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Mechanics of Aeronautical Solids Materials and

Structures

Christophe Bouvet

First published 2017 in Great Britain and the United States by ISTE Ltd and John Wiley amp Sons Inc

Apart from any fair dealing for the purposes of research or private study or criticism or review as permitted under the Copyright Designs and Patents Act 1988 this publication may only be reproduced stored or transmitted in any form or by any means with the prior permission in writing of the publishers or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address

ISTE Ltd John Wiley amp Sons Inc 27-37 St Georgersquos Road 111 River Street London SW19 4EU Hoboken NJ 07030 UK USA

wwwistecouk wwwwileycom

copy ISTE Ltd 2017 The rights of Christophe Bouvet to be identified as the author of this work have been asserted by him in accordance with the Copyright Designs and Patents Act 1988

Library of Congress Control Number 2016962010 British Library Cataloguing-in-Publication Data A CIP record for this book is available from the British Library ISBN 978-1-78630-115-4

Contents

Foreword ix

Preface xi

Introduction xiii

Chapter 1 Stress 1

11 Notion of stress 1 111 External forces 1 112 Internal cohesive forces 2 113 Normal stress shear stress 2

12 Properties of the stress vector 3 121 Boundary conditions 3 122 Torsor of internal forces 5 123 Reciprocal actions 8 124 Cauchy reciprocal theorem 9

13 Stress matrix 11 131 Notation 11 132 Invariants of the stress tensor 13 133 Relation between the stress matrix and the stress vector 15 134 Principal stresses and principal directions 18

14 Equilibrium equation 21 15 Mohrrsquos circle 23

Chapter 2 Strain 27

21 Notion of strain 27 211 Displacement vector 27

vi Mechanics of Aeronautical Solids Materials and Structures

212 Unit strain 28 213 Angular distortion 30

22 Strain matrix 33 221 Definition of the strain matrix 33 222 Principal strains and principal directions 37 223 Volume expansion 39 224 Invariants of strain tensor 40 225 Compatibility condition 40

23 Strain measurement strain gage 41

Chapter 3 Behavior Law 43

31 A few definitions 43 32 Tension test 43

321 Brittle materials 44 322 Ductile materials 45 323 Particular cases 46

33 Shear test 46 331 Brittle materials 47 332 Ductile materials 48

34 General rule 48 341 Linear elasticity 48

35 Anisotropic materials example of a composite 53 351 Elasticity 53

36 Thermoelasticity 54

Chapter 4 Resolution Methods 59

41 Assessment 59 42 Displacement method 61 43 Stress method 61 44 Finite element method 62

Chapter 5 Work-energy Theorem Principle of Finite Element Method 63

51 Work-energy theorem 63 511 Hypotheses 63 512 Strain energy 64 513 Work of external forces 65 514 Strain energy 66 515 Energy minimization Ritz method 68

Contents vii

52 Finite element method 69 521 General principle of finite element method 69 522 Example of the three-node triangular element 74

53 Application triangle with plate finite element using Catia 80

Chapter 6 Sizing Criteria of an Aeronautical Structure 83

61 Introduction 83 62 Experimental determination of a sizing criterion 85 63 Normal stress or principal stress criterion brittle material 87 64 Stress or maximum shear energy criterion ductile material 91

641 Tresca criterion 91 642 Von Mises criterion 93 643 Rupture of a ductile material 96

65 Maximum shear criterion with friction compression of brittle materials 99 66 Anisotropic criterion example of the composite 105

Chapter 7 Plasticity 109

71 Introduction 109 72 Plastic instability necking true stress and true strain 111 73 Plastic behavior law RambergndashOsgood law 116 74 Example of an elasticndashplastic calculation plate with open hole in tension 118

Chapter 8 Physics of Aeronautical Structure Materials 127

81 Introduction 127 82 Aluminum 2024 130 83 Carbonepoxy composite T300914 135 84 Polymers 140

Chapter 9 Exercises 151

91 Rosette analysis 151 92 Pure shear 154 93 Compression of an elastic solid 154 94 Gravity dam 155 95 Shear modulus 156 96 Modulus of a composite 157 97 Torsional cylinder 158 98 Plastic compression 160 99 Bi-material beam tension 162

viii Mechanics of Aeronautical Solids Materials and Structures

910 Beam thermal expansion 164 911 Cube under shear stress 165 912 Spherical reservoir under pressure 166 913 Plastic bending 169 914 Disc under radial tension 171 915 Bending beam resolution by the Ritz method 173 916 Stress concentration in open hole 174 917 Bending beam 178

Chapter 10 Solutions to Exercises 183

101 Rosette analysis 183 102 Pure shear 191 103 Compression of an elastic solid 192 104 Gravity dam 196 105 Shear modulus 201 106 Modulus of a composite 203 107 Torsional cylinder 206 108 Plastic compression 212 109 Bi-material beam tension 215 1010 Beam thermal expansion 225 1011 Cube under shear stress 231 1012 Spherical reservoir under pressure 235 1013 Plastic bending 240 1014 Disc under radial tension 245 1015 Bending beam resolution by the Ritz method 252 1016 Stress concentration in open hole 256 1017 Bending beam 259

Appendix 273

Bibliography 279

Index 281

Foreword

This book follows a long-standing tradition of mechanical engineering tuition which is already a century old and comes from the Toulouse mechanical engineering scientific community now merged into the Institut Cleacutement Ader (wwwinstitut-clement-aderorg) Just as all of its illustrious predecessors this book is very timely and illustrates the specificity and originality of the approaches that we have developed which have both a high scientific standing and a quasi-permanent connection with the aeronautical industry This publication provides the reader with the necessary knowledge and techniques to calculate structures and decompartmentalize disciplines and fields The aeronautical engineer will find all of the helpful information he or she needs within these pages the basis of continuum mechanics the finite element method and knowledge of materials metals and composites both within linear and non-linear fields The information is presented in an extremely clear and educational manner The reader may draw on an impressive series of exercises with detailed corrections something which is not so commonly found

Bruno CASTANIEacute INSA Toulouse

Institut Cleacutement Adler

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

First published 2017 in Great Britain and the United States by ISTE Ltd and John Wiley amp Sons Inc

Apart from any fair dealing for the purposes of research or private study or criticism or review as permitted under the Copyright Designs and Patents Act 1988 this publication may only be reproduced stored or transmitted in any form or by any means with the prior permission in writing of the publishers or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address

ISTE Ltd John Wiley amp Sons Inc 27-37 St Georgersquos Road 111 River Street London SW19 4EU Hoboken NJ 07030 UK USA

wwwistecouk wwwwileycom

copy ISTE Ltd 2017 The rights of Christophe Bouvet to be identified as the author of this work have been asserted by him in accordance with the Copyright Designs and Patents Act 1988

Library of Congress Control Number 2016962010 British Library Cataloguing-in-Publication Data A CIP record for this book is available from the British Library ISBN 978-1-78630-115-4

Contents

Foreword ix

Preface xi

Introduction xiii

Chapter 1 Stress 1

11 Notion of stress 1 111 External forces 1 112 Internal cohesive forces 2 113 Normal stress shear stress 2

12 Properties of the stress vector 3 121 Boundary conditions 3 122 Torsor of internal forces 5 123 Reciprocal actions 8 124 Cauchy reciprocal theorem 9

13 Stress matrix 11 131 Notation 11 132 Invariants of the stress tensor 13 133 Relation between the stress matrix and the stress vector 15 134 Principal stresses and principal directions 18

14 Equilibrium equation 21 15 Mohrrsquos circle 23

Chapter 2 Strain 27

21 Notion of strain 27 211 Displacement vector 27

vi Mechanics of Aeronautical Solids Materials and Structures

212 Unit strain 28 213 Angular distortion 30

22 Strain matrix 33 221 Definition of the strain matrix 33 222 Principal strains and principal directions 37 223 Volume expansion 39 224 Invariants of strain tensor 40 225 Compatibility condition 40

23 Strain measurement strain gage 41

Chapter 3 Behavior Law 43

31 A few definitions 43 32 Tension test 43

321 Brittle materials 44 322 Ductile materials 45 323 Particular cases 46

33 Shear test 46 331 Brittle materials 47 332 Ductile materials 48

34 General rule 48 341 Linear elasticity 48

35 Anisotropic materials example of a composite 53 351 Elasticity 53

36 Thermoelasticity 54

Chapter 4 Resolution Methods 59

41 Assessment 59 42 Displacement method 61 43 Stress method 61 44 Finite element method 62

Chapter 5 Work-energy Theorem Principle of Finite Element Method 63

51 Work-energy theorem 63 511 Hypotheses 63 512 Strain energy 64 513 Work of external forces 65 514 Strain energy 66 515 Energy minimization Ritz method 68

Contents vii

52 Finite element method 69 521 General principle of finite element method 69 522 Example of the three-node triangular element 74

53 Application triangle with plate finite element using Catia 80

Chapter 6 Sizing Criteria of an Aeronautical Structure 83

61 Introduction 83 62 Experimental determination of a sizing criterion 85 63 Normal stress or principal stress criterion brittle material 87 64 Stress or maximum shear energy criterion ductile material 91

641 Tresca criterion 91 642 Von Mises criterion 93 643 Rupture of a ductile material 96

65 Maximum shear criterion with friction compression of brittle materials 99 66 Anisotropic criterion example of the composite 105

Chapter 7 Plasticity 109

71 Introduction 109 72 Plastic instability necking true stress and true strain 111 73 Plastic behavior law RambergndashOsgood law 116 74 Example of an elasticndashplastic calculation plate with open hole in tension 118

Chapter 8 Physics of Aeronautical Structure Materials 127

81 Introduction 127 82 Aluminum 2024 130 83 Carbonepoxy composite T300914 135 84 Polymers 140

Chapter 9 Exercises 151

91 Rosette analysis 151 92 Pure shear 154 93 Compression of an elastic solid 154 94 Gravity dam 155 95 Shear modulus 156 96 Modulus of a composite 157 97 Torsional cylinder 158 98 Plastic compression 160 99 Bi-material beam tension 162

viii Mechanics of Aeronautical Solids Materials and Structures

910 Beam thermal expansion 164 911 Cube under shear stress 165 912 Spherical reservoir under pressure 166 913 Plastic bending 169 914 Disc under radial tension 171 915 Bending beam resolution by the Ritz method 173 916 Stress concentration in open hole 174 917 Bending beam 178

Chapter 10 Solutions to Exercises 183

101 Rosette analysis 183 102 Pure shear 191 103 Compression of an elastic solid 192 104 Gravity dam 196 105 Shear modulus 201 106 Modulus of a composite 203 107 Torsional cylinder 206 108 Plastic compression 212 109 Bi-material beam tension 215 1010 Beam thermal expansion 225 1011 Cube under shear stress 231 1012 Spherical reservoir under pressure 235 1013 Plastic bending 240 1014 Disc under radial tension 245 1015 Bending beam resolution by the Ritz method 252 1016 Stress concentration in open hole 256 1017 Bending beam 259

Appendix 273

Bibliography 279

Index 281

Foreword

This book follows a long-standing tradition of mechanical engineering tuition which is already a century old and comes from the Toulouse mechanical engineering scientific community now merged into the Institut Cleacutement Ader (wwwinstitut-clement-aderorg) Just as all of its illustrious predecessors this book is very timely and illustrates the specificity and originality of the approaches that we have developed which have both a high scientific standing and a quasi-permanent connection with the aeronautical industry This publication provides the reader with the necessary knowledge and techniques to calculate structures and decompartmentalize disciplines and fields The aeronautical engineer will find all of the helpful information he or she needs within these pages the basis of continuum mechanics the finite element method and knowledge of materials metals and composites both within linear and non-linear fields The information is presented in an extremely clear and educational manner The reader may draw on an impressive series of exercises with detailed corrections something which is not so commonly found

Bruno CASTANIEacute INSA Toulouse

Institut Cleacutement Adler

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Contents

Foreword ix

Preface xi

Introduction xiii

Chapter 1 Stress 1

11 Notion of stress 1 111 External forces 1 112 Internal cohesive forces 2 113 Normal stress shear stress 2

12 Properties of the stress vector 3 121 Boundary conditions 3 122 Torsor of internal forces 5 123 Reciprocal actions 8 124 Cauchy reciprocal theorem 9

13 Stress matrix 11 131 Notation 11 132 Invariants of the stress tensor 13 133 Relation between the stress matrix and the stress vector 15 134 Principal stresses and principal directions 18

14 Equilibrium equation 21 15 Mohrrsquos circle 23

Chapter 2 Strain 27

21 Notion of strain 27 211 Displacement vector 27

vi Mechanics of Aeronautical Solids Materials and Structures

212 Unit strain 28 213 Angular distortion 30

22 Strain matrix 33 221 Definition of the strain matrix 33 222 Principal strains and principal directions 37 223 Volume expansion 39 224 Invariants of strain tensor 40 225 Compatibility condition 40

23 Strain measurement strain gage 41

Chapter 3 Behavior Law 43

31 A few definitions 43 32 Tension test 43

321 Brittle materials 44 322 Ductile materials 45 323 Particular cases 46

33 Shear test 46 331 Brittle materials 47 332 Ductile materials 48

34 General rule 48 341 Linear elasticity 48

35 Anisotropic materials example of a composite 53 351 Elasticity 53

36 Thermoelasticity 54

Chapter 4 Resolution Methods 59

41 Assessment 59 42 Displacement method 61 43 Stress method 61 44 Finite element method 62

Chapter 5 Work-energy Theorem Principle of Finite Element Method 63

51 Work-energy theorem 63 511 Hypotheses 63 512 Strain energy 64 513 Work of external forces 65 514 Strain energy 66 515 Energy minimization Ritz method 68

Contents vii

52 Finite element method 69 521 General principle of finite element method 69 522 Example of the three-node triangular element 74

53 Application triangle with plate finite element using Catia 80

Chapter 6 Sizing Criteria of an Aeronautical Structure 83

61 Introduction 83 62 Experimental determination of a sizing criterion 85 63 Normal stress or principal stress criterion brittle material 87 64 Stress or maximum shear energy criterion ductile material 91

641 Tresca criterion 91 642 Von Mises criterion 93 643 Rupture of a ductile material 96

65 Maximum shear criterion with friction compression of brittle materials 99 66 Anisotropic criterion example of the composite 105

Chapter 7 Plasticity 109

71 Introduction 109 72 Plastic instability necking true stress and true strain 111 73 Plastic behavior law RambergndashOsgood law 116 74 Example of an elasticndashplastic calculation plate with open hole in tension 118

Chapter 8 Physics of Aeronautical Structure Materials 127

81 Introduction 127 82 Aluminum 2024 130 83 Carbonepoxy composite T300914 135 84 Polymers 140

Chapter 9 Exercises 151

91 Rosette analysis 151 92 Pure shear 154 93 Compression of an elastic solid 154 94 Gravity dam 155 95 Shear modulus 156 96 Modulus of a composite 157 97 Torsional cylinder 158 98 Plastic compression 160 99 Bi-material beam tension 162

viii Mechanics of Aeronautical Solids Materials and Structures

910 Beam thermal expansion 164 911 Cube under shear stress 165 912 Spherical reservoir under pressure 166 913 Plastic bending 169 914 Disc under radial tension 171 915 Bending beam resolution by the Ritz method 173 916 Stress concentration in open hole 174 917 Bending beam 178

Chapter 10 Solutions to Exercises 183

101 Rosette analysis 183 102 Pure shear 191 103 Compression of an elastic solid 192 104 Gravity dam 196 105 Shear modulus 201 106 Modulus of a composite 203 107 Torsional cylinder 206 108 Plastic compression 212 109 Bi-material beam tension 215 1010 Beam thermal expansion 225 1011 Cube under shear stress 231 1012 Spherical reservoir under pressure 235 1013 Plastic bending 240 1014 Disc under radial tension 245 1015 Bending beam resolution by the Ritz method 252 1016 Stress concentration in open hole 256 1017 Bending beam 259

Appendix 273

Bibliography 279

Index 281

Foreword

This book follows a long-standing tradition of mechanical engineering tuition which is already a century old and comes from the Toulouse mechanical engineering scientific community now merged into the Institut Cleacutement Ader (wwwinstitut-clement-aderorg) Just as all of its illustrious predecessors this book is very timely and illustrates the specificity and originality of the approaches that we have developed which have both a high scientific standing and a quasi-permanent connection with the aeronautical industry This publication provides the reader with the necessary knowledge and techniques to calculate structures and decompartmentalize disciplines and fields The aeronautical engineer will find all of the helpful information he or she needs within these pages the basis of continuum mechanics the finite element method and knowledge of materials metals and composites both within linear and non-linear fields The information is presented in an extremely clear and educational manner The reader may draw on an impressive series of exercises with detailed corrections something which is not so commonly found

Bruno CASTANIEacute INSA Toulouse

Institut Cleacutement Adler

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

vi Mechanics of Aeronautical Solids Materials and Structures

212 Unit strain 28 213 Angular distortion 30

22 Strain matrix 33 221 Definition of the strain matrix 33 222 Principal strains and principal directions 37 223 Volume expansion 39 224 Invariants of strain tensor 40 225 Compatibility condition 40

23 Strain measurement strain gage 41

Chapter 3 Behavior Law 43

31 A few definitions 43 32 Tension test 43

321 Brittle materials 44 322 Ductile materials 45 323 Particular cases 46

33 Shear test 46 331 Brittle materials 47 332 Ductile materials 48

34 General rule 48 341 Linear elasticity 48

35 Anisotropic materials example of a composite 53 351 Elasticity 53

36 Thermoelasticity 54

Chapter 4 Resolution Methods 59

41 Assessment 59 42 Displacement method 61 43 Stress method 61 44 Finite element method 62

Chapter 5 Work-energy Theorem Principle of Finite Element Method 63

51 Work-energy theorem 63 511 Hypotheses 63 512 Strain energy 64 513 Work of external forces 65 514 Strain energy 66 515 Energy minimization Ritz method 68

Contents vii

52 Finite element method 69 521 General principle of finite element method 69 522 Example of the three-node triangular element 74

53 Application triangle with plate finite element using Catia 80

Chapter 6 Sizing Criteria of an Aeronautical Structure 83

61 Introduction 83 62 Experimental determination of a sizing criterion 85 63 Normal stress or principal stress criterion brittle material 87 64 Stress or maximum shear energy criterion ductile material 91

641 Tresca criterion 91 642 Von Mises criterion 93 643 Rupture of a ductile material 96

65 Maximum shear criterion with friction compression of brittle materials 99 66 Anisotropic criterion example of the composite 105

Chapter 7 Plasticity 109

71 Introduction 109 72 Plastic instability necking true stress and true strain 111 73 Plastic behavior law RambergndashOsgood law 116 74 Example of an elasticndashplastic calculation plate with open hole in tension 118

Chapter 8 Physics of Aeronautical Structure Materials 127

81 Introduction 127 82 Aluminum 2024 130 83 Carbonepoxy composite T300914 135 84 Polymers 140

Chapter 9 Exercises 151

91 Rosette analysis 151 92 Pure shear 154 93 Compression of an elastic solid 154 94 Gravity dam 155 95 Shear modulus 156 96 Modulus of a composite 157 97 Torsional cylinder 158 98 Plastic compression 160 99 Bi-material beam tension 162

viii Mechanics of Aeronautical Solids Materials and Structures

910 Beam thermal expansion 164 911 Cube under shear stress 165 912 Spherical reservoir under pressure 166 913 Plastic bending 169 914 Disc under radial tension 171 915 Bending beam resolution by the Ritz method 173 916 Stress concentration in open hole 174 917 Bending beam 178

Chapter 10 Solutions to Exercises 183

101 Rosette analysis 183 102 Pure shear 191 103 Compression of an elastic solid 192 104 Gravity dam 196 105 Shear modulus 201 106 Modulus of a composite 203 107 Torsional cylinder 206 108 Plastic compression 212 109 Bi-material beam tension 215 1010 Beam thermal expansion 225 1011 Cube under shear stress 231 1012 Spherical reservoir under pressure 235 1013 Plastic bending 240 1014 Disc under radial tension 245 1015 Bending beam resolution by the Ritz method 252 1016 Stress concentration in open hole 256 1017 Bending beam 259

Appendix 273

Bibliography 279

Index 281

Foreword

This book follows a long-standing tradition of mechanical engineering tuition which is already a century old and comes from the Toulouse mechanical engineering scientific community now merged into the Institut Cleacutement Ader (wwwinstitut-clement-aderorg) Just as all of its illustrious predecessors this book is very timely and illustrates the specificity and originality of the approaches that we have developed which have both a high scientific standing and a quasi-permanent connection with the aeronautical industry This publication provides the reader with the necessary knowledge and techniques to calculate structures and decompartmentalize disciplines and fields The aeronautical engineer will find all of the helpful information he or she needs within these pages the basis of continuum mechanics the finite element method and knowledge of materials metals and composites both within linear and non-linear fields The information is presented in an extremely clear and educational manner The reader may draw on an impressive series of exercises with detailed corrections something which is not so commonly found

Bruno CASTANIEacute INSA Toulouse

Institut Cleacutement Adler

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Contents vii

52 Finite element method 69 521 General principle of finite element method 69 522 Example of the three-node triangular element 74

53 Application triangle with plate finite element using Catia 80

Chapter 6 Sizing Criteria of an Aeronautical Structure 83

61 Introduction 83 62 Experimental determination of a sizing criterion 85 63 Normal stress or principal stress criterion brittle material 87 64 Stress or maximum shear energy criterion ductile material 91

641 Tresca criterion 91 642 Von Mises criterion 93 643 Rupture of a ductile material 96

65 Maximum shear criterion with friction compression of brittle materials 99 66 Anisotropic criterion example of the composite 105

Chapter 7 Plasticity 109

71 Introduction 109 72 Plastic instability necking true stress and true strain 111 73 Plastic behavior law RambergndashOsgood law 116 74 Example of an elasticndashplastic calculation plate with open hole in tension 118

Chapter 8 Physics of Aeronautical Structure Materials 127

81 Introduction 127 82 Aluminum 2024 130 83 Carbonepoxy composite T300914 135 84 Polymers 140

Chapter 9 Exercises 151

91 Rosette analysis 151 92 Pure shear 154 93 Compression of an elastic solid 154 94 Gravity dam 155 95 Shear modulus 156 96 Modulus of a composite 157 97 Torsional cylinder 158 98 Plastic compression 160 99 Bi-material beam tension 162

viii Mechanics of Aeronautical Solids Materials and Structures

910 Beam thermal expansion 164 911 Cube under shear stress 165 912 Spherical reservoir under pressure 166 913 Plastic bending 169 914 Disc under radial tension 171 915 Bending beam resolution by the Ritz method 173 916 Stress concentration in open hole 174 917 Bending beam 178

Chapter 10 Solutions to Exercises 183

101 Rosette analysis 183 102 Pure shear 191 103 Compression of an elastic solid 192 104 Gravity dam 196 105 Shear modulus 201 106 Modulus of a composite 203 107 Torsional cylinder 206 108 Plastic compression 212 109 Bi-material beam tension 215 1010 Beam thermal expansion 225 1011 Cube under shear stress 231 1012 Spherical reservoir under pressure 235 1013 Plastic bending 240 1014 Disc under radial tension 245 1015 Bending beam resolution by the Ritz method 252 1016 Stress concentration in open hole 256 1017 Bending beam 259

Appendix 273

Bibliography 279

Index 281

Foreword

This book follows a long-standing tradition of mechanical engineering tuition which is already a century old and comes from the Toulouse mechanical engineering scientific community now merged into the Institut Cleacutement Ader (wwwinstitut-clement-aderorg) Just as all of its illustrious predecessors this book is very timely and illustrates the specificity and originality of the approaches that we have developed which have both a high scientific standing and a quasi-permanent connection with the aeronautical industry This publication provides the reader with the necessary knowledge and techniques to calculate structures and decompartmentalize disciplines and fields The aeronautical engineer will find all of the helpful information he or she needs within these pages the basis of continuum mechanics the finite element method and knowledge of materials metals and composites both within linear and non-linear fields The information is presented in an extremely clear and educational manner The reader may draw on an impressive series of exercises with detailed corrections something which is not so commonly found

Bruno CASTANIEacute INSA Toulouse

Institut Cleacutement Adler

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

viii Mechanics of Aeronautical Solids Materials and Structures

910 Beam thermal expansion 164 911 Cube under shear stress 165 912 Spherical reservoir under pressure 166 913 Plastic bending 169 914 Disc under radial tension 171 915 Bending beam resolution by the Ritz method 173 916 Stress concentration in open hole 174 917 Bending beam 178

Chapter 10 Solutions to Exercises 183

101 Rosette analysis 183 102 Pure shear 191 103 Compression of an elastic solid 192 104 Gravity dam 196 105 Shear modulus 201 106 Modulus of a composite 203 107 Torsional cylinder 206 108 Plastic compression 212 109 Bi-material beam tension 215 1010 Beam thermal expansion 225 1011 Cube under shear stress 231 1012 Spherical reservoir under pressure 235 1013 Plastic bending 240 1014 Disc under radial tension 245 1015 Bending beam resolution by the Ritz method 252 1016 Stress concentration in open hole 256 1017 Bending beam 259

Appendix 273

Bibliography 279

Index 281

Foreword

This book follows a long-standing tradition of mechanical engineering tuition which is already a century old and comes from the Toulouse mechanical engineering scientific community now merged into the Institut Cleacutement Ader (wwwinstitut-clement-aderorg) Just as all of its illustrious predecessors this book is very timely and illustrates the specificity and originality of the approaches that we have developed which have both a high scientific standing and a quasi-permanent connection with the aeronautical industry This publication provides the reader with the necessary knowledge and techniques to calculate structures and decompartmentalize disciplines and fields The aeronautical engineer will find all of the helpful information he or she needs within these pages the basis of continuum mechanics the finite element method and knowledge of materials metals and composites both within linear and non-linear fields The information is presented in an extremely clear and educational manner The reader may draw on an impressive series of exercises with detailed corrections something which is not so commonly found

Bruno CASTANIEacute INSA Toulouse

Institut Cleacutement Adler

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Foreword

This book follows a long-standing tradition of mechanical engineering tuition which is already a century old and comes from the Toulouse mechanical engineering scientific community now merged into the Institut Cleacutement Ader (wwwinstitut-clement-aderorg) Just as all of its illustrious predecessors this book is very timely and illustrates the specificity and originality of the approaches that we have developed which have both a high scientific standing and a quasi-permanent connection with the aeronautical industry This publication provides the reader with the necessary knowledge and techniques to calculate structures and decompartmentalize disciplines and fields The aeronautical engineer will find all of the helpful information he or she needs within these pages the basis of continuum mechanics the finite element method and knowledge of materials metals and composites both within linear and non-linear fields The information is presented in an extremely clear and educational manner The reader may draw on an impressive series of exercises with detailed corrections something which is not so commonly found

Bruno CASTANIEacute INSA Toulouse

Institut Cleacutement Adler

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Preface

This volume on the mechanics of solids and materials as well as aeronautical structures aims to give an overview of the necessary notions for structure sizing within the aeronautics field It begins by establishing all of the classic notions of mechanics stress strain behavior law and sizing criteria Also covered are notions that are specific to aeronautics with a particular emphasis on the notion of limit loads and ultimate loads

Different problem-solving methods particularly the finite element method are then introduced The methods are not classically presented and instead energy minimization is drawn on in order to minimize the number of equations all while remaining within a framework that we may comprehend ldquowith their handsrdquo

The book then addresses the subject of plasticity showcasing its influence on structure sizing and especially the advantages it has for sizing criteria

Finally the physics of the two main materials in aeronautics namely aluminum and composite materials is discussed so as to shed light on the sizing criteria outlined in the previous chapters

The corrected exercises help the student to test their understanding of the different topics

What is so original about this book is that from the outset it places itself within the field of aeronautics Sizing criteria are indeed rather specific to this field Nevertheless the notions discussed remain valid for the majority of industrial fields in Mechanical Engineering and Finite Elements these notions in fact remain the same

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

xii Mechanics of Aeronautical Solids Materials and Structures

Another original aspect of this work is that it consolidates basic continuum mechanics with a very succinct description of finite elements and a description of the material aspect of the main materials used in aeronautical structures that being aluminum and composites This publication is therefore a summary of the basic knowledge deemed necessary for the (ldquoAirbusrdquo) engineer working within research departments The book is simultaneously aimed at both students who are beginning their training and also engineers already working in the field who desire a summary of the basic theories

Lastly the publication aims to limit the amount of formulas provided as much as possible in order to highlight the significance of the physical Any readers who may be interested in demonstrations are advised to refer to more specific and theoretical works such as [COI 01 DUV 98 GER 73 HEA 77 KHA 95 LEM 96 MIR 03 SAL 01 UGU 03] and [THU 97] etc

Christophe BOUVET January 2017

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Introduction

I1 Outlining the problem

Let us consider a solid S that is subjected to imposed displacements and external forces

S

Fext

uimp = 0

Figure I1 Outlining the problem For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The aim of the mechanics of deformable solids is to study the internal state of the material (notion of stress) and the way in which it becomes deformed (notion of strain) [FRA 12 SAL 01 LEM 96]

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

xiv Mechanics of Aeronautical Solids Materials and Structures

In mechanics a mechanical piece or system may be designed

ndash to prevent it from breaking

ndash to prevent it from becoming permanently deformed

ndash to prevent it from becoming too deformed or

ndash for any another purposes

A solid shall be deemed a continuous medium meaning that it shall be regarded as a continuous set of material points with a mass representing the state of matter that is surrounded by an infinitesimal volume

Mechanics of deformable solids enables the study of cohesive forces (notion of stress) at a point M like the forces exerted on the small volume surrounding it called a Representative Elementary Volume (REV) For metals the REV is typically within the range of a tenth of a millimeter

The matter in this REV must be seen as continuous and homogeneous

ndash if it is too small the matter cannot be as seen homogeneous atomic piling inclusion within matter grains etc (for example for concrete the REV is within the range of 10 cm)

ndash if it is too big the state of the cohesive forces in its center will no longer represent the REV state

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

1

Stress

11 Notion of stress

111 External forces

There are three types of external forces

ndash concentrated forces this is a force exerted on a point (in Newton units noted as N) In practice this force does not actually exist It is just a model If we were to apply a force to a point that has zero surface the contact pressure would be infinite and the deformation of the solid would therefore induce a non-zero contact surface Nevertheless it can still be imagined for studying problems with a very concentrated contact type load between balls The results will thus yield an infinite stress and will need to be interpreted accordingly

ndash surface forces which will be noted as Fext for the rest of this volume (in Pascal units it is noted Pa) This type of force includes contact forces between two solids as well as the pressure of a fluid Practically any concentrated force can be seen as a surface force distributed onto a small contact surface

ndash volume forces which will be noted as fv for the rest of this book (in Nm3) Examples of volume forces are forces of gravity electromagnetic forces etc

Incidentally in this book you will notice that vectors are underlined once and matrices (or tensors of rank 2) which you will come across further on are underlined twice

Mechanics of Aeronautical Solids Materials and Structures copy ISTE Ltd 2017 Published by ISTE Ltd and John Wiley amp Sons Inc

First Edition Christophe Bouvet

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

2 Mechanics of Aeronautical Solids Materials and Structures

112 Internal cohesive forces

We wish to study the cohesive forces of the solid S at point M and which is in equilibrium under the action of external forces The solid is cut into two parts E1 and E2 by a plane with a normal vector n passing through M The part E1 is in equilibrium under the action of the external forces on E1 and the cohesive force of E2 on E1

E2

Fext

uimp = 0

E1

n

F21

uimp = 0

E1

n

ΔF

M ΔS

Figure 11 Principle of internal cohesive forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let ΔS be the surface around M and ΔF be the cohesive force of 2 on 1 exerting on ΔS then the stress vector at the point M associated with the facet with a normal vector n is called

( ) lim0

F d FM nS S dS

σ Δ= =Δ rarr Δ

[11]

The unit is Nm2 or Pa and we generally use MPa or Nmm2

Physically the stress notion is fairly close to the notion of pressure that can be found in everyday life (the unit is even the same) but as we will see further on pressure is but only one particular example of stress

113 Normal stress shear stress

We define the different stresses as

ndash normal stress the projection of σ (M n) onto n noted as σ

ndash shear stress the projection of σ (M n) onto the plane with normal n noted as τ

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Stress 3

dF

M dS

n

t

σ

τ

Figure 12 Decomposition of a stress vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Thus σ represents the cohesive forces perpendicular to the facet meaning the tractioncompression and τ the forces tangential to the facet meaning the shear In a physical sense the pressure found in our everyday lives is simply a normal compression stress

We then definitely have

( )M n n tσ σ τ= + [12]

NOTEndash n and t must be unit vectors

And conversely

( )( )

M n n

M n t

σ σ

τ σ

⎧ =⎪⎨

=⎪⎩ [13]

12 Properties of the stress vector

121 Boundary conditions

If n is an external normal then

( )M n Fextextσ = [14]

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

4 Mechanics of Aeronautical Solids Materials and Structures

Fext

S

next

M

Figure 13 External force and associated normal vector For a color version of this figure see wwwistecoukbouvetaeronauticalzip

NOTEndash Fext is in MPa and a normal external vector is always moving from the matter towards the exterior

So Fext can be seen as a stress vector exerted on S particularly if the surface is a free surface

( ) 0M nextσ = [15]

These relations are important as they translate the stress boundary conditions on the structure In order for this to be the solution to the problem (see Chapter 3) these relations are part of a group of conditions that are needed to verify a stress field

EXAMPLE TANK UNDER PRESSUREndash

next

Pressure p

σ(Mnext)

Figure 14 Tank under pressure

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Stress 5

For every point on the internal wall of the tank we find

( ) M n p next extσ = minus [16]

With the external normal vector moving towards the center of the circle from where the normal and shear stresses are

( )( )

0

M n n pext extM n text

σ σ

τ σ

⎧ = = minus⎪⎨

= =⎪⎩ [17]

Given that the normal stress is negative and the shear stress is zero the material is subjected to pure compression The first relation shows that the physical notion of pressure is simply a normal stress of compression hence the minus sign before the pressure

122 Torsor of internal forces

E2

Fext

uimp = 0

E1

n

Section S

G

Figure 15 Set of internal forces For a color version of this figure see wwwistecoukbouvetaeronauticalzip

The torsor of internal forces of 2 on 1 at G the center of gravity of S is

2121 ( )21

RcohTM GG⎧ ⎫

= ⎨ ⎬⎩ ⎭

[18]

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

6 Mechanics of Aeronautical Solids Materials and Structures

At first sight the torsor notion may seem primitive but it enables us to simply consolidate the force with the moment Should the notion of torsor bother you you may settle for referring to it in plainer language as force and moment However you should not forget that when speaking about internal forces between 2 parts of a solid it needs to be remembered that there is a force (in N) and a moment (in Nmm) The ambiguity comes from the term ldquoforcerdquo which is used for a force (in the common everyday sense of the word) and as a whole force + moment

Let us now seek to link this set of internal forces to the previously discussed stress vector We then have

( ) ( ) 21d F M M n dSσ= [19]

therefore

( )

( )

( ) 21 21

( )( ) ( ) ( )21 2121

R d F M M n dSSM SMM G d M G GM d F M

M S M SGM M n dSS

σ

σ

= sum =⎧ intintisin⎪

⎪= sum = sum and⎨

isin isin⎪⎪ = andintint⎩

[110]

These relations are somewhat (or very) complex but physically they simply translate the fact that if we add up all of the stress vectors on section S then we will obtain the force of part E2 on part E1 Lastly we should not forget that when we add up the stress vectors we will obtain not only a force but also a moment (which obviously depends on the point at which it is calculated)

These relations can also be written on an external surface as

( )

( )

σ

σ

⎧ = =⎪⎨

= and = and⎪⎩

intint intintintint intint

11

( ) ext ext

ext ext

ext ext extS S

ext ext extS S

R M n dS F dS

M G GM M n dS GM F dS [111]

These relations are important because in practice although we know the resultant Rext1 or Mext1 we do not generally know Fext In fact an external force is practically applied via the intermediary of a beam a screed a jack etc and the applied resulting force (or the moment) is known but the way in which it is divided is unknown

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Stress 7

EXAMPLE TRACTIONndash

x

y F

F

x

y σ0

Uniform stress

σ0

Sy

Figure 16 Tensile test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In a tensile test we know that the resultant of the forces applied to Sy is worth F

( )( )

0

M y dS FSy

GM M y dSSy

σ

σ

⎧ =intint⎪⎨

and =⎪intint⎩

[112]

However in order to deduce that

( ) 0FM y y

Syσ σ= = [113]

we must add a homogeneity hypothesis of the force applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

EXAMPLE BENDINGndash

x

y

MM σ Stress

distribution

x

y

σ

Sx

Figure 17 Bending test For a color version of this figure see wwwistecoukbouvetaeronauticalzip

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

8 Mechanics of Aeronautical Solids Materials and Structures

In a pure bending test we know that the resultant of the forces applied to Sx is worth Mz

( )( )

σ

σ

⎧ =intint⎪⎨

and =intint⎪⎩

0 M x dSSxGM M x dS M zSx

[114]

However by deducing that on Sx

( ) MM x y xIz

σ minus= [115]

This formula is a classic example of the mechanics of material which we will discuss (and demonstrate) again when doing the exercises Should you need to you can read a more detailed publication such as [AGA 08 BAM 08 CHE 08 DEL 08 DUP 09] etc

Obviously with the moment of inertia

2 I y dSz Sx= intint [116]

we must add a linear distribution hypothesis of the stress applied which remains to be verified Incidentally we can demonstrate that the two previous integrals are verified with this stress vector

123 Reciprocal actions

E2

Fext

uimp = 0

E1

n M dS

Figure 18 Reciprocal actions For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Stress 9

According to the Law of Reciprocal Action we have

= minus21 12dF dF [117]

Yet

( ) ( )( ) ( )

21 12

d F M M n dS

d F M M n dS

σσ

=⎧⎪⎨

= minus⎪⎩ [118]

Hence

( ) ( )σ σ= minus minus M n M n [119]

This can be translated by the fact that a fine slice of matter of surface dS which has a normal vector +n on one side and ndashn on the other is at equilibrium under the action of the two opposing forces σ(M n) dS and σ(M minusn) dS Evidently it is very much at equilibrium

124 Cauchy reciprocal theorem

Figure 19 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

Let us put this in 2D in order to make the demonstration easier

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

10 Mechanics of Aeronautical Solids Materials and Structures

A square is assumed to be infinitely small therefore the stresses are assumed to be constant everywhere in the cube hence we have

( ) ( )

( ) ( )( ) ( )

( ) ( )

1

2

1

2

M x M x x yxx xy

M y M y y xyy yx

M x M x x yxx xy

M y M y y xyy yx

σ σ σ τ

σ σ σ τ

σ σ σ τ

σ σ σ τ

⎧ = = +⎪⎪ = = +⎪⎨

minus = minus = minus minus⎪⎪⎪ minus = minus = minus minus⎩

[120]

In the notation of τxy the first ldquoxrdquo corresponds to the direction of the facet meaning the normal vector on the cutting plane in question and the second index ldquoyrdquo represents the direction of the stress

The equilibrium equation on the square is written as

sum = 0cohText cube [121]

which for the force equation induces the following

( ) ( ) ( ) ( ) 01 2 1 1M x dy dz M y dx dz M x dy dz M y dx dzσ σ σ σ+ + minus + minus = [122]

This is an automatically verified equation For the moment equation in M at the center of the square the below is induced

( ) ( )( ) ( )

1 21 2

01 21 2

MM M x dy dz MM M y dx dz

MM M x dy dz MM M y dx dz

σ σ

σ σ

and + and

+ and minus + and minus = [123]

where the Cauchy reciprocity theorem is

xy yxτ τ= [124]

It can be shown in the same way in 3D

xy yx

xz zxyz zy

τ τ

τ ττ τ

=⎧⎪

=⎨⎪ =⎩

[125]

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Stress 11

EXAMPLE TORSION OF A WELDED TUBEndash

C

τ

Longitudinal welding C

τ

τ

τ

τ

τ

τ

τxy

τxy

τxy

τxy x

y

τxy

τxy

τxy

τxy

τxy

τxy

τxy

τxy

Figure 110 Torsion of a welded tube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

A torsion moment is applied to a tube comprised of a rolled and welded sheet It can be shown (see exercises) that it has been subjected to a homogeneous circumferential shear stress The Cauchy reciprocity theorem then induces the welding to be stressed by a longitudinal shear τ equal to the circumferential shear stress All that remains then is the sizing of the welding so that it can withstand this shearing applied force

13 Stress matrix

131 Notation

There is a stress vector on the facet with the normal vector x

( ) M x x tσ σ τ σ τ= + = + [126]

σ (M y)

y

z

x

M

σ (M x)

σ (M z)

Figure 111 Stress vectors on the faces of a unit cube For a color version of this figure see wwwistecoukbouvetaeronauticalzip

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

12 Mechanics of Aeronautical Solids Materials and Structures

and t can be broken down in accordance with y and z hence

( ) M x x y zxx xy xzσ σ τ τ= + + [127]

Once again in the notation of τxy the first index ldquoxrdquo corresponds to the direction of the facet and the second index ldquoyrdquo represents the direction of the stress

In the same way for the faces of the cube with normal vectors y and z we have

( )( )( )

σ σ τ τ

σ τ σ τ

σ τ τ σ

⎧ = + +⎪⎪ = + +⎨⎪

= + +⎪⎩

M x x y zxx xy xz

M y x y zyx yy yzM z x y zzx zy zz

[128]

σ(M) is what is referred to as the stress matrix

( )( )

xx yx zx

xy yy zy

xz yz zz x y z

Mσ τ τ

σ τ σ ττ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

[129]

Evidently the expression of this matrix depends on the coordinate

Moreover it is symmetrical in accordance with the reciprocity of the stresses

( )( )

σ τ τσ τ σ τ

τ τ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

xx xy xz

xy yy yz

xz yz zz x y z

M [130]

The very notion itself of the stress matrix is important as it assumes that at one point and in the small volume (REV) surrounding it the state of the internal forces of matter is entirely represented by this matrix This is in contrast to the stress vector which only gives the internal force for one single facet

EXAMPLE DRAWING THE STRESSES IN 2Dndash

A small square is subjected to a 2D-stress tensor

( )( )

σ τσ

τ σ⎡ ⎤

= ⎢ ⎥⎣ ⎦

xx xy

xy yy x y

M [131]

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

Stress 13

These faces are therefore subjected to the following forces

x

y

Mσxx

σ (Mx)τxy

σxx

σ (M-x) τxy

σyy σ (My)

τxy

σyy

σ (M-y)

τxy

Figure 112 Stress vectors on the faces of a square For a color version of this figure see wwwistecoukbouvetaeronauticalzip

This drawing may be simple but it is paramount for interpreting the stress tensor

You obviously would have noticed that the projection of σ(M x) on y is equal to that of σ(M y) on x which is evidently due to the symmetry of the stress tensor

I will leave it up to you to do the same drawing in 3D

132 Invariants of the stress tensor

The stress tensor possesses three elementary invariants We classically use

ndash The hydrostatic pressure

( )13 3

x y zp traceσ σ σ

σ+ +

= = [132]

It is named thus because when we apply a uniform pressure to a cube in all directions we obtain

0 00 00 0

pp

minus⎡ ⎤⎢ ⎥= minus⎢ ⎥⎢ ⎥minus⎣ ⎦

[133]

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants

14 Mechanics of Aeronautical Solids Materials and Structures

xII

xIII

xI

σIII = -p

σII = -p σII = -p

σI = -p

σI = -p

σIII = -p

Figure 113 Hydrostatic pressure For a color version of this figure see wwwistecoukbouvetaeronauticalzip

In this case the hydrostatic pressure is evidently equal to the pressure p in its absolute value

ndash The Von Mises equivalent stress

( ) ( ) ( )2 2 23 1 ( ) ( ) ( )2 2 I II II III III Idev devσ σ σ σ σ σ σ σ σ= = minus + minus + minus [134]

With the stress deviator which is written as

( ) ( )1 3

dev trace Iσ σ σ= minus [135]

This stress is very important for estimating the beginning of a ductile materialrsquos plasticity (see Chapters 3 and 4)

ndash The determinant

( )3 detI σ= [136]

These invariants are very important for writing the fracture yield or damage criteria Given that a criterion of this type is indeed representative of the state of matter it must not depend on the coordinate at which we write the stress matrix and therefore it can be written based on these invariants