Aircraft Structural Analysis

We may now completely describe the state of stress at a point O in a body by specifying components of shear and direct stresses on the faces of an element of sideδx, δy, and δz, formed a

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Copyright © 2010, T H G Megson Published by Elsevier Ltd All rights reserved.

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British Library Cataloguing in Publication Data

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Library of Congress Cataloging-in-Publication Data

Megson, T.H.G (Thomas Henry Gordon)

An introduction to aircraft structural analysis / T.H.G Megson.

p cm.

Rev ed of: Aircraft structures for engineering students / T.H.G Megson 4th ed 2007.

Includes bibliographical references and index.

ISBN 978-1-85617-932-4 (alk paper)

1 Airframes 2 Structural analysis (Engineering) I Title.

TL671.6.M36 2010

629.134’31–dc22

2009050354

For information on all Butterworth-Heinemann publications

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Printed in the United States of America

10 11 12 13 14 10 9 8 7 6 5 4 3 2 1

Preface vii

PART A FUNDAMENTALS OF STRUCTURAL ANALYSIS CHAPTER 1 Basic Elasticity 3

1.1 Stress 3

1.2 Notation for Forces and Stresses 5

1.3 Equations of Equilibrium 7

1.4 Plane Stress 9

1.5 Boundary Conditions 9

1.6 Determination of Stresses on Inclined Planes 10

1.7 Principal Stresses 14

1.8 Mohr’s Circle of Stress 16

1.9 Strain 20

1.10 Compatibility Equations 24

1.11 Plane Strain 25

1.12 Determination of Strains on Inclined Planes 25

1.13 Principal Strains 27

1.14 Mohr’s Circle of Strain 28

1.15 Stress–Strain Relationships 28

1.16 Experimental Measurement of Surface Strains 37

Problems 41

CHAPTER 2 Two-Dimensional Problems in Elasticity 45

2.1 Two-Dimensional Problems 45

2.2 Stress Functions 47

2.3 Inverse and Semi-Inverse Methods 48

2.4 St Venant’s Principle 53

2.5 Displacements 54

2.6 Bending of an End-Loaded Cantilever 55

Problems 60

CHAPTER 3 Torsion of Solid Sections 65

3.1 Prandtl Stress Function Solution 65

3.2 St Venant Warping Function Solution 75

3.3 The Membrane Analogy 77

3.4 Torsion of a Narrow Rectangular Strip 79

Problems 82

CHAPTER 4 Virtual Work and Energy Methods 85

4.1 Work 85

4.2 Principle of Virtual Work 86

4.3 Applications of the Principle of Virtual Work 99

Problems 107

CHAPTER 5 Energy Methods 111

5.1 Strain Energy and Complementary Energy 111

5.2 The Principle of the Stationary Value of the Total Complementary Energy 113

iii

5.3 Application to Deflection Problems 114

5.4 Application to the Solution of Statically Indeterminate Systems 122

5.5 Unit Load Method 138

5.6 Flexibility Method 141

5.7 Total Potential Energy 147

5.8 The Principle of the Stationary Value of the Total Potential Energy 148

5.9 Principle of Superposition 151

5.10 The Reciprocal Theorem 151

5.11 Temperature Effects 156

Problems 158

CHAPTER 6 Matrix Methods 169

6.1 Notation 170

6.2 Stiffness Matrix for an Elastic Spring 171

6.3 Stiffness Matrix for Two Elastic Springs in Line 172

6.4 Matrix Analysis of Pin-jointed Frameworks 176

6.5 Application to Statically Indeterminate Frameworks 183

6.6 Matrix Analysis of Space Frames 183

6.7 Stiffness Matrix for a Uniform Beam 185

6.8 Finite Element Method for Continuum Structures 193

Problems 211

CHAPTER 7 Bending of Thin Plates 219

7.1 Pure Bending of Thin Plates 219

7.2 Plates Subjected to Bending and Twisting 223

7.3 Plates Subjected to a Distributed Transverse Load 227

7.4 Combined Bending and In-Plane Loading of a Thin Rectangular Plate 236

7.5 Bending of Thin Plates Having a Small Initial Curvature 240

7.6 Energy Method for the Bending of Thin Plates 241

Problems 250

CHAPTER 8 Columns 253

8.1 Euler Buckling of Columns 253

8.2 Inelastic Buckling 259

8.3 Effect of Initial Imperfections 263

8.4 Stability of Beams under Transverse and Axial Loads 266

8.5 Energy Method for the Calculation of Buckling Loads in Columns 270

8.6 Flexural–Torsional Buckling of Thin-Walled Columns 274

Problems 287

CHAPTER 9 Thin Plates 293

9.1 Buckling of Thin Plates 293

9.2 Inelastic Buckling of Plates 296

9.3 Experimental Determination of Critical Load for a Flat Plate 298

9.4 Local Instability 299

9.5 Instability of Stiffened Panels 300

9.6 Failure Stress in Plates and Stiffened Panels 302

9.7 Tension Field Beams 304

Problems 320

Contents v

PART B ANALYSIS OF AIRCRAFT STRUCTURES

CHAPTER 10 Materials 327

10.1 Aluminum Alloys 327

10.2 Steel 329

10.3 Titanium 330

10.4 Plastics 331

10.5 Glass 331

10.6 Composite Materials 331

10.7 Properties of Materials 333

Problems 349

CHAPTER 11 Structural Components of Aircraft 351

11.1 Loads on Structural Components 351

11.2 Function of Structural Components 354

11.3 Fabrication of Structural Components 359

11.4 Connections 363

Problems 370

CHAPTER 12 Airworthiness 373

12.1 Factors of Safety-Flight Envelope 373

12.2 Load Factor Determination 375

CHAPTER 13 Airframe Loads 379

13.1 Aircraft Inertia Loads 379

13.2 Symmetric Maneuver Loads 386

13.3 Normal Accelerations Associated with Various Types of Maneuver 391

13.4 Gust Loads 393

Problems 399

CHAPTER 14 Fatigue 403

14.1 Safe Life and Fail-Safe Structures 403

14.2 Designing Against Fatigue 404

14.3 Fatigue Strength of Components 405

14.4 Prediction of Aircraft Fatigue Life 409

14.5 Crack Propagation 414

Problems 420

CHAPTER 15 Bending of Open and Closed, Thin-Walled Beams 423

15.1 Symmetrical Bending 424

15.2 Unsymmetrical Bending 433

15.3 Deflections due to Bending 441

15.4 Calculation of Section Properties 456

15.5 Applicability of Bending Theory 466

15.6 Temperature Effects 466

Problems 471

CHAPTER 16 Shear of Beams 479

16.1 General Stress, Strain, and Displacement Relationships for Open and Single Cell Closed Section Thin-Walled Beams 479

16.2 Shear of Open Section Beams 483

16.3 Shear of Closed Section Beams 488

Problems 496

CHAPTER 17 Torsion of Beams 503

17.1 Torsion of Closed Section Beams 503

17.2 Torsion of Open Section Beams 514

Problems 521

CHAPTER 18 Combined Open and Closed Section Beams 529

18.1 Bending 529

18.2 Shear 529

18.3 Torsion 533

Problems 534

CHAPTER 19 Structural Idealization 537

19.1 Principle 537

19.2 Idealization of a Panel 538

19.3 Effect of Idealization on the Analysis of Open and Closed Section Beams 541

19.4 Deflection of Open and Closed Section Beams 553

Problems 556

CHAPTER 20 Wing Spars and Box Beams 561

20.1 Tapered Wing Spar 561

20.2 Open and Closed Section Beams 565

20.3 Beams Having Variable Stringer Areas 571

Problems 574

CHAPTER 21 Fuselages 577

21.1 Bending 577

21.2 Shear 578

21.3 Torsion 581

21.4 Cutouts in Fuselages 584

Problems 585

CHAPTER 22 Wings 587

22.1 Three-Boom Shell 587

22.2 Bending 588

22.3 Torsion 590

22.4 Shear 594

22.5 Shear Center 599

22.6 Tapered Wings 600

22.7 Deflections 603

22.8 Cutouts in Wings 605

Problems 613

CHAPTER 23 Fuselage Frames and Wing Ribs 619

23.1 Principles of Stiffener/Web Construction 619

23.2 Fuselage Frames 625

23.3 Wing Ribs 626

Problems 630

Index 633

During my experience of teaching aircraft structures, I have felt the need for a textbook written ically for students of aeronautical engineering Although there have been a number of excellent bookswritten on the subject, they are now either out of date or too specialized in content to fulfill the require-

specif-ments of an undergraduate textbook With that in mind, I wrote Aircraft Structures for Engineering

Students, the text on which this one is based Users of that text have supplied many useful comments to

the publisher, including comments that a briefer version of the book might be desirable, particularly forprograms that do not have the time to cover all the material in the “big” book That feedback, along with

a survey done by the publisher, resulted in this book, An Introduction to Aircraft Structural Analysis,

designed to meet the needs of more time-constrained courses

Much of the content of this book is similar to that of Aircraft Structures for Engineering Students, but

the chapter on “Vibration of Structures” has been removed since this is most often covered in a separatestandalone course The topic of Aeroelasticity has also been removed, leaving detailed treatment to thegraduate-level curriculum The section on “Structural Loading and Discontinuities” remains in the bigbook but not this “intro” one While these topics help develop a deeper understanding of load transferand constraint effects in aircraft structures, they are often outside the scope of an undergraduate text.The reader interested in learning more on those topics should refer to the “big” book In the interest ofsaving space, the appendix on “Design of a Rear Fuselage” is available for download from the book’s

companion Web site Please visit www.elsevierdirect.com and search on “Megson” to find the Web site

and the downloadable content

Supplementary materials, including solutions to end-of-chapter problems, are available for registered

instructors who adopt this book as a course text Please visit www.textbooks.elsevier.com for information

and to register for access to these resources

The help of Tom Lacy, Associate Professor of Mechanical and Aerospace Engineering at sippi State University, is gratefully acknowledged in the development of this book

Missis-T.H.G Megson

Supporting material accompanying this book

A full set of worked solutions for this book are available for teaching purposes

Please visit www.textbooks.elsevier.com and follow the registration instructions to access this

material, which is intended for use by lecturers and tutors

vii

Consider the arbitrarily shaped, three-dimensional body shown in Fig 1.1 The body is in equilibrium

under the action of externally applied forces P1, P2, , and is assumed to comprise a continuous anddeformable material so that the forces are transmitted throughout its volume It follows that at anyinternal point O, there is a resultant forceδP The particle of material at O subjected to the force δP is

in equilibrium so that there must be an equal but opposite forceδP (shown dotted in Fig 1.1) acting on the particle at the same time If we now divide the body by any plane nn containing O, then these two

Fig 1.1

Internal force at a point in an arbitrarily shaped body

Copyright © 2010, T H G Megson Published by Elsevier Ltd All rights reserved.

3

Fig 1.2

Internal force components at the point O

forcesδP may be considered uniformly distributed over a small area δA of each face of the plane at the corresponding point O, as in Fig 1.2 The stress at O is then defined by the equation

Stress= lim

δA→0

δP

The directions of the forcesδP in Fig 1.2 are such that they produce tensile stresses on the faces

of the plane nn It must be realized here that while the direction of δP is absolute, the choice of plane

is arbitrary so that although the direction of the stress at O will always be in the direction of δP, its

magnitude depends on the actual plane chosen, since a different plane will have a different inclinationand therefore a different value for the areaδA This may be more easily understood by reference to the bar in simple tension in Fig 1.3 On the cross-sectional plane mm, the uniform stress is given by P /A, while on the inclined plane mm, the stress is of magnitude P /A In both cases, the stresses are parallel

to the direction of P.

Generally, the direction ofδP is not normal to the area δA, in which case it is usual to resolve δP

into two components: one,δP n, normal to the plane and the other,δP s, acting in the plane itself (seeFig 1.2) Note that in Fig 1.2 the plane containingδP is perpendicular to δA The stresses associated with these components are a normal or direct stress defined as

1.2 Notation for Forces and Stresses 5

Fig 1.3

Values of stress on different planes in a uniform bar

The resultant stress is computed from its components by the normal rules of vector addition, namely

Resultant stress=σ2+ τ2

Generally, however, as just indicated, we are interested in the separate effects ofσ and τ

However, to be strictly accurate, stress is not a vector quantity, for, in addition to magnitude and

direction, we must specify the plane on which the stress acts Stress is therefore a tensor, with its

complete description depending on the two vectors of force and surface of action

It is usually convenient to refer the state of stress at a point in a body to an orthogonal set of axes

Oxyz In this case, we cut the body by planes parallel to the direction of the axes The resultant force

δP acting at the point O on one of these planes may then be resolved into a normal component and two

in-plane components as shown in Fig 1.4, thereby producing one component of direct stress and twocomponents of shear stress

The direct stress component is specified by reference to the plane on which it acts, but the stresscomponents require a specification of direction in addition to the plane We therefore allocate a singlesubscript to direct stress to denote the plane on which it acts and two subscripts to shear stress, thefirst specifying the plane and the second direction Therefore, in Fig 1.4, the shear stress componentsareτzx andτzy acting on the z plane and in the x and y directions, respectively, while the direct stress

component isσz

We may now completely describe the state of stress at a point O in a body by specifying components

of shear and direct stresses on the faces of an element of sideδx, δy, and δz, formed at O by the cutting

planes as indicated in Fig 1.5

Fig 1.4

Components of stress at a point in a body

The sides of the element are infinitesimally small so that the stresses may be assumed to be formly distributed over the surface of each face On each of the opposite faces, there will be, to a firstsimplification, equal but opposite stresses

uni-We shall now define the directions of the stresses in Fig 1.5 as positive so that normal stressesdirected away from their related surfaces are tensile and positive, and opposite compressive stressesare negative Shear stresses are positive when they act in the positive direction of the relevant axis in aplane on which the direct tensile stress is in the positive direction of the axis If the tensile stress is inthe opposite direction, then positive shear stresses are in directions opposite to the positive directions

of the appropriate axes

Two types of external forces may act on a body to produce the internal stress system we have already

discussed Of these, surface forces such as P1, P2, , or hydrostatic pressure are distributed over thesurface area of the body The surface force per unit area may be resolved into components parallel to

our orthogonal system of axes, and these are generally given the symbols X, Y , and Z The second force system derives from gravitational and inertia effects, and the forces are known as body forces These

are distributed over the volume of the body, and the components of body force per unit volume are

designated X, Y , and Z.

acting on the z plane isσz , then the direct stress acting on the z +δz plane is, from the first two terms of

a Taylor’s series expansion,σz+(∂σz /∂z)δz We now investigate the equilibrium of an element at some

internal point in an elastic body where the stress system is obtained by the method just described

In Fig 1.6, the element is in equilibrium under forces corresponding to the stresses shown and thecomponents of body forces (not shown) Surface forces acting on the boundary of the body, althoughcontributing to the production of the internal stress system, do not directly feature in the equilibriumequations

Taking moments about an axis through the center of the element parallel to the z axis

We see, therefore, that a shear stress acting on a given plane (τxy,τxz,τyz) is always accompanied by

an equal complementary shear stress (τyx,τzx,τzy) acting on a plane perpendicular to the given planeand in the opposite sense

Now considering the equilibrium of the element in the x direction

The equations of equilibrium must be satisfied at all interior points in a deformable body under a

three-dimensional force system

Most aircraft structural components are fabricated from thin metal sheet so that stresses across the

thickness of the sheet are usually negligible Assuming, say, that the z axis is in the direction of the

thickness, then the three-dimensional case of Section 1.3 reduces to a two-dimensional case in which

σz,τxz, andτyz are all zero This condition is known as plane stress; the equilibrium equations then

The equations of equilibrium (1.5) (and also (1.6) for a two-dimensional system) satisfy the requirements

of equilibrium at all internal points of the body Equilibrium must also be satisfied at all positions on

the boundary of the body where the components of the surface force per unit area are X, Y , and Z The

triangular element of Fig 1.7 at the boundary of a two-dimensional body of unit thickness is then inequilibrium under the action of surface forces on the elemental length AB of the boundary and internalforces on internal faces AC and CB

Summation of forces in the x direction gives

X δs − σ x δy − τ yx δx + X1

2δxδy = 0

Fig 1.7

Stresses on the faces of an element at the boundary of a two-dimensional body

which, by taking the limit asδx approaches zero, becomes

X= σx

dy

ds+ τyx

dx ds The derivatives dy/ds and dx/ds are the direction cosines l and m of the angles that a normal to AB makes with the x and y axes, respectively It follows that

The complex stress system of Fig 1.6 is derived from a consideration of the actual loads applied to abody and is referred to a predetermined, though arbitrary, system of axes The values of these stressesmay not give a true picture of the severity of stress at that point, so it is necessary to investigate the

1.6 Determination of Stresses on Inclined Planes 11

Fig 1.8

(a) Stresses on a two-dimensional element; (b) stresses on an inclined plane at the point

state of stress on other planes on which the direct and shear stresses may be greater We shall restrictthe analysis to the two-dimensional system of plane stress defined in Section 1.4

Figure 1.8(a) shows a complex stress system at a point in a body referred to axes Ox, Oy All

stresses are positive as defined in Section 1.2 The shear stressesτxyandτyxwere shown to be equal inSection 1.3 We now, therefore, designate them bothτxy The element of sideδx,δy and of unit thickness

is small, so stress distributions over the sides of the element may be assumed to be uniform Body forcesare ignored, since their contribution is a second-order term

Suppose that we want to find the state of stress on a plane AB inclined at an angleθ to the vertical.The triangular element EDC formed by the plane and the vertical through E is in equilibrium under theaction of the forces corresponding to the stresses shown in Fig 1.8(b), whereσnandτ are the directand shear components of the resultant stress on AB Then, resolving forces in a direction perpendicular

Example 1.1

A cylindrical pressure vessel has an internal diameter of 2 m and is fabricated from plates 20 mm thick

If the pressure inside the vessel is 1.5 N/mm2and, in addition, the vessel is subjected to an axial tensileload of 2500 kN, calculate the direct and shear stresses on a plane inclined at an angle of 60◦to the axis

of the vessel Calculate also the maximum shear stress

The expressions for the longitudinal and circumferential stresses produced by the internal pressuremay be found in any text on stress analysis and are

A rectangular element in the wall of the pressure vessel is then subjected to the stress system shown in

Fig 1.9 Note that there are no shear stresses acting on the x and y planes; in this case,σxandσythen

form a biaxial stress system.

The direct stress,σn, and shear stress,τ, on the plane AB that makes an angle of 60◦with the axis of

the vessel may be found from first principles by considering the equilibrium of the triangular elementABC or by direct substitution in Eqs (1.8) and (1.9) Note that in the latter case,θ =30◦andτxy=0.Then,

1.6 Determination of Stresses on Inclined Planes 13

The negative sign forτ indicates that the shear stress is in the direction BA and not in AB

From Eq (1.9) whenτxy=0,

τ = (σx− σy)(sin2θ)/2 (i)The maximum value ofτ therefore occurs when sin2θ is a maximum—that is, when sin2θ =1 and

θ =45◦ Then, substituting the values ofσxandσyin Eq (i),

τmax= (57.4 − 75)/2 = −8.8N/mm2

Example 1.2

A cantilever beam of solid, circular cross section supports a compressive load of 50 kN applied to itsfree end at a point 1.5 mm below a horizontal diameter in the vertical plane of symmetry together with

a torque of 1200 Nm (Fig 1.10) Calculate the direct and shear stresses on a plane inclined at 60◦to the

axis of the cantilever at a point on the lower edge of the vertical plane of symmetry

The direct loading system is equivalent to an axial load of 50 kN together with a bending moment

of 50×103×1.5=75000N/mm in a vertical plane Therefore, at any point on the lower edge of thevertical plane of symmetry, there are compressive stresses due to the axial load and bending momentwhich act on planes perpendicular to the axis of the beam and are given, respectively, by Eqs (1.2) and(15.9):

Fig 1.11

Stress system on two-dimensional element of the beam of Example 1.2

The stress system acting on a two-dimensional rectangular element at the point is shown in Fig 1.11.Note that since the element is positioned at the bottom of the beam, the shear stress due to the torque is

in the direction shown and is negative (see Fig 1.8)

Againσnandτ may be found from first principles or by direct substitution in Eqs (1.8) and (1.9).Note thatθ =30◦,σy=0, and τxy=−28.3N/mm2, the negative sign is arising from the fact that it is inthe opposite direction toτxyas in Fig 1.8

(acting in the direction AB)

Different answers would have been obtained if the plane AB had been chosen on the opposite side

tan 2θ = 2τxy

σx− σy

(1.10)

1.7 Principal Stresses 15

Two solutions,θ and θ +π/2, are obtained from Eq (1.10) so that there are two mutually dicular planes on which the direct stress is either a maximum or a minimum Further, by comparingEqs (1.9) and (1.10), it will be observed that these planes correspond to those on which there is no

perpen-shear stress The direct stresses on these planes are called principal stresses, and the planes themselves are called principal planes.

From Eq (1.10),

sin 2θ = 2τxy

(σx− σy)2+ 4τ2

xy

cos 2θ = σx− σy

(σx− σy)2+ 4τ2

σI=σx+ σy

2 +12

(σx− σy)2+ 4τ2

and

σII=σx+ σy

2 −12

(σx− σy)2+ 4τ2

whereσIis the maximum or major principal stress andσIIis the minimum or minor principal stress.

Note thatσI is algebraically the greatest direct stress at the point, whileσIIis algebraically the least.Therefore, whenσIIis negative—that is, compressive—it is possible forσIIto be numerically greaterthanσI

The maximum shear stress at this point in the body may be determined in an identical manner From

Eq (1.9),

dτ

dθ = (σx− σy)cos2θ + 2τxysin 2θ = 0giving

tan 2θ = −(σx− σy)

2τxy

(1.13)

Equations (1.14) and (1.15) give the maximum shear stress at the point in the body in the plane of

the given stresses For a three-dimensional body supporting a two-dimensional stress system, this is not

necessarily the maximum shear stress at the point

Since Eq (1.13) is the negative reciprocal of Eq (1.10), then the angles 2θ given by these twoequations differ by 90◦, or the planes of maximum shear stress are inclined at 45◦ to the principal

planes

The state of stress at a point in a deformable body may be determined graphically by Mohr’s circle of

stress.

In Section 1.6, the direct and shear stresses on an inclined plane were given by

σn= σxcos2θ + σysin2θ + τxysin 2θ (Eq (1.8))and

τ =(σx− σy)

2 sin 2θ − τxycos 2θ (Eq (1.9))respectively The positive directions of these stresses and the angle θ are defined in Fig 1.12(a).Equation (1.8) may be rewritten in the form

σn=σx

2(1 + cos2θ) +σy

2 (1 − cos2θ) + τxysin 2θ

1.8 Mohr’s Circle of Stress 17

xyand having its center at the point((σx−σy)/2, 0)

The circle is constructed by locating the points Q1(σx,τxy) and Q2(σy,−τxy) referred to axes Oστ asshown in Fig 1.12(b) The center of the circle then lies at C the intersection of Q1Q2and the Oσ axis;clearly C is the point ((σx−σy)/2, 0), and the radius of the circle is 1

2

(σx−σy)2+4τ2

σn= σxcos2θ + σysin2θ + τxysin 2θ

as in Eq (1.8) Similarly, it may be shown that

as in Eq (1.9) Note that the construction of Fig 1.12(b) corresponds to the stress system of Fig 1.12(a)

so that any sign reversal must be allowed for Also, the Oσ and Oτ axes must be constructed to thesame scale, or the equation of the circle is not represented

The maximum and minimum values of the direct stress—that is, the major and minor principalstressesσIandσII—occur when N and Qcoincide with B and A, respectively Thus,

σ1= OC + radius of circle

=(σx+ σy)

2 + CP21+ P1Q21or

σI=(σx+ σy)

2 +1

2

(σx− σy)2+ 4τ2

xy

The principal planes are then given by 2θ =β(σI) and 2θ =β +π(σII)

Also the maximum and minimum values of shear stress occur when Qcoincides with D and E at

the upper and lower extremities of the circle

At these points, QN is equal to the radius of the circle which is given by

CQ1=

(σx− σy)2

xy as before The planes of maximum and minimum shear stressesare given by 2θ =β +π/2 and 2θ =β +3π/2, these being inclined at 45◦to the principal planes.

Example 1.3

Direct stresses of 160 N/mm2(tension) and 120 N/mm2(compression) are applied at a particular point in

an elastic material on two mutually perpendicular planes The principal stress in the material is limited

1.8 Mohr’s Circle of Stress 19

to 200 N/mm2(tension) Calculate the allowable value of shear stress at the point on the given planes.Determine also the value of the other principal stress and the maximum value of shear stress at the point.Verify your answer using Mohr’s circle

The stress system at the point in the material may be represented as shown in Fig 1.13 by consideringthe stresses to act uniformly over the sides of a triangular element ABC of unit thickness Supposethat the direct stress on the principal plane AB isσ For horizontal equilibrium of the element,

σABcosθ = σxBC+ τxyACwhich simplifies to

τxytanθ = σ − σx (i)Considering vertical equilibrium gives

σ ABsinθ = σyAC+ τxyBCor

τxycotθ = σ − σy (ii)Hence, from the product of Eqs (i) and (ii),

τ2

xy= (σ − σx)(σ − σy)Now substituting the valuesσx=160N/mm2,σy=−120N/mm2, andσ =σ1= 200N/mm2, we have

Fig 1.14

Solution of Example 1.3 using Mohr’s circle of stress

The numerical solutions of Eq (iii) corresponding to the given values ofσx,σy, andτxyare the principalstresses at the point, namely

The solution is rapidly verified from Mohr’s circle of stress (Fig 1.14) From the arbitrary origin O,

OP1and OP2are drawn to representσx=160N/mm2andσy=−120N/mm2 The midpoint C of P1P2

is then located OB=σ1=200N/mm2is marked out, and the radius of the circle is then CB OA is therequired principal stress Perpendiculars P1Q1and P2Q2to the circumference of the circle are equal to

±τxy(to scale), and the radius of the circle is the maximum shear stress

The external and internal forces described in the previous sections cause linear and angular

displace-ments in a deformable body These displacedisplace-ments are generally defined in terms of strain Longitudinal

1.9 Strain 21

or direct strains are associated with direct stresses σ and relate to changes in length, while shear

strains define changes in angle produced by shear stresses These strains are designated, with

appro-priate suffixes, by the symbols ε and γ , respectively, and have the same sign as the associatedstresses

Consider three mutually perpendicular line elements OA, OB, and OC at a point O in a deformablebody Their original or unstrained lengths areδx,δy, and δz, respectively If, now, the body is subjected

to forces that produce a complex system of direct and shear stresses at O, such as that in Fig 1.6, thenthe line elements deform to the positions OA, OB, and OCas shown in Fig 1.15.

The coordinates of O in the unstrained body are (x, y, z) so that those of A, B, and C are (x +δx,y,z), (x, y +δy,z), and (x,y,z +δz) The components of the displacement of O to Oparallel to the x, y, and

z axes are u, v, and w These symbols are used to designate these displacements throughout the book

and are defined as positive in the positive directions of the axes We again use the first two terms of aTaylor’s series expansion to determine the components of the displacements of A, B, and C Thus, the

displacement of A in a direction parallel to the x axis is u +(∂u/∂x)δx The remaining components are

found in an identical manner and are shown in Fig 1.15

We now define direct strain in more quantitative terms If a line element of length L at a point in a

body suffers a change in lengthL, then the longitudinal strain at that point in the body in the direction

Fig 1.15

Displacement of line elements OA, OB, and OC

of the line element is

ε = lim

L→0

L

L

The change in length of the element OA is (OA−OA) so that the direct strain at O in the x direction

is obtained from the equation

εx=OA− OA

OA =OAδx− δx (1.16)Now,

The shear strain at a point in a body is defined as the change in the angle between two mutually

perpendicular lines at the point Therefore, if the shear strain in the xz plane isγxz, then the anglebetween the displaced line elements OAand OCin Fig 1.15 isπ/2−γxzradians

But for small displacements, the derivatives of u, v, and w are small compared with l so that, as we are

concerned here with actual length rather than change in length, we may use the approximations

1.10 COMPATIBILITY EQUATIONS

In Section 1.9, we expressed the six components of strain at a point in a deformable body in terms of the

three components of displacement at that point, u, v, and w We have supposed that the body remains continuous during the deformation so that no voids are formed It follows that each component, u, v, and w, must be a continuous, single-valued function or, in quantitative terms,

u = f1(x,y,z) v = f2(x,y,z) w = f3(x,y,z)

If voids were formed, then displacements in regions of the body separated by the voids would be

expressed as different functions of x, y, and z The existence, therefore, of just three single-valued functions for displacement is an expression of the continuity or compatibility of displacement which

we have presupposed

Since the six strains are defined in terms of three displacement functions, then they must bear somerelationship to each other and cannot have arbitrary values These relationships are found as follows.Differentiatingγxy from Eq (1.20) with respect to x and y gives

∂y∂z

∂u

∂x

1.12 Determination of Strains on Inclined Planes 25

Substituting from Eqs (1.18) and (1.21) and rearranging,

Equations (1.21) through (1.26) are the six equations of strain compatibility which must be satisfied in

the solution of three-dimensional problems in elasticity

Although we have derived the compatibility equations and the expressions for strain for the generalthree-dimensional state of strain, we shall be mainly concerned with the two-dimensional case described

in Section 1.4 The corresponding state of strain, in which it is assumed that particles of the body suffer

displacements in one plane only, is known as plane strain We shall suppose that this plane is, as for plane stress, the xy plane Then,εz,γxz, andγyzbecome zero, and Eqs (1.18) and (1.20) reduce to

εx=∂u ∂x εy=∂v ∂y (1.27)and

γxy=∂v ∂x+∂u ∂y (1.28)Further, by substitutingεz=γxz=γyz=0 in the six equations of compatibility and noting that εx,εy,andγxy are now purely functions of x and y, we are left with Eq (1.21), namely

as the only equation of compatibility in the two-dimensional or plane strain case

Having defined the strain at a point in a deformable body with reference to an arbitrary system ofcoordinate axes, we may calculate direct strains in any given direction and the change in the angle(shear strain) between any two originally perpendicular directions at that point We shall consider thetwo-dimensional case of plane strain described in Section 1.11

Fig 1.16

(a) Stress system on rectangular element; (b) distorted shape of element due to stress system in (a)

An element in a two-dimensional body subjected to the complex stress system of Fig 1.16(a) distortsinto the shape shown in Fig 1.16(b) In particular, the triangular element ECD suffers distortion to theshape ECD with corresponding changes in the length FC and angle EFC Suppose that the known

direct and shear strains associated with the given stress system areεx,εy, andγxy(the actual relationshipswill be investigated later) and that we want to find the direct strainεnin a direction normal to the plane

ED and the shear strainγ produced by the shear stress acting on the plane ED

2(ED)2εn+π/2= 2(CD)2εx+ 2(CE)2εy− 2(CE)(CD)γxy

Dividing by 2(ED)2gives

εn+π/2= εxsin2θ + εycos2θ − cosθ sinθγxy (1.30)

(CE)2(1 + εy)2=(CF)2(1 + εn)2+ (FE)2(1 + εn+π/2)2

− 2(CF)(FE)(1 + εn)(1 + εn+π/2)sinγ (1.33)All the strains are assumed to be small so that their squares and higher powers may be ignored Further,sinγ ≈γ and Eq (1.33) becomes

(CE)2(1 + 2εy) = (CF)2(1 + 2εn) + (FE)2(1 + 2εn+π/2) − 2(CF)(FE)γ

From Fig 1.16(a), (CE)2=(CF)2+(FE)2and the preceding equation simplifies to

2(CE)2εy= 2(CF)2εn+ 2(FE)2εn+π/2− 2(CF)(FE)γDividing by 2(CE)2and transposing,

γ =εnsin2θ + εn+π/2cos2θ − εy

sinθ cosθSubstitution ofεn+π/2andεnfrom Eqs (1.30) and (1.31) yields

Therefore, at a point in a deformable body, there are two mutually perpendicular planes on whichthe shear strainγ is zero and normal to which the direct strain is a maximum or minimum These strains

are the principal strains at that point and are given (from comparison with Eqs (1.11) and (1.12)) by

εI=εx+ εy

2 +12

(εx− εy)2+ γ2

and

εII=εx+ εy

2 −12

(εx− εy)2+ γ2

If the shear strain is zero on these planes, it follows that the shear stress must also be zero, and wededuce, from Section 1.7, that the directions of the principal strains and principal stresses coincide Therelated planes are then determined from Eq (1.10) or from

(εx− εy)2+ γ2

We now apply the arguments of Section 1.13 to the Mohr’s circle of stress described in Section 1.8

A circle of strain, analogous to that shown in Fig 1.12(b), may be drawn whenσx,σy, and so on arereplaced byεx,εy, and so on, as specified in Section 1.13 The horizontal extremities of the circlerepresent the principal strains, the radius of the circle, half the maximum shear strain, and so on

In the preceding sections, we have developed, for a three-dimensional deformable body, three equations

of equilibrium (Eqs (1.5)) and six strain–displacement relationships (Eqs (1.18) and (1.20)) Fromthe latter, we eliminated displacements, thereby deriving six auxiliary equations relating strains Thesecompatibility equations are an expression of the continuity of displacement which we have assumed

as a prerequisite of the analysis At this stage, therefore, we have obtained nine independent equationstoward the solution of the three-dimensional stress problem However, the number of unknowns totals

15, comprising six stresses, six strains, and three displacements An additional six equations are thereforenecessary to obtain a solution

1.15 Stress–Strain Relationships 29

So far we have made no assumptions regarding the force–displacement or stress–strain relationship

in the body This will, in fact, provide us with the required six equations, but before these are derived,

it is worthwhile to consider some general aspects of the analysis

The derivation of the equilibrium, strain–displacement, and compatibility equations does not involveany assumption as to the stress–strain behavior of the material of the body It follows that these basicequations are applicable to any type of continuous, deformable body no matter how complex its behavior

under stress In fact, we shall consider only the simple case of linearly elastic isotropic materials for

which stress is directly proportional to strain and whose elastic properties are the same in all directions

A material possessing the same properties at all points is said to be homogeneous.

Particular cases arise where some of the stress components are known to be zero, and the number

of unknowns may then be no greater than the remaining equilibrium equations that have not identicallyvanished The unknown stresses are then found from the conditions of equilibrium alone, and the

problem is said to be statically determinate For example, the uniform stress in the member supporting

a tensile load P in Fig 1.3 is found by applying one equation of equilibrium and a boundary condition.

This system is therefore statically determinate

Statically indeterminate systems require the use of some, if not all, of the other equations involving

strain–displacement and stress–strain relationships However, whether the system is statically minate or not, stress–strain relationships are necessary to determine deflections The role of the sixauxiliary compatibility equations will be discussed when actual elasticity problems are formulated inChapter 2

deter-We now proceed to investigate the relationship of stress and strain in a three-dimensional, linearlyelastic, isotropic body

Experiments show that the application of a uniform direct stress, sayσx, does not produce any sheardistortion of the material and that the direct strainεxis given by the equation

in whichν is a constant termed Poisson’s ratio.

For a body subjected to direct stressesσx,σy, and σz, the direct strains are from Eqs (1.40) and

(1.41) and the principle of superposition (see Chapter 5, Section 5.9)

Equations (1.42) may be transposed to obtain expressions for each stress in terms of the strains Theprocedure adopted may be any of the standard mathematical approaches and gives

σx= E

1− ν2(εx+ νεy) (1.46)

σy= E

1− ν2(εy+ νεx) (1.47)Suppose now that at some arbitrary point in a material, there are principal strainsεI andεIIcor-responding to principal stresses σI andσII If these stresses (and strains) are in the direction of the

coordinate axes x and y, respectively, thenτxy=γxy=0, and from Eq (1.34), the shear strain on anarbitrary plane at the point inclined at an angleθ to the principal planes is

γ = (εI− εII)sin2θ (1.48)Using the relationships of Eqs (1.42) and substituting in Eq (1.48), we have

γ = 1

E[(σI− νσII) − (σII− νσI)]sin2θor

γ =(1 + ν)

E (σI− σII)sin2θ (1.49)Using Eq (1.9) and noting that for this particular caseτxy=0,σx=σI, andσy=σII,

2τ = (σI− σII)sin2θfrom which we may rewrite Eq (1.49) in terms ofτ as

γ =2(1 + ν)

The term E/2(1 +ν) is a constant known as the modulus of rigidity G Hence,

γ = τ/G

to a linearly elastic isotropic body.

For the case of plane stress, they simplify to

Changes in the linear dimensions of a strained body may lead to a change in volume Suppose that

a small element of a body has dimensionsδx,δy, and δz When subjected to a three-dimensional stress system, the element sustains a volumetric strain e (change in volume/unit volume) equal to

as a prerequisite of the analysis At this stage, therefore, we have obtained nine independent equationstoward the solution... equations, but before these are derived,

it is worthwhile to consider some general aspects of the analysis

The derivation of the equilibrium, strain–displacement, and compatibility equations